Draw A Generic Engineering Stress-Strain Diagram For A Ductile Metal

Table of contents :
Cover
Title Page
Copyright Page
Dedication
About the Authors
Brief Contents
Contents
Preface
Acknowledgments
Part 1 Basics
Chapter 1 Introduction to Mechanical Engineering Design
1–1 Design
1–2 Mechanical Engineering Design
1–3 Phases and Interactions of the Design Process
1–4 Design Tools and Resources
1–5 The Design Engineer's Professional Responsibilities
1–6 Standards and Codes
1–7 Economics
1–8 Safety and Product Liability
1–9 Stress and Strength
1–10 Uncertainty
1–11 Design Factor and Factor of Safety
1–12 Reliability and Probability of Failure
1–13 Relating Design Factor to Reliability
1–14 Dimensions and Tolerances
1–15 Units
1–16 Calculations and Significant Figures
1–17 Design Topic Interdependencies
1–18 Power Transmission Case Study Specifications
Problems
Chapter 2 Materials
2–1 Material Strength and Stiffness
2–2 The Statistical Significance of Material Properties
2–3 Plastic Deformation and Cold Work
2–4 Cyclic Stress-Strain Properties
2–5 Hardness
2–6 Impact Properties
2–7 Temperature Effects
2–8 Numbering Systems
2–9 Sand Casting
2–10 Shell Molding
2–11 Investment Casting
2–12 Powder-Metallurgy Process
2–13 Hot-Working Processes
2–14 Cold-Working Processes
2–15 The Heat Treatment of Steel
2–16 Alloy Steels
2–17 Corrosion-Resistant Steels
2–18 Casting Materials
2–19 Nonferrous Metals
2–20 Plastics
2–21 Composite Materials
2–22 Materials Selection
Problems
Chapter 3 Load and Stress Analysis
3–1 Equilibrium and Free-Body Diagrams
3–2 Shear Force and Bending Moments in Beams
3–3 Singularity Functions
3–4 Stress
3–5 Cartesian Stress Components
3–6 Mohr's Circle for Plane Stress
3–7 General Three-Dimensional Stress
3–8 Elastic Strain
3–9 Uniformly Distributed Stresses
3–10 Normal Stresses for Beams in Bending
3–11 Shear Stresses for Beams in Bending
3–12 Torsion
3–13 Stress Concentration
3–14 Stresses in Pressurized Cylinders
3–15 Stresses in Rotating Rings
3–16 Press and Shrink Fits
3–17 Temperature Effects
3–18 Curved Beams in Bending
3–19 Contact Stresses
3–20 Summary
Problems
Chapter 4 Deflection and Stiffness
4–1 Spring Rates
4–2 Tension, Compression, and Torsion
4–3 Deflection Due to Bending
4–4 Beam Deflection Methods
4–5 Beam Deflections by Superposition
4–6 Beam Deflections by Singularity Functions
4–7 Strain Energy
4–8 Castigliano's Theorem
4–9 Deflection of Curved Members
4–10 Statically Indeterminate Problems
4–11 Compression Members—General
4–12 Long Columns with Central Loading
4–13 Intermediate-Length Columns with Central Loading
4–14 Columns with Eccentric Loading
4–15 Struts or Short Compression Members
4–16 Elastic Stability
4–17 Shock and Impact
Problems
Part 2 Failure Prevention
Chapter 5 Failures Resulting from Static Loading
5–1 Static Strength
5–2 Stress Concentration
5–3 Failure Theories
5–4 Maximum-Shear-Stress Theory for Ductile Materials
5–5 Distortion-Energy Theory for Ductile Materials
5–6 Coulomb-Mohr Theory for Ductile Materials
5–7 Failure of Ductile Materials Summary
5–8 Maximum-Normal-Stress Theory for Brittle Materials
5–9 Modifications of the Mohr Theory for Brittle Materials
5–10 Failure of Brittle Materials Summary
5–11 Selection of Failure Criteria
5–12 Introduction to Fracture Mechanics
5–13 Important Design Equations
Problems
Chapter 6 Fatigue Failure Resulting from Variable Loading
6–1 Introduction to Fatigue
6–2 Chapter Overview
6–3 Crack Nucleation and Propagation
6–4 Fatigue-Life Methods
6–5 The Linear-Elastic Fracture Mechanics Method
6–6 The Strain-Life Method
6–7 The Stress-Life Method and the S-N Diagram
6–8 The Idealized S-N Diagram for Steels
6–9 Endurance Limit Modifying Factors
6–10 Stress Concentration and Notch Sensitivity
6–11 Characterizing Fluctuating Stresses
6–12 The Fluctuating-Stress Diagram
6–13 Fatigue Failure Criteria
6–14 Constant-Life Curves
6–15 Fatigue Failure Criterion for Brittle Materials
6–16 Combinations of Loading Modes
6–17 Cumulative Fatigue Damage
6–18 Surface Fatigue Strength
6–19 Road Maps and Important Design Equations for the Stress-Life Method
Problems
Part 3 Design of Mechanical Elements
Chapter 7 Shafts and Shaft Components
7–1 Introduction
7–2 Shaft Materials
7–3 Shaft Layout
7–4 Shaft Design for Stress
7–5 Deflection Considerations
7–6 Critical Speeds for Shafts
7–7 Miscellaneous Shaft Components
7–8 Limits and Fits
Problems
Chapter 8 Screws, Fasteners, and the Design of Nonpermanent Joints
8–1 Thread Standards and Definitions
8–2 The Mechanics of Power Screws
8–3 Threaded Fasteners
8–4 Joints—Fastener Stiffness
8–5 Joints—Member Stiffness
8–6 Bolt Strength
8–7 Tension Joints—The External Load
8–8 Relating Bolt Torque to Bolt Tension
8–9 Statically Loaded Tension Joint with Preload
8–10 Gasketed Joints
8–11 Fatigue Loading of Tension Joints
8–12 Bolted and Riveted Joints Loaded in Shear
Problems
Chapter 9 Welding, Bonding, and the Design of Permanent Joints
9–1 Welding Symbols
9–2 Butt and Fillet Welds
9–3 Stresses in Welded Joints in Torsion
9–4 Stresses in Welded Joints in Bending
9–5 The Strength of Welded Joints
9–6 Static Loading
9–7 Fatigue Loading
9–8 Resistance Welding
9–9 Adhesive Bonding
Problems
Chapter 10 Mechanical Springs
10–1 Stresses in Helical Springs
10–2 The Curvature Effect
10–3 Deflection of Helical Springs
10–4 Compression Springs
10–5 Stability
10–6 Spring Materials
10–7 Helical Compression Spring Design for Static Service
10–8 Critical Frequency of Helical Springs
10–9 Fatigue Loading of Helical Compression Springs
10–10 Helical Compression Spring Design for Fatigue Loading
10–11 Extension Springs
10–12 Helical Coil Torsion Springs
10–13 Belleville Springs
10–14 Miscellaneous Springs
10–15 Summary
Problems
Chapter 11 Rolling-Contact Bearings
11–1 Bearing Types
11–2 Bearing Life
11–3 Bearing Load Life at Rated Reliability
11–4 Reliability versus Life—The Weibull Distribution
11–5 Relating Load, Life, and Reliability
11–6 Combined Radial and Thrust Loading
11–7 Variable Loading
11–8 Selection of Ball and Cylindrical Roller Bearings
11–9 Selection of Tapered Roller Bearings
11–10 Design Assessment for Selected Rolling-Contact Bearings
11–11 Lubrication
11–12 Mounting and Enclosure
Problems
Chapter 12 Lubrication and Journal Bearings
12–1 Types of Lubrication
12–2 Viscosity
12–3 Petroff's Equation
12–4 Stable Lubrication
12–5 Thick-Film Lubrication
12–6 Hydrodynamic Theory
12–7 Design Variables
12–8 The Relations of the Variables
12–9 Steady-State Conditions in Self-Contained Bearings
12–10 Clearance
12–11 Pressure-Fed Bearings
12–12 Loads and Materials
12–13 Bearing Types
12–14 Dynamically Loaded Journal Bearings
12–15 Boundary-Lubricated Bearings
Problems
Chapter 13 Gears—General
13–1 Types of Gears
13–2 Nomenclature
13–3 Conjugate Action
13–4 Involute Properties
13–5 Fundamentals
13–6 Contact Ratio
13–7 Interference
13–8 The Forming of Gear Teeth
13–9 Straight Bevel Gears
13–10 Parallel Helical Gears
13–11 Worm Gears
13–12 Tooth Systems
13–13 Gear Trains
13–14 Force Analysis—Spur Gearing
13–15 Force Analysis—Bevel Gearing
13–16 Force Analysis—Helical Gearing
13–17 Force Analysis—Worm Gearing
Problems
Chapter 14 Spur and Helical Gears
14–1 The Lewis Bending Equation
14–2 Surface Durability
14–3 AGMA Stress Equations
14–4 AGMA Strength Equations
14–5 Geometry Factors I and J (ZI and YJ)
14–6 The Elastic Coefficient Cp (ZE)
14–7 Dynamic Factor Kv
14–8 Overload Factor Ko
14–9 Surface Condition Factor Cf (ZR)
14–10 Size Factor Ks
14–11 Load-Distribution Factor Km (KH)
14–12 Hardness-Ratio Factor CH (ZW)
14–13 Stress-Cycle Factors YN and ZN
14–14 Reliability Factor KR (YZ)
14–15 Temperature Factor KT (Yθ)
14–16 Rim-Thickness Factor KB
14–17 Safety Factors SF and SH
14–18 Analysis
14–19 Design of a Gear Mesh
Problems
Chapter 15 Bevel and Worm Gears
15–1 Bevel Gearing—General
15–2 Bevel-Gear Stresses and Strengths
15–3 AGMA Equation Factors
15–4 Straight-Bevel Gear Analysis
15–5 Design of a Straight-Bevel Gear Mesh
15–6 Worm Gearing—AGMA Equation
15–7 Worm-Gear Analysis
15–8 Designing a Worm-Gear Mesh
15–9 Buckingham Wear Load
Problems
Chapter 16 Clutches, Brakes, Couplings, and Flywheels
16–1 Static Analysis of Clutches and Brakes
16–2 Internal Expanding Rim Clutches and Brakes
16–3 External Contracting Rim Clutches and Brakes
16–4 Band-Type Clutches and Brakes
16–5 Frictional-Contact Axial Clutches
16–6 Disk Brakes
16–7 Cone Clutches and Brakes
16–8 Energy Considerations
16–9 Temperature Rise
16–10 Friction Materials
16–11 Miscellaneous Clutches and Couplings
16–12 Flywheels
Problems
Chapter 17 Flexible Mechanical Elements
17–1 Belts
17–2 Flat- and Round-Belt Drives
17–3 V Belts
17–4 Timing Belts
17–5 Roller Chain
17–6 Wire Rope
17–7 Flexible Shafts
Problems
Chapter 18 Power Transmission Case Study
18–1 Design Sequence for Power Transmission
18–2 Power and Torque Requirements
18–3 Gear Specification
18–4 Shaft Layout
18–5 Force Analysis
18–6 Shaft Material Selection
18–7 Shaft Design for Stress
18–8 Shaft Design for Deflection
18–9 Bearing Selection
18–10 Key and Retaining Ring Selection
18–11 Final Analysis
Problems
Part 4 Special Topics
Chapter 19 Finite-Element Analysis
19–1 The Finite-Element Method
19–2 Element Geometries
19–3 The Finite-Element Solution Process
19–4 Mesh Generation
19–5 Load Application
19–6 Boundary Conditions
19–7 Modeling Techniques
19–8 Thermal Stresses
19–9 Critical Buckling Load
19–10 Vibration Analysis
19–11 Summary
Problems
Chapter 20 Geometric Dimensioning and Tolerancing
20–1 Dimensioning and Tolerancing Systems
20–2 Definition of Geometric Dimensioning and Tolerancing
20–3 Datums
20–4 Controlling Geometric Tolerances
20–5 Geometric Characteristic Definitions
20–6 Material Condition Modifiers
20–7 Practical Implementation
20–8 GD&T in CAD Models
20–9 Glossary of GD&T Terms
Problems
Appendixes
A Useful Tables
B Answers to Selected Problems
Index

Shigley' s

Mechanical Engineering Design

Eleventh Edition

Richard G. Budynas J. Keith Nisbett

Shigley's Mechanical Engineering Design

E O

D C

2

2

A xy

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Shigley's Mechanical Engineering Design Eleventh Edition Richard G. Budynas Professor Emeritus, Kate Gleason College of Engineering, Rochester Institute of Technology

J. Keith Nisbett Associate Professor of Mechanical Engineering, Missouri University of Science and Technology

SHIGLEY'S MECHANICAL ENGINEERING DESIGN, ELEVENTH EDITION Published by McGraw-Hill Education, 2 Penn Plaza, New York, NY 10121. Copyright © 2020 by McGraw-Hill Education. All rights reserved. Printed in the United States of America. Previous editions © 2015, 2011, and 2008. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of McGraw-Hill Education, including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning. Some ancillaries, including electronic and print components, may not be available to customers outside the United States. This book is printed on acid-free paper. 1 2 3 4 5 6 7 8 9 LWI 21 20 19 ISBN 978-0-07-339821-1 (bound edition) MHID 0-07-339821-7 (bound edition) ISBN 978-1-260-40764-8 (loose-leaf edition) MHID 1-260-40764-0 (loose-leaf edition) Product Developers: Tina Bower and Megan Platt Marketing Manager: Shannon O'Donnell Content Project Managers: Jane Mohr, Samantha Donisi-Hamm, and Sandy Schnee Buyer: Laura Fuller Design: Matt Backhaus Content Licensing Specialist: Beth Cray Cover Image: Courtesy of Dee Dehokenanan Compositor: Aptara®, Inc. All credits appearing on page or at the end of the book are considered to be an extension of the copyright page.

Library of Congress Cataloging-in-Publication Data

Names: Budynas, Richard G. (Richard Gordon), author. | Nisbett, J. Keith,   author. | Shigley, Joseph Edward. Mechanical engineering design. Title: Shigley's mechanical engineering design / Richard G. Budynas,  Professor Emeritus, Kate Gleason College of Engineering, Rochester Institute of Technology, J. Keith Nisbett, Associate Professor of Mechanical Engineering, Missouri University of Science and Technology. Other titles: Mechanical engineering design Description: Eleventh edition. ∣ New York, NY : McGraw-Hill Education, [2020]   ∣ Includes index. Identifiers: LCCN 2018023098 ∣ ISBN 9780073398211 (alk. paper) ∣ ISBN   0073398217 (alk. paper) Subjects: LCSH: Machine design. Classification: LCC TJ230 .S5 2020 | DDC 621.8/15--dc23 LC record available at  https://lccn.loc.gov/2018023098 The Internet addresses listed in the text were accurate at the time of publication. The inclusion of a website does not indicate an endorsement by the authors or McGraw-Hill Education, and McGraw-Hill Education does not guarantee the accuracy of the information presented at these sites.

mheducation.com/highered

Dedication To my wife, Joanne. I could not have accomplished what I have without your love and support.  Richard G. Budynas To my colleague and friend, Dr. Terry Lehnhoff, who encouraged me early in my teaching career to pursue opportunities to improve the presentation of machine design topics.  J. Keith Nisbett

Dedication to Joseph Edward Shigley Joseph Edward Shigley (1909–1994) is undoubtedly one of the most well-known and respected contributors in machine design education. He authored or coauthored eight books, including Theory of Machines and Mechanisms (with John J. Uicker, Jr.), and Applied Mechanics of Materials. He was coeditor-in-chief of the well-known Standard Handbook of Machine Design. He began Machine Design as sole author in 1956, and it evolved into Mechanical Engineering Design, setting the model for such textbooks. He contributed to the first five editions of this text, along with coauthors Larry Mitchell and Charles Mischke. Uncounted numbers of students across the world got their first taste of machine design with Shigley's textbook, which has literally become a classic. Nearly every mechanical engineer for the past half century has referenced terminology, equations, or procedures as being from "Shigley." McGraw-Hill is honored to have worked with Professor Shigley for more than 40 years, and as a tribute to his lasting contribution to this textbook, its title officially reflects what many have already come to call it—Shigley's Mechanical Engineering Design. Having received a bachelor's degree in Electrical and Mechanical Engineering from Purdue University and a master of science in Engineering Mechanics from the University of Michigan, Professor Shigley pursued an academic career at Clemson College from 1936 through 1954. This led to his position as professor and head of Mechanical Design and Drawing at Clemson College. He joined the faculty of the Department of Mechanical Engineering of the University of Michigan in 1956, where he remained for 22 years until his retirement in 1978. Professor Shigley was granted the rank of Fellow of the American Society of Mechanical Engineers in 1968. He received the ASME Mechanisms Committee Award in 1974, the Worcester Reed Warner Medal for outstanding contribution to the permanent literature of engineering in 1977, and the ASME Machine Design Award in 1985. Joseph Edward Shigley indeed made a difference. His legacy shall continue.

vi

About the Authors Richard G. Budynas is Professor Emeritus of the Kate Gleason College of Engineering at Rochester Institute of Technology. He has more than 50 years experience in teaching and practicing mechanical engineering design. He is the author of a McGraw-Hill textbook, Advanced Strength and Applied Stress Analysis, Second Edition; and coauthor of a McGraw-Hill reference book, Roark's Formulas for Stress and Strain, Eighth Edition. He was awarded the BME of Union College, MSME of the University of Rochester, and the PhD of the University of Massachusetts. He is a licensed Professional Engineer in the state of New York. J. Keith Nisbett is an Associate Professor and Associate Chair of Mechanical Engineering at the Missouri University of Science and Technology. He has more than 30 years of experience with using and teaching from this classic textbook. As demonstrated by a steady stream of teaching awards, including the Governor's Award for Teaching Excellence, he is devoted to finding ways of communicating concepts to the students. He was awarded the BS, MS, and PhD of the University of Texas at Arlington.

vii

Brief Contents Preface xv

Part 1 Basics 2 1 Introduction to Mechanical Engineering Design  3 2 Materials 41 3 Load and Stress Analysis  93 4 Deflection and Stiffness  173

Part 2 Failure Prevention  240 5 Failures Resulting from Static Loading  241 6 Fatigue Failure Resulting from Variable Loading  285

Part 3 Design of Mechanical Elements  372 7 Shafts and Shaft Components  373 8 Screws, Fasteners, and the Design of Nonpermanent Joints  421 9 Welding, Bonding, and the Design of Permanent Joints  485 10 Mechanical Springs  525 11 Rolling-Contact Bearings  575 12 Lubrication and Journal Bearings  623 13 Gears—General  681 14 Spur and Helical Gears  739 15 Bevel and Worm Gears  791 16 Clutches, Brakes, Couplings, and Flywheels  829 17 Flexible Mechanical Elements  881 18 Power Transmission Case Study  935

viii

Brief Contents     ix

Part 4 Special Topics  954 19 Finite-Element Analysis  955 20 Geometric Dimensioning and Tolerancing  977

Appendixes A Useful Tables  1019 B Answers to Selected Problems  1075 Index  1081

Contents Preface  xv Part

1

Basics 2 Chapter

1

Introduction to Mechanical Engineering Design  3 1–1 Design 4 1–2 Mechanical Engineering Design  5 1–3 Phases and Interactions of the Design

Process 5 1–4 Design Tools and Resources  8 1–5 The Design Engineer's Professional Responsibilities 10 1–6 Standards and Codes  12 1–7 Economics 13 1–8 Safety and Product Liability  15 1–9 Stress and Strength  16 1–10 Uncertainty 16 1–11 Design Factor and Factor of Safety  18 1–12 Reliability and Probability of Failure  20 1–13 Relating Design Factor to Reliability  24 1–14 Dimensions and Tolerances  27 1–15 Units 31 1–16 Calculations and Significant Figures  32 1–17 Design Topic Interdependencies  33 1–18 Power Transmission Case Study Specifications 34 Problems 36 Chapter

2

Plastic Deformation and Cold Work  50 Cyclic Stress-Strain Properties  57 Hardness 61 Impact Properties  62 Temperature Effects  63 Numbering Systems  64 Sand Casting  66 Shell Molding  66 Investment Casting  67 Powder-Metallurgy Process  67 Hot-Working Processes  67 Cold-Working Processes  68 The Heat Treatment of Steel  69 Alloy Steels  72 Corrosion-Resistant Steels  73 Casting Materials  73 Nonferrous Metals  75 Plastics 78 Composite Materials  80 Materials Selection  81

Problems 87 Chapter

3

Load and Stress Analysis  93

Materials 41

2–1 2–2

Material Strength and Stiffness  42

The Statistical Significance of Material Properties 48

x

2–3 2–4 2–5 2–6 2–7 2–8 2–9 2–10 2–11 2–12 2–13 2–14 2–15 2–16 2–17 2–18 2–19 2–20 2–21 2–22

3–1 3–2 3–3 3–4 3–5 3–6 3–7 3–8 3–9 3–10

Equilibrium and Free-Body Diagrams  94 Shear Force and Bending Moments in Beams 97 Singularity Functions  98 Stress 101 Cartesian Stress Components  101 Mohr's Circle for Plane Stress  102 General Three-Dimensional Stress  108 Elastic Strain  109 Uniformly Distributed Stresses  110 Normal Stresses for Beams in Bending 111

Contents     xi

3–11 3–12 3–13 3–14 3–15 3–16 3–17 3–18 3–19 3–20

Shear Stresses for Beams in Bending  116 Stress Concentration  132 Stresses in Pressurized Cylinders  135 Stresses in Rotating Rings  137 Press and Shrink Fits  139 Temperature Effects  140 Curved Beams in Bending  141 Contact Stresses  145 Summary 149

4

Deflection and Stiffness  173

4–1 4–2 4–3 4–4 4–5 4–6 4–7 4–8 4–9 4–10 4–11 4–12 4–13

Spring Rates  174

4–14 4–15 4–16 4–17

Columns with Eccentric Loading  212

Tension, Compression, and Torsion  175 Deflection Due to Bending  176 Beam Deflection Methods  179 Beam Deflections by Superposition  180 Beam Deflections by Singularity Functions  182 Strain Energy  188 Castigliano's Theorem  190 Deflection of Curved Members  195 Statically Indeterminate Problems  201 Compression Members—General  207 Long Columns with Central Loading  207 Intermediate-Length Columns with Central Loading 210 Struts or Short Compression Members  215 Elastic Stability  217 Shock and Impact  218

Problems 220

Part

2

Failure Prevention  240 Chapter

5

Failures Resulting from Static Loading  241

5–1 5–2 5–3

5–4

Maximum-Shear-Stress Theory for Ductile Materials 247

5–5

Distortion-Energy Theory for Ductile Materials 249

5–6

Coulomb-Mohr Theory for Ductile Materials 255

5–7 Failure of Ductile Materials Summary  258 5–8 Maximum-Normal-Stress Theory for Brittle

5–9 Modifications of the Mohr Theory for Brittle

Torsion 123

Problems 150 Chapter

Static Strength  244 Stress Concentration  245 Failure Theories  247

Materials 262 Materials 263

5–10 5–11 5–12 5–13

Failure of Brittle Materials Summary  265 Selection of Failure Criteria  266 Introduction to Fracture Mechanics  266 Important Design Equations  275

Problems 276 Chapter

6

Fatigue Failure Resulting from Variable Loading  285

6–1 6–2 6–3 6–4 6–5

Introduction to Fatigue  286

Chapter Overview  287

6–6 6–7

The Strain-Life Method  299

6–8 6–9 6–10 6–11 6–12 6–13 6–14 6–15

The Idealized S-N Diagram for Steels  304

Crack Nucleation and Propagation  288 Fatigue-Life Methods  294 The Linear-Elastic Fracture Mechanics Method 295 The Stress-Life Method and the S-N Diagram 302 Endurance Limit Modifying Factors  309 Stress Concentration and Notch Sensitivity  320 Characterizing Fluctuating Stresses  325 The Fluctuating-Stress Diagram  327 Fatigue Failure Criteria  333 Constant-Life Curves  342 Fatigue Failure Criterion for Brittle Materials 345

6–16 Combinations of Loading Modes  347 6–17 Cumulative Fatigue Damage  351 6–18 Surface Fatigue Strength  356 6–19 Road Maps and Important Design Equations for the Stress-Life Method  359

Problems 363

xii      Mechanical Engineering Design

Part

3

Design of Mechanical Elements  372 Chapter

7

9–6 9–7 9–8 9–9

Introduction 374 Shaft Materials  374 Shaft Layout  375 Shaft Design for Stress  380 Deflection Considerations  391 Critical Speeds for Shafts  395 Miscellaneous Shaft Components  400 Limits and Fits  406

Problems 411 Chapter

8

Screws, Fasteners, and the Design of Nonpermanent Joints  421

Fatigue Loading  505 Resistance Welding  507 Adhesive Bonding  508

Problems 516

Shafts and Shaft Components  373 7–1 7–2 7–3 7–4 7–5 7–6 7–7 7–8

Static Loading  502

Chapter

10

Mechanical Springs  525 10–1 10–2 10–3 10–4 10–5 10–6 10–7

Stresses in Helical Springs  526

10–8 10–9

Critical Frequency of Helical Springs  542

The Curvature Effect  527 Deflection of Helical Springs  528 Compression Springs  528 Stability 529 Spring Materials  531 Helical Compression Spring Design for Static Service 535 Fatigue Loading of Helical Compression Springs 543

10–10 Helical Compression Spring Design for

8–1 8–2 8–3 8–4 8–5 8–6 8–7 8–8 8–9

Thread Standards and Definitions  422

Statically Loaded Tension Joint with Preload 452

Chapter

Gasketed Joints  456

Rolling-Contact Bearings  575

Fatigue Loading of Tension Joints  456

11–1 11–2 11–3 11–4

Bearing Types  576

11–5 11–6 11–7 11–8

Relating Load, Life, and Reliability  583

8–10 8–11 8–12

The Mechanics of Power Screws  426 Threaded Fasteners  434 Joints—Fastener Stiffness  436 Joints—Member Stiffness  437 Bolt Strength  443 Tension Joints—The External Load  446 Relating Bolt Torque to Bolt Tension  448

Bolted and Riveted Joints Loaded in Shear  463

Problems 471 Chapter

9

Welding, Bonding, and the Design of Permanent Joints  485 9–1 9–2 9–3 9–4 9–5

Welding Symbols  486 Butt and Fillet Welds  488 Stresses in Welded Joints in Torsion  492 Stresses in Welded Joints in Bending  497 The Strength of Welded Joints  499

Fatigue Loading  547

10–11 10–12 10–13 10–14 10–15

Extension Springs  550 Helical Coil Torsion Springs  557 Belleville Springs  564 Miscellaneous Springs  565 Summary 567

Problems 567

11

Bearing Life  579 Bearing Load Life at Rated Reliability  580 Reliability versus Life—The Weibull Distribution 582 Combined Radial and Thrust Loading  585 Variable Loading  590 Selection of Ball and Cylindrical Roller Bearings 593

11–9 Selection of Tapered Roller Bearings  596 11–10 Design Assessment for Selected Rolling-Contact Bearings  604

Contents     xiii

11–11 Lubrication 608 11–12 Mounting and Enclosure  609

13–16 13–17

Problems 613

Problems 724

Chapter

12

Lubrication and Journal Bearings  623 12–1 12–2 12–3 12–4 12–5 12–6 12–7 12–8 12–9

Types of Lubrication  624

12–10 12–11 12–12 12–13 12–14

Clearance 653

12–15

Boundary-Lubricated Bearings  670

Viscosity 625 Petroff's Equation  627 Stable Lubrication  632 Thick-Film Lubrication  633 Hydrodynamic Theory  634 Design Variables  639 The Relations of the Variables  640 Steady-State Conditions in Self-Contained Bearings 649 Pressure-Fed Bearings  655 Loads and Materials  661 Bearing Types  662 Dynamically Loaded Journal Bearings 663

Problems 677 Chapter

13

Gears—General 681

13–1 13–2 13–3 13–4 13–5 13–6 13–7 13–8 13–9 13–10 13–11 13–12 13–13 13–14 13–15

Chapter

Force Analysis—Helical Gearing  716 Force Analysis—Worm Gearing  719

14

Spur and Helical Gears  739 14–1 14–2 14–3 14–4 14–5

The Lewis Bending Equation  740

14–6 14–7 14–8 14–9 14–10 14–11 14–12 14–13 14–14 14–15 14–16 14–17 14–18 14–19

The Elastic Coefficient Cp (ZE) 761

Surface Durability  749 AGMA Stress Equations  751 AGMA Strength Equations  752 Geometry Factors I and J (ZI and YJ) 757 Dynamic Factor Kv 763 Overload Factor Ko 764 Surface Condition Factor Cf (ZR) 764 Size Factor Ks 765 Load-Distribution Factor Km (KH) 765 Hardness-Ratio Factor CH (ZW) 767 Stress-Cycle Factors YN and ZN 768 Reliability Factor KR (YZ) 769 Temperature Factor KT (Yθ) 770 Rim-Thickness Factor KB 770 Safety Factors SF and SH 771 Analysis 771 Design of a Gear Mesh  781

Problems 786

Types of Gears  682 Nomenclature 683 Conjugate Action  684 Involute Properties  685 Fundamentals 686

Chapter

15

Bevel and Worm Gears  791 15–1 15–2 15–3 15–4 15–5

Bevel Gearing—General  792

Worm Gearing—AGMA Equation  814

Force Analysis—Spur Gearing  710

15–6 15–7 15–8 15–9

Force Analysis—Bevel Gearing  713

Problems 826

Contact Ratio  689 Interference 690 The Forming of Gear Teeth  693 Straight Bevel Gears  695 Parallel Helical Gears  696 Worm Gears  700 Tooth Systems  701 Gear Trains  703

Bevel-Gear Stresses and Strengths  794 AGMA Equation Factors  797 Straight-Bevel Gear Analysis  808 Design of a Straight-Bevel Gear Mesh 811 Worm-Gear Analysis  818 Designing a Worm-Gear Mesh  822 Buckingham Wear Load  825

xiv      Mechanical Engineering Design

Chapter

16

Clutches, Brakes, Couplings, and Flywheels  829

18–10 18–11

Key and Retaining Ring Selection  950 Final Analysis  953

Problems 953

4

16–1 16–2

Static Analysis of Clutches and Brakes  831

16–3

External Contracting Rim Clutches and Brakes 844

Chapter

16–4 16–5 16–6 16–7 16–8 16–9 16–10 16–11 16–12

Band-Type Clutches and Brakes  847

Finite-Element Analysis 955

Internal Expanding Rim Clutches and Brakes 836

Frictional-Contact Axial Clutches  849 Disk Brakes  852 Cone Clutches and Brakes  856 Energy Considerations  858 Temperature Rise  860 Friction Materials  863 Miscellaneous Clutches and Couplings  866 Flywheels 868

Problems 873 Chapter

17

Flexible Mechanical Elements  881

Belts 882

17–1 17–2 17–3 17–4 17–5 17–6 17–7

Flat- and Round-Belt Drives  885 V Belts  900 Timing Belts  908 Roller Chain  909 Wire Rope  917 Flexible Shafts  926

Problems 927 Chapter

18

Power Transmission Case Study  935 18–1

Design Sequence for Power Transmission 937

18–2 18–3 18–4 18–5 18–6 18–7 18–8 18–9

Power and Torque Requirements  938 Gear Specification  938 Shaft Layout  945

Part

Special Topics

19–1 19–2 19–3 19–4 19–5 19–6 19–7 19–8 19–9 19–10 19–11

Chapter

Element Geometries  959 The Finite-Element Solution Process  961 Mesh Generation  964 Load Application  966 Boundary Conditions  967 Modeling Techniques  967 Thermal Stresses  970 Critical Buckling Load  972 Vibration Analysis  973 Summary 974

20

20–1 20–2

Dimensioning and Tolerancing Systems  978

20–3 20–4 20–5 20–6 20–7 20–8 20–9

Datums 983

Definition of Geometric Dimensioning and Tolerancing 979 Controlling Geometric Tolerances  989 Geometric Characteristic Definitions  992 Material Condition Modifiers  1002 Practical Implementation  1004 GD&T in CAD Models  1009 Glossary of GD&T Terms  1010

Problems 1012

Shaft Material Selection  947

Bearing Selection  949

The Finite-Element Method  957

Geometric Dimensioning and Tolerancing  977

Shaft Design for Deflection  948

19

Problems 975

Force Analysis  947 Shaft Design for Stress  948

954

Appendixes

A Useful Tables  1019 B Answers to Selected P ­ roblems  1075

Index  1081

Preface Objectives This text is intended for students beginning the study of mechanical engineering design. The focus is on blending fundamental development of concepts with practical specification of components. Students of this text should find that it inherently directs them into familiarity with both the basis for decisions and the standards of industrial components. For this reason, as students transition to practicing engineers, they will find that this text is indispensable as a reference text. The objectives of the text are to: ∙ Cover the basics of machine design, including the design process, engineering mechanics and materials, failure prevention under static and variable loading, and characteristics of the principal types of mechanical elements. ∙ Offer a practical approach to the subject through a wide range of real-world applications and examples. ∙ Encourage readers to link design and analysis. ∙  Encourage readers to link fundamental concepts with practical component ­specification.

New to This Edition Enhancements and modifications to the eleventh edition are described in the following summaries: ∙ Chapter 6, Fatigue Failure Resulting from Variable Loading, has received a complete update of its presentation. The goals include clearer explanations of underlying mechanics, streamlined approach to the stress-life method, and updates consistent with recent research. The introductory material provides a greater appreciation of  the processes involved in crack nucleation and propagation. This allows the strain-life method and the linear-elastic fracture mechanics method to be given proper context within the coverage, as well as to add to the understanding of the factors driving the data used in the stress-life method. The overall methodology of the stress-life approach remains the same, though with expanded explanations and improvements in the presentation. ∙ Chapter 2, Materials, includes expanded coverage of plastic deformation, strainhardening, true stress and true strain, and cyclic stress-strain properties. This information provides a stronger background for the expanded discussion in Chapter 6 of the mechanism of crack nucleation and propagation. ∙ Chapter 12, Lubrication and Journal Bearings, is improved and updated. The chapter contains a new section on dynamically loaded journal bearings, including the mobility method of solution for the journal dynamic orbit. This includes new examples and end-of-chapter problems. The design of big-end connecting rod bearings, used in automotive applications, is also introduced.

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xvi      Mechanical Engineering Design

∙ Approximately 100 new end-of-chapter problems are implemented. These are focused on providing more variety in the fundamental problems for first-time exposure to the topics. In conjunction with the web-based parameterized problems available through McGraw-Hill Connect Engineering, the ability to assign new problems each semester is ever stronger. The following sections received minor but notable improvements in presentation: Section 3–8 Elastic Strain Section 3–11 Shear Stresses for Beams in Bending Section 3–14 Stresses in Pressurized Cylinders Section 3–15 Stresses in Rotating Rings Section 4–12 Long Columns with Central Loading Section 4–13 Intermediate-Length Columns with Central Loading Section 4–14 Columns with Eccentric Loading

Section Section Section Section Section Section Section Section

7–4 Shaft Design for Stress 8–2 The Mechanics of Power Screws 8–7 Tension Joints—The External Load 13–5 Fundamentals 16–4 Band-Type Clutches and Brakes 16–8 Energy Considerations 17–2 Flat- and Round-Belt Drives 17–3 V Belts

In keeping with the well-recognized accuracy and consistency within this text, minor improvements and corrections are made throughout with each new edition. Many of these are in response to the diligent feedback from the community of users.

Instructor Supplements Additional media offerings available at www.mhhe.com/shigley include: ∙ Solutions manual.  The instructor's manual contains solutions to most end-of-chapter nondesign problems. ∙ PowerPoint® slides.  Slides outlining the content of the text are provided in PowerPoint format for instructors to use as a starting point for developing lecture presentation materials. The slides include all figures, tables, and equations from the text. ∙ C.O.S.M.O.S.  A complete online solutions manual organization system that allows instructors to create custom homework, quizzes, and tests using end-of-chapter problems from the text.

Acknowledgments The authors would like to acknowledge those who have contributed to this text for over 50 years and eleven editions. We are especially grateful to those who provided input to this eleventh edition: Steve Boedo, Rochester Institute of Technology: Review and update of Chapter 12, Lubrication and Journal Bearings. Lokesh Dharani, Missouri University of Science and Technology: Review and advice regarding the coverage of fracture mechanics and fatigue. Reviewers of This and Past Editions Kenneth Huebner, Arizona State Gloria Starns, Iowa State Tim Lee, McGill University Robert Rizza, MSOE Richard Patton, Mississippi State University Stephen Boedo, Rochester Institute of Technology

Om Agrawal, Southern Illinois University Arun Srinivasa, Texas A&M Jason Carey, University of Alberta Patrick Smolinski, University of Pittsburgh Dennis Hong, Virginia Tech

List of Symbols This is a list of common symbols used in machine design and in this book. Specialized use in a subject-matter area often attracts fore and post subscripts and superscripts. To make the table brief enough to be useful, the symbol kernels are listed. See Table 14–1 for spur and helical gearing symbols, and Table 15–1 for bevel-gear symbols. A Area, coefficient a Distance B Coefficient, bearing length Bhn Brinell hardness b Distance, fatigue strength exponent, Weibull shape parameter, width C Basic load rating, bolted-joint constant, center distance, coefficient of variation, column end condition, correction factor, specific heat capacity, spring index, radial clearance c Distance, fatigue ductility exponent, radial clearance COV Coefficient of variation D Diameter, helix diameter d Diameter, distance E Modulus of elasticity, energy, error e Distance, eccentricity, efficiency, Naperian logarithmic base F Force, fundamental dimension force f Coefficient of friction, frequency, function fom Figure of merit G Torsional modulus of elasticity g Acceleration due to gravity, function H Heat, power HB Brinell hardness HRC Rockwell C-scale hardness h Distance, film thickness hCR Combined overall coefficient of convection and radiation heat transfer I Integral, linear impulse, mass moment of inertia, second moment of area i Index i Unit vector in x-direction J Mechanical equivalent of heat, polar second moment of area, geometry factor j Unit vector in the y-direction K Service factor, stress-concentration factor, stress-augmentation factor, torque coefficient k Marin endurance limit modifying factor, spring rate k Unit vector in the z-direction L Length, life, fundamental dimension length ℒ Life in hours

xvii

xviii      Mechanical Engineering Design

l Length M Fundamental dimension mass, moment M Moment vector, mobility vector m Mass, slope, strain-strengthening exponent N Normal force, number, rotational speed, number of cycles n Load factor, rotational speed, factor of safety nd Design factor P Force, pressure, diametral pitch PDF Probability density function p Pitch, pressure, probability Q First moment of area, imaginary force, volume q Distributed load, notch sensitivity R Radius, reaction force, reliability, Rockwell hardness, stress ratio, reduction in area R Vector reaction force r Radius r Distance vector S Sommerfeld number, strength s Distance, sample standard deviation, stress T Temperature, tolerance, torque, fundamental dimension time T Torque vector t Distance, time, tolerance U Strain energy u Strain energy per unit volume V Linear velocity, shear force v Linear velocity W Cold-work factor, load, weight w Distance, gap, load intensity X Coordinate, truncated number x Coordinate, true value of a number, Weibull parameter Y Coordinate y Coordinate, deflection Z Coordinate, section modulus, viscosity z Coordinate, dimensionless transform variable for normal distributions α Coefficient, coefficient of linear thermal expansion, end-condition for springs, thread angle β Bearing angle, coefficient Δ Change, deflection δ Deviation, elongation ϵ Eccentricity ratio ε Engineering strain ε˜ True or logarithmic strain ε˜f True fracture strain ε′f Fatigue ductility coefficient Γ Gamma function, pitch angle γ Pitch angle, shear strain, specific weight λ Slenderness ratio for springs μ Absolute viscosity, population mean ν Poisson ratio ω Angular velocity, circular frequency

List of Symbols     xix

ϕ Angle, wave length ψ Slope integral ρ Radius of curvature, mass density σ Normal stress σa Alternating stress, stress amplitude σar Completely reversed alternating stress σm Mean stress σ0 Nominal stress, strength coefficient or strain-strengthening coefficient σ′f Fatigue strength coefficient σ˜ True stress σ˜f True fracture strength σ′ Von Mises stress σˆ Standard deviation τ Shear stress θ Angle, Weibull characteristic parameter ¢ Cost per unit weight $ Cost

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Shigley's Mechanical Engineering Design

E O

D C

2

2

A xy

x G

part

Courtesy of Dee Dehokenanan

Basics Chapter 1

Introduction to Mechanical Engineering Design  3

Chapter 2 Materials  41 Chapter 3 Load and Stress Analysis  93 Chapter 4 Deflection and Stiffness  173

1

1

Introduction to Mechanical Engineering Design

©Monty Rakusen/Getty Images

Chapter Outline 1–1

Design  4

1–10

Uncertainty  16

1–2

Mechanical Engineering Design   5

1–11

Design Factor and Factor of Safety   18

1–3  Phases and Interactions of the Design Process  5

1–12

Reliability and Probability of Failure   20

1–13

Relating Design Factor to Reliability   24

1–4

1–14

Dimensions and Tolerances   27

1–15

Units  31

1–16

Calculations and Significant Figures   32

1–17

Design Topic Interdependencies   33

Design Tools and Resources   8

1–5  The Design Engineer's Professional Responsibilities  10 1–6

Standards and Codes   12

1–7

Economics  13

1–8

Safety and Product Liability   15

1–9

Stress and Strength   16

1–18  Power Transmission Case Study Specifications  34 3

4      Mechanical Engineering Design

Mechanical design is a complex process, requiring many skills. Extensive relationships need to be subdivided into a series of simple tasks. The complexity of the process requires a sequence in which ideas are introduced and iterated. We first address the nature of design in general, and then mechanical engineering design in particular. Design is an iterative process with many interactive phases. Many resources exist to support the designer, including many sources of information and an abundance of computational design tools. Design engineers need not only develop competence in their field but they must also cultivate a strong sense of responsibility and professional work ethic. There are roles to be played by codes and standards, ever-present economics, safety, and considerations of product liability. The survival of a mechanical component is often related through stress and strength. Matters of uncertainty are ever-present in engineering design and are typically addressed by the design factor and factor of safety, either in the form of a deterministic (absolute) or statistical sense. The latter, statistical approach, deals with a design's reliability and requires good statistical data. In mechanical design, other considerations include dimensions and tolerances, units, and calculations. This book consists of four parts. Part 1, Basics, begins by explaining some differences between design and analysis and introducing some fundamental notions and approaches to design. It continues with three chapters reviewing material properties, stress analysis, and stiffness and deflection analysis, which are the principles necessary for the remainder of the book. Part 2, Failure Prevention, consists of two chapters on the prevention of failure of mechanical parts. Why machine parts fail and how they can be designed to prevent failure are difficult questions, and so we take two chapters to answer them, one on preventing failure due to static loads, and the other on preventing fatigue failure due to time-varying, cyclic loads. In Part 3, Design of Mechanical Elements, the concepts of Parts 1 and 2 are applied to the analysis, selection, and design of specific mechanical elements such as shafts, fasteners, weldments, springs, rolling contact bearings, film bearings, gears, belts, chains, and wire ropes. Part 4, Special Topics, provides introductions to two important methods used in mechanical design, finite element analysis and geometric dimensioning and tolerancing. This is optional study material, but some sections and examples in Parts 1 to 3 demonstrate the use of these tools. There are two appendixes at the end of the book. Appendix A contains many useful tables referenced throughout the book. Appendix B contains answers to selected end-of-chapter problems.

1–1 Design To design is either to formulate a plan for the satisfaction of a specified need or to solve a specific problem. If the plan results in the creation of something having a physical reality, then the product must be functional, safe, reliable, competitive, usable, manufacturable, and marketable. Design is an innovative and highly iterative process. It is also a decision-making process. Decisions sometimes have to be made with too little information, occasionally with just the right amount of information, or with an excess of partially contradictory information. Decisions are sometimes made tentatively, with the right reserved to

Introduction to Mechanical Engineering Design     5

adjust as more becomes known. The point is that the engineering designer has to be personally comfortable with a decision-making, problem-solving role. Design is a communication-intensive activity in which both words and pictures are used, and written and oral forms are employed. Engineers have to communicate effectively and work with people of many disciplines. These are important skills, and an engineer's success depends on them. A designer's personal resources of creativeness, communicative ability, and problemsolving skill are intertwined with the knowledge of technology and first principles. Engineering tools (such as mathematics, statistics, computers, graphics, and languages) are combined to produce a plan that, when carried out, produces a product that is functional, safe, reliable, competitive, usable, manufacturable, and marketable, regardless of who builds it or who uses it.

1–2 Mechanical Engineering Design Mechanical engineers are associated with the production and processing of energy and with providing the means of production, the tools of transportation, and the ­techniques of automation. The skill and knowledge base are extensive. Among the disciplinary bases are mechanics of solids and fluids, mass and momentum transport, manufacturing processes, and electrical and information theory. Mechanical engineering design involves all the disciplines of mechanical engineering. Real problems resist compartmentalization. A simple journal bearing involves fluid flow, heat transfer, friction, energy transport, material selection, thermomechanical treatments, statistical descriptions, and so on. A building is environmentally controlled. The heating, ventilation, and air-conditioning considerations are sufficiently specialized that some speak of heating, ventilating, and air-conditioning design as if it is separate and distinct from mechanical engineering design. Similarly, internalcombustion engine design, turbomachinery design, and jet-engine design are sometimes considered discrete entities. Here, the leading string of words preceding the word design is merely a product descriptor. Similarly, there are phrases such as machine design, machine-element design, machine-component design, systems design, and fluid-power design. All of these phrases are somewhat more focused examples of mechanical engineering design. They all draw on the same bodies of knowledge, are similarly organized, and require similar skills.

1–3  Phases and Interactions of the Design Process What is the design process? How does it begin? Does the engineer simply sit down at a desk with a blank sheet of paper and jot down some ideas? What happens next? What factors influence or control the decisions that have to be made? Finally, how does the design process end? The complete design process, from start to finish, is often outlined as in Figure 1–1. The process begins with an identification of a need and a decision to do something about it. After many iterations, the process ends with the presentation of the plans for satisfying the need. Depending on the nature of the design task, several design phases may be repeated throughout the life of the product, from inception to termination. In the next several subsections, we shall examine these steps in the design process in detail. Identification of need generally starts the design process. Recognition of the need and phrasing the need often constitute a highly creative act, because the need may be

6      Mechanical Engineering Design

Figure 1–1

Identification of need

The phases in design, acknowledging the many feedbacks and iterations.

Definition of problem

Synthesis

Analysis and optimization

Evaluation Iteration Presentation

only a vague discontent, a feeling of uneasiness, or a sensing that something is not right. The need is often not evident at all; recognition can be triggered by a particular adverse circumstance or a set of random circumstances that arises almost simultaneously. For example, the need to do something about a food-packaging machine may be indicated by the noise level, by a variation in package weight, and by slight but perceptible variations in the quality of the packaging or wrap. There is a distinct difference between the statement of the need and the definition of the problem. The definition of problem is more specific and must include all the specifications for the object that is to be designed. The specifications are the input and output quantities, the characteristics and dimensions of the space the object must occupy, and all the limitations on these quantities. We can regard the object to be designed as something in a black box. In this case we must specify the inputs and outputs of the box, together with their characteristics and limitations. The specifications define the cost, the number to be manufactured, the expected life, the range, the operating temperature, and the reliability. Specified characteristics can include the speeds, feeds, temperature limitations, maximum range, expected variations in the variables, dimensional and weight limitations, and more. There are many implied specifications that result either from the designer's particular environment or from the nature of the problem itself. The manufacturing processes that are available, together with the facilities of a certain plant, constitute restrictions on a designer's freedom, and hence are a part of the implied specifications. It may be that a small plant, for instance, does not own cold-working machinery. Knowing this, the designer might select other metal-processing methods that can be performed in the plant. The labor skills available and the competitive situation also constitute implied constraints. Anything that limits the designer's freedom of choice is a constraint. Many materials and sizes are listed in supplier's catalogs, for instance, but these are not all easily available and shortages frequently occur. Furthermore, inventory economics requires that a manufacturer stock a minimum number of materials and sizes. An example of a specification is given in Section 1–18. This example is for a case study of a power transmission that is presented throughout this text. The synthesis of a scheme connecting possible system elements is sometimes called the invention of the concept or concept design. This is the first and most important

Introduction to Mechanical Engineering Design     7

step in the synthesis task. Various schemes must be proposed, investigated, and quantified in terms of established metrics.1 As the fleshing out of the scheme progresses, analyses must be performed to assess whether the system performance is satisfactory or better, and, if satisfactory, just how well it will perform. System schemes that do not survive analysis are revised, improved, or discarded. Those with potential are optimized to determine the best performance of which the scheme is capable. Competing schemes are compared so that the path leading to the most competitive product can be chosen. Figure 1–1 shows that synthesis and analysis and optimization are intimately and iteratively related. We have noted, and we emphasize, that design is an iterative process in which we proceed through several steps, evaluate the results, and then return to an earlier phase of the procedure. Thus, we may synthesize several components of a system, analyze and optimize them, and return to synthesis to see what effect this has on the remaining parts of the system. For example, the design of a system to transmit power requires attention to the design and selection of individual components (e.g., gears, bearings, shaft). However, as is often the case in design, these components are not independent. In order to design the shaft for stress and deflection, it is necessary to know the applied forces. If the forces are transmitted through gears, it is necessary to know the gear specifications in order to determine the forces that will be transmitted to the shaft. But stock gears come with certain bore sizes, requiring knowledge of the necessary shaft diameter. Clearly, rough estimates will need to be made in order to proceed through the process, refining and iterating until a final design is obtained that is satisfactory for each individual component as well as for the overall design specifications. Throughout the text we will elaborate on this process for the case study of a power transmission design. Both analysis and optimization require that we construct or devise abstract models of the system that will admit some form of mathematical analysis. We call these models mathematical models. In creating them it is our hope that we can find one that will simulate the real physical system very well. As indicated in Figure 1–1, evaluation is a significant phase of the total design process. Evaluation is the final proof of a successful design and usually involves the testing of a prototype in the laboratory. Here we wish to discover if the design really satisfies the needs. Is it reliable? Will it compete successfully with similar products? Is it economical to manufacture and to use? Is it easily maintained and adjusted? Can a profit be made from its sale or use? How likely is it to result in product-liability lawsuits? And is insurance easily and cheaply obtained? Is it likely that recalls will be needed to replace defective parts or systems? The project designer or design team will need to address a myriad of engineering and non-engineering questions. Communicating the design to others is the final, vital presentation step in the design process. Undoubtedly, many great designs, inventions, and creative works have been lost to posterity simply because the originators were unable or unwilling to properly explain their accomplishments to others. Presentation is a selling job. The engineer, when presenting a new solution to administrative, management, or supervisory persons, is attempting to sell or to prove to them that their solution is a better one. Unless this can be done successfully, the time and effort spent on obtaining the

1

An excellent reference for this topic is presented by Stuart Pugh, Total Design—Integrated Methods for Successful Product Engineering, Addison-Wesley, 1991. A description of the Pugh method is also provided in Chapter 8, David G. Ullman, The Mechanical Design Process, 3rd ed., McGraw-Hill, New York, 2003.

8      Mechanical Engineering Design

solution have been largely wasted. When designers sell a new idea, they also sell themselves. If they are repeatedly successful in selling ideas, designs, and new solutions to management, they begin to receive salary increases and promotions; in fact, this is how anyone succeeds in his or her profession. Design Considerations Sometimes the strength required of an element in a system is an important factor in the determination of the geometry and the dimensions of the element. In such a situation we say that strength is an important design consideration. When we use the expression design consideration, we are referring to some characteristic that influences the design of the element or, perhaps, the entire system. Usually quite a number of such characteristics must be considered and prioritized in a given design situation. Many of the important ones are as follows (not necessarily in order of importance):  1 Functionality  2 Strength/stress  3 Distortion/deflection/stiffness  4 Wear  5 Corrosion  6 Safety  7 Reliability  8 Manufacturability  9 Utility 10 Cost 11 Friction 12 Weight 13 Life

14 Noise 15 Styling 16 Shape 17 Size 18 Control 19 Thermal properties 20 Surface 21 Lubrication 22 Marketability 23 Maintenance 24 Volume 25 Liability 26 Remanufacturing/resource recovery

Some of these characteristics have to do directly with the dimensions, the material, the processing, and the joining of the elements of the system. Several characteristics may be interrelated, which affects the configuration of the total system.

1–4  Design Tools and Resources Today, the engineer has a great variety of tools and resources available to assist in the solution of design problems. Inexpensive microcomputers and robust computer software packages provide tools of immense capability for the design, analysis, and simulation of mechanical components. In addition to these tools, the engineer always needs technical information, either in the form of basic science/engineering behavior or the characteristics of specific off-the-shelf components. Here, the resources can range from science/engineering textbooks to manufacturers' brochures or catalogs. Here too, the computer can play a major role in gathering information.2 Computational Tools Computer-aided design (CAD) software allows the development of three-dimensional (3-D) designs from which conventional two-dimensional orthographic views with automatic dimensioning can be produced. Manufacturing tool paths can be generated 2

An excellent and comprehensive discussion of the process of "gathering information" can be found in Chapter 4, George E. Dieter, Engineering Design, A Materials and Processing Approach, 3rd ed., McGraw-Hill, New York, 2000.

Introduction to Mechanical Engineering Design     9

from the computer 3-D models, and in many cases, parts can be created directly from the 3-D database using rapid prototyping additive methods referred to as 3-D printing or STL (stereolithography). Another advantage of a 3-D database is that it allows rapid and accurate calculation of mass properties such as mass, location of the center of gravity, and mass moments of inertia. Other geometric properties such as areas and distances between points are likewise easily obtained. There are a great many CAD software packages available such as CATIA, AutoCAD, NX, MicroStation, SolidWorks, and Creo, to name only a few.3 The term computer-aided engineering (CAE) generally applies to all computerrelated engineering applications. With this definition, CAD can be considered as a subset of CAE. Some computer software packages perform specific engineering analysis and/or simulation tasks that assist the designer, but they are not considered a tool for the creation of the design that CAD is. Such software fits into two categories: engineering-based and non-engineering-specific. Some examples of engineering-based software for mechanical engineering applications—software that might also be integrated within a CAD system—include finite-element analysis (FEA) programs for analysis of stress and deflection (see Chapter 19), vibration, and heat transfer (e.g., ALGOR, ANSYS, MSC/NASTRAN, etc.); computational fluid dynamics (CFD) programs for fluid-flow analysis and simulation (e.g., CFD++, Star-CCM+, Fluent, etc.); and programs for simulation of dynamic force and motion in mechanisms (e.g., ADAMS, LMS Virtual.Lab Motion, Working Model, etc.). Examples of non-engineering-specific computer-aided applications include software for word processing, spreadsheet software (e.g., Excel, Quattro-Pro, Google Sheets, etc.), and mathematical solvers (e.g., Maple, MathCad, MATLAB, Mathematica, TKsolver, etc.). Your instructor is the best source of information about programs that may be available to you and can recommend those that are useful for specific tasks. One caution, however: Computer software is no substitute for the human thought process. You are the driver here; the computer is the vehicle to assist you on your journey to a solution. Numbers generated by a computer can be far from the truth if you entered incorrect input, if you misinterpreted the application or the output of the program, if the program contained bugs, etc. It is your responsibility to assure the validity of the results, so be careful to check the application and results carefully, perform benchmark testing by submitting problems with known solutions, and monitor the software company and user-group newsletters. Acquiring Technical Information We currently live in what is referred to as the information age, where information is generated at an astounding pace. It is difficult, but extremely important, to keep abreast of past and current developments in one's field of study and occupation. The reference in footnote 2 provides an excellent description of the informational resources available and is highly recommended reading for the serious design engineer. Some sources of information are: ∙ Libraries (community, university, and private). Engineering dictionaries and encyclopedias, textbooks, monographs, handbooks, indexing and abstract services, journals, translations, technical reports, patents, and business sources/brochures/catalogs. 3

The commercial softwares mentioned in this section are but a few of the many that are available and are by no means meant to be endorsements by the authors.

10      Mechanical Engineering Design

∙ Government sources. Departments of Defense, Commerce, Energy, and Transportation; NASA; Government Printing Office; U.S. Patent and Trademark Office; National Technical Information Service; and National Institute for Standards and Technology. ∙ Professional societies. American Society of Mechanical Engineers, Society of Manufacturing Engineers, Society of Automotive Engineers, American Society for Testing and Materials, and American Welding Society. ∙ Commercial vendors. Catalogs, technical literature, test data, samples, and cost information. ∙ Internet. The computer network gateway to websites associated with most of the categories previously listed.4 This list is not complete. The reader is urged to explore the various sources of information on a regular basis and keep records of the knowledge gained.

1–5  The Design Engineer's Professional Responsibilities In general, the design engineer is required to satisfy the needs of customers (management, clients, consumers, etc.) and is expected to do so in a competent, responsible, ethical, and professional manner. Much of engineering course work and practical experience focuses on competence, but when does one begin to develop engineering responsibility and professionalism? To start on the road to success, you should start to develop these characteristics early in your educational program. You need to cultivate your professional work ethic and process skills before graduation, so that when you begin your formal engineering career, you will be prepared to meet the challenges. It is not obvious to some students, but communication skills play a large role here, and it is the wise student who continuously works to improve these skills—even if it is not a direct requirement of a course assignment! Success in engineering (achievements, promotions, raises, etc.) may in large part be due to competence but if you cannot communicate your ideas clearly and concisely, your technical proficiency may be compromised. You can start to develop your communication skills by keeping a neat and clear journal/logbook of your activities, entering dated entries frequently. (Many companies require their engineers to keep a journal for patent and liability concerns.) Separate journals should be used for each design project (or course subject). When starting a project or problem, in the definition stage, make journal entries quite frequently. Others, as well as yourself, may later question why you made certain decisions. Good chronological records will make it easier to explain your decisions at a later date. Many engineering students see themselves after graduation as practicing engineers designing, developing, and analyzing products and processes and consider the need of good communication skills, either oral or writing, as secondary. This is far from the truth. Most practicing engineers spend a good deal of time communicating with others, writing proposals and technical reports, and giving presentations and interacting with engineering and non-engineering support personnel. You have the time now to sharpen your communication skills. When given an assignment to write or 4

Some helpful Web resources, to name a few, include www.globalspec.com, www.engnetglobal.com, www.efunda.com, www.thomasnet.com, and www.uspto.gov.

Introduction to Mechanical Engineering Design     11

make any presentation, technical or nontechnical, accept it enthusiastically, and work on improving your communication skills. It will be time well spent to learn the skills now rather than on the job. When you are working on a design problem, it is important that you develop a systematic approach. Careful attention to the following action steps will help you to organize your solution processing technique. ∙ Understand the problem. Problem definition is probably the most significant step in the engineering design process. Carefully read, understand, and refine the problem statement. ∙ Identify the knowns. From the refined problem statement, describe concisely what information is known and relevant. ∙ Identify the unknowns and formulate the solution strategy. State what must be determined, in what order, so as to arrive at a solution to the problem. Sketch the component or system under investigation, identifying known and unknown parameters. Create a flowchart of the steps necessary to reach the final solution. The steps may require the use of free-body diagrams; material properties from tables; equations from first principles, textbooks, or handbooks relating the known and unknown parameters; experimentally or numerically based charts; specific computational tools as discussed in Section 1–4; etc. ∙ State all assumptions and decisions. Real design problems generally do not have unique, ideal, closed-form solutions. Selections, such as the choice of materials, and heat treatments, require decisions. Analyses require assumptions related to the modeling of the real components or system. All assumptions and decisions should be identified and recorded. ∙ Analyze the problem. Using your solution strategy in conjunction with your decisions and assumptions, execute the analysis of the problem. Reference the sources of all equations, tables, charts, software results, etc. Check the credibility of your results. Check the order of magnitude, dimensionality, trends, signs, etc. ∙ Evaluate your solution. Evaluate each step in the solution, noting how changes in strategy, decisions, assumptions, and execution might change the results, in positive or negative ways. Whenever possible, incorporate the positive changes in your final solution. ∙ Present your solution. Here is where your communication skills are important. At this point, you are selling yourself and your technical abilities. If you cannot skillfully explain what you have done, some or all of your work may be misunderstood and unaccepted. Know your audience. As stated earlier, all design processes are interactive and iterative. Thus, it may be necessary to repeat some or all of the aforementioned steps more than once if less than satisfactory results are obtained. In order to be effective, all professionals must keep current in their fields of endeavor. The design engineer can satisfy this in a number of ways by: being an active member of a professional society such as the American Society of Mechanical Engineers (ASME), the Society of Automotive Engineers (SAE), and the Society of Manufacturing Engineers (SME); attending meetings, conferences, and seminars of societies, manufacturers, universities, etc.; taking specific graduate courses or programs at universities; regularly reading technical and professional journals; etc. An engineer's education does not end at graduation.

12      Mechanical Engineering Design

The design engineer's professional obligations include conducting activities in an ethical manner. Reproduced here is the Engineers' Creed from the National Society of Professional Engineers (NSPE):5 As a Professional Engineer I dedicate my professional knowledge and skill to the advancement and betterment of human welfare. I pledge: To give the utmost of performance; To participate in none but honest enterprise; To live and work according to the laws of man and the highest standards of professional conduct; To place service before profit, the honor and standing of the profession before personal advantage, and the public welfare above all other considerations. In humility and with need for Divine Guidance, I make this pledge.

1–6  Standards and Codes A standard is a set of specifications for parts, materials, or processes intended to achieve uniformity, efficiency, and a specified quality. One of the important purposes of a standard is to limit the multitude of variations that can arise from the arbitrary creation of a part, material, or process. A code is a set of specifications for the analysis, design, manufacture, and construction of something. The purpose of a code is to achieve a specified degree of safety, efficiency, and performance or quality. It is important to observe that safety codes do not imply absolute safety. In fact, absolute safety is impossible to obtain. Sometimes the unexpected event really does happen. Designing a building to withstand a 120 mi/h wind does not mean that the designers think a 140 mi/h wind is impossible; it simply means that they think it is highly improbable. All of the organizations and societies listed here have established specifications for standards and safety or design codes. The name of the organization provides a clue to the nature of the standard or code. Some of the standards and codes, as well as addresses, can be obtained in most technical libraries or on the Internet. The organizations of interest to mechanical engineers are: Aluminum Association (AA) American Bearing Manufacturers Association (ABMA) American Gear Manufacturers Association (AGMA) American Institute of Steel Construction (AISC) American Iron and Steel Institute (AISI) American National Standards Institute (ANSI) American Society of Heating, Refrigerating and Air-Conditioning Engineers (ASHRAE) American Society of Mechanical Engineers (ASME) American Society of Testing and Materials (ASTM) American Welding Society (AWS) 5

Adopted by the National Society of Professional Engineers, June 1954. "The Engineer's Creed." Reprinted by permission of the National Society of Professional Engineers. NSPE also publishes a much more extensive Code of Ethics for Engineers with rules of practice and professional obligations. For the current revision, July 2007 (at the time of this book's printing), see the website www.nspe.org/Ethics/CodeofEthics/index.html.

Introduction to Mechanical Engineering Design     13

ASM International British Standards Institution (BSI) Industrial Fasteners Institute (IFI) Institute of Transportation Engineers (ITE) Institution of Mechanical Engineers (IMechE) International Bureau of Weights and Measures (BIPM) International Federation of Robotics (IFR) International Standards Organization (ISO) National Association of Power Engineers (NAPE) National Institute for Standards and Technology (NIST) Society of Automotive Engineers (SAE)

1–7 Economics The consideration of cost plays such an important role in the design decision process that we could easily spend as much time in studying the cost factor as in the study of the entire subject of design. Here we introduce only a few general concepts and simple rules. First, observe that nothing can be said in an absolute sense concerning costs. Materials and labor usually show an increasing cost from year to year. But the costs of processing the materials can be expected to exhibit a decreasing trend because of the use of automated machine tools and robots. The cost of manufacturing a single product will vary from city to city and from one plant to another because of overhead, labor, taxes, and freight differentials and the inevitable slight manufacturing variations. Standard Sizes The use of standard or stock sizes is a first principle of cost reduction. An engineer who specifies an AISI 1020 bar of hot-rolled steel 53 mm square has added cost to the product, provided that a bar 50 or 60 mm square, both of which are preferred sizes, would do equally well. The 53-mm size can be obtained by special order or by rolling or machining a 60-mm square, but these approaches add cost to the product. To ensure that standard or preferred sizes are specified, designers must have access to stock lists of the materials they employ. A further word of caution regarding the selection of preferred sizes is necessary. Although a great many sizes are usually listed in catalogs, they are not all readily available. Some sizes are used so infrequently that they are not stocked. A rush order for such sizes may add to the expense and delay. Thus you should also have access to a list such as those in Table A–17 for preferred inch and millimeter sizes. There are many purchased parts, such as motors, pumps, bearings, and fasteners, that are specified by designers. In the case of these, too, you should make a special effort to specify parts that are readily available. Parts that are made and sold in large quantities usually cost somewhat less than the odd sizes. The cost of rolling bearings, for example, depends more on the quantity of production by the bearing manufacturer than on the size of the bearing. Large Tolerances Among the effects of design specifications on costs, tolerances are perhaps most significant. Tolerances, manufacturing processes, and surface finish are interrelated and influence the producibility of the end product in many ways. Close tolerances

14      Mechanical Engineering Design

Figure 1–2

Costs, %

Cost versus tolerance/machining process. (Source: From Ullman, David G., The Mechanical Design Process, 3rd ed., McGraw-Hill, New York, 2003.)

400 380 360 340 320 300 280 260 240 220 200 180 160 140 120 100 80 60 40 20 ± 0.030

Material: steel

± 0.015

± 0.010

± 0.005

± 0.003

± 0.001

± 0.0005 ± 0.00025

± 0.063

± 0.025

± 0.012

± 0.006

Semifinish turn

Finish turn

Grind

Hone

Nominal tolerances (inches) ± 0.75

± 0.50

± 0.50

± 0.125

Nominal tolerance (mm) Rough turn

Machining operations

may necessitate additional steps in processing and inspection or even render a part completely impractical to produce economically. Tolerances cover dimensional variation and surface-roughness range and also the variation in mechanical properties resulting from heat treatment and other processing operations. Because parts having large tolerances can often be produced by machines with higher production rates, costs will be significantly smaller. Also, fewer such parts will be rejected in the inspection process, and they are usually easier to assemble. A plot of cost versus tolerance/machining process is shown in Figure 1–2, and illustrates the drastic increase in manufacturing cost as tolerance diminishes with finer machining processing. Breakeven Points Sometimes it happens that, when two or more design approaches are compared for cost, the choice between the two depends on a set of conditions such as the quantity of production, the speed of the assembly lines, or some other condition. There then occurs a point corresponding to equal cost, which is called the breakeven point. As an example, consider a situation in which a certain part can be manufactured at the rate of 25 parts per hour on an automatic screw machine or 10 parts per hour on a hand screw machine. Let us suppose, too, that the setup time for the automatic is 3 h and that the labor cost for either machine is $20 per hour, including overhead. Figure 1–3 is a graph of cost versus production by the two methods. The breakeven point for this example corresponds to 50 parts. If the desired production is greater than 50 parts, the automatic machine should be used.

Introduction to Mechanical Engineering Design     15

Figure 1–3

140

100 Cost, $

A breakeven point.

Breakeven point

120

Automatic screw machine

80 60 Hand screw machine

40 20 0

0

20

40

60 Production

80

100

Cost Estimates There are many ways of obtaining relative cost figures so that two or more designs can be roughly compared. A certain amount of judgment may be required in some instances. For example, we can compare the relative value of two automobiles by comparing the dollar cost per pound of weight. Another way to compare the cost of one design with another is simply to count the number of parts. The design having the smaller number of parts is likely to cost less. Many other cost estimators can be used, depending upon the application, such as area, volume, horsepower, torque, capacity, speed, and various performance ratios.6

1–8  Safety and Product Liability The strict liability concept of product liability generally prevails in the United States. This concept states that the manufacturer of an article is liable for any damage or harm that results because of a defect. And it doesn't matter whether the manufacturer knew about the defect, or even could have known about it. For example, suppose an article was manufactured, say, 10 years ago. And suppose at that time the article could not have been considered defective on the basis of all technological knowledge then available. Ten years later, according to the concept of strict liability, the manufacturer is still liable. Thus, under this concept, the plaintiff needs only to prove that the article was defective and that the defect caused some damage or harm. Negligence of the manufacturer need not be proved. The best approaches to the prevention of product liability are good engineering in analysis and design, quality control, and comprehensive testing procedures. Advertising managers often make glowing promises in the warranties and sales literature for a product. These statements should be reviewed carefully by the engineering staff to eliminate excessive promises and to insert adequate warnings and instructions for use.

6

For an overview of estimating manufacturing costs, see Chapter 11, Karl T. Ulrich and Steven D. Eppinger, Product Design and Development, 3rd ed., McGraw-Hill, New York, 2004.

16      Mechanical Engineering Design

1–9  Stress and Strength The survival of many products depends on how the designer adjusts the maximum stresses in a component to be less than the component's strength at critical locations. The designer must allow the maximum stress to be less than the strength by a sufficient margin so that despite the uncertainties, failure is rare. In focusing on the stress-strength comparison at a critical (controlling) location, we often look for "strength in the geometry and condition of use." Strengths are the magnitudes of stresses at which something of interest occurs, such as the proportional limit, 0.2 percent-offset yielding, or fracture (see Section 2–1). In many cases, such events represent the stress level at which loss of function occurs. Strength is a property of a material or of a mechanical element. The strength of an element depends on the choice, the treatment, and the processing of the material. Consider, for example, a shipment of springs. We can associate a strength with a specific spring. When this spring is incorporated into a machine, external forces are applied that result in load-induced stresses in the spring, the magnitudes of which depend on its geometry and are independent of the material and its processing. If the spring is removed from the machine undamaged, the stress due to the external forces will return to zero. But the strength remains as one of the properties of the spring. Remember, then, that strength is an inherent property of a part, a property built into the part because of the use of a particular material and process. Various metalworking and heat-treating processes, such as forging, rolling, and cold forming, cause variations in the strength from point to point throughout a part. The spring cited previously is quite likely to have a strength on the outside of the coils different from its strength on the inside because the spring has been formed by a cold winding process, and the two sides may not have been deformed by the same amount. Remember, too, therefore, that a strength value given for a part may apply to only a particular point or set of points on the part. In this book we shall use the capital letter S to denote strength, with appropriate subscripts to denote the type of strength. Thus, Sy is a yield strength, Su an ultimate strength, Ssy a shear yield strength, and Se an endurance strength. In accordance with accepted engineering practice, we shall employ the Greek letters σ (sigma) and τ (tau) to designate normal and shear stresses, respectively. Again, various subscripts will indicate some special characteristic. For example, σ1 is a principal normal stress, σy a normal stress component in the y direction, and σr a normal stress component in the radial direction. Stress is a state property at a specific point within a body, which is a function of load, geometry, temperature, and manufacturing processing. In an elementary course in mechanics of materials, stress related to load and geometry is emphasized with some discussion of thermal stresses. However, stresses due to heat treatments, molding, assembly, etc. are also important and are sometimes neglected. A review of stress analysis for basic load states and geometry is given in Chapter 3.

1–10 Uncertainty Uncertainties in machinery design abound. Examples of uncertainties concerning stress and strength include ∙ Composition of material and the effect of variation on properties. ∙ Variations in properties from place to place within a bar of stock. ∙ Effect of processing locally, or nearby, on properties.

Introduction to Mechanical Engineering Design     17

∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙

Effect of nearby assemblies such as weldments and shrink fits on stress conditions. Effect of thermomechanical treatment on properties. Intensity and distribution of loading. Validity of mathematical models used to represent reality. Intensity of stress concentrations. Influence of time on strength and geometry. Effect of corrosion. Effect of wear. Uncertainty as to the length of any list of uncertainties.

Engineers must accommodate uncertainty. Uncertainty always accompanies change. Material properties, load variability, fabrication fidelity, and validity of mathematical models are among concerns to designers. There are mathematical methods to address uncertainties. The primary techniques are the deterministic and stochastic methods. The deterministic method establishes a design factor based on the absolute uncertainties of a loss-of-function parameter and a maximum allowable parameter. Here the parameter can be load, stress, deflection, etc. Thus, the design factor nd is defined as nd =

loss-of-function parameter maximum allowable parameter

(1–1)

If the parameter is load (as would be the case for column buckling), then the maximum allowable load can be found from

Maximum allowable load =

loss-of-function load nd

(1–2)

EXAMPLE 1–1 Consider that the maximum load on a structure is known with an uncertainty of ±20 percent, and the load causing failure is known within ±15 percent. If the load causing failure is nominally 2000 lbf, determine the design factor and the maximum allowable load that will offset the absolute uncertainties. Solution To account for its uncertainty, the loss-of-function load must increase to 1∕0.85, whereas the maximum allowable load must decrease to 1∕1.2. Thus to offset the absolute uncertainties the design factor, from Equation (1–1), should be Answer

nd =

1∕0.85 = 1.4 1∕1.2

From Equation (1–2), the maximum allowable load is found to be Answer

Maximum allowable load =

2000 = 1400 lbf 1.4

Stochastic methods are based on the statistical nature of the design parameters and focus on the probability of survival of the design's function (that is, on reliability). This is discussed further in Sections 1–12 and 1–13.

18      Mechanical Engineering Design

1–11  Design Factor and Factor of Safety A general approach to the allowable load versus loss-of-function load problem is the deterministic design factor method, and sometimes called the classical method of design. The fundamental equation is Equation (1–1) where nd is called the design factor. All loss-of-function modes must be analyzed, and the mode leading to the smallest design factor governs. After the design is completed, the actual design factor may change as a result of changes such as rounding up to a standard size for a cross section or using offthe-shelf components with higher ratings instead of employing what is calculated by using the design factor. The factor is then referred to as the factor of safety, n. The factor of safety has the same definition as the design factor, but it generally differs numerically. Because stress may not vary linearly with load (see Section 3–19), using load as the loss-of-function parameter may not be acceptable. It is more common then to express the design factor in terms of a stress and a relevant strength. Thus Equation (1–1) can be rewritten as loss-of-function strength S nd = = (1–3) allowable stress σ(or τ) The stress and strength terms in Equation (1–3) must be of the same type and units. Also, the stress and strength must apply to the same critical location in the part. EXAMPLE 1–2 A rod with a cross-sectional area of A and loaded in tension with an axial force of P = 2000 lbf undergoes a stress of σ = P∕A. Using a material strength of 24 kpsi and a design factor of 3.0, determine the minimum diameter of a solid circular rod. Using Table A–17, select a preferred fractional diameter and determine the rod's factor of safety. Solution Since A = πd 2∕4, σ = P∕A, and from Equation (1–3), σ = S∕nd, then

σ=

Solving for d yields d=(

Answer

4Pnd 1∕2 4(2000)3 1∕2 = ( π(24 000) ) = 0.564 in πS )

From Table A–17, the next higher preferred size is equation developed above, the factor of safety n is n=

Answer

P P S = 2 = n A πd ∕4 d

5 8

in = 0.625 in. Thus, when nd is replaced with n in the

πSd 2 π(24 000)0.6252 = = 3.68 4P 4(2000)

Thus, rounding the diameter has increased the actual design factor. It is tempting to offer some recommendations concerning the assignment of the design factor for a given application.7 The problem in doing so is with the evaluation 7

If the reader desires some examples of assigning design factor values see David G. Ullman, The Mechanical Design Process, 4th ed., McGraw-Hill, New York, 2010, Appendix C.

Introduction to Mechanical Engineering Design     19

of the many uncertainties associated with the loss-of-function modes. The reality is, the designer must attempt to account for the variance of all the factors that will affect the results. Then, the designer must rely on experience, company policies, and the many codes that may pertain to the application (e.g., the ASME Boiler and Pressure Vessel Code) to arrive at an appropriate design factor. An example might help clarify the intricacy of assigning a design factor.

EXAMPLE 1–3 A vertical round rod is to be used to support a hanging weight. A person will place the weight on the end without dropping it. The diameter of the rod can be manufactured within ±1 percent of its nominal dimension. The support ends can be centered within ±1.5 percent of the nominal diameter dimension. The weight is known within ±2 percent of the nominal weight. The strength of the material is known within ±3.5 percent of the nominal strength value. If the designer is using nominal values and the nominal stress equation, σnom = P∕A (as in the previous example), determine what design factor should be used so that the stress does not exceed the strength. Solution There are two hidden factors to consider here. The first, due to the possibility of eccentric loading, the maximum stress is not σ = P∕A (review Chapter 3). Second, the person may not be placing the weight onto the rod support end gradually, and the load application would then be considered dynamic. Consider the eccentricity first. With eccentricity, a bending moment will exist giving an additional stress of σ = 32M∕(πd 3) (see Section 3–10). The bending moment is given by M = Pe, where e is the eccentricity. Thus, the maximum stress in the rod is given by

σ=

P 32Pe P 32Pe + = 2 + A πd 3 πd ∕4 πd 3

(1)

Since the eccentricity tolerance is expressed as a function of the diameter, we will write the eccentricity as a percentage of d. Let e = ked, where ke is a constant. Thus, Equation (1) is rewritten as

σ=

32Pked 4P 4P + = 2 (1 + 8ke ) 2 3 πd πd πd

(2)

Applying the tolerances to achieve the maximum the stress can reach gives

σmax =

4P(1 + 0.02)

π[d(1 − 0.01)] = 1.166σnom

2

[1 + 8(0.015) ] = 1.166 (

4P πd 2 )

(3)

Suddenly applied loading is covered in Section 4–17. If a weight is dropped from a height, h, from the support end, the maximum load in the rod is given by Equation (4–59) which is

hk 1∕2 F = W + W (1 + ) W

where F is the force in the rod, W is the weight, and k is the rod's spring constant. Since the person is not dropping the weight, h = 0, and with W = P, then F = 2P. This assumes the person is not gradually placing the weight on, and there is no damping in the rod. Thus, Equation (3) is modified by substituting 2P for P and the maximum stress is

σmax = 2(1.166) σnom = 2.332 σnom

20      Mechanical Engineering Design

The minimum strength is

Smin = (1 − 0.035) Snom = 0.965 Snom

Equating the maximum stress to the minimum strength gives

2.332 σnom = 0.965 Snom

From Equation (1–3), the design factor using nominal values should be nd =

Answer

Snom 2.332 = = 2.42 σnom 0.965

Obviously, if the designer takes into account all of the uncertainties in this example and accounts for all of the tolerances in the stress and strength in the calculations, a design factor of one would suffice. However, in practice, the designer would probably use the nominal geometric and strength values with the simple σ = P∕A calculation. The designer would probably not go through the calculations given in the example and would assign a design factor. This is where the experience factor comes in. The designer should make a list of the loss-of-function modes and estimate a factor, ni, for each. For this example, the list would be Loss-of-Function

Estimated Accuracy

ni

Geometry dimensions

Good tolerances

1.05

Stress calculation   Dynamic load   Bending

Not gradual loading Slight possibility

2.0* 1.1

Well known

1.05

Strength data *Minimum

Each term directly affects the results. Therefore, for an estimate, we evaluate the product of each term

nd = ∏ ni = 1.05(2.0) (1.1)(1.05) = 2.43

1–12  Reliability and Probability of Failure In these days of greatly increasing numbers of liability lawsuits and the need to conform to regulations issued by governmental agencies such as EPA and OSHA, it is very important for the designer and the manufacturer to know the reliability of their product. The reliability method of design is one in which we obtain the distribution of stresses and the distribution of strengths and then relate these two in order to achieve an acceptable success rate. The statistical measure of the probability that a mechanical element will not fail in use is called the reliability of that element and as we will see, is related to the probability of failure, pf. Probability of Failure The probability of failure, pf, is obtained from the probability density function (PDF), which represents the distribution of events within a given range of values. A number of standard discrete and continuous probability distributions are commonly applicable to engineering problems. The two most important continuous probability distributions

Introduction to Mechanical Engineering Design     21 f (x)

Figure 1–4

f(x)

The shape of the normal distribution curve: (a) small σˆ; (b) large σˆ.

x

μ

x

μ (b)

(a)

Figure 1–5 f(z)

Transformed normal distribution function of Table A–10.

Φ(zα) α 0

z

for our use in this text are the Gaussian (normal) distribution and the Weibull distribution. We will describe the normal distribution in this section and in Section 2–2. The Weibull distribution is widely used in rolling-contact bearing design and will be described in Chapter 11. The continuous Gaussian (normal) distribution is an important one whose probability density function (PDF) is expressed in terms of its mean, μx, and its standard deviation8 σˆx as

f(x) =

1 x − μx 2 exp[ − ( 2 σˆx ) ] σˆx √2π 1

(1–4)

Plots of Equation (1–4) are shown in Figure 1–4 for small and large standard deviations. The bell-shaped curve is taller and narrower for small values of σˆ and shorter and broader for large values of σˆ. Note that the area under each curve is unity. That is, the probability of all events occurring is one (100 percent). To obtain values of pf , integration of Equation (1–4) is necessary. This can come easily from a table if the variable x is placed in dimensionless form. This is done using the transform

z=

x − μx σˆx

(1–5)

The integral of the transformed normal distribution is tabulated in Table A–10, where α is defined, and is shown in Figure 1–5. The value of the normal density function is used so often, and manipulated in so many equations, that it has its own particular symbol, Φ(z). The transform variant z has a mean value of zero and a standard deviation of unity. In Table A–10, the probability of an observation less than z is Φ(z) for negative values of z and 1 − Φ(z) for positive values of z. 8

The symbol σ is normally used for the standard deviation. However, in this text σ is used for stress. Consequently, we will use σˆ for the standard deviation.

22      Mechanical Engineering Design

EXAMPLE 1–4 In a shipment of 250 connecting rods, the mean tensile strength is found to be S = 45 kpsi and has a standard deviation of σˆS = 5 kpsi. (a) Assuming a normal distribution, how many rods can be expected to have a strength less than S = 39.5 kpsi? (b) How many are expected to have a strength between 39.5 and 59.5 kpsi? Solution (a) Substituting in Equation (1–5) gives the transform z variable as

z39.5 =

x − μx S − S 39.5 − 45 = = = −1.10 σˆx σˆS 5

The probability that the strength is less than 39.5 kpsi can be designated as F(z) = Φ(z39.5) = Φ(−1.10). Using Table A–10, and referring to Figure 1–6, we find Φ(z39.5) = 0.1357. So the number of rods having a strength less than 39.5 kpsi is, NΦ(z39.5 ) = 250(0.1357) = 33.9 ≈ 34 rods

Answer

because Φ(z39.5) represents the proportion of the population N having a strength less than 39.5 kpsi.

Figure 1–6 f (z)

–1.1 z39.5

0

+2.9 z59.5

z

(b) Corresponding to S = 59.5 kpsi, we have

z59.5 =

59.5 − 45 = 2.90 5

Referring again to Figure 1–6, we see that the probability that the strength is less than 59.5 kpsi is F(z) = Φ(z59.5) = Φ(2.90). Because the z variable is positive, we need to find the value complementary to unity. Thus, from Table A–10

Φ(2.90) = 1 − Φ(−2.90) = 1 − 0.001 87 = 0.998 13

The probability that the strength lies between 39.5 and 59.5 kpsi is the area between the ordinates at z39.5 and z59.5 in Figure 1–6. This probability is found to be

p = Φ(z59.5 ) − Φ(z39.5 ) = Φ(2.90) − Φ(−1.10) = 0.998 13 − 0.1357 = 0.862 43

Therefore the number of rods expected to have strengths between 39.5 and 59.5 kpsi is Answer

Np = 250(0.862) = 215.5 ≈ 216 rods

Introduction to Mechanical Engineering Design     23

Events typically arise as discrete distributions, which can be approximated by continuous distributions. Consider N samples of events. Let xi be the value of an event (i = 1, 2, . . . k) and fi is the class frequency or number of times the event xi occurs within the class frequency range. The discrete mean, x, and standard deviation, defined as sx, are given by

1 k ∑ fi xi N i=1

x=

(1–6)

k

∑ fi x 2i − N x 2 i=1

sx =

N−1

(1–7)

EXAMPLE 1–5 Five tons of 2-in round rods of 1030 hot-rolled steel have been received for workpiece stock. Nine standardgeometry tensile test specimens have been machined from random locations in various rods. In the test report, the ultimate tensile strength was given in kpsi. The data in the ranges 62–65, 65–68, 68–71, and 71–74 kpsi is given in histographic form as follows: Sut (kpsi)

63.5 66.5 69.5 72.5 2 2 3 2

f

where the values of Sut are the midpoints of each range. Find the mean and standard deviation of the data. Solution Table 1–1 provides a tabulation of the calculations for the solution. Table 1–1 Class Midpoint x, kpsi

Class Frequency f

63.5

 2

127

8 064.50

66.5

 2

133

8 844.50

69.5

 3

208.5

14 480.75

72.5

 2

145

10 513.50

613.5

41 912.25

Σ  9

Extension fx

f x2

From Equation (1–6), Answer

x=

From Equation (1–7),

1 k 1 ∑ fi xi = (613.5) = 68.16667 = 68.2 kpsi N i=1 9 k

Answer

sx =

∑ fi x2i − N x2 i=1

N−1 = 3.39 kpsi

=√

41 912.25 − 9(68.166672 ) 9−1

24      Mechanical Engineering Design

Reliability The reliability R can be expressed by

R = 1 − pf

(1–8)

where pf is the probability of failure, given by the number of instances of failures per total number of possible instances. The value of R falls in the range 0 ≤ R ≤ 1. A reliability of R = 0.90 means there is a 90 percent chance that the part will perform its proper function without failure. The failure of 6 parts out of every 1000 manufactured, pf = 6∕1000, might be considered an acceptable failure rate for a certain class of products. This represents a reliability of R = 1 − 6∕1000 = 0.994 or 99.4 percent. In the reliability method of design, the designer's task is to make a judicious selection of materials, processes, and geometry (size) so as to achieve a specific reliability goal. Thus, if the objective reliability is to be 99.4 percent, as shown, what combination of materials, processing, and dimensions is needed to meet this goal? If a mechanical system fails when any one component fails, the system is said to be a series system. If the reliability of component i is Ri in a series system of n components, then the reliability of the system is given by n

R = ∏ Ri

(1–9)

i=1

For example, consider a shaft with two bearings having reliabilities of 95 percent and 98 percent. From Equation (1–9), the overall reliability of the shaft system is then R = R1R2 = 0.95(0.98) = 0.93 or 93 percent. Analyses that lead to an assessment of reliability address uncertainties, or their estimates, in parameters that describe the situation. Stochastic variables such as stress, strength, load, or size are described in terms of their means, standard deviations, and distributions. If bearing balls are produced by a manufacturing process in which a diameter distribution is created, we can say upon choosing a ball that there is uncertainty as to size. If we wish to consider weight or moment of inertia in rolling, this size uncertainty can be considered to be propagated to our knowledge of weight or inertia. There are ways of estimating the statistical parameters describing weight and inertia from those describing size and density. These methods are variously called propagation of error, propagation of uncertainty, or propagation of dispersion. These methods are integral parts of analysis or synthesis tasks when probability of failure is involved. It is important to note that good statistical data and estimates are essential to perform an acceptable reliability analysis. This requires a good deal of testing and validation of the data. In many cases, this is not practical and a deterministic approach to the design must be undertaken.

1–13  Relating Design Factor to Reliability Reliability is the statistical probability that machine systems and components will perform their intended function satisfactorily without failure. Stress and strength are statistical in nature and very much tied to the reliability of the stressed component. Consider the probability density functions for stress and strength, σ and S, shown in

Introduction to Mechanical Engineering Design     25

f (s), f (σ)

Figure 1–7 σ

Plots of density functions showing how the interference of S and σ is used to explain the stress margin m. (a) Stress and strength distributions. (b) Distribution of interference; the reliability R is the area of the density function for m > 0; the interference is the area (1 − R).

S

μs

μσ Stress

(a)

f (m)

m (1 – R) R –∞

+∞ μm 0 Stress margin (b)

Figure 1–7a. The mean values of stress and strength are σ = μσ and S = μS , respectively. Here, the "average" design factor is

nd =

μS μσ

(a)

The margin of safety for any value of stress σ and strength S is defined as

m = S − σ

(b)

The average of the margin of safety is m = μS − μσ . However, for the overlap of the distributions shown by the shaded area in Figure 1–7a, the stress exceeds the strength. Here, the margin of safety is negative, and these parts are expected to fail. This shaded area is called the interference of σ and S. Figure 1–7b shows the distribution of m, which obviously depends on the distributions of stress and strength. The reliability that a part will perform without failure, R, is the area of the margin of safety distribution for m > 0. The interference is the area, 1 − R, where parts are expected to fail. Assuming that σ and S each have a normal distribution, the stress margin m will also have a normal distribution. Reliability is the probability p that m > 0. That is,

R = p(S > σ) = p(S − σ > 0) = p(m > 0)

(c)

To find the probability that m > 0, we form the z variable of m and substitute m = 0. Noting that μm = μS − μσ, and9 σˆm = (σˆS2 + σˆσ2) 1∕2 , use Equation (1–5) to write

9

z=

m − μm 0 − μm μm μS − μσ = =− =− 2 σˆm σˆm σˆm (σˆS + σˆσ2) 1∕2

(1–10)

Note: If a and b are normal distributions, and c = a ± b, then c is a normal distribution with a mean of μc = μa ± μb, and a standard deviation of σˆc = (σˆa2 + σˆb2) 1∕2 . Tabular results for means and standard deviations for simple algebraic operations can be found in R. G. Budynas and J. K. Nisbett, Shigley's Mechanical Engineering Design, 9th ed., McGraw-Hill, New York, 2011, Table 20-6, p. 993.

26      Mechanical Engineering Design

Comparing Figure 1–7b with Table A–10, we see that R = 1 − Φ(z)  z ≤ 0   = Φ(z)     z > 0

(d)

To relate to the design factor, nd = μS∕μσ , divide each term on the right side of Equation (1–10) by μσ and rearrange as shown:

z=−

=−

μS −1 μσ

[ μ2σ

σˆS2

+

=−

μ2σ ] σˆσ2

1∕2

nd − 1

[ μ2σ μ2S + μ2σ ] μ2S σˆS2

σˆσ2

1∕2

nd − 1

[ μ2σ μ2S σˆS2 μ2S

=−

2 [ nd

+

μ2σ ] σˆσ2

1∕2

(e)

nd − 1 σˆS2 μ2S

+

μ2σ ] σˆσ2

1∕2

Introduce the terms CS = σˆS∕μS and Cσ = σˆσ∕μσ , called the coefficients of variance for strength and stress, respectively. Equation (e) is then rewritten as

z=−

nd − 1 √nd2 C S2 + Cσ2

(1–11)

Squaring both sides of Equation (1–11) and solving for nd results in

nd =

1 ± √1 − (1 − z2C 2S ) (1 − z2Cσ2 ) 1 − z2C 2S

(1–12)

The plus sign is associated with R > 0.5, and the minus sign with R ≤ 0.5. Equation (1–12) is remarkable in that it relates the design factor nd to the reliability goal R (through z) and the coefficients of variation of the strength and stress. EXAMPLE 1–6 A round cold-drawn 1018 steel rod has 0.2 percent mean yield strength Sy = 78.4 kpsi with a standard deviation of 5.90 kpsi. The rod is to be subjected to a mean static axial load of P = 50 kip with a standard deviation of 4.1 kip. Assuming the strength and load have normal distributions, what value of the design factor nd corresponds to a reliability of 0.999 against yielding? Determine the corresponding diameter of the rod. Solution For strength, CS = σˆS∕μS = 5.90∕78.4 = 0.0753. For stress,

σ=

P 4P = A πd 2

Since the tolerance on the diameter will be an order of magnitude less than that of the load or strength, the diameter will be treated deterministically. Thus, statistically, the stress is linearly proportional to the load, and Cσ = CP = σˆP∕μP = 4.1∕50 = 0.082. From Table A–10, for R = 0.999, z = −3.09. Then, Equation (1–12) gives Answer

nd =

1 + √1 − [1 − (−3.09) 2 (0.0753) 2 ][1 − (−3.09) 2 (0.082) 2 ] 1 − (−3.09) 2 (0.0753) 2

= 1.416

Introduction to Mechanical Engineering Design     27

The diameter is found deterministically from

σ=

Sy 4P = 2 nd πd

Solving for d gives Answer

d=√

4Pnd 4(50)(1.416) =√ = 1.072 in π(78.4) πSy

1–14 Dimensions and Tolerances Part of a machine designer's task is to specify the parts and components necessary for a machine to perform its desired function. Early in the design process, it is usually sufficient to work with nominal dimensions to determine function, stresses, deflections, and the like. However, eventually it is necessary to get to the point of specificity that every component can be purchased and every part can be manufactured. For a part to be manufactured, its essential shape, dimensions, and tolerances must be communicated to the manufacturers. This is usually done by means of a machine drawing, which may either be a multiview drawing on paper, or digital data from a CAD file. Either way, the drawing usually represents a legal document between the parties involved in the design and manufacture of the part. It is essential that the part be defined precisely and completely so that it can only be interpreted in one way. The designer's intent must be conveyed in such a way that any manufacturer can make the part and/or component to the satisfaction of any inspector. Common Dimensioning Terminology Before going further, we will define a few terms commonly used in dimensioning. ∙ Nominal size. The size we use in speaking of an element. For example, we may specify a 112 -in pipe or a 12 -in bolt. Either the theoretical size or the actual measured size may be quite different. The theoretical size of a 112 -in pipe is 1.900 in for the outside diameter. And the diameter of the 12 -in bolt, say, may actually measure 0.492 in. ∙ Limits. The stated maximum and minimum dimensions. ∙ Tolerance. The difference between the two limits. ∙ Bilateral tolerance. The variation in both directions from the basic dimension. That is, the basic size is between the two limits, for example, 1.005 ± 0.002 in. The two parts of the tolerance need not be equal. ∙ Unilateral tolerance. The basic dimension is taken as one of the limits, and variation is permitted in only one direction, for example, 1.005 +0.004 −0.000 in ∙ Clearance. A general term that refers to the mating of cylindrical parts such as a bolt and a hole. The word clearance is used only when the internal member is smaller than the external member. The diametral clearance is the measured difference in the two diameters. The radial clearance is the difference in the two radii. ∙ Interference. The opposite of clearance, for mating cylindrical parts in which the internal member is larger than the external member (e.g., press-fits).

28      Mechanical Engineering Design

∙ Allowance. The minimum stated clearance or the maximum stated interference for mating parts. ∙ Fit. The amount of clearance or interference between mating parts. See Section 7–8 for a standardized method of specifying fits for cylindrical parts, such as gears and bearings onto a shaft. ∙ GD&T. Geometric Dimensioning and Tolerancing (GD&T) is a comprehensive system of symbols, rules, and definitions for defining the nominal (theoretically perfect) geometry of parts and assemblies, along with the allowable variation in size, location, orientation, and form of the features of a part. See Chapter 20 for an overview of GD&T. Choice of Tolerances The choice of tolerances is the designer's responsibility and should not be made arbitrarily. Tolerances should be selected based on a combination of considerations including functionality, fit, assembly, manufacturing process ability, quality control, and cost. While there is need for balancing these considerations, functionality must not be compromised. If the functionality of the part or assembly cannot be achieved with a reasonable balance of the other considerations, the entire design may need to be reconsidered. The relationship of tolerances to functionality is usually associated with the need to assemble multiple parts. For example, the diameter of a shaft does not generally need a tight tolerance, except for the portions that must fit with components like bearings or gears. The bearings need a particular press fit in order to function properly. Section 7–8 addresses this issue in detail. Manufacturing methods evolve over time. The manufacturer is free to use any manufacturing process, as long as the final part meets the specifications. This allows the manufacturer to take advantage of available materials and tools, and to specify the least expensive manufacturing methods. Excessive precision on the part of the designer may seem like an easy way to achieve functionality, but it is actually a poor design choice in that it limits the manufacturing options and drives up the cost. In a competitive manufacturing environment, the designer must embrace the idea that less expensive manufacturing methods should be selected, even though the parts may be less than perfect. Since tight tolerances usually correlate to higher production costs, as shown in Figure 1–2, the designer should generally be thinking in terms of loosening the tolerances as much as possible, while still achieving the desired functionality. Choice of Dimensions Dimensioning a part is a designer's responsibility, since the choice of which dimensions to specify can make a difference in the functionality of the part. A properly dimensioned part will include just enough information, with no extraneous information that can lead to confusion or multiple interpretations. For example, the part shown in Figure 1–8a is over-specified in its length dimensions. Note, in machine drawings, the units for the dimensions are typically specified in an overall note on the drawing, and are not shown with the dimensions. If all the dimensions were theoretically perfect, there would be no inconsistency in the over-specified dimensions. But in reality every dimension can only be manufactured to some less-than-perfect level of accuracy. Suppose every dimension in Figure 1–8a is specified with a tolerance of +∕− 1. It would be possible to manufacture the part such that some dimensions were within the specified tolerance, while forcing related redundant dimensions to be out of tolerance.

Introduction to Mechanical Engineering Design     29 150 100 50

Figure 1–8

149 102

50

50

51

51

(a)

47

(b) 150 ± 1

50 ± 1

50 ± 1

50 ± 1

50 ± 1

50 ± 1

(b)

(a)

150 ± 1 150 ± 1 50 ± 1

100 ± 1 50 ± 1

(c)

Example of over-specified dimensions. (a) Five nominal dimensions specified. (b) With +∕− 1 tolerances, two dimensions are incompatible.

50 ± 1

(d)

For example, in Figure 1–8b, three of the dimensions are within the +∕− 1 tolerance, but they force the other two dimensions to be out of tolerance. In this example, only three length dimensions should be specified. The designer should determine which three are most important to the functioning and assembly of the part. Figure 1–9 shows four different choices of how the length dimensions might be specified for the same part. None of them are incorrect, but they are not all equivalent in terms of satisfying a particular function. For example, if the two holes are to mate with a pair of corresponding features from another part, the distance between the holes is critical. The choice of dimensions in Figure 1–9c would not be a good choice in this case. Even if the part is manufactured within the specified tolerances of +∕− 1, the distance between the holes could range anywhere from 47 to 53, an effective tolerance of +∕− 3. Choosing dimensions as shown in Figure 1–9a or 1–9b would serve the purpose better to limit the dimension between the holes to a tolerance of +∕− 1. For a different application, the distance of the holes to one or both edges might be important, while the overall length might be critical for another application. The point is, the designer should make this determination, not the manufacturer. Tolerance Stack-up Note that while there are always choices of which dimensions to specify, the cumulative effect of the individual specified tolerances must be allowed to accumulate somewhere. This is known as tolerance stack-up. Figure 1–9a shows an example of chain

Figure 1–9 Examples of choice of which dimensions to specify.

30      Mechanical Engineering Design

dimensioning, in which several dimensions are specified in series such that the tolerance stack-up can become large. In this example, even though the individual tolerances are all +∕− 1, the total length of the part has an implied tolerance of +∕− 3 due to the tolerance stack-up. A common method of minimizing a large tolerance stack-up is to dimension from a common baseline, as shown in Figure 1–9d. The tolerance stack-up issue is also pertinent when several parts are assembled. A gap or interference will occur, and will depend on the dimensions and tolerances of the individual parts. An example will demonstrate the point.

EXAMPLE 1–7 A shouldered screw contains three hollow right circular cylindrical parts on the screw before a nut is tightened against the shoulder. To sustain the function, the gap w must equal or exceed 0.003 in. The parts in the assembly depicted in Figure 1–10 have dimensions and tolerances as follows: a = 1.750 ± 0.003 in  b = 0.750 ± 0.001 in c = 0.120 ± 0.005 in  d = 0.875 ± 0.001 in Figure 1–10

a

An assembly of three cylindrical sleeves of lengths b, c, and d on a shoulder bolt shank of length a. The gap w is of interest. b

c

d

w

All parts except the part with the dimension d are supplied by vendors. The part containing the dimension d is made in-house. (a) Estimate the mean and tolerance on the gap w. (b) What basic value of d will assure that w ≥ 0.003 in? Solution (a) The mean value of w is given by Answer

w = a − b − c − d = 1.750 − 0.750 − 0.120 − 0.875 = 0.005 in

For equal bilateral tolerances, the tolerance of the gap is Answer

tw =

∑ t = 0.003 + 0.001 + 0.005 + 0.001 = 0.010 in all

Then, w = 0.005 ± 0.010 in, and wmax = w + tw = 0.005 + 0.010 = 0.015 in wmin = w − tw = 0.005 − 0.010 = −0.005 in Thus, both clearance and interference are possible. (b) If wmin is to be 0.003 in, then, w = wmin + tw = 0.003 + 0.010 = 0.013 in. Thus, Answer

d = a − b − c − w = 1.750 − 0.750 − 0.120 − 0.013 = 0.867 in

Introduction to Mechanical Engineering Design     31

The previous example represented an absolute tolerance system. Statistically, gap dimensions near the gap limits are rare events. Using a statistical tolerance system, the probability that the gap falls within a given limit is determined. This probability deals with the statistical distributions of the individual dimensions. For example, if the distributions of the dimensions in the previous example were normal and the tolerances, t, were given in terms of standard deviations of the dimension distribution, the standard deviation of the gap w would be tw = Σ t2 . However, this assumes a all

normal distribution for the individual dimensions, a rare occurrence. To find the distribution of w and/or the probability of observing values of w within certain limits requires a computer simulation in most cases. Monte Carlo computer simulations can be used to determine the distribution of w by the following approach: 1 Generate an instance for each dimension in the problem by selecting the value of each dimension based on its probability distribution. 2 Calculate w using the values of the dimensions obtained in step 1. 3 Repeat steps 1 and 2 N times to generate the distribution of w. As the number of trials increases, the reliability of the distribution increases.

1–15 Units In the symbolic units equation for Newton's second law, F = ma, F = MLT −2

(1–13)

F stands for force, M for mass, L for length, and T for time. Units chosen for any three of these quantities are called base units. The first three having been chosen, the fourth unit is called a derived unit. When force, length, and time are chosen as base units, the mass is the derived unit and the system that results is called a gravitational system of units. When mass, length, and time are chosen as base units, force is the derived unit and the system that results is called an absolute system of units. In some English-speaking countries, the U.S. customary foot-pound-second system (fps) and the inch-pound-second system (ips) are the two standard gravitational systems most used by engineers. In the fps system the unit of mass is

M=

2 FT 2 (pound-force)(second) = = lbf · s2/ft = slug L foot

(1–14)

Thus, length, time, and force are the three base units in the fps gravitational system. The unit of force in the fps system is the pound, more properly the pound-force. We shall often abbreviate this unit as lbf; the abbreviation lb is permissible however, since we shall be dealing only with the U.S. customary gravitational system. In some branches of engineering it is useful to represent 1000 lbf as a kilopound and to abbreviate it as kip. Note: In Equation (1–14) the derived unit of mass in the fps gravitational system is the lbf · s2/ft and is called a slug; there is no abbreviation for slug. The unit of mass in the ips gravitational system is

M=

2 FT 2 (pound-force)(second) = = lbf · s2/in L inch

The mass unit lbf · s2/in has no official name.

(1–15)

32      Mechanical Engineering Design

The International System of Units (SI) is an absolute system. The base units are the meter, the kilogram (for mass), and the second. The unit of force is derived by using Newton's second law and is called the newton. The units constituting the newton (N) are

F=

ML (kilogram)(meter) = = kg · m/s2 = N T2 (second) 2

(1–16)

The weight of an object is the force exerted upon it by gravity. Designating the weight as W and the acceleration due to gravity as g, we have

W = mg

(1–17)

In the fps system, standard gravity is g = 32.1740 ft/s2. For most cases this is rounded off to 32.2. Thus the weight of a mass of 1 slug in the fps system is W = mg = (1 slug)(32.2 ft/s2 ) = 32.2 lbf In the ips system, standard gravity is 386.088 or about 386 in/s2. Thus, in this system, a unit mass weighs W = (1 lbf · s2/in) (386 in/s2 ) = 386 lbf With SI units, standard gravity is 9.806 or about 9.81 m/s. Thus, the weight of a 1-kg mass is W = (1 kg)(9.81 m/s2 ) = 9.81 N A series of names and symbols to form multiples and submultiples of SI units has been established to provide an alternative to the writing of powers of 10. Table A–1 includes these prefixes and symbols. Numbers having four or more digits are placed in groups of three and separated by a space instead of a comma. However, the space may be omitted for the special case of numbers having four digits. A period is used as a decimal point. These recommendations avoid the confusion caused by certain European countries in which a comma is used as a decimal point, and by the English use of a centered period. Examples of correct and incorrect usage are as follows: 1924 or 1 924 but not 1,924 0.1924 or 0.192 4 but not 0.192,4 192 423.618 50 but not 192,423.61850 The decimal point should always be preceded by a zero for numbers less than unity.

1–16  Calculations and Significant Figures The discussion in this section applies to real numbers, not integers. The accuracy of a real number depends on the number of significant figures describing the number. Usually, but not always, three or four significant figures are necessary for engineering accuracy. Unless otherwise stated, no less than three significant figures should be used in your calculations. The number of significant figures is usually inferred by the number of figures given (except for leading zeros). For example, 706, 3.14, and 0.002 19 are assumed to be numbers with three significant figures. For trailing zeros, a little more clarification is necessary. To display 706 to four significant figures insert a trailing zero and display either 706.0, 7.060 × 102, or 0.7060 × 103. Also, consider

Introduction to Mechanical Engineering Design     33

a number such as 91 600. Scientific notation should be used to clarify the accuracy. For three significant figures express the number as 91.6 × 103. For four significant figures express it as 91.60 × 103. Computers and calculators display calculations to many significant figures. However, you should never report a number of significant figures of a calculation any greater than the smallest number of significant figures of the numbers used for the calculation. Of course, you should use the greatest accuracy possible when performing a calculation. For example, determine the circumference of a solid shaft with a diameter of d = 0.40 in. The circumference is given by C = πd. Since d is given with two significant figures, C should be reported with only two significant figures. Now if we used only two significant figures for π our calculator would give C = 3.1 (0.40) = 1.24 in. This rounds off to two significant figures as C = 1.2 in. However, using π = 3.141 592 654 as programmed in the calculator, C = 3.141 592 654 (0.40) = 1.256 637 061 in. This rounds off to C = 1.3 in, which is 8.3 percent higher than the first calculation. Note, however, since d is given with two significant figures, it is implied that the range of d is 0.40 ± 0.005. This means that the calculation of C is only accurate to within ±0.005∕0.40 = ±0.0125 = ±1.25%. The calculation could also be one in a series of calculations, and rounding each calculation separately may lead to an accumulation of greater inaccuracy. Thus, it is considered good engineering practice to make all calculations to the greatest accuracy possible and report the results within the accuracy of the given input.

1–17  Design Topic Interdependencies One of the characteristics of machine design problems is the interdependencies of the various elements of a given mechanical system. For example, a change from a spur gear to a helical gear on a drive shaft would add axial components of force, which would have implications on the layout and size of the shaft, and the type and size of the bearings. Further, even within a single component, it is necessary to consider many different facets of mechanics and failure modes, such as excessive deflection, static yielding, fatigue failure, contact stress, and material characteristics. However, in order to provide significant attention to the details of each topic, most machine design textbooks focus on these topics separately and give end-of-chapter problems that relate only to that specific topic. To help the reader see the interdependence between the various design topics, this textbook presents many ongoing and interdependent problems in the end-of-chapter problem sections. Each row of Table 1–2 shows the problem numbers that apply to the same mechanical system that is being analyzed according to the topics being presented in that particular chapter. For example, in the second row, Problems 3–41, 5–78, and 5–79 correspond to a pin in a knuckle joint that is to be analyzed for stresses in Chapter 3 and then for static failure in Chapter 5. This is a simple example of interdependencies, but as can be seen in the table, other systems are analyzed with as many as 10 separate problems. It may be beneficial to work through some of these continuing sequences as the topics are covered to increase your awareness of the various interdependencies. In addition to the problems given in Table 1–2, Section 1–18 describes a power transmission case study where various interdependent analyses are performed throughout the book, when appropriate in the presentation of the topics. The final results of the case study are then presented in Chapter 18.

34      Mechanical Engineering Design

Table 1–2  Problem Numbers for Linked End-of-Chapter Problems* 3–1

4–50

4–81

3–41

5–78

5–79

3–79

4–23

4–29

4–35

5–50

6–37

7–12

11–16

3–80

4–24

4–30

4–36

5–51

6–38

7–13

11–17

3–81

4–25

4–31

4–37

5–52

6–39

7–14

11–18

3–82

4–26

4–32

4–38

5–53

6–40

7–15

11–19

3–83

4–27

4–33

4–39

5–54

6–41

7–16

7–24

7–25

7–39

11–29

11–30

13–44

14–37

3–84

4–28

4–34

4–40

5–55

6–42

7–17

7–26

7–27

7–40

11–31

11–32

13–45

14–38

3–85

5–56

6–43

7–18

11–47

13–48

3–87

5–57

6–44

7–19

11–48

13–48

3–88

5–58

6–45

7–20

11–20

13–46

14–39

3–90

5–59

6–46

7–21

11–21

13–47

14–40

3–91

4–41

4–78

5–60

6–47

3–92

5–61

6–48

3–93

5–62

6–49

3–94

5–63

6–50

3–95

4–43

4–80

5–64

5–67

3–96

5–65

6–52

3–97

5–66

6–53

3–98

5–67

6–51

*Each row corresponds to the same mechanical component repeated for a different design concept.

1–18  Power Transmission Case Study Specifications We will consider a case study incorporating the many facets of the design process for a power transmission speed reducer throughout this textbook. The problem will be introduced here with the definition and specification for the product to be designed. Further details and component analysis will be presented in subsequent chapters. Chapter 18 provides an overview of the entire process, focusing on the design sequence, the interaction between the component designs, and other details pertinent to transmission of power. It also contains a complete case study of the power transmission speed reducer introduced here. Many industrial applications require machinery to be powered by engines or electric motors. The power source usually runs most efficiently at a narrow range of rotational speed. When the application requires power to be delivered at a slower speed than supplied by the motor, a speed reducer is introduced. The speed reducer should transmit the power from the motor to the application with as little energy loss as practical, while reducing the speed and consequently increasing the torque. For example, assume that a company wishes to provide off-the-shelf speed reducers in various capacities and speed ratios to sell to a wide variety of target applications.

Introduction to Mechanical Engineering Design     35

The marketing team has determined a need for one of these speed reducers to satisfy the following customer requirements. Design Requirements Power to be delivered: 20 hp Input speed: 1750 rev/min Output speed: 85 rev/min Targeted for uniformly loaded applications, such as conveyor belts, blowers, and generators Output shaft and input shaft in-line Base mounted with 4 bolts Continuous operation 6-year life, with 8 hours/day, 5 days/wk Low maintenance Competitive cost Nominal operating conditions of industrialized locations Input and output shafts standard size for typical couplings In reality, the company would likely design for a whole range of speed ratios for each power capacity, obtainable by interchanging gear sizes within the same overall design. For simplicity, in this case study we will consider only one speed ratio. Notice that the list of customer requirements includes some numerical specifics, but also includes some generalized requirements, e.g., low maintenance and competitive cost. These general requirements give some guidance on what needs to be considered in the design process, but are difficult to achieve with any certainty. In order to pin down these nebulous requirements, it is best to further develop the customer requirements into a set of product specifications that are measurable. This task is usually achieved through the work of a team including engineering, marketing, management, and customers. Various tools may be used (see footnote 1) to prioritize the requirements, determine suitable metrics to be achieved, and to establish target values for each metric. The goal of this process is to obtain a product specification that identifies precisely what the product must satisfy. The following product specifications provide an appropriate framework for this design task. Design Specifications Power to be delivered: 20 hp Power efficiency: >95% Steady state input speed: 1750 rev/min Maximum input speed: 2400 rev/min Steady-state output speed: 82–88 rev/min Usually low shock levels, occasional moderate shock Input and output shafts extend 4 in outside gearbox Input and output shaft diameter tolerance: ±0.001 in Input and output shafts in-line: concentricity ±0.005 in, alignment ±0.001 rad Maximum allowable loads on input shaft: axial, 50 lbf; transverse, 100 lbf Maximum allowable loads on output shaft: axial, 50 lbf; transverse, 500 lbf Maximum gearbox size: 14-in × 14-in base, 22-in height Base mounted with 4 bolts Mounting orientation only with base on bottom 100% duty cycle

36      Mechanical Engineering Design

Maintenance schedule: lubrication check every 2000 hours; change of lubrication every 8000 hours of operation; gears and bearing life >12 000 hours; infinite shaft life; gears, bearings, and shafts replaceable Access to check, drain, and refill lubrication without disassembly or opening of gasketed joints Manufacturing cost per unit: σ3), so always order your principal stresses. Do this in any computer code you generate and you'll always generate τmax.

3–8  Elastic Strain Normal strain ε is defined and discussed in Section 2–1 for the tensile specimen and is given by Equation (2–2) as ε = δ∕l, where δ is the total elongation of the bar within the length l. Hooke's law for the tensile specimen is given by Equation (2–3) as σ = Eε (3–17) where the constant E is called Young's modulus or the modulus of elasticity. When a material is placed in tension, there exists not only an axial strain, but also negative strain (contraction) perpendicular to the axial strain. Assuming a linear, homogeneous, isotropic material, this lateral strain is proportional to the axial strain. If the axial direction is x, then the lateral strains are εy = εz = −νεx. The constant of proportionality ν is called Poisson's ratio, which is about 0.3 for most structural metals. See Table A–5 for values of ν for common materials. 1

For development of this equation and further elaboration of three-dimensional stress transformations see: Richard G. Budynas, Advanced Strength and Applied Stress Analysis, 2nd ed., McGraw-Hill, New York, 1999, pp. 46–78. 2 Note the difference between this notation and that for a shear stress, say, τxy. The use of the shilling mark is not accepted practice, but it is used here to emphasize the distinction.

110      Mechanical Engineering Design

If the axial stress is in the x direction, then from Equation (3–17) εx =

σx E

εy = εz = −ν

σx E

(3–18)

For a stress element undergoing σx, σy, and σz simultaneously, the normal strains are given by 1 εx = [σx − ν(σy + σz )] E 1 εy = [σy − ν(σx + σz )] (3–19) E 1 εz = [σz − ν(σx + σy )] E Shear strain γ is the change in a right angle of a stress element when subjected to pure shear stress, and Hooke's law for shear is given by

τ = Gγ

(3–20)

where the constant G is the shear modulus of elasticity or modulus of rigidity. It can be shown for a linear, isotropic, homogeneous material, the three elastic constants are related to each other by E = 2G(1 + ν) (3–21) A state of strain, called plane strain, occurs when all strains in a given direction are zero. For example, if all the strains in the z direction were zero, then εz = γyz = γzx = 0. As an example of plane strain, consider the beam shown in Figure 3–16. If the beam thickness is very thin in the z direction (approaching zero thickness in the limit), then the state of stress will be plane stress. If the beam is very wide in the z direction (approaching infinite width in the limit), then the state of strain will be plane strain. The strain-stress equations, Equation (3–19), can then be modified for plane strain (see Problem 3–31).

3–9  Uniformly Distributed Stresses The assumption of a uniform distribution of stress is frequently made in design. The result is then often called pure tension, pure compression, or pure shear, depending upon how the external load is applied to the body under study. The word simple is sometimes used instead of pure to indicate that there are no other complicating effects. The tension rod is typical. Here a tension load F is applied through pins at the ends of the bar. The assumption of uniform stress means that if we cut the bar at a section remote from the ends and remove one piece, we can replace its effect by applying a uniformly distributed force of magnitude σA to the cut end. So the stress σ is said to be uniformly distributed. It is calculated from the equation

σ=

F A

(3–22)

This assumption of uniform stress distribution requires that:

∙ The bar be straight and of a homogeneous material ∙ The line of action of the force contains the centroid of the section ∙ The section be taken remote from the ends and from any discontinuity or abrupt change in cross section

Load and Stress Analysis     111

For simple compression, Equation (3–22) is applicable with F normally being considered a negative quantity. Also, a slender bar in compression may fail by buckling, and this possibility must be eliminated from consideration before Equation (3–22) is used.3 Another type of loading that assumes a uniformly distributed stress is known as direct shear. This occurs when there is a shearing action with no bending. An example is the action on a piece of sheet metal caused by the two blades of tin snips. Bolts and pins that are loaded in shear often have direct shear. Think of a cantilever beam with a force pushing down on it. Now move the force all the way up to the wall so there is no bending moment, just a force trying to shear the beam off the wall. This is direct shear. Direct shear is usually assumed to be uniform across the cross section, and is given by

τ=

V A

(3–23)

where V is the shear force and A is the area of the cross section that is being sheared. The assumption of uniform stress is not accurate, particularly in the vicinity where the force is applied, but the assumption generally gives acceptable results.

3–10  Normal Stresses for Beams in Bending The equations for the normal bending stresses in straight beams are based on the following assumptions. ∙ The beam is subjected to pure bending. This means that the shear force is zero, and that no torsion or axial loads are present (for most engineering applications it is assumed that these loads affect the bending stresses minimally). ∙ The material is isotropic and homogeneous. ∙ The material obeys Hooke's law. ∙ The beam is initially straight with a cross section that is constant throughout the beam length. ∙ The beam has an axis of symmetry in the plane of bending. ∙ The proportions of the beam are such that it would fail by bending rather than by crushing, wrinkling, or sidewise buckling. ∙ Plane cross sections of the beam remain plane during bending. In Figure 3–13 we visualize a portion of a straight beam acted upon by a positive bending moment M shown by the curved arrow showing the physical action of the moment together with a straight, double-headed, arrow indicating the moment vector. The x axis Figure 3–13

y

Straight beam in positive bending.

M

z

M 3

See Section 4–11.

x

112      Mechanical Engineering Design

Figure 3–14

y

Compression

Bending stresses according to Equation (3–24).

c y

Neutral axis, Centroidal axis x

Tension

is coincident with the neutral axis of the section, and the xz plane, which contains the neutral axes of all cross sections, is called the neutral plane. Elements of the beam coincident with this plane have zero bending stress. The location of the neutral axis with respect to the cross section is coincident with the centroidal axis of the cross section. The bending stress varies linearly with the distance from the neutral axis, y, and is given by My σx = − (3–24) I where I is the second-area moment about the z axis. That is,

I = y2dA

(3–25)

The stress distribution given by Equation (3–24) is shown in Figure 3–14. The maximum magnitude of the bending stress will occur where y has the greatest magnitude. Designating σmax as the maximum magnitude of the bending stress, and c as the maximum magnitude of y Mc σmax = (3–26a) I Equation (3–24) can still be used to ascertain whether σmax is tensile or compressive. Equation (3–26a) is often written as M σmax = (3–26b) Z where Z = I∕c is called the section modulus. EXAMPLE 3–5 A beam having a T section with the dimensions shown in Figure 3–15 is subjected to a bending moment of 1600 N · m, about the negative z axis, that causes tension at the top surface. Locate the neutral axis and find the maximum tensile and compressive bending stresses. Solution Dividing the T section into two rectangles, numbered 1 and 2, the total area is A = 12(75) + 12(88) = 1956 mm2. Summing the area moments of these rectangles about the top edge, where the moment arms of areas 1 and 2 are 6 mm and (12 + 88∕2) = 56 mm respectively, we have

1956c1 = 12(75) (6) + 12(88)(56)

and hence c1 = 32.99 mm. Therefore c2 = 100 − 32.99 = 67.01 mm.

Load and Stress Analysis     113

Figure 3–15

y

Dimensions in millimeters.

75 12

1 c1

z 100

2 c2

12

Next we calculate the second moment of area of each rectangle about its own centroidal axis. Using Table A–18, we find for the top rectangle

I1 =

1 3 1 bh = (75)123 = 1.080 × 10 4 mm4 12 12

For the bottom rectangle, we have

I2 =

1 (12)883 = 6.815 × 10 5 mm4 12

We now employ the parallel-axis theorem to obtain the second moment of area of the composite figure about its own centroidal axis. This theorem states Iz = Ica + Ad 2 where Ica is the second moment of area about its own centroidal axis and Iz is the second moment of area about any parallel axis a distance d removed. For the top rectangle, the distance is and for the bottom rectangle,

d1 = 32.99 − 6 = 26.99 mm

d2 = 67.01 −

88 = 23.01 mm 2

Using the parallel-axis theorem for both rectangles, we now find that

I = [1.080 × 104 + 12(75)26.992] + [6.815 × 105 + 12(88)23.012] = 1.907 × 106 mm4

Finally, the maximum tensile stress, which occurs at the top surface, is found to be Answer

σ=

Mc1 1600(32.99)10−3 = = 27.68(106 ) Pa = 27.68 MPa I 1.907(10−6 )

Similarly, the maximum compressive stress at the lower surface is found to be Answer

σ=−

Mc2 1600(67.01)10−3 =− = −56.22(106 ) Pa = −56.22 MPa I 1.907(10−6 )

114      Mechanical Engineering Design

Two-Plane Bending Quite often, in mechanical design, bending occurs in both xy and xz planes. Considering cross sections with one or two planes of symmetry only, the bending stresses are given by

σx = −

Mz y My z + Iz Iy

(3–27)

where the first term on the right side of the equation is identical to Equation (3–24), My is the bending moment in the xz plane (moment vector in y direction), z is the distance from the neutral y axis, and Iy is the second area moment about the y axis. For noncircular cross sections, Equation (3–27) is the superposition of stresses caused by the two bending moment components. The maximum tensile and compressive bending stresses occur where the summation gives the greatest positive and negative stresses, respectively. For solid circular cross sections, all lateral axes are the same and the plane containing the moment corresponding to the vector sum of Mz and My contains the maximum bending stresses. For a beam of diameter d the maximum distance from the neutral axis is d∕2, and from Table A–18, I = πd 4∕64. The maximum bending stress for a solid circular cross section is then 2 2 1∕2 Mc (M y + M z ) (d∕2) 32 σm = = = 3 (M 2y + M 2z ) 1∕2 4 I πd ∕64 πd

(3–28)

EXAMPLE 3–6 As shown in Figure 3–16a, beam OC is loaded in the xy plane by a uniform load of 50 lbf/in, and in the xz plane by a concentrated force of 100 lbf at end C. The beam is 8 in long. Figure 3–16 (a) Beam loaded in two planes; (b) loading and bending-moment diagrams in xy plane; (c) loading and bending-moment diagrams in xz plane.

y

y 50 lbf/in

A

O 1600 lbf-in 400 lbf

50 lbf/in

O B

x

C

Mz (lbf-in)

z C

1.5 in

100 lbf 0.75 in (a)

100 lbf 800 lbf-in O

C

z

100 lbf

My (lbf-in) 800 0

x (c)

0

x

x

x

–1600 (b)

Load and Stress Analysis     115

(a) For the cross section shown determine the maximum tensile and compressive bending stresses and where they act. (b) If the cross section was a solid circular rod of diameter, d = 1.25 in, determine the magnitude of the maximum bending stress. Solution (a) The reactions at O and the bending-moment diagrams in the xy and xz planes are shown in Figures 3–16b and c, respectively. The maximum moments in both planes occur at O where

1 (Mz ) O = − (50)82 = −1600 lbf-in  (My ) O = 100(8) = 800 lbf-in 2

The second moments of area in both planes are

Iz =

1 1 (0.75)1.53 = 0.2109 in4   Iy = (1.5)0.753 = 0.05273 in4 12 12

The maximum tensile stress occurs at point A, shown in Figure 3–16a, where the maximum tensile stress is due to both moments. At A, yA = 0.75 in and zA = 0.375 in. Thus, from Equation (3–27) Answer

(σx ) A = −

−1600(0.75) 800(0.375) + = 11 380 psi = 11.38 kpsi 0.2109 0.05273

The maximum compressive bending stress occurs at point B, where yB = −0.75 in and zB = −0.375 in. Thus Answer

(σx ) B = −

−1600(−0.75) 800(−0.375) + = −11 380 psi = −11.38 kpsi 0.2109 0.05273

(b) For a solid circular cross section of diameter, d = 1.25 in, the maximum bending stress at end O is given by Equation (3–28) as Answer

σm =

32 [8002 + (−1600) 2]1∕2 = 9329 psi = 9.329 kpsi π(1.25) 3

Beams with Asymmetrical Sections4 The bending stress equations, given by Equations (3–24) and (3–27), can also be applied to beams having asymmetrical cross sections, provided the planes of bending coincide with the area principal axes of the section. The method for determining the orientation of the area principal axes and the values of the corresponding principal second-area moments can be found in any statics book. If a section has an axis of symmetry, that axis and its perpendicular axis are the area principal axes. For example, consider a beam in bending, using an equal leg angle as shown in Table A–6. Equation (3–27) cannot be used if the bending moments are resolved about axis 1–1 and/or axis 2–2. However, Equation (3–27) can be used if the moments are resolved about axis 3–3 and its perpendicular axis (let us call it, say, axis 4–4). Note, for this cross section, axis 4–4 is an axis of symmetry. Table A–6 is a standard table, and for brevity, does not directly give all the information needed to use it. The orientation of the area principal axes and the values of I2–2, I3–3, and I4–4 are not given because they can be determined as follows. Since the legs are equal, the principal 4

For further discussion, see Section 5.3, Richard G. Budynas, Advanced Strength and Applied Stress Analysis, 2nd ed., McGraw-Hill, New York, 1999.

116      Mechanical Engineering Design

axes are oriented ±45° from axis 1–1, and I2–2 = I1–1. The second-area moment I3–3 is given by

I3−3 = A(k3−3 ) 2

(a)

where k3–3 is called the radius of gyration. The sum of the second-area moments for a cross section is invariant, so I1–1 + I2–2 = I3–3 + I4–4. Thus, I4–4 is given by

I4−4 = 2 I1−1 − I3−3

(b)

1 4

where I2–2 = I1–1. For example, consider a 3 × 3 × angle. Using Table A–6 and Equations (a) and (b), I3–3 = 1.44 (0.592)2 = 0.505 in4, and I4–4 = 2 (1.24) − 0.505 = 1.98 in4.

3–11  Shear Stresses for Beams in Bending Most beams have both shear forces and bending moments present. It is only occasionally that we encounter beams subjected to pure bending, that is to say, beams having zero shear force. The flexure formula is developed on the assumption of pure bending. This is done, however, to eliminate the complicating effects of shear force in the development of the formula. For engineering purposes, the flexure formula is valid no matter whether a shear force is present or not. For this reason, we shall utilize the same normal bending-stress distribution [Equations (3–24) and (3–26)] when shear forces are also present. In Figure 3–17a we show a beam segment of constant cross section subjected to a shear force V and a bending moment M at x. Because of external loading and V, the shear force and bending moment change with respect to x. At x + dx the shear force and bending moment are V + dV and M + dM, respectively. Considering forces in the x direction only, Figure 3–17b shows the stress distribution σx due to the bending moments. If dM is positive, with the bending moment increasing, the stresses on the right face, for a given value of y, are larger in magnitude than the stresses on the left face. If we further isolate the element by making a slice at y = y1 (see Figure 3–17b), the net force in the x direction will be directed to the left with a value of

c

y1

(dM)y dA I

as shown in the rotated view of Figure 3–17c. For equilibrium, a shear force on the bottom face, directed to the right, is required. This shear force gives rise to a shear stress τ, where, if assumed uniform, the force is τb dx. Thus

τb dx =

c

y1

(dM)y dA I

(a)

The term dM∕I can be removed from within the integral and b dx placed on the right side of the equation; then, from Equation (3–3) with V = dM∕dx, Equation (a) becomes

τ=

V Ib

c

∫ y dA

(3–29)

y1

In this equation, the integral is the first moment of the area A′ with respect to the neutral axis (see Figure 3–17c). This integral is usually designated as Q. Thus c

Q=

∫ y dA = y ′A′ y1

(3–30)

Load and Stress Analysis     117 w (x)

y

σx = –

My I

σx = – y1

V

M + dM

M

c x

x

V + dV x

My dMy + I I

dx

dx

(b)

(a) A' y

dx τ

b

F=

τ

c y1

dM y I

Figure 3–17 Beam section isolation. Note: Only forces shown in x direction on dx element in (b).

x (c)

where, for the isolated area y1 to c, y′ is the distance in the y direction from the neutral plane to the centroid of the area A′. With this, Equation (3–29) can be written as

τ=

VQ Ib

(3–31)

This stress is known as the transverse shear stress. It is always accompanied with bending stress. In using this equation, note that b is the width of the section at y = y1. Also, I is the second moment of area of the entire section about the neutral axis. Because cross shears are equal, and area A′ is finite, the shear stress τ given by Equation (3–31) and shown on area A′ in Figure 3–17c occurs only at y = y1. The shear stress on the lateral area varies with y, normally maximum at y = 0 (where y′A′ is maximum) and zero at the outer fibers of the beam where A′ = 0. The shear stress distribution in a beam depends on how Q∕b varies as a function of y1. Here we will show how to determine the shear stress distribution for a beam with a rectangular cross section and provide results of maximum values of shear stress for other standard cross sections. Figure 3–18 shows a portion of a beam with a rectangular cross section, subjected to a shear force V and a bending moment M. As a result of the bending moment, a normal stress σ is developed on a cross section such as A–A, which is in compression above the neutral axis and in tension below. To investigate the shear stress at a distance y1 above the neutral axis, we select an element of area dA at a distance y above the neutral axis. Then, dA = b dy, and so Equation (3–30) becomes

Q=

c

y1

y dA = b

c

y dy =

y1

by 2 c b = (c2 − y 21 ) ⎹ 2 y1 2

(b)

Substituting this value for Q into Equation (3–31) gives

τ=

V 2 (c − y21 ) 2I

(3–32)

This is the general equation for shear stress in a rectangular beam. To learn something about it, let us make some substitutions. From Table A–18, the second moment of area for a rectangular section is I = bh3∕12; substituting h = 2c and A = bh = 2bc gives

I=

Ac 2 3

(c)

118      Mechanical Engineering Design

Figure 3–18

y

y

y A

Transverse shear stresses in a rectangular beam.

dy

σ

b

dA

τmax =

3V 2A

V

M

x

O

z

c

y y 1 O

x

h

A (b)

(a)

(c)

y dx τ

c y1

τ

x

(d )

If we now use this value of I for Equation (3–32) and rearrange, we get

τ=

y 21 3V 1 − 2A ( c2 )

(3–33)

We note that the maximum shear stress exists when y1 = 0, which is at the bending neutral axis. Thus

τmax =

3V 2A

(3–34)

for a rectangular section. As we move away from the neutral axis, the shear stress decreases parabolically until it is zero at the outer surfaces where y1 = ±c, as shown in Figure 3–18c. Horizontal shear stress is always accompanied by vertical shear stress of the same magnitude, and so the distribution can be diagrammed as shown in Figure 3–18d. Figure 3–18c shows that the shear τ on the vertical surfaces varies with y. We are almost always interested in the horizontal shear, τ in Figure 3–18d, which is nearly uniform over dx with constant y = y1. The maximum horizontal shear occurs where the vertical shear is largest. This is usually at the neutral axis but may not be if the width b is smaller somewhere else. Furthermore, if the section is such that b can be minimized on a plane not horizontal, then the horizontal shear stress occurs on an inclined plane. For example, with tubing, the horizontal shear stress occurs on a radial plane and the corresponding "vertical shear" is not vertical, but tangential. The distributions of transverse shear stresses for several commonly used cross sections are shown in Table 3–2. The profiles represent the VQ∕Ib relationship, which is a function of the distance y from the neutral axis. For each profile, the formula for the maximum value at the neutral axis is given. Note that the expression given for the I beam is a commonly used approximation that is reasonable for a standard I beam with a thin web. Also, the profile for the I beam is idealized. In reality the transition from the web to the flange is quite complex locally, and not simply a step change.

Load and Stress Analysis     119

Table 3–2  Formulas for Maximum Transverse Shear Stress from VQ∕Ib Beam Shape

Formula

τavc =

V A

τmax =

Beam Shape

3V 2A

Rectangular

Formula τavc =

V A

τmax =

2V A

τmax ≈

V Aweb

Hollow, thin-walled round τavc =

V A

τmax =

4V 3A

Aweb

Structural I beam (thin-walled)

Circular

It is significant to observe that the transverse shear stress in each of these common cross sections is maximum on the neutral axis, and zero on the outer surfaces. Since this is exactly the opposite of where the bending and torsional stresses have their maximum and minimum values, the transverse shear stress is often not critical from a design perspective. Let us examine the significance of the transverse shear stress, using as an example a cantilever beam of length L, with rectangular cross section b × h, loaded at the free end with a transverse force F. At the wall, where the bending moment is the largest, at a distance y from the neutral axis, a stress element will include both bending stress and transverse shear stress. In Section 5–4 it will be shown that a good measure of the combined effects of multiple stresses on a stress element is the maximum shear stress. The bending stress is given by σ = My∕I, where, for this case, M = FL, I = bh3∕12, and h = 2c. This gives

σ=

y y My 12FLy 3F L 3F L = = · 8( )( ) = · 4( )( ) 3 c I 2bh h 2c 2bh h bh

(d)

The shear stress given by Equation (3–33), with V = F and A = bh, is

τ=

y 2 3F 1 − (c) ] 2bh [

(e)

Substituting Equations (d) and (e) into Equation (3–14) we obtain a general equation for the maximum shear stress in a cantilever beam with a rectangular cross section. The result is

σ 2 3F √4(L∕h) 2 (y∕c) 2 + [1 − (y∕c) 2]2 τmax = √ ( ) + τ2 = 2 2bh

(f )

Defining a normalized maximum shear stress, τmax as τmax divided by the maximum shear stress due to bending, (σ∕2)∣y=c = 6FLc∕(bh3 ) = 3FL∕(bh2 ), Equation ( f ) can be rewritten as

τmax =

1 √4(L∕h) 2 (y∕c) 2 + [1 − (y∕c) 2 ] 2 2(L∕h)

(g)

To investigate the significance of transverse shear stress, we plot τmax as a function of L∕h for several values of y∕c, as shown in Figure 3–19. Since F and b appear

120      Mechanical Engineering Design

Figure 3–19

y⁄c = 1

τmax

Plot of dimensionless maximum shear stress, τmax, for a rectangular cantilever beam, combining the effects of bending and transverse shear stresses.

1

y⁄c = 2/3

0.5

y⁄c = 1/3

0

y⁄c = 0 0

0.5

1

1.5

2 L/h

2.5

3

3.5

4

only as linear multipliers outside the radical, they will only serve to scale the plot in the vertical direction without changing any of the relationships. Notice that at the neutral axis where y∕c = 0, τmax is constant for any length beam, since the bending stress is zero at the neutral axis and the transverse shear stress is independent of L. On the other hand, on the outer surface where y∕c = 1, τmax increases linearly with L∕h because of the bending moment. For y∕c between zero and one, τmax is nonlinear for low values of L∕h, but behaves linearly as L∕h increases, displaying the dominance of the bending stress as the moment arm increases. We can see from the graph that the critical stress element (the largest value of τmax) will always be either on the outer surface (y∕c = 1) or at the neutral axis (y∕c = 0), and never between. Thus, for the rectangular cross section, the transition between these two locations occurs at L∕h = 0.5 where the line for y∕c = 1 crosses the horizontal line for y∕c = 0. The critical stress element is either on the outer surface where the transverse shear is zero, or if L∕h is small enough, it is on the neutral axis where the bending stress is zero. The conclusions drawn from Figure 3–19 are generally similar for any cross section that does not increase in width farther away from the neutral axis. This notably includes solid round cross sections, but not I beams or channels. Care must be taken with I beams and channels that have thin webs that extend far enough from the neutral axis that the bending and shear may both be significant on the same stress element (See Example 3–7). For any common cross section beam, if the beam length to height ratio is greater than 10, the transverse shear stress is generally considered negligible compared to the bending stress at any point within the cross section.

EXAMPLE 3–7 A simply supported beam, 12 in long, is to support a load of 488 lbf acting 3 in from the left support, as shown in Figure 3–20a. The beam is an I beam with the cross-sectional dimensions shown. To simplify the calculations, assume a cross section with square corners, as shown in Figure 3–20c. Points of interest are labeled (a, b, c, and d) at distances y from the neutral axis of 0 in, 1.240− in, 1.240+ in, and 1.5 in (Figure 3–20c). At the critical axial location along the beam, find the following information. (a) Determine the profile of the distribution of the transverse shear stress, obtaining values at each of the points of interest. (b) Determine the bending stresses at the points of interest. (c) Determine the maximum shear stresses at the points of interest, and compare them.

Load and Stress Analysis     121

Figure 3–20

y

3 in

488 lbf

9 in

0.260 in

3.00 in

x

O

0.170 in

2.33 in R1 = 366 lbf

R2 = 122 lbf (a)

y 488 lbf x

O 366 lbf

122 lbf 366 lbf

d c

0.260 in

O

y τd

τc

τb

b

1.24 in

τa τ

a

1.08 in O (c)

(b)

(d)

Solution First, we note that the transverse shear stress is not likely to be negligible in this case since the beam length to height ratio is much less than 10, and since the thin web and wide flange will allow the transverse shear to be large. The loading, shear-force, and bending-moment diagrams are shown in Figure 3–20b. The critical axial location is at x = 3− where the shear force and the bending moment are both maximum. (a) We obtain the area moment of inertia I by evaluating I for a solid 3.0-in × 2.33-in rectangular area, and then subtracting the two rectangular areas that are not part of the cross section.

I=

(2.33)(3.00) 3 (1.08)(2.48) 3 − 2[ = 2.50 in4 ] 12 12

Finding Q at each point of interest using Equation (3–30) gives Qa =

( ∑ y′A′ ⎹ )

y=1.5 y=0

= (1.24 +

0.260 1.24 [(2.33)(0.260)] + ( [(1.24)(0.170)] = 0.961 in3 2 ) 2 )

[(2.33) (0.260)] = 0.830 in3 ( ∑y′A′ )⎹ y=1.24 = (1.24 + 0.260 2 ) y=1.5

Qb = Qc =

y=1.5

Qd =

( ∑y′A′ )⎹ y=1.5 = (1.5)(0) = 0 in3

Applying Equation (3–31) at each point of interest, with V and I constant for each point, and b equal to the width of the cross section at each point, shows that the magnitudes of the transverse shear stresses are

122      Mechanical Engineering Design

Answer

τa =

VQa (366)(0.961) = = 828 psi Iba (2.50) (0.170)

τb =

VQb (366)(0.830) = = 715 psi Ibb (2.50) (0.170)

τc =

VQc (366) (0.830) = = 52.2 psi Ibc (2.50)(2.33)

τd =

VQd (366) (0) = = 0 psi Ibd (2.50) (2.33)

The magnitude of the idealized transverse shear stress profile through the beam depth will be as shown in Figure 3–20d. (b) The bending stresses at each point of interest are Mya (1098)(0) = = 0 psi I 2.50

Answer

σa =

σb = σc = −

σd = −

Myb (1098) (1.24) =− = −545 psi I 2.50

Myd (1098)(1.50) =− = −659 psi I 2.50

(c) Now at each point of interest, consider a stress element that includes the bending stress and the transverse shear stress. The maximum shear stress for each stress element can be determined by Mohr's circle, or analytically by Equation (3–14) with σy = 0, Thus, at each point

σ 2 τmax = √ ( ) + τ 2 2

τmax, a = √0 + (828) 2 = 828 psi

−545 2 τmax,b = √ ( + (715) 2 = 765 psi 2 )

−545 2 τmax,c = √ ( + (52.2) 2 = 277 psi 2 )

−659 2 τmax,d = √ ( + 0 = 330 psi 2 )

Answer Interestingly, the critical location is at point a where the maximum shear stress is the largest, even though the bending stress is zero. The next critical location is at point b in the web, where the thin web thickness dramatically increases the transverse shear stress compared to points c or d. These results are counterintuitive, since both points a and b turn out to be more critical than point d, even though the bending stress is maximum at point d. The thin web and wide flange increase the impact of the transverse shear stress. If the beam length to height ratio were increased, the critical point would move from point a to point b, since the transverse shear stress at point a would remain constant, but the bending stress at point b would increase. The designer should be particularly alert to the possibility of the critical stress element not being on the outer surface with cross sections that get wider farther from the neutral axis, particularly in cases with thin web sections and wide flanges. For rectangular and circular cross sections, however, the maximum bending stresses at the outer surfaces will dominate, as was shown in Figure 3–19.

Load and Stress Analysis     123

3–12  Torsion Any moment vector that is collinear with an axis of a mechanical part is called a torque vector, because the moment causes the part to be twisted about that axis. A bar subjected to such a moment is said to be in torsion. As shown in Figure 3–21, the torque T applied to a bar can be designated by drawing arrows on the surface of the bar to indicate direction or by drawing torquevector arrows along the axes of twist of the bar. Torque vectors are the hollow arrows shown on the x axis in Figure 3–21. Note that they conform to the right-hand rule for vectors. The angle of twist, in radians, for a solid round bar is

θ=

Tl GJ

(3–35)

where T = torque l = length G = modulus of rigidity J = polar second moment of area Shear stresses develop throughout the cross section. For a round bar in torsion, these stresses are proportional to the radius ρ and are given by

τ=

Tρ J

(3–36)

Designating r as the radius to the outer surface, we have

τmax =

Tr J

(3–37)

The assumptions used in the analysis are: ∙ The bar is acted upon by a pure torque, and the sections under consideration are remote from the point of application of the load and from a change in diameter. ∙ The material obeys Hooke's law. ∙ Adjacent cross sections originally plane and parallel remain plane and parallel after twisting, and any radial line remains straight. Figure 3–21 T

l

A

y dx B T C B'

θ C'

z

r

ρ O x

124      Mechanical Engineering Design

The last assumption depends upon the axisymmetry of the member, so it does not hold true for noncircular cross sections. Consequently, Equations (3–35) through (3–37) apply only to circular sections. For a solid round section, J=

πd 4 32

(3–38)

where d is the diameter of the bar. For a hollow round section,

J=

π (do4 − di4 ) 32

(3–39)

where the subscripts o and i refer to the outside and inside diameters, respectively. There are some applications in machinery for noncircular cross-section members and shafts where a regular polygonal cross section is useful in transmitting torque to a gear or pulley that can have an axial change in position. Because no key or keyway is needed, the possibility of a lost key is avoided. The development of equations for stress and deflection for torsional loading of noncircular cross sections can be obtained from the mathematical theory of elasticity. In general, the shear stress does not vary linearly with the distance from the axis, and depends on the specific cross section. In fact, for a rectangular section bar the shear stress is zero at the corners where the distance from the axis is the largest. The maximum shearing stress in a rectangular b × c section bar occurs in the middle of the longest side b and is of the magnitude

τmax =

T T 1.8 ≈ 2 (3 + 2 ) (3–40) b∕c αbc bc

where b is the width (longer side) and c is the thickness (shorter side). They can not be interchanged. The parameter α is a factor that is a function of the ratio b∕c as shown in the following table.5 The angle of twist is given by

θ=

Tl βbc3G

(3–41)

where β is a function of b∕c, as shown in the table. b∕c 1.00 1.50 1.75 2.00 2.50 3.00 4.00 6.00 8.00 10 ∞ α 0.208 0.231 0.239 0.246 0.258 0.267 0.282 0.299 0.307 0.313 0.333 β 0.141 0.196 0.214 0.228 0.249 0.263 0.281 0.299 0.307 0.313 0.333

Equation (3–40) is also approximately valid for equal-sided angles; these can be considered as two rectangles, each of which is capable of carrying half the torque.6 It is often necessary to obtain the torque T from a consideration of the power and speed of a rotating shaft. For convenience when U.S. Customary units are used, three forms of this relation are 5

H=

FV 2πTn Tn = = 33 000 33 000(12) 63 025

(3–42)

S. Timoshenko, Strength of Materials, Part I, 3rd ed., D. Van Nostrand Company, New York, 1955, p. 290. For other sections see W. C. Young, R. G. Budynas, and A. M. Sadegh, Roark's Formulas for Stress and Strain, 8th ed., McGraw-Hill, New York, 2012. 6

Load and Stress Analysis     125

where H T n F V

= = = = =

power, hp torque, lbf · in shaft speed, rev/min force, lbf velocity, ft/min

When SI units are used, the equation is

H = Tω

(3–43)

where H = power, W T = torque, N · m ω = angular velocity, rad/s The torque T corresponding to the power in watts is given approximately by H T = 9.55 n

(3–44)

where n is in revolutions per minute. EXAMPLE 3–8 Figure 3–22 shows a crank loaded by a force F = 300 lbf that causes twisting and bending of a 34 -indiameter shaft fixed to a support at the origin of the reference system. In actuality, the support may be an inertia that we wish to rotate, but for the purposes of a stress analysis we can consider this a statics problem. (a) Draw separate free-body diagrams of the shaft AB and the arm BC, and compute the values of all forces, moments, and torques that act. Label the directions of the coordinate axes on these diagrams. (b) For member BC, compute the maximum bending stress and the maximum shear stress associated with the applied torsion and transverse loading. Indicate where these stresses act. (c) Locate a stress element on the top surface of the shaft at A, and calculate all the stress components that act upon this element. (d) Determine the maximum normal and shear stresses at A. Figure 3–22

y

1.5 in F

C

A

3 4

in dia. B

1 4

in

1 2 1 14

in

z 4 in 5 in x

in dia.

126      Mechanical Engineering Design

Solution (a) The two free-body diagrams are shown in Figure 3–23. The results are At At At At

end end end end

C of arm BC: B of arm BC: B of shaft AB: A of shaft AB:

F F F F

= = = =

−300j lbf, TC = −450k lbf · in 300j lbf, M1 = 1200i lbf · in, T1 = 450k lbf · in −300j lbf, T2 = −1200i lbf · in, M2 = −450k lbf · in 300j lbf, MA = 1950k lbf · in, TA = 1200i lbf · in

(b) For arm BC, the bending moment will reach a maximum near the shaft at B. If we assume this is 1200 lbf · in, then the bending stress for a rectangular section will be σ=

Answer

6(1200) M 6M = = = 18 400 psi = 18.4 kpsi I∕c bh2 0.25(1.25) 2

Of course, this is not exactly correct, because at B the moment is actually being transferred into the shaft, probably through a weldment. For the torsional stress, use Equation (3–40). Thus τT =

T 1.8 450 1.8 3+ = 3+ = 19 400 psi = 19.4 kpsi 2 ( 2 ( ) b∕c 1.25∕0.25 ) bc 1.25(0.25 )

This stress occurs on the outer surfaces at the middle of the 114 -in side. In addition to the torsional shear stress, a transverse shear stress exists at the same location and in the same direction on the visible side of BC, hence they are additive. On the opposite side, they subtract. The transverse shear stress, given by Equation (3–34), with V = F, is

τV =

3(300) 3F = = 1 440 psi = 1 .44 kpsi 2A 2(1.25)(0.25)

Figure 3–23 F

y

TC

C

4 in B M1

F

T1

x

z y

MA

TA A z

F

5 in

F

M2

B T2 x

Load and Stress Analysis     127

Adding this to τT gives the maximum shear stress of τmax = τT + τV = 19.4 + 1.44 = 20.84 kpsi

Answer

This stress occurs on the outer-facing surface at the middle of the 114 -in side. (c) For a stress element at A, the bending stress is tensile and is Answer

σx =

M 32M 32(1950) = = = 47 100 psi = 47.1 kpsi I∕c πd 3 π(0.75) 3

The torsional stress is Answer

τxz =

−T −16T −16(1200) = = = −14 500 psi = −14.5 kpsi J∕c πd 3 π(0.75) 3

where the reader should verify that the negative sign accounts for the direction of τxz. (d) Point A is in a state of plane stress where the stresses are in the xz plane. Thus, the principal stresses are given by Equation (3–13) with subscripts corresponding to the x, z axes. Answer The maximum normal stress is then given by

σ1 =

=

σx + σz σx − σz 2 + √( + τ2xz 2 2 ) 47.1 + 0 47.1 − 0 2 2 + √( ) + (−14.5) = 51.2 kpsi 2 2

Answer The maximum shear stress at A occurs on surfaces different from the surfaces containing the principal stresses or the surfaces containing the bending and torsional shear stresses. The maximum shear stress is given by Equation (3–14), again with modified subscripts, and is given by

τ1 = √ (

σx − σz 2 47.1 − 0 2 2 2 + τ = xz √( ) + (−14.5) = 27.7 kpsi 2 ) 2

EXAMPLE 3–9 The 1.5-in-diameter solid steel shaft shown in Figure 3–24a is simply supported at the ends. Two pulleys are keyed to the shaft where pulley B is of diameter 4.0 in and pulley C is of diameter 8.0 in. Considering bending and torsional stresses only, determine the locations and magnitudes of the greatest tensile, compressive, and shear stresses in the shaft. Solution Figure 3–24b shows the net forces, reactions, and torsional moments on the shaft. Although this is a threedimensional problem and vectors might seem appropriate, we will look at the components of the moment vector by performing a two-plane analysis. Figure 3–24c shows the loading in the xy plane, as viewed down the z axis, where bending moments are actually vectors in the z direction. Thus we label the moment diagram as Mz versus x. For the xz plane, we look down the y axis, and the moment diagram is My versus x as shown in Figure 3–24d.

128      Mechanical Engineering Design

y y

A z

10 in

A

800 lbf

200 lbf

10 in

z

B

200 lbf

10 in

1600 lbf ⋅ in B 10 in

1200 lbf

C 10 in

1000 lbf

600 lbf

x

100 lbf

x 400 lbf

(b)

y

1200 lbf

600 lbf B

C

200 lbf Mz (1bf · in)

D

400 lbf

(a)

A

10 in

C

D

500 lbf

1600 lbf ⋅ in

D

A

x

800 lbf

400 lbf

4000

B

z

My (1bf · in)

2000 x

O

C

D

x

400 lbf 8000 4000 x

O

(c)

(d) Location: at B (x = 10+) 8000 lbf · in

8246 lbf · in

F Max. compression and shear

β

2000 lbf · in

β = tan–1 8000 = 76° 2000

E Max. tension and shear

Figure 3–24

(e)

The net moment on a section is the vector sum of the components. That is,

M = √My2 + M 2z

At point B,

MB = √20002 + 80002 = 8246 lbf · in

At point C,

MC = √40002 + 40002 = 5657 lbf · in

(1)

Load and Stress Analysis     129

Thus the maximum bending moment is 8246 lbf · in and the maximum bending stress at pulley B is

σ=

M d∕2

=

4

πd ∕64

32M 32(8246) = = 24 890 psi = 24.89 kpsi πd 3 π(1.53 )

The maximum torsional shear stress occurs between B and C and is

τ=

T d∕2 4

πd ∕32

=

16T 16(1600) = = 2414 psi = 2.414 kpsi πd3 π(1.53 )

The maximum bending and torsional shear stresses occur just to the right of pulley B at points E and F as shown in Figure 3–24e. At point E, the maximum tensile stress will be σ1 given by Answer

σ1 =

σ σ 2 24.89 24.89 2 + √ ( ) + τ2 = + √( + 2.4142 = 25.12 kpsi 2 2 2 2 )

At point F, the maximum compressive stress will be σ2 given by Answer

σ2 =

−σ −σ 2 −24.89 −24.89 2 − √ ( ) + τ2 = − √( + 2.4142 = −25.12 kpsi 2 2 2 2 )

The extreme shear stress also occurs at E and F and is ±σ 2 ±24.89 2 τ1 = √ ( ) + τ2 = √ ( + 2.4142 = 12.68 kpsi 2 2 )

Answer

Closed Thin-Walled Tubes (t l∕2, then we would set dyAB∕dx = 0. For more complicated problems, plotting the equations using numerical data is the simplest approach to finding the maximum deflection. Sometimes it may not be obvious that we can use superposition with the tables at hand, as demonstrated in the next example.

EXAMPLE 4–3 Consider the beam in Figure 4–4a and determine the deflection equations using superposition. Solution For region AB we can superpose beams 7 and 10 of Table A–9 to obtain wx Fax 2 Answer yAB = (2lx 2 − x 3 − l 3 ) + (l − x 2 ) 24EI 6EI l For region BC, how do we represent the uniform load? Considering the uniform load only, the beam deflects as shown in Figure 4–4b. Region BC is straight since there is no bending moment due to w. The slope of the beam at B is θB and is obtained by taking the derivative of y given in the table with respect to x and setting x = l. Thus, dy d wx w = [ (2lx 2 − x 3 − l 3 )] = (6lx 2 − 4x 3 − l 3 ) dx dx 24EI 24EI

Substituting x = l gives

θB =

w wl 3 (6ll 2 − 4l 3 − l 3 ) = 24EI 24EI

The deflection in region BC due to w is θB(x − l), and adding this to the deflection due to F, in BC, yields F(x − l) wl 3 Answer yBC = (x − l) + [(x − l) 2 − a(3x − l)] 24EI 6EI y

Figure 4–4

y a

l w B

A

R2

R1 (a)

F C

w

B

A x

l x (b)

θB

yBC = θB(x – l) x C

(a) Beam with uniformly distributed load and overhang force; (b) deflections due to uniform load only.

182      Mechanical Engineering Design

EXAMPLE 4–4 Figure 4–5a shows a cantilever beam with an end load. Normally we model this problem by considering the left support as rigid. After testing the rigidity of the wall it was found that the translational stiffness of the wall was kt force per unit vertical deflection, and the rotational stiffness was kr moment per unit angular (radian) deflection (see Figure 4–5b). Determine the deflection equation for the beam under the load F. Solution Here we will superpose the modes of deflection. They are: (1) translation due to the compression of spring kt, (2) rotation of the spring kr, and (3) the elastic deformation of beam 1 given in Table A–9. The force in spring kt is R1 = F, giving a deflection from Equation (4–2) of F y1 = − kt

(1)

The moment in spring kr is M1 = Fl. This gives a clockwise rotation of θ = Fl∕kr. Considering this mode of deflection only, the beam rotates rigidly clockwise, leading to a deflection equation of

y2 = −

Fl x kr

(2)

Finally, the elastic deformation of beam 1 from Table A–9 is

y3 =

Fx2 (x − 3l) 6EI

(3)

Adding the deflections from each mode yields Fx2 F Fl Answer y= (x − 3l) − − x 6EI kt kr

Figure 4–5

kr

y l

F

F

x x

kt

M1 R1

R1 (a)

(b)

4–6  Beam Deflections by Singularity Functions Introduced in Section 3–3, singularity functions are excellent for managing discontinuities, and their application to beam deflection is a simple extension of what was presented in the earlier section. They are easy to program, and as will be seen later, they can greatly simplify the solution of statically indeterminate problems. The following examples illustrate the use of singularity functions to evaluate deflections of statically determinate beam problems.

Deflection and Stiffness     183

EXAMPLE 4–5 Consider beam 6 of Table A–9, which is a simply supported beam having a concentrated load F not in the center. Develop the deflection equations using singularity functions. Solution First, write the load intensity equation from the free-body diagram, q = R1⟨x⟩−1 − F⟨x − a⟩−1 + R2⟨x − l⟩−1

(1)

Integrating Equation (1) twice results in

V = R1⟨x⟩0 − F⟨x − a⟩0 + R2⟨x − l ⟩0

M = R1⟨x⟩1 − F⟨x − a⟩1 + R2⟨x − l⟩1 (3)

(2)

Recall that as long as the q equation is complete, integration constants are unnecessary for V and M; therefore, they are not included up to this point. From statics, setting V = M = 0 for x slightly greater than l yields R1  = Fb∕l and R2 = Fa∕l. Thus Equation (3) becomes

M=

Fb Fa ⟨x⟩1 − F ⟨x − a⟩1 + ⟨x − l⟩1 l l

Integrating Equations (4–12) and (4–13) as indefinite integrals gives

EI

dy Fb 2 F Fa = ⟨x⟩ − ⟨x − a⟩2 + ⟨x − l⟩2 + C1 dx 2l 2 2l

EI y =

Fb F Fa ⟨x⟩3 − ⟨x − a⟩3 + ⟨x − l⟩3 + C1 x + C2 6l 6 6l

Note that the first singularity term in both equations always exists, so ⟨x⟩2 = x2 and ⟨x⟩3 = x3 . Also, the last singularity term in both equations does not exist until x = l, where it is zero, and since there is no beam for x > l we can drop the last term. Thus,

EI

dy Fb 2 F = x − ⟨x − a⟩2 + C1 dx 2l 2

EI y =

(4)

Fb 3 F x − ⟨x − a⟩3 + C1x + C2 6l 6

(5)

The constants of integration C1 and C2 are evaluated by using the two boundary conditions y = 0 at x = 0 and y = 0 at x = l. The first condition, substituted into Equation (5), gives C2 = 0 (recall that ⟨0 − a⟩3 = 0). The second condition, substituted into Equation (5), yields

0=

Fb 3 F Fbl 2 Fb3 l − (l − a) 3 + C1l = − + C1l 6l 6 6 6

Solving for C1 gives Fb 2 (l − b2 ) 6l Finally, substituting C1 and C2 in Equation (5) and simplifying produces

C1 = −

y=

F [bx(x2 + b2 − l 2 ) − l ⟨x − a⟩3] 6EI l

(6)

184      Mechanical Engineering Design

Comparing Equation (6) with the two deflection equations for beam 6 in Table A–9, we note that the use of singularity functions enables us to express the deflection equation with a single discontinuous equation. For x ≤ a, the singularity function in Equation (6) does not exist, and we see that Equation (6) perfectly agrees with yAB from Table A–9. However, for x ≥ a, Equation (6) appears different from yBC. This is because Equation (6) uses a, b, and l, whereas yBC only uses a and l. Substituting b = l − a into Equation (6) and expanding, we get

y=−

Fa 3 (x − 3lx2 + 2l2x + a2x − la2 ) 6EI l

Expanding yBC from Table A–9 yields the same results.

EXAMPLE 4–6 Determine the deflection equation for the simply supported beam with the load distribution shown in Figure 4–6. Solution This is a good beam to add to our table for later use with superposition. The load intensity equation for the beam is q = R1⟨x⟩−1 − w⟨x⟩0 + w⟨x − a⟩0 + R2⟨x − l⟩−1

(1)

where the w⟨x − a⟩0 is necessary to "turn off" the uniform load at x = a. From statics, the reactions are

R1 =

wa (2l − a) 2l

R2 =

wa2 2l

(2)

For simplicity, we will retain the form of Equation (1) for integration and substitute the values of the reactions in later. Two integrations of Equation (1) reveal V = R1⟨x⟩0 − w⟨x⟩1 + w⟨x − a⟩1 + R2⟨x − l⟩0

M = R1⟨x⟩1 −

w w ⟨x⟩2 + ⟨x − a⟩2 + R2⟨x − l⟩1 2 2

(3) (4)

As in the previous example, singularity functions of order zero or greater starting at x = 0 can be replaced by normal polynomial functions. Also, once the reactions are determined, singularity functions starting at the extreme right end of the beam can be omitted. Thus, Equation (4) can be rewritten as

M = R1x −

w 2 w x + ⟨x − a⟩2 2 2

y

Figure 4–6

l a w B

A R1

C R2

x

(5)

Deflection and Stiffness     185

Integrating two more times for slope and deflection gives

EI

EI y =

dy R1 2 w 3 w = x − x + ⟨ x − a⟩3 + C1 dx 2 6 6 R1 3 w 4 w x − x + ⟨x − a⟩4 + C1x + C2 6 24 24

(6) (7)

The boundary conditions are y = 0 at x = 0 and y = 0 at x = l. Substituting the first condition in Equation (7) shows C2 = 0. For the second condition 0=

R1 3 w 4 w l − l + (l − a) 4 + C1l 6 24 24

Solving for C1 and substituting into Equation (7) yields

EI y =

R1 w w w x(x 2 − l 2 ) − x(x 3 − l 3 ) − x(l − a) 4 + ⟨x − a⟩4 6 24 24l 24

Finally, substitution of R1 from Equation (2) and simplifying results gives Answer

y=

w [2ax(2l − a)(x 2 − l 2 ) − xl(x3 − l 3 ) − x(l − a) 4 + l⟨x − a⟩4] 24EI l

As stated earlier, singularity functions are relatively simple to program, as they are omitted when their arguments are negative, and the ⟨ ⟩ brackets are replaced with ( ) parentheses when the arguments are positive.

EXAMPLE 4–7 The steel step shaft shown in Figure 4–7a is mounted in bearings at A and F. A pulley is centered at C where a total radial force of 600 lbf is applied. Using singularity functions evaluate the shaft displacements at 12 -in increments. Assume the shaft is simply supported. Solution The reactions are found to be R1 = 360 lbf and R2 = 240 lbf. Ignoring R2, using singularity functions, the moment equation is M = 360x − 600⟨x − 8⟩1

(1)

This is plotted in Figure 4–7b. For simplification, we will consider only the step at D. That is, we will assume section AB has the same diameter as BC and section EF has the same diameter as DE. Since these sections are short and at the supports, the size reduction will not add much to the deformation. We will examine this simplification later. The second area moments for BC and DE are

IBC =

π 1.54 = 0.2485 in4 64

IDE =

π 1.754 = 0.4604 in4 64

186      Mechanical Engineering Design

Figure 4–7

600 lbf

y

Dimensions in inches.

1.500

1.000 A B

C

1.750

D

1.000

E F

x 0.5 (a)

R1

8

8.5

19.5

M

2880 lbf-in

M/I

a

R2 20

2760 lbf-in

(b) b c d

(c)

A plot of M∕I is shown in Figure 4–7c. The values at points b and c, and the step change are M 2760 3 ( I )b = 0.2485 = 11 106.6 lbf/in

M 2760 3 ( I )c = 0.4604 = 5 994.8 lbf/in

M Δ ( ) = 5 994.8 − 11 106.6 = −5 111.8 lbf/in3 I

The slopes for ab and cd, and the change are 2760 − 2880 = −965.8 lbf/in4 0.2485(0.5)

mab =

Δm = −521.3 − (−965.8) = 444.5 lbf/in4

mcd =

−5 994.8 = −521.3 lbf/in4 11.5

Dividing Equation (1) by IBC and, at x = 8.5 in, adding a step of −5 111.8 lbf/in3 and a ramp of slope 444.5 lbf/in4, gives

M = 1 448.7x − 2 414.5⟨x − 8⟩1 − 5 111.8⟨x − 8.5⟩0 + 444.5⟨x − 8.5⟩1 I

(2)

dy = 724.35x 2 − 1207.3⟨x − 8⟩2 − 5 111.8⟨x − 8.5⟩1 + 222.3⟨x − 8.5⟩2 + C1 dx

(3)

Integration gives

E

Integrating again yields

Ey = 241.5x 3 − 402.4⟨x − 8⟩3 − 2 555.9⟨x − 8.5⟩2 + 74.08⟨x − 8.5⟩3 + C1x + C2

(4)

At x = 0, y = 0. This gives C2 = 0 (remember, singularity functions do not exist until the argument is positive). At x = 20 in, y = 0, and

0 = 241.5(20) 3 − 402.4(20 − 8)3 − 2 555.9(20 − 8.5)2 + 74.08(20 − 8.5)3 + C1 (20)

Deflection and Stiffness     187

Solving, gives C1 = −50 565 lbf/in2. Thus, Equation (4) becomes, with E = 30(10)6 psi,

y=

1 (241.5x3 − 402.4⟨x − 8⟩3 − 2 555.9⟨x − 8.5⟩2 30(106 ) + 74.08⟨x − 8.5⟩ − 50 565x) 3

(5)

When using a spreadsheet, program the following equations:

y=

1 (241.5x3 − 50 565x) 6 30(10 )

0 ≤ x ≤ 8 in

y=

1 [241.5x3 − 402.4(x − 8) 3 − 50 565x] 6 30(10 )

8 ≤ x ≤ 8.5 in

y=

1 [241.5x3 − 402.4(x − 8) 3 − 2 555.9(x − 8.5) 2 6 30(10 ) + 74.08(x − 8.5) 3 − 50 565x]

8.5 ≤ x ≤ 20 in

The following table results. x

y

x

y

x

y

x

y

x

y

0

−0.000000

4.5

−0.006851

9

−0.009335

13.5

−0.007001

18

−0.002377

0.5

−0.000842

5

−0.007421

9.5

−0.009238

14

−0.006571

18.5

−0.001790

1

−0.001677

5.5

−0.007931

10

−0.009096

14.5

−0.006116

19

−0.001197

1.5

−0.002501

6

−0.008374

10.5

−0.008909

15

−0.005636

19.5

−0.000600

2

−0.003307

6.5

−0.008745

11

−0.008682

15.5

−0.005134

20

0.000000

2.5

−0.004088

7

−0.009037

11.5

−0.008415

16

−0.004613

3

−0.004839

7.5

−0.009245

12

−0.008112

16.5

−0.004075

3.5

−0.005554

8

−0.009362

12.5

−0.007773

17

−0.003521

4

−0.006227

8.5

−0.009385

13

−0.007403

17.5

−0.002954

where x and y are in inches. We see that the greatest deflection is at x = 8.5 in, where y = −0.009385 in. Substituting C1 into Equation (3) the slopes at the supports are found to be θA = 1.686(10−3) rad = 0.09657 deg, and θF = 1.198(10−3) rad = 0.06864 deg. You might think these to be insignificant deflections, but as you will see in Chapter 7, on shafts, they are not. A finite-element analysis was performed for the same model and resulted in y∣ x=8.5 in = −0.009380 in  θA = −0.09653°  θF = 0.06868° Virtually the same answer save some round-off error in the equations. If the steps of the bearings were incorporated into the model, more equations result, but the process is the same. The solution to this model is y∣ x=8.5 in = −0.009387 in  θA = −0.09763°  θF = 0.06973° The largest difference between the models is of the order of 1.5 percent. Thus the simplification was  justified.

188      Mechanical Engineering Design

In Section 4–9, we will demonstrate the usefulness of singularity functions in solving statically indeterminate problems.

4–7  Strain Energy The external work done on an elastic member in deforming it is transformed into strain, or potential, energy. If the member is deformed a distance y, and if the forcedeflection relationship is linear, this energy is equal to the product of the average force and the deflection, or U=

F F2 y= 2 2k

(4–15)

This equation is general in the sense that the force F can also mean torque, or moment, provided, of course, that consistent units are used for k. By substituting appropriate expressions for k, strain-energy formulas for various simple loadings may be obtained. For tension and compression, for example, we employ Equation (4–4) and obtain

F 2l 2AE

 tension and compression F2  U= dx  2AE

or

U=

(4–16) (4–17)

where the first equation applies when all the terms are constant throughout the length, and the more general integral equation allows for any of the terms to vary through the length. Similarly, from Equation (4–7), the strain energy for torsion is given by or

U=

T 2l 2GJ

U=

 torsion

T2  dx  2GJ

(4–18) (4–19)

To obtain an expression for the strain energy due to direct shear, consider the element with one side fixed in Figure 4–8a. The force F places the element in pure shear, and the work done is U = Fδ∕2. Since the shear strain is γ = δ∕l = τ∕G = F∕AG, we have

F 2l 2AG

 direct shear F2  U= dx  2AG

or

U=

Figure 4–8

O

δ

ρ

F dθ

γ

A

F l

F

ds

B

dx (a) Pure shear element

(b) Beam bending element

(4–20) (4–21)

Deflection and Stiffness     189

The strain energy stored in a beam or lever by bending may be obtained by referring to Figure 4–8b. Here AB is a section of the elastic curve of length ds having a radius of curvature ρ. The strain energy stored in this element of the beam is dU = (M∕2)dθ. Since ρ dθ = ds, we have

M ds 2ρ

dU =

(a)

We can eliminate ρ by using Equation (4–8), ρ = EI∕M. Thus

M 2 ds 2EI

dU =

(b)

For small deflections, ds ≈ dx. Then, for the entire beam

U=

2

M dx ∫ dU = ∫ 2EI

(c)

The integral equation is commonly needed for bending, where the moment is typically a function of x. Summarized to include both the integral and nonintegral form, the strain energy for bending is

M 2l 2EI

 bending M2  U= dx  2EI

or

U=

(4–22) (4–23)

Equations (4–22) and (4–23) are exact only when a beam is subject to pure bending. Even when transverse shear is present, these equations continue to give quite good results, except for very short beams. The strain energy due to shear loading of a beam is a complicated problem. An approximate solution can be obtained by using Equation (4–20) with a correction factor whose value depends upon the shape of the cross section. If we use C for the correction factor and V for the shear force, then the strain energy due to shear in bending is or

CV 2l 2AG

 transverse shear CV 2  U= dx  2AG U=

Values of the factor C are listed in Table 4–1. Table 4–1  Strain-Energy Correction Factors for Transverse Shear Beam Cross-Sectional Shape Rectangular Circular Thin-walled tubular, round Box sections† Structural sections† †

Factor C 1.2 1.11 2.00 1.00 1.00

Use area of web only. Source: Richard G. Budynas, Advanced Strength and Applied Stress Analysis, 2nd ed., McGraw-Hill, New York, 1999. Copyright © 1999 The McGraw-Hill Companies.

(4–24) (4–25)

190      Mechanical Engineering Design

EXAMPLE 4–8 A cantilever beam with a round cross section has a concentrated load F at the end, as shown in Figure 4–9a. Find the strain energy in the beam. x

Figure 4–9

F

l ymax

F M V (b)

(a)

Solution To determine what forms of strain energy are involved with the deflection of the beam, we break into the beam and draw a free-body diagram to see the forces and moments being carried within the beam. Figure 4–9b shows such a diagram in which the transverse shear is V = −F, and the bending moment is M = −Fx. The variable x is simply a variable of integration and can be defined to be measured from any convenient point. The same results will be obtained from a free-body diagram of the right-hand portion of the beam with x measured from the wall. Using the free end of the beam usually results in reduced effort since the ground reaction forces do not need to be determined. For the transverse shear, using Equation (4–24) with the correction factor C = 1.11 from Table 4–1, and noting that V is constant through the length of the beam,

Ushear =

CV 2l 1.11F 2l = 2AG 2AG

For the bending, since M is a function of x, Equation (4–23) gives

Ubend =

M 2dx 1 = 2EI 2EI

l 0

(−Fx) 2dx =

F 2l 3 6EI

The total strain energy is Answer

U = Ubend + Ushear =

F 2l 3 1.11F 2l + 6EI 2AG

Note, except for very short beams, the shear term (of order l) is typically small compared to the bending term (of order l3). This will be demonstrated in the next example.

4–8  Castigliano's Theorem A most unusual, powerful, and often surprisingly simple approach to deflection analysis is afforded by an energy method called Castigliano's theorem. It is a unique way of analyzing deflections and is even useful for finding the reactions of indeterminate structures. Castigliano's theorem states that when forces act on elastic systems subject to small displacements, the displacement corresponding to any force, in the direction of the force, is equal to the partial derivative of the total strain energy with respect to that force. The terms force and displacement in this statement are broadly

Deflection and Stiffness     191

interpreted to apply equally to moments and angular displacements. Mathematically, the theorem of Castigliano is ∂U δi = (4–26) ∂Fi where δi is the displacement of the point of application of the force Fi in the direction of Fi. For rotational displacement Equation (4–26) can be written as ∂U (4–27) ∂Mi where θi is the rotational displacement, in radians, of the beam where the moment Mi exists and in the direction of Mi. As an example, apply Castigliano's theorem using Equations (4–16) and (4–18) to get the axial and torsional deflections. The results are

θi =

∂ F 2l Fl = ∂F ( 2AE ) AE ∂ T 2l Tl θ= = ∂T ( 2GJ ) GJ

δ=

(a) (b)

Compare Equations (a) and (b) with Equations (4–3) and (4–5). EXAMPLE 4–9 The cantilever of Example 4–8 is a carbon steel bar 10 in long with a 1-in diameter and is loaded by a force F = 100 lbf. (a) Find the maximum deflection using Castigliano's theorem, including that due to shear. (b) What error is introduced if shear is neglected? Solution (a) From Example 4–8, the total energy of the beam is

U=

F 2l 3 1.11F 2l + 6EI 2AG

(1)

Then, according to Castigliano's theorem, the deflection of the end is We also find that

ymax =

∂U Fl 3 1.11Fl = + ∂F 3EI AG

(2)

πd 4 π(1) 4 = = 0.0491 in4 64 64 πd 2 π(1) 2 A= = = 0.7854 in2 4 4 I=

Substituting these values, together with F = 100 lbf, l = 10 in, E = 30 Mpsi, and G = 11.5 Mpsi, in Equation (3) gives Answer ymax = 0.022 63 + 0.000 12 = 0.022 75 in Note that the result is positive because it is in the same direction as the force F. Answer (b) The error in neglecting shear for this problem is (0.02275 − 0.02263)∕0.02275 = 0.0053 = 0.53 percent.

192      Mechanical Engineering Design

The relative contribution of transverse shear to beam deflection decreases as the length-to-height ratio of the beam increases, and is generally considered negligible for l∕d > 10. Note that the deflection equations for the beams in Table A–9 do not include the effects of transverse shear. Castigliano's theorem can be used to find the deflection at a point even though no force or moment acts there. The procedure is as follows: 1 Set up the equation for the total strain energy U by including the energy due to a fictitious force or moment Q acting at the point whose deflection is to be found. 2 Find an expression for the desired deflection δ, in the direction of Q, by taking the derivative of the total strain energy with respect to Q. 3 Since Q is a fictitious force, solve the expression obtained in step 2 by setting Q equal to zero. Thus, the displacement at the point of application of the fictitious force Q is

δ=

∂U ∂Q⎹ Q=0

(4–28)

In cases where integration is necessary to obtain the strain energy, it is more efficient to obtain the deflection directly without explicitly finding the strain energy, by moving the partial derivative inside the integral. For the example of the bending case,

δi =

∂U ∂ M ∂ M = dx = dx = ∂Fi ∂Fi ( 2EI ) ∂Fi ( 2EI )

2

2

∂M ∂Fi 1 ∂M dx = M dx 2EI EI ( ∂Fi )

2M

This allows the derivative to be taken before integration, simplifying the mathematics. This method is especially helpful if the force is a fictitious force Q, since it can be set to zero as soon as the derivative is taken. The expressions for the common cases in Equations (4–17), (4–19), and (4–23) are rewritten as

∂U 1 ∂F = F dx    tension and compression ∂Fi AE ( ∂Fi )

∫ ∂U 1 ∂T θ = = ∫ T dx  torsion ∂M GJ ( ∂M ) ∂U 1 ∂M δ = = ∫ (M dx   bending ∂F EI ∂F ) δi = i

i

(4–30)

i

i

i

(4–29)

(4–31)

i

EXAMPLE 4–10 Using Castigliano's method, determine the deflections of points A and B due to the force F applied at the end of the step shaft shown in Figure 4–10. The second area moments for sections AB and BC are I1 and 2I1, respectively. Solution To avoid the need to determine the support reaction forces, define the origin of x at the left end of the beam as shown. For 0 ≤ x ≤ l, the bending moment is

M = −Fx

(1)

Deflection and Stiffness     193

Figure 4–10

y l/2

x

A

l/2

I1

2I1

B

F

C

Q

Since F is at A and in the direction of the desired deflection, the deflection at A from Equation (4–31) is

δA =

∂U = ∂F

l 0

1 ∂M M dx EI ( ∂F )

(2)

Substituting Equation (1) into Equation (2), noting that I = I1 for 0 ≤ x ≤ l∕2, and I = 2I1 for l∕2 ≤ x ≤ l, we get 1 1 δA = [ (−Fx)(−x) dx + E 0 I1 Answer 1 Fl 3 7Fl 3 3 Fl 3 = [ + = E 24I1 48I1 ] 16 EI1

l∕2

l

l∕2

1 (−Fx) (−x) dx ] 2I1

which is positive, as it is in the direction of F. For B, a fictitious force Q is necessary at the point. Assuming Q acts down at B, and x is as before, the moment equation is

M = −Fx         0 ≤ x ≤ l∕2

l M = −Fx − Q (x − )   l∕2 ≤ x ≤ l 2

(3)

For Equation (4–31), we need ∂M∕∂Q. From Equation (3),

∂M = 0       0 ≤ x ≤ l∕2 ∂Q

∂M l = − (x − )   l∕2 ≤ x ≤ l ∂Q 2

Once the derivative is taken, Q can be set to zero, so Equation (4–31) becomes

δB = [

=

l 0

1 EI1

1 ∂M M dx EI ( ∂Q ) ]Q=0

l∕2 0

(−Fx)(0)dx +

1 E(2I1 )

l (−Fx) [ − (x − )] dx 2 l∕2

l

Evaluating the last integral gives F x 3 lx 2 l 5 Fl 3 Answer δB = − = 2EI1 ( 3 4 )⎹ l∕2 96 EI1 which again is positive, in the direction of Q.

(4)

194      Mechanical Engineering Design

EXAMPLE 4–11 For the wire form of diameter d shown in Figure 4–11a, determine the deflection of point B in the direction of the applied force F (neglect the effect of transverse shear). Solution Figure 4–11b shows free body diagrams where the body has been broken in each section, and internal balancing forces and moments are shown. The sign convention for the force and moment variables is positive in the directions shown. With energy methods, sign conventions are arbitrary, so use a convenient one. In each section, the variable x is defined with its origin as shown. The variable x is used as a variable of integration for each section independently, so it is acceptable to reuse the same variable for each section. For completeness, the transverse shear forces are included, but the effects of transverse shear on the strain energy (and deflection) will be neglected. Element BC is in bending only so from Equation (4–31),5 ∂UBC 1 = ∂F EI

a

0

(Fx) (x) dx =

Fa 3 3EI

(1)

Element CD is in bending and in torsion. The torsion is constant so Equation (4–30) can be written as ∂U ∂T l = (T ) ∂Fi ∂Fi GJ

G

Figure 4–11

c F

D

B a

b

C

(a)

MDG2 = Fb F F

F

MCD = Fx

B

B

C

C VBC = F

D MDG1 = Fa

a

a x

x

B

TCD = Fa

x

MBC = Fx

F

b

VCD = F

(b)

5

It is very tempting to mix techniques and try to use superposition also, for example. However, some subtle things can occur that you may visually miss. It is highly recommended that if you are using Castigliano's theorem on a problem, you use it for all parts of the problem.

Deflection and Stiffness     195

where l is the length of the member. So for the torsion in member CD, Fi = F, T = Fa, and l = b. Thus, ∂UCD b Fa2b = (Fa) (a) = ( ∂F )torsion GJ GJ

(2)

For the bending in CD, ∂UCD 1 ( ∂F )bending = EI

b 0

(Fx) (x) dx =

Fb3 3EI

(3)

Member DG is axially loaded and is bending in two planes. The axial loading is constant, so Equation (4–29) can be written as ∂U ∂F l = (F ) ∂Fi ∂Fi AE

where l is the length of the member. Thus, for the axial loading of DG, Fi = F, l = c, and ∂UDG Fc ( ∂F )axial = AE

(4)

The bending moments in each plane of DG are constant along the length, with MDG2 = Fb and MDG1 = Fa. Considering each one separately in the form of Equation (4–31) gives

∂UDG 1 ( ∂F )bending = EI

=

c

c

∫ (Fb)(b) dx + EI1 ∫ (Fa)(a) dx 0

0

2

2

Fc (a + b ) EI

(5)

Adding Equations (1) to (5), noting that I = πd4∕64, J = 2I, A = πd2∕4, and G = E∕[2(1 + ν)], we find that the deflection of B in the direction of F is Answer

(δB ) F =

4F [16(a3 + b3 ) + 48c(a2 + b2 ) + 48(1 + ν)a2b + 3cd 2] 3πEd 4

Now that we have completed the solution, see if you can physically account for each term in the result using an independent method such as superposition.

4–9  Deflection of Curved Members Machine frames, springs, clips, fasteners, and the like frequently occur as curved shapes. The determination of stresses in curved members has already been described in Section 3–18. Castigliano's theorem is particularly useful for the analysis of deflections in curved parts too.6 Consider, for example, the curved frame of Figure 4–12a. We are interested in finding the deflection of the frame due to F and in the direction of F. Unlike straight beams, the bending moment 6

For more solutions than are included here, see Joseph E. Shigley, "Curved Beams and Rings," Chapter 38 in Joseph E. Shigley, Charles R. Mischke, and Thomas H. Brown, Jr. (eds.), Standard Handbook of Machine Design, 3rd ed., McGraw-Hill, New York, 2004.

196      Mechanical Engineering Design

Figure 4–12 (a) Curved bar loaded by force F. R = radius to centroidal axis of section; h = section thickness. (b) Diagram showing forces acting on section taken at angle θ. Fr = V = shear component of F; Fθ is component of F normal to section; M is moment caused by force F.

Fr

M F

h R θ

θ

F (b)

(a)

and axial force are coupled for curved beams, creating an additional energy term.7 The energy due to the moment alone is

U1 =

2

M dθ ∫ 2AeE

(4–32)

In this equation, the eccentricity e is

e = R − rn

(4–33)

where rn is the radius of the neutral axis as defined in Section 3–18 and shown in Figure 3–35. Analogous to Equation (4–17), the strain energy component due to the axial force Fθ alone is

U2 =

F 2θ R dθ 2AE

(4–34)

The additional coupling term between M and Fθ is

U3 = −

∫ MFAEdθ θ

(4–35)

The negative sign of Equation (4–35) can be appreciated by referring to both parts of Figure 4–12. Note that the moment M tends to decrease the angle dθ. On the other hand, Fθ tends to increase dθ. Thus U3 is negative. If Fθ had been acting in the opposite direction, then both M and Fθ would tend to decrease the angle dθ. The fourth and last term is the transverse shear energy due to Fr. Adapting Equation (4–25) gives

U4 =

CF 2r R dθ 2AG

(4–36)

where C is the correction factor of Table 4–1. Combining the four terms gives the total strain energy

7

U=

2

2 θ

2 r

MF dθ R dθ CF R dθ − ∫ + ∫ ∫ M2AeEdθ + ∫ F2AE AE 2AG θ

See Richard G. Budynas, Advanced Strength and Applied Stress Analysis, 2nd ed., Section 6.7, McGraw-Hill, New York, 1999.

(4–37)

Deflection and Stiffness     197

The deflection produced by the force F can now be found. It is

δ=

∂U = ∂F

F R ∂F M ∂M ∫ AeE ( ∂F ) dθ + ∫ AE ( ∂F ) dθ θ

θ

) CF R ∂F dθ + ∫ dθ ∫ AE1 ∂ (MF ∂F AG ( ∂F ) θ

r

r

(4–38)

This equation is general and may be applied to any section of a thick-walled circular curved beam with application of appropriate limits of integration. For the specific curved beam in Figure 4–12b, the integrals are evaluated from 0 to π. Also, for this case we find M = FR sin θ

∂M = R sin θ ∂F

Fθ = F sin θ

∂Fθ = sin θ ∂F

   MFθ = F 2R sin2 θ    Fr = F cos θ

   ∂ (MFθ ) = 2FR sin2 θ ∂F ∂Fr = cos θ ∂F

Substituting these into Equation (4–38) and factoring yields

δ=

FR 2 AeE

π

π

2

0

2

0

π

FR sin θ dθ ∫ sin θ dθ + AE ∫ sin θ dθ − 2FR AE ∫ +

=

2

CFR AG

0 π

2

∫ cos θ dθ 2

0

2

πFR πFR πFR πCFR πFR πFR πCFR + − + = − + 2AeE 2AE AE 2AG 2AeE 2AE 2AG

(4–39)

Because the first term contains the square of the radius, the second two terms will be small if the frame has a large radius. For curved sections in which the radius is significantly larger than the thickness, say R∕h > 10, the effect of the eccentricity is negligible, so that the strain energies can be approximated directly from Equations (4–17), (4–23), and (4–25) with a substitution of R dθ for dx. Further, as R increases, the contributions to deflection from the normal force and tangential force becomes negligibly small compared to the bending component. Therefore, an approximate result can be obtained for a thin circular curved member as 2

U≈

M R dθ ∫ 2EI

δ=

∂U ≈ ∂F

  R∕h > 10

(4–40)

R dθ  R∕h > 10 ∫ EI1 (M ∂M ∂F )

(4–41)

198      Mechanical Engineering Design

EXAMPLE 4–12 The cantilevered hook shown in Figure 4–13a is formed from a round steel wire with a diameter of 2 mm. The hook dimensions are l = 40 and R = 50 mm. A force P of 1 N is applied at point C. Use Castigliano's theorem to estimate the deflection at point D at the tip. Solution Since l∕d and R∕d are significantly greater than 10, only the contributions due to bending will be ­considered. To obtain the vertical deflection at D, a fictitious force Q will be applied there. Freebody  diagrams are shown in Figures 4–13b, c, and d, with breaks in sections AB, BC, and CD, respectively. The normal and shear forces, N and V respectively, are shown but are considered negligible in the deflection analysis. For section AB, with the variable of integration x defined as shown in Figure 4–13b, summing moments about the break gives an equation for the moment in section AB,

MAB = P(R + x) + Q(2R + x)

(1)

∂MAB∕∂Q = 2R + x

(2)

Since the derivative with respect to Q has been taken, we can set Q equal to zero. From Equation (4–31), inserting Equations (1) and (2),

∂MAB 1 1 (δD ) AB = [ (MAB ∂Q ) dx ]Q=0 = EI EI 0

P EI

=

l

l

∫ (2R 0

2

+ 3Rx + x 2 ) dx =

Figure 4–13

l 0

P(R + x)(2R + x)dx

P 3 1 2R 2l + l 2R + l 3) EI ( 2 3

(3)

l

A

D

B

R P

C (a)

VAB

MAB

Q

Q

x

D

D

B

R P

Q

NBC VBC

θ

P

MBC

VCD NCD

C

C

(b)

(c)

D

θ

R

MCD

(d)

Deflection and Stiffness     199

For section BC, with the variable of integration θ defined as shown in Figure 4–13c, summing moments about the break gives the moment equation for section BC.

MBC = Q(R + R sin θ) + PR sin θ

(4)

∂MBC∕∂Q = R(1 + sin θ)

(5)

From Equation (4–41), inserting Equations (4) and (5) and setting Q = 0, we get (δD ) BC = [

=

π∕2

0

∂MBC 1 R MBC R dθ ] = EI ( ∂Q ) EI Q=0

PR 3 π 1+ ) EI ( 4

π∕2

0

(PR sin θ)[R(1 + sin θ) ] dθ

(6)

Noting that the break in section CD contains nothing but Q, and after setting Q = 0, we can conclude that there is no actual strain energy contribution in this section. Combining terms from Equations (3) and (6) to get the total vertical deflection at D,

δD = (δD ) AB + (δD ) BC =

=

P 3 2 1 3 PR3 π 2 2R l + l R + l + 1+ ) EI ( 2 3 ) EI ( 4

P (1.785R 3 + 2R 2l + 1.5 Rl 2 + 0.333l 3 ) EI

(7)

Substituting values, and noting I = πd4∕64, and E = 207 GPa for steel, we get Answer  δD =

1 [1.785(0.053 ) + 2(0.052 )0.04 + 1.5(0.05)0.042 + 0.333(0.043 )] 207(10 )[π(0.0024 )∕64] 9

= 3.47(10−3 ) m = 3.47 mm

The general result expressed in Equation (4–39) is useful in sections that are uniform and in which the centroidal locus is circular. The bending moment is largest where the material is farthest from the load axis. Strengthening requires a larger second area moment I. A variable-depth cross section is attractive, but it makes the integration to a closed form very difficult. However, if you are seeking results, numerical integration with computer assistance is helpful.

EXAMPLE 4–13 Deflection in a Variable-Cross-Section Punch-Press Frame Consider the steel C frame depicted in Figure 4–14a in which the centroidal radius is 32 in, the cross section at the ends is 2 in × 2 in, and the depth varies sinusoidally with an amplitude of 2 in. The load is 1000 lbf. It follows that C = 1.2, G = 11.5(106) psi, E = 30(106) psi. The outer and inner radii are

Rout = 33 + 2 sin θ  Rin = 31 − 2 sin θ

200      Mechanical Engineering Design

Figure 4–14 (a) A steel punch press has a C frame with a varying-depth rectangular cross section depicted. The cross section varies sinusoidally from 2 in × 2 in at θ = 0° to 2 in × 6 in at θ = 90°, and back to 2 in × 2 in at θ = 180°. Of immediate interest to the designer is the deflection in the load axis direction under the load. (b) Finite-element model.

32-in R

1000 lbf

θ

1000 lbf

1000 lbf

(a)

(b)

The remaining geometrical terms are h = Rout − Rin = 2(1 + 2 sin θ) A = bh = 4(1 + 2 sin θ) rn =

2(1 + 2 sin θ) h = ln(Rout∕Rin ) ln[(33 + 2 sin θ)∕(31 − 2 sin θ)]

e = R − rn = 32 − rn Note that M = FR sin θ

∂M∕∂F = R sin θ

Fθ = F sin θ

∂Fθ∕∂F = sin θ

2

   2

MFθ = F R sin θ    ∂M Fθ∕∂F = 2FR sin2 θ Fr = F cos θ

   ∂Fr∕∂F = cos θ

Substitution of the terms into Equation (4–38) yields three integrals

δ = I1 + I2 + I3

(1)

where the integrals are π

sin2 θ dθ 2(1 + 2 sin θ) 0 32 − (1 + 2 sin θ) 33 + 2 sin θ ln ( 31 − 2 sin θ )

I1 = 8.5333(10−3 )

I2 = −2.6667(10−4 )

I3 = 8.3478(10−4 )

[

π

sin2 θ dθ 1 + 2 sin θ 0

(3)

(4)

π

cos2 θ dθ 1 + 2 sin θ 0

]

(2)

Deflection and Stiffness     201

The integrals may be evaluated in a number of ways: by a program using Simpson's rule integration,8 by a program using a spreadsheet, or by mathematics software. Using MathCad and checking the results with Excel gives the integrals as I1 = 0.076 615, I2 = −0.000 159, and I3 = 0.000 773. Substituting these into Equation (1) gives δ = 0.077 23 in

Answer

Finite-element (FE) programs are also very accessible. Figure 4–14b shows a simple half-model, using symmetry, of the press consisting of 216 plane-stress (2-D) elements. Creating the model and analyzing it to obtain a solution took minutes. Doubling the results from the FE analysis yielded δ = 0.07790 in, a less than 1 percent variation from the results of the numerical integration.

4–10  Statically Indeterminate Problems A system is overconstrained when it has more unknown support (reaction) forces and/ or moments than static equilibrium equations. Such a system is said to be statically indeterminate and the extra constraint supports are called redundant supports. In addition to the static equilibrium equations, a deflection equation is required for each redundant support reaction in order to obtain a solution. For example, consider a beam in bending with a wall support on one end and a simple support on the other, such as beam 12 of Table A–9. There are three support reactions and only two static equilibrium equations are available. This beam has one redundant support. To solve for the three unknown support reactions we use the two equilibrium equations and one additional deflection equation. For another example, consider beam 15 of Table A–9. This beam has a wall on both ends, giving rise to two redundant supports requiring two deflection equations in addition to the equations from statics. The purpose of redundant supports is to provide additional safety and reduce deflection. A simple example of a statically indeterminate problem is furnished by the nested helical springs in Figure 4–15a. When this assembly is loaded by the compressive force F, it deforms through the distance δ. What is the compressive force in each spring? Only one equation of static equilibrium can be written. It is

∑ F = F − F1 − F2 = 0

(a)

which simply says that the total force F is resisted by a force F1 in spring 1 plus the force F2 in spring 2. Since there are two unknowns and only one static equilibrium equation, the system is statically indeterminate. To write another equation, note the deformation relation in Figure 4–15b. The two springs have the same deformation. Thus, we obtain the second equation as

δ1 = δ2 = δ

(b)

If we now substitute Equation (4–2) in Equation (b), we have

8

F1 F2 = k1 k2

See Case Study 4, p. 203, J. E. Shigley and C. R. Mischke, Mechanical Engineering Design, 6th ed., McGraw-Hill, New York, 2001.

(c)

202      Mechanical Engineering Design

Figure 4–15

F δ

k1

k2

(a) F1

F2 δ

k1

k2

(b)

Now we solve Equation (c) for F1 and substitute the result in Equation (a). This gives

F−

k1 F2 − F2 = 0 k2

or

F2 =

k2 F k1 + k2

(d)

Substituting F2 into Equation (c) gives F1 = k1F∕(k1 + k2) and so δ = δ1 = δ2 = F∕(k1 + k2). Thus, for two springs in parallel, the overall spring constant is k  =  F∕δ = k1 + k2. In the spring example, obtaining the necessary deformation equation was very straightforward. However, for other situations, the deformation relations may not be as easy. A more structured approach may be necessary. Here we will show two basic procedures for general statically indeterminate problems. Procedure 1  1 Choose the redundant reaction(s). There may be alternative choices (see Example 4–14).  2 Write the equations of static equilibrium for the remaining reactions in terms of the applied loads and the redundant reaction(s) of step 1.  3 Write the deflection equation(s) for the point(s) at the locations of the redundant reaction(s) of step 1 in terms of the applied loads and the redundant reaction(s) of step 1. Normally the deflection(s) is (are) zero. If a redundant reaction is a moment, the corresponding deflection equation is a rotational deflection equation.  4 The equations from steps 2 and 3 can now be solved to determine the reactions. In step 3 the deflection equations can be solved in any of the standard ways. Here we will demonstrate the use of superposition and Castigliano's theorem on a beam problem.

Deflection and Stiffness     203

EXAMPLE 4–14 The indeterminate beam 11 of Appendix Table A–9 is reproduced in Figure 4–16. Determine the reactions using procedure 1. Solution The reactions are shown in Figure 4–16b. Without R2 the beam is a statically determinate cantilever beam. Without M1 the beam is a statically determinate simply supported beam. In either case, the beam has only one redundant support. We will first solve this problem using superposition, choosing R2 as the redundant reaction. For the second solution, we will use Castigliano's theorem with M1 as the redundant reaction. Solution 1 1 Choose R2 at B to be the redundant reaction. 2 Using static equilibrium equations solve for R1 and M1 in terms of F and R2. This results in

R1 = F − R2   M1 =

Fl − R2l 2

(1)

3  Write the deflection equation for point B in terms of F and R2. Using superposition of beam 1 of Table A–9 with F = −R2, and beam 2 of Table A–9 with a = l∕2, the deflection of B, at x = l, is

δB = −

F(l∕2) 2 l R2 l 2 R2 l 3 5Fl 3 (l − 3l) + − 3l = ) 3EI − 48EI = 0 6EI 6EI ( 2

(2)

4 Equation (2) can be solved for R2 directly. This yields R2 =

Answer

5F 16

(3)

Next, substituting R2 into Equations (1) completes the solution, giving 11F 3Fl   M1 = 16 16

R1 =

Answer

(4)

Note that the solution agrees with what is given for beam 11 in Table A–9. Solution 2 1 Choose M1 at O to be the redundant reaction. 2 Using static equilibrium equations solve for R1 and R2 in terms of F and M1. This results in

R1 = y

F M1 F M1 +   R2 = − 2 l 2 l

(5)

y

Figure 4–16 F

l l 2

F A

O (a)

B

A B

x

x

O M1

R1

xˆ (b)

R2

204      Mechanical Engineering Design

3  Since M1 is the redundant reaction at O, write the equation for the angular deflection at point O. From Castigliano's theorem this is

∂U ∂M1

θO =

(6)

We can apply Equation (4–31), using the variable x as shown in Figure 4–16b. However, simpler terms can be found by using a variable xˆ that starts at B and is positive to the left. With this and the expression for R2 from Equation (5) the moment equations are

M=(

F M1 − xˆ 2 l )

M=(

l   0 ≤ xˆ ≤ 2

(7)

F M1 l l − xˆ − F (xˆ − )   ≤ xˆ ≤ l 2 l ) 2 2

(8)

For both equations ∂M xˆ =− ∂M1 l

(9)

Substituting Equations (7) to (9) in Equation (6), using the form of Equation (4–31) where Fi = M1, gives

θO =

∂U 1 = ∂M1 EI {

l∕2

0

F M1 xˆ ( 2 − l ) xˆ (− l ) dxˆ +

F M1 l xˆ [( 2 − l ) xˆ − F (xˆ − 2 )](− l ) dxˆ} = 0 l∕2

l

Canceling 1∕EIl, and combining the first two integrals, simplifies this quite readily to

F M1 (2 − l )

l

l (xˆ − 2 ) xˆ dxˆ = 0 l∕2

∫ xˆ dxˆ − F ∫ 2

0

l

Integrating gives

F M1 l 3 F 3 l 3 Fl 2 l 2 − − l − + l − (2 (2) ] (2) ] = 0 l )3 3[ 4[

which reduces to

3Fl 16

(10)

11F 5F   R2 = 16 16

(11)

M1 =

4 Substituting Equation (10) into (5) results in

R1 =

which again agrees with beam 11 of Table A–9.

For some problems even procedure 1 can be a task. Procedure 2 eliminates some tricky geometric problems that would complicate procedure 1. We will describe the procedure for a beam problem.

Deflection and Stiffness     205

Procedure 2 1 Write the equations of static equilibrium for the beam in terms of the applied loads and unknown restraint reactions. 2 Write the deflection equation for the beam in terms of the applied loads and unknown restraint reactions. 3 Apply boundary conditions to the deflection equation of step 2 consistent with the restraints. 4 Solve the equations from steps 1 and 3.

EXAMPLE 4–15 The rods AD and CE shown in Figure 4–17a each have a diameter of 10 mm. The second-area moment of beam ABC is I = 62.5(103) mm4. The modulus of elasticity of the material used for the rods and beam is E  = 200 GPa. The threads at the ends of the rods are single-threaded with a pitch of 1.5 mm. The nuts are first snugly fit with bar ABC horizontal. Next the nut at A is tightened one full turn. Determine the resulting tension in each rod and the deflections of points A and C. Solution There is a lot going on in this problem; a rod shortens, the rods stretch in tension, and the beam bends. Let's try the procedure! 1 The free-body diagram of the beam is shown in Figure 4–17b. Summing forces, and moments about B, gives

FB − FA − FC = 0

(1)

4FA − 3FC = 0

(2)

2 Using singularity functions, we find the moment equation for the beam is M = −FAx + FB⟨x − 0.2⟩1

where x is in meters. Integration yields

EI

dy FA FB = − x2 + ⟨x − 0.2⟩2 + C1 dx 2 2

EI y = −

FA 3 FB x + ⟨x − 0.2⟩3 + C1x + C2 6 6

(3)

The term EI = 200(109) 62.5(10−9) = 1.25(104) N · m2.

200

150

FA

B

A

C

200

A

150 B

x FB

600

800 D E (a)

(b) Free-body diagram of beam ABC

FC

Figure 4–17

C

Dimensions in mm.

206      Mechanical Engineering Design

3  The upward deflection of point A is (Fl∕AE)AD − Np, where the first term is the elastic stretch of AD, N is the number of turns of the nut, and p is the pitch of the thread. Thus, the deflection of A in meters is

yA =

FA (0.6) − (1)(0.0015) π (0.010) 2 (200)(109 ) 4

(4)

= 3.8197(10−8 )FA − 1.5(10−3 )

The upward deflection of point C is (Fl∕AE)CE, or

yC =

FC (0.8) π (0.010) 2 (200) (109 ) 4

= 5.093(10−8 )FC

(5)

Equations (4) and (5) will now serve as the boundary conditions for Equation (3). At x = 0, y = yA. Substituting Equation (4) into (3) with x = 0 and EI = 1.25 (104), noting that the singularity function is zero for x = 0, gives −4.7746(10−4 )FA + C2 = −18.75

(6)

At x = 0.2 m, y = 0, and Equation (3) yields −1.3333(10−3 )FA + 0.2C1 + C2 = 0

(7)

At x = 0.35 m, y = yC. Substituting Equation (5) into (3) with x = 0.35 m and EI = 1.25 (104) gives −7.1458(10−3 )FA + 5.625(10−4 )FB − 6.3662(10−4 )FC + 0.35C1 + C2 = 0

(8)

Equations (1), (2), (6), (7), and (8) are five equations in FA, FB, FC, C1, and C2. Written in matrix form, they are

[

−1 4 −4.7746(10−4 ) −1.3333(10−3 ) −7.1458(10−3 )

1 0 0 0 5.625(10−4 )

−1 −3 0 0 −6.3662(10−4 )

0 0 0 0.2 0.35

0 0 1 1 1

]{ } { FA FB FC C1 C2

=

0 0 −18.75 0 0

Solving these equations yields Answer

FA = 2988 N     FB = 6971 N      FC = 3983 N

C1 = 106.54 N · m2    C2 = −17.324 N · m3

Equation (3) can be reduced to

y = −(39.84x3 − 92.95⟨x − 0.2⟩3 − 8.523x + 1.386)(10−3 )

Answer

At x = 0, y = yA = −1.386(10−3 ) m = −1.386 mm

Answer

At x = 0.35m, y = yC = −[39.84(0.35) 3 − 92.95(0.35 − 0.2) 3 − 8.523(0.35)

+ 1.386](10−3 ) = 0.203(10−3 ) m = 0.203 mm

}

Deflection and Stiffness     207

Note that we could have easily incorporated the stiffness of the support at B if we were given a spring constant.

4–11  Compression Members—General The analysis and design of compression members can differ significantly from that of members loaded in tension or in torsion. If you were to take a long rod or pole, such as a meterstick, and apply gradually increasing compressive forces at each end, very small axial deflections would happen at first, but then the stick would bend (buckle), and very quickly bend so much as to possibly fracture. Try it. The other extreme would occur if you were to saw off, say, a 5-mm length of the meterstick and perform the same experiment on the short piece. You would then observe that the failure exhibits itself as a mashing of the specimen, that is, a simple compressive failure. For these reasons it is convenient to classify compression members according to their length and according to whether the loading is central or eccentric. The term column is applied to all such members except those in which failure would be by simple or pure compression. Columns can be categorized then as:

1 2 3 4

Long columns with central loading Intermediate-length columns with central loading Columns with eccentric loading Struts or short columns with eccentric loading

Classifying columns as above makes it possible to develop methods of analysis and design specific to each category. Furthermore, these methods will also reveal whether or not you have selected the category appropriate to your particular problem. The four sections that follow correspond, respectively, to the four categories of columns listed above.

4–12  Long Columns with Central Loading Figure 4–18 shows long columns with differing end (boundary) conditions. If the axial force P shown acts along the centroidal axis of the column, simple compression of the member occurs for low values of the force. However, under certain conditions, when P reaches a specific value, the column becomes unstable and bending as shown in Figure 4–18 develops rapidly. This force is determined by writing the bending deflection equation for the column, resulting in a differential equation where when the boundary conditions are applied, results in the critical load for unstable bending.9 The critical force for the pin-ended column of Figure 4–18a is given by

Pcr =

π2 EI l2

(4–42)

which is called the Euler column formula. Equation (4–42) can be extended to apply to other end-conditions by writing

Pcr =

Cπ2 EI l2

(4–43)

where the constant C depends on the end conditions as shown in Figure 4–18. 9

See F. P. Beer, E. R. Johnston, Jr., and J. T. DeWolf, Mechanics of Materials, 5th ed., McGraw-Hill, New York, 2009, pp. 610–613.

208      Mechanical Engineering Design

Figure 4–18

P

(a) Both ends rounded or pivoted; (b) both ends fixed; (c) one end free and one end fixed; (d) one end rounded and pivoted, and one end fixed.

P

P P

y A

l 2

l

0.707l

l 4

l

l 4

l A

B

x (a) C = 1

(b) C = 4

(c) C =

1 4

(d) C = 2

Using the relation I = Ak2, where A is the area and k the radius of gyration, enables us to rearrange Equation (4–43) into the more convenient form

Pcr Cπ2E = A (l∕k) 2

(4–44)

where l∕k is called the slenderness ratio. This ratio, rather than the actual column length, will be used in classifying columns according to length categories. The quantity Pcr∕A in Equation (4–44) is the critical unit load. It is the load per unit area necessary to place the column in a condition of unstable equilibrium. In this state any small crookedness of the member, or slight movement of the support or load, will cause the column to begin to collapse. The unit load has the same units as strength, but this is the strength of a specific column, not of the column material. Doubling the length of a member, for example, will have a drastic effect on the value of Pcr∕A but no effect at all on, say, the yield strength Sy of the column material itself. Equation (4–44) shows that the critical unit load depends only upon the end conditions, the modulus of elasticity, and the slenderness ratio. Thus a column obeying the Euler formula made of high-strength alloy steel is no stronger than one made of low-carbon steel, since E is the same for both. The factor C is called the end-condition constant, and it may have any one of the theoretical values 14 , 1, 2, and 4, depending upon the manner in which the load is applied. In practice it is difficult, if not impossible, to fix the column ends so that the factor C = 2 or C = 4 would apply. Even if the ends are welded, some deflection will occur. Because of this, some designers never use a value of C greater than unity. However, if liberal factors of safety are employed, and if the column load is accurately known, then a value of C not exceeding 1.2 for both ends fixed, or for one end rounded and one end fixed, is not unreasonable, since it supposes only partial fixation. Of course, the value C = 14 must always be used for a column having one end fixed and one end free. These recommendations are summarized in Table 4–2. When Equation (4–44) is solved for various values of the unit load Pcr∕A in terms of the slenderness ratio l∕k, we obtain the curve PQR shown in Figure 4–19. Since the yield strength of the material has the same units as the unit load, the horizontal line through Sy and Q has been added to the figure. This would appear to make the figure Sy QR cover the entire range of compression problems from the shortest to the longest compression member. Thus it would appear that any

Deflection and Stiffness     209

Table 4–2  End-Condition Constants for Euler Columns [to Be Used with Equation (4–43)] End-Condition Constant C Column End Conditions

Theoretical Value

Conservative Value

Recommended Value*

Fixed-free

1 4

1 4

1 4

Rounded-rounded

1

1

1

Fixed-rounded

2

1

1.2

Fixed-fixed

4

1

1.2

*To be used only with liberal factors of safety when the column load is accurately known.

Figure 4–19

P

Q

Pcr A

Sy Unit load

Euler curve plotted using Equation (4–43) with C = 1.

Parabolic curve T Euler curve l kQ

l k

R

1

Slenderness ratio

l k

compression member having an l∕k value less than (l∕k)Q should be treated as a pure compression member while all others are to be treated as Euler columns. Unfortunately, this is not true. In the actual design of a member that functions as a column, the designer will be aware of the end conditions shown in Figure 4–18, and will endeavor to configure the ends, using bolts, welds, or pins, for example, so as to achieve the required ideal end conditions. In spite of these precautions, the result, following manufacture, is likely to contain defects such as initial crookedness or load eccentricities. The existence of such defects and the methods of accounting for them will usually involve a factor-of-safety approach or a stochastic analysis. These methods work well for long columns and for simple compression members. However, tests show numerous failures for columns with slenderness ratios below and in the vicinity of point Q, as shown in the shaded area in Figure 4–19. These have been reported as occurring even when near-perfect geometric specimens were used in the testing procedure. A column failure is always sudden, total, unexpected, and hence dangerous. There is no advance warning. A beam will bend and give visual warning that it is overloaded, but not so for a column. For this reason neither simple compression methods nor the Euler column equation should be used when the slenderness ratio is near (l∕k)Q. Then what should we do? The usual approach is to choose some point T on the Euler curve of Figure 4–19. If the slenderness ratio is specified as (l∕k)1 corresponding to point T, then use the Euler equation only when the actual

210      Mechanical Engineering Design

slenderness ratio is greater than (l∕k)1. Otherwise, use one of the methods in the sections that follow. See Examples 4–17 and 4–18. Most designers select point T such that Pcr∕A = Sy∕2. Using Equation (4–43), we find the corresponding value of (l∕k)1 to be

l 2π2 CE 1∕2 = ( k )1 ( Sy )

(4–45)

Euler Equations for Specific Cross Sections Equations can be developed for simple round or rectangular cross sections. Round Cross Sections Given a column of diameter d, the area is A = πd2∕4 and the radius of gyration is k = (I∕A)1∕2 = [(πd4∕64)∕(πd2∕4)]1∕2 = d∕4. Substituting these into Equation (4–44) gives Pcr

πd2∕4 Solving for d yields

=

Cπ2E [l∕(d/4) ] 2

64Pcrl2 1∕4 d=( 3 π CE )

(4–46)

Rectangular Cross Sections  Consider a column of cross section h × b with the restriction that h ≤ b. If the end conditions are the same for buckling in both directions, then buckling will occur about the axis of the least thickness. Therefore, I = bh3∕12, A = bh, and k2 = I∕A = h2∕12. Substituting these into Equation (4–44) gives

Pcr Cπ2E = bh [l∕(h∕ √12)] 2

Solving for b or h yields

b=

12Pcrl2

π2CEh3

12Pcrl2 1∕3 h=( 2 π CEb )

(4–47a)

(4–47b)

Note, however, that rectangular columns do not generally have the same end conditions in both directions.

4–13  Intermediate-Length Columns with Central Loading Over the years there have been a number of column formulas proposed and used for the range of l∕k values for which the Euler formula is not suitable. Many of these are based on the use of a single material; others, on a so-called safe unit load rather than the critical value. Most of these formulas are based on the use of a linear relationship between the slenderness ratio and the unit load. The parabolic or J. B. Johnson formula now seems to be the preferred one among designers in the machine, automotive, aircraft, and structural-steel construction fields.

Deflection and Stiffness     211

The general form of the parabolic formula is Pcr l 2 = a − b( ) A k

(a)

where a and b are constants that are evaluated by fitting a parabola to the Euler curve of Figure 4–19 as shown by the dashed line ending at T. If the parabola is begun at Sy, then a = Sy. If point T is selected as previously noted, then Equation (4–45) gives the value of (l∕k)1 and the constant b is found to be b=(

1 ) 2π CE 2

Sy

(b)

Upon substituting the known values of a and b into Equation (a), we obtain, for the parabolic equation,

Sy l 2 1 Pcr l l = Sy − (    ≤ ( ) A 2π k ) CE k k 1

(4–48)

Parabolic Equations for Specific Cross Sections The J. B. Johnson Equations can be developed for simple round or rectangular cross sections. Round Cross Sections  Given a column of diameter d, the area is A = πd2∕4 and the radius of gyration is k = (I∕A)1∕2 = [(πd4∕64)∕(πd2∕4)]1∕2 = d∕4. Substituting these into Equation (4–48) gives

2 Sy 1 = Sy − ( 2π(d∕4) ) CE (πd ∕4)

Pcr 2

Solving for d yields Syl2 1∕2 Pcr d = 2( + 2 ) (4–49) πSy π CE

Rectangular Cross Sections Consider a column of cross section h × b with the restriction that h ≤ b. If the end conditions are the same for buckling in both directions, then buckling will occur about the axis of the least thickness. Therefore, I = bh3∕12, A = bh, and k2 = I∕A = h2∕12. Substituting these into Equation (4–48) gives

2 Sy Pcr 1 = Sy − ( ) bh 2π(h∕ √12) CE

Solving for b or h yields

b=

h=

Pcr

hSy(1 −

π2CEh2 ) 3l2Sy

2 Pcr 2 3l Sy 1∕2 Pcr + [( + 2bSy 2bSy ) π2CE ]

(4–50a)

(4–50b)

Note, however, that rectangular columns do not generally have the same end conditions in both directions.

212      Mechanical Engineering Design

4–14  Columns with Eccentric Loading We have noted before that deviations from an ideal column, such as load eccentricities or crookedness, are likely to occur during manufacture and assembly. Though these deviations are often quite small, it is still convenient to have a method of dealing with them. Frequently, too, problems occur in which load eccentricities are unavoidable. Figure 4–20a shows a column in which the line of action of the column forces is separated from the centroidal axis of the column by the eccentricity e. From Figure 4–20b, M = −P(e + y). Substituting this into Equation (4–12), d2y∕dx2 = M∕EI, results in the differential equation d 2y

dx 2

+

P Pe y=− EI EI

(a)

The solution of Equation (a), using the boundary conditions, y = 0 at x = 0 and l is l P P P y = e [ tan ( √ ) sin (√ x) + cos (√ x) − 1] 2 EI EI EI

(b)

By substituting x = l∕2 in Equation (b) and using a trigonometric identity, we obtain P l δ = e [ sec (√ − 1] EI 2 )

(4–51)

The magnitude of the maximum bending moment also occurs at midspan and is l P Mmax = P(e + δ) = Pe sec ( √ ) 2 EI

(4–52)

The magnitude of the maximum compressive stress at midspan is found by superposing the axial component and the bending component. This gives

σc =

P Mc P Mc + = + 2 A I A Ak

(c)

Substituting Mmax from Equation (4–52) yields

σc =

P ec l P 1 + 2 sec ( √ )] [ A 2k EA k

x

Figure 4–20 Notation for an eccentrically loaded column.

P

A

x

P

l

δ

M y x

y

O

P

y Pe

e (a)

P (b)

(4–53)

Deflection and Stiffness     213

Figure 4–21 2

Unit load P/A

ec/k = 0.1

Comparison of secant and Euler equations for steel with Sy = 40 kpsi.

Sy

0 .3

0 .6 1.0

0

50

Euler's curve

100

150

200

250

Slenderness ratio l/k

By imposing the compressive yield strength Syc as the maximum value of σc, we can write Equation (4–53) in the form Syc P = A 1 + (ec∕k2 ) sec[(l∕2k) √P∕AE]

(4–54)

This is called the secant column formula. The term ec∕k2 is called the eccentricity ratio. Figure 4–21 is a plot of Equation (4–54) for a steel having a compressive (and tensile) yield strength of 40 kpsi. Note how the P∕A contours asymptotically approach the Euler curve as l∕k increases. Equation (4–54) cannot be solved explicitly for the load P. Design charts, in the fashion of Figure 4–21, can be prepared for a single material if much column design is to be done. Otherwise, a root-finding technique using numerical methods must be used. EXAMPLE 4–16 Specify the diameter of a round column 1.5 m long that is to carry a maximum load estimated to be 22 kN. Use a design factor nd = 4 and consider the ends as pinned (rounded). The column material selected has a minimum yield strength of 500 MPa and a modulus of elasticity of 207 GPa. Solution We shall design the column for a critical load of Pcr = nd P = 4(22) = 88 kN Then, using Equation (4–46) with C = 1 (see Table 4–2) gives d=(

π3CE )

64Pcr l 2

1∕4

=[

π3 (1)207(109 ) ]

64(88)103 (1.5) 2

1∕4

= 0.0375 m = 37.5 mm

Table A–17 shows that the preferred size is 40 mm. The slenderness ratio for this size is 1.5(103 ) l l = = = 150 k d∕4 40∕4

214      Mechanical Engineering Design

To be sure that this is an Euler column, we use Equation (4–45) and obtain

2π2 (1)207(109 ) 1∕2 l 2π2CE 1∕2 = = = 90.4 ( k )1 ( Sy ) [ 500(106 ) ]

where l∕k > (l∕k)1 indicates that it is indeed an Euler column. So select d = 40 mm

Answer

EXAMPLE 4–17 Repeat Example 4–16 for a fixed-fixed column with a rectangular cross section with b = 4h. Round up the results to the next higher whole millimeter. Solution From Example 4–16, Pcr = 88 kN. From Table 4–1, we select the recommended value of C = 1.2. Setting b = 4h into Equation (4–47a) and solving for h gives

h=(

π2CE )

3Pcrl2

1∕4

=[

π2 (1.2) (207)109 ] 3(88)103 (1.5) 2

1∕4

= 0.0222 m = 22.2 mm

Rounding up the next higher whole millimeter gives h = 23 mm and b = 92 mm. The slenderness ratio for this size is √12(1.5) √12 l l = = = 226 k h 0.023

From Equation (4–45) and using the recommended value of C = 1.2 from Table 4–2 2π2 (1.2)207(109 ) 1∕2 l 2π2CE 1∕2 = = = 90.40 ( k )1 ( Sy ) [ ] 500(106 ) Since (l∕k) > (l∕k)1, the column is an Euler column. So select h = 23 mm, b = 92 mm

Answer

EXAMPLE 4–18 Repeat Example 4–16 with l = 375 mm. Solution Following the same procedure as Example 4–16, 64Pcrl2 1∕4 64(88)103 (0.375) 2 1∕4 d=( 3 = [ π3 (1)207(109 ) ] = 0.0187 m = 18.7 mm π CE ) From Table A–17, d = 20 mm. l 4l 4(375) = = = 75 k d 20

Deflection and Stiffness     215

From Example 4–16, (l∕k)1 = 90.4. Since (l∕k) < (l∕k)1, the column is not a Euler column and the parabolic equation, Equation (4–49), must be used. Thus, Syl2 1∕2 Pcr 88(103 ) 500(106 )(0.375) 2 1∕2 d = 2( + 2 ) = 2[ + πSy π CE π2 (1)207(109 ) ] π(500)106 = 0.0190 m = 19.0 mm From Table A–17 select d = 20 mm

Answer

The answer is not substantially different from what was obtained from the Euler formula since we were close to point T of Figure 4–19.

EXAMPLE 4–19 Choose a set of dimensions for a rectangular link that is to carry a maximum compressive load of 5000 lbf. The material selected has a minimum yield strength of 75 kpsi and a modulus of elasticity E = 30 Mpsi. Use a design factor of 4 and an end condition constant C = 1 for buckling in the weakest direction, and design for (a) a length of 15 in, and (b) a length of 8 in with a minimum thickness of 12 in. Solution (a) Using Equation (4–45), we find the limiting slenderness ratio to be 2π2 (1) (30)(106 ) 1∕2 l 2π2CE 1∕2 = = ( k )1 ( Sy ) [ ] = 88.9 75(10) 3 By using Pcr = ndP = 4(5000) = 20 000 lbf, Equations (4–47a) and (4–50a) are solved, using various values of h, to form Table 4–3. The table shows that a cross section of 58 by 34 in, which is marginally suitable, gives the least area. (b) An approach similar to that in part (a) is used with l = 8 in. All trial computations are found to be in the J. B. Johnson region of l∕k values. A minimum area occurs when the section is a near square. Thus a cross section of 12 by 34 in is found to be suitable and safe. Table 4–3  Table Generated to Solve Example 4–19, part (a) h

b

A

0.375

3.46

1.298

0.500

1.46 0.730

0.625

0.76 0.475

0.5625 1.03 0.579

l/k

Type

139

Euler

104 Euler

Eq. No. (4–47a) (4–47a)

83 Johnson (4–50a) 92 Euler

(4–47a)

4–15  Struts or Short Compression Members A short bar loaded in pure compression by a force P acting along the centroidal axis will shorten in accordance with Hooke's law, until the stress reaches the elastic limit of the material. At this point, permanent set is introduced and usefulness as a machine member may be at an end. If the force P is increased still more, the material either

216      Mechanical Engineering Design

becomes "barrel-like" or fractures. When there is eccentricity in the loading, the elastic limit is encountered at smaller loads. A strut is a short compression member such as the one shown in Figure 4–22. The magnitude of the maximum compressive stress in the x direction at point B in an intermediate section is the sum of a simple component P∕A and a flexural component Mc∕I; that is,

P x e

B

l c

y

P

Figure 4–22 Eccentrically loaded strut.

σc =

P Mc P PecA P ec + = + = (1 + 2 ) A I A IA A k

(4–55)

where k = (I∕A)1∕2 and is the radius of gyration, c is the coordinate of point B, and e is the eccentricity of loading. Note that the length of the strut does not appear in Equation (4–55). In order to use the equation for design or analysis, we ought, therefore, to know the range of lengths for which the equation is valid. In other words, how long is a short member? The difference between the secant formula Equation (4–54) and Equation (4–55) is that the secant equation, unlike Equation (4–55), accounts for an increased bending moment due to bending deflection. Thus the secant equation shows the eccentricity to be magnified by the bending deflection. This difference between the two formulas suggests that one way of differentiating between a "secant column" and a strut, or short compression member, is to say that in a strut, the effect of bending deflection must be limited to a certain small percentage of the eccentricity. If we decide that the limiting percentage is to be 1 percent of e, then, from Equation (4–44), the limiting slenderness ratio turns out to be

l AE 1∕2 = 0.282 ( k )2 (P)

(4–56)

This equation then gives the limiting slenderness ratio for using Equation (4–55). If the actual slenderness ratio is greater than (l∕k)2, then use the secant formula; otherwise, use Equation (4–55).

EXAMPLE 4–20 Figure 4–23a shows a workpiece clamped to a milling machine table by a bolt tightened to a tension of 2000 lbf. The clamp contact is offset from the centroidal axis of the strut by a distance e = 0.10 in, as shown in part b of the figure. The strut, or block, is steel, 1 in square and 4 in long, as shown. Determine the maximum compressive stress in the block. Solution First we find A = bh = 1(1) = 1 in2, I = bh3∕12 = 1(1)3∕12 = 0.0833 in4, k2 = I∕A = 0.0833∕1 = 0.0833 in2, and l∕k = 4∕(0.0833)1∕2 = 13.9. Equation (4–56) gives the limiting slenderness ratio as 1(30)(106 ) 1∕2 l AE 1∕2 = 0.282 = 0.282 ( k )2 (P) [ 1000 ] = 48.8

Deflection and Stiffness     217

Thus the block could be as long as l = 48.8k = 48.8(0.0833) 1∕2 = 14.1 in

before it need be treated by using the secant formula. So Equation (4–55) applies and the maximum compressive stress is Answer

σc =

0.1(0.5) P ec 1000 1 + 2) = 1+ = 1600 psi [ A( 1 0.0833 ] k P = 1000 lbf

1-in square

4 in

P (a)

0.10 in

(b)

4–16  Elastic Stability Section 4–12 presented the conditions for the unstable behavior of long, slender columns. Elastic instability can also occur in structural members other than columns. Compressive loads/stresses within any long, thin structure can cause structural instabilities (buckling). The compressive stress may be elastic or inelastic and the instability may be global or local. Global instabilities can cause catastrophic failure, whereas local instabilities may cause permanent deformation and function failure but not a catastrophic failure. The buckling discussed in Section 4–12 was global instability. However, consider a wide flange beam in bending. One flange will be in compression, and if thin enough, can develop localized buckling in a region where the bending moment is a maximum. Localized buckling can also occur in the web of the beam, where transverse shear stresses are present at the beam centroid. Recall, for the case of pure shear stress τ, a stress transformation will show that at 45°, a compressive stress of σ = −τ exists. If the web is sufficiently thin where the shear force V is a maximum, localized buckling of the web can occur. For this reason, additional support in the form of bracing is typically applied at locations of high shear forces.10 Thin-walled beams in bending can buckle in a torsional mode as illustrated in Figure 4–24. Here a cantilever beam is loaded with a lateral force, F. As F increases from zero, the end of the beam will deflect in the negative y direction normally according to the bending equation, y = −FL3∕(3EI). However, if the beam is long enough and the ratio of b∕h is sufficiently small, there is a critical value of F for which the beam will collapse in a twisting mode as shown. This is due to the compression in the bottom fibers of the beam that cause the fibers to buckle sideways (z direction). 10

See C. G. Salmon, J. E. Johnson, and F. A. Malhas, Steel Structures: Design and Behavior, 5th ed., Prentice Hall, Upper Saddle River, NJ, 2009.

Figure 4–23 A strut that is part of a workpiece clamping assembly.

218      Mechanical Engineering Design y

Figure 4–24 Torsional buckling of a thinwalled beam in bending. z

h

z y

x b

F

Figure 4–25 Finite-element representation of flange buckling of a channel in compression.

There are a great many other examples of unstable structural behavior, such as thin-walled pressure vessels in compression or with outer pressure or inner vacuum, thin-walled open or closed members in torsion, thin arches in compression, frames in compression, and shear panels. Because of the vast array of applications and the complexity of their analyses, further elaboration is beyond the scope of this book. The intent of this section is to make the reader aware of the possibilities and potential safety issues. The key issue is that the designer should be aware that if any unbraced part of a structural member is thin, and/or long, and in compression (directly or indirectly), the possibility of buckling should be investigated.11 For unique applications, the designer may need to revert to a numerical solution such as using finite elements. Depending on the application and the finite-element code available, an analysis can be performed to determine the critical loading (see Figure 4–25).

4–17  Shock and Impact Impact refers to the collision of two masses with initial relative velocity. In some cases it is desirable to achieve a known impact in design; for example, this is the case in the design of coining, stamping, and forming presses. In other cases, impact occurs because of excessive deflections, or because of clearances between parts, and in these cases it is desirable to minimize the effects. The rattling of mating gear teeth in their tooth spaces is an impact problem caused by shaft deflection and the clearance between the teeth. This impact causes gear noise and fatigue failure of the tooth surfaces. The clearance space between a cam and follower or between a journal and its bearing may result in crossover impact and also cause excessive noise and rapid fatigue failure. Shock is a more general term that is used to describe any suddenly applied force or disturbance. Thus the study of shock includes impact as a special case. Figure 4–26 represents a highly simplified mathematical model of an automobile in collision with a rigid obstruction. Here m1 is the lumped mass of the engine. The displacement, velocity, and acceleration are described by the coordinate x1 and its time derivatives. The lumped mass of the vehicle less the engine is denoted by m2, and its motion by the coordinate x2 and its derivatives. Springs k1, k2, and k3 represent the linear and nonlinear stiffnesses of the various structural elements that compose the vehicle. Friction and damping can and should be included, but is not shown in this 11

See S. P. Timoshenko and J. M. Gere, Theory of Elastic Stability, 2nd ed., McGraw-Hill, New York, 1961. See also, Z. P. Bazant and L. Cedolin, Stability of Structures, Oxford University Press, New York, 1991.

Deflection and Stiffness     219

Figure 4–26

x2

x1 k1

Two-degree-of-freedom mathematical model of an automobile in collision with a rigid obstruction.

k2 m1

m2

k3

model. The determination of the spring rates for such a complex structure will almost certainly have to be performed experimentally. Once these values—the k's, m's, damping and frictional coefficients—are obtained, a set of nonlinear differential equations can be written and a computer solution obtained for any impact velocity. For sake of illustration, assuming the springs to be linear, isolate each mass and write their equations of motion. This results in  m x1 + k1x1 + k2 (x1 − x2 ) = 0 (4–57)  m x2 + k3x2 − k2 (x1 − x2 ) = 0 The analytical solution of the Equation (4–57) pair is harmonic and is studied in a course on mechanical vibrations.12 If the values of the m's and k's are known, the solution can be obtained easily using a program such as MATLAB. Suddenly Applied Loading A simple case of impact is illustrated in Figure 4–27a. Here a weight W falls a distance h and impacts a cantilever of stiffness EI and length l. We want to find the maximum deflection and the maximum force exerted on the beam due to the impact. Figure 4–27b shows an abstract model of the system considering the beam as a simple spring. For beam 1 of Table A–9, we find the spring rate to be k = F∕y = 3EI∕l3. The beam mass and damping can be accounted for, but for this example will be considered negligible. If the beam is considered massless, there is no momentum transfer, only energy. If the maximum deflection of the spring (beam) is considered to be δ, the drop of the weight is h + δ, and the loss of potential energy is W(h + δ). The resulting increase in potential (strain) energy of the spring is 12 kδ2 . Thus, for energy conservation, 12 kδ2 = W(h + δ). Rearranging this gives

δ2 − 2

W W δ − 2 h = 0 k k

(a)

Solving for δ yields

δ=

W W 2hk 1∕2 ± 1 + k k ( W)

(b)

The negative solution is possible only if the weight "sticks" to the beam and vibrates between the limits of Equation (b). Thus, the maximum deflection is

12

δ=

W W 2hk 1∕2 + 1 + k k ( W)

(4–58)

See William T. Thomson and Marie Dillon Dahleh, Theory of Vibrations with Applications, 5th ed., Prentice Hall, Upper Saddle River, NJ, 1998.

220      Mechanical Engineering Design

Figure 4–27 (a) A weight free to fall a distance h to free end of a beam. (b) Equivalent spring model.

W

W

h

h

EI, l

k

(a)

(b)

The maximum force acting on the beam is now found to be F = kδ = W + W(1 +

2hk 1∕2 W)

(4–59)

Note, in this equation, that if h = 0, then F = 2W. This says that when the weight is released while in contact with the spring but is not exerting any force on the spring, the largest force is double the weight. Most systems are not as ideal as those explored here, so be wary about using these relations for nonideal systems.

PROBLEMS Problems marked with an asterisk (*) are linked to problems in other chapters, as summarized in Table 1–2 of Section 1–17.

  4–1 The figure shows a torsion bar OA fixed at O, simply supported at A, and connected

to a cantilever AB. The spring rate of the torsion bar is kT, in newton-meters per radian, and that of the cantilever is kl, in newtons per meter. What is the overall spring rate based on the deflection y at point B?

F

O

Problem 4–1

B

L l

A

y R

  4–2 For Problem 4–1, if the simple support at point A were eliminated and the cantilever spring rate of OA is given by kL, determine the overall spring rate of the bar based on the deflection of point B.

  4–3 A torsion-bar spring consists of a prismatic bar, usually of round cross section, that is

twisted at one end and held fast at the other to form a stiff spring. An engineer needs a stiffer one than usual and so considers building in both ends and applying the torque somewhere in the central portion of the span, as shown in the figure. This effectively creates two springs in parallel. If the bar is uniform in diameter, that is, if d = d1 = d2, (a) determine how the spring rate and the end reactions depend on the location x at which the torque is applied, (b) determine the spring rate, the end reactions, and the maximum shear stress, if d = 0.5 in, x = 5 in, l = 10 in, T = 1500 lbf · in, and G = 11.5 Mpsi.

Deflection and Stiffness     221 d2 T d1

Problem 4–3

l x

  4–4 An engineer is forced by geometric considerations to apply the torque on the spring

of Problem 4–3 at the location x = 0.4l. For a uniform-diameter spring, this would cause one leg of the span to be underutilized when both legs have the same diameter. For optimal design the diameter of each leg should be designed such that the maximum shear stress in each leg is the same. This problem is to redesign the spring of part (b) of Problem 4–3. Using x = 0.4l, l = 10 in, T = 1500 lbf · in, and G = 11.5 Mpsi, design the spring such that the maximum shear stresses in each leg are equal and the spring has the same spring rate (angle of twist) as part (b) of Problem 4–3. Specify d1, d2, the spring rate k, and the torque and the maximum shear stress in each leg.

  4–5 A bar in tension has a circular cross section and includes a tapered portion of length l, as shown.

l

(a) For the tapered portion, use Equation (4–3) in the form of δ = show that δ=

∫ [F∕(AE)] dx to 0

4 Fl π d1 d2 E

(b) Determine the elongation of each portion if d1 = 0.5 in, d2 = 0.75 in, l = l1 = l2 = 2.0 in, E = 30 Mpsi, and F = 1000 lbf. y Problem 4–5

F

dl l1

x

d2

l

l2

F

  4–6 Instead of a tensile force, consider the bar in Problem 4–5 to be loaded by a torque T. l

(a) Use Equation (4–5) in the form of θ = twist of the tapered portion is

θ=

∫ [T∕(GJ )] dx to show that the angle of 0

32 Tl (d 21 + d1d2 + d 22 ) 3π Gd13d 32

(b) Using the same geometry as in Problem 4–5b with T = 1500 lbf · in and G = 11.5 Mpsi, determine the angle of twist in degrees for each portion.

  4–7 When a vertically suspended hoisting cable is long, the weight of the cable itself

contributes to the elongation. If a 500-ft steel cable has an effective diameter of 0.5 in and lifts a load of 5000 lbf, determine the total elongation and the percent of the total elongation due to the cable's own weight.

222      Mechanical Engineering Design

4–8

Derive the equations given for beam 2 in Table A–9 using statics and the doubleintegration method.

4–9

Derive the equations given for beam 5 in Table A–9 using statics and the doubleintegration method.

4–10 The figure shows a cantilever consisting of steel angles size 100 × 100 × 12 mm mounted back to back. Using superposition, find the deflection at B and the maximum stress in the beam. y 3m

2.5 kN

2m

Problem 4–10

1 kN/m O

4–11

x

B

A

A simply supported beam loaded by two forces is shown in the figure. Select a pair of structural steel channels mounted back to back to support the loads in such a way that the deflection at midspan will not exceed 12 in and the maximum stress will not exceed 15 kpsi. Use superposition. y

450 lbf 300 lbf Problem 4–11 6 ft

O

10 ft

4 ft A

C

x

B

4–12 Using superposition, find the deflection of the steel shaft at A in the figure. Find the deflection at midspan. By what percentage do these two values differ? y 24 in

15 in 340 lbf

Problem 4–12

150 lbf/ft

B

O

x

A 1.5 in-dia. shaft

4–13 A rectangular steel bar supports the two overhanging loads shown in the figure. Using superposition, find the deflection at the ends and at the center. y 300

300

500

400 N

Problem 4–13 Dimensions in millimeters.

400 N A

B

x C

O Bar, b = 6, h = 32

Deflection and Stiffness     223

4–14 An aluminum tube with outside diameter of 2 in and inside diameter of 1.5 in is cantilevered and loaded as shown. Using the formulas in Appendix Table A–9 and superposition, find the deflection at B. y 200 lbf

300 lbf Problem 4–14 2 ft

2 ft

O

x

A

B

4–15 The cantilever shown in the figure consists of two structural-steel channels size 3 in,

5.0  lbf/ft. Using superposition, find the deflection at A. Include the weight of the  channels. y 60 in

Problem 4–15

150 lbf

5 lbf/in

x

A

O

4–16 Using superposition for the bar shown, determine the minimum diameter of a steel shaft for which the maximum deflection is 2 mm. y 250

Problem 4–16 Dimensions in millimeters.

250

250

250

375 N

550 N

375 N D

O A

B

x

C

4–17 A simply supported beam has a concentrated moment MA applied at the left support

and a concentrated force F applied at the free end of the overhang on the right. Using superposition, determine the deflection equations in regions AB and BC. y a

l

Problem 4–17 MA

B

A R1

F C

x

R2

4–18 Calculating beam deflections using superposition is quite convenient provided you

have a comprehensive table to refer to. Because of space limitations, this book provides a table that covers a great deal of applications, but not all possibilities. Take for example, Problem 4–19, which follows this problem. Problem 4–19 is not directly

224      Mechanical Engineering Design

solvable from Table A–9, but with the addition of the results of this problem, it is. For the beam shown, using statics and double integration, show that

R1 =

MAB =

yAB =

wa wa2 w wa2 (2l − a)   R2 =   VAB = [2l(a − x) − a2]  VBC = − 2l 2l 2l 2l wx wa2 (2al − a2 − lx)   MBC = (l − x) 2l 2l wx w [2ax 2 (2l − a) − lx3 − a2 (2l − a) 2]  yBC = yAB + (x − a) 4 24EI l 24EI

y l Problem 4–18

a w A

B

C

R1

x

R2

4–19 Using the results of Problem 4–18, use superposition to determine the deflection equations for the three regions of the beam shown. y l b a

Problem 4–19

w D

A

B

C

R1

x

R2

4–20 Like Problem 4–18, this problem provides another beam to add to Table A–9. For the simply supported beam shown with an overhanging uniform load, use statics and double integration to show that wa2 wa wa2   R2 = (2l + a)   VAB = −   VBC = w(l + a − x) 2l 2l 2l

R1 =

MAB = −

yAB =

w wa2 x  MBC = − (l + a − x) 2 2 2l

wa2x 2 w (l − x 2 )   yBC = − [(l + a − x) 4 − 4a2 (l − x) (l + a) − a4] 12EI l 24EI

y a w

l

Problem 4–20

R1 A

C

B

x

R2

4–21 Consider the uniformly loaded simply supported steel beam with an overhang as shown. The second-area moment of the beam is I = 0.05 in4. Use superposition (with Table A–9 and the results of Problem 4–20) to determine the reactions and the deflection equations of the beam. Plot the deflections. w = 100 lbf/in Problem 4–21

C

A y

y 10 in

B 4 in

Deflection and Stiffness     225

4–22 Illustrated is a rectangular steel bar with simple supports at the ends and loaded by a

force F at the middle; the bar is to act as a spring. The ratio of the width to the thickness is to be about b = 10h, and the desired spring scale is 1800 lbf/in. (a) Find a set of cross-section dimensions, using preferred fractional sizes from Table A–17. (b) What deflection would cause a permanent set in the spring if this is estimated to occur at a normal stress of 60 kpsi? F A

b

Problem 4–22 A

h

3 ft

Section A–A

4–23* For the steel countershaft specified in the table, find the deflection and slope of the to shaft at point A. Use superposition with the deflection equations in Table A–9. Assume 4–28* the bearings constitute simple supports. Problem Number

Problem Number Defining Shaft

4–23*

3–79

4–24*

3–80

4–25* 4–26* 4–27* 4–28*

3–81 3–82 3–83 3–84

4–29* For the steel countershaft specified in the table, find the slope of the shaft at each to bearing. Use superposition with the deflection equations in Table A–9. Assume the 4–34* bearings constitute simple supports. Problem Number

Problem Number Defining Shaft

4–29*

3–79

4–30*

3–80

4–31* 4–32* 4–33* 4–34*

3–81 3–82 3–83 3–84

4–35* For the steel countershaft specified in the table, assume the bearings have a maximum to slope specification of 0.06° for good bearing life. Determine the minimum shaft 4–40* diameter. Problem Number

Problem Number Defining Shaft

4–35*

3–79

4–36*

3–80

4–37* 4–38* 4–39* 4–40*

3–81 3–82 3–83 3–84

226      Mechanical Engineering Design

4–41* The cantilevered handle in the figure is made from mild steel that has been welded at

the joints. For Fy = 200 lbf, Fx = Fz = 0, determine the vertical deflection (along the y axis) at the tip. Use superposition. See the discussion on Equation (3–41) for the twist in the rectangular cross section in section BC. y

2 in C

A 1-in dia. Problem 4–41

1 4

Fy

in

1 12

B

in

z

Fz

D

3 -in 4

dia.

Fx

5 in 6 in x

4–42 For the cantilevered handle in Problem 4–41, let Fx = −150 lbf, Fy = 0 lbf, Fz = −100 lbf. Find the deflection at the tip along the x axis.

4–43* The cantilevered handle in Problem 3–95 is made from mild steel. Let Fy = 250 lbf, Fx = Fz = 0. Determine the angle of twist in bar OC, ignoring the fillets but including the changes in diameter along the 13-in effective length. Compare the angle of twist if the bar OC is simplified to be all of uniform 1-in diameter. Use superposition to determine the vertical deflection (along the y axis) at the tip, using the simplified bar OC.

4–44 A flat-bed trailer is to be designed with a curvature such that when loaded to capacity the trailer bed is flat. The load capacity is to be 3000 lbf/ft between the axles, which are 25 ft apart, and the second-area moment of the steel structure of the bed is I = 485 in4. Determine the equation for the curvature of the unloaded bed and the maximum height of the bed relative to the axles.

4–45 The designer of a shaft usually has a slope constraint imposed by the bearings used. This limit will be denoted as ξ. If the shaft shown in the figure is to have a uniform diameter d except in the locality of the bearing mounting, it can be approximated as a uniform beam with simple supports. Show that the minimum diameters to meet the slope constraints at the left and right bearings are, respectively, dL =⎹

32Fb (l 2 − b2 ) 1∕4 32Fa(l 2 − a2 ) 1∕4   d = R ⎹ 3π Elξ ⎹ 3π Elξ ⎹

F a

b

l Problem 4–45

y F θ

x

Deflection and Stiffness     227

4–46 A steel shaft is to be designed so that it is supported by roller bearings. The basic

geometry is shown in the figure from Problem 4–45, with l = 300 mm, a = 100 mm, and F = 3 kN. The allowable slope at the bearings is 0.001 mm/mm without bearing life penalty. For a design factor of 1.28, what uniform-diameter shaft will support the load without penalty? Determine the maximum deflection of the shaft.

4–47 If the diameter of the steel beam shown is 1.25 in, determine the deflection of the beam at x = 8 in.

y 150 lbf

5 A

Problem 4–47 Dimensions in inches.

z

15 10 B

250 lbf

x

4–48 For the beam of Problem 4–47, plot the magnitude of the displacement of the beam

in 0.1-in increments. Approximate the maximum displacement and the value of x where it occurs.

4–49 Shown in the figure is a uniform-diameter shaft with bearing shoulders at the ends; the shaft is subjected to a concentrated moment M = 1000 lbf · in. The shaft is of carbon

a

steel and has a = 4 in and l = 10 in. The slope at the ends must be limited to 0.002 rad. Find a suitable diameter d.

B l

4–50* The figure shows an aluminum beam OB with rectangular cross section, pinned to the to ground at one end, and supported by a round steel rod with hooks formed on the ends. 4–53 A load is applied as shown. Use superposition to determine the vertical deflection at point B.

2 in

1 -in 2

100 lbf

1 -in 4

dia.

C

1 -in 2

12 in

thick

12 in

2 in

B 6 in

7 in 6 in

12 in

220

12 thick A

4000 N

12 in

4000 N

A B

300

200

C

6 dia.

220

12 thick

50

D

O 150

B

Problem 4–51

C

6 dia.

thick

D

O

Problem 4–50*

50

1 -in 4

A

A O

Problem 4–49

100 lbf

C

dia.

D B

O 150

300

Problem 4–52

Problem 4–53

All dimensions in mm.

All dimensions in mm.

200

b MB

228      Mechanical Engineering Design

4–54 Solve Problem 4–50 for the vertical deflection at point A. 4–55 Solve Problem 4–51 for the vertical deflection at point A. 4–56 Solve Problem 4–52 for the vertical deflection at point A. 4–57 Solve Problem 4–53 for the vertical deflection at point A. 4–58 The figure illustrates a stepped torsion-bar spring OA with an actuating cantilever AB.

Both parts are of carbon steel. Use superposition and find the spring rate k corresponding to a force F acting at B. y

O

18 mm 12 mm

Problem 4–58

C

0.2 m F

A

8 mm

x

0.4 m

z 0.2 m

B

4–59 Consider the simply supported beam 5 with a center load in Appendix A–9. Determine the

deflection equation if the stiffness of the left and right supports are k1 and k2, respectively.

4–60 Consider the simply supported beam 10 with an overhanging load in Appendix A–9. Determine the deflection equation if the stiffness of the left and right supports are k1 and k2, respectively.

4–61 Prove that for a uniform-cross-section beam with simple supports at the ends loaded by a single concentrated load, the location of the maximum deflection will never be outside the range of 0.423l ≤ x ≤ 0.577l regardless of the location of the load along the beam. The importance of this is that you can always get a quick estimate of ymax by using x = l∕2.

4–62 Solve Problem 4–10 using singularity functions. Use statics to determine the reactions. 4–63 Solve Problem 4–11 using singularity functions. Use statics to determine the reactions. 4–64 Solve Problem 4–12 using singularity functions. Use statics to determine the reactions. 4–65 Solve Problem 4–21 using singularity functions to determine the deflection equation of the beam. Use statics to determine the reactions.

4–66 Solve Problem 4–13 using singularity functions. Since the beam is symmetric, only write the equation for half the beam and use the slope at the beam center as a boundary condition. Use statics to determine the reactions.

4–67 Solve Problem 4–17 using singularity functions. Use statics to determine the reactions. 4–68 Solve Problem 4–19 using singularity functions to determine the deflection equation of the beam. Use statics to determine the reactions.

4–69 Using singularity functions, write the deflection equation for the steel beam shown.

Since the beam is symmetric, write the equation for only half the beam and use the slope at the beam center as a boundary condition. Plot your results and determine the maximum deflection.

Deflection and Stiffness     229 w = 180 lbf/in

1.375-in dia.

1.375-in dia.

1.75-in dia.

Problem 4–69 3 in

10 in

3 in

4–70 Determine the deflection equation for the cantilever beam shown using singularity functions. Evaluate the deflections at B and C and compare your results with Example 4–10. y l/2 Problem 4–70

l/2

2I1

A

I1

B

x

C

F

4–71 Use Castigliano's theorem to verify the maximum deflection for the uniformly loaded beam 7 of Appendix Table A–9. Neglect shear.

4–72 Use Castigliano's theorem to verify the maximum deflection for the uniformly loaded cantilever beam 3 of Appendix Table A–9. Neglect shear.

4–73 Solve Problem 4–15 using Castigliano's theorem. 4–74 The beam shown in the figure is pinned to the ground at points A and B, and loaded

by a force P at point C. (a) Using Castigliano's method, derive an expression for the vertical deflection at D. (b) Using Table A–9, determine the vertical deflection at D, and compare your results with part (a). P

Problem 4–74

B

C D

A l/2

l/2

a

4–75 Solve Problem 4–58 using Castigliano's theorem. 4–76 Determine the deflection at midspan for the beam of Problem 4–69 using Castigliano's theorem.

4–77 Using Castigliano's theorem, determine the deflection of point B in the direction of the force F for the steel bar shown.

15 in O Problem 4–77

1 -in 2

dia.

A 7 in B

4 3

F = 15 lbf

230      Mechanical Engineering Design

4–78* Solve Problem 4–41 using Castigliano's theorem. Since Equation (4–18) for torsional

strain energy was derived from the angular displacement for circular cross sections, it is not applicable for section BC. You will need to obtain a new strain energy equation for the rectangular cross section from Equations (4–18) and (3–41).

4–79 Solve Problem 4–42 using Castigliano's theorem. 4–80* The cantilevered handle in Problem 3–95 is made from mild steel. Let Fy = 250 lbf

and Fx = Fz = 0. Using Castigliano's theorem, determine the vertical deflection (along the y axis) at the tip. Repeat the problem with shaft OC simplified to a uniform diameter of 1 in for its entire length. What is the percent error from this simplification?

4–81* Solve Problem 4–50 using Castigliano's theorem. 4–82 Solve Problem 4–51 using Castigliano's theorem. 4–83 Solve Problem 4–52 using Castigliano's theorem. 4–84 Solve Problem 4–53 using Castigliano's theorem. 4–85 Solve Problem 4–54 using Castigliano's theorem. 4–86 Solve Problem 4–55 using Castigliano's theorem. 4–87 The figure shows a rectangular member OB, made from 0.5-cm thick aluminum

plate, pinned to the ground at one end and supported by a coil spring at point A. The spring has been measured to deflect 1 mm with an applied force of 10 kN. Find the deflection at point B. Use Castigliano's method directly without using superposition or indirect geometric extrapolations. Organize your work in the ­following manner: (a) Determine the component of the deflection at B due to the energy in section OA. (b) Determine the component of the deflection at B due to the energy in section AB. (c) Determine the component of the deflection at B due to the energy in section AC. You may find it helpful to consider the definition of strain energy in Equation (4–15), and apply it to the spring. (d) Determine the total deflection at B due to all of the components combined.

C 300 N Problem 4–87

0.5-cm thick A B

O 25 cm

3 cm

25 cm

4–88 The part shown in the figure is made from cold-drawn AISI 1020 steel. Use Castigliano's and method to directly find the deflection of point D in the y direction. Organize your 4–89 work in the following manner: (a) Determine the component of the deflection at D due to the energy in section AB. (b) Determine the component of the deflection at D due to the energy in section BC. (c) Determine the component of the deflection at D due to the energy in section BD. (d) Determine the total deflection at D due to all of the components combined.

Deflection and Stiffness     231 y

y

A

0.25 × 1.25 in

1-in dia.

500 lbf

A

C

8 × 35

25 dia.

C

300 lbf B

z

450 N

B

z 5 in

200 lbf 6 in

D

125

x

300 N

4 in

D

Problem 4–88

150

x 100

Problem 4–89 Dimensions in mm

4–90 For Problem 4–88, determine the displacement of point D in the z direction. 4–91 For Problem 4–89, determine the displacement of point D in the z direction. 4–92 The steel curved bar shown has a rectangular cross section with a radial height h = 6

mm, and a thickness b = 4 mm. The radius of the centroidal axis is R = 40 mm. A force P = 10 N is applied as shown. Find the vertical deflection at B. Use Castigliano's method for a curved flexural member, and since R∕h < 10, do not neglect any of the terms.

A C

R P

4–93 Repeat Problem 4–92 to find the vertical deflection at A. 4–94 For the curved steel beam shown, F = 6.7 kips. Determine the relative deflection of the applied forces.

B Problem 4–92

3 in

Problem 4–94

2 in

0.375 in

F F

A

A

0.75 in

1.5 in 0.375 in Section A–A

4 in

4–95 A steel piston ring has a mean diameter of 70 mm, a radial height h = 4.5 mm, and

a thickness b = 3 mm. The ring is assembled using an expansion tool that separates the split ends a distance δ by applying a force F as shown. Use Castigliano's theorem and determine the force F needed to expand the split ends a distance δ = 1 mm.

4–96 For the steel wire form shown, use Castigliano's method to determine the horizontal reaction forces at A and B and the deflection at C. C

Problem 4–96

40 mm A

h = 4.5 mm F + δ F

2-mm dia.

30 N

Problem 4–95

B

232      Mechanical Engineering Design

4–97 The part shown is formed from a 18 -in diameter steel wire, with R = 2.5 in and l = 2 in. to A force is applied with P = 1 lbf. Use Castigliano's method to estimate the horizontal 4–100 deflection at point A. Justify any components of strain energy that you choose to neglect. D

R

C C B R

B

A

C

B

B

R l

P

P

l

P

R

l P

A Problem 4–97

l

A

C

A

Problem 4–98

Problem 4–99

Problem 4–100

4–101 Repeat Problem 4–97 for the vertical deflection at point A.

A

4–102 Repeat Problem 4–98 for the vertical deflection at point A. 8 cm

2-mm dia.

4–103 Repeat Problem 4–99 for the vertical deflection at point A. 4–104 Repeat Problem 4–100 for the vertical deflection at point A.

B

E 4 cm

C

D

4–105 A hook is formed from a 2-mm-diameter steel wire and fixed firmly into the ceiling as shown. A 1-kg mass is hung from the hook at point D. Use Castigliano's theorem to determine the vertical deflection of point D.

4–106 The figure shows a rectangular member OB, made from 14 -in-thick aluminum plate,

pinned to the ground at one end, and supported by a 12 -in-diameter round steel rod that is formed into an arc and pinned to the ground at C. A load of 100 lbf is applied at B. Use Castigliano's theorem to determine the vertical deflection at point B. Justify any choices to neglect any components of strain energy.

Problem 4–105

1 -in 2

Problem 4–106

1 -in 4

thick

C

dia.

100 lbf 10 in

2 in A

B

O 10 in

10 in

10 in

4–107 Repeat Problem 4–106 for the vertical deflection at point A. 4–108 For the wire form shown, determine the deflection of point A in the y direction.

Assume R∕h > 10 and consider the effects of bending and torsion only. The wire is steel with E = 200 GPa, ν = 0.29, and has a diameter of 6 mm. Before application of the 250-N force the wire form is in the xz plane where the radius R is 80 mm.

Deflection and Stiffness     233 y

x Problem 4–108

R

z

90° A

250 N

4–109 A 100-ft cable is made using a 12-gauge (0.1055-in) steel wire and three strands of 10-gauge (0.1019-in) copper wire. Find the deflection of the cable and the stress in each wire if the cable is subjected to a tension of 400 lbf.

4–110 The figure shows a steel pressure cylinder of diameter 5 in that uses six SAE grade

4 steel bolts having a grip of 10 in. These bolts have a proof strength (see Chapter 8) of 65 kpsi. Suppose the bolts are tightened to 75 percent of this strength. (a) Find the tensile stress in the bolts and the compressive stress in the cylinder walls. (b) Repeat part (a), but assume now that a fluid under a pressure of 500 psi is introduced into the cylinder. Six

1 2

-in grade 4 bolts

t = 14 in

Problem 4–110

lc = 9 in

D = 5 in

lb = 10 in

y B

4–111 A torsion bar of length L consists of a round core of stiffness (GJ)c and a shell of stiffness (GJ)s. If a torque T is applied to this composite bar, what percentage of the total torque is carried by the shell?

600 mm

60 mm

W

10 mm thick A

400 mm

4–112 A rectangular aluminum bar 10 mm thick and 60 mm wide is welded to fixed supports at the ends, and the bar supports a load W = 4 kN, acting through a pin as shown. Find the reactions at the supports and the deflection of point A.

x O Problem 4–112

234      Mechanical Engineering Design

4–113 Solve Problem 4–112 using Castigliano's method and procedure 1 from Section 4–10. 4–114 An aluminum step bar is loaded as shown. (a) Verify that end C deflects to the rigid

wall, and (b) determine the wall reaction forces, the stresses in each member, and the deflection of B. A

0.75-in dia.

Problem 4–114

0.5-in dia.

B

3 kip

(Not drawn to scale)

C

2 kip

8 in

5 in 0.005 in

4–115 The steel shaft shown in the figure is subjected to a torque of 200 lbf · in applied at point A. Find the torque reactions at O and B; the angle of twist at A, in degrees; and the shear stress in sections OA and AB. y 200 lbf ·in

1 2

-in dia.

Problem 4–115

x O

A

B

4 in

6 in

4–116 Repeat Problem 4–115 with the diameters of section OA being 0.5 in and section AB being 0.75 in.

4–117 The figure shows a 12 - by 1-in rectangular steel bar welded to fixed supports at each

end. The bar is axially loaded by the forces FA = 12 kip and FB = 6 kip acting on pins at A and B. Assuming that the bar will not buckle laterally, find the reactions at the fixed supports, the stress in section AB, and the deflection of point A. Use procedure 1 from Section 4–10. y 10 in

20 in

A

Problem 4–117 1 in

FA O

15 in B

C

FB 1 2

x

in thick

4–118 For the beam shown, determine the support reactions using superposition and ­procedure 1 from Section 4–10. w

Problem 4–118

B

A

C

a l

Deflection and Stiffness     235

4–119 Solve Problem 4–118 using Castigliano's theorem and procedure 1 from Section 4–10. 4–120 Consider beam 13 in Table A–9, but with flexible supports. Let w = 500 lbf/ft,

l = 2 ft, E = 30 Mpsi, and I = 0.85 in4. The support at the left end has a translational spring constant of k1 = 1.5(106) lbf/in and a rotational spring constant of k2 = 2.5(106) lbf · in. The right support has a translational spring constant of k3 = 2.0(106) lbf/in. Using procedure 2 of Section 4–10, determine the reactions at the supports and the deflection at the midpoint of the beam.

4–121 The steel beam ABCD shown is simply supported at A and supported at B and D

by steel cables, each having an effective diameter of 0.5 in. The second area moment of the beam is I = 1.2 in4. A force of 5 kips is applied at point C. Using procedure 2 of Section 4–10 determine the stresses in the cables and the deflections of B, C, and D.

E

F 38 in

A

Problem 4–121

B

C

D 5 kips

16 in

16 in

16 in

4–122 The steel beam ABCD shown is simply supported at C as shown and supported at

B and D by shoulder steel bolts, each having a diameter of 8 mm. The lengths of BE and DF are 50 mm and 65 mm, respectively. The beam has a second area moment of 21(103) mm4. Prior to loading, the members are stress-free. A force of 2 kN is then applied at point A. Using procedure 2 of Section 4–10, determine the stresses in the bolts and the deflections of points A, B, and D.

2 kN

E

A

B

D

C

Problem 4–122

F 75 mm

75 mm

75 mm

4–123 Repeat Example 4–15 except consider the ground support at point B to be elastic with a spring constant of kB = 40 MN/m.

4–124 A thin ring is loaded by two equal and opposite forces F in part a of the figure. A

free-body diagram of one quadrant is shown in part b. This is a statically indeterminate problem, because the moment MA cannot be found by statics. (a) Find the maximum bending moment in the ring due to the forces F, and (b) find the increase in the

236      Mechanical Engineering Design

diameter of the ring along the y axis. Assume that the radius of the ring is large so that Equation (4–41) can be used.

y y F B

B

ds dθ R A

Problem 4–124 C

O

θ x

A

O

x MA

F 2

D F (a)

(b)

4–125 A round tubular column has outside and inside diameters of D and d, respectively,

and a diametral ratio of K = d∕D. Show that buckling will occur when the outside diameter is 1∕4 64Pcr l 2 D=[ 3 4 ] π CE(1 − K )

4–126 For the conditions of Problem 4–125, show that buckling according to the parabolic formula will occur when the outside diameter is

1∕2 Sy l 2 Pcr D = 2[ + πSy (1 − K 2 ) π2CE(1 + K 2 ) ]

4–127 Link 2, shown in the figure, is 25 mm wide, has 12-mm-diameter bearings at the

ends, and is cut from low-carbon steel bar stock having a minimum yield strength of 165 MPa. The end-condition constants are C = 1 and C = 1.2 for buckling in and out of the plane of the drawing, respectively. (a) Using a design factor nd = 4, find a suitable thickness for the link. (b) Are the bearing stresses at O and B of any significance?

y

1 Problems 4–127 and 4–128

O

2

x A

3 B

900 mm

800 N

500 mm C 750 mm

Deflection and Stiffness     237

4–128 Link 2, shown in the figure, is 25 mm wide and 11 mm thick. It is made from low-

carbon steel with Sy = 165 MPa. The pin joints are constructed with sufficient size and fit to provide good resistance to out-of-plane bending. Use Table 4–2 for recommended values for C. Determine the following for link 2. (a) Axial force (b) Yielding factor of safety (c) In-plane buckling factor of safety (d) Out-of-plane buckling factor of safety Hint: Be sure to check Euler versus Johnson for both parts (b) and (c), as the (l∕k)1 point is different for each case.

4–129 Link OB is 20 mm wide and 10 mm thick, and is made from low-carbon steel with

Sy = 200 MPa. The pin joints are constructed with sufficient size and fit to provide good resistance to out-of-plane bending. Determine the factor of safety for out-ofplane buckling.

Problem 4–129

A

O

400 mm

y B

800 mm B

1200 N

F = 270 1bf 3

4–130 Link 3, shown schematically in the figure, acts as a brace to support the 270-lbf

load. For buckling in the plane of the figure, the link may be regarded as pinned at both ends. For out-of-plane buckling, the ends are fixed. Select a suitable material and a method of manufacture, such as forging, casting, stamping, or machining, for casual applications of the brace in oil-field machinery. Specify the dimensions of the cross section as well as the ends so as to obtain a strong, safe, well-made, and economical brace.

4–131 The hydraulic cylinder shown in the figure has a 2-in bore and is to operate at a pres-

sure of 1500 psi. With the clevis mount shown, the piston rod should be sized as a column with both ends rounded for any plane of buckling. The rod is to be made of forged AISI 1030 steel without further heat treatment. d

Problem 4–131

2 in

(a) Use a design factor nd = 2.5 and select a preferred size for the rod diameter if the column length is 50 in. (b) Repeat part (a) but for a column length of 16 in. (c) What factor of safety actually results for each of the cases above?

4–132 The figure shows a schematic drawing of a vehicular jack that is to be designed to

support a maximum mass of 300 kg based on the use of a design factor nd = 3.50. The opposite-handed threads on the two ends of the screw are cut to allow the link angle θ

3 ft

2

O

60

A

1 Problem 4–130

x

238      Mechanical Engineering Design

to vary from 15 to 70°. The links are to be machined from AISI 1010 hot-rolled steel bars. Each of the four links is to consist of two bars, one on each side of the central bearings. The bars are to be 350 mm long and have a bar width of w = 30 mm. The pinned ends are to be designed to secure an end-condition constant of at least C = 1.4 for out-of-plane buckling. Find a suitable preferred thickness and the resulting factor of safety for this thickness. W

Problem 4–132 θ

w

4–133 If drawn, a figure for this problem would resemble that for Problem 4–110. A strut

that is a standard hollow right circular cylinder has an outside diameter of 3 in and a wall thickness of 14 in and is compressed between two circular end plates held by four bolts equally spaced on a bolt circle of 4.5-in diameter. All four bolts are handtightened, and then bolt A is tightened to a tension of 1500 lbf and bolt C, diagonally opposite, is tightened to a tension of 9000 lbf. The strut axis of symmetry is coincident with the center of the bolt circles. Find the maximum compressive load, the eccentricity of loading, and the largest compressive stress in the strut.

4–134 Design link CD of the hand-operated toggle press shown in the figure. Specify the

cross-section dimensions, the bearing size and rod-end dimensions, the material, and the method of processing. F A B L l

Problem 4–134 L = 9 in, l = 3 in, θmin = 0°.

C

θ l D

Deflection and Stiffness     239

4–135 Find the maximum values of the spring force and deflection of the impact system

shown in the figure if W = 30 lbf, k = 100 lbf/in, and h = 2 in. Ignore the mass of the spring and solve using energy conservation. W y k

Problem 4–135

h

4–136 As shown in the figure, the weight W1 strikes W2 from a height h. If W1 = 40 N,

W2 = 400 N, h = 200 mm, and k = 32 kN/m, find the maximum values of the spring force and the deflection of W2. Assume that the impact between W1 and W2 is inelastic, ignore the mass of the spring, and solve using energy conservation. h

W1

Problem 4–136

W2 y k

4–137 Part a of the figure shows a weight W mounted between two springs. If the free end

of spring k1 is suddenly displaced through the distance x = a, as shown in part b, determine the maximum displacement y of the weight. Let W = 5 lbf, k1 = 10 lbf/in, k2 = 20 lbf/in, and a = 0.25 in. Ignore the mass of each spring and solve using energy conservation. x

y k1

k2 W

Problem 4–137

a t

x (a)

(b)

E O

D C

2

2

A xy

x G

part

Courtesy of Dee Dehokenanan

Failure Prevention Chapter 5 Failures Resulting from Static Loading  241 Chapter 6 F  atigue Failure Resulting from Variable Loading  285

2

5

Failures Resulting from Static Loading

©Aroon Phukeed/123RF

Chapter Outline 5–1

Static Strength   244

5–2

Stress Concentration   245

5–3

Failure Theories   247

5–4  Maximum-Shear-Stress Theory for Ductile Materials  247 5–5  Distortion-Energy Theory for Ductile Materials  249 5–6  Coulomb-Mohr Theory for Ductile Materials  255 5–7

Failure of Ductile Materials Summary   258

5–8  Maximum-Normal-Stress Theory for Brittle Materials  262 5–9  Modifications of the Mohr Theory for Brittle Materials   263 5–10  Failure of Brittle Materials Summary  265 5–11

Selection of Failure Criteria   266

5–12  Introduction to Fracture Mechanics  266 5–13

Important Design Equations   275 241

242      Mechanical Engineering Design

In Chapter 1 we learned that strength is a property or characteristic of a mechanical element. This property results from the material identity, the treatment and processing incidental to creating its geometry, and the loading, and it is at the controlling or critical location. In addition to considering the strength of a single part, we must be cognizant that the strengths of the mass-produced parts will all be somewhat different from the others in the collection or ensemble because of variations in dimensions, machining, forming, and composition. Descriptors of strength are necessarily statistical in nature, involving parameters such as mean, standard deviations, and distributional identification. A static load is a stationary force or couple applied to a member. To be stationary, the force or couple must be unchanging in magnitude, point or points of application, and direction. A static load can produce axial tension or compression, a shear load, a bending load, a torsional load, or any combination of these. To be considered static, the load cannot change in any manner. In this chapter we consider the relations between strength and static loading in order to make the decisions concerning material and its treatment, fabrication, and geometry for satisfying the requirements of functionality, safety, reliability, competitiveness, usability, manufacturability, and marketability. How far we go down this list is related to the scope of the examples. "Failure" is the first word in the chapter title. Failure can mean a part has separated into two or more pieces; has become permanently distorted, thus ruining its geometry; has had its reliability downgraded; or has had its function compromised, whatever the reason. A designer speaking of failure can mean any or all of these possibilities. In this chapter our attention is focused on the predictability of permanent distortion or separation. In strength-sensitive situations the designer must separate mean stress and mean strength at the critical location sufficiently to accomplish his or her purposes. Figures 5–1 to 5–5 are photographs of several failed parts. The photographs exemplify the need of the designer to be well-versed in failure prevention. Toward this end we shall consider one-, two-, and three-dimensional stress states, with and without stress concentrations, for both ductile and brittle materials.

Figure 5–1 (a) Failure of a truck drive-shaft spline due to corrosion fatigue. Note that it was necessary to use clear tape to hold the pieces in place. (b) Direct end view of failure. (For permission to reprint Figures 5–1 through 5–5, the authors are grateful for the personal photographs of Larry D. Mitchell, coauthor of Mechanical Engineering Design, 4th ed., McGraw-Hill, New York, 1983.)

Failures Resulting from Static Loading     243

Figure 5–2 Impact failure of a lawn-mower blade driver hub. The blade impacted a surveying pipe marker.

Figure 5–3 Failure of an overhead-pulley retaining bolt on a weightlifting machine. A manufacturing error caused a gap that forced the bolt to take the entire moment load.

Figure 5–4 Chain test fixture that failed in one cycle. To alleviate complaints of excessive wear, the manufacturer decided to case-harden the material. (a) Two halves showing fracture; this is an excellent example of brittle fracture initiated by stress concentration. (b) Enlarged view of one portion to show cracks induced by stress concentration at the support-pin holes.

244      Mechanical Engineering Design

Figure 5–5 Valve-spring failure caused by spring surge in an oversped engine. The fractures exhibit the classic 45° shear failure.

5–1  Static Strength Ideally, in designing any machine element, the engineer should have available the results of a great many strength tests of the particular material chosen. These tests should be made on specimens having the same heat treatment, surface finish, and size as the element the engineer proposes to design; and the tests should be made under exactly the same loading conditions as the part will experience in service. This means that if the part is to experience a bending load, it should be tested with a bending load. If it is to be subjected to combined bending and torsion, it should be tested under combined bending and torsion. If it is made of heat-treated AISI 1040 steel drawn at 500°C with a ground finish, the specimens tested should be of the same material prepared in the same manner. Such tests will provide very useful and precise information. Whenever such data are available for design purposes, the engineer can be assured of doing the best possible job of engineering. The cost of gathering such extensive data prior to design is justified if failure of the part may endanger human life or if the part is manufactured in sufficiently large quantities. Refrigerators and other appliances, for example, have very good reliabilities because the parts are made in such large quantities that they can be thoroughly tested in advance of manufacture. The cost of making these tests is very low when it is divided by the total number of parts manufactured. You can now appreciate the following four design categories: 1 Failure of the part would endanger human life, or the part is made in extremely large quantities; consequently, an elaborate testing program is justified during design. 2 The part is made in large enough quantities that a moderate series of tests is feasible. 3 The part is made in such small quantities that testing is not justified at all; or the design must be completed so rapidly that there is not enough time for testing. 4 The part has already been designed, manufactured, and tested and found to be unsatisfactory. Analysis is required to understand why the part is unsatisfactory and what to do to improve it.

Failures Resulting from Static Loading     245

More often than not it is necessary to design using only published values of yield strength, ultimate strength, percentage reduction in area, and percentage elongation, such as those listed in Appendix A. How can one use such meager data to design against both static and dynamic loads, two- and three-dimensional stress states, high and low temperatures, and very large and very small parts? These and similar questions will be addressed in this chapter and those to follow, but think how much better it would be to have data available that duplicate the actual design situation.

5–2  Stress Concentration Stress concentration (see Section 3–13) is a highly localized effect. In some instances it may be due to a surface scratch. If the material is ductile and the load static, the design load may cause yielding in the critical location in the notch. This yielding can involve strain strengthening of the material and an increase in yield strength at the small critical notch location. Since the loads are static and the material is ductile, that part can carry the loads satisfactorily with no general yielding. In these cases the designer sets the geometric (theoretical) stress-concentration factor Kt to unity. The rationale can be expressed as follows. The worst-case scenario is that of an idealized non–strain-strengthening material shown in Figure 5–6. The stress-strain curve rises linearly to the yield strength Sy, then proceeds at constant stress, which is equal to Sy. Consider a filleted rectangular bar as depicted in Figure A–15–5, where the cross-section area of the small shank is 1 in2. If the material is ductile, with a yield point of 40 kpsi, and the theoretical stress-concentration factor (SCF) Kt is 2, ∙ A load of 20 kip induces a nominal tensile stress of 20 kpsi in the shank as depicted at point A in Figure 5–6. At the critical location in the fillet the stress is 40 kpsi, and the SCF is K = σmax∕σnom = 40∕20 = 2. ∙ A load of 30 kip induces a nominal tensile stress of 30 kpsi in the shank at point B. The fillet stress is still 40 kpsi (point D), and the SCF K = σmax∕σnom = Sy∕σ = 40∕30 = 1.33. ∙ At a load of 40 kip the induced tensile stress (point C) is 40 kpsi in the shank. At the critical location in the fillet, the stress (at point E) is 40 kpsi. The SCF K = σmax∕σnom = Sy∕σ = 40∕40 = 1. Figure 5–6 An idealized stress-strain curve. The dashed line depicts a strain-strengthening material.

Tensile stress σ, kpsi

50

Sy

C D

E

B

A

0

Tensile strain, ε

246      Mechanical Engineering Design

For materials that strain-strengthen, the critical location in the notch has a higher Sy. The shank area is at a stress level a little below 40 kpsi, is carrying load, and is very near its failure-by-general-yielding condition. This is the reason designers do not apply Kt in static loading of a ductile material loaded elastically, instead setting Kt = 1. When using this rule for ductile materials with static loads, be careful to assure yourself that the material is not susceptible to brittle fracture (see Section 5–12) in the environment of use. The usual definition of geometric (theoretical) stressconcentration factor for normal stress Kt and shear stress Kts is given by Equation pair (3–48) as

σmax = Kt σnom

(a)

τmax = Ktsτnom

(b)

Since your attention is on the stress-concentration factor, and the definition of σnom or τnom is given in the graph caption or from a computer program, be sure the value of nominal stress is appropriate for the section carrying the load. As shown in Figure 2–2b, brittle materials do not exhibit a plastic range. The stress-concentration factor given by Equation (a) or (b) could raise the stress to a level to cause fracture to initiate at the stress raiser, and initiate a catastrophic failure of the member. An exception to this rule is a brittle material that inherently contains microdiscontinuity stress concentration, worse than the macrodiscontinuity that the designer has in mind. Sand molding introduces sand particles, air, and water vapor bubbles. The grain structure of cast iron contains graphite flakes (with little strength), which are literally cracks introduced during the solidification process. When a tensile test on a cast iron is performed, the strength reported in the literature includes this stress concentration. In such cases Kt or Kts need not be applied. An important source of stress-concentration factors is R. E. Peterson, who compiled them from his own work and that of others.1 Peterson developed the style of presentation in which the stress-concentration factor Kt is multiplied by the nominal stress σnom to estimate the magnitude of the largest stress in the locality. His approximations were based on photoelastic studies of two-dimensional strips (Hartman and Levan, 1951; Wilson and White, 1973), with some limited data from threedimensional photoelastic tests of Hartman and Levan. A contoured graph was included in the presentation of each case. Filleted shafts in tension were based on two-dimensional strips. Table A–15 provides many charts for the theoretical stress-concentration factors for several fundamental load conditions and geometry. Additional charts are also available from Peterson.2 Finite element analysis (FEA) can also be applied to obtain stress-concentration factors. Improvements on Kt and Kts for filleted shafts were reported by Tipton, Sorem, and Rolovic.3 1

R. E. Peterson, "Design Factors for Stress Concentration," Machine Design, vol. 23, no. 2, February 1951; no. 3, March 1951; no. 5, May 1951; no. 6, June 1951; no. 7, July 1951. 2 Walter D. Pilkey and Deborah Pilkey, Peterson's Stress-Concentration Factors, 3rd ed, John Wiley & Sons, New York, 2008. 3 S. M. Tipton, J. R. Sorem Jr., and R. D. Rolovic, "Updated Stress-Concentration Factors for Filleted Shafts in Bending and Tension," Trans. ASME, Journal of Mechanical Design, vol. 118, September 1996, pp. 321–327.

Failures Resulting from Static Loading     247

5–3  Failure Theories Section 5–1 illustrated some ways that loss of function is manifested. Events such as distortion, permanent set, cracking, and rupturing are among the ways that a machine element fails. Testing machines appeared in the 1700s, and specimens were pulled, bent, and twisted in simple loading processes. If the failure mechanism is simple, then simple tests can give clues. Just what is simple? The tension test is uniaxial (that's simple) and elongations are largest in the axial direction, so strains can be measured and stresses inferred up to "failure." Just what is important: a critical stress, a critical strain, a critical energy? In the next several sections, we shall show failure theories that have helped answer some of these questions. Unfortunately, there is no universal theory of failure for the general case of material properties and stress state. Instead, over the years several hypotheses have been formulated and tested, leading to today's accepted practices. Being accepted, we will characterize these "practices" as theories as most designers do. Structural metal behavior is typically classified as being ductile or brittle, although under special situations, a material normally considered ductile can fail in a brittle manner (see Section 5–12). Ductile materials are normally classified such that εf ≥ 0.05 and have an identifiable yield strength that is often the same in compression as in tension (Syt = Syc = Sy). Brittle materials, εf < 0.05, do not exhibit an identifiable yield strength, and are typically classified by ultimate tensile and compressive strengths, Sut and Suc, respectively (where Suc is given as a positive quantity). The generally accepted theories are: Ductile materials (yield criteria) ∙ Maximum shear stress (MSS), Section 5–4 ∙ Distortion energy (DE), Section 5–5 ∙ Ductile Coulomb-Mohr (DCM), Section 5–6 Brittle materials (fracture criteria) ∙ Maximum normal stress (MNS), Section 5–8 ∙ Brittle Coulomb-Mohr (BCM), Section 5–9 ∙ Modified Mohr (MM), Section 5–9 It would be inviting if we had one universally accepted theory for each material type, but for one reason or another, they are all used. Later, we will provide rationales for selecting a particular theory. First, we will describe the bases of these theories and apply them to some examples.

5–4  Maximum-Shear-Stress Theory for Ductile Materials The maximum-shear-stress (MSS) theory predicts that yielding begins whenever the maximum shear stress in any element equals or exceeds the maximum shear stress in a tension-test specimen of the same material when that specimen begins to yield. The MSS theory is also referred to as the Tresca or Guest theory. Many theories are postulated on the basis of the consequences seen from tensile tests. As a strip of a ductile material is subjected to tension, slip lines (called Lüder lines) form at approximately 45° with the axis of the strip. These slip lines are the

248      Mechanical Engineering Design

beginning of yield, and when loaded to fracture, fracture lines are also seen at angles approximately 45° with the axis of tension. Since the shear stress is maximum at 45° from the axis of tension, it makes sense to think that this is the mechanism of failure. It will be shown in the next section that there is a little more going on than this. However, it turns out the MSS theory is an acceptable but conservative predictor of failure; and since engineers are conservative by nature, it is quite often used. Recall that for simple tensile stress, σ = P∕A, and the maximum shear stress occurs on a surface 45° from the tensile surface with a magnitude of τmax = σ∕2. So the maximum shear stress at yield is τmax = Sy∕2. For a general state of stress, three principal stresses can be determined and ordered such that σ1 ≥ σ2 ≥ σ3. The maximum shear stress is then τmax = (σ1 − σ3)∕2 (see Figure 3–12). Thus, for a general state of stress, the maximum-shear-stress theory predicts yielding when

τmax =

σ1 − σ3 Sy ≥ 2 2

or

σ1 − σ3 ≥ Sy

(5–1)

Note that this implies that the yield strength in shear is given by

Ssy = 0.5Sy

(5–2)

which, as we will see later is about 15 percent low (conservative). For design purposes, Equation (5–1) can be modified to incorporate a factor of safety, n. Thus,

τmax =

Sy 2n

or

σ1 − σ3 =

Sy n

(5–3)

Plane stress is a very common state of stress in design. However, it is extremely important to realize that plane stress is a three-dimensional state of stress. Plane stress transformations in Section 3–6 are restricted to the in-plane stresses only, where the in-plane principal stresses are given by Equation (3–13) and labeled as σ1 and σ2. It is true that these are the principal stresses in the plane of analysis, but out of plane there is a third principal stress and it is always zero for plane stress. This means that if we are going to use the convention of ordering σ1 ≥ σ2 ≥ σ3 for three-dimensional analysis, upon which Equation (5–1) is based, we cannot arbitrarily call the in-plane principal stresses σ1 and σ2 until we relate them with the third principal stress of zero. To illustrate the MSS theory graphically for plane stress, we will first label the principal stresses given by Equation (3–13) as σA and σB, and then order them with the zero principal stress according to the convention σ1 ≥ σ2 ≥ σ3. Assuming that σA ≥ σB, there are three cases to consider when using Equation (5–1) for plane stress: Case 1: σA ≥ σB ≥ 0. For this case, σ1 = σA and σ3 = 0. Equation (5–1) reduces to a yield condition of

σA ≥ Sy

(5–4)

Case 2: σA ≥ 0 ≥ σB. Here, σ1 = σA and σ3 = σB, and Equation (5–1) becomes

σA − σB ≥ Sy

(5–5)

Case 3: 0 ≥ σA ≥ σB. For this case, σ1 = 0 and σ3 = σB, and Equation (5–1) gives

σB ≤ −Sy

(5–6)

Equations (5–4) to (5–6) are represented in Figure 5–7 by the three lines indicated in the σA, σB plane. The remaining unmarked lines are cases for σB ≥ σA, which completes the stress yield envelope but are not normally used. The maximum-shear-stress theory predicts yield if a stress state is outside the shaded region bordered by the stress yield

Failures Resulting from Static Loading     249 σB

Figure 5–7

Case 1

Sy

b Load line

gi

on

a

re

O

Sy

No ny i

eld

–Sy

σA

Case 2 –Sy Case 3

envelope. In Figure 5–7, suppose point a represents the stress state of a critical stress element of a member. If the load is increased, it is typical to assume that the principal stresses will increase proportionally along the line from the origin through point a. Such a load line is shown. If the stress situation increases along the load line until it crosses the stress failure envelope, such as at point b, the MSS theory predicts that the stress element will yield. The factor of safety guarding against yield at point a is given by the ratio of strength (distance to failure at point b) to stress (distance to stress at point a), that is n = Ob∕Oa. Note that the first part of Equation (5–3), τmax = Sy∕2n, is sufficient for design purposes provided the designer is careful in determining τmax. For plane stress, Equation (3–14) does not always predict τmax. However, consider the special case when one normal stress is zero in the plane, say σx and τxy have values and σy = 0. It can be easily shown that this is a Case 2 problem, and the shear stress determined by Equation (3–14) is τmax. Shaft design problems typically fall into this category where a normal stress exists from bending and/or axial loading, and a shear stress arises from torsion.

5–5  Distortion-Energy Theory for Ductile Materials The distortion-energy theory predicts that yielding occurs when the distortion strain energy per unit volume reaches or exceeds the distortion strain energy per unit volume for yield in simple tension or compression of the same material. The distortion-energy (DE) theory originated from the observation that ductile materials stressed hydrostatically (equal principal stresses) exhibited yield strengths greatly in excess of the values given by the simple tension test. Therefore it was postulated that yielding was not a simple tensile or compressive phenomenon at all, but, rather, that it was related somehow to the angular distortion of the stressed element. To develop the theory, note, in Figure 5–8a, the unit volume subjected to any threedimensional stress state designated by the stresses σ1, σ2, and σ3. The stress state shown in Figure 5–8b is one of hydrostatic normal stresses due to the stresses σav acting in each of the same principal directions as in Figure 5–8a. The formula for σav is simply

σav =

σ1 + σ2 + σ3 3

(a)

Thus, the element in Figure 5–8b undergoes pure volume change, that is, no angular distortion. If we regard σav as a component of σ1, σ2, and σ3, then this component can be subtracted from them, resulting in the stress state shown in Figure 5–8c. This element is subjected to pure angular distortion, that is, no volume change.

The maximum-shear-stress (MSS) theory yield envelope for plane stress, where σA and σB are the two nonzero principal stresses.

250      Mechanical Engineering Design

Figure 5–8 (a) Element with triaxial stresses; this element undergoes both volume change and angular distortion. (b) Element under hydrostatic normal stresses undergoes only volume change. (c) Element has angular distortion without volume change.

σ1 σ3

σ2 – σav

σav

σ2

=

σav

(a) Triaxial stresses

σ1 – σav σ3 – σav

σav

σ1 > σ2 > σ3

+

(b) Hydrostatic component

(c) Distortional component

The strain energy per unit volume for simple tension is u = 12 εσ. For the element of Figure 5–8a the strain energy per unit volume is u = 12 [ε1σ1 + ε2σ2 + ε3σ3]. Substituting Equation (3–19) for the principal strains gives

u=

1 2 [σ1 + σ22 + σ23 − 2ν(σ1σ2 + σ2σ3 + σ3σ1 )] 2E

(b)

The strain energy for producing only volume change uv can be obtained by substituting σav for σ1, σ2, and σ3 in Equation (b). The result is

uv =

3σ2av (1 − 2ν) 2E

(c)

If we now substitute the square of Equation (a) in Equation (c) and simplify the expression, we get

uv =

1 − 2ν 2 (σ1 + σ22 + σ23 + 2σ1σ2 + 2σ2σ3 + 2σ3σ1 ) 6E

(5–7)

Then the distortion energy is obtained by subtracting Equation (5–7) from Equation (b). This gives

ud = u − uv =

1 + ν (σ1 − σ2 ) 2 + (σ2 − σ3 ) 2 + (σ3 − σ1 ) 2 ] 3E [ 2

(5–8)

Note that the distortion energy is zero if σ1 = σ2 = σ3. For the simple tensile test, at yield, σ1 = Sy and σ2 = σ3 = 0, and from Equation (5–8) the distortion energy is

ud =

1+ν 2 Sy 3E

(5–9)

So for the general state of stress given by Equation (5–8), yield is predicted if Equation (5–8) equals or exceeds Equation (5–9). This gives

(σ1 − σ2 ) 2 + (σ2 − σ3 ) 2 + (σ3 − σ1 ) 2 1∕2 [ ] ≥ Sy 2

(5–10)

If we had a simple case of tension σ, then yield would occur when σ ≥ Sy. Thus, the left of Equation (5–10) can be thought of as a single, equivalent, or effective stress for the entire general state of stress given by σ1, σ2, and σ3. This effective stress is usually called the von Mises stress, σ′, named after Dr. R. von Mises, who contributed to the theory. Thus Equation (5–10), for yield, can be written as

σ′ ≥ Sy

(5–11)

Failures Resulting from Static Loading     251 σB

Figure 5–9 The distortion-energy (DE) theory yield envelope for plane stress states. This is a plot of points obtained from Equation (5–13) with σ′ = Sy.

–Sy

N on yi el d

re gi on

Sy

Sy

σA

Pure shear load line (σA = –σB = τ) –Sy

DE MSS

where the von Mises stress is σ′ = [

(σ1 − σ2 ) 2 + (σ2 − σ3 ) 2 + (σ3 − σ1 ) 2 1∕2 ] 2

(5–12)

For plane stress, the von Mises stress can be represented by the principal stresses σA, σB, and zero. Then from Equation (5–12), we get σ′ = (σ2A − σAσB + σ2B ) 1∕2

(5–13)

Equation (5–13) is a rotated ellipse in the σA, σB plane, as shown in Figure 5–9 with σ′ = Sy. The dotted lines in the figure represent the MSS theory, which can be seen to be more restrictive, hence, more conservative.4 Using xyz components of three-dimensional stress, the von Mises stress can be written as

σ′ =

1 √2

[(σx − σy ) 2 + (σy − σz ) 2 + (σz − σx ) 2 + 6(τ2xy + τ2yz + τ2zx )]1∕2

(5–14)

and for plane stress,

σ′ = (σ2x − σx σy + σ2y + 3τ2xy ) 1∕2

(5–15)

The distortion-energy theory is also called:

∙ The von Mises or von Mises–Hencky theory ∙ The shear-energy theory ∙ The octahedral-shear-stress theory Understanding octahedral shear stress will shed some light on why the MSS is conservative. Consider an isolated element in which the normal stresses on each surface are equal to the hydrostatic stress σav. There are eight surfaces symmetric to the principal directions that contain this stress. This forms an octahedron as shown in Figure 5–10. The shear stresses on these surfaces are equal and are called the 4

The three-dimensional equations for DE and MSS can be plotted relative to three-dimensional σ1, σ2, σ3, coordinate axes. The failure surface for DE is a circular cylinder with an axis inclined at 45° from each principal stress axis, whereas the surface for MSS is a hexagon inscribed within the cylinder. See Arthur P. Boresi and Richard J. Schmidt, Advanced Mechanics of Materials, 6th ed., John Wiley & Sons, New York, 2003, Section 4.4.

252      Mechanical Engineering Design

octahedral shear stresses (Figure 5–10 shows only one of the octahedral surfaces labeled). Through coordinate transformations the octahedral shear stress is given by5 σ2

σ3

Figure 5–10 Octahedral surfaces.

1 [(σ1 − σ2 ) 2 + (σ2 − σ3 ) 2 + (σ3 − σ1 ) 2]1∕2 3

(5–16)

Under the name of the octahedral-shear-stress theory, failure is assumed to occur whenever the octahedral shear stress for any stress state equals or exceeds the octahedral shear stress for the simple tension-test specimen at failure. As before, on the basis of the tensile test results, yield occurs when σ1 = Sy and σ2 = σ3 = 0. From Equation (5–16) the octahedral shear stress under this condition is

σav

τoct

τoct =

σ1

τoct =

√2

3

Sy

(5–17)

When, for the general stress case, Equation (5–16) is equal or greater than Equation (5–17), yield is predicted. This reduces to (σ1 − σ2 ) 2 + (σ2 − σ3 ) 2 + (σ3 − σ1 ) 2 1∕2 [ ] ≥ Sy 2

(5–18)

which is identical to Equation (5–10), verifying that the maximum-octahedral-shearstress theory is equivalent to the distortion-energy theory. The model for the MSS theory ignores the contribution of the normal stresses on the 45° surfaces of the tensile specimen. However, these stresses are P∕2A, and not the hydrostatic stresses which are P∕3A. Herein lies the difference between the MSS and DE theories. The mathematical manipulation involved in describing the DE theory might tend to obscure the real value and usefulness of the result. The equations given allow the most complicated stress situation to be represented by a single quantity, the von Mises stress, which then can be compared against the yield strength of the material through Equation (5–11). This equation can be expressed as a design equation by

σ′ =

Sy n

(5–19)

The distortion-energy theory predicts no failure under hydrostatic stress and agrees well with all data for ductile behavior. Hence, it is the most widely used theory for ductile materials and is recommended for design problems unless otherwise specified. One final note concerns the shear yield strength. Consider a case of pure shear τxy, where for plane stress σx = σy = 0. For yield, Equation (5–11) with Equation (5–15) gives (3τ2xy ) 1∕2 = Sy

or

τxy =

Sy √3

= 0.577Sy

(5–20)

Thus, the shear yield strength predicted by the distortion-energy theory is

Ssy = 0.577Sy

(5–21)

which as stated earlier, is about 15 percent greater than the 0.5 Sy predicted by the MSS theory. For pure shear, τxy the principal stresses from Equation (3–13) are σA  = −σB = τxy. The load line for this case is in the third quadrant at an angle of 45° from the σA, σB axes shown in Figure 5–9. 5

For a derivation, see Arthur P. Boresi, op. cit., pp. 36–37.

Failures Resulting from Static Loading     253

EXAMPLE 5–1 A hot-rolled steel has a yield strength of Syt = Syc = 100 kpsi and a true strain at fracture of εf = 0.55. Estimate the factor of safety for the following principal stress states: (a) σx = 70 kpsi, σy = 70 kpsi, τxy = 0 kpsi (b) σx = 60 kpsi, σy = 40 kpsi, τxy = −15 kpsi (c) σx = 0 kpsi, σy = 40 kpsi, τxy = 45 kpsi (d) σx = −40 kpsi, σy = −60 kpsi, τxy = 15 kpsi (e) σ1 = 30 kpsi, σ2 = 30 kpsi, σ3 = 30 kpsi Solution Since εf > 0.05 and Syt and Syc are equal, the material is ductile and both the distortion-energy (DE) theory and maximum-shear-stress (MSS) theory apply. Both will be used for comparison. Note that cases a to d are plane stress states. (a) Since there is no shear stress on this stress element, the normal stresses are equal to the principal stresses. The ordered principal stresses are σA = σ1 = 70, σB = σ2 = 70, σ3 = 0 kpsi. DE  From Equation (5–13), σ′ = [702 − 70(70) + 702]1∕2 = 70 kpsi

From Equation (5–19),

n=

Answer

Sy σ′

=

100 = 1.43 70

MSS  Noting that the two nonzero principal stresses are equal, τmax will be from the largest Mohr's circle, which will incorporate the third principal stress at zero. From Equation (3–16),

τmax =

σ1 − σ3 70 − 0 = = 35 kpsi 2 2

From Equation (5–3), n=

Answer

Sy∕2 100∕2 = = 1.43 τmax 35

(b) From Equation (3–13), the nonzero principal stresses are

σA, σB =

60 + 40 60 − 40 2 ± √( + (−15) 2 = 68.0, 32.0 kpsi 2 2 )

The ordered principal stresses are σA = σ1 = 68.0, σB = σ2 = 32.0, σ3 = 0 kpsi. DE Answer

σ′ = [682 − 68(32) + 322]1∕2 = 59.0 kpsi n=

Sy σ′

=

100 = 1.70 59.0

254      Mechanical Engineering Design

MSS  Noting that the two nonzero principal stresses are both positive, τmax will be from the largest Mohr's circle which will incorporate the third principal stress at zero. From Equation (3–16),

τmax =

σ1 − σ3 68.0 − 0 = = 34.0 kpsi 2 2 n=

Answer

Sy∕2 100∕2 = = 1.47 τmax 34.0

(c) This time, we shall obtain the factors of safety directly from the xy components of stress. DE  From Equation (5–15), σ′ = (σ2x − σx σy + σ2y + 3τ2xy ) 1∕2 = [(402 + 3(45) 2]1∕2 = 87.6 kpsi n=

Answer

Sy σ′

=

100 = 1.14 87.6

MSS  Taking care to note from a quick sketch of Mohr's circle that one nonzero principal stress will be positive while the other one will be negative, τmax can be obtained from the extreme-value shear stress given by Equation (3–14) without finding the principal stresses. τmax = √ (

σx − σy 2

2 ) + τxy = √ ( 2

n=

Answer

0 − 40 2 + 452 = 49.2 kpsi 2 )

Sy∕2 100∕2 = = 1.02 τmax 49.2

For graphical comparison purposes later in this problem, the nonzero principal stresses can be obtained from Equation (3–13) to be 69.2 kpsi and −29.2 kpsi. (d) From Equation (3–13), the nonzero principal stresses are σA, σB =

−40 − (−60) 2 −40 + (−60) 2 + √( ) + (15) = −32.0, −68.0 kpsi 2 2

The ordered principal stresses are σ1 = 0, σA = σ2 = −32.0, σB = σ3 = −68.0 kpsi. σ′ = [(−32) 2 − (−32) (−68) + (−68) 2]1∕2 = 59.0 kpsi

DE

n=

Answer

Sy σ′

=

100 = 1.70 59.0

MSS  From Equation (3–16), τmax =

σ1 − σ3 0 − (−68.0) = = 34.0 kpsi 2 2 n=

Answer

Sy∕2 100∕2 = = 1.47 τmax 34.0

(e) The ordered principal stresses are σ1 = 30, σ2 = 30, σ3 = 30 kpsi. DE  From Equation (5–12), σ′ = [

(30 − 30) 2 + (30 − 30) 2 + (30 − 30) 2 1∕2 ] = 0 kpsi 2

Failures Resulting from Static Loading     255

n=

Answer

Sy σ′

=

100 →∞ 0

MSS  From Equation (5–3), n=

Answer

Sy 100 = →∞ σ1 − σ3 30 − 30

A tabular summary of the factors of safety is included for comparisons. (a)

(b)

(c)

(d)

(e)

DE

1.43

1.70

1.14

1.70

MSS

1.43

1.47

1.02

1.47

Since the MSS theory is on or within the boundary of the DE theory, it will always predict a factor of safety equal to or less than the DE theory, as can be seen in the table. For each case, except case (e), the coordinates and load lines in the σA, σB plane are shown in Figure 5–11. Case (e) is not plane stress. Note that the load line for case (a) is the only plane stress case given in which the two theories agree, thus giving the same factor of safety. σB

Figure 5–11

(a)

Load lines for Example 5–1.

Sy

(b)

σB σA

–Sy

Sy

σA (c)

–Sy

DE MSS Load lines

(d )

5–6  Coulomb-Mohr Theory for Ductile Materials Not all materials have compressive strengths equal to their corresponding tensile values. For example, the yield strength of magnesium alloys in compression may be as little as 50 percent of their yield strength in tension. The ultimate strength of gray cast irons in compression varies from 3 to 4 times greater than the ultimate tensile strength. So, in this section, we are primarily interested in those theories that can be used to predict failure for materials whose strengths in tension and compression are not equal. Historically, the Mohr theory of failure dates to 1900, a date that is relevant to its presentation. There were no computers, just slide rules, compasses, and French curves. Graphical procedures, common then, are still useful today for visualization. The idea of Mohr is based on three "simple" tests: tension, compression, and shear, to yielding if the material can yield, or to rupture. It is easier to define shear yield strength as Ssy than it is to test for it.

256      Mechanical Engineering Design τ A Mohr failure curve

Coulomb-Mohr failure line

B C

–Sc

D

B3

E

St

τ

B2 B1

σ

O –Sc

Figure 5–12 Three Mohr circles, one for the uniaxial compression test, one for the test in pure shear, and one for the uniaxial tension test, are used to define failure by the Mohr hypothesis. The strengths Sc and St are the compressive and tensile strengths, respectively; they can be used for yield or ultimate strength.

σ3 C C 3 2

σ1 C1

St

σ

Figure 5–13 Mohr's largest circle for a general state of stress.

The practical difficulties aside, Mohr's hypothesis was to use the results of tensile, compressive, and torsional shear tests to construct the three circles of Figure 5–12 defining a failure envelope tangent to the three circles, depicted as curve ABCDE in the figure. The argument amounted to the three Mohr circles describing the stress state in a body (see Figure 3–12) growing during loading until one of them became tangent to the failure envelope, thereby defining failure. Was the form of the failure envelope straight, circular, or quadratic? A compass or a French curve defined the failure envelope. A variation of Mohr's theory, called the Coulomb-Mohr theory or the internalfriction theory, assumes that the boundary BCD in Figure 5–12 is straight. With this assumption only the tensile and compressive strengths are necessary. Consider the conventional ordering of the principal stresses such that σ1 ≥ σ2 ≥ σ3. The largest circle connects σ1 and σ3, as shown in Figure 5–13. The centers of the circles in Figure 5–13 are C1, C2, and C3. Triangles OBiCi are similar, therefore

B2C2 − B1C1 B3C3 − B1C1 = OC2 − OC1 OC3 − OC1

or,

B2C2 − B1C1 B3C3 − B1C1 = C1C2 C1C3

where B1C1 = St∕2, B2C2 = (σ1 − σ3)∕2, and B3C3 = Sc∕2, are the radii of the right, center, and left circles, respectively. The distance from the origin to C1 is St∕2, to C3 is Sc∕2, and to C2 (in the positive σ direction) is (σ1 + σ3)∕2. Thus, σ1 − σ3 St Sc St − − 2 2 2 2 = St σ1 + σ3 St Sc − + 2 2 2 2

Failures Resulting from Static Loading     257 σB

Figure 5–14

N on fa ilu re

–Sc

re gi on

St

σA

St

–Sc

Canceling the 2 in each term, cross-multiplying, and simplifying reduces this equation to

σ1 σ3 − = 1 St Sc

(5–22)

where either yield strength or ultimate strength can be used. For plane stress, when the two nonzero principal stresses are σA ≥ σB, we have a situation similar to the three cases given for the MSS theory, Equations (5–4) to (5–6). That is, the failure conditions are Case 1: σA ≥ σB ≥ 0. For this case, σ1 = σA and σ3 = 0. Equation (5–22) reduces to

σA ≥ St

(5–23)

Case 2: σA ≥ 0 ≥ σB. Here, σ1 = σA and σ3 = σB, and Equation (5–22) becomes

σA σB − ≥ 1 St Sc

(5–24)

Case 3: 0 ≥ σA ≥ σB. For this case, σ1 = 0 and σ3 = σB, and Equation (5–22) gives

σB ≤ −Sc

(5–25)

A plot of these cases, together with the normally unused cases corresponding to σB ≥ σA, is shown in Figure 5–14. For design equations, incorporating the factor of safety n, divide all strengths by n. For example, Equation (5–22) as a design equation can be written as

σ1 σ3 1 − = St Sc n

(5–26)

Since for the Coulomb-Mohr theory we do not need the torsional shear strength circle we can deduce it from Equation (5–22). For pure shear τ, σ1 = −σ3 = τ. The torsional yield strength occurs when τmax = Ssy. Substituting σ1 = −σ3 = Ssy into Equation (5–22) and simplifying gives

Ssy =

Syt Syc Syt + Syc

(5–27)

Plot of the Coulomb-Mohr theory failure envelope for plane stress states.

258      Mechanical Engineering Design

EXAMPLE 5–2 A 25-mm-diameter shaft is statically torqued to 230 N · m. It is made of cast 195-T6 aluminum, with a yield strength in tension of 160 MPa and a yield strength in compression of 170 MPa. It is machined to final diameter. Estimate the factor of safety of the shaft. Solution The maximum shear stress is given by τ=

16(230) 16T = = 75(106 ) N/m2 = 75 MPa πd 3 π[25(10−3 )]3

The two nonzero principal stresses are 75 and −75 MPa, making the ordered principal stresses σ1 = 75, σ2 = 0, and σ3 = −75 MPa. From Equation (5–26), for yield, n=

Answer

1 1 = = 1.10 σ1∕Syt − σ3∕Syc 75∕160 − (−75)∕170

Alternatively, from Equation (5–27),

Ssy =

Syt Syc Syt + Syc

=

160(170) = 82.4 MPa 160 + 170

and τmax = 75 MPa. Thus, n=

Answer

Ssy 82.4 = = 1.10 τmax 75

5–7  Failure of Ductile Materials Summary Having studied some of the various theories of failure, we shall now evaluate them and show how they are applied in design and analysis. In this section we limit our studies to materials and parts that are known to fail in a ductile manner. Materials that fail in a brittle manner will be considered separately because these require different failure theories. To help decide on appropriate and workable theories of failure, Marin6 collected data from many sources. Some of the data points used to select failure theories for ductile materials are shown in Figure 5–15.7 Mann also collected many data for copper and nickel alloys; if shown, the data points for these would be mingled with those already diagrammed. Figure 5–15 shows that either the maximum-shear-stress theory or the distortion-energy theory is acceptable for design and analysis of materials that would fail in a ductile manner. The selection of one or the other of these two theories is something that you, the engineer, must decide. For design purposes the maximum-shear-stress theory is easy, quick to use, and conservative. If the problem is to learn why a part failed, then the 6

Joseph Marin was one of the pioneers in the collection, development, and dissemination of material on the failure of engineering elements. He has published many books and papers on the subject. Here the reference used is Joseph Marin, Engineering Materials, Prentice Hall, Englewood Cliffs, N.J., 1952. 7 Note that some data in Figure 5–15 are displayed along the top horizontal boundary where σB ≥ σA. This is often done with failure data to thin out congested data points by plotting on the mirror image of the line σB = σA.

Failures Resulting from Static Loading     259 σ2 /Sc

Oct. shear

Yielding (Sc = Sy )

1.0

Ni-Cr-Mo steel AISI 1023 steel 2024-T4 Al 3S-H Al

Max. shear –1.0 0

1.0

Figure 5–15 Experimental data superposed on failure theories. (Source: Adapted from Fig. 7.11, p. 257, Mechanical Behavior of Materials, 2nd ed., N. E. Dowling, Prentice Hall, Englewood Cliffs, N.J., 1999. Modified to show only ductile failures.)

σ1 /Sc

–1.0

distortion-energy theory may be the best to use; Figure 5–15 shows that the plot of the distortion-energy theory passes closer to the central area of the data points, and thus is generally a better predictor of failure. However, keep in mind that though a failure curve passing through the center of the experimental data is typical of the data, its reliability from a statistical standpoint is about 50 percent. For design purposes, a larger factor of safety may be warranted when using such a failure theory. For ductile materials with unequal yield strengths, Syt in tension and Syc in compression, the Mohr theory is the best available. However, the theory requires the results from three separate modes of tests, graphical construction of the failure locus, and fitting the largest Mohr's circle to the failure locus. The alternative to this is to use the Coulomb-Mohr theory, which requires only the tensile and compressive yield strengths and is easily dealt with in equation form. EXAMPLE 5–3 This example illustrates the use of a failure theory to determine the strength of a mechanical element or component. The example may also clear up any confusion existing between the phrases strength of a machine part, strength of a material, and strength of a part at a point. A certain force F applied at D near the end of the 15-in lever shown in Figure 5–16, which is quite similar to a socket wrench, results in certain stresses in the cantilevered bar OABC. This bar (OABC) is of AISI 1035 steel, forged and heat-treated so that it has a minimum (ASTM) yield strength of 81 kpsi. We presume that this component would be of no value after yielding. Thus the force F required to initiate yielding can be regarded as the strength of the component part. Find this force. Solution We will assume that lever DC is strong enough and hence not a part of the problem. A 1035 steel, heat-treated, will have a reduction in area of 50 percent or more and hence is a ductile material at normal temperatures. This also means that stress concentration at shoulder A need not be considered. A stress element at A on the

260      Mechanical Engineering Design

y

Figure 5–16 2 in

O A 12 in z

1 12 -in dia. 1 8

B -in rad.

2 in C 1-in dia.

15 in F

1 12 -in dia.

x

D

top surface will be subjected to a tensile bending stress and a torsional stress. This point, on the 1-in-diameter section, is the weakest section, and governs the strength of the assembly. The two stresses are M 32M 32(14F) σx = = = = 142.6F I∕c πd 3 π(13 ) Tr 16T 16(15F ) τzx = = = = 76.4F J πd 3 π(13 ) Employing the distortion-energy theory, we find, from Equation (5–15), that σ′ = (σ2x + 3τ2zx ) 1∕2 = [(142.6F) 2 + 3(76.4F) 2]1∕2 = 194.5F Equating the von Mises stress to Sy, we solve for F and get Answer

F=

Sy 194.5

=

81 000 = 416 lbf 194.5

In this example the strength of the material at point A is Sy = 81 kpsi. The strength of the assembly or component is F = 416 lbf. Let us apply the MSS theory for comparison. For a point undergoing plane stress with only one nonzero normal stress and one shear stress, the two nonzero principal stresses will have opposite signs, and hence the maximum shear stress is obtained from the Mohr's circle between them. From Equation (3–14) σx 2 142.6F 2 τmax = √ ( ) + τ2zx = √ ( + (76.4F) 2 = 104.5F 2 2 ) Setting this equal to Sy∕2, from Equation (5–3) with n = 1, and solving for F, we get 81 000∕2 = 388 lbf 104.5 which is about 7 percent less than found for the DE theory. As stated earlier, the MSS theory is more conservative than the DE theory. F=

Failures Resulting from Static Loading     261

EXAMPLE 5–4 The cantilevered tube shown in Figure 5–17 is to be made of 2014 aluminum alloy treated to obtain a specified minimum yield strength of 276 MPa. We wish to select a stock-size tube from Table A–8 using a design factor nd = 4. The bending load is F = 1.75 kN, the axial tension is P = 9.0 kN, and the torsion is T = 72 N · m. What is the realized factor of safety? Solution The critical stress element is at point A on the top surface at the wall, where the bending moment is the largest, and the bending and torsional stresses are at their maximum values. The critical stress element is shown in Figure 5–17b. Since the axial stress and bending stress are both in tension along the x axis, they are additive for the normal stress, giving

σx =

P Mc 9 120(1.75)(do∕2) 9 105do + = + = + A I A I A I

(1)

where, if millimeters are used for the area properties, the stress is in gigapascals. The torsional stress at the same point is

τzx =

Tr 72(do∕2) 36do = = J J J

(2)

For accuracy, we choose the distortion-energy theory as the design basis. The von Mises stress from Equation (5–15) is σ′ = (σ2x + 3τ2zx ) 1∕2

(3)

On the basis of the given design factor, the goal for σ′ is

σ′ ≤

Sy 0.276 = = 0.0690 GPa nd 4

(4)

where we have used gigapascals in this relation to agree with Equations (1) and (2). Figure 5–17

y

12

0m

m

τzx A F z

σx

P T x (a)

z (b)

x

262      Mechanical Engineering Design

Programming Equations (1) to (3) on a spreadsheet and entering metric sizes from Table A–8 reveals that a 42 × 5-mm tube is satisfactory. The von Mises stress is found to be σ′= 0.06043 GPa for this size. Thus the realized factor of safety is n=

Answer

Sy σ′

=

0.276 = 4.57 0.06043

For the next size smaller, a 42 × 4-mm tube, σ′ = 0.07105 GPa giving a factor of safety of

n=

Sy σ′

=

0.276 = 3.88 0.07105

5–8  Maximum-Normal-Stress Theory for Brittle Materials The maximum-normal-stress (MNS) theory states that failure occurs whenever one of the three principal stresses equals or exceeds the strength. Again we arrange the principal stresses for a general stress state in the ordered form σ1 ≥ σ2 ≥ σ3. This theory then predicts that failure occurs whenever

σ1 ≥ Sut

or

σ3 ≤ −Suc

(5–28)

where Sut and Suc are the ultimate tensile and compressive strengths, respectively, given as positive quantities. For plane stress, with the principal stresses given by Equation (3–13), with σA ≥ σB, Equation (5–28) can be written as

σA ≥ Sut

or

σB ≤ −Suc

(5–29)

which is plotted in Figure 5–18. As before, the failure criteria equations can be converted to design equations. We can consider two sets of equations where σA ≥ σB as

σA =

Sut n

or

σB

Figure 5–18 Graph of maximum-normalstress (MNS) theory failure envelope for plane stress states.

Sut

–Suc

e ur ail

re

on gi

Sut

f

n No

– Suc

σA

σB = −

Suc n

(5–30)

Failures Resulting from Static Loading     263

As will be seen later, the maximum-normal-stress theory is not very good at predicting failure in the second and fourth quadrants of the σA, σB plane. Thus, we will not recommend the theory for use. It has been included here mainly for historical reasons.

5–9  Modifications of the Mohr Theory for Brittle Materials We will discuss two modifications of the Mohr theory for brittle materials: the BrittleCoulomb-Mohr (BCM) theory and the modified Mohr (MM) theory. The equations provided for the theories will be restricted to plane stress and be of the design type incorporating the factor of safety. The Coulomb-Mohr theory was discussed earlier in Section 5–6 with Equations (5–23) to (5–25). Written as design equations for a brittle material, they are: Brittle-Coulomb-Mohr

σA =

Sut n

σA ≥ σB ≥ 0

σA σB 1 − = Sut Suc n

σB = −

Suc n

(5–31a)

σA ≥ 0 ≥ σB

(5–31b)

0 ≥ σA ≥ σB

(5–31c)

On the basis of observed data for the fourth quadrant, the modified Mohr theory expands the fourth quadrant with the solid lines shown in the second and fourth quadrants of Figure 5–19 (where the factor of safety, n, is set to one).

max. normal

–700

omb

Coul

–500

Figure 5–19

300

Biaxial fracture data of gray cast iron compared with various failure criteria. (Source: Data points from Grassi, R. C., and Cornet, I., "Fracture of Gray Cast Iron Tubes under Biaxial Stress," J. of Applied Mechanics, vol. 16, June 1949, 178–182.)

Sut

mod. Mohr –Suc

σB , MPa

–300

hr

- Mo

100 Sut

–100

0

100

–100 –Sut –300

–500

–Suc –700

300

σA, MPa

264      Mechanical Engineering Design

Modified Mohr

σA =

Sut n

σA ≥ σB ≥ 0 σA ≥ 0 ≥ σB

(Suc − Sut )σA σB 1 − = Suc Sut Suc n

σB = −

Suc n

(5–32a)

σB and ⎹ ⎹ ≤ 1 σA

σA ≥ 0 ≥ σB and

σB

⎹ σ⎹ A

0 ≥ σA ≥ σB

> 1

(5–32b)

(5–32c)

Data are still outside this extended region. The straight line introduced by the modified Mohr theory, for σA ≥ 0 ≥ σB and ∣σB∕σA∣ > 1, can be replaced by a parabolic relation which can more closely represent some of the data.8 However, this introduces a nonlinear equation for the sake of a minor correction, and will not be presented here. EXAMPLE 5–5 Consider the wrench in Example 5–3, Figure 5–16, as made of cast iron, machined to dimension. The force F required to fracture this part can be regarded as the strength of the component part. If the material is ASTM grade 30 cast iron, find the force F with (a) Coulomb-Mohr failure model. (b) Modified Mohr failure model. Solution We assume that the lever DC is strong enough, and not part of the problem. Since grade 30 cast iron is a brittle material and cast iron, the stress-concentration factors Kt and Kts are set to unity. From Table A–24, the tensile ultimate strength is 31 kpsi and the compressive ultimate strength is 109 kpsi. The stress element at A on the top surface will be subjected to a tensile bending stress and a torsional stress. This location, on the 1-in-diameter section fillet, is the weakest location, and it governs the strength of the assembly. The normal stress σx and the shear stress at A are given by 32(14F) M 32M = Kt 3 = (1) = 142.6F I∕c πd π(1) 3

σx = Kt

τxy = Kts

16(15F) Tr 16T = Kts 3 = (1) = 76.4F J πd π(1) 3

From Equation (3–13) the nonzero principal stresses σA and σB are

σA, σB =

142.6F + 0 142.6F − 0 2 2 ± √( ) + (76.4F) = 175.8F, −33.2F 2 2

This puts us in the fourth-quadrant of the σA, σB plane.

8

See J. E. Shigley, C. R. Mischke, and R. G. Budynas, Mechanical Engineering Design, 7th ed., McGraw-Hill, New York, 2004, p. 275.

Failures Resulting from Static Loading     265

(a) For BCM, Equation (5–31b) applies with n = 1 for failure. σA σB (−33.2F) 175.8F − = − =1 Sut Suc 31(103 ) 109(103 )

Solving for F yields

F = 167 lbf

Answer

(b) For MM, the slope of the load line is ∣σB∕σA∣ = 33.2∕175.8 = 0.189 < 1. Obviously, Equation (5–32a) applies. σA 175.8F = =1 Sut 31(103 )

F = 176 lbf

Answer

As one would expect from inspection of Figure 5–19, Coulomb-Mohr is more conservative.

5–10  Failure of Brittle Materials Summary We have identified failure or strength of brittle materials that conform to the usual meaning of the word brittle, relating to those materials whose true strain at fracture is 0.05 or less. We also have to be aware of normally ductile materials that for some reason may develop a brittle fracture or crack if used below the transition temperature. Figure 5–20 shows data for a nominal grade 30 cast iron taken under biaxial stress conditions, with several brittle failure hypotheses shown, superposed. We note the following: ∙ In the first quadrant the data appear on both sides and along the failure curves of maximum-normal-stress, Coulomb-Mohr, and modified Mohr. All failure curves are the same, and data fit well. σB

Modified Mohr

Figure 5–20

–Sut 30 Sut

–120

– Suc –90

–60

–30

30

ASTM No. 30 C.I. Sut = 31 kpsi, Suc = 109 kpsi

–30

C

σA

–90

D

A

σB σA = –1

σB –60

B

σA

–Sut

Coulomb-Mohr

Maximum-normal-stress

Sut

–120

–150

–Suc

A plot of experimental data points obtained from tests on cast iron. Shown also are the graphs of three failure theories of possible usefulness for brittle materials. Note points A, B, C, and D. To avoid congestion in the first quadrant, points have been plotted for σA > σB as well as for the opposite sense. (Source: Data from Charles F. Walton (ed.), Iron Castings Handbook, Iron Founders' Society, 1971, pp. 215, 216, Cleveland, Ohio.)

266      Mechanical Engineering Design

Figure 5–21

Brittle behavior

Failure theory selection flowchart.

< 0.05

No

Mod. Mohr (MM) Eq. (5–32)

Conservative?

Yes

Ductile behavior

εf

≤ 0.05

No

Yes

Syt ≈ Syc?

Brittle Coulomb-Mohr Ductile Coulomb-Mohr (BCM) (DCM) Eq. (5–31) Eq. (5–26)

No

Conservative?

Distortion-energy (DE) Eqs. (5–15) and (5–19)

Yes

Maximum shear stress (MSS) Eq. (5–3)

∙ In the fourth quadrant the modified Mohr theory represents the data best, whereas the maximum-normal-stress theory does not. ∙ In the third quadrant the points A, B, C, and D are too few to make any suggestion concerning a fracture locus.

5–11  Selection of Failure Criteria For ductile behavior the preferred criterion is the distortion-energy theory, although some designers also apply the maximum-shear-stress theory because of its simplicity and conservative nature. In the rare case when Syt ≠ Syc , the ductile Coulomb-Mohr method is employed. For brittle behavior, the original Mohr hypothesis, constructed with tensile, compression, and torsion tests, with a curved failure locus is the best hypothesis we have. However, the difficulty of applying it without a computer leads engineers to choose modifications, namely, Coulomb Mohr, or modified Mohr. Figure 5–21 provides a summary flowchart for the selection of an effective procedure for analyzing or predicting failures from static loading for brittle or ductile behavior. Note that the maximum-normal-stress theory is excluded from Figure 5–21 as the other theories better represent the experimental data.

5–12  Introduction to Fracture Mechanics The idea that cracks exist in parts even before service begins, and that cracks can grow during service, has led to the descriptive phrase "damage-tolerant design." The focus of this philosophy is on crack growth until it becomes critical, and the part is removed from service. The analysis tool is linear elastic fracture mechanics (LEFM). Inspection and maintenance are essential in the decision to retire parts before cracks reach catastrophic size. Where human safety is concerned, periodic inspections for cracks are mandated by codes and government ordinance.

Failures Resulting from Static Loading     267

We shall now briefly examine some of the basic ideas and vocabulary needed for the potential of the approach to be appreciated. The intent here is to make the reader aware of the dangers associated with the sudden brittle fracture of so-called ductile materials. The topic is much too extensive to include in detail here and the reader is urged to read further on this complex subject.9 The use of elastic stress-concentration factors provides an indication of the average load required on a part for the onset of plastic deformation, or yielding. These factors are also useful for analysis of the loads on a part that will cause fatigue fracture. However, stress-concentration factors are limited to structures for which all dimensions are precisely known, particularly the radius of curvature in regions of high stress concentration. When there exists a crack, flaw, inclusion, or defect of unknown small radius in a part, the elastic stress-concentration factor approaches infinity as the root radius approaches zero, thus rendering the stress-concentration factor approach useless. Furthermore, even if the radius of curvature of the flaw tip is known, the high local stresses there will lead to local plastic deformation surrounded by a region of elastic deformation. Elastic stress-concentration factors are no longer valid for this situation, so analysis from the point of view of stress-concentration factors does not lead to criteria useful for design when very sharp cracks are present. By combining analysis of the gross elastic changes in a structure or part that occur as a sharp brittle crack grows with measurements of the energy required to produce new fracture surfaces, it is possible to calculate the average stress (if no crack were present) that will cause crack growth in a part. Such calculation is possible only for parts with cracks for which the elastic analysis has been completed, and for materials that crack in a relatively brittle manner and for which the fracture energy has been carefully measured. The term relatively brittle is rigorously defined in the test procedures,10 but it means, roughly, fracture without yielding occurring throughout the fractured cross section. Thus glass, hard steels, strong aluminum alloys, and even low-carbon steel below the ductile-to-brittle transition temperature can be analyzed in this way. Fortunately, ductile materials blunt sharp cracks, as we have previously discovered, so that fracture occurs at average stresses of the order of the yield strength, and the designer is prepared for this condition. The middle ground of materials that lie between "relatively brittle" and "ductile" is now being actively analyzed, but exact design criteria for these materials are not yet available. Quasi-Static Fracture Many of us have had the experience of observing brittle fracture, whether it is the breaking of a cast-iron specimen in a tensile test or the twist fracture of a piece of blackboard chalk. It happens so rapidly that we think of it as instantaneous, that is, the cross section simply parting. Fewer of us have skated on a frozen pond in the spring, with no one near us, heard a cracking noise, and stopped to observe. The noise 9

References on brittle fracture include:   H. Tada, P. C. Paris, and G. R. Irwin, The Stress Analysis of Cracks Handbook, 3rd ed., ASME Press, New York, 2000.   D. Broek, Elementary Engineering Fracture Mechanics, 4th ed., Martinus Nijhoff, London, 1985.   D. Broek, The Practical Use of Fracture Mechanics, Kluwar Academic Publishers, London, 1988.   David K. Felbeck and Anthony G. Atkins, Strength and Fracture of Engineering Solids, 2nd ed., Prentice Hall, Englewood Cliffs, N.J., 1995.  Kåre Hellan, Introduction to Fracture Mechanics, McGraw-Hill, New York, 1984. 10 BS 5447:1977 and ASTM E399-78.

268      Mechanical Engineering Design

Figure 5–22

y σ

b x a

σ

is due to cracking. The cracks move slowly enough for us to see them run. The phenomenon is not instantaneous, since some time is necessary to feed the crack energy from the stress field to the crack for propagation. Quantifying these things is important to understanding the phenomenon "in the small." In the large, a static crack may be stable and will not propagate. Some level of loading can render the crack unstable, and the crack propagates to fracture. The foundation of fracture mechanics was first established by Griffith in 1921 using the stress field calculations for an elliptical flaw in a plate developed by Inglis in 1913. For the infinite plate loaded by an applied uniaxial stress σ in Figure 5–22, the maximum stress occurs at (±a, 0) and is given by

a (σy ) max = (1 + 2 )σ b

(5–33)

Note that when a = b, the ellipse becomes a circle and Equation (5–33) gives a stressconcentration factor of 3. This agrees with the well-known result for an infinite plate with a circular hole (see Table A–15–1). For a fine crack, b∕a → 0, and Equation (5–33) predicts that (σy ) max → ∞ . However, on a microscopic level, an infinitely sharp crack is a hypothetical abstraction that is physically impossible, and when plastic deformation occurs, the stress will be finite at the crack tip. Griffith showed that the crack growth occurs when the energy release rate from applied loading is greater than the rate of energy for crack growth. Crack growth can be stable or unstable. Unstable crack growth occurs when the rate of change of the energy release rate relative to the crack length is equal to or greater than the rate of change of the crack growth rate of energy. Griffith's experimental work was restricted to brittle materials, namely glass, which pretty much confirmed his surface energy hypothesis. However, for ductile materials, the energy needed to perform plastic work at the crack tip is found to be much more crucial than surface energy. Crack Modes and the Stress Intensity Factor Three distinct modes of crack propagation exist, as shown in Figure 5–23. A tensile stress field gives rise to mode I, the opening crack propagation mode, as shown in Figure 5–23a. This mode is the most common in practice. Mode II is the sliding mode, is due to in-plane shear, and can be seen in Figure 5–23b. Mode III is the tearing mode, which arises from out-of-plane shear, as shown in Figure 5–23c. Combinations of these modes can also occur. Since mode I is the most common and important mode, the remainder of this section will consider only this mode.

Failures Resulting from Static Loading     269 y σ

dx r θ a

(a) Mode I

(b) Mode II

(c) Mode III

σ

Figure 5–23

Figure 5–24

Crack propagation modes.

Mode I crack model.

Consider a mode I crack of length 2a in the infinite plate of Figure 5–24. By using complex stress functions, it has been shown that the stress field on a dx dy element in the vicinity of the crack tip is given by

a θ θ 3θ σx = σ √ cos (1 − sin sin ) 2r 2 2 2

(5–34a)

a θ θ 3θ σy = σ √ cos (1 + sin sin ) 2r 2 2 2

(5–34b)

a θ θ 3θ τxy = σ √ sin cos cos 2r 2 2 2

(5–34c)

0 σz = { ν(σx + σy )

(for plane stress) (for plane strain)

(5–34d)

where plane stress and plane strain are defined in Sections 3–5 and 3–8, respectively. The stress σy near the tip, with θ = 0, is

a σy ∣ θ=0 = σ √ 2r

(a)

As with the elliptical crack, we see that σy∣θ=0 → ∞ as r → 0, and again the concept of an infinite stress concentration at the crack tip is inappropriate. The quantity σy∣θ=0 √2r = σ √a, however, does remain constant as r → 0. It is common practice to define a factor K called the stress intensity factor given by

K = σ √πa

(b)

where the units are MPa √m or kpsi √in. Since we are dealing with a mode I crack, Equation (b) is written as

KI = σ √πa

(5–35)

The stress intensity factor is not to be confused with the static stress-concentration factors Kt and Kts defined in Sections 3–13 and 5–2.

dy x

270      Mechanical Engineering Design

Thus Equations (5–34) can be rewritten as

σx =

σy =

τxy =

0 σz = { ν(σx + σy )

KI θ θ 3θ cos (1 − sin sin ) √2πr 2 2 2

(5–36a)

cos

θ θ 3θ 1 + sin sin ) 2( 2 2

(5–36b)

sin

θ θ 3θ cos cos 2 2 2

(5–36c)

(for plane stress) (for plane strain)

(5–36d)

KI √2πr

KI √2πr

The stress intensity factor is a function of geometry, size and shape of the crack, and the type of loading. For various load and geometric configurations, Equation (5–35) can be written as

K I = βσ √πa

(5–37)

where β is the stress intensity modification factor. Tables for β are available in the literature for basic configurations.11 Figures 5–25 to 5–30 present a few examples of β for mode I crack propagation. Fracture Toughness When the magnitude of the mode I stress intensity factor reaches a critical value, KIc, crack propagation initiates. The critical stress intensity factor KIc is a material property that depends on the material, crack mode, processing of the material, temperature, loading rate, and the state of stress at the crack site (such as plane stress versus plane strain). The critical stress intensity factor KIc is also called the fracture toughness of the material. The fracture toughness for plane strain is normally lower than that for plane stress. For this reason, the term KIc is typically defined as the mode I, plane strain fracture toughness. Fracture toughness KIc for engineering metals lies in the range 20 ≤ KIc ≤ 200 MPa · √m; for engineering polymers and ceramics, 1 ≤ KIc ≤ 5 MPa · √m. For a 4340 steel, where the yield strength due to heat treatment ranges from 800 to 1600 MPa, KIc decreases from 190 to 40 MPa · √m. Table 5–1 gives some approximate typical room-temperature values of KIc for several materials. As previously noted, the fracture toughness depends on many factors and the table is meant only to convey some typical magnitudes of KIc. For an actual application, it is recommended that the material specified for the application be 11

See, for example:   H. Tada, P. C. Paris, and G. R. Irwin, The Stress Analysis of Cracks Handbook, 3rd ed., ASME Press, New York, 2000.   G. C. Sib, Handbook of Stress Intensity Factors for Researchers and Engineers, Institute of Fracture and Solid Mechanics, Lehigh University, Bethlehem, Pa., 1973.   Y. Murakami, ed., Stress Intensity Factors Handbook, Pergamon Press, Oxford, U.K., 1987.   W. D. Pilkey, Formulas for Stress, Strain, and Structural Matrices, 2nd ed. John Wiley & Sons, New York, 2005.

Failures Resulting from Static Loading     271

Table 5–1  Values of KIc for Some Engineering Materials at Room Temperature Material

KIc, MPa √m

Sy, MPa

 2024

26

455

 7075

24

495

 7178

33

490

Aluminum

Titanium  Ti-6AL-4V

115

910

 Ti-6AL-4V

55

1035

Steel  4340

99

860

 4340

60

1515

 52100

14

2070

7.0

2.2

A A σ

σ 6.0

h

2.0

a

2a A

A

B

1.8

b

h

5.0

d

σ

2b β 4.0 β 1.6

σ

0.4 1.4

d ̸b = 1.0

B

3.0

h b̸ = 0.5

0.2 B 0.4

1.2

1.0 2.0

0.2

1.0 1.0

0

0.2

0.4 a d̸ ratio

0.6

0

0.2

0.8

Figure 5–25 Off-center crack in a plate in longitudinal tension; solid curves are for the crack tip at A; dashed curves are for the tip at B.

0.4 a ̸ b ratio

0.6

0.8

Figure 5–26 Plate loaded in longitudinal tension with a crack at the edge; for the solid curve there are no constraints to bending; the dashed curve was obtained with bending constraints added.

272      Mechanical Engineering Design 3 σ 2.0

h

a

2a M

M

2

F 1.8 a F 2

r r = 0.25 b

2b

β

h

r = 0.5 b

σ

l

l

F 2

1 r =0 b

1.6 β

0

Pure bending 1.4

0

0.2

0.4 a ̸ b ratio

0.6

0.8

Figure 5–28 Plate in tension containing a circular hole with two cracks.

l =4 h

1.2

l =2 h 3.4

1.0

a 0

0.2

0.4 a ̸ h ratio

0.6

0.8

3.0 pi

ri

Figure 5–27

ro

Beams of rectangular cross section having an edge crack.

2.6

β 2.2 4.0

σ

ri r̸ o = 0

a

a

3.0

1.8

ri r̸ o = 0.9

0.75

0.1 β

0.4

σ 2.0

0.8

ro

ri

1.0 1.0

0

0.2

0.35

1.4

0.4 a (r ̸ o – ri ) ratio

0.6

0

0.2

0.8

Figure 5–29 A cylinder loading in axial tension having a radial crack of depth a extending completely around the circumference of the cylinder.

0.4 a (r ̸ o – ri ) ratio

0.6

Figure 5–30 Cylinder subjected to internal pressure pi, having a radial crack in the longitudinal direction of depth a. Use Equation (3–50) for the tangential stress at r = r0.

0.8

Failures Resulting from Static Loading     273

certified using standard test procedures [see the American Society for Testing and Materials (ASTM) standard E399]. One of the first problems facing the designer is that of deciding whether the conditions exist, or not, for a brittle fracture. Low-temperature operation, that is, operation below room temperature, is a key indicator that brittle fracture is a possible failure mode. Tables of transition temperatures for various materials have not been published, possibly because of the wide variation in values, even for a single material. Thus, in many situations, laboratory testing may give the only clue to the possibility of a brittle fracture. Another key indicator of the possibility of fracture is the ratio of the yield strength to the ultimate strength. A high ratio of Sy∕Su indicates there is only a small ability to absorb energy in the plastic region and hence there is a likelihood of brittle fracture. The strength-to-stress ratio KIc∕KI can be used as a factor of safety as

n=

KIc KI

(5–38)

EXAMPLE 5–6 A steel ship deck plate is 30 mm thick and 12 m wide. It is loaded with a nominal uniaxial tensile stress of 50 MPa. It is operated below its ductile-to-brittle transition temperature with KIc equal to 28.3 MPa. If a 65-mm-long central transverse crack is present, estimate the tensile stress at which catastrophic failure will occur. Compare this stress with the yield strength of 240 MPa for this steel. Solution For Figure 5–25, with d = b, 2a = 65 mm and 2b = 12 m, so that d∕b = 1 and a∕d = 65∕12(103) = 0.00542. Since a∕d is so small, β = 1, so that

KI = σ √πa = 50 √π(32.5 × 10−3 ) = 16.0 MPa √m

From Equation (5–38),

n=

KIc 28.3 = = 1.77 KI 16.0

The stress at which catastrophic failure occurs is Answer

σc =

KIc 28.3 σ= (50) = 88.4 MPa KI 16.0

The yield strength is 240 MPa, and catastrophic failure occurs at 88.4∕240 = 0.37, or at 37 percent of yield. The factor of safety in this circumstance is KIc∕KI = 28.3∕16 = 1.77 and not 240∕50 = 4.8.

EXAMPLE 5–7 A plate of width 1.4 m and length 2.8 m is required to support a tensile force in the 2.8-m direction of 4.0 MN. Inspection procedures will detect only through-thickness edge cracks larger than 2.7 mm. The two Ti-6AL-4V alloys in Table 5–1 are being considered for this application, for which the safety factor must be 1.3 and minimum weight is important. Which alloy should be used?

274      Mechanical Engineering Design

Solution (a) We elect first to estimate the thickness required to resist yielding. Because σ = P∕wt, we have t = P∕wσ. For the weaker alloy, we have, from Table 5–1, Sy = 910 MPa. Thus,

σall =

Sy 910 = = 700 MPa n 1.3

Thus,

t=

4.0(10) 3 P = = 4.08 mm or greater wσall 1.4(700)

For the stronger alloy, we have, from Table 5–1,

σall =

1035 = 796 MPa 1.3

and so the thickness is t=

Answer

4.0(10) 3 P = = 3.59 mm or greater wσall 1.4(796)

(b) Now let us find the thickness required to prevent crack growth. Using Figure 5–26, we have h 2.8∕2 = =1 b 1.4

a 2.7 = = 0.001 93 b 1.4(103 )

Corresponding to these ratios we find from Figure 5–26 that β ≈ 1.1, and KI = 1.1σ √πa. n=

KIc 115 √103 = , KI 1.1σ √πa

σ=

KIc 1.1n √πa

From Table 5–1, KIc = 115 MPa √m for the weaker of the two alloys. Solving for σ with n = 1 gives the fracture stress

σ=

115 1.1 √π(2.7 × 10−3 )

= 1135 MPa

which is greater than the yield strength of 910 MPa, and so yield strength is the basis for the geometry decision. For the stronger alloy Sy = 1035 MPa, with n = 1 the fracture stress is

σ=

KIc 55 = = 542.9 MPa nKI 1(1.1) √π(2.7 × 10−3 )

which is less than the yield strength of 1035 MPa. The thickness t is t=

4.0(103 ) P = = 6.84 mm or greater wσall 1.4(542.9∕1.3)

This example shows that the fracture toughness KIc limits the geometry when the stronger alloy is used, and so a thickness of 6.84 mm or larger is required. When the weaker alloy is used the geometry is limited by the yield strength, giving a thickness of only 4.08 mm or greater. Thus the weaker alloy leads to a thinner and lighter weight choice since the failure modes differ.

Failures Resulting from Static Loading     275

5–13  Important Design Equations The following equations are provided as a summary. Note for plane stress: The principal stresses in the following equations that are labeled σA and σB represent the principal stresses determined from the two-dimensional Equation (3–13). Maximum Shear Theory σ1 − σ3 Sy = 2 2n

(5–3)

(σ1 − σ2 ) 2 + (σ2 − σ3 ) 2 + (σ3 − σ1 ) 2 1∕2 ] 2

(5–12)

τmax =

Distortion-Energy Theory Von Mises stress σ′ = [

σ′ =

1 √2

[(σx − σy ) 2 + (σy − σz ) 2 + (σz − σx ) 2 + 6(τ2xy + τ2yz + τ2zx )]1∕2 (5–14)

Plane stress σ′ = (σ2A − σAσB + σ2B ) 1∕2

σ′ =

(σ2x

− σx σy +

σ2y

+

(5–13)

3τ2xy ) 1∕2

(5–15)

Yield design equation

σ′ =

Sy n

(5–19)

Shear yield strength

Ssy = 0.577 Sy

(5–21)

σ1 σ3 1 − = St Sc n

(5–26)

Coulomb-Mohr Theory

where St is tensile yield (ductile) or ultimate tensile (brittle), and St is compressive yield (ductile) or ultimate compressive (brittle) strengths. Modified Mohr (Plane Stress) Sut σA = σA ≥ σB ≥ 0 n σA ≥ 0 ≥ σB and

(Suc − Sut )σA σB 1 − = Suc Sut Suc n

σB = −

Suc n

(5–32a)

σB ≤1 ⎹ σ⎹ A

σA ≥ 0 ≥ σB and 0 ≥ σA ≥ σB

σB

⎹ σ⎹ A

> 1

(5–32b) (5–32c)

276      Mechanical Engineering Design

Failure Theory Flowchart Figure 5–21 Brittle behavior

< 0.05

No

Conservative?

Mod. Mohr (MM) Eq. (5–32)

Ductile behavior

≤ 0.05

εf

Yes

No

Yes

Syt ≈ Syc?

Brittle Coulomb-Mohr Ductile Coulomb-Mohr (BCM) (DCM) Eq. (5–31) Eq. (5–26)

No

Conservative?

Distortion-energy (DE) Eqs. (5–15) and (5–19)

Fracture Mechanics

KI = βσ √πa

Yes

Maximum shear stress (MSS) Eq. (5–3)

(5–37)

where β is found in Figures 5–25 to 5–30

n=

where KIc is found in Table 5–1

KIc KI

(5–38)

PROBLEMS Problems marked with an asterisk (*) are linked to problems in other chapters, as summarized in Table 1–2 of Section 1–17.

5–1

A ductile hot-rolled steel bar has a minimum yield strength in tension and compression of 350 MPa. Using the distortion-energy and maximum-shear-stress theories determine the factors of safety for the following plane stress states: (a) σx = 100 MPa, σy = 100 MPa (b) σx = 100 MPa, σy = 50 MPa (c) σx = 100 MPa, τxy = −75 MPa (d) σx = −50 MPa, σy = −75 MPa, τxy = −50 MPa (e) σx = 100 MPa, σy = 20 MPa, τxy = −20 MPa

5–2

Repeat Problem 5–1 with the following principal stresses obtained from Equation (3–13): (a) σA = 100 MPa, σB = 100 MPa (b) σA = 100 MPa, σB = −100 MPa (c) σA = 100 MPa, σB = 50 MPa (d) σA = 100 MPa, σB = −50 MPa (e) σA = −50 MPa, σB = −100 MPa

Failures Resulting from Static Loading     277

5–3

Repeat Problem 5–1 for a bar of AISI 1030 hot-rolled steel and: (a) σx = 25 kpsi, σy = 15 kpsi (b) σx = 15 kpsi, σy = −15 kpsi (c) σx = 20 kpsi, τxy = −10 kpsi (d) σx = −12 kpsi, σy = 15 kpsi, τxy = −9 kpsi (e) σx = −24 kpsi, σy = −24 kpsi, τxy = −15 kpsi

5–4

Repeat Problem 5–1 for a bar of AISI 1015 cold-drawn steel with the following principal stresses obtained from Equation (3–13): (a) σA = 30 kpsi, σB = 30 kpsi (b) σA = 30 kpsi, σB = −30 kpsi (c) σA = 30 kpsi, σB = 15 kpsi (d) σA = −30 kpsi, σB = −15 kpsi (e) σA = −50 kpsi, σB = 10 kpsi

5–5

Repeat Problem 5–1 by first plotting the failure loci in the σA, σB plane to scale; then, for each stress state, plot the load line and by graphical measurement estimate the factors of safety.

5–6

Repeat Problem 5–3 by first plotting the failure loci in the σA, σB plane to scale; then, for each stress state, plot the load line and by graphical measurement estimate the factors of safety.

5–7 An AISI 1018 steel has a yield strength, Sy = 295 MPa. Using the distortion-energy to theory for the given state of plane stress, (a) determine the factor of safety, (b) plot 5–11 the failure locus, the load line, and estimate the factor of safety by graphical measurement. Problem Number

σx (MPa)

σy (MPa)

τxy (MPa)

5–7

75

−35

0

5–8

−100

30

0

5–9

100

0

−25

5–10

−30

−65

40

5–11

−80

30

−10

5–12 A ductile material has the properties Syt = 60 kpsi and Syc = 75 kpsi. Using the duc-

tile Coulomb-Mohr theory, determine the factor of safety for the states of plane stress given in Problem 5–3.

5–13 Repeat Problem 5–12 by first plotting the failure loci in the σA, σB plane to scale; then for each stress state, plot the load line and by graphical measurement estimate the factor of safety.

5–14 An AISI 4142 steel Q&T at 800°F exhibits Syt = 235 kpsi, Syc = 285 kpsi, and εf = 0.07. to For the given state of plane stress, (a) determine the factor of safety, (b) plot the failure 5–18 locus and the load line, and estimate the factor of safety by graphical measurement. Problem Number

σx (kpsi)

σy (kpsi)

τxy (kpsi)

5–14

150

−50

5–15

−150

50

0 0

5–16

125

0

−75

5–17

−80

−125

50

5–18

125

80

−75

278      Mechanical Engineering Design

5–19 A brittle material has the properties Sut = 30 kpsi and Suc = 90 kpsi. Using the brittle

Coulomb-Mohr and modified-Mohr theories, determine the factor of safety for the following states of plane stress. (a) σx = 25 kpsi, σy = 15 kpsi (b) σx = 15 kpsi, σy = −15 kpsi (c) σx = 20 kpsi, τxy = −10 kpsi (d) σx = −15 kpsi, σy = 10 kpsi, τxy = −15 kpsi (e) σx = −20 kpsi, σy = −20 kpsi, τxy = −15 kpsi

5–20 Repeat Problem 5–19 by first plotting the failure loci in the σA, σB plane to scale; then

for each stress state, plot the load line and by graphical measurement estimate the factor of safety.

5–21 For an ASTM 30 cast iron, (a) find the factors of safety using the BCM and MM to theories, (b) plot the failure diagrams in the σA, σB plane to scale and locate the coordinates 5–25 of the stress state, and (c) estimate the factors of safety from the two theories by graphical measurements along the load line. Problem Number

σx (kpsi)

σy (kpsi)

τxy (kpsi)

5–21

15

10

0

5–22

15

−50

0

5–23

15

0

−10

5–24

−10

−25

−10

5–25

−35

13

−10

5–26 A cast aluminum 195-T6 exhibits Sut = 36 kpsi, Suc = 35 kpsi, and εf = 0.045. For to the given state of plane stress, (a) using the Coulomb-Mohr theory, determine the 5–30 factor of safety, (b) plot the failure locus and the load line, and estimate the factor of safety by graphical measurement. Problem Number

σx (kpsi)

σy (kpsi)

τxy (kpsi)

5–26

15

−10

0

5–27

−15

10

0

5–28

12

0

−8

5–29

−10

−15

10

5–30

15

8

−8

5–31 Repeat Problems 5–26 to 5–30 using the modified-Mohr theory. to Problem number 5–31 5–32 5–33 5–34 5–35 5–35 Repeat problem 5–26 5–27 5–28 5–29 5–30

5–36 This problem illustrates that the factor of safety for a machine element depends on the

particular point selected for analysis. Here you are to compute factors of safety, based upon the distortion-energy theory, for stress elements at A and B of the member shown in the figure. This bar is made of AISI 1006 cold-drawn steel and is loaded by the forces F = 0.55 kN, P = 4.0 kN, and T = 25 N · m.

Failures Resulting from Static Loading     279 y

10 0m

m

A B Problem 5–36 z

F 15-mm dia.

T

P x

5–37 For the beam in Problem 3–45 determine the minimum yield strength that should be

considered to obtain a minimum factor of safety of 2 based on the distortion-energy theory.

5–38 A 1020 CD steel shaft is to transmit 20 hp while rotating at 1750 rpm. Determine the minimum diameter for the shaft to provide a minimum factor of safety of 3 based on the maximum-shear-stress theory.

5–39 A 30-mm-diameter shaft, made of AISI 1018 HR steel, transmits 10 kW of power while rotating at 200 rev/min. Assume any bending moments present in the shaft to be negligibly small compared to the torque. Determine the static factor of safety based on (a) the maximum-shear-stress failure theory. (b) the distortion-energy failure theory.

5–40 A 20-mm-diameter steel shaft, made of AISI 1035 HR steel, transmits power while rotating at 400 rev/min. Assume any bending moments in the shaft to be relatively small compared to the torque. Determine how much power, in units of kW, the shaft can transmit with a static factor of safety of 1.5 based on (a) the maximum-shear-stress theory. (b) distortion-energy theory.

5–41 The short cantilevered peg in Problem 3–47 is made from AISI 1040 cold-drawn steel. Evaluate the state of stress at points A, B, and C, determine the critical case, and evaluate the minimum factor of safety based on (a) the maximum-shear-stress theory. (b) the distortion-energy theory.

5–42 The shaft ABD in Problem 3–53 is made from AISI 1040 CD steel. Based on failure at the wall at A, determine the factor of safety using (a) the maximum-shear-stress theory. (b) the distortion-energy theory.

5–43 The shaft ABD in Problem 3–54 is made from AISI 1020 CD steel. Based on failure at the wall at A, determine the factor of safety using (a) the maximum-shear-stress theory. (b) the distortion-energy theory.

5–44 The shaft AB in Problem 3–55 is made from AISI 1035 CD steel. Based on failure at the wall at A, determine the factor of safety using (a) the maximum-shear-stress theory. (b) the distortion-energy theory.

280      Mechanical Engineering Design

5–45 The input shaft, AB, of the gear reducer in Problem 3–101 is made of AISI 1040 CD steel. Determine the factor of safety of the shaft using (a) the maximum-shear-stress theory. (b) the distortion-energy theory.

5–46 The output shaft, CD, of the gear reducer in Problem 3–102 is made of AISI 1040 CD steel. Determine the factor of safety of the shaft using (a) the maximum-shear-stress theory. (b) the distortion-energy theory.

5–47 A round shaft supports a transverse load of F = 15  000 lbf and carries a torque of

T = 7 kip · in, as shown in the figure. The shaft does not rotate. The shaft is machined from AISI 4140 steel, quenched and tempered at 400°F. Document the location of the critical stress element and how you determined that element to be critical. Do not assume transverse shear stress is negligible without proving it. Then, (a) determine the minimum factor of safety based on yielding according to the maximumshear-stress theory. (b) determine the minimum factor of safety based on yielding according to the distortion-energy theory. 1.1-in dia. 1.5-in dia.

A

F

B

T

1.3-in dia.

RA

Problem 5–47

C D

1 in

T

1 in RD

1 in

5–48 A 1-in, constant diameter shaft, is loaded with forces at A and B as shown, with ground

reaction forces at O and C. The shaft also transmits a torque of 1500 lbf · in throughout the length of the shaft. The shaft has a tensile yield strength of 50 kpsi. Determine the minimum static factor of safety using (a) the maximum-shear-stress failure theory. (b) the distortion-energy failure theory. y

460 lbf 12 in

Problem 5–48

O

1500 lbf-in

18 in A

RO

575 lbf

B

C

x

10 in RC

5–49 Cantilevered rod OA is 0.5 m long, and made from AISI 1010 hot-rolled steel. A constant force and torque are applied as shown. Determine the minimum diameter, d, for the rod that will achieve a minimum static factor of safety of 2 (a) using the maximum-shear-stress failure theory. (b) using the distortion-energy failure theory.

Failures Resulting from Static Loading     281 y

150 N

Problem 5–49 O

A x 25 N-m

z

5–50* For the problem specified in the table, build upon the results of the original problem to to determine the minimum factor of safety for yielding. Use both the maximum-shear5–66* stress theory and the distortion-energy theory, and compare the results. The material is 1018 CD steel.

Problem Number

Original Problem Number

5–50* 5–51* 5–52* 5–53* 5–54* 5–55* 5–56* 5–57* 5–58* 5–59* 5–60* 5–61* 5–62* 5–63* 5–64* 5–65* 5–66*

3–79 3–80 3–81 3–82 3–83 3–84 3–85 3–87 3–88 3–90 3–91 3–92 3–93 3–94 3–95 3–96 3–97

5–67* Build upon the results of Problems 3–95 and 3–98 to compare the use of a low-strength,

ductile material (1018 CD) in which the stress-concentration factor can be ignored to a high-strength but more brittle material (4140 Q&T @ 400°F) in which the stressconcentration factor should be included. For each case, determine the factor of safety for yielding using the distortion-energy theory.

5–68 Using F = 416 lbf, design the lever arm CD of Figure 5–16 by specifying a suitable size and material.

5–69 A spherical pressure vessel is formed of 16-gauge (0.0625-in) cold-drawn AISI 1020 sheet steel. If the vessel has a diameter of 15 in, use the distortion-energy theory to estimate the pressure necessary to initiate yielding. What is the estimated bursting pressure?

5–70 An AISI 1040 cold-drawn steel tube has an outside diameter of 50 mm and an inside

diameter of 42 mm. The tube is 150 mm long and is capped at both ends. Determine the maximum allowable internal pressure for a static factor of safety of 2 for the tube walls (a) based on the maximum-shear-stress theory. (b) based on the distortion-energy theory.

282      Mechanical Engineering Design

5–71 An AISI 1040 cold-drawn steel tube has an outside diameter of 50 mm and an inside diameter of 42 mm. The tube is 150 mm long, and is capped on both ends. An internal pressure of 40 MPa is applied. Determine the static factor of safety using (a) the maximum-shear-stress theory. (b) the distortion-energy theory.

5–72 This problem illustrates that the strength of a machine part can sometimes be measured

in units other than those of force or moment. For example, the maximum speed that a flywheel can reach without yielding or fracturing is a measure of its strength. In this problem you have a rotating ring made of hot-forged AISI 1020 steel; the ring has a 6-in inside diameter and a 10-in outside diameter and is 0.5 in thick. Using the distortionenergy theory, determine the speed in revolutions per minute that would cause the ring to yield. At what radius would yielding begin? [Note: The maximum radial stress occurs at r = (ro ri ) 1∕2 ; see Equation (3–55).]

5–73 A light pressure vessel is made of 2024-T3 aluminum alloy tubing with suitable end closures. This cylinder has a 321 -in OD, a 0.065-in wall thickness, and ν = 0.334. The purchase order specifies a minimum yield strength of 46 kpsi. Using the ­distortion-energy theory, determine the factor of safety if the pressure-release valve is set at 500 psi.

5–74 A cold-drawn AISI 1015 steel tube is 300 mm OD by 200 mm ID and is to be subjected

to an external pressure caused by a shrink fit. Using the distortion-energy theory, determine the maximum pressure that would cause the material of the tube to yield.

5–75 What speed would cause fracture of the ring of Problem 5–72 if it were made of grade 30 cast iron?

5–76 The figure shows a shaft mounted in bearings at A and D and having pulleys at B and C.

The forces shown acting on the pulley surfaces represent the belt tensions. The shaft is to be made of AISI 1035 CD steel. Using a conservative failure theory with a design factor of 2, determine the minimum shaft diameter to avoid yielding.

300 lbf

50 lbf

y

Problem 5–76 8-in dia.

z A

x

6-in dia.

B

59 lbf 392 lbf D C 6 in

8 in

8 in

5–77 By modern standards, the shaft design of Problem 5–76 is poor because it is so long. Suppose it is redesigned by halving the length dimensions. Using the same material and design factor as in Problem 5–76, find the new shaft diameter.

5–78* Build upon the results of Problem 3–41 to determine the factor of safety for yielding based on the distortion-energy theory for each of the simplified models in parts c, d, and e of the figure for Problem 3–41. The pin is machined from AISI 1018 hot-rolled steel. Compare the three models from a designer's perspective in terms of accuracy, safety, and modeling time.

Failures Resulting from Static Loading     283

5–79* For the clevis pin of Problem 3–41, redesign the pin diameter to provide a factor of safety of 2.5 based on a conservative yielding failure theory, and the most conservative loading model from parts c, d, and e of the figure for Problem 3–41. The pin is machined from AISI 1018 hot-rolled steel.

5–80 A split-ring clamp-type shaft collar is shown in the figure. The collar is 50 mm OD

by 25 mm ID by 12 mm wide. The screw is designated as M 6 × 1. The relation between the screw tightening torque T, the nominal screw diameter d, and the tension in the screw Fi is approximately T = 0.2 Fid. The shaft is sized to obtain a close running fit. Find the axial holding force Fx of the collar as a function of the coefficient of friction and the screw torque.

5–81 Suppose the collar of Problem 5–80 is tightened by using a screw torque of 20 N · m.

The collar material is AISI 1035 steel heat-treated to a minimum tensile yield strength of 450 MPa. (a) Estimate the tension in the screw. (b) By relating the tangential stress to the hoop tension, find the internal pressure of the shaft on the ring. (c) Find the tangential and radial stresses in the ring at the inner surface. (d) Determine the maximum shear stress and the von Mises stress. (e) What are the factors of safety based on the maximum-shear-stress and the distortionenergy theories?

5–82 In Problem 5–80, the role of the screw was to induce the hoop tension that produces

the clamping. The screw should be placed so that no moment is induced in the ring. Just where should the screw be located?

5–83 A tube has another tube shrunk over it. The specifications are: Inner Member

Outer Member

ID

1.250 ± 0.003 in

2.001 ± 0.0004 in

OD

2.002 ± 0.0004 in

3.000 ± 0.004 in

Both tubes are made of a plain carbon steel. (a) Find the nominal shrink-fit pressure and the von Mises stresses at the fit surface. (b) If the inner tube is changed to solid shafting with the same outside dimensions, find the nominal shrink-fit pressure and the von Mises stresses at the fit surface.

5–84 Two steel tubes have the specifications: Inner Tube

Outer Tube

ID

20 ± 0.050 mm

39.98 ± 0.008 mm

OD

40 ± 0.008 mm

65 ± 0.10 mm

These are shrink-fitted together. Find the nominal shrink-fit pressure and the von Mises stress in each body at the fit surface.

5–85 Repeat Problem 5–84 for maximum shrink-fit conditions. 5–86 A solid steel shaft has a gear with ASTM grade 20 cast-iron hub (E = 14.5 Mpsi) shrink-

fitted to it. The shaft diameter is 2.001 ± 0.0004 in. The specifications for the gear hub are 2.000

+0.0004 in −0.0000

A

Problem 5–80

284      Mechanical Engineering Design

ID with an OD of 4.00 ± 321 in. Using the midrange values and the modified Mohr theory, estimate the factor of safety guarding against fracture in the gear hub due to the shrink fit.

5–87 Two steel tubes are shrink-fitted together where the nominal diameters are 40, 45, and

50 mm. Careful measurement before fitting determined the diametral interference between the tubes to be 0.062 mm. After the fit, the assembly is subjected to a torque of 900 N · m and a bending-moment of 675 N · m. Assuming no slipping between the cylinders, analyze the outer cylinder at the inner and outer radius. Determine the factor of safety using distortion energy with Sy = 415 MPa.

5–88 Repeat Problem 5–87 for the inner tube. 5–89 For the problem given in the table, the specifications for the press fit of two cylinders to are given in the original problem from Chapter 3. If both cylinders are hot-rolled AISI 5–94 1040 steel, determine the minimum factor of safety for the outer cylinder based on the distortion-energy theory. Problem Number

Original Problem Number 3–124 3–125 3–126 3–127 3–128 3–129

5–89 5–90 5–91 5–92 5–93 5–94

5–95 For Equations (5–36) show that the principal stresses are given by

σ1 =

σ2 =

KI √2πr

KI √2πr

cos

θ θ 1 + sin ) 2( 2

cos

θ θ 1 − sin 2( 2)

0 σ3 =  θ  2  √ νKI cos 2  πr

(plane stress) (plane strain)

5–96 Use the results of Problem 5–95 for plane strain near the tip with θ = 0 and ν = 13 . If the yield strength of the plate is Sy, what is σ1 when yield occurs? (a) Use the distortion-energy theory. (b) Use the maximum-shear-stress theory. Using Mohr's circles, explain your answer.

5–97 A plate 100 mm wide, 200 mm long, and 12 mm thick is loaded in tension in the

direction of the length. The plate contains a crack as shown in Figure 5–26 with the crack length of 16 mm. The material is steel with KIc = 80 MPa · √m, and Sy = 950 MPa. Determine the maximum possible load that can be applied before the plate (a) yields, and (b) has uncontrollable crack growth.

5–98 A cylinder subjected to internal pressure pi has an outer diameter of 14 in and a 1-in

wall thickness. For the cylinder material, KIc = 72 kpsi · √in, Sy = 170 kpsi, and Sut = 192 kpsi. If the cylinder contains a radial crack in the longitudinal direction of depth 0.5 in determine the pressure that will cause uncontrollable crack growth.

6

Fatigue Failure Resulting from Variable Loading

From ASM Handbook, Vol. 12: Fractography, 2nd printing, 1992, ASM International, Materials Park, OH 44073-0002, fig. 50, p. 120. Reprinted by permission of ASM International®, www.asminternational.org.

Chapter Outline 6–1

Introduction to Fatigue   286

6–2

Chapter Overview   287

6–3  Crack Nucleation and Propagation   288 6–4

Fatigue-Life Methods   294

6–5  The Linear-Elastic Fracture Mechanics Method  295 6–6

The Strain-Life Method   299

6–7  The Stress-Life Method and the S-N Diagram  302 6–8  The Idealized S-N Diagram for Steels  304 6–9

Endurance Limit Modifying Factors   309

6–10  Stress Concentration and Notch Sensitivity  320 6–11  Characterizing Fluctuating Stresses   325 6–12

The Fluctuating-Stress Diagram   327

6–13

Fatigue Failure Criteria   333

6–14

Constant-Life Curves   342

6–15 Fatigue Failure Criterion for Brittle Materials  345 6–16

Combinations of Loading Modes   347

6–17

Cumulative Fatigue Damage   351

6–18

Surface Fatigue Strength   356

6–19

 oad Maps and Important Design Equations R for the Stress-Life Method   359

285

286      Mechanical Engineering Design

In Chapter 5 we considered the analysis and design of parts subjected to static loading. In this chapter we shall examine the failure mechanisms of parts subjected to fluctuating loading conditions.

6–1  Introduction to Fatigue Prior to the nineteenth century, engineering design was based primarily on static loading. Speeds were relatively slow, loads were light, and factors of safety were large. About the time of the development of the steam engine, this changed. Shafts were expected to run faster, accumulating high repetitions of dynamically cycling stresses. An example is a particular stress element on the surface of a rotating shaft subjected to the action of bending loads that cycles between tension and compression with each revolution of the shaft. In the days of waterwheel-powered machinery, a shaft might experience only a few thousand cycles in a day. The steam engine could rotate at several hundred revolutions per minute, accumulating hundreds of thousands of cycles in a single day. As material manufacturing processes improved, steels with higher strengths and greater uniformity became available, leading to the confidence to design for higher stresses and lower static factors of safety. Increasingly, machine members failed suddenly and often dramatically, under the action of fluctuating stresses; yet a careful analysis revealed that the maximum stresses were well below the yield strength of the material. The most distinguishing characteristic of these failures was that the stresses had been repeated a large number of times. This led to the notion that the part had simply become "tired" from the repeated cycling, hence the origin of the term fatigue failure. However, posttesting of failed parts indicated that the material still had its original properties. In actuality, fatigue failure is not a consequence of changed material properties but from a crack initiating and growing when subjected to many repeated cycles of stress. Some of the first notable fatigue failures involved railroad axles in the mid-1800s and led to investigations by Albert Wöhler, who is credited with deliberately studying and articulating some of the basic principles of fatigue failure, as well as design strategies to avoid fatigue failure. The next 150 years saw a steady stream of research to both understand the underlying mechanisms of fatigue and to develop theoretical modeling techniques for practical design use. There have been times in which some postulated theories gained traction, only to later be discarded as false paths. The understanding of fatigue failure has been somewhat elusive, perhaps to be expected due to the fact that the failures are a result of many factors combining, seemingly with a high level of randomness included. With persistence, the field of study has stabilized. With the advancement of technology to examine microstructures at extreme magnification, much of the physical mechanics of the problem has been well established. More recent work in the last few decades has focused on the theoretical modeling to incorporate both the mechanics as well as the empirical evidence. Historically, even as progress was made, dramatic failures continued to prove the need for continued study. A few notable examples that can be readily researched for details include the following: Versailles railroad axle (1842), Liberty ships (1943), multiple de Havilland Comet crashes (1954), Kielland oil platform collapse (1980), Aloha B737 accident (1988), DC10 Sioux City accident (1989), MD-88 Pensacola engine failure (1996), Eschede railway accident (1998), GE CF6 engine failure (2016).

Fatigue Failure Resulting from Variable Loading     287

6–2  Chapter Overview In this chapter, we will present the concepts pertinent to understanding the mechanisms of fatigue failure, as well as a practical approach to estimating fatigue life. The study of fatigue includes the overlaying of several complexities. As an aid to organizing these complexities, a brief outline of the chapter is given here. I. Crack Nucleation and Propagation (Section 6–3) This section describes the stages of crack nucleation, propagation, and fracture. The description starts with macroscale observations, then moves to examining the reaction of the material at a microscopic level to cyclical strain loading. It provides the conceptual background to understanding why and how fatigue fracture happens. II. Fatigue-Life Methods (Section 6–4) This section describes and contrasts the three major approaches to predicting the fatigue life of a part, namely, the linear-elastic fracture mechanics method, the strain-life method, and the stress-life method. The purpose of this section is to briefly introduce these topics to provide a big picture of the methods to be described in the following sections. III. The Linear-Elastic Fracture Mechanics Method (Section 6–5) The linear-elastic fracture mechanics method assumes a crack exists and uses principles of fracture mechanics to model the growth of the crack. This section is useful for a general understanding of crack growth. It is intended as an overview and is not essential for understanding the remainder of the chapter. IV. The Strain-Life Method (Section 6–6) The strain-life method uses strain testing to analyze the effects of local yielding at notches. It is particularly useful for situations with high stress and plastic strain, and therefore low life expectancy. Local plastic strain is the driving mechanism for fatigue failures, so a general understanding of the concepts in this section is worthwhile. The method is complex enough that this section does not attempt to cover all of the details. It is intended as an overview and is not essential for understanding the remainder of the chapter. V. The Stress-Life Method in Detail (Sections 6–7 through 6–17) The stress-life method is fully developed in the remainder of the chapter as a practical approach to estimating the fatigue life of a part. The coverage is organized into three categories of loading: completely reversed, fluctuating, and combined modes. A. Completely Reversed Loading (Sections 6–7 through 6–10) Completely reversed loading is the simplest type of stress cycling between equal amplitudes of tension and compression. The majority of fatigue experimentation uses this type of loading, so the fundamentals are first addressed with this loading. These sections represent the heart of the stress-life method. i. The Stress-Life Method and the S-N Diagram (Section 6–7) ii. The Idealized S-N Diagram for Steels (Section 6–8) iii. Endurance Limit Modifying Factors (Section 6–9) iv. Stress Concentration and Notch Sensitivity (Section 6–10)

288      Mechanical Engineering Design

B. Fluctuating Loading (Sections 6–11 through 6–15) Generalized fluctuating stress levels are introduced to model the effect of nonzero mean stress on the fatigue life. Several well-known models are introduced to represent the experimental data. i. Characterizing Fluctuating Stresses (Section 6–11) ii. The Fluctuating-Stress Diagram (Section 6–12) iii. Fatigue Failure Criteria (Section 6–13) iv. Constant-Life Curves (Section 6–14) v. Fatigue Failure Criterion for Brittle Materials (Section 6–15) C. Combinations of Loading Modes (Section 6–16) A procedure based on the distortion-energy theory is presented for analyzing combined fluctuating stress states, such as combined bending and torsion. D. Cumulative Fatigue Damage (Section 6–17) The fluctuating stress levels on a machine part may be time-varying. Methods are provided to assess the fatigue damage on a cumulative basis. IV. Surface Fatigue (Section 6–18) Surface fatigue is discussed as an important special case when contact stress is cyclic. It is useful background for the study of fatigue failure in gears and bearings. This section is not essential for a first approach to the fatigue methods. V. Road Maps and Important Design Equations for the Stress-Life Method (Section  6–19) The important equations and procedures for the stress-life method are summarized in a concise manner.

6–3  Crack Nucleation and Propagation Fatigue failure is due to crack nucleation and propagation. A fatigue crack will initiate at a location that experiences repeated applications of locally high stress (and thus high strain). This is often at a discontinuity in the material, such as any of the following: ∙ Geometric changes in cross section, keyways, holes, and the like where stress concentrations occur, as discussed in Sections 3–13 and 5–2. ∙ Manufacturing imperfections such as stamp marks, tool marks, scratches, and burrs; poor joint design; improper assembly; and other fabrication faults. ∙ Composition of the material itself as processed by rolling, forging, casting, extrusion, drawing, heat treatment, and more. Microscopic and submicroscopic surface and subsurface discontinuities arise, such as inclusions of foreign material, alloy segregation, voids, hard precipitated particles, and crystal discontinuities. A fatigue failure is characterized by three stages. Stage I is the initiation of one or more microcracks due to cyclic plastic deformation followed by crystallographic propagation extending from three to ten grains from the origin. Stage I cracks are not normally discernible to the naked eye. Stage II progresses from microcracks to macrocracks forming parallel plateau-like fracture surfaces separated by longitudinal ridges. The plateaus are generally smooth and normal to the direction of maximum tensile stress. These surfaces are often characterized by dark and light bands referred to as beach marks, as seen in Figure 6–1. During cyclic loading, these cracked

Fatigue Failure Resulting from Variable Loading     289

Figure 6–1 Fatigue failure of a bolt due to repeated unidirectional bending. The failure started at the thread root at A, propagated across most of the cross section shown by the beach marks at B, before final fast fracture at C. (From ASM Handbook, Vol. 12: Fractography, 2nd printing, 1992, ASM International, Materials Park, OH 44073-0002, fig. 50, p. 120. Reprinted by permission of ASM International®, www.asminternational.org.)

Figure 6–2 Schematics of fatigue fracture surfaces produced in smooth and notched components with round cross sections under various loading conditions and nominal stress levels. (From ASM Metals Handbook, Vol. 11: Failure Analysis and Prevention, 1986, ASM International, Materials Park, OH 44073-0002, fig. 18, p. 111. Reprinted by permission of ASM International®, www.asminternational.org.)

surfaces open and close, rubbing together, and the beach mark appearance depends on the changes in the level or frequency of loading and the corrosive nature of the environment. Stage III occurs during the final stress cycle when the material of the remaining cross section cannot support the loads, resulting in a sudden, fast fracture. A stage III fracture can be brittle, ductile, or a combination of both. There is a good deal that can be observed visually from the fracture patterns of a fatigue failure. Quite often the beach marks, if they exist, and possible patterns in the stage III fracture called chevron lines, point toward the origins of the initial cracks. Figure 6–2 shows representations of typical failure surfaces of round cross sections under differing load conditions and levels of stress concentration. Note that the level of stress is evident from the amount of area for the stage III final fracture. The sharper stress concentrations tend to advance the stage II crack growth into a more aggressive and prominent leading edge. Figures 6–3 to 6–8 show a few examples of fatigue failure surfaces. Several of these examples are from the ASM Metals Handbook, which is a 21-volume major reference source in the study of fatigue failure. Crack Nucleation Crack nucleation occurs in the presence of localized plastic strain. Plastic strain involves the breaking of a limited number of atomic bonds by the movement of

290      Mechanical Engineering Design

Figure 6–3 Fatigue fracture of an AISI 4320 drive shaft. The fatigue failure initiated at the end of the keyway at points B and progressed to final rupture at C. The final rupture zone is small, indicating that loads were low. (From ASM Handbook, Vol. 12: Fractography, 2nd printing, 1992, ASM International, Materials Park, OH 44073-0002, fig. 51, p. 120. Reprinted by permission of ASM International®, www.asminternational.org.)

Figure 6–5 Fatigue fracture surface of a forged connecting rod of AISI 8640 steel. The fatigue crack origin is at the left edge, at the flash line of the forging, but no unusual roughness of the flash trim was indicated. The fatigue crack progressed halfway around the oil hole at the left, indicated by the beach marks, before final fast fracture occurred. Note the pronounced shear lip in the final fracture at the right edge. (From ASM Handbook, Vol. 12: Fractography, 2nd printing, 1992, ASM International, Materials Park, OH 44073-0002, fig. 523, p. 332. Reprinted by permission of ASM International®, www.asminternational.org.)

Figure 6–4 Fatigue fracture surface of an AISI 8640 pin. Sharp corners of the mismatched grease holes provided stress concentrations that initiated two fatigue cracks indicated by the arrows. (From ASM Handbook, Vol. 12: Fractography, 2nd printing, 1992, ASM International, Materials Park, OH 44073-0002, fig. 520, p. 331. Reprinted by permission of ASM International®, www.asminternational.org.)

Fatigue Failure Resulting from Variable Loading     291

Figure 6–6 Fatigue fracture surface of a 200-mm (8-in) diameter piston rod of an alloy steel steam hammer used for forging. This is an example of a fatigue fracture caused by pure tension where surface stress concentrations are absent and a crack may initiate anywhere in the cross section. In this instance, the initial crack formed at a forging flake slightly below center, grew outward symmetrically, and ultimately produced a brittle fracture without warning. (From ASM Handbook, Vol. 12: Fractography, 2nd printing, 1992, ASM International, Materials Park, OH 44073-0002, fig. 570, p. 342. Reprinted by permission of ASM International®, www.asminternational.org.)

Medium-carbon steel (ASTM A186) 30 dia

Web

Fracture Fracture Tread

Flange (1 of 2)

(a) Coke-oven-car wheel

Figure 6–7 Fatigue failure of an ASTM A186 steel double-flange trailer wheel caused by stamp marks. (a) Coke-oven-car wheel showing position of stamp marks and fractures in the rib and web. (b) Stamp mark showing heavy impression and fracture extending along the base of the lower row of numbers. (c) Notches, indicated by arrows, created from the heavily indented stamp marks from which cracks initiated along the top at the fracture surface. (From ASM Metals Handbook, Vol. 11: Failure Analysis and Prevention, 1986, ASM International, Materials Park, OH 44073-0002, fig. 51, p. 130. Reprinted by permission of ASM International®, www.asminternational.org.)

292      Mechanical Engineering Design

Figure 6–8 Aluminum alloy 7075-T73 landing-gear torque-arm assembly redesign to eliminate fatigue fracture at a lubrication hole. (a) Arm configuration, original and improved design (dimensions given in inches). (b) Fracture surface where arrows indicate multiple crack origins. (Photo: From ASM Metals Handbook, Vol. 11: Failure Analysis and Prevention, 1986, ASM International, Materials Park, OH 44073-0002, fig 23, p. 114. Reprinted by permission of ASM International ®, www.asminternational.org.)

4.94

Aluminum alloy 7075-T73 Rockwell B 85.5 25.5 10.200

Lug (1 of 2)

Fracture A Primary-fracture surface

Lubrication hole

1.750-in.-dia bushing, 0.090-in. wall

Lubrication hole

1 in Secondary fracture

3.62 dia

Improved design

Original design Detail A (a)

(b)

Figure 6–9

Static normal stress

Formation of persistent slip bands leading to crack nucleation. (a) Static stress causes slip planes 45° from load direction. (b) Increased static stress causes parallel slip planes to bunch into slip bands. (c) Cyclic stress causes slipping in both directions, leaving persistent slip bands protruding past the grain boundary. (d) Continued cyclic stress causes pronounced extrusions and intrusions of the persistent slip bands, which are prime locations for microcracks to nucleate.

Cyclic normal stress

Intrusion Slip plane

Extrusion Grain

Grain boundary

Persistent slip bands

Slip bands (a)

(b)

(c)

(d)

dislocations, allowing atoms in crystal planes to slip past one another (Figure 6–9a). The slip planes prefer movement within a grain of the material in a direction requiring the least energy. Within a grain, then, slip planes will tend to be parallel to one another, and will bunch together to form slip bands (Figure 6–9b). Grains that happen to be preferentially oriented with respect to the macro-level normal stress direction will be the first to form slip planes. This preferential orientation will usually be along the plane of maximum shear stress at a 45° angle to the loading direction. When the slip bands reach the edge of a grain, and especially at the surface of the material, they can extrude very slightly past the edge of the grain boundary and are then called persistent slip bands (Figure 6–9c). So far, this explanation is consistent with static plastic strain, and is the same mechanism that leads to strain hardening, as explained in Section 2–3. However, the properties of a material subject to cyclic loading, as described in Section 2–4, plays a dominant role at this point. Repeated cycling of stresses can lead to cyclic hardening or cyclic softening, depending on the original condition of the material. The cyclic stress-strain properties of the material are of more significance in dictating the plastic yield strength than the monotonic stress-strain properties. Strain hardening and cyclic hardening will resist the slip planes, slowing or halting the crack nucleation.

Fatigue Failure Resulting from Variable Loading     293

Continued cyclic loading of sufficient level eventually causes the persistent slip bands to slide back and forth with respect to one another, forming both extrusions and intrusions at the grain boundary on the order of 1 to 10 microns (Figure 6–9d). This leaves tiny steps in the surface that act as stress concentrations that are prone to nucleating a microcrack. Conditions that can accelerate crack nucleation include residual tensile stresses, elevated temperatures, temperature cycling, and a corrosive environment. Microcrack nucleation is much more likely to occur at the free surface of a part. The stresses are often highest at the outer part of the cross section; stress concentrations are often at the surface; surface roughness creates local stress concentrations; oxidation and corrosion at the surface accelerate the process; the plastic deformation necessary for persistent slip bands to form has less resistance at the surface. Crack Propagation After a microcrack nucleates, it begins stage I crack growth by progressively breaking the bonds between slip planes across a single grain. The growth rate is very slow, on the order of 1 nm per stress cycle. When a microcrack reaches the grain boundary, it may halt, or eventually it may transfer into the adjacent grain, especially if that grain is also preferentially oriented such that its maximum shear plane is near 45° to the direction of loading. After repeating this process across approximately three to ten grains, the crack is sufficiently large to form a stress concentration at its tip that forms a tensile plastic zone. If several microcracks are in near vicinity to one another, they may join together, increasing the size of the tensile plastic zone. At this point, the crack is vulnerable to being "opened" by a tensile normal stress. This causes the crack to change from growing along slip planes to growing in a direction perpendicular to the direction of the applied load, beginning stage II crack growth. Figure 6–10 shows a schematic of the stage I and II transcrystalline growth of a crack. A compressive normal stress does not open the crack, so the change from stage I to stage II may not be as distinct. Later we will see that compressive stresses have much less effect on fatigue life than tensile stress. Figure 6–11 shows a representative graph of the crack nucleation, growth, and fracture on a plot of cycling stress amplitude versus the fatigue life in number of cycles. The graph shows that at higher stress levels a crack initiates quickly, and most of the fatigue life is used in growing the crack. This lengthy crack growth at high Stage I

Stage II

Cycling stress amplitude

Free surface

Persistent slip bands

Crack growth region Final fracture Slip band region Microcrack nucleation

Figure 6–10

Loading direction

Schematic of stages I (shear mode) and II (tensile mode) transcrytalline microscopic fatigue crack growth.

Number of cycles

Figure 6–11 Crack nucleation and growth as a portion of total fatigue life.

294      Mechanical Engineering Design

stress levels is well modeled by methods of fracture mechanics, described in Section 6–5. At lower stress levels, a large fraction of the fatigue life is spent to nucleate a crack, followed by a quick crack growth. If the stress level is low enough, it is possible that a crack never nucleates, or that a nucleated crack never grows to fracture. This last phenomenon is characteristic of some materials, including steels, and is one of the early discoveries of Wöhler that allowed for designing for long or infinite life.

6–4  Fatigue-Life Methods The three major fatigue-life methods used in design and analysis are the stress-life method, the strain-life method, and the linear-elastic fracture mechanics (LEFM) method. These methods attempt to predict the life in number of cycles to failure, N, for a specific level of loading. Broadly speaking, the strain-life method focuses on crack nucleation (stage I), the LEFM method focuses on crack propagation (stage II), and the stress-life method merges all three stages together with an empirical view to estimating fatigue life based on comparison to experimental test specimens. When cyclic loads are relatively low, stresses and strains are mostly elastic, and long lives are achieved (say, greater than about 104), the domain is referred to as high-cycle fatigue. On the other hand, for high cyclic loads with mostly plastic stresses and strains, and short lives, the domain is low-cycle fatigue. The stress-life method estimates the life to complete fracture, ignoring the details of crack initiation and propagation. It is based on nominal stress levels only, applying stress concentration factors at notches, with no accounting for local plastic strain. Consequently, it is not useful for conditions with high stresses, plastic strains, and low cycles (i.e., the low-cycle fatigue domain). It is almost entirely based on empirical data, with little theoretical basis, with the goal of developing a stress-life diagram, a plot of the cycling stress levels versus the number of cycles before failure. Though it is the least accurate approach, it is the most traditional method, since it is the easiest to implement for a wide range of design applications, has ample supporting data, and represents high-cycle applications adequately. The strain-life method involves more detailed analysis of the plastic deformation at localized regions where both elastic and plastic strains are considered for life estimates. Like the stress-life method, this method is based on test specimens. This time, the testing is strain-based, taking into account the cyclic characteristics of material properties at the localized level in the vicinity of notches that are assumed to be where cracks nucleate. The method requires a significant amount of material property information from cyclic stress-strain and strain-life curves. A key tool is the strain-life diagram, a plot of both elastic and plastic strain levels versus the number of cycles (or strain reversals) before a crack nucleates. This method is especially good for low-cycle fatigue applications where the strains are high, but also works for high-cycle life. The fracture mechanics method assumes a crack is already present and detected. It is then employed to predict crack growth with respect to stress intensity. This approach deals directly with predicting the growth of a crack versus the number of cycles, an approach that is unique from the other two approaches, which do not address the actual crack. The method is useful for predicting fatigue life when the stress levels are high and a large fraction of the fatigue life is spent in the slow growth of a crack, as indicated in the discussion of Figure 6–11. All three methods have their place in fatigue design. It is not that one is better than the others. Each one is the prime tool for the application and domain in which it is strong, but cumbersome or completely incapable for applications and domain in which it is weak. For applications that need to monitor the actual growth rate of a crack, LEFM is the prime tool. For the low-cycle domain in the presence of a notch, the strain-life

Fatigue Failure Resulting from Variable Loading     295

method is optimal. For high-cycle domain, both the strain-life and stress-life methods apply. The strain-life method is more accurate but requires significantly more overhead. From a practical perspective, the stress-life method is by far the easiest for beginning engineers. It requires less conceptual mastery, less need to acquire material properties, and less mathematical analysis. It allows rough life estimates to be made in the highcycle domain with minimal effort. It provides a path to observe the factors that have been proven to affect fatigue life. In this text, the focus is on first exposure to fatigue, so we focus on the stress-life method. The brief coverage of LEFM and strain-life is to encourage familiarity with the concepts, and to provide a good starting point for digging deeper when greater investment is warranted for the strengths of these methods. Fatigue Design Criteria As the understanding of fatigue failure mechanisms has developed, design philosophies have also evolved to provide strategies for safe designs. There are four design approaches, each with an appropriate application for use. The infinite-life design is the oldest strategy, originating in the mid-1800s when Wöhler discovered that steels have a stress level, referred to as a fatigue limit or endurance limit, below which fatigue cracks do not ever grow. This design approach uses materials for which this characteristic is true and designs for stresses that never exceed the endurance limit. This approach is most suitable for applications that need to sustain over a million stress cycles before the end of service life is reached. Because the endurance limit is significantly below the yield strength, this approach requires very low stress levels, such that plastic strain is almost entirely avoided. Consequently, this approach is often used with the stress-life method. The safe-life design criterion designs for a finite life, typically less than a million cycles. This approach is appropriate for applications that experience a limited number of stress cycles. It is not possible to be precise with estimating fatigue life, as the actual failures exhibit a large scatter for the same operating conditions. Significant safety margins are used. Both the strain-life and stress-life approach are used for safelife designs, though the safe-life approach is not very accurate for lower numbers of cycles where the stress levels are high enough to produce locally plastic strains. When warranted, test data from actual simulated operating conditions on the actual parts can reduce the necessary factor of safety. Standardized parts like bearings are tested sufficiently to specify a reliability for a given load and life. The fail-safe design criterion requires that the overall design of the system is such that if one part fails, the system does not fail. It uses load paths, load transfer between members, crack stoppers, and scheduled inspections. The philosophy originated in the aircraft industry that could not tolerate the added weight from large factors of safety, nor the potentially catastrophic consequences from low factors of safety. This approach does not attempt to entirely prevent crack initiation and growth. The damage-tolerant design criterion is a refinement of the fail-safe criterion. It assumes the existence of a crack, due to material processing, manufacturing, or nucleation. It uses the linear-elastic fracture mechanics method to predict the growth of a hypothetical crack, in order to dictate an inspection and replacement schedule. Materials that exhibit slow crack growth and high fracture toughness are best for this criterion.

6–5  The Linear-Elastic Fracture Mechanics Method Fracture mechanics is the field of mechanics that studies the propagation of cracks. Linear-elastic fracture mechanics is an analytical approach to evaluating the stress field at the tip of a crack, with the assumptions that the material is isotropic and linear elastic, and that the plastic deformation at the tip of a crack is small compared to the size

296      Mechanical Engineering Design

of the crack. The stress field is evaluated at the crack tip using the theory of elasticity. When the stresses near the crack tip exceed the material fracture toughness, the crack is predicted to grow. The fundamentals of fracture mechanics are developed in Section 5–12, where it is applied to quasi-static loading situations and brittle materials. Here, we will extend the concepts for dynamically loaded applications. The fracture mechanics approach is useful first of all for studying and understanding the fracture mechanism. Secondly, for a certain class of problems it is effective in predicting fatigue life. In particular, it works well for situations in which most of the fatigue life consists of a slow propagation of a crack. From Figure 6–11, we can see that this is often the case for high stresses in which the crack nucleates quickly. In fact, the method is often used when it is appropriate to assume the crack exists from the beginning, either in the form of a prior flaw or a quickly nucleated crack at a sharp stress concentration. This is the approach used for the damage-tolerant design criterion (Section 6–4), and is prominent in the aircraft industry. Fatigue cracks nucleate and grow when stresses vary and there is some tension in each stress cycle. Consider the stress to be fluctuating between the limits of σmin and σmax, where the stress range is defined as Δσ = σmax − σmin. From Equation (5–37) the stress intensity is given by KI = βσ √πa. Thus, for Δσ, the stress intensity range per cycle is

ΔKI = β(σmax − σmin ) √πa = βΔσ √πa

(6–1)

To develop fatigue strength data, a number of specimens of the same material are tested at various levels of Δσ. Cracks nucleate at or very near a free surface or large discontinuity. Assuming an initial crack length of ai, crack growth as a function of the number of stress cycles N will depend on Δσ, that is, ΔKI. For ΔKI below some threshold value (ΔKI)th a crack will not grow. Figure 6–12 represents the crack length a as a function of N for three stress levels (Δσ)3 > (Δσ)2 > (Δσ)1, where (ΔKI)3 > (ΔKI)2 > (ΔKI)1 for a given crack size. Notice the effect of the higher stress range in Figure 6–12 in the production of longer cracks at a particular cycle count. When the rate of crack growth per cycle, da∕dN in Figure 6–12, is plotted as shown in Figure 6–13, the data from all three stress range levels superpose to give a sigmoidal curve. A group of similar curves can be generated by changing the stress ratio R = σmin∕σmax of the experiment. Three unique regions of crack development are observable from the curve. Region I is known as the near threshold region. If the stress intensity factor range is less than a threshold value (ΔKI)th, the crack does not grow. However, this threshold level is very low, so it is not usually a practical option to design to stay below this level. Figure 6–12

(Δσ)3 Crack length a

The increase in crack length a from an initial length of ai as a function of cycle count for three stress ranges, (Δσ)3 > (Δσ)2 > (Δσ)1.

(Δσ)2

(Δσ)1 da

a dN ai

Log N Stress cycles N

Fatigue Failure Resulting from Variable Loading     297

Figure 6–13

Log da dN Region I

Region II

Low-speed propagation

Stable propagation

Crack growth rate versus stress intensity factor range, showing three regions of crack development.

Region III High-speed unstable propagation

Increasing stress ratio R No propagation

(ΔKI)th

KIc

Log ΔKI

Above this threshold, the crack growth rate, though still very small, begins to increase rapidly. Region II is characterized by stable crack propagation with a linear relationship between crack growth and stress intensity factor range. In region III, the crack growth rate is very high and rapidly accelerates to instability. When ΔKI exceeds the critical stress intensity factor ΔKIc (also known as the fracture toughness, and defined in Section 5–12), then there is a sudden, complete fracture of the remaining cross section. Here we present a simplified procedure for estimating the remaining life of a cyclically stressed part after discovery of a crack. This requires the assumption that plane strain conditions prevail.1 Assuming a crack is discovered early in region II, the crack growth in region II of Figure 6–13 can be approximated by the Paris equation, which is of the form da = C(ΔKI ) m dN

(6–2)

where C and m are empirical material constants and ΔKI is given by Equation (6–1). Representative, but conservative, values of C and m for various classes of steels are listed in Table 6–1. Substituting Equation (6–1) and integrating gives

Nf

0

dN = Nf =

1 C

af

ai

da (βΔσ √πa) m

(6–3)

Here ai is the initial crack length, af is the final crack length corresponding to failure, and Nf is the estimated number of cycles to produce a failure after the initial crack is formed. Note that β may vary in the integration variable (e.g., see Figures 5–25 to 5–30). If this should happen, then Reemsnyder suggests the use of numerical integration.2 1

Recommended references are: Dowling, op. cit.; J. A. Collins, Failure of Materials in Mechanical Design, John Wiley & Sons, New York, 1981; R. Stephens, A. Fatemi, R. Stephens, and H. Fuchs, Metal Fatigue in Engineering, 2nd ed., John Wiley & Sons, New York, 2001; and Harold S. Reemsnyder, "Constant Amplitude Fatigue Life Assessment Models," SAE Trans. 820688, vol. 91, Nov. 1983. 2 Op. cit.

298      Mechanical Engineering Design

Table 6–1  Conservative Values of Factor C and Exponent m in Equation (6–2) for Various Forms of Steel (R = σmin∕σmax ≈ 0) C,

Material

m/cycle (MPa √m)

C,

m

−12

Ferritic-pearlitic steels

6.89(10

)

1.36(10

Austenitic stainless steels

5.61(10−12)

(kpsi √in) m

)

m

−10

3.00

−9

6.60(10 )

2.25

3.00(10−10)

3.25

3.60(10

−10

Martensitic steels

in/cycle )

Source: Barsom, J. M. and Rolfe, S. T., Fatigue and Fracture Control in Structures, 2nd ed., Prentice Hall, Upper Saddle River, NJ, 1987, 288–291.

The following example is highly simplified with β constant in order to give some understanding of the procedure. Normally, one uses fatigue crack growth computer programs with more comprehensive theoretical models to solve these problems.

EXAMPLE 6–1 The bar shown in Figure 6–14 is subjected to a repeated moment 0 ≤ M ≤ 1200 lbf · in. The bar is AISI 4430 steel with Sut = 185 kpsi, Sy = 170 kpsi, and KIc = 73 kpsi √in. Material tests on various specimens of this material with identical heat treatment indicate worst-case constants of C = 3.8(10−11 ) (in/cycle)∕(kpsi √in) m and m = 3.0. As shown, a nick of size 0.004 in has been discovered on the bottom of the bar. Estimate the number of cycles of life remaining. Solution The stress range Δσ is always computed by using the nominal (uncracked) area. Thus I bh2 0.25(0.5) 2 = = = 0.01042 in3 c 6 6

Therefore, before the crack initiates, the stress range is

Δσ =

ΔM 1200 = = 115.2(103 ) psi = 115.2 kpsi I∕c 0.01042

which is below the yield strength. As the crack grows, it will eventually become long enough such that the bar will completely yield or undergo a brittle fracture. For the ratio of Sy∕Sut it is highly unlikely that the bar will reach complete yield. For brittle fracture, designate the crack length as af. If β = 1, then from Equation (5–37) with KI = KIc, we approximate af as

af =

KIc 2 1 1 73 2 ≈ = 0.1278 in π ( βσmax ) π ( 115.2 )

Figure 6–14

1 4

M

M Nick

in 1 2

in

Fatigue Failure Resulting from Variable Loading     299

From Figure 5–27, we compute the ratio af∕h as af

h

=

0.1278 = 0.256 0.5

Thus, af∕h varies from near zero to approximately 0.256. From Figure 5–27, for this range β is nearly constant at approximately 1.07. We will assume it to be so, and re-evaluate af as

af =

2 1 73 = 0.112 in π ( 1.07(115.2) )

Thus, from Equation (6–3), the estimated remaining life is Answer

Nf =

1 C

=−

af

ai

da 1 = (βΔσ √πa) m 3.8(10−11 )

0.112

da √πa]3 [1.07(115.2) 0.004

5.047(103 ) 0.112 = 65(103 ) cycles √a ⎹ 0.004

6–6  The Strain-Life Method A fatigue failure almost always begins at a local discontinuity such as a notch, crack, or other area of stress concentration. When the stress at the discontinuity exceeds the elastic limit, plastic strain occurs. If a fatigue fracture is to occur, cyclic plastic strains must exist. The strain-life method is based on evaluation of the plastic and elastic strains in the localized regions where a crack begins, usually at a notch. The method is particularly useful for situations involving local yielding, which is often the case when stresses and strains are high and there is a relatively short life expectancy. Because the strain-life method also includes elastic strains, it is also applicable for long life expectancy, when the stresses and strains are lower and mostly elastic. The method is therefore comprehensive, and is widely viewed as the best method for serious work when the fatigue life needs to be predicted with reasonable reliability. The method requires a relatively high learning curve and a significant investment to obtain the necessary strain-based material behaviors. In this section, the goal is to present the fundamentals to allow the reader to be conversant and to have a framework for further study.3 Unlike the fracture mechanics approach, the strain-life method does not specifically analyze the crack growth but predicts life based on comparison to the experimentally measured behavior of test specimens. The assumption is that the life of the notched part will be the same as the life of a small unnotched specimen cycled to the same strains as the material at the notch root. The test specimens are loaded in tension to a given strain level, followed by loading in compression to the same strain level. The number of load reversals is used in the evaluations, which is twice the number of complete cycles that is used for the stress-life method.

3

N. Dowling, Mechanical Behavior of Materials, 4th ed., Pearson Education, Upper Saddle River, N.J., 2013, chapter 14.

300      Mechanical Engineering Design

Figure 6–15

σ˜

Δε˜e

A stable cyclic hysteresis loop.

Δε˜e

Δε˜p

2

2 ε˜pa

ε˜pa

E

σ˜a

Δσ˜

ε˜

σ˜a

E

Δε˜e

Δε˜p Δε˜

Initially, the strain cycling may lead to changes in the monotonic material properties, such as cyclic hardening or softening, as described in Section 2–4. With relatively few cycles, though, the material response settles into a stable cyclic hysteresis loop, such as in Figure 6–15, and fully described in Section 2–4. The total true strain amplitude Δε̃∕2 is resolved into elastic and plastic components, Δε̃e∕2 and Δε̃p∕2, respectively, as defined in Figure 6–15. This means that a strain-life fatigue test produces two data points, elastic and plastic, for each strain reversal. These strain amplitudes are plotted on a log-log scale versus the number of strain reversals 2N, such as for the example in Figure 6–16. Both elastic and plastic components have a linear relationship with strain reversals on the log-log scale. The equations for these two lines are well known and named, and are given below, followed by definitions of the terms.

10 0

A strain-life plot for hot-rolled SAE 1020 steel. The total strainlife curve is the sum of the plastic-strain and elastic-strain lines (on a log-log scale).

10–1

Strain amplitude, Δε˜/2

Figure 6–16

–2

10

ε'f

c 1.0

σ'f E b

10–3

10– 4 100

Total strain

Plastic strain 1.0 Elastic strain

101

10 2

10 3

10 4

Reversals to failure, 2N

10 5

106

Fatigue Failure Resulting from Variable Loading     301

Plastic-strain Manson-Coffin equation:  Elastic-strain Basquin equation:

Δε̃p 2

= ε′f (2N) c

Δε̃e σ′f = (2N) b 2 E

(6–4) (6–5)

∙ Fatigue ductility coefficient ε′f is the ordinate intercept (at 1 reversal, 2N = 1) of the plastic-strain line. It is approximately equal to the true fracture strain. ∙ Fatigue strength coefficient σ ′f is approximately equal to the true fracture strength. σ f′∕E is the ordinate intercept of the elastic-strain line. ∙ Fatigue ductility exponent c is the slope of the plastic-strain line. ∙ Fatigue strength exponent b is the slope of the elastic-strain line. All four of these parameters are considered empirical material properties, with some representative values tabulated in Table A–23. The point of specimen strain testing is to obtain these parameters. Cyclic strain testing is carried out to obtain stable cyclic hysteresis loops, from which elastic and plastic strain data points are obtained. This is repeated over a broad range of strain amplitudes to generate enough data points to plot both the lines for the strain-life plot. This can be rather expensive, and sometimes approximations are used, such as those identified in the aforementioned parameter definitions. The total strain amplitude is the sum of elastic and plastic components, giving

Δε̃ Δε̃e Δε̃p = + 2 2 2

(6–6)

Therefore, the total-strain amplitude is

Δε̃ σ′f = (2N) b + ε′f (2N) c 2 E

(6–7)

This equation is termed the strain-life relation and is the basis of the strain-life method. For high strain amplitudes, the strain-life curve approaches the plastic-strain Manson-Coffin line. At low strain amplitudes, the curve approaches the elastic-strain Basquin equation. In the stress-life method, a similar relationship is developed between stress and life, plotted on the S-N diagram, though it only includes an elastic term. The Basquin equation is essentially the same as the stress-life line. This makes sense, because at low strain levels the strain is almost entirely elastic, in which case stress and strain are linearly related. The uniqueness and beauty of the strain-life curve is that it includes both elastic and plastic influences on the life. The strain-life relation is based on completely reversed tensile and compressive strain of equal amplitudes. There is much research and discussion regarding the effect of nonzero mean strain. Strain-controlled cycling with a mean strain results in a relaxation of the mean stress for large strains, due to plastic deformation. The stress, then, tends to move toward centering around zero mean stress, even as the strain continues to cycle with a mean strain. In this case, the mean strain has little effect. But if the strain level is not sufficiently high, the mean stress does not completely relax, which has a detrimental effect on fatigue life for positive mean stress, and positive effect for negative mean stress. Dowling provides a good discussion of modeling the mean stress effects.4 4

N. Dowling, "Mean Stress Effects in Strain-Life Fatigue," Fatigue Fract. Eng. Mater. Struct., vol. 32, pp. 1004–1019, 2009.

302      Mechanical Engineering Design

6–7  The Stress-Life Method and the S-N Diagram The stress-life method relies on studies of test specimens subjected to controlled cycling between two stress levels, known as constant amplitude loading. Figure 6–17a shows a general case of constant amplitude loading between a minimum and maximum stress. Load histories of actual parts are often much more diverse, with variable amplitude loading, but many cases can be reasonably modeled with the constant amplitude approach. This is especially true for rotating equipment that experiences repetitive loading with each revolution. Testing with constant amplitude loading also provides a controlled environment to study the nature of fatigue behavior and material fatigue properties. The constant amplitude stress situation is characterized by the following ­terminology: σmin = minimum stress σmax = maximum stress σm = mean stress, or midrange stress σa = alternating stress, or stress amplitude σr = stress range The following relations are evident from Figure 6–17a,

σmax − σmin 2 σmax + σmin σm = 2

σa =

(6–9)

σa

σr

Stress

(6–8)

Stress

Stress

The mean stress can have positive or negative values,σ while the alternatingσa stress is max always a positive magnitude representing the amplitude of the stress that alternatesσ m above and below the mean stress. σmin Two special cases are common enough to warrant special attention. Figure 6–17b shows what is called a repeated stress, in which the stress cycles between a minimum Time stress of zero to a maximum stress. Figure 6–17c shows a completely(a)reversed stress, in which the stress alternates between equal magnitudes of tension and compression, σa shall add the subscript r to the alternating stress, that with a mean stress of zero. We σr is σar , when it is advantageous to clarify that an alternating stress is completely σa reversed. Most then the a σmaxfatigue testing is done with completely reversed σstresses; σr modifying effect of nonzero mean stress is considered separately. σm σmax σ σ σmin m One of the first and most widely used fatigue-testing devices is the R.a R. Moore high-speed rotating-beam machine. This machine subjectsσminthe test specimen to pure = 0 Time Time

(b)

(a) σa

σm

σr

σmax

σmin

σa

σa σa σmin = 0

Time

Time

Stress

Constant amplitude loading. (a) General; (b) Repeated, with σmin = 0; (c) Completely reversed, with σm = 0 σa

σmax

σa σmin = 0

σm Time

Stress

σr

Time σa

σm = 0

σm = 0 (c)

Figure 6–17 σa

σa

σm

(b)

(a)

Time

Stress

σmax

Stress

Stress

σr σa

σr

σr

Fatigue Failure Resulting from Variable Loading     303

Figure 6–18

7 3 16 in

0.30 in

a

F

F

Test-specimen geometry for the R. R. Moore rotating-beam machine.

a 9 78 in R.

F

F

bending by loading as shown in Figure 6–18. A quick sketch of transverse shear and bending moment diagrams will show that throughout the center section of the specimen the transverse shear is zero and the bending moment is constant. The test specimen is very carefully machined and polished, with a final polishing in an axial direction to avoid circumferential scratches. The goal is to eliminate as many extraneous effects as possible to study just the fatigue behavior of the material. Other fatiguetesting machines are available for applying fluctuating or reversed axial stresses, torsional stresses, or combined stresses to the test specimens. The S-N Diagram For the rotating-beam test, each revolution of the specimen causes a stress element on the surface to cycle between equal magnitudes of tension and compression, thus achieving completely reversed stress cycling. A large number of tests are necessary because of the statistical nature of fatigue. Many specimens are tested to failure at each level of completely reversed stress. A plot of the completely reversed alternating stress versus the life in cycles to failure is made on a semi-log or log-log scale, such as in Figure 6–19. This plot is called a Wöhler curve, a stress-life diagram, or an S-N diagram. The curve typically passes through the median of the test data for each stress level. A value of completely reversed stress on the ordinate is referred to as a fatigue strength Sf when accompanied by a statement of the number of cycles N to which it corresponds. Fatigue failure with less than 1000 cycles is generally classified as low-cycle fatigue, and fatigue failure with greater than 1000 cycles as high-cycle fatigue, as indicated in Figure 6–19. Low-cycle fatigue is associated with high stresses and Low cycle

Figure 6–19

High cycle Finite life

A characteristic S-N diagram.

Infinite life

Fatigue strength Sf

Sut

Se

100

101

102

103

104

105

Number of stress cycles, N

106

107

108

109

304      Mechanical Engineering Design

Figure 6–20 Peak alternating bending stress S, kpsi (log)

S-N bands for representative aluminum alloys.

80 70 60 50 40 35 30 25

Wrought

20 18 16 14 12

Permanent mold cast

10

Sand cast

8 7 6 5 103

104

105

106 Life N, cycles (log)

107

108

109

strains, often in the plastic range, at least locally. Consequently, low-cycle fatigue is best modeled with the strain-life approach. Because the stress-life approach is best suited for lower stresses, which are associated with the high-cycle region, a healthy factor of safety with regard to static yielding, including any stress concentration, will usually suffice to avoid failure in the low-cycle region. In the case of ferrous metals and alloys, a knee occurs in the S-N diagram. Reversed stresses below this level do not cause fatigue failure, no matter how great the number of cycles. Accordingly, Figure 6–19 distinguishes a finite-life region and an infinite-life region. The boundary between these regions depends on the specific material, but it lies somewhere between 106 and 107 cycles for steels. The fatigue strength corresponding to the knee is called the endurance limit Se, or the fatigue limit. Many materials, particularly nonferrous metals and plastics, do not exhibit an endurance limit. For example, Figure 6–20 shows regions for the S-N curves for most common aluminum alloys. For materials without an endurance limit, the fatigue strength Sf is reported at a specific number of cycles, commonly N = 5(108) cycles of reversed stress (see Table A–24). The endurance limit in steels is thought to be associated with the presence of a solute such as carbon or nitrogen that pins dislocations at small strains, thus preventing the slip mechanism that leads to the formation of microcracks. Aluminum alloys lack these dislocation-pinning solutes, so that even small levels of strain eventually lead to dislocation slipping, and therefore an absence of an endurance limit. Care must be taken when using an endurance limit in design applications because it can disappear if the dislocation pinning is overcome by such things as periodic overloads or high temperatures. A corrosive environment can also interact with the fatigue process to make a material behave as if it does not have an endurance limit.

6–8  The Idealized S-N Diagram for Steels The S-N diagram represents many data points that have a wide spread due to the somewhat chaotic nature of fatigue. Care must be taken to constantly be aware of this scatter, realizing that fatigue is not as clean and predictable as many other facets of engineering. However, for preliminary and prototype design and for some failure analysis as well, a simple idealized version of the S-N diagram is useful. In this section, we will develop such a diagram for steels. A similar approach can be used for other materials.

Fatigue Failure Resulting from Variable Loading     305 Low cycle

Figure 6–21

High cycle Infinite life

Finite life

An idealized S-N diagram for steels.

Sut

Fatigue strength Sf

fSut

Sf = aN b

Sf

Se

100

101

102

103

104 N

105

106

107

108

109

Number of stress cycles, N

Figure 6–21 shows an idealized S-N diagram for steels, in which the median failure curve in the finite-life region is represented by two lines. In the low-cycle region (N < 1000 cycles), the line has a relatively low slope and runs between the ultimate strength to some fraction f of the ultimate strength, where f is generally between 0.8 and 0.9. Between 103 and 106 cycles, a line of steeper slope represents the median failure curve. At 106 cycles, the endurance limit is reached, and the failure curve becomes horizontal. To work with this idealized S-N diagram, we will need to determine two points: f Sut at 103 cycles and Se at 106 cycles. With these points, the relationship between fatigue strength and life can be approximated for the finite life region, and infinite life can be predicted for stress levels less than the endurance limit. Estimating the Endurance Limit Many fatigue tests are run at ever lower stress levels to determine the endurance limit for a particular material. This is very time consuming since close to a million revolutions is necessary for each data point. Consequently, it is desirable for purposes of obtaining an idealized S-N diagram to have a quick means of estimating the endurance limit, ideally based on readily available published data. Plotting the experimentally determined endurance limit versus ultimate strength for a large quantity of ferrous metals shows there is a reasonably strong correlation, as shown in Figure 6–22. The plot indicates that the endurance limit generally ranges between 40 to 60 percent of the ultimate strength up to about 200 kpsi. At that point, the scatter increases and the slope flattens. For steels, simplifying our observation of Figure 6–22, we will estimate the endurance limit as

 0.5Sut  S′e =  100 kpsi   700 MPa

Sut ≤ 200 kpsi (1400 MPa) Sut > 200 kpsi Sut > 1400 MPa

(6–10)

where Sut is the minimum tensile strength. The prime mark on S′e in this equation refers to the rotating-beam specimen itself. We wish to reserve the unprimed symbol Se for the endurance limit of an actual machine element subjected to any kind of loading. Soon we shall learn that the two strengths may be quite different.

306      Mechanical Engineering Design

Figure 6–22

140

Graph of endurance limits versus tensile strengths from actual test results for a large number of wrought irons and steels. Ratios of S′e∕Sut of 0.60, 0.50, and 0.40 are shown by the solid and dashed lines. Note also the horizontal dashed line for S′e = 105 kpsi. Points shown having a tensile strength greater than 210 kpsi have a mean endurance limit of S′e = 105 kpsi and a standard deviation of 13.5 kpsi. (Collated from data compiled by H. J. Grover, S. A. Gordon, and L. R. Jackson in Fatigue of Metals and Structures, Bureau of Naval Weapons Document NAVWEPS 00-25-534, 1960; and from Fatigue Design Handbook, SAE, 1968, p. 42.)

Carbon steels Alloy steels Wrought irons

120

Endurance limit S 'e , kpsi

S 'e = Su

0.5

0.6

0.4 105 kpsi

100 80 60 40 20 0

0

20

40

60

80

100

120

140

160

180

200

220

240

260

280

300

Tensile strength Sut , kpsi

Steels treated to give different microstructures have different S′e∕Sut ratios. It appears that the more ductile microstructures have a higher ratio. Martensite has a very brittle nature and is highly susceptible to fatigue-induced cracking; thus the ratio is low. When designs include detailed heat-treating specifications to obtain specific micro structures, it is possible to use an estimate of the endurance limit based on test data for the particular microstructure; such estimates are much more reliable and indeed should be used. The endurance limits for various classes of cast irons, polished or machined, are given in Table A–24. Aluminum alloys do not have an endurance limit. The fatigue strengths of some aluminum alloys at 5(108) cycles of reversed stress are given in Table A–24. Estimating the Fatigue Strength at 103 Cycles The next step in determining the fatigue line in the high-cycle region between 103 to 106 cycles is to estimate the fatigue strength fSut at 103 cycles. Many references have settled on a value of f of 0.9. Others, noting that some experimental data indicate a lower value is warranted, opt for a more conservative value of f equal to 0.8. The difference is not terribly significant, as it represents a small change on the low-cycle end of a line that is primarily of interest at higher cycles. However, it has been observed that for steels, f is generally a little lower for higher strength materials. A relationship between f and Sut has been developed based on the elastic strain line in the strain-life approach to fatigue analysis.5 The resulting relationship for steels is plotted in Figure 6–23 and expressed by the curve-fit equations

f = 1.06 − 2.8(10 −3 )Sut + 6.9(10 −6 )S2ut   70 < Sut < 200 kpsi

f = 1.06 − 4.1(10 −4 )Sut + 1.5(10 −7 )S2ut   500 < Sut < 1400 MPa

5

(6–11)

R. G. Budynas and J. K. Nisbett, Shigley's Mechanical Engineering Design, 10th ed., McGraw-Hill Education, New York, 2015, p. 292.

Fatigue Failure Resulting from Variable Loading     307

Figure 6–23

Sut , MPa f

0.9

500

600

700

800

900

1000

1100

1200

1300

0.88 0.86 0.84 0.82 0.80 0.78 0.76 70

80

90

100 110 120 130 140 150 160 170 180 190 200 Sut , kpsi

Values for f can be estimated from the plot or equations, noting that they are not experimentally based and are only intended to provide a better estimate than using a fixed value. For values of Sut lower than the limit given, f = 0.9 is recommended. The High-Cycle S-N Line The relationship between fatigue strength and life in the high-cycle finite-life region, that is between 103 and 106 cycles, is approximately linear on the log-log scale. It thus can be represented on a normal scale by a power function known as Basquin's equation, Sf = aNb

(6–12)

where Sf is the fatigue strength correlating to a life N in cycles to failure. The constants a and b are the ordinate intercept and the slope of the line in log-log coordinates, which can be readily recognized by taking the logarithm of both sides of Equation (6–12), giving log Sf = b log N + log a. To obtain expressions for a and b, substitute into Equation (6–12) for (N, Sf) the two known points (103, fSut) and (106, Se). Solving for a and b, ( f Sut ) 2 Se

a=

(6–13)

f Sut 1 b = − log( ) 3 Se

(6–14)

Equation (6–12) can be solved for the life in cycles correlating to a completely reversed stress, replacing Sf with σar,

N=(

σar 1/b a)

(6–15)

The typical S-N diagram is only applicable for completely reversed loading. For general fluctuating loading situations, the effect of mean stress must be accounted for (see Section 6–11). This will lead to an equivalent completely reversed stress that is

Fatigue strength fraction, f, of Sut at 103 cycles for steels, with Se = S′e = 0.5Sut at 106 cycles.

308      Mechanical Engineering Design

considered to be equally as damaging as the actual fluctuating stress (see Section 6–14), and which can therefore be used with the S-N diagram and Equation (6–15). Basquin's equation is commonly encountered in the research literature in terms of load reversals (two reversals per cycle) in the form of σar = σ′f (2N) b

(6–16)

where σar is the alternating stress (completely reversed), N is the number of cycles, b is the fatigue strength exponent, and is the slope of the line, and σ′f is the fatigue strength coefficient. This equation is equivalent to the strain-based version of Equation (6–5) used in the strain-life method, and the parameters are fully discussed there. Parameters b and σ′f are empirically determined material properties, based on completely reversed loading of unnotched specimens, with some representative values given in Table A–23. For other situations where the parameters are not available, Equations (6–12), (6–13), and (6–14) provide a means to estimate the high-cycle S-N line.

EXAMPLE 6–2 Given a 1050 HR steel, estimate (a) the rotating-beam endurance limit at 106 cycles. (b) the endurance strength of a polished rotating-beam specimen corresponding to 104 cycles to failure. (c) the expected life of a polished rotating-beam specimen under a completely reversed stress of 55 kpsi. Solution (a) From Table A–20, Sut = 90 kpsi. From Equation (6–10), S′e = 0.5(90) = 45 kpsi

Answer

(b) From Figure 6–23, or Equation (6–11), for Sut = 90 kpsi, f ≈ 0.86. From Equation (6–13), a=

[0.86(90)]2 = 133.1 kpsi 45

From Equation (6–14), 0.86(90) 1 b = − log [ = −0.0785 3 45 ] Thus, Equation (6–12) is Sf = 133.1 N −0.0785 Answer For 104 cycles to failure, Sf = 133.1(104)−0.0785 = 65 kpsi (c) From Equation (6–15), with σar = 55 kpsi, Answer

N=(

55 1∕−0.0785 = 77 500 = 7.8(104 ) cycles 133.1 )

Keep in mind that these are only estimates, thus the rounding of the results to fewer significant figures.

Fatigue Failure Resulting from Variable Loading     309

6–9  Endurance Limit Modifying Factors The rotating-beam specimen used in the laboratory to determine endurance limits is prepared very carefully and tested under closely controlled conditions. It is unrealistic to expect the endurance limit of a mechanical or structural member to match the values obtained in the laboratory. Some differences include ∙ Material: composition, basis of failure, variability ∙ Manufacturing: method, heat treatment, fretting corrosion, surface condition, stress concentration ∙ Environment: corrosion, temperature, stress state, relaxation times ∙ Design: size, shape, life, stress state, speed, fretting, galling Marin6 identified factors that quantified the effects of surface condition, size, loading, temperature, and miscellaneous items. Marin proposed that correction factors for each effect be applied as multipliers to adjust the endurance limit. The multiplicative combination of the various effects has not been thoroughly tested and proven, particularly in capturing the actual impact of interactions between different effects. However, limited testing has shown it to be reasonable for the rough approximations expected from the stress-life approach. Marin's equation is written as

Se = ka k b k c k d k e S′e (6–17)

where  ka = surface factor kb = size factor kc = load factor kd = temperature factor ke = reliability factor S′e = rotary-beam test specimen endurance limit Se = endurance limit at the critical location of a machine part in the geometry and condition of use When endurance tests of parts are not available, estimations are made by applying Marin factors to the endurance limit. This will effectively lower the high-cycle end of the S-N line, while not moving the low-cycle end of the line. Accordingly, the modifying effects are applied proportionately through the finite life region. This seems reasonable and consistent with limited experimental data. Surface Factor ka The surface factor is to account for the effect on the endurance limit for actual parts which rarely have, or can maintain, a surface finish as smooth as the polished test specimen. Stresses are often highest at the surface. Surface roughness puts many localized stress concentrations in the area of high stress, often leading to localized plastic strain at the roots of surface imperfections. This creates an environment that is prone to crack initiation. If materials were perfectly homogeneous, elastic, and isotropic, the effect of surface roughness could be predicted from an analysis of geometric stress concentration associated with the surface profile. However, with any real material subject to real manufacturing operations, the surface roughness only adds 6

Joseph Marin, Mechanical Behavior of Engineering Materials, Prentice Hall, Englewood Cliffs, N.J., 1962, p. 224.

310      Mechanical Engineering Design

Figure 6–24 1.0

Polished Ground

0.9 0.8

Machined or cold drawn

0.7 Surface factor ka

Trends for surface factor ka for steels. (Generated from data from C. J. Noll and C. Lipson, "Allowable Working Stresses," Society for Experimental Stress Analysis, vol. 3, no. 2, 1946, p. 29.)

Sut , GPa 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7

0.6 Hot rolled

0.5 0.4 0.3 As forged

0.2 0.1 0 60

80

100 120 140 160 180 200 220 240 260 Sut , kpsi

stress raisers to those already present in the existing microstructure due to its complicated history of such things as metallurgical treatment, cold working, residual stresses from manufacturing operations, etc. Consequently, from a practical perspective, a surface factor based on purely surface roughness can be misleading when compared to a test specimen that also includes a history of manufacturing processes. Lipson and Noll collected data from many studies, organizing them into several common commercial surface finishes (ground, machined, hot-rolled, and as-forged).7 A commonly used adaptation of their data is shown in Figure 6–24. It is clear from the curves that the surface effect is significant, and that it is more detrimental for higher-strength materials. Bringing critical stress locations to at least a machined finish is certainly worth consideration as a cost-effective substantial improvement in fatigue life. Several observations regarding the data for each category is warranted. The ground surface category includes any surface finish that affects the endurance limit by no more than approximately 10 percent (except for higher strengths), and that has a maximum profilometer reading of 100 micro-inches. This includes ground, honed, and lapped finishes. The test data for higher strengths is significantly scattered. In recognition of this, the declining curve shown for higher strengths is generally conservative and not typical of all the data points. The machined category includes rough and finish machining operations, as well as unmachined cold-drawn surfaces. The data collected for this curve was very limited for higher strengths, so the curve is extrapolated for strengths above about 160 kpsi. The hot-rolled class represents surface conditions typical of a hot-rolled manufacturing process and includes surface irregularities, scale defects, oxide, and partial surface decarburization. It is significant to note that this is not entirely a factor of the surface roughness, but also includes metallurgical conditions. Similar to the hot-rolled category, the as-forged category is heavily influenced by metallurgical conditions, notably that the surface layer is significantly decarburized, 7

C. J. Noll and C. Lipson, "Allowable Working Stresses," Society for Experimental Stress Analysis, vol. 3, no. 2, 1946, p. 29. Reproduced by O. J. Horger (ed.), Metals Engineering Design ASME Handbook, McGraw-Hill, New York, 1953, p. 102.

Fatigue Failure Resulting from Variable Loading     311

Table 6–2  Curve Fit Parameters for Surface Factor, Equation (6–18) Factor a Surface Finish

Exponent b

Sut , kpsi

Sut , MPa

Ground

1.21

1.38

−0.067

Machined or cold-drawn

2.00

3.04

−0.217

Hot-rolled

11.0

38.6

−0.650

As-forged

12.7

54.9

−0.758

which is probably as much or more significant than the actual surface roughness. When machining a forged part to improve its surface factor, it is important to machine to a depth that will remove the decarburized layer. More recent investigations by McKelvey and others have noted that the forging process has seen significant improvements since the Lipson and Noll data was collected in the 1940s.8 Consequently, based on more recent experimental data, McKelvey recommends a surface factor for the as-forged surface that is at least as high as the hot-rolled curve shown in Figure 6–24. The curves in Figure 6–24 are only intended to capture the broad tendencies. The data came from many studies, gathered under a variety of conditions. Lipson and Noll indicated that they attempted to compensate for other factors, such as the differences in specimen sizes included in the studies. In general, the curves are thought to represent the lower bounds of the spread of the data, and are therefore likely to be conservative compared to what might be proven by specific part testing. For convenience, the curves of Figure 6–24 are fitted with a power curve equation ka = aSbut

(6–18)

where Sut is the minimum tensile strength, in units of kpsi or MPa, and a and b are curve fit parameters tabulated in Table 6–2. The power curve form of Equation (6–18) is not the optimal choice for fitting the shape of the curves, but it is suitable for its simplicity and is sufficient for the level of scatter in the original data. The fitting parameters in Table 6–2 have been updated from previous editions of this text to better fit the curves, and giving less weighting to strengths above 200 kpsi (1400 MPa), where there was little test data available.

EXAMPLE 6–3 A steel has a minimum ultimate strength of 520 MPa and a machined surface. Estimate ka. Solution From Table 6–2, a = 3.04 and b = −0.217. Then, from Equation (6–18) Answer

8

ka = 3.04(520) −0.217 = 0.78

S. A. McKelvey and A. Fatemi, "Surface Finish Effect on Fatigue Behavior of Forged Steel," International Journal of Fatigue, vol. 36, 2012, pp. 130–145.

312      Mechanical Engineering Design

Size Factor kb The endurance limit of specimens loaded in bending and torsion has been observed to decrease slightly as the specimen size increases. One explanation of this is that a larger specimen size leads to a greater volume of material experiencing the highest stress levels, thus a higher probability of a crack initiating. However, the size effect is not present in axially loaded specimens. This leads to a further hypothesis that the presence of a stress gradient is somehow involved in the size effect. A size factor has been evaluated by Mischke for bending and torsion of round rotating bars, using a compilation of data from several sources.9 The data has significant scatter, but curve-fit equations representing the lower edge of the data (and therefore conservative for design purposes) can be given by

 (d∕0.3) −0.107 = 0.879d −0.107  0.91d −0.157 kb =  −0.107 = 1.24d −0.107  (d∕7.62) −0.157  1.51d

0.3 ≤ d ≤ 2 in 2 < d ≤ 10 in 7.62 ≤ d ≤ 51 mm 51 < d ≤ 254 mm

(6–19)

For d less than 0.3 inches (7.62 mm), the data is quite scattered. Unless more specific data is available to warrant a higher value, a value of kb = 1 is recommended. For axial loading there is no size effect, so

kb = 1

(6–20)

but see kc. Equation (6–19) applies to round rotating bars in bending and torsion, in which the highly stressed volume is around the outer circumference. If a round bar is not rotating, the highly stressed volume is the same for torsion (all the way around the circumference), but is much less for bending (e.g., just on opposite sides of the cross section). Kuguel introduced a critical volume theory in which the volume of material experiencing a stress above 95 percent of the maximum stress is considered to be critical.10 The method employs an equivalent diameter de obtained by equating the volume of material stressed at and above 95 percent of the maximum stress to the same volume in the rotating-beam specimen. When these two volumes are equated, the lengths cancel, and so we need only consider the areas. For a rotating round section, the 95 percent stress area is the area in a ring having an outside diameter d and an inside diameter of 0.95d. So, designating the 95 percent stress area A0.95σ, we have

A0.95σ =

π 2 [d − (0.95d) 2] = 0.0766d 2 4

(6–21)

This equation is also valid for a rotating hollow round. For nonrotating solid or hollow rounds, the 95 percent stress area is twice the area outside of two parallel chords having a spacing of 0.95d, where d is the diameter, that is

9

A0.95σ = 0.01046d 2

(6–22)

Charles R. Mischke, "Prediction of Stochastic Endurance Strength," Trans. of ASME, Journal of Vibration, Acoustics, Stress, and Reliability in Design, vol. 109, no. 1, January 1987, Table 3. 10 R. Kuguel, "A Relation between Theoretical Stress-Concentration Factor and Fatigue Notch Factor Deduced from the Concept of Highly Stressed Volume," Proc. ASTM, vol. 61, 1961, pp. 732–748.

Fatigue Failure Resulting from Variable Loading     313

With de in Equation (6–21), setting Equations (6–21) and (6–22) equal to each other enables us to solve for the effective diameter. This gives

de = 0.370d

(6–23)

as the effective size of a round corresponding to a nonrotating solid or hollow round. A rectangular section of dimensions h × b has A0.95σ = 0.05hb. Using the same approach as before, de = 0.808(hb)1∕2

(6–24)

Table 6–3 provides A0.95σ areas of common structural shapes undergoing nonrotating bending. Table 6–3  A0.95σ Areas of Common Nonrotating Structural Shapes Loaded in Bending A0.95σ = 0.01046d 2

d

de = 0.370d

b 2

A0.95σ = 0.05hb h

1

de = 0.808 √hb

1 2 a 1

b

2

0.10atf A0.95σ = { 0.05ba

2

tf > 0.025a

   axis 1-1 axis 2-2

tf

1

a

b

2

1 x tf

2

0.05ab axis 1-1 A0.95σ = {    0.052xa + 0.1tf (b − x) axis 2-2

1

EXAMPLE 6–4 A steel shaft loaded in bending is 32 mm in diameter, abutting a filleted shoulder 38 mm in diameter. Estimate the Marin size factor kb if the shaft is used in (a) A rotating mode. (b) A nonrotating mode.

314      Mechanical Engineering Design

Solution (a) From Equation (6–19) d −0.107 32 −0.107 kb = ( = = 0.86 ( 7.62 ) 7.62 )

Answer (b) From Table 6–3,

de = 0.37d = 0.37(32) = 11.84 mm

From Equation (6–19), 11.84 −0.107 kb = ( = 0.95 7.62 )

Answer

Loading Factor kc Estimates of endurance limit, such as that given in Equation (6–10), are typically obtained from testing with completely reversed bending. With axial or torsional loading, fatigue tests indicate different relationships between the endurance limit and the ultimate strength for each type of loading.11 These differences can be accounted for with a load factor to adjust the endurance limit obtained from bending. Though the load factor is actually a function of the ultimate strength, the variation is minor, so it is appropriate to specify average values of the load factor as

1 kc =  0.85  0.59

bending axial torsion

(6–25)

Note that the load factor for torsion is very close to the prediction from the distortion energy theory, Equation (5–21), where for ductile materials the shear strength is 0.577 times the normal strength. This implies that the load factor for torsion is mainly accounting for the difference in shear strength versus normal strength. Therefore, use the torsion load factor only for pure torsional fatigue loading. When torsion is combined with other loading, such as bending, set kc = 1, and manage the loading situation by using the effective von Mises stress, as described in Section 6–16. Temperature Factor kd Fatigue life predictions can be complicated at temperatures significantly below or above room temperature, due to complex interactions between a variety of other time-dependent and material-dependent processes.12 We shall note some of the issues and trends here, but will provide only very limited quantitative recommendations. 11

H. J. Grover, S. A. Gordon, and L. R. Jackson, Fatigue of Metals and Structures, Bureau of Naval Weapons, Document NAVWEPS 00-2500435, 1960; R. G. Budynas and J. K. Nisbett, Shigley's Mechanical Engineering Design, 9th ed., McGraw-Hill, New York, 2011, pp. 332–333. 12 A good overview is provided by R. I. Stephens, A. Fatemi, R. R. Stephens, and H. O. Fuchs, Metal Fatigue in Engineering, 2nd ed., John Wiley & Sons, 2001, pp. 364–391.

Fatigue Failure Resulting from Variable Loading     315

As operating temperatures decrease below room temperature, many materials, particularly steels, show an increase in ultimate and yield strength. However, the ductility decreases, as evidenced by a decrease in the plastic zone between the yield and ultimate strengths. The effect on fatigue is sometimes positive in that the higher ultimate strength also evidences a higher endurance limit. But if strain levels at stress concentrations are high, there may be a greater notch sensitivity due to the reduced ductility. The lower fracture toughness and lower ductility can lead to a smaller crack length necessary before rapid crack growth to fracture. For steels, the overall fatigue performance effect is often improved at lower temperatures, especially in the highcycle region (and thus low stress and low strain). In the low-cycle region, steels often experience a slight improvement for smooth specimens, but little positive effect, and possibly detrimental effect, for notched specimens. Any impact loading during the dynamic fatigue cycling can have a significant detrimental effect, especially in the presence of sharp notches. In general, then, by avoiding sharp notches and impact loading, the regular fatigue life prediction methods are not unreasonable for low temperatures, especially for the high-cycle region. High temperature fatigue is primarily a concern for temperatures above about 40 percent of the absolute (Kelvin) melting temperature. At these temperatures, thermally activated processes are involved, including oxidation, creep, relaxation, and metallurgical aspects. Also, time-dependent factors that are of little importance at lower temperatures begin to play an important role, such as frequency of cycling, wave shape, creep, and relaxation. Thermally induced cycling stresses can be significant in applications with frequent start-up and shut-down, or any other form of nonuniform or transient heating. When temperatures exceed about 50 percent of the absolute melting temperature, creep becomes a predominant factor, and the stress-life approach is no longer reasonable. Crack nucleation and growth rates are accelerated due to high temperature environmental oxidation. Freshly exposed surfaces due to local plasticity are rapidly oxidized, with grain boundaries being particularly attacked. This leads to a greater incidence of intercrystalline crack growth, rather than the usual transcrystalline growth, which can have a faster crack growth rate. At high temperatures, metals lose the distinct endurance limit, such that the fatigue strength continues to decline with cycles to failure. Further, most metals exhibit a lower long-life fatigue strength (e.g., at 108 cycles) with increased temperatures, with the notable exception of an increase for mild carbon steel in the 200° to 400°C range. For steels operating in steady temperatures less than about 40 percent of the absolute melting point, which is about 380°C (720°F), the primary fatigue life mechanism is probably just the temperature effect on the strength properties. Accordingly, for moderate temperature variations from room temperature, it is recommended to simply utilize the appropriate ultimate strength, and if available, the endurance limit or fatigue strength, for the operating temperature, with kd equal to unity.13 If the temperature-specific ultimate strength is not available, the graph from Figure 2–17 can be utilized (for steels), represented by the curve-fit polynomials

ST∕SRT = 0.98 + 3.5(10 −4 )TF − 6.3(10 −7 )TF2

ST∕SRT = 0.99 + 5.9(10 −4 )TC − 2.1(10 −6 )TC2

13

(6–26)

See, for example, W. F. Gale and T. C. Totemeier (eds.), Smithells Metals Reference Book, 8th ed., Elsevier, 2004, pp. 22-138 to 22-140, where endurance limits are tabulated for several steels from 100° to 650°C.

316      Mechanical Engineering Design

where ST and SRT are the ultimate strengths at the operating temperature and room temperature, respectively, and TF and TC are the operating temperature in degrees Fahrenheit and Celsius, respectively. Equation (6–26) is for steels, and should be limited to the range 20°C (70°F) to 550°C (1000°F). The other factors discussed earlier in this section should be considered as potentially significant for temperatures above about 380°C (720°F). In the case where the endurance limit is known (e.g., by testing, or tabulated material data) at room temperature, it can be adjusted for the operating temperature by applying

kd = ST∕SRT

(6–27)

where values from Equation (6–26) can be used for steels. Again, though, if the endurance limit is available or being estimated based on the ultimate strength at the operating temperature, it needs no further adjustment and kd should be set to unity. EXAMPLE 6–5 A 1035 steel has a tensile strength of 80 kpsi and is to be used for a part that operates in a steady temperature of 750°F. Estimate the endurance limit at the operating temperature if (a) only the tensile strength at room temperature is known. (b) the room-temperature endurance limit for the material is found by test to be (S′e)70° = 39 kpsi. Solution (a) Estimate the tensile strength at the operating temperature from Equation (6–26),

(ST∕SRT ) 750° = 0.98 + 3.5(10 −4 ) (750) − 6.3(10 −7 )(750) 2 = 0.89 (Sut ) 750° = (ST∕SRT ) 750° (Sut ) 70° = 0.89(80) = 71.2 kpsi

Thus, From Equation (6–10),

(Se ) 750° = 0.5(Sut ) 750° = 0.5(71.2) = 35.6 kpsi

Answer

and use kd = 1 since this is already adjusted for the operating temperature. (b) Since the endurance limit is known at room temperature, apply the temperature factor to adjust it to the operating temperature. From Equation (6–27), kd = (ST∕SRT ) 750° = 0.89

(Se ) 750° = kd (S′e ) 70° = 0.89(39) = 35 kpsi

Answer

Reliability Factor ke The reliability factor accounts only for the scatter in the endurance limit fatigue data and is not part of a complete stochastic analysis. The presentation in this text is strictly of a deterministic nature. For a more in-depth stochastic discussion see the previous edition of this text,14 based on work by Mischke.15 A true reliability for fatigue life can only be reasonably attempted with actual part testing. 14

R. G. Budynas and J. K. Nisbett, Shigley's Mechanical Engineering Design, 9th ed., McGraw-Hill Education, New York, 2011, p. 330. 15 C. R. Mischke, "Prediction of Stochastic Endurance Strength," Trans. ASME, Journal of Vibration, Acoustics, Stress, and Reliability in Design, vol. 109, no. 1, January 1987, pp. 113–122.

Fatigue Failure Resulting from Variable Loading     317

Table 6–4  Reliability Factors ke Corresponding to 8 Percent Standard Deviation of the Endurance Limit Reliability, %

Transformation Variate za

Reliability Factor ke

50

0

1.000

90

1.288

0.897

95

1.645

0.868

99

2.326

0.814

99.9

3.091

0.753

99.99

3.719

0.702

Most endurance strength data are reported as mean values. In Figure 6–22 and Equation (6–10), the endurance limit is experimentally determined to be related to the ultimate strength by S′e∕Sut = 0.5. This corresponds roughly to a line through the middle of the scattered data. The reliability factor allows the slope of the line to be adjusted to increase the reliability of data points being included above the line. In turn, this adjusted (lower) endurance limit will account for a higher reliability on the S-N diagram of actual failure points being above the predictions of the highcycle line. Data presented by Haugen and Wirching16 show standard deviations of endurance strengths of less than 8 percent. Thus the reliability modification factor to account for this can be written as

ke = 1 − 0.08 z a

(6–28)

where za is defined by Equation (1–5) and values for any desired reliability can be determined from Table A–10. Table 6–4 gives reliability factors for some standard specified reliabilities. Miscellaneous Effects Any number of additional factors could be included as a multiplier in the Marin equation. Several other effects are known to exist, but they are difficult to quantify in the absence of specific testing. Some of these effects are briefly discussed here. Further investigation of pertinent literature is warranted when dealing with any of these issues. Residual stresses may either improve the endurance limit or affect it adversely. Generally, if the residual stress in the surface of the part is compression, the endurance limit is improved. Fatigue failures appear to be tensile failures, or at least to be caused by tensile stress, and so anything that reduces tensile stress will also reduce the possibility of a fatigue failure. Operations such as shot peening, hammering, and cold rolling build compressive stresses into the surface of the part and improve the endurance limit significantly. Of course, the material must not be worked to exhaustion. The endurance limits of parts that are made from rolled or drawn sheets or bars, as well as parts that are forged, may be affected by the so-called directional characteristics of the operation. Rolled or drawn parts, for example, have an endurance limit in the transverse direction that may be 10 to 20 percent less than the endurance limit in the longitudinal direction. 16

E. B. Haugen and P. H. Wirsching, "Probabilistic Design," Machine Design, vol. 47, no. 12, 1975, pp. 10–14.

318      Mechanical Engineering Design

Figure 6–25

Se (case)

The failure of a case-hardened part in bending or torsion. In this example, failure occurs in the core.

σ or τ Case

Core

Se (core)

Parts that are case-hardened may fail at the surface or at the maximum core radius, depending upon the stress gradient. Figure 6–25 shows the typical triangular stress distribution of a bar under bending or torsion. Also plotted as a heavy line in this figure are the endurance limits Se for the case and core. For this example the endurance limit of the core rules the design because the figure shows that the stress σ or τ, whichever applies, at the outer core radius, is appreciably larger than the core endurance limit. Corrosion It is to be expected that parts that operate in a corrosive atmosphere will have a lowered fatigue resistance. This is true, and it is partly due to the roughening or pitting of the surface by the corrosive material. But the problem is not so simple as the one of finding the endurance limit of a specimen that has been corroded. The reason for this is that the corrosion and the stressing occur at the same time, such that the corrosion has opportunity to interact with and accelerate the crack growth mechanism. Basically, this means that in time any part will fail when subjected to repeated stressing in a corrosive atmosphere. There is no fatigue limit. Thus the designer's problem is to attempt to minimize the factors that affect the fatigue life; these are: ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙

Mean or static stress Alternating stress Electrolyte concentration Dissolved oxygen in electrolyte Material properties and composition Temperature Cyclic frequency Fluid flow rate around specimen Local crevices

Electrolytic Plating Metallic coatings, such as chromium plating, nickel plating, or cadmium plating, reduce the endurance limit by as much as 50 percent. In some cases the reduction by coatings has been so severe that it has been necessary to eliminate the plating process. Zinc plating does not affect the fatigue strength. Anodic oxidation of light alloys reduces bending endurance limits by as much as 39 percent but has no effect on the torsional endurance limit.

Fatigue Failure Resulting from Variable Loading     319

Metal Spraying Metal spraying results in surface imperfections that can initiate cracks. Limited tests show reductions of 14 percent in the fatigue strength. Cyclic Frequency If, for any reason, the fatigue process becomes time-dependent, then it also becomes frequency-dependent. Under normal conditions, fatigue failure is independent of frequency. But when corrosion or high temperatures, or both, are encountered, the cyclic rate becomes important. The slower the frequency and the higher the temperature, the higher the crack propagation rate and the shorter the life at a given stress level. Frettage Corrosion The phenomenon of frettage corrosion is the result of microscopic motions of tightly fitting parts or structures. Bolted joints, bearing-race fits, wheel hubs, and any set of tightly fitted parts are examples. The process involves surface discoloration, pitting, and eventual fatigue. A frettage factor depends upon the material of the mating pairs and ranges from 0.24 to 0.90. EXAMPLE 6–6 A 1080 hot-rolled steel bar has been machined to a diameter of 1 in. It is to be placed in reversed axial loading for 70 000 cycles to failure in an operating environment of 650°F. Using ASTM minimum properties, and a reliability for the endurance limit estimate of 99 percent, estimate the endurance limit and fatigue strength at 70 000 cycles. Solution From Table A–20, Sut = 112 kpsi at 70°F. Since the rotating-beam specimen endurance limit is not known at room temperature, we determine the ultimate strength at the elevated temperature first, using Equation (6–26), (ST∕SRT ) 650° = 0.98 + 3.5(10−4 ) (650) − 6.3(10−7 )(650) 2 = 0.94 The ultimate strength at 650°F is then (Sut ) 650° = (ST∕SRT ) 650° (Sut ) 70° = 0.94(112) = 105 kpsi The rotating-beam specimen endurance limit at 650°F is then estimated from Equation (6–10) as S′e = 0.5(105) = 52.5 kpsi Next, we determine the Marin factors. For the machined surface, Equation (6–18) with Table 6–2 gives ka = aSbut = 2.0(105) −0.217 = 0.73 For axial loading, from Equation (6–20), the size factor kb = 1, and from Equation (6–25) the loading factor is kc = 0.85. The temperature factor kd = 1, since we accounted for the temperature in modifying the ultimate strength and consequently the endurance limit. For 99 percent reliability, from Table 6–4, ke = 0.814. The endurance limit for the part is estimated by Equation (6–17) as Answer

Se = ka kb kc kd ke S′e = 0.73(1)(0.85)(1)(0.814)52.5 = 26.5 kpsi

320      Mechanical Engineering Design

For the fatigue strength at 70 000 cycles we need to construct the S-N equation. From Equation (6–11), or we could use Figure 6–23, f = 1.06 − 2.8(10−3 )(105) + 6.9(10−6 ) (105) 2 = 0.84

From Equation (6–13),

a=

( f Sut ) 2 [0.84(105)]2 = = 293.6 kpsi Se 26.5

and Equation (6–14) f Sut 0.84(105) 1 1 b = − log( = − log[ = −0.1741 3 Se ) 3 26.5 ] Finally, for the fatigue strength at 70 000 cycles, Equation (6–12) gives Sf = a N b = 293.6(70 000) −0.1741 = 42.1 kpsi

Answer

6–10  Stress Concentration and Notch Sensitivity In Section 3–13 it was pointed out that the existence of irregularities or discontinuities, such as holes, grooves, or notches, in a part increases the theoretical stresses significantly in the immediate vicinity of the discontinuity. Equation (3–48) defined a stress-concentration factor Kt (or Kts), which is used with the nominal stress to obtain the maximum resulting stress very local to the irregularity or defect. The theoretical stress-concentration factor Kt is defined for static loading conditions. Under dynamic loading conditions leading to fatigue, it turns out that the fatigue strength of a notched specimen is often not affected as much as would be expected from applying the stressconcentrated maximum stress. Consequently, for fatigue purposes, a fatigue stressconcentration factor, Kf (also known as the fatigue notch factor), is defined based on the fatigue strengths of notch-free versus notched specimens at long-life conditions (e.g., the endurance limit at 106 cycles) under completely reversed loading, such that

Kf =

Fatigue strength of notch-free specimen Fatigue strength of notched specimen

(6–29)

Thus, Kf is a reduced version of Kt , taking into account the sensitivity of the actual part to the stress concentrating effects of a notch in a fatigue situation, and is used in place of Kt as a stress increaser of the nominal stress,

σmax = Kf σ0

or

τmax = Kfsτ0

(6–30)

The reason for the reduced sensitivity is complex and not fully understood, and seems to include the interaction of several different phenomena. Perhaps the most fundamental explanation is that the notch stress affecting the fatigue life is not the maximum stress at the notch, but rather an average stress acting over a finite volume of material adjacent to the notch. Additionally, a small crack initiating at a notch will be growing into a region with rapidly decreasing stress levels. Consequently, the stress gradient plays a part. A large notch radius will have less stress gradient, such that the concentrated stress reduces gradually for distances further from the notch. In this case, the average stress near the notch is nearly as high as the maximum stress at the notch,

Fatigue Failure Resulting from Variable Loading     321

so Kf will be nearly as large as Kt. On the other hand, a small (sharp) notch radius will have a large stress gradient, such that the average stress near the notch is significantly lower than the maximum stress. This leads to an apparent lower sensitivity to the notch. This behavior might be expected due to the fact that at the very small scale, the material is not homogeneous, but is made up of a discrete microstructure of grains which have the effect of equalizing the stress over a small finite distance. In general, materials that are soft, ductile, and low strength with a large grain structure are less sensitive to notches than materials that are hard, brittle, and high strength with a small grain structure. Another factor for notch sensitivity applies particularly when stresses are high enough to cause localized plastic strain at the root of a crack, effectively blunting the tip and reducing the maximum stress to a level that is less than predicted by Ktσ0. To quantify the sensitivity of materials to notches, a notch sensitivity q is defined as Kf − 1

q=

or

Kt − 1

qs =

Kfs − 1 Kts − 1

(6–31)

where q is usually between zero and unity. Equation (6–31) shows that if q = 0, then Kf = 1, and the material has no sensitivity to notches at all. On the other hand, if q = 1, then Kf = Kt, and the material has full notch sensitivity. In analysis or design work, find Kt first, from the geometry of the part. Then specify the material, find q, and solve for Kf from the equation

Kf = 1 + q(Kt − 1)

or

Kfs = 1 + qs (Kts − 1)

(6–32)

Notch sensitivities for specific materials are obtained experimentally. Published experimental values are limited, but some values are available for steels and aluminum. Trends for notch sensitivity as a function of notch radius and ultimate strength are shown in Figure 6–26 for reversed bending or axial loading, and Figure 6–27 for reversed torsion. In using these charts it is well to know that the actual test results from which the curves were derived exhibit a large amount of scatter. Because of this scatter it is always safe to use Kf = Kt if there is any doubt about the true value of q. Also, note that q is not far from unity for large notch radii. Notch radius r, mm 1.0

0

0.5 S ut

si 00 kp =2

2.0

2.5

3.0

3.5

4.0

(0.7)

0

(0.4)

0

10

0.6

1.5 (1.4 GPa)

(1.0)

15

0.8 Notch sensitivity q

1.0

60

0.4

Steels Alum. alloy

0.2

0

0

0.02

0.04

0.06 0.08 0.10 Notch radius r, in

0.12

0.14

0.16

Figure 6–26 Notch-sensitivity charts for steels and UNS A92024-T wrought aluminum alloys subjected to reversed bending or reversed axial loads. For larger notch radii, use the values of q corresponding to the r = 0.16-in (4-mm) ordinate. Source: Sines, George and Waisman, J. L. (eds.), Metal Fatigue, McGraw-Hill, New York, 1969.

322      Mechanical Engineering Design

Figure 6–27

Notch radius r, mm

Notch-sensitivity curves for materials in reversed torsion. For larger notch radii, use the values of qs corresponding to r = 0.16 in (4 mm).

1.0

0

1.0

1.5

2.0

2.5

3.0

3.5

4.0

psi (1.4 GPa) 200 k = S ut 0) (1. 0 7) 15 (0. 100 4) (0. 60

0.8 Notch sensitivity qs

0.5

0.6

0.4

Steels Alum. alloy

0.2

0

0

0.02

0.04

0.06

0.08

0.10

0.12

0.14

0.16

Notch radius r, in

Figures 6–26 and 6–27 have as their basis the work of Neuber17 and Kuhn,18 in which the notch sensitivity is described as a function of the notch radius and a material characteristic length dimension a. This characteristic length is roughly several times the size of a single microstructure grain, and can be thought of as near the size of the material's natural internal imperfections. It is often shown in the form of the Neuber constant √a. The relationship is 1 q= (6–33) √a 1+ √r where r is the notch radius. For convenience, we can combine Equations (6–32) and (6–33), giving Kt − 1 Kf = 1 + (6–34) 1 + √a∕ √r The Neuber constant is experimentally determined for each material. For steels, it correlates with the ultimate strength, and can be represented with the following curve-fit equations in both U.S. customary and SI units. Bending or axial: √a √a

= 0.246 − 3.08(10−3 )Sut + 1.51(10−5 )S2ut − 2.67(10−8 )S3ut −3

−6

= 1.24 − 2.25(10 )Sut + 1.60(10

)S2ut

−10

− 4.11(10

)S3ut

50 ≤ Sut ≤ 250 kpsi 340 ≤ Sut ≤ 1700 MPa   (6–35)

Torsion: √a √a

17

= 0.190 − 2.51(10−3 )Sut + 1.35(10−5 )S2ut − 2.67(10−8 )S3ut −3

−6

= 0.958 − 1.83(10 )Sut + 1.43(10

)S2ut −

−10

4.11(10

)S3ut

50 ≤ Sut ≤ 220 kpsi 340 ≤ Sut ≤ 1500 MPa   (6–36)

H. Neuber, Theory of Notch Stresses, J. W. Edwards, Ann Arbor, Mich., 1946. P. Kuhn and H. F. Hardrath, An Engineering Method for Estimating Notch-size Effect in Fatigue Tests on Steel. Technical Note 2805, NACA, Washington, D.C., October 1952. 18

Fatigue Failure Resulting from Variable Loading     323

For the first of each pair of equations, Sut is in kpsi, the Neuber constant √a has units of (inch)1∕2, and when used in Equation (6–33), r is in inches. In the second of each pair of equations, Sut is in MPa, the Neuber constant √a has units of (mm)1∕2, and when used in Equation (6–33), r is in mm. Equation (6–33) used in conjunction with Equations (6–35) or (6–36) is equivalent to Figures (6–26) and (6–27). As with the graphs, the results from the curve fit equations provide only approximations to the experimental data. The notch sensitivity of cast irons is very low, varying from 0 to about 0.20, depending upon the tensile strength. To be on the conservative side, it is recommended that the value q = 0.20 be used for all grades of cast iron. EXAMPLE 6–7 A steel shaft in bending has an ultimate strength of 690 MPa and a shoulder with a fillet radius of 3 mm connecting a 32-mm diameter with a 38-mm diameter. Estimate Kf using: (a) Figure 6–26. (b) Equations (6–34) and (6–35). Solution From Figure A–15–9, using D∕d = 38∕32 = 1.1875, r∕d = 3∕32 = 0.093 75, we read the graph to find Kt = 1.65. (a) From Figure 6–26, for Sut = 690 MPa and r = 3 mm, q = 0.84. Thus, from Equation (6–32) Answer

Kf = 1 + q(Kt − 1) = 1 + 0.84(1.65 − 1) = 1.55

(b) From Equation (6–35) with Sut = 690 MPa, with r = 3 mm gives Answer

Kf = 1 +

√a

= 0.314 √mm. Substituting this into Equation (6–34)

Kt − 1 1.65 − 1 =1+ = 1.55 0.314 1 + √a∕r 1+ √3

The fatigue stress-concentration factor was defined in Equation (6–29) based on its effect at long-life conditions, such as at 106 cycles. At shorter lives, experimental evidence indicates that reduced amounts of Kf are appropriate, particularly for ductile materials. In fact, for ductile materials with static loading (less than 1 cycle), it is standard practice to ignore the stress concentration effect, which we can now understand as using a reduced value of Kf equal to unity. At 103 cycles, high strength or brittle materials need nearly the full amount of Kf , while low strength and ductile materials need a concentrating effect that is between 1 and the full amount of Kf . To simplify the matter, a conservative approach will be used in this text, in which the full amount of Kf will be applied even for the low-cycle region. In fact, since the stress-life approach does not attempt to account for localized plastic strain, we will recommend that even the maximum localized stresses at a notch, as represented by Kf σ0, be kept below yielding. Some designers use 1∕Kf as a Marin factor to reduce Se, primarily as a convenient method of applying Kf to only the high-cycle end of the S-N curve. As has just been explained, this is only appropriate for very ductile materials for which stress concentration is not needed in the low-cycle region. Of course, for simple loading, infinite life problems, it makes no difference whether Se is reduced

324      Mechanical Engineering Design

by dividing it by Kf or the nominal stress is multiplied by Kf . Furthermore, in Section 6–16, when we consider combining loads, there generally are multiple fatigue stress-concentration factors occurring at a point (e.g., Kf for bending and Kfs for torsion). Here, it is only practical to modify the nominal stresses. To be consistent in this text, we will exclusively use the fatigue stress-concentration factor as a multiplier of the nominal stress. EXAMPLE 6–8 Figure 6–28a shows a rotating shaft simply supported in ball bearings at A and D and loaded by a nonrotating force F of 6.8 kN. The shaft is machined from AISI 1050 cold-drawn steel. Estimate the life of the part. Solution From Figure 6–28b we learn that failure will probably occur at B rather than at C or at the point of maximum moment. Point B has a smaller cross section, a higher bending moment, and a higher stress-concentration factor than C, and the location of maximum moment has a larger size and no stress-concentration factor. We shall solve the problem by first estimating the strength at point B and comparing this strength with the stress at the same point. From Table A–20 we find Sut = 690 MPa and Sy = 580 MPa. The endurance limit S′e is estimated as S′e = 0.5(690) = 345 MPa From Equation (6–18) and Table 6–2, ka = 3.04(690) −0.217 = 0.74 From Equation (6–19), kb = (32∕7.62) −0.107 = 0.86 Since kc = kd = ke = 1, Se = 0.74(0.86)345 = 220 MPa Figure 6–28

A

6.8 kN

B

(a) Shaft drawing showing all dimensions in millimeters; all fillets 3-mm radius. (b) Bending-moment diagram.

250

75

C 100

125

10

10

32

30

D

35

38

30 R2

R1 (a) Mmax

MB

MC

A

B

C (b)

D

Fatigue Failure Resulting from Variable Loading     325

To find the geometric stress-concentration factor Kt we enter Figure A–15–9 with D∕d = 38∕32 = 1.1875 and r∕d = 3∕32 = 0.093 75 and read Kt = 1.65. From Equation (6–35a) with Sut = 690 MPa, √a = 0.314 √mm. Substituting this into Equation (6–34) gives Kt − 1 1.65 − 1 =1+ = 1.55 √ 1 + a∕r 1 + 0.314∕√3

Kf = 1 +

The next step is to estimate the bending stress at point B. The bending moment is MB = R1 x =

225(6.8) 225F 250 = 250 = 695.5 N · m 550 550

Just to the left of B the section modulus is I∕c = πd3∕32 = π323∕32 = 3.217 (103 ) mm3. The reversing bending stress is, assuming infinite life, σar = Kf

MB 695.5 = 1.55 (10) −6 = 335.1(106 ) Pa = 335.1 MPa I∕c 3.217

This stress is greater than Se and less than Sy. This means we have both finite life and no yielding on the first cycle. For finite life, we will need to use Equation (6–15). The ultimate strength, Sut = 690 MPa. From Figure 6–23, f = 0.85. From Equation (6–13) a=

( f Sut ) 2 [0.85(690)]2 = = 1564 MPa Se 220

and from Equation (6–14) f Sut 0.85(690) 1 1 b = − log( = − log[ = −0.1419 ) 3 Se 3 220 ] From Equation (6–15), Answer

N=(

σar 1∕b 335.1 −1∕0.1419 = = 52(103 ) cycles ( 1564 ) a)

6–11  Characterizing Fluctuating Stresses In Section 6–7 the concept of fluctuating stresses was introduced to define terminology. But so far the focus has been on completely reversed stress situations, with zero mean stress. This has allowed the focus to be on the data predominately generated by completely reversed testing. The next step is to examine the effect of nonzero mean stresses. To do so, we will first consider some general issues characterizing fluctuating stresses and reiterate some of the definitions that will be used in the following sections. Figure 6–29 illustrates some of the various stress-time traces that occur. Fluctuating stresses in machinery often take the form of a sinusoidal pattern because of the nature of rotating machinery. However, other patterns are common as well. Very irregular or random stress is extremely difficult to model. Simplifications to the stress pattern are sometimes necessary to apply the stress-life method. Tiny ripples in an otherwise large stress variation, such as in Figure 6–29a, can be ignored. Irregularities in the shape or frequency of a wave pattern, such as in Figure 6–29b, do not have any impact. In general, the important thing is the magnitudes of the peaks. For variations in the

326      Mechanical Engineering Design

Stress

Time

Time

(a)

(c)

σa Stress

σr

Stress

Some stress-time relations: (a) fluctuating stress with highfrequency ripple; (b and c) nonsinusoidal fluctuating stress; (d) sinusoidal fluctuating stress.

Stress

Figure 6–29

Time

σa

σmax

σm σmin (b)

Time

O (d)

peaks, such as in Figure 6–29c, reasonable life estimates may be possible by ignoring the intermediate peaks. To study the effect of mean stress on the fatigue life predictions of the S-N diagram, we will focus on the general fluctuating case shown in Figure 6–29d, along with the following definitions. σmin = minimum stress σmax = maximum stress σa = alternating stress, or stress amplitude

σm = mean stress, or midrange stress σr = stress range

The following relations are evident from Figure 6–29d. These were defined in Section 6–7 and are repeated here for convenience:

σa =⎹

σm =

σmax − σmin ⎹ 2

σmax + σmin 2

(6–8) (6–9)

In addition, the stress ratio is defined as

R=

σmin σmax

(6–37)

The stress ratio can have values between −1 and +1, and is commonly used to represent with a single value the nature of the stress pattern. For example, R = −1 is completely reversed, R = 0 is repeated load, R = 1 is steady. Application of Fatigue Stress-Concentration Factor While we are defining terms, we will take a moment to address the terminology and the method of applying the fatigue stress-concentration factor Kf to the stresses. The stress terms already defined, such as σa and σm, are to be understood to refer to the stress at the location under scrutiny, including any stress concentration. When it is helpful to be

Fatigue Failure Resulting from Variable Loading     327

clear that a stress is referring to a nominal stress, without the inclusion of stress concentration, it will be shown with an additional subscript, as in σa0 and σm0. Therefore, in the presence of a notch, σa = Kf σa0, and when no notch is present, Kf = 1 and σa = σa0. When the mean stress is high enough to induce localized notch yielding, the first-cycle local yielding produces plastic strain and strain strengthening. This is occurring at the location where fatigue crack nucleation and growth are most likely. The material properties (Sy and Sut) are new and difficult to quantify. We recommend as a design criterion for the stress-life method that the maximum stress at a notch, including the fatigue stress concentration, not exceed the yield strength. In other words, design to avoid plastic yielding at a notch. This recommendation is because the stress-life method does not address plastic strain and is not particularly suitable for predicting life when plastic strain is included. Consequently, in this text, we will apply the fatigue stress-concentration factor to both the alternating and mean stresses, as well as to the maximum stress when checking for yielding at a notch. Thus,

σa = Kf σa0 (6–38)

σm = Kf σm0 (6–39)

There are other strategies for handling the stress concentration when localized yielding occurs. They mostly take into account that the peak stress is actually reduced when the localized yielding occurs, so the full amount of the concentration factor is not accurate, and can be reduced. This is usually done only for the mean stress. Some of these approaches include the nominal mean stress method, the residual stress method,19 and Dowling's method.20

6–12  The Fluctuating-Stress Diagram The S-N diagram is normally generated with completely reversed stresses. It is possible to generate S-N diagrams with loading situations that include a mean stress. Figure 6–30 shows a characteristic family of S-N curves, generated as usual by plotting alternating stresses versus fatigue life, but with each curve corresponding to a Figure 6–30

Alternating stress, σa

Completely reversed, σm = 0

Set of (σm, σa) points having the same life

σm increasing 103

104

105

106

107

Number of stress cycles, N 19

R. C. Juvinall and K. M. Marshek, Fundamentals of Machine Component Design, 4th ed., Wiley, New York, 2006, Section 8.11; M. E. Dowling, Mechanical Behavior of Materials, 3rd ed., Prentice Hall, Upper Saddle River, N.J., 2007, Sections 10.2–10.6. 20 Dowling, op. cit., pp. 486–487.

Characteristic family of S-N curves for increasing levels of mean stress.

328      Mechanical Engineering Design

different mean stress. The top curve is for zero mean stress and corresponds to the usual line on the completely reversed S-N diagram. Though details would be different for each material tested, the curves show the usual characteristic that higher mean stresses reduce the endurance limit, and in general reduce the life expectancy for a given alternating stress. Historically, there have been many ways of plotting the data obtained for general fluctuating stresses, with such names as Goodman diagram, modified-Goodman diagram, master fatigue diagram, and Haigh diagram.21 Today, probably the simplest and most common means of displaying the fluctuating-stress data is on a plot of alternating stress versus mean stress. We will call this σa − σm set of axes a fluctuating-stress diagram. From the family of S-N curves in Figure 6–30, taking sets of points correlating to the same value of life, combinations of σa and σm can be determined for each life, and plotted as constant-life curves on a fluctuatingstress diagram, as shown in Figure 6–31. Of course, the curves are attempts to fit the scattered data points. Specific equations to "fit" the data are discussed in Section 6–14. This diagram depicts the effect of mean stress on fatigue life, for a specific material from which the curves were generated. Among other things, it is clear that for any desired life, an increase in mean stress must be accompanied by a decrease in alternating stress. Now, from Figure 6–31, focus in on the data points that correlate to the constantlife curve of 106 cycles, shown in Figure 6–32, with many more data points shown to indicate the scatter of data. For steels, with the idealized assumption that the endurance limit corresponds to a life of 106 cycles, these data points reflect the predicted boundary between finite life and infinite life. If it is desired to achieve infinite life, the combination of σa and σm must be inside an envelope bounded by these data points. A straight line between the endurance limit and the ultimate strength is a commonly used design criterion to conservatively represent the infinite life boundary. The line is well-known and is called the modified-Goodman line. As

104

Alternating stress, σa

Alternating stress, σa

N = 103

105

Se

6

10

Se

Sa σa O

Mean stress, σm

Sut

Figure 6–31 Constant-life curves on the fluctuating-stress diagram.

21

Load line

B

Goodman line

A σm

Sm

Sut

Mean stress, σm

Figure 6–32 Fluctuating-stress diagram showing the Goodman line as an infinite-life fatigue criterion.

W. Schutz, "A History of Fatigue," Engineering Fracture Mechanics, vol. 54, no. 2, 1996, pp. 263–300.

Fatigue Failure Resulting from Variable Loading     329

time has passed since John Goodman's life (1862–1935), very little reference is ever made to the original Goodman line, so for simplicity, we will refer to the modifiedGoodman line as simply the Goodman line. The Goodman line connects two points: the endurance limit when the mean stress is zero (completely reversed) and the ultimate strength when the alternating stress is zero (steady stress). The equation for the Goodman line is σa σm + = 1 Se Sut

(6–40)

To utilize the Goodman line as a design criterion, define a load-line from the origin, through a given stress point (σm, σa). The default assumption is that if the stress situation is to increase, it will do so along the load line, maintaining the same slope, that is, the ratio of σa∕σm. A factor of safety with respect to achieving infinite life can then be defined as the proportion of the distance along the load line from the origin to the failure criterion that has been achieved by the load stress. In other words, referring to Figure 6–32, define a strength as distance OB, a stress as distance OA, and the factor of safety as strength∕stress = OB∕OA. This can be done by solving for coordinates (Sm, Sa) at point B at the intersection of the failure line and the load line. Then the factor of safety is nf = OB∕OA = Sa∕σa = Sm∕σm. Or, for this load line from the origin, it turns out that the factor of safety can be found mathematically by simply applying a fatigue life design factor nf to each of the stresses in Equation (6–40) and solving for the factor. That is,

(nf σa ) Se

+

(nf σm ) Sut

σa σm −1 nf = ( + ) Se Sut

= 1 σm ≥ 0

(a) (6–41)

This factor of safety based on the Goodman line is commonly used for design situations in which some conservatism is desired (though the level of conservatism is not quantified). In using it, remember that it is being applied to a specific stress element of a specific part made from a specific material. The stresses should both include any applicable fatigue concentration factor, and the endurance limit should be either obtained from actual testing of the part, or estimated and adjusted with appropriate Marin factors. A factor of safety greater than 1 predicts infinite life, and a factor of safety less than 1 predicts a finite life. So far, the presentation of the Goodman line has been based on hypothetical data from a single material (as in Figures 6–30 and 6–31). To evaluate how well a curve fits the data, it is useful to normalize the fluctuating-stress diagram to allow data from different materials and different testing programs to be plotted on the same plot. Figure 6–33 shows such a plot for a variety of steels, where the ordinate axis is normalized by dividing the alternating stress by the endurance limit, and the abscissa is normalized by dividing the mean stress by the material's ultimate strength. The plot also includes the negative mean stress range, where it is evident that the failure points do not indicate a reduction in alternating stresses for increasing

330      Mechanical Engineering Design

Figure 6–33

1.0 Amplitude ratio σa /Se

Plot of fatigue failures for mean stresses in both tensile and compressive regions. Normalizing the data by using the ratio of steady strength component to tensile strength σm∕Sut, steady strength component to compressive strength σm∕Suc and strength amplitude component to endurance limit σa∕Se enables a plot of experimental results for a variety of steels. [Data source: Thomas J. Dolan, "Stress Range," Section 6.2 in O. J. Horger (ed.), ASME Handbook—Metals Engineering Design, McGraw-Hill, New York, 1953.]

1.2 A

B

0.8 0.6 0.4 0.2 C –1.2

–1.0

–0.8

–0.6

–0.4

–0.2

0

Compression σm /Suc

0.2

0.4

0.6

0.8

1.0

Tension σm /Sut Mean ratio

compressive mean stresses. In fact, tests actually indicate an increase in alternating stress for many materials. This is an interesting phenomenon—that compressive mean stress is not detrimental to fatigue life, and is in fact often beneficial. There are several thoughts as to why this is so, some of which are discussed in Section 6–3 in the crack propagation section. From the conglomeration of test data such as in Figure  6–33, it is confirmed that the Goodman line is a reasonable conservative design tool for positive mean stress. For negative mean stress, a similarly conservative design criterion is a horizontal line from the endurance limit. A fatigue factor of safety based on such a line is

nf =

Se σa

σm < 0

(6–42)

It is necessary to check for yielding. So far, since we are only considering a uniaxial stress situation, we can simply compare the maximum stress to the yield strength. The factor of safety guarding against first-cycle yield is given by

ny =

Sy Sy = σmax σa + ∣σm∣

(6–43)

where the absolute value of the mean stress allows the equation to apply for both positive and negative mean stress. It is helpful to plot the yield condition on the fluctuating-stress diagram to see how it compares with the fatigue criterion. Setting ny equal to unity, and plotting the equation σa + ∣σm∣ = Sy produces two diagonal lines as shown in Figure 6–34. The yield lines are known as Langer lines. Figure 6–34 provides a summary of the design space for both infinite-life fatigue and yielding. A fluctuating-stress state outside the bounds of the yield lines predicts yielding. Notice there is a zone on the positive mean stress side where the yield line crosses inside the fatigue line. This zone indicates it is possible to yield in the first cycle, even when infinite fatigue life is predicted. For this reason, it is always wise to check for yielding, even on a fatigue-loading situation, especially when the mean stress is large and the alternating stress is small.

Fatigue Failure Resulting from Variable Loading     331

Figure 6–34

σa ny =

Sy

ny =

σa + ∣σm∣

S nf = σe a

N=∞

σa + ∣σm∣ N tan λ

(8–3)

This relation states that self-locking is obtained whenever the coefficient of thread friction is equal to or greater than the tangent of the thread lead angle. An expression for efficiency is also useful in the evaluation of power screws. If we let f = 0 in Equation (8–1), we obtain

T0 =

Fl 2π

(g)

which, since thread friction has been eliminated, is the torque required only to raise the load. The thread efficiency is thus defined as

e=

T0 Fl = TR 2πTR

(8–4)

The preceding equations have been developed for square threads where the normal thread loads are parallel to the axis of the screw. In the case of Acme or other threads, the normal thread load is inclined to the axis because of the thread angle 2α and the lead angle λ. Since lead angles are small, this inclination can be neglected and only the effect of the thread angle (Figure 8–7a) considered. The effect of the angle α is to increase the frictional force by the wedging action of the threads. Therefore the frictional terms in Equation (8–1) must be divided by cos α. For raising the load, or for tightening a screw or bolt, this yields

TR =

Fdm l + π f dm sec α 2 ( πdm − f l sec α )

(8–5)

Screws, Fasteners, and the Design of Nonpermanent Joints     429

Figure 8–7

dc

α

F⁄ 2

F cos α

F

(a) Normal thread force is increased because of angle α; (b) thrust collar has frictional diameter dc.

F⁄ 2 Collar Nut

2α =

Thread angle F⁄ 2

F⁄ 2 (a)

(b)

In using Equation (8–5), remember that it is an approximation because the effect of the lead angle has been neglected. For power screws, the Acme thread is not as efficient as the square thread, because of the additional friction due to the wedging action, but it is often preferred because it is easier to machine and permits the use of a split nut, which can be adjusted to take up for wear. Usually a third component of torque must be applied in power-screw applications. When the screw is loaded axially, a thrust or collar bearing must be employed between the rotating and stationary members in order to carry the axial component. Figure 8–7b shows a typical thrust collar in which the load is assumed to be concentrated at the mean collar diameter dc. If fc is the coefficient of collar friction, the torque required is

Tc =

F f c dc 2

(8–6)

For large collars, the torque should probably be computed in a manner similar to that employed for disk clutches (see Section 16–5). Nominal body stresses in power screws can be related to thread parameters as follows. The maximum nominal shear stress τ in torsion of the screw body can be expressed as

τ=

16T πd r3

(8–7)

The compressive axial stress σ in the body of the screw due to load F is

σ=−

F 4F = − 2 A πd r

(8–8)

in the absence of column action. For a short column the J. B. Johnson buckling formula is given by Equation (4–48), which is

Sy l 2 1 F = S − y ( A )crit ( 2π k ) CE

(8–9)

430      Mechanical Engineering Design

Figure 8–8

dm

y

Geometry of square thread useful in finding bending and transverse shear stresses at the thread root.

F

σy τyz

z

σx

τzx

x

y Ff x

p/2

p/2

T F

Nominal thread stresses in power screws can be related to thread parameters as follows. The bearing stress, σB, is from the force F pressing into the surface area of the thread, as in Figure 8–8, giving

σB = −

F 2F =− πdm nt p∕2 πdm nt p

(8–10)

where nt is the number of engaged threads. The bending stress at the root of the thread, in the x direction, is found from Z=

2 I bh2 (πdrnt ) (p∕2) π = = = dr nt p2 c 6 6 24

M=

Fp 4

so

σx =

M Fp 24 6F = = Z 4 πdr nt p2 πdr nt p

(8–11)

Due to the length to height ratio of the thread (see Figure 3–19) the transverse shear stress at the center of the thread root is not critical. The actual stress situation where the root of the thread connects to the main screw body is complex and not easily modeled analytically. However, an estimate of the critical stress can be obtained by evaluating a stress element at the outer radius of the screw body adjacent to the root of the thread, as shown in Figure 8–8. The expanded element in Figure 8–8 shows a rotated 3-D representation of the stresses at the root. This stress element experiences all of the screw body stresses, as well as supporting the bending stress from the adjacent stress element in the thread. In addition, isolating the root of the thread of Figure 8–8 from the main screw body also exposes a tangential shear stress that is transferred from the thread to the screw body due to friction on the thread. This shear is directly related to the torsion and is on the x face in the negative z direction, hence τzx. The shear area, As, is the circumference of the thread at the root diameter times the total thread

Screws, Fasteners, and the Design of Nonpermanent Joints     431

thickness. The shear force, τzx As, acting at the root radius, rr, balances the torsion, T. Thus, the tangential shear stress is

τzx = −

T T 4T =− =− 2 Asrr [πdrnt (p/2)] (dr/2) πdr ntp

(8–12)

Thus, the inset of Figure 8–8 shows the stress element at the outer radius of the screw body, with the stress equations summarized as 6F πdr nt p

σx =

σy = −

σz = 0    τzx =

4F πd 2r

τxy = 0 τyz =

16T πd 3r 4T πd2r ntp

The von Mises stress, σ′, for this stress element is found by substituting these stresses into Equation (5–14). The screw-thread form is complicated from an analysis viewpoint. The equations above assume all engaged threads are equally sharing the load, which turns out to be a weak assumption. A power screw lifting a load is in compression and its thread pitch is shortened by elastic deformation. Its engaging nut is in tension and its thread pitch is lengthened. The engaged threads cannot share the load equally. Some experiments show that the first engaged thread carries 0.38 of the load, the second 0.25, the third 0.18, and the seventh is free of load. In estimating thread stresses by the equations above, substituting 0.38F for F and setting nt to 1 will give the largest level of stresses in the thread-nut combination. EXAMPLE 8–1 A square-thread power screw has a major diameter of 32 mm and a pitch of 4 mm with double threads, and it is to be used in an application similar to that in Figure 8–4. The given data include f = fc = 0.08, dc = 40 mm, and F = 6.4 kN per screw. (a) Find the thread depth, thread width, pitch diameter, minor diameter, and lead. (b) Find the torque required to raise and lower the load. (c) Find the efficiency during lifting the load. (d) Find the body stresses, torsional and compressive. (e) Find the bearing stress on the first thread. ( f ) Find the thread bending stress at the root of the first thread. (g) Determine the von Mises stress at the critical stress element where the root of the first thread interfaces with the screw body. Solution (a) From Figure 8–3a the thread depth and width are the same and equal to half the pitch, or 2 mm. Also Answer

dm = d − p∕2 = 32 − 4∕2 = 30 mm dr = d − p = 32 − 4 = 28 mm l = np = 2(4) = 8 mm

432      Mechanical Engineering Design

(b) Using Equations (8–1) and (8–6), the torque required to turn the screw against the load is TR = = Answer

Ffc dc Fdm l + πf dm + 2 ( πdm − f l ) 2 6.4(30) 8 + π(0.08)(30) 6.4(0.08)40 + [ ] 2 π(30) − 0.08(8) 2

= 15.94 + 10.24 = 26.18 N · m

Using Equations (8–2) and (8–6), we find the load-lowering torque is Ffc dc Fdm πf d m − l + 2 ( πdm + f l ) 2

TL =

=

Answer

= −0.466 + 10.24 = 9.77 N · m

6.4(30) π(0.08)30 − 8 6.4(0.08) (40) + [ ] 2 π(30) + 0.08(8) 2

The minus sign in the first term indicates that the screw alone is not self-locking and would rotate under the action of the load except for the fact that the collar friction is present and must be overcome, too. Thus the torque required to rotate the screw "with" the load is less than is necessary to overcome collar friction alone. (c) The overall efficiency in raising the load is e=

Answer

6.4(8) Fl = = 0.311 2πTR 2π(26.18)

(d) The body shear stress τ due to torsional moment TR at the outside of the screw body is τ=

Answer

16TR πd 3r

=

16(26.18)(103 ) π(283 )

= 6.07 MPa

The axial nominal normal stress σ is Answer

σ=−

4(6.4)103 4F =− = −10.39 MPa 2 πd r π(282 )

(e) The bearing stress σB is, with one thread carrying 0.38F, Answer

σB = −

2(0.38F) 2(0.38)(6.4)103 =− = −12.9 MPa πd m (1)p π(30) (1)(4)

( f ) The thread-root bending stress σb with one thread carrying 0.38F is Answer

σb =

6(0.38F) 6(0.38) (6.4)103 = = 41.5 MPa πdr (1)p π(28) (1)4

(g) The tangential shear stress given by Equation (8–12) with one thread carrying 0.38 T,

τzx = −

4(0.38T) πd2r (1)p

=−

4(0.38)26.18(103 ) π(282 )4

= −4.04 MPa

Screws, Fasteners, and the Design of Nonpermanent Joints     433

The 3-D stresses for Figure 8–8 are

σx = 41.5 MPa    τxy = 0

σy = −10.39 MPa   τyz = 6.07 MPa

σz = 0       τzx = −4.04 MPa

For the von Mises stress, Equation (5–14) can be written as Answer

σ′ =

1 √2

{[41.5 − (−10.39)] 2 + (−10.39 − 0) 2 + (0 − 41.5) 2 + 6(6.07) 2 + 6(−4.04) 2 }

= 49.2 MPa

Ham and Ryan1 showed that the coefficient of friction in screw threads is independent of axial load, practically independent of speed, decreases with heavier lubricants, shows little variation with combinations of materials, and is best for steel on bronze. Sliding coefficients of friction in power screws are about 0.10–0.15. Table 8–4 shows safe bearing pressures on threads, to protect the moving surfaces from abnormal wear. Table 8–5 shows the coefficients of sliding friction for common material pairs. Table 8–6 shows coefficients of starting and running friction for common material pairs. Table 8–4  Screw Bearing Pressure pb Screw Material

Nut Material

Safe pb, psi

Notes

Steel

Bronze

2500–3500

Low speed

Steel

Bronze

1600–2500

≤10 fpm

Cast iron

1800–2500

≤8 fpm

Bronze

800–1400

20–40 fpm

Cast iron

600–1000

20–40 fpm

Bronze

150–240

≥50 fpm

Steel Steel

Source: Data from H. A. Rothbart and T. H. Brown, Jr., Mechanical Design Handbook, 2nd ed., McGraw-Hill, New York, 2006.

Table 8–5  Coefficients of Friction f for Threaded Pairs

Combination

Nut Material

Running

Starting

Steel

Bronze

Brass

Cast Iron

Soft steel on cast iron

0.12

0.17

Steel, dry

0.15–0.25

0.15–0.23

0.15–0.19

0.15–0.25

Hard steel on cast iron

0.09

0.15

Steel, machine oil

0.11–0.17

0.10–0.16

0.10–0.15

0.11–0.17

Soft steel on bronze

0.08

0.10

Bronze

0.08–0.12

0.04–0.06

0.06–0.09

Hard steel on bronze

0.06

0.08

Screw Material

Source: Data from H. A. Rothbart and T. H. Brown, Jr., Mechanical Design Handbook, 2nd ed., McGraw-Hill, New York, 2006.

1

Table 8–6  Thrust-Collar Friction Coefficients

Source: Data from H. A. Rothbart and T. H. Brown, Jr., Mechanical Design Handbook, 2nd ed., McGraw-Hill, New York, 2006.

Ham and Ryan, An Experimental Investigation of the Friction of Screw-threads, Bulletin 247, University of Illinois Experiment Station, Champaign-Urbana, Ill., June 7, 1932.

434      Mechanical Engineering Design

8–3  Threaded Fasteners Figure 8–9 is a drawing of a standard hexagon-head bolt. Points of stress concentration are at the fillet, at the start of the threads (runout), and at the thread-root fillet in the plane of the nut when it is present. See Table A–29 for dimensions. The diameter of the washer face is the same as the width across the flats of the hexagon. The thread length of inch-series bolts, where d is the nominal diameter, is 2d + 14 in LT = { 2d + 12 in

L ≤ 6 in L > 6 in

(8–13)

and for metric bolts is

 2d + 6  L T =  2d + 12   2d + 25

L ≤ 125 125 < L ≤ 200 L > 200

d ≤ 48

(8–14)

where the dimensions are in millimeters. The ideal bolt length is one in which only one or two threads project from the nut after it is tightened. Bolt holes may have burrs or sharp edges after drilling. These could bite into the fillet and increase stress concentration. Therefore, washers must always be used under the bolt head to prevent this. They should be of hardened steel and loaded onto the bolt so that the rounded edge of the stamped hole faces the washer face of the bolt. Sometimes it is necessary to use washers under the nut too. The purpose of a bolt is to clamp two or more parts together. The clamping load stretches or elongates the bolt; the load is obtained by twisting the nut until the bolt has elongated almost to the elastic limit. If the nut does not loosen, this bolt tension remains as the preload or clamping force. When tightening, the mechanic should, if possible, hold the bolt head stationary and twist the nut; in this way the bolt shank will not feel the thread-friction torque. The head of a hexagon-head cap screw is slightly thinner than that of a hexagonhead bolt. Dimensions of hexagon-head cap screws are listed in Table A–30. Hexagonhead cap screws are used in the same applications as bolts and also in applications in which one of the clamped members is threaded. Three other common cap-screw head styles are shown in Figure 8–10. A variety of machine-screw head styles are shown in Figure 8–11. Inch-series machine screws are generally available in sizes from No. 0 to about 38 in. Several styles of hexagonal nuts are illustrated in Figure 8–12; their dimensions are given in Table A–31. The material of the nut must be selected carefully to match that of the bolt. During tightening, the first thread of the nut tends to take the entire load; but yielding occurs, with some strengthening due to the cold work that takes place, and the load is eventually divided over about three nut threads. For this reason you should never reuse nuts; in fact, it can be dangerous to do so. Figure 8–9

H

Hexagon-head bolt; note the washer face, the fillet under the head, the start of threads, and the chamfer on both ends. Bolt lengths are always measured from below the head.

Approx.

R 30°

1 64

in

W

Screws, Fasteners, and the Design of Nonpermanent Joints     435

Figure 8–10

A

A

A

80 to 82°

H

H

H

Typical cap-screw heads: (a) fillister head; (b) flat head; (c) hexagonal socket head. Cap screws are also manufactured with hexagonal heads similar to the one shown in Figure 8–9, as well as a variety of other head styles. This illustration uses one of the conventional methods of representing threads.

d

d

d

L

L LT

L LT

LT

(a)

(b)

(c)

A

A

d

H

80 to 82°

Figure 8–11

H

L

(a) Round head

L

80 to 82°

(b) Flat head

A

A

d

H

d

d H

L

(c) Fillister head

L

(d) Oval head

±3° 5°

A

A

d

d

R H

L

L

(e) Truss head

( f ) Binding head

d W

d W

H

L

(g) Hex head (trimmed)

H (h) Hex head (upset)

L

Types of heads used on machine screws.

436      Mechanical Engineering Design

Figure 8–12

W

Hexagonal nuts: (a) end view, general; (b) washer-faced regular nut; (c) regular nut chamfered on both sides; (d) jam nut with washer face; (e) jam nut chamfered on both sides.

H

1 Approx. 64 in

30° (a)

H

H

30° (b)

Approx.

in

H

30°

30°

(c)

1 64

(d)

(e)

8–4  Joints—Fastener Stiffness When a connection is desired that can be disassembled without destructive methods and that is strong enough to resist external tensile loads, moment loads, and shear loads, or a combination of these, then the simple bolted joint using hardened-steel washers is a good solution. Such a joint can also be dangerous unless it is properly designed and assembled by a trained mechanic. A section through a tension-loaded bolted joint is illustrated in Figure 8–13. Notice the clearance space provided by the bolt holes. Notice, too, how the bolt threads extend into the body of the connection. As noted previously, the purpose of the bolt is to clamp the two, or more, parts together. Twisting the nut stretches the bolt to produce the clamping force. This clamping force is called the pretension or bolt preload. It exists in the connection after the nut has been properly tightened no matter whether the external tensile load P is exerted or not. Of course, since the members are being clamped together, the clamping force that produces tension in the bolt induces compression in the members. Figure 8–14 shows another tension-loaded connection. This joint uses cap screws threaded into one of the members. An alternative approach to this problem (of not using a nut) would be to use studs. A stud is a rod threaded on both ends. The stud is screwed into the lower member first; then the top member is positioned and fastened P

P

l

P

l

P

Figure 8–13 A bolted connection loaded in tension by the forces P. Note the use of two washers. Note how the threads extend into the body of the connection. This is usual and is desired. l is the grip of the connection.

Figure 8–14 Section of cylindrical pressure vessel. Hexagon-head cap screws are used to fasten the cylinder head to the body. Note the use of an O-ring seal. l is the effective grip of the connection (see Table 8–7).

Screws, Fasteners, and the Design of Nonpermanent Joints     437

down with hardened washers and nuts. The studs are regarded as permanent, and so the joint can be disassembled merely by removing the nut and washer. Thus the threaded part of the lower member is not damaged by reusing the threads. The spring rate is a limit as expressed in Equation (4–1). For an elastic member such as a bolt, as we learned in Equation (4–2), it is the ratio between the force applied to the member and the deflection produced by that force. We can use Equation (4–4) and the results of Problem 4–1 to find the stiffness constant of a fastener in any bolted connection. The grip l of a connection is the total thickness of the clamped material. In Figure 8–13 the grip is the sum of the thicknesses of both members and both washers. In Figure 8–14 the effective grip is given in Table 8–7. The stiffness of the portion of a bolt or screw within the clamped zone will generally consist of two parts, that of the unthreaded shank portion and that of the threaded portion. Thus the stiffness constant of the bolt is equivalent to the stiffnesses of two springs in series. Using the results of Problem 4–1, we find 1 1 1 = + k k1 k2

or

k=

k1k2 k1 + k2

(8–15)

for two springs in series. From Equation (4–4), the spring rates of the threaded and unthreaded portions of the bolt in the clamped zone are, respectively, where  At lt Ad ld

kt = = = = =

At E lt

kd =

Ad E ld

(8–16)

tensile-stress area (Tables 8–1, 8–2) length of threaded portion of grip major-diameter area of fastener length of unthreaded portion in grip

Substituting these stiffnesses in Equation (8–15) gives

kb =

Ad A t E Ad l t + At ld

(8–17)

where kb is the estimated effective stiffness of the bolt or cap screw in the clamped zone. For short fasteners, the one in Figure 8–14, for example, the unthreaded area is small and so the first of the expressions in Equation (8–16) can be used to find kb. For long fasteners, the threaded area is relatively small, and so the second expression in Equation (8–16) can be used. Table 8–7 is useful.

8–5  Joints—Member Stiffness In the previous section, we determined the stiffness of the fastener in the clamped zone. In this section, we wish to study the stiffnesses of the members in the clamped zone. Both of these stiffnesses must be known in order to learn what happens when the assembled connection is subjected to an external tensile loading. There may be more than two members included in the grip of the fastener. All together these act like compressive springs in series, and hence the total spring rate of the members is

1 1 1 1 1 = + + +…+ km k1 k2 k3 ki

(8–18)

438      Mechanical Engineering Design

Table 8–7  Suggested Procedure for Finding Fastener Stiffness lt

ld

h t1

t H t

t2

d

d

lt

LT l

l

LT

ld

L

L (a)

(b)

Given fastener diameter d and pitch p in mm or number of threads per inch Washer thickness:  t  from Table A–32 or A–33 Nut thickness [Figure (a) only]: H  from Table A–31 Grip length:    For Figure (a):  l = thickness of all material squeezed between face of bolt and face of nut h + t2∕2,   For Figure (b):  l = { h + d∕2,

t2 < d t2 ≥ d

Fastener length (round up using Table A–17*):    For Figure (a):  L > l + H    For Figure (b):  L > h + 1.5d Threaded length LT:  Inch series:

2d + 14 in, LT = { 2d + 12 in,

Metric series:

 2d + 6 mm,  L T =  2d + 12 mm,   2d + 25 mm,

Length of unthreaded portion in grip: Length of threaded portion in grip: Area of unthreaded portion: Area of threaded portion: Fastener stiffness:

L ≤ 6 in L > 6 in L ≤ 125 mm, d ≤ 48 mm 125 < L ≤ 200 mm L > 200 mm

ld = L − LT lt = l − ld Ad = πd 2∕4 At from Table 8–1 or 8–2 A d At E kb = Ad lt + At ld

*Bolts and cap screws may not be available in all the preferred lengths listed in Table A–17. Large fasteners may not be available in fractional inches or in millimeter lengths ending in a nonzero digit. Check with your bolt supplier for availability.

Screws, Fasteners, and the Design of Nonpermanent Joints     439

Figure 8–15

D

x

α y

dw t d

l 2

y

x

t dx

d x

(a)

(b)

If one of the members is a soft gasket, its stiffness relative to the other members is usually so small that for all practical purposes the others can be neglected and only the gasket stiffness used. If there is no gasket, the stiffness of the members is rather difficult to obtain, except by experimentation, because the compression region spreads out between the bolt head and the nut and hence the area is not uniform. There are, however, some cases in which this area can be determined. Ito2 has used ultrasonic techniques to determine the pressure distribution at the member interface. The results show that the pressure stays high out to about 1.5 bolt radii. The pressure, however, falls off farther away from the bolt. Thus Ito suggests the use of Rotscher's pressure-cone method for stiffness calculations with a variable cone angle. This method is quite complicated, and so here we choose to use a simpler approach using a fixed cone angle. Figure 8–15 illustrates the general cone geometry using a half-apex angle α. An angle α = 45° has been used, but Little3 reports that this overestimates the clamping stiffness. When loading is restricted to a washer-face annulus (hardened steel, cast iron, or aluminum), the proper apex angle is smaller. Osgood4 reports a range of 25°  ≤ α ≤ 33° for most combinations. In this book we shall use α = 30° except in cases in which the material is insufficient to allow the frusta to exist. Referring now to Figure 8–15b, the contraction of an element of the cone of thickness dx subjected to a compressive force P is, from Equation (4–3),

dδ =

P dx EA

(a)

The area of the element is D 2 d 2 A = π(r 2o − r 2i ) = π[(x tan α + ) − ( ) ] 2 2

= π (x tan α +

D+d D−d x tan α + 2 )( 2 )

(b)

Substituting this in Equation (a) and integrating gives a total contraction of 2

δ=

P πE

t

dx ∫ [x tan α + (D + d)∕2][x tan α + (D − d)∕2] 0

Y. Ito, J. Toyoda, and S. Nagata, "Interface Pressure Distribution in a Bolt-Flange Assembly," ASME paper no. 77-WA/DE-11, 1977. 3 R. E. Little, "Bolted Joints: How Much Give?" Machine Design, November 9, 1967. 4 C. C. Osgood, "Saving Weight on Bolted Joints," Machine Design, October 25, 1979.

(c)

Compression of a member with the equivalent elastic properties represented by a frustum of a hollow cone. Here, l represents the grip length.

440      Mechanical Engineering Design

Using a table of integrals, we find the result to be

δ=

(2t tan α + D − d) (D + d) P ln πEd tan α (2t tan α + D + d) (D − d)

(d)

Thus the spring rate or stiffness of this frustum is

k=

P = δ

πEd tan α (2t tan α + D − d)(D + d) ln (2t tan α + D + d)(D − d)

(8–19)

With α = 30°, this becomes

k=

0.5774π Ed (1.155t + D − d) (D + d) ln (1.155t + D + d) (D − d)

(8–20)

Equation (8–20), or (8–19), must be solved separately for each frustum in the joint. Then individual stiffnesses are assembled to obtain km using Equation (8–18). If the members of the joint have the same Young's modulus E with symmetrical frusta back to back, then they act as two identical springs in series. From Equation (8–18) we learn that km = k∕2. Using the grip as l = 2t and dw as the diameter of the washer face, from Equation (8–19) we find the spring rate of the members to be

km =

π Ed tan α (l tan α + dw − d)(dw + d) 2 ln (l tan α + dw + d)(dw − d)

(8–21)

The diameter of the washer face is about 50 percent greater than the fastener diameter for standard hexagon-head bolts and cap screws. Thus we can simplify Equation (8–21) by letting dw = 1.5d. If we also use α = 30°, then Equation (8–21) can be written as

km =

0.5774π Ed 0.5774l + 0.5d 2 ln (5 0.5774l + 2.5d )

(8–22)

It is easy to program the numbered equations in this section, and you should do so. The time spent in programming will save many hours of formula plugging. To see how good Equation (8–21) is, solve it for km∕Ed: km = Ed

π tan α (l tan α + dw − d)(dw + d) 2 ln [ (l tan α + dw + d)(dw − d) ]

Earlier in the section use of α = 30° was recommended for hardened steel, cast iron, or aluminum members. Wileman, Choudury, and Green5 conducted a finite element study of this problem. The results, which are depicted in Figure 8–16, agree with the α = 30° recommendation, coinciding exactly at the aspect ratio d∕l = 0.4. Additionally, they offered an exponential curve-fit of the form 5

km = A exp (Bd∕l) Ed

(8–23)

J. Wileman, M. Choudury, and I. Green, "Computation of Member Stiffness in Bolted Connections," Trans. ASME, J. Mech. Design, vol. 113, December 1991, pp. 432–437.

Screws, Fasteners, and the Design of Nonpermanent Joints     441 3.4

Figure 8–16

3.2

The dimensionless plot of stiffness versus aspect ratio of the members of a bolted joint, showing the relative accuracy of methods of Rotscher, Mischke, and Motosh, compared to a finite-element analysis (FEA) conducted by Wileman, Choudury, and Green.

3.0

Dimensionless stiffness, k m ⁄ Ed

2.8 2.6 2.4 2.2 2.0 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4

0.1

0.3

0.5

0.7

0.9

1.1

1.3

1.5

1.7

1.9

Aspect ratio, d ⁄ l FEA

Mischke 45°

Rotscher

Mischke 30°

Motosh

Table 8–8  Stiffness Parameters of Various Member Materials† Elastic Modulus

Poisson Ratio

GPa

Mpsi

Steel

0.291

207

30.0

0.787 15

0.628 73

Aluminum

0.334

71

10.3

0.796 70

0.638 16

Copper

0.326

119

17.3

0.795 68

0.635 53

Gray cast iron

0.211

100

14.5

0.778 71

0.616 16

0.789 52

0.629 14

Material Used

General expression

A

B

Source: Data from J. Wileman, M. Choudury, and I. Green, "Computation of Member Stiffness in Bolted Connections," Trans. ASME, J. Mech. Design, vol. 113, December 1991, pp. 432–437.

with constants A and B defined in Table 8–8. Equation (8–23) offers a simple calculation for member stiffness km. However, it is very important to note that the entire joint must be made up of the same material. For departure from these conditions, Equation (8–20) remains the basis for approaching the problem. EXAMPLE 8–2 As shown in Figure 8–17a, two plates are clamped by washer-faced 12 in-20 UNF × 112 in SAE grade 5 bolts each with a standard 12 N steel plain washer. (a) Determine the member spring rate km if the top plate is steel and the bottom plate is gray cast iron. (b) Using the method of conical frusta, determine the member spring rate km if both plates are steel. (c) Using Equation (8–23), determine the member spring rate km if both plates are steel. Compare the results with part (b). (d) Determine the bolt spring rate kb.

442      Mechanical Engineering Design

Figure 8–17 Dimensions in inches.

1.437 0.75

0.095

1 2

0.6725 0.0775

3 4

0.6725

1.527 (b)

(a)

Solution From Table A–32, the thickness of a standard 12 N plain washer is 0.095 in. (a) As shown in Figure 8–17b, the frusta extend halfway into the joint the distance 1 (0.5 + 0.75 + 0.095) = 0.6725 in 2 The distance between the joint line and the dotted frusta line is 0.6725 − 0.5 − 0.095 = 0.0775 in. Thus, the top frusta consist of the steel washer, steel plate, and 0.0775 in of the cast iron. Since the washer and top plate are both steel with E = 30(106) psi, they can be considered a single frustum of 0.595 in thick. The outer diameter of the frustum of the steel member at the joint interface is 0.75 + 2(0.595) tan 30° = 1.437 in. The outer diameter at the midpoint of the entire joint is 0.75 + 2(0.6725) tan 30° = 1.527 in. Using Equation (8–20), the spring rate of the steel is k1 =

0.5774π(30)(106 )0.5 = 30.80(106 ) lbf/in [1.155(0.595) + 0.75 − 0.5](0.75 + 0.5) ln { [1.155(0.595) + 0.75 + 0.5](0.75 − 0.5)}

From Tables 8–8 or A–5, for gray cast iron, E = 14.5 Mpsi. Thus for the upper cast-iron frustum k2 =

0.5774π(14.5)(106 )0.5 = 285.5(106 ) lbf/in [1.155(0.0775) + 1.437 − 0.5](1.437 + 0.5) ln { [1.155(0.0775) + 1.437 + 0.5](1.437 − 0.5)}

For the lower cast-iron frustum k3 =

0.5774π(14.5)(106 )0.5 = 14.15(106 ) lbf/in [1.155(0.6725) + 0.75 − 0.5](0.75 + 0.5) ln { [1.155(0.6725) + 0.75 + 0.5](0.75 − 0.5)}

The three frusta are in series, so from Equation (8–18) 1 1 1 1 = + + 6 6 km 30.80(10 ) 285.5(10 ) 14.15(106 ) Answer This results in km = 9.378 (106) lbf/in. (b) If the entire joint is steel, Equation (8–22) with l = 2(0.6725) = 1.345 in gives

Screws, Fasteners, and the Design of Nonpermanent Joints     443

km =

Answer

0.5774π(30.0)(106 )0.5 = 14.64(106 ) lbf/in. 0.5774(1.345) + 0.5(0.5) 2 ln {5 [ 0.5774(1.345) + 2.5(0.5) ]}

(c) From Table 8–8, A = 0.787 15, B = 0.628 73. Equation (8–23) gives Answer

km = 30(106 )(0.5) (0.787 15) exp[0.628 73(0.5)∕1.345] = 14.92(106 ) lbf/in

For this case, the difference between the results for Equations (8–22) and (8–23) is less than 2 percent. (d) Following the procedure of Table 8–7, the threaded length of a 0.5-in bolt is LT = 2(0.5) + 0.25 = 1.25 in. The length of the unthreaded portion is ld = 1.5 − 1.25 = 0.25 in. The length of the unthreaded portion in grip is lt = 1.345 − 0.25 = 1.095 in. The major diameter area is Ad = (π∕4)(0.52) = 0.196 3 in2. From Table 8–2, the tensile-stress area is At = 0.159 9 in2. From Equation (8–17) Answer

kb =

0.196 3(0.159 9)30(106 ) = 3.69(106 ) lbf/in 0.196 3(1.095) + 0.159 9(0.25)

8–6  Bolt Strength In the specification standards for bolts, the strength is specified by stating SAE or ASTM minimum quantities, the minimum proof strength, or minimum proof load, and the minimum tensile strength. The proof load is the maximum load (force) that a bolt can withstand without acquiring a permanent set. The proof strength is the quotient of the proof load and the tensile-stress area. The proof strength thus corresponds roughly to the proportional limit and corresponds to 0.0001-in permanent set in the fastener (first measurable deviation from elastic behavior). Tables 8–9, 8–10, and 8–11 provide minimum strength specifications for steel bolts. The values of the mean proof strength, the mean tensile strength, and the corresponding standard deviations are not part of the specification codes, so it is the designer's responsibility to obtain these values, perhaps by laboratory testing, if designing to a reliability specification. The SAE specifications are found in Table 8–9. The bolt grades are numbered according to the tensile strengths, with decimals used for variations at the same strength level. Bolts and screws are available in all grades listed. Studs are available in grades 1, 2, 4, 5, 8, and 8.1. Grade 8.1 is not listed. ASTM specifications are listed in Table 8–10. ASTM threads are shorter because ASTM deals mostly with structures; structural connections are generally loaded in shear, and the decreased thread length provides more shank area. Specifications for metric fasteners are given in Table 8–11. Specifications-grade bolts usually bear a manufacturer's mark or logo, in addition to the grade marking, on the bolt head. Such marks confirm that the bolt meets or exceeds specifications. If such marks are missing, assume the bolt strength is unregulated, or is relatively low and not intended for engineering applications. Bolts in fatigue axial loading fail at the fillet under the head, at the thread runout, and at the first thread engaged in the nut. If the bolt has a standard shoulder under the head, it has a value of Kf from 2.1 to 2.3, and this shoulder fillet is protected from scratching or scoring by a washer. If the thread runout has a 15° or less half-cone angle, the stress is higher at the first engaged thread in the nut. Bolts

444      Mechanical Engineering Design

Table 8–9  SAE Specifications for Steel Bolts SAE Grade No.

Size Range Inclusive, in

1 1 1   4 –12

Minimum Minimum Minimum Proof Tensile Yield Strength,* Strength,* Strength,* kpsi kpsi kpsi 36

Low or medium carbon

1 3 2   55 74 57 4 –4 7 1   – 33 60 36 1 8 2

Low or medium carbon

1 1 4   4 –12

33

65

60

Material

115

100

1 5   85 120 92 4 –1 1 18 –112 74 105 81

1 5.2   4 –1

Head Marking

Medium carbon, cold-drawn

Medium carbon, Q&T

Low-carbon martensite, Q&T

85

120

92

1 1 7   4 –12

105

133

115

Medium-carbon alloy, Q&T

1 1 8   4 –12

120

150

130

Medium-carbon alloy, Q&T

1 8.2   4 –1

120

150

130

Low-carbon martensite, Q&T

*Minimum strengths are strengths exceeded by 99 percent of fasteners.

are sized by examining the loading at the plane of the washer face of the nut. This is the weakest part of the bolt if and only if the conditions above are satisfied (washer protection of the shoulder fillet and thread runout ≤15°). Inattention to this requirement has led to a record of 15 percent fastener fatigue failure under the head, 20 percent at thread runout, and 65 percent where the designer is focusing attention. It does little good to concentrate on the plane of the nut washer face if it is not the weakest location. Nuts are graded so that they can be mated with their corresponding grade of bolt. The purpose of the nut is to have its threads deflect to distribute the load of the bolt more evenly to the nut. The nut's properties are controlled in order to accomplish this. The grade of the nut should be the grade of the bolt.

Screws, Fasteners, and the Design of Nonpermanent Joints     445

Table 8–10  ASTM Specifications for Steel Bolts ASTM Size Minimum Minimum Minimum Desig- Range, Proof Tensile Yield nation Inclusive, Strength,* Strength,* Strength,* No. in kpsi kpsi kpsi 1 1 A307   4 –12

33

60

36

A325,   12 –1 85 120 92 type 1 1 1 18 –12 74 105 81

Material

Head Marking

Low carbon

Medium carbon, Q&T A325

A325,   12 –1 85 120 92 Low-carbon, martensite, type 2 Q&T 1 1 18 –12 74 105 81

A325

A325,   12 –1 85 120 92 Weathering steel, type 3 Q&T 1 1 18 –12 74 105 81

A325

A354,   14 –221 105 125 109 grade BC 3 24 – 4 95 115 99

Alloy steel, Q&T

A354,   14 – 4 grade BD

Alloy steel, Q&T

120

150

130

85

120

92

1 A449   4 –1

118 –112

134 –3 55 90 58

A490,   12 –121 type 1

BC

Medium-carbon, Q&T

74 105 81

120

150

130

A490,   12 –121 120 150 130 type 3

*Minimum strengths are strengths exceeded by 99 percent of fasteners.

Alloy steel, Q&T A490

Weathering steel, Q&T

A490

446      Mechanical Engineering Design

Table 8–11  Metric Mechanical-Property Classes for Steel Bolts, Screws, and Studs Minimum Minimum Minimum Size Proof Tensile Yield Property Range, Strength,* Strength,* Strength,* Class Inclusive MPa MPa MPa

4.6

M5–M36

225

400

240

Material

Head Marking

Low or medium carbon 4.6

4.8

M1.6–M16

310

420

340

Low or medium carbon 4.8

5.8

M5–M24

380

520

420

Low or medium carbon 5.8

8.8

M16–M36

600

830

660

Medium carbon, Q&T 8.8

9.8

M1.6–M16

650

900

720

Medium carbon, Q&T 9.8

10.9 M5–M36 830 1040 940 Low-carbon martensite, Q&T

12.9

M1.6–M36

970

1220

1100

10.9

Alloy, Q&T 12.9

*Minimum strengths are strengths exceeded by 99 percent of fasteners.

8–7  Tension Joints—The External Load Let us now consider what happens when an external tensile load P is applied to a bolted connection, as shown in Figure 8–18. The nomenclature used is: Fi = preload Ptotal = Total external tensile load applied to the joint P = external tensile load per bolt Pb = portion of P taken by bolt Pm = portion of P taken by members Fb = Pb + Fi = resultant bolt load Fm = Pm − Fi = resultant load on members C = fraction of external load P carried by bolt 1 − C = fraction of external load P carried by members N = Number of bolts in the joint

Screws, Fasteners, and the Design of Nonpermanent Joints     447

If N bolts equally share the total external load, then P = Ptotal∕N

(a)

In Figure 8–18a, a bolted joint is shown where the nut is in contact with the members, but not yet tightened to introduce a preload. The members and the bolt can be modeled as springs in parallel. The figure schematically represents the members with a large stiff spring and the bolt with a low-stiffness spring, each starting in an undeformed equilibrium condition. In Figure 8–18b, the nut is tightened to introduce a preload Fi to the joint. In doing so, the stiff spring of the members is compressed by an amount δm = Fi∕km , while the soft spring of the bolt is stretched by an amount δb = Fi∕kb . Figure 8–19 shows a plot of these forcedeflection characteristics, where the linear spring stiffness relationships are evident. Note that the ordinate ∣F∣ represents the absolute value of the force in each member, where the force in the bolt is positive, and the force in the members is negative. As both the bolt and the members experience the preload Fi, the member loading moves up its stiffness line to point a, while the bolt loading moves to point b. Since the member stiffness is usually much greater than the bolt stiffness, this results in δm being significantly smaller than δb . When an external tensile load P is applied to the members (not directly to the bolt), the bolt elongates further by Δδb , whereas the members' deflection decreases by Δδm , as shown in Figure 8–18c. By inspection of the figure, it can be realized that these deflections must be equal. From Figure 8–19, observe that the external load P is split into two portions, Pb applied to the bolt and Pm applied to the members, apportioned as necessary to ensure that Δδb = Δδm . Thus,

Δδb =

Pb Pm = Δδm = kb km

(b)

km Pb kb

(c)

or,

Pm =

P

∣F∣

Member

Bolt

km

d

Fb a

Pb

Fi Pm c

b P

Fm

Δδm

δm

P Δδb

δb

Δδm

O

δm (a) No preload

kb

(b) Preload

(c) External load

Figure 8–18 Spring model of the effect of preload and external tensile load to a bolted joint.

Δδb

+

δb

Figure 8–19 Force-deflection characteristics of an externally loaded bolted member with an initial bolt preload force, Fi. The external load is P.

δ

448      Mechanical Engineering Design

Table 8–12  An example of Bolt and Member Stiffnesses. Steel members clamped using a

1 2

in-13 NC steel bolt. C =

kb kb + km

Stiffnesses, M lbf/in kb

km

C

1−C

2

2.57

12.69

0.168

0.832

3

1.79

11.33

0.136

0.864

4

1.37

10.63

0.114

0.886

Bolt Grip, in

Since P = Pb + Pm, then, Pm = P − Pb. Substituting this into Equation (c) and rearranging we find

Pb =

kb P = CP kb + km

(d)

and

Pm = P − Pb = (1 − C)P

(e)

where

C=

kb kb + km

(f )

is called the stiffness constant of the joint. The resultant bolt load is

Fb = Pb + Fi = CP + Fi

Fm < 0

(8–24)

and the resultant load on the connected members is

Fm = Pm − Fi = (1 − C)P − Fi

Fm < 0

(8–25)

Of course, these results are valid only as long as some clamping load remains in the members; this is indicated by the qualifier in the equations. Returning to Figure 8–19, Equation 8–24 represents the preloading of the bolt to point b, followed by the movement to point d due to the application of Pb to the bolt. Similarly, Equation 8–25 represents the compressive preloading of the members to point a, followed by the movement to point c due to the application of Pm to the members. Table 8–12 is included to provide some information on the relative values of the stiffnesses encountered. The grip contains only two members, both of steel, and no washers. The ratios C and 1 − C are the coefficients of P in Equations (8–24) and (8–25), respectively. They describe the proportion of the external load taken by the bolt and by the members, respectively. In all cases, the members take over 80 percent of the external load. Think how important this is when fatigue loading is present. Note also that making the grip longer causes the members to take an even greater percentage of the external load.

8–8  Relating Bolt Torque to Bolt Tension Having learned that a high preload is very desirable in important bolted connections, we must next consider means of ensuring that the preload is actually developed when the parts are assembled.

Screws, Fasteners, and the Design of Nonpermanent Joints     449

If the overall length of the bolt can actually be measured with a micrometer when it is assembled, the bolt elongation due to the preload Fi can be computed using the formula δ = Fil∕(AE). Then the nut is simply tightened until the bolt elongates through the distance δ. This ensures that the desired preload has been attained. The elongation of a screw cannot usually be measured, because the threaded end is often in a blind hole. It is also impractical in many cases to measure bolt elongation. In such cases the wrench torque required to develop the specified preload must be estimated. Then torque wrenching, pneumatic-impact wrenching, or the turn-ofthe-nut method may be used. The torque wrench has a built-in dial that indicates the proper torque. With impact wrenching, the air pressure is adjusted so that the wrench stalls when the proper torque is obtained, or in some wrenches, the air automatically shuts off at the desired torque. The turn-of-the-nut method requires that we first define the meaning of snugtight. The snug-tight condition is the tightness attained by a few impacts of an impact wrench, or the full effort of a person using an ordinary wrench. When the snug-tight condition is attained, all additional turning develops useful tension in the bolt. The turn-of-the-nut method requires that you compute the fractional number of turns necessary to develop the required preload from the snug-tight condition. For example, for heavy hexagonal structural bolts, the turn-of-the-nut specification states that the nut should be turned a minimum of 180° from the snug-tight condition under optimum conditions. Problems 8–26 to 8–28 illustrate the method further. Although the coefficients of friction may vary widely, we can obtain a good estimate of the torque required to produce a given preload by combining Equations (8–5) and (8–6):

T=

Fi fc dc Fi dm l + π f dm sec α + 2 ( πdm − f l sec α ) 2

(a)

where dm is the average of the major and minor diameters. Since tan λ = l∕πdm, we divide the numerator and denominator of the first term by πdm and get

T=

Fi fc dc Fi dm tan λ + f sec α + 2 ( 1 − f tan λ sec α ) 2

(b)

The diameter of the washer face of a hexagonal nut is the same as the width across flats and equal to 112 times the nominal size. Therefore the mean collar diameter is dc = (d + 1.5d)∕2 = 1.25d. Equation (b) can now be arranged to give

tan λ + f sec α dm T = [( )( + 0.625 fc ]Fi d 2d 1 − f tan λ sec α )

(c)

We now define a torque coefficient K as the term in brackets, and so

K=(

tan λ + f sec α dm + 0.625 fc 2d )( 1 − f tan λ sec α )

(8–26)

Equation (c) can now be written

T = KFi d

(8–27)

450      Mechanical Engineering Design

The coefficient of friction depends upon the surface smoothness, accuracy, and degree of lubrication. On the average, both f and fc are about 0.15. The interesting fact about Equation (8–26) is that K ≈ 0.20 for f = fc = 0.15 no matter what size bolts are employed and no matter whether the threads are coarse or fine. Blake and Kurtz have published results of numerous tests of the torquing of bolts.6 By subjecting their data to a statistical analysis, we can learn something about the distribution of the torque coefficients and the resulting preload. Blake and Kurtz determined the preload in quantities of unlubricated and lubricated bolts of size 12 in-20 UNF when torqued to 800 lbf · in. This corresponds roughly to an M12 × 1.25 bolt torqued to 90 N · m. The statistical analyses of these two groups of bolts, converted to SI units, are displayed in Tables 8–13 and 8–14. We first note that both groups have about the same mean preload, 34 kN. The unlubricated bolts have a standard deviation of 4.9 kN. The lubricated bolts have a standard deviation of 3 kN. The means obtained from the two samples are nearly identical, approximately 34 kN; using Equation (8–27), we find, for both samples, K = 0.208. Bowman Distribution, a large manufacturer of fasteners, recommends the values shown in Table 8–15. In this book we shall use these values and use K = 0.2 when the bolt condition is not stated. Table 8–13  Distribution of Preload Fi for 20 Tests of Unlubricated Bolts Torqued to 90 N · m 23.6 27.6 28.0 29.4 30.3 30.7 32.9 33.8 33.8 33.8 34.7 35.6 35.6 37.4 37.8 37.8 39.2 40.0 40.5 42.7 Mean value Fi = 34.3 kN. Standard deviation, σˆ = 4.91 kN.

Table 8–14  Distribution of Preload Fi for 10 Tests of Lubricated Bolts Torqued to 90 N · m 30.3 32.5 32.5 32.9 32.9 33.8 34.3 34.7 37.4 40.5 Mean value, Fi = 34.18 kN. Standard deviation, σˆ = 2.88 kN.

Table 8–15  Torque Factors K for Use with Equation (8–27) Bolt Condition

6

K

Nonplated, black finish

0.30

Zinc-plated

0.20

Lubricated

0.18

Cadmium-plated

0.16

With Bowman Anti-Seize

0.12

With Bowman-Grip nuts

0.09

J. C. Blake and H. J. Kurtz, "The Uncertainties of Measuring Fastener Preload," Machine Design, vol. 37, September 30, 1965, pp. 128–131.

Screws, Fasteners, and the Design of Nonpermanent Joints     451

EXAMPLE 8–3 A 34 in-16 UNF × 212 in SAE grade 5 bolt is subjected to a load P of 6 kip in a tension joint. The initial bolt tension is Fi = 25 kip. The bolt and joint stiffnesses are kb = 6.50 and km = 13.8 Mlbf/in, respectively. (a) Determine the preload and service load stresses in the bolt. Compare these to the SAE minimum proof strength of the bolt. (b) Specify the torque necessary to develop the preload, using Equation (8–27). (c) Specify the torque necessary to develop the preload, using Equation (8–26) with f = fc = 0.15. Solution From Table 8–2, At = 0.373 in2. (a) The preload stress is σi =

Answer

Fi 25 = = 67.02 kpsi At 0.373

The stiffness constant is C=

kb 6.5 = = 0.320 kb + km 6.5 + 13.8

From Equation (8–24), the stress under the service load is σb =

Answer

Fb CP + Fi P = = C + σi At At At

= 0.320

6 + 67.02 = 72.17 kpsi 0.373

From Table 8–9, the SAE minimum proof strength of the bolt is Sp = 85 kpsi. The preload and service load stresses are respectively 21 and 15 percent less than the proof strength. (b) From Equation (8–27), the torque necessary to achieve the preload is T = KFi d = 0.2(25) (103 )(0.75) = 3750 lbf · in

Answer

(c) The minor diameter can be determined from the minor area in Table 8–2. Thus dr = √4Ar∕π = √4(0.351)∕π = 0.6685 in. Thus, the mean diameter is dm = (0.75 + 0.6685)∕2 = 0.7093 in. The lead angle is λ = tan−1

l 1 1 = tan−1 = tan−1 = 1.6066° πdm πd m N π(0.7093)(16)

For α = 30°, Equation (8–26) gives tan 1.6066° + 0.15(sec 30°) 0.7093 T = {[ + 0.625(0.15)}25(103 ) (0.75) ][ 2(0.75) 1 − 0.15(tan 1.6066°)(sec 30°) ] = 3551 lbf · in which is 5.3 percent less than the value found in part (b).

452      Mechanical Engineering Design

8–9  Statically Loaded Tension Joint with Preload Equations (8–24) and (8–25) represent the forces in a bolted joint with preload. The tensile stress in the bolt can be found as in Example 8–3 as

σb =

Fb CP + Fi = At At

(a)

Thus, the yielding factor of safety guarding against the static stress exceeding the proof strength is

np =

Sp Sp = σb (CP + Fi )∕At

(b)

or

np =

Sp At CP + Fi

(8–28)

Since it is common to load a bolt close to the proof strength, the yielding factor of safety is often not much greater than unity. Another indicator of yielding that is sometimes used is a load factor, which is applied only to the load P as a guard against overloading. Applying such a load factor, nL, to the load P in Equation (a), and equating it to the proof strength gives

CnLP + Fi = Sp (c) At

Solving for the load factor gives

nL =

Sp At − Fi CP

(8–29)

It is also essential for a safe joint that the external load be smaller than that needed to cause the joint to separate. If separation does occur, then the entire external load will be imposed on the bolt. Let P0 be the value of the external load that would cause joint separation. At separation, Fm = 0 in Equation (8–25), and so

(1 − C) P0 − Fi = 0

(d)

Let the factor of safety against joint separation be

n0 =

P0 P

(e)

Substituting P0 = n0P in Equation (d), we find

n0 =

Fi P(1 − C)

(8–30)

as a load factor guarding against joint separation. Figure 8–20 is the stress-strain diagram of a good-quality bolt material. Notice that there is no clearly defined yield point and that the diagram progresses smoothly up to fracture, which corresponds to the tensile strength. This means that no matter how much preload is given the bolt, it will retain its load-carrying capacity. This is what keeps the bolt tight and determines the joint strength. The pretension is the "muscle" of the joint,

Screws, Fasteners, and the Design of Nonpermanent Joints     453 Sut

Typical stress-strain diagram for bolt materials showing proof strength Sp, yield strength Sy, and ultimate tensile strength Sut.

Sy

Stress

Sp

Strain

and its magnitude is determined by the bolt strength. If the full bolt strength is not used in developing the pretension, then money is wasted and the joint is weaker. Good-quality bolts can be preloaded into the plastic range to develop more strength. Some of the bolt torque used in tightening produces torsion, which increases the principal tensile stress. However, this torsion is held only by the friction of the bolt head and nut; in time it relaxes and lowers the bolt tension slightly. Thus, as a rule, a bolt will either fracture during tightening, or not at all. Above all, do not rely too much on wrench torque; it is not a good indicator of preload. Actual bolt elongation should be used whenever possible—especially with fatigue loading. In fact, if high reliability is a requirement of the design, then preload should always be determined by bolt elongation. Russell, Burdsall & Ward Inc. (RB&W) recommendations for preload are 60 kpsi for SAE grade 5 bolts for nonpermanent connections, and that A325 bolts (equivalent to SAE grade 5) used in structural applications be tightened to proof load or beyond (85 kpsi up to a diameter of 1 in).7 Bowman8 recommends a preload of 75 percent of proof load, which is about the same as the RB&W recommendations for reused bolts. In view of these guidelines, it is recommended for both static and fatigue loading that the following be used for preload: 0.75Fp Fi = { 0.90Fp

for nonpermanent connections, reused fasteners for permanent connections

(8–31)

where Fp is the proof load, obtained from the equation

Fp = At Sp

(8–32)

Here Sp is the proof strength obtained from Tables 8–9 to 8–11. For other materials, an approximate value is Sp = 0.85Sy. Be very careful not to use a soft material in a threaded fastener. For high-strength steel bolts used as structural steel connectors, if advanced tightening methods are used, tighten to yield. 7

Russell, Burdsall & Ward Inc., Helpful Hints for Fastener Design and Application, Mentor, Ohio, 1965, p. 42. Bowman Distribution–Barnes Group, Fastener Facts, Cleveland, 1985, p. 90.

8

Figure 8–20

454      Mechanical Engineering Design

You can see that the RB&W recommendations on preload are in line with what we have encountered in this chapter. The purposes of development were to give the reader the perspective to appreciate Equations (8–31) and a methodology with which to handle cases more specifically than the recommendations.

EXAMPLE 8–4 Figure 8–21 is a cross section of a grade 25 cast-iron pressure vessel. A total of N bolts are to be used to resist a separating force of 36 kip. (a) Determine kb, km, and C. (b) Find the number of bolts required for a load factor of 2 where the bolts may be reused when the joint is taken apart. (c) With the number of bolts obtained in part (b), determine the realized load factor for overload, the yielding factor of safety, and the load factor for joint separation. Solution (a) The grip is l = 1.50 in. From Table A–31, the nut thickness is of 112 in gives a bolt length of L=

35 64

in. Adding two threads beyond the nut

35 2 + 1.50 + = 2.229 in 64 11

From Table A–17 the next fraction size bolt is L = 214 in. From Equation (8–13), the thread length is LT = 2(0.625) + 0.25 = 1.50 in. Thus, the length of the unthreaded portion in the grip is ld = 2.25 − 1.50 = 0.75 in. The threaded length in the grip is lt = l − ld = 0.75 in. From Table 8–2, At = 0.226 in2. The majordiameter area is Ad = π(0.625)2∕4 = 0.3068 in2. The bolt stiffness is then kb =

Answer

Ad At E 0.3068(0.226)(30) = Ad lt + At ld 0.3068(0.75) + 0.226(0.75)

= 5.21 Mlbf/in

From Table A–24, for no. 25 cast iron we will use E = 14 Mpsi. The stiffness of the members, from Equation (8–22), is

Figure 8–21

5 8

in-11 UNC × 2 14 in grade 5 finished hex head bolt No. 25 CI 3 4

in

3 4

in

Screws, Fasteners, and the Design of Nonpermanent Joints     455

Answer

km =

0.5774π(14) (0.625) 0.5774πEd = 0.5774(1.5) + 0.5(0.625) 0.5774l + 0.5d 2 ln (5 2 ln[ 5 0.5774l + 2.5d ) 0.5774(1.5) + 2.5(0.625) ]

= 8.95 Mlbf/in

If you are using Eq. (8–23), from Table 8–8, A = 0.778 71 and B = 0.616 16, and

k m = EdA exp(Bd∕l)

= 14(0.625) (0.778 71) exp[0.616 16(0.625)∕1.5]

= 8.81 Mlbf/in

which is only 1.6 percent lower than the previous result. From the first calculation for km, the stiffness constant C is C=

Answer

kb 5.21 = = 0.368 k b + k m 5.21 + 8.95

(b) From Table 8–9, Sp = 85 kpsi. Then, using Equations (8–31) and (8–32), we find the recommended preload to be Fi = 0.75A t Sp = 0.75(0.226)(85) = 14.4 kip For N bolts, Equation (8–29) can be written

nL =

Sp A t − Fi C(Ptotal∕N)

or N=

CnL Ptotal 0.368(2) (36) = = 5.52 Sp At − Fi 85(0.226) − 14.4

Answer Six bolts should be used to provide the specified load factor. (c) With six bolts, the load factor actually realized is nL =

Answer

85(0.226) − 14.4 = 2.18 0.368(36∕6)

From Equation (8–28), the yielding factor of safety is Answer

np =

Sp At C(Ptotal∕N) + Fi

=

85(0.226) = 1.16 0.368(36∕6) + 14.4

From Equation (8–30), the load factor guarding against joint separation is Answer

n0 =

Fi 14.4 = = 3.80 (Ptotal∕N) (1 − C) (36∕6) (1 − 0.368)

(1)

456      Mechanical Engineering Design

8–10  Gasketed Joints If a full gasket of area Ag is present in the joint, the gasket pressure p is found by dividing the force in the member by the gasket area per bolt. Thus, for N bolts,

p=−

Fm Ag∕N

(a)

With a load factor n, Equation (8–25) can be written as

Fm = (1 − C) nP − Fi

(b)

Substituting this into Equation (a) gives the gasket pressure as

p = [Fi − nP(1 − C)]

N Ag

(8–33)

In full-gasketed joints uniformity of pressure on the gasket is important. To maintain adequate uniformity of pressure adjacent bolts should not be placed more than six nominal diameters apart on the bolt circle. To maintain wrench clearance, bolts should be placed at least three diameters apart. A rough rule for bolt spacing around a bolt circle is

3≤

πDb ≤ 6 (8–34) Nd

where Db is the diameter of the bolt circle and N is the number of bolts.

8–11  Fatigue Loading of Tension Joints Tension-loaded bolted joints subjected to fatigue action can be analyzed directly by the methods of Chapter 6. Table 8–16 lists average fatigue stress-concentration factors for the fillet under the bolt head and also at the beginning of the threads on the bolt shank. These are already corrected for notch sensitivity. Designers should be aware that situations may arise in which it would be advisable to investigate these factors more closely, since they are only average values. Peterson9 observes that the distribution of typical bolt failures is about 15 percent under the head, 20 percent at the end of the thread, and 65 percent in the thread at the nut face. Use of rolled threads is the predominant method of thread-forming in screw fasteners. In thread-rolling, the amount of cold work and strain-strengthening is unknown to the designer; therefore, fully corrected (including Kf) axial endurance strength is reported in Table 8–17. Since Kf is included as an endurance strength reducer in Table 8–17, it should not be applied as a stress increaser when using Table 8–16  Fatigue Stress-Concentration Factors Kf for Threaded Elements SAE Grade

9

Metric Grade

Rolled Threads

Cut Threads

Fillet

0 to 2

3.6 to 5.8

2.2

2.8

2.1

4 to 8

6.6 to 10.9

3.0

3.8

2.3

W. D. Pilkey and D. F. Pilkey, Peterson's Stress-Concentration Factors, 3rd ed., John Wiley & Sons, New York, 2008, p. 411.

Screws, Fasteners, and the Design of Nonpermanent Joints     457

Table 8–17  Fully Corrected Endurance Strengths for Bolts and Screws with Rolled Threads* Grade or Class SAE 5

Size Range 1 4 –1

Endurance Strength 18.6 kpsi

in

118 –121 in

16.3 kpsi

SAE 7

1 1 4 –12

in

20.6 kpsi

SAE 8

1 1 4 –12

in

23.2 kpsi

ISO 8.8

M16–M36

129 MPa

ISO 9.8

M1.6–M16

140 MPa

ISO 10.9

M5–M36

162 MPa

ISO 12.9

M1.6–M36

190 MPa

*Repeatedly applied, axial loading, fully corrected, including Kf as a strength reducer.

values from this table. For cut threads, the methods of Chapter 6 are useful. Anticipate that the endurance strengths will be considerably lower. For a general case with a constant preload, and an external load on a per bolt basis fluctuating between Pmin and Pmax, a bolt will experience fluctuating forces such that

Fbmin = CPmin + Fi

(a)

Fbmax = CPmax + Fi

(b)

The alternating stress experienced by a bolt is

σa =

(Fb max − Fbmin )∕2 (CPmax + Fi ) − (CPmin + Fi ) = At 2At

σa =

C(Pmax − Pmin ) 2At

(8–35)

The midrange stress experienced by a bolt is (Fbmax + Fbmin )∕2 (CPmax + Fi ) + (CPmin + Fi ) = At 2At

σm =

C(Pmax + Pmin ) Fi σm = + 2At At

(8–36)

A load line typically experienced by a bolt is shown in Figure 8–22, where the stress starts from the preload stress and increases with a constant slope of σa∕(σm − σi). The Goodman failure line is also shown in Figure 8–22. The fatigue factor of safety can be found by intersecting the load line and the Goodman line to find the intersection point (Sm, Sa). The load line is given by Load line:

Sa =

σa (S − σi ) σm − σi m

(a)

The Goodman line, rearranging Equation (6–40), is Goodman line:

Sa = Se −

Se Sm (b) Sut

458      Mechanical Engineering Design

Figure 8–22 Se

Load line Alternating stress σa

Fluctuating-stress diagram showing a Goodman failure line and a commonly used load line for a constant preload and a fluctuating load.

C

Sa B

σa

D Sm

A σi =

Fi At

σm

Sut

Steady stress σm

Equating Equations (a) and (b), solving for Sm, then substituting Sm back into Equation (b) yields Seσa (Sut − σi ) Sa = (c) Sut σa + Se (σm − σi ) The fatigue factor of safety is given by

nf =

Sa σa

(8–37)

Substituting Equation (c) into Equation (8–37) gives

nf =

Se (Sut − σi ) Sut σa + Se (σm − σi )

(8–38)

The same approach can be used for the other failure curves, though for any of the nonlinear failure curves the algebra is a bit more tedious to put in equation form such as Equation (8–38). An easier approach would be to solve in stages numerically, first Sm, then Sa, and finally nf. Special Case of Repeated Loading Often, the type of fatigue loading encountered in the analysis of bolted joints is one in which the externally applied load fluctuates between zero and some maximum force P. This would be the situation in a pressure cylinder, for example, where a pressure either exists or does not exist. For such cases, Equations (8–35) and (8–36) can be simplified by setting Pmax = P and Pmin = 0, resulting in

σa =

CP 2At

(8–39)

σm =

CP Fi + 2At At

(8–40)

Note that Equation (8–40) can be viewed as the sum of the alternating stress and the preload stress. If the preload is considered to be constant, the load line relationship between the alternating and midrange stresses can be treated as

σm = σa + σi

(8–41)

Screws, Fasteners, and the Design of Nonpermanent Joints     459

This load line has a slope of unity, and is a special case of the load line shown in Figure 8–22. With the simplifications in the algebra, we can now proceed as before to obtain the fatigue factor of safety using a few of the typical failure criteria, duplicated here from Equations (6–40), (6–47), and (6–51), in terms of the intersection point (Sm, Sa). Goodman: Sa Sm + = 1 Se Sut

(8–42)

Sa Sm 2 + ( ) = 1 Se Sut

(8–43)

Sa 2 Sm 2 + ( Se ) ( Sp ) = 1

(8–44)

Gerber: ASME-elliptic:

Now if we intersect Equation (8–41) and each of Equations (8–42) to (8–44) to solve for Sa, and apply Equation (8–37), we obtain fatigue factors of safety for each failure criteria in a repeated loading situation. Goodman:

nf =

Se (Sut − σi ) σa (Sut + Se )

(8–45)

Gerber:

nf =

1 [Sut √S 2ut + 4Se (Se + σi ) − S 2ut − 2σi Se] (8–46) 2σa Se

ASME-elliptic:

nf =

Se σa (S 2p

+ S 2e )

(Sp √S 2p + S 2e − σ 2i − σi Se ) (8–47)

Note that Equations (8–45) to (8–47) are only applicable for repeated loads. If Kf is being applied to the stresses, rather than to Se, be sure to apply it to both σa and σm. Otherwise, the slope of the load line will not remain 1 to 1. If desired, σa from Equation (8–39) and σi = Fi∕At can be directly substituted into any of Equations (8–45) to (8–47). If we do so for the Goodman criteria in Equation (8–45), we obtain 2Se (Sut At − Fi ) nf = (8–48) CP(Sut + Se ) when preload Fi is present. With no preload, C = 1, Fi = 0, and Equation (8–48) becomes 2Se Sut At nf 0 = (8–49) P(Sut + Se ) Preload is beneficial for resisting fatigue when nf∕nf 0 is greater than unity. For Goodman, Equations (8–48) and (8–49) with nf∕nf 0 ≥ 1 puts an upper bound on the preload Fi of

Fi ≤ (1 − C)Sut A t

(8–50)

460      Mechanical Engineering Design

If this cannot be achieved, and nf is unsatisfactory, use the Gerber or ASME-elliptic criterion to obtain a less conservative assessment. If the design is still not satisfactory, additional bolts and/or a different size bolt may be called for. Preload Considerations Bolts loosen, as they are friction devices, and cyclic loading and vibration as well as other effects allow the fasteners to lose tension with time. How does one fight loosening? Within strength limitations, the higher the preload the better. A rule of thumb is that preloads of 60 percent of proof load rarely loosen. If more is better, how much more? Well, not enough to create reused fasteners as a future threat. Alternatively, fastener-locking schemes can be employed. After solving for the fatigue factor of safety, you should also check the possibility of yielding, using the proof strength

np =

Sp σm + σa

(8–51)

which is equivalent to Equation (8–28). EXAMPLE 8–5 Figure 8–23 shows a connection using cap screws. The joint is subjected to a fluctuating force whose maximum value is 5 kip per screw. The required data are: cap screw, 58 in-11 UNC, SAE 5; hardened-steel washer, tw = 161 in thick; steel cover plate, t1 = 58 in, Es = 30 Mpsi; and cast-iron base, t2 = 58 in, Eci = 16 Mpsi. (a) Find kb, km, and C using the assumptions given in the caption of Figure 8–23. (b) Find all factors of safety and explain what they mean. Solution (a) For the symbols of Figures 8–15 and 8–23, h = t1 + tw = 0.6875 in, l = h + d∕2 = 1 in, and D2 = 1.5d = 0.9375 in. The joint is composed of three frusta; the upper two frusta are steel and the lower one is cast iron. For the upper frustum: t = l∕2 = 0.5 in, D = 0.9375 in, and E = 30 Mpsi. Using these values in Equation (8–20) gives k1 = 46.46 Mlbf/in. For the middle frustum: t = h − l∕2 = 0.1875 in and D = 0.9375 + 2(l − h) tan 30° = 1.298 in. With these and Es = 30 Mpsi, Equation (8–20) gives k2 = 197.43 Mlbf/in.

Figure 8–23 Pressure-cone frustum member model for a cap screw. For this model the significant sizes are h + t2∕2 t2 < d l={ h + d∕2 t2 ≥ d D1 = dw + l tan α = 1.5d + 0.577l D2 = dw = 1.5d where l = effective grip. The solutions are for α = 30° and dw = 1.5d.

D1

l

l 2

t1 t2

d D2

h

Screws, Fasteners, and the Design of Nonpermanent Joints     461

The lower frustum has D = 0.9375 in, t = l − h = 0.3125 in, and Eci = 16 Mpsi. The same equation yields k3 = 32.39 Mlbf/in. Substituting these three stiffnesses into Equation (8–18) gives km =17.40 Mlbf/in. The cap screw is short and threaded all the way. Using l = 1 in for the grip and At = 0.226 in2 from Table 8–2, we find the stiffness to be kb = AtE∕l = 6.78 Mlbf/in. Thus the joint constant is Answer

C=

kb 6.78 = = 0.280 kb + km 6.78 + 17.40

(b) Equation (8–30) gives the preload as Fi = 0.75Fp = 0.75At Sp = 0.75(0.226) (85) = 14.4 kip where from Table 8–9, Sp = 85 kpsi for an SAE grade 5 cap screw. Using Equation (8–28), we obtain the load factor as the yielding factor of safety is Answer

np =

Sp At CP + Fi

=

85(0.226) = 1.22 0.280(5) + 14.4

This is the traditional factor of safety, which compares the maximum bolt stress to the proof strength. Using Equation (8–29), Answer

nL =

Sp At − Fi CP

=

85(0.226) − 14.4 = 3.44 0.280(5)

This factor is an indication of the overload on P that can be applied without exceeding the proof strength. Next, using Equation (8–30), we have Answer

n0 =

Fi 14.4 = = 4.00 P(1 − C) 5(1 − 0.280)

If the force P gets too large, the joint will separate and the bolt will take the entire load. This factor guards against that event. For fatigue, refer to Figure 8–24, where we will show a comparison of the Goodman and Gerber criteria. The intersection of the load line L with the respective failure lines at points C, D, and E defines a set of strengths Sa and Sm at each intersection. Point B represents the stress state σm, σa. Point A is the preload stress σi. Therefore the load line begins at A and makes an angle having a unit slope. This angle is 45° only when both stress axes have the same scale. The factors of safety are found by dividing the distances AC, AD, and AE by the distance AB. Note that this is the same as dividing Sa for each theory by σa. The quantities shown in the caption of Figure 8–24 are obtained as follows: Point A σi =

Fi 14.4 = = 63.72 kpsi At 0.226

Point B CP 0.280(5) = = 3.10 kpsi 2At 2(0.226)

σa =

σm = σa + σi = 3.10 + 63.72 = 66.82 kpsi

462      Mechanical Engineering Design

Figure 8–24

L

E

Sa

D

Sa Sa

C

Sp σa

B A

Stress amplitude σa

Fluctuating-stress diagram for preloaded bolts, drawn to scale, showing the Goodman line, the Gerber curve, and the Langer proof-strength line, with an exploded view of the area of interest. The strengths used are Sp = 85 kpsi, Se = 18.6 kpsi, and Sut = 120 kpsi. The coordinates are A, σi = 63.72 kpsi; B, σa = 3.10 kpsi, σm = 66.82 kpsi; C, Sa = 7.55 kpsi, Sm = 71.29 kpsi; D, Sa = 10.64 kpsi, Sm = 74.36 kpsi; E, Sa = 11.32 kpsi, Sm = 75.04 kpsi.

60

σi

σm

Sm

Sm

Sm

70

80

Sp

90

Proofstrength line Gerber line L

Se

Goodman line

σi

Sp

Sut

Steady stress component σm

Point C This is the Goodman criterion. From Table 8–17, we find Se = 18.6 kpsi. Then, using Eq. (8–45), the factor of safety is found to be Answer

nf =

Se (Sut − σi ) 18.6(120 − 63.72) = = 2.44 σa (Sut + Se ) 3.10(120 + 18.6)

Point D This is on the proof-strength line where

Sm + Sa = Sp

(1)

In addition, the horizontal projection of the load line AD is

Sm = σi + Sa

(2)

Solving Equations (1) and (2) simultaneously results in

Sa =

Sp − σi 2

=

85 − 63.72 = 10.64 kpsi 2

The factor of safety resulting from this is Sa 10.64 = = 3.43 σa 3.10 which, of course, is identical to the result previously obtained by using Equation (8–29). A similar analysis of a fatigue diagram could have been done using yield strength instead of proof strength. Though the two strengths are somewhat related, proof strength is a much better and more positive indicator of a fully loaded bolt than is the yield strength. It is also worth remembering that proof-strength values are specified in design codes for bolts; yield strengths are not. We found nf = 2.44 on the basis of fatigue and the Goodman line, and np = 3.43 on the basis of proof strength. Thus the danger of failure is by fatigue, not by overproof loading. These two factors should always be compared to determine where the greatest danger lies. Answer

np =

Screws, Fasteners, and the Design of Nonpermanent Joints     463

Point E For the Gerber criterion, from Eq. (8–46), the safety factor is Answer

nf =

1 [Sut √Sut2 + 4Se (Se + σi ) − Sut2 − 2σi Se] 2σaSe

1 [120 √1202 + 4(18.6)(18.6 + 63.72) − 1202 − 2(63.72) (18.6)] 2(3.10)(18.6) = 3.65 =

which is greater than np = 3.43 and contradicts the conclusion earlier that the danger of failure is fatigue. Figure 8–24 clearly shows the conflict where point D lies between points C and E. Again, the conservative nature of the Goodman criterion explains the discrepancy and the designer must form his or her own conclusion.

8–12  Bolted and Riveted Joints Loaded in Shear10 Riveted and bolted joints loaded in shear are treated exactly alike in design and analysis. Figure 8–25a shows a riveted connection loaded in shear. Let us now study the various means by which this connection might fail. Figure 8–25b shows a failure by bending of the rivet or of the riveted members. The bending moment is approximately M = Ft∕2, where F is the shearing force and Figure 8–25

(a)

(b)

(e) 10

(c)

(f)

(d )

(g)

The design of bolted and riveted connections for boilers, bridges, buildings, and other structures in which danger to human life is involved is strictly governed by various construction codes. When designing these structures, the engineer should refer to the American Institute of Steel Construction Handbook, the American Railway Engineering Association specifications, or the Boiler Construction Code of the American Society of Mechanical Engineers.

Modes of failure in shear loading of a bolted or riveted connection: (a) shear loading; (b) bending of rivet; (c) shear of rivet; (d) tensile failure of members; (e) bearing of rivet on members or bearing of members on rivet; ( f ) shear tear-out; (g) tensile tear-out.

464      Mechanical Engineering Design

t is the grip of the rivet, that is, the total thickness of the connected parts. The bending stress in the members or in the rivet is, neglecting stress concentration,

σ=

M I∕c

(8–52)

where I∕c is the section modulus for the weakest member or for the rivet or rivets, depending upon which stress is to be found. The calculation of the bending stress in this manner is an assumption, because we do not know exactly how the load is distributed to the rivet or the relative deformations of the rivet and the members. Although this equation can be used to determine the bending stress, it is seldom used in design; instead its effect is compensated for by an increase in the factor of safety. In Figure 8–25c failure of the rivet by pure shear is shown; the stress in the rivet is

τ=

F A

(8–53)

where A is the cross-sectional area of all the rivets in the group. It may be noted that it is standard practice in structural design to use the nominal diameter of the rivet rather than the diameter of the hole, even though a hot-driven rivet expands and nearly fills up the hole. Rupture of one of the connected members or plates by pure tension is illustrated in Figure 8–25d. The tensile stress is

σ=

F A

(8–54)

where A is the net area of the plate, that is, the area reduced by an amount equal to the area of all the rivet holes. For brittle materials and static loads and for either ductile or brittle materials loaded in fatigue, the stress-concentration effects must be included. It is true that the use of a bolt with an initial preload and, sometimes, a rivet will place the area around the hole in compression and thus tend to nullify the effects of stress concentration, but unless definite steps are taken to ensure that the preload does not relax, it is on the conservative side to design as if the full stressconcentration effect were present. The stress-concentration effects are not considered in structural design, because the loads are static and the materials ductile. In calculating the area for Equation (8–54), the designer should, of course, use the combination of rivet or bolt holes that gives the smallest area. Figure 8–25e illustrates a failure by crushing of the rivet or plate. Calculation of this stress, which is usually called a bearing stress, is complicated by the distribution of the load on the cylindrical surface of the rivet. The exact values of the forces acting upon the rivet are unknown, and so it is customary to assume that the components of these forces are uniformly distributed over the projected contact area of the rivet. This gives for the stress

F σ=− A

(8–55)

where the projected area for a single rivet is A = td. Here, t is the thickness of the thinnest plate and d is the rivet or bolt diameter. Edge shearing, or tearing, of the margin is shown in Figure 8–25f and g, respectively. In structural practice this failure is avoided by spacing the rivets at

Screws, Fasteners, and the Design of Nonpermanent Joints     465

least 112 diameters away from the edge. Bolted connections usually are spaced an even greater distance than this for satisfactory appearance, and hence this type of failure may usually be neglected. In a rivet joint, the rivets all share the load in shear, bearing in the rivet, bearing in the member, and shear in the rivet. Other failures are participated in by only some of the joint. In a bolted joint, shear is taken by clamping friction, and bearing does not exist. When bolt preload is lost, one bolt begins to carry the shear and bearing until yielding slowly brings other fasteners in to share the shear and bearing. Finally, all participate, and this is the basis of most bolted-joint analysis if loss of bolt preload is complete. The usual analysis involves ∙ ∙ ∙ ∙ ∙ ∙ ∙

Bearing in the bolt (all bolts participate) Bearing in members (all holes participate) Shear of bolt (all bolts participate eventually) Distinguishing between thread and shank shear Edge shearing and tearing of member (edge bolts participate) Tensile yielding of member across bolt holes Checking member capacity

EXAMPLE 8–6 Two 1- by 4-in 1018 cold-rolled steel bars are butt-spliced with two 12 - by 4-in 1018 cold-rolled splice plates using four 34 in-16 UNF grade 5 bolts as depicted in Figure 8–26. For a design factor of nd = 1.5 estimate the static load F that can be carried if the bolts lose preload. Solution From Table A–20, minimum strengths of Sy = 54 kpsi and Sut = 64 kpsi are found for the members, and from Table 8–9 minimum strengths of Sp = 85 kpsi, Sy = 92 kpsi, and Sut = 120 kpsi for the bolts are found. 1

1 2 in

1

1

1 2 in

1 2 in

Figure 8–26

1

1 2 in

1

1 4 in F

1

4 in

1 2 in

F

1

1 4 in (a) 1 2

3 4

in

1in

F

1 2

in-16 UNF SAE grade 5

F

in (b)

466      Mechanical Engineering Design

F∕2 is transmitted by each of the splice plates, but since the areas of the splice plates are half those of the center bars, the stresses associated with the plates are the same. So for stresses associated with the plates, the force and areas used will be those of the center plates. Bearing in bolts, all bolts loaded:

σ=

Sy F = n 2td d

F=

2td Sy 2(1)( 34 )92 = = 92 kip nd 1.5

Bearing in members, all bolts active:

σ=

(Sy ) mem F = nd 2td

F=

2td(Sy ) mem 2(1) ( 34 )54 = = 54 kip nd 1.5

Shear of bolt, all bolts active: If the bolt threads do not extend into the shear planes for four shanks: Sy F = 0.577 2 nd 4πd ∕4

τ=

F = 0.577πd 2

Sy 92 = 0.577π(0.75) 2 = 62.5 kip nd 1.5

If the bolt threads extend into a shear plane:

τ=

Sy F = 0.577 n 4Ar d

F=

0.577(4)Ar Sy 0.577(4)0.351(92) = = 49.7 kip nd 1.5

Edge shearing of member at two margin bolts: From Figure 8–27,

τ=

0.577(Sy ) mem F = nd 4at

F=

4at0.577(Sy ) mem 4(1.125)(1)0.577(54) = = 93.5 kip nd 1.5

Tensile yielding of members across bolt holes: (Sy ) mem F = 3 nd [4 − 2( 4 )]t

σ=

[4 − 2( 34 )]t(Sy ) mem [4 − 2( 34 )](1)54 F= = = 90 kip nd 1.5

Screws, Fasteners, and the Design of Nonpermanent Joints     467

On the basis of bolt shear, the limiting value of the force is 49.7 kip, assuming the threads extend into a shear plane. However, it would be poor design to allow the threads to extend into a shear plane. So, assuming a good design based on bolt shear, the limiting value of the force is 62.5 kip. For the members, the bearing stress limits the load to 54 kip. Figure 8–27 Edge shearing of member.

Bolt d

a

Shear Joints with Eccentric Loading In the previous example, the load distributed equally to the bolts since the load acted along a line of symmetry of the fasteners. The analysis of a shear joint undergoing eccentric loading requires locating the center of relative motion between the two members. In Figure 8–28 let A1 to A5 be the respective cross-sectional areas of a group of five pins, or hot-driven rivets, or tight-fitting shoulder bolts. Under this assumption the rotational pivot point lies at the centroid of the cross-sectional area pattern of the pins, rivets, or bolts. Using statics, we learn that the centroid G is located by the coordinates x and y, where xi and yi are the distances to the ith area center: A1x1 + A2 x2 + A3 x3 + A 4 x 4 + A5 x5 x= = A1 + A2 + A3 + A 4 + A5

y=

A1 y1 + A2 y2 + A3 y3 + A 4 y 4 + A5 y5 = A1 + A2 + A3 + A 4 + A5

∑1n A i x i ∑n1 Ai

(8–56)

∑ Ai yi ∑n1 Ai n 1

In many instances the centroid can be located by symmetry. Figure 8–28

y

Centroid of pins, rivets, or bolts.

A3

A2 A4

G A1 _ y A5 O

x _ x

468      Mechanical Engineering Design

Figure 8–29

w lbf ⁄ in M1

(a) Beam bolted at both ends with distributed load; (b) free-body diagram of beam; (c) enlarged view of bolt group centered at O showing primary and secondary resultant shear forces.

O

M2

V1

V2 (b) FA'

w lbf ⁄ in

O

FB'

F B"

A F A"

+

B rB

rA O

Beam FC'

rC

rD

FD'

(a) C

F D"

D F C" (c)

An example of eccentric loading of fasteners is shown in Figure 8–29. This is a portion of a machine frame containing a beam subjected to the action of a bending load. In this case, the beam is fastened to vertical members at the ends with specially prepared load-sharing bolts. You will recognize the schematic representation in Figure 8–29b as a statically indeterminate beam with both ends fixed and with moment and shear reactions at each end. For convenience, the centers of the bolts at the left end of the beam are drawn to a larger scale in Figure 8–29c. Point O represents the centroid of the group, and it is assumed in this example that all the bolts are of the same diameter. Note that the forces shown in Figure 8–29c are the resultant forces acting on the pins with a net force and moment equal and opposite to the reaction loads V1 and M1 acting at O. The total load taken by each bolt will be calculated in three steps. In the first step the shear V1 is divided equally among the bolts so that each bolt takes F′ = V1∕n, where n refers to the number of bolts in the group and the force F′ is called the direct load, or primary shear. It is noted that an equal distribution of the direct load to the bolts assumes an absolutely rigid member. The arrangement of the bolts or the shape and size of the members sometimes justifies the use of another assumption as to the division of the load. The direct loads F′n are shown as vectors on the loading diagram (Figure 8–29c). The moment load, or secondary shear, is the additional load on each bolt due to the moment M1. If rA, rB, rC, etc., are the radial distances from the centroid to the center of each bolt, the moment and moment loads are related as follows:

M1 = F″A rA + F″B rB + F″C rC + …

(a)

where the F″ are the moment loads. The force taken by each bolt depends upon its radial distance from the centroid; that is, the bolt farthest from the centroid takes the greatest load, while the nearest bolt takes the smallest. This is because the moment loads are caused by bearing pressure of the bolt holes against the bolts as the members rotate slightly around the centroid with respect to each other, with the amount of displacement at each bolt proportional to its radius from the centroid. We can therefore write

F″A F″B F″C = = rA rB rC

(b)

Screws, Fasteners, and the Design of Nonpermanent Joints     469

where again, the diameters of the bolts are assumed equal. If not, then one replaces F″ in Equation (b) with the shear stresses τ″ = 4F″∕πd2 for each bolt. Solving Equations (a) and (b) simultaneously, we obtain M1rn F″n = 2 (8–57) 2 r A + r B + rC2 + … where the subscript n refers to the particular bolt whose load is to be found. Each moment load is a force vector perpendicular to the radial line from the centroid to the bolt center. In the third step the direct and moment loads are added vectorially to obtain the resultant load on each bolt. Since all the bolts or rivets are usually the same size, only that bolt having the maximum load need be considered. When the maximum load is found, the strength may be determined by using the various methods already described. EXAMPLE 8–7 Shown in Figure 8–30 is a 15- by 200-mm rectangular steel bar cantilevered to a 250-mm steel channel using four tightly fitted bolts located at A, B, C, and D. Assume the bolt threads do not extend into the joint. For the F = 16 kN load shown find (a) The resultant load on each bolt (b) The maximum shear stress in each bolt (c) The maximum bearing stress (d) The critical bending stress in the bar Solution (a) Point O, the centroid of the bolt group in Figure 8–30, is found by symmetry. If a free-body diagram of the beam were constructed, the shear reaction V would pass through O and the moment reactions M would be about O. These reactions are V = 16 kN M = 16(300 + 50 + 75) = 6800 N · m In Figure 8–31, the bolt group has been drawn to a larger scale and the reactions and resultants are shown. The distance from the centroid to the center of each bolt is r = √ (60) 2 + (75) 2 = 96.0 mm

250

Figure 8–30

10

M16 × 2 bolts C

Dimensions in millimeters.

15

F = 16 kN

B 60 O

D

75

75

200

60

A

50

300

470      Mechanical Engineering Design

Figure 8–31

y

FC" FC

B

C FC'

FB' rB

rC

F B" FB

x

O F D"

M

V

rA

rD

FD D

A FA'

FD' F A" FA

The resultants are found as follows. The primary shear load per bolt is

F′ =

V 16 = = 4 kN n 4

Since the rn are equal, the secondary shear forces are equal, and Equation (8–57) becomes

F″ =

Mr M 6800 = = = 17.7 kN 2 4r 4(96.0) 4r

The primary and secondary shear forces are plotted to scale in Figure 8–31 and the resultants obtained by using the parallelogram rule. The magnitudes are found by measurement (or analysis) to be Answer

FA = FB = 21.0 kN

Answer

FC = FD = 14.8 kN

(b) Bolts A and B are critical because they carry the largest shear load. The problem stated to assume that the bolt threads are not to extend into the joint. This would require special bolts. If standard nuts and bolts were used, the bolts would need to be 45 mm long with a thread length of LT = 38 mm. Thus the unthreaded portion of the bolt is 45 − 38 = 7 mm long. This is less than the 15 mm for the plate in Figure 8–30, and the bolts would tend to shear along the minor diameter at a stress of τ = F∕As = 21.0(10)3∕144 = 146 MPa. Using bolts not extending into the joint, or shoulder bolts, is preferred. For this example, the body area of each bolt is A =  π(162)∕4 = 201.1 mm2, resulting in a shear stress of Answer

τ=

F 21.0(10) 3 = = 104 MPa A 201.1

Screws, Fasteners, and the Design of Nonpermanent Joints     471

(c) The channel is thinner than the bar, and so the largest bearing stress is due to the pressing of the bolt against the channel web. The bearing area is Ab = td = 10(16) = 160 mm2. Thus the bearing stress is σ=−

Answer

21.0(10) 3 F =− = −131 MPa Ab 160

(d) The critical bending stress in the bar is assumed to occur in a section parallel to the y axis and through bolts A and B. At this section the bending moment is

M = 16(300 + 50) = 5600 N · m

The second moment of area through this section is obtained as follows: I = I bar − 2(I holes + d 2A)

=

15(200) 3 15(16) 3 − 2[ + (60) 2 (15) (16) ] = 8.26(10) 6 mm4 12 12

Then σ=

Answer

Mc 5600(100) = (10) 3 = 67.8 MPa I 8.26(10) 6

PROBLEMS   8–1 A power screw is 25 mm in diameter and has a thread pitch of 5 mm.

(a) Find the thread depth, the thread width, the mean and root diameters, and the lead, provided square threads are used. (b) Repeat part (a) for Acme threads.

  8–2 Using the information in the footnote of Table 8–1, show that the tensile-stress area is

At =

π (d − 0.938 194p) 2 4

  8–3 Show that for zero collar friction the efficiency of a square-thread screw is given by the equation

e = tan λ

1 − f tan λ tan λ + f

Plot a curve of the efficiency for lead angles up to 45°. Use f = 0.08.

  8–4 A single-threaded power screw is 25 mm in diameter with a pitch of 5 mm. A vertical

load on the screw reaches a maximum of 5 kN. The coefficients of friction are 0.06 for the collar and 0.09 for the threads. The frictional diameter of the collar is 45 mm. Find the overall efficiency and the torque to "raise" and "lower" the load.

  8–5 The machine shown in the figure can be used for a tension test but not for a compression test. Why? Can both screws have the same hand?

472      Mechanical Engineering Design

Motor

Bearings

Worm

Spur gears

[

Problem 8–5 Bronze bushings

2 's C.I.

Collar bearing

B C

Foot

1

3 2 in

Problem 8–7

A

2 [ 's

8–6

The press shown for Problem 8–5 has a rated load of 5000 lbf. The twin screws have Acme threads, a diameter of 2 in, and a pitch of 14 in. Coefficients of friction are 0.05 for the threads and 0.08 for the collar bearings. Collar diameters are 3.5 in. The gears have an efficiency of 95 percent and a speed ratio of 60:1. A slip clutch, on the motor shaft, prevents overloading. The full-load motor speed is 1720 rev/min. (a) When the motor is turned on, how fast will the press head move? (b) What should be the horsepower rating of the motor?

8–7

For the C clamp shown, a force is applied at the end of the 38 -in diameter handle. The screw is a 34 in-6 Acme thread (see Figure 8–3), and is 10 in long overall, with a maximum of 8 in possible in the clamping region. The handle and screw are both made from cold-drawn AISI 1006 steel. The coefficients of friction for the screw and the collar are 0.15. The collar, which in this case is the anvil striker's swivel joint, has a friction diameter of 1 in. It is desired that the handle will yield before the screw will fail. Check this by the following steps. (a) Determine the maximum force that can be applied to the end of the handle to reach the point of yielding of the handle. (b) Using the force from part (a), determine the clamping force. (c) Using the force from part (a), determine the factor of safety for yielding at the interface of the screw body and the base of the first engaged thread, assuming the first thread carries 38 percent of the total clamping force. (d) Determine a factor of safety for buckling of the screw.

8–8

The C clamp shown in the figure for Problem 8–7 uses a 34 in-6 Acme thread. The frictional coefficients are 0.15 for the threads and for the collar. The collar, which in this case is the anvil striker's swivel joint, has a friction diameter of 1 in. Calculations are to be based on a maximum force of 8 lbf applied to the handle at a radius of 312 in from the screw centerline. Find the clamping force.

8–9

Find the power required to drive a 1.5-in power screw having double square threads with a pitch of 14 in. The nut is to move at a velocity of 2 in/s and move a load of F = 2.2 kips. The frictional coefficients are 0.10 for the threads and 0.15 for the collar. The frictional diameter of the collar is 2.25 in.

Screws, Fasteners, and the Design of Nonpermanent Joints     473

8–10 A single square-thread power screw has an input power of 3 kW at a speed of 1 rev/s.

The screw has a diameter of 40 mm and a pitch of 8 mm. The frictional coefficients are 0.14 for the threads and 0.09 for the collar, with a collar friction radius of 50 mm. Find the axial resisting load F and the combined efficiency of the screw and collar.

8–11

An M14 × 2 hex-head bolt with a nut is used to clamp together two 15-mm steel plates. (a) Determine a suitable length for the bolt, rounded up to the nearest 5 mm. (b) Determine the bolt stiffness. (c) Determine the stiffness of the members.

8–12 Repeat Problem 8–11 with the addition of one 14R metric plain washer under the nut. 8–13 Repeat Problem 8–11 with one of the plates having a threaded hole to eliminate the nut. 8–14 A 2-in steel plate and a 1-in cast-iron plate are compressed with one bolt and nut. The bolt is 12 in-13 (a) Determine (b) Determine (c) Determine

UNC. a suitable length for the bolt, rounded up to the nearest the bolt stiffness. the stiffness of the members.

8–15 Repeat Problem 8–14 with the addition of one

1 4

in.

1 2

N American Standard plain washer under the head of the bolt, and another identical washer under the nut.

8–16 Repeat Problem 8–14 with the cast-iron plate having a threaded hole to eliminate the nut. 8–17 Two identical aluminum plates are each 2 in thick, and are compressed with one bolt and nut. Washers are used under the head of the bolt and under the nut. Washer properties: steel; ID = 0.531 in; OD = 1.062 in; thickness = 0.095 in Nut properties: steel; height = 167 in Bolt properties: 12 in-13 UNC grade 8 Plate properties: aluminum; E = 10.3 Mpsi; Su = 47 kpsi; Sy = 25 kpsi (a) Determine a suitable length for the bolt, rounded up to the nearest 14 in. (b) Determine the bolt stiffness. (c) Determine the stiffness of the members.

8–18 Repeat Problem 8–17 with no washer under the head of the bolt, and two washers stacked under the nut.

8–19 A 30-mm thick AISI 1020 steel plate is sandwiched between two 10-mm thick 2024-T3 aluminum plates and compressed with a bolt and nut with no washers. The bolt is M10 × 1.5, property class 5.8. (a) Determine a suitable length for the bolt, rounded up to the nearest 5 mm. (b) Determine the bolt stiffness. (c) Determine the stiffness of the members.

8–20 Repeat Problem 8–19 with the bottom aluminum plate replaced by one that is 20 mm thick. 8–21 Repeat Problem 8–19 with the bottom aluminum plate having a threaded hole to eliminate the nut.

8–22 Two 20-mm steel plates are to be clamped together with a bolt and nut. Specify a coarse thread metric bolt to provide a joint constant C of approximately 0.2.

8–23 A 2-in steel plate and a 1-in cast-iron plate are to be compressed with one bolt and nut. Specify a UNC bolt to provide a joint constant C of approximately 0.2.

474      Mechanical Engineering Design 3 4

in-16 UNF grade 5

8–24 An aluminum bracket with a 12 -in thick flange is to be clamped to a steel column with

a 34 -in wall thickness. A cap screw passes through a hole in the bracket flange, and threads into a tapped hole through the column wall. Specify a UNC cap screw to provide a joint constant C of approximately 0.25.

1.125 in

8–25 An M14 × 2 hex-head bolt with a nut is used to clamp together two 20-mm steel

10 in

plates. Compare the results of finding the overall member stiffness by use of Equations (8–20), (8–22), and (8–23).

in-16 UNF series SAE grade 5 bolt has a 34 -in ID steel tube 10 in long, clamped between washer faces of bolt and nut by turning the nut snug and adding one-third of a turn. The tube OD is the washer-face diameter dw = 1.5d = 1.5(0.75) = 1.125 in = OD. (a) Determine the bolt stiffness, the tube stiffness, and the joint constant C. (b) When the one-third turn-of-nut is applied, what is the initial tension Fi in the bolt?

8–26 A

Problem 8–26

3 4

8–27 From your experience with Problem 8–26, generalize your solution to develop a turnof-nut equation

Nt = where  Nt θ N Fi kb, km

= = = = =

kb + km θ = Fi N 360° ( kb km )

turn of the nut, in rotations, from snug tight turn of the nut in degrees number of thread/in (1∕p where p is pitch) initial preload spring rates of the bolt and members, respectively

Use this equation to find the relation between torque-wrench setting T and turn-of-nut Nt. ("Snug tight" means the joint has been tightened to perhaps half the intended preload to flatten asperities on the washer faces and the members. Then the nut is loosened and retightened finger tight, and the nut is rotated the number of degrees indicated by the equation. Properly done, the result is competitive with torque wrenching.)

8–28 RB&W11 recommends turn-of-nut from snug fit to preload as follows: 1∕3 turn for bolt grips of 1–4 diameters, 1∕2 turn for bolt grips 4–8 diameters, and 2∕3 turn for

grips of 8–12 diameters. These recommendations are for structural steel fabrication (permanent joints), producing preloads of 100 percent of proof strength and beyond. Machinery fabricators with fatigue loadings and possible joint disassembly have much smaller turns-of-nut. The RB&W recommendation enters the nonlinear plastic deformation zone. For Example 8–4, use Equation (8–27) with K = 0.2 to estimate the torque necessary to establish the desired preload. Then, using the results from Problem 8–27, determine the turn of the nut in degrees. How does this compare with the RB&W recommendations?

8–29 For a bolted assembly with six bolts, the stiffness of each bolt is kb = 3 Mlbf/in and the stiffness of the members is km = 12 Mlbf/in per bolt. An external load of 80 kips

is applied to the entire joint. Assume the load is equally distributed to all the bolts. It has been determined to use 12 in-13 UNC grade 8 bolts with rolled threads. Assume the bolts are preloaded to 75 percent of the proof load. (a) Determine the yielding factor of safety. (b) Determine the overload factor of safety. (c) Determine the factor of safety based on joint separation.

11

Russell, Burdsall & Ward, Inc., Metal Forming Specialists, Mentor, Ohio.

Screws, Fasteners, and the Design of Nonpermanent Joints     475

8–30 For the bolted assembly of Problem 8–29, it is desired to find the range of torque that

a mechanic could apply to initially preload the bolts without expecting failure once the joint is loaded. Assume a torque coefficient of K = 0.2. (a) Determine the maximum bolt preload that can be applied without exceeding the proof strength of the bolts. (b) Determine the minimum bolt preload that can be applied while avoiding joint separation. (c) Determine the value of torque in units of lbf · ft that should be specified for preloading the bolts if it is desired to preload to the midpoint of the values found in parts (a) and (b).

8–31 For a bolted assembly with eight bolts, the stiffness of each bolt is kb = 1.0 MN/mm

and the stiffness of the members is km = 2.6 MN/mm per bolt. The joint is subject to occasional disassembly for maintenance and should be preloaded accordingly. Assume the external load is equally distributed to all the bolts. It has been determined to use M6 × 1 class 5.8 bolts with rolled threads. (a) Determine the maximum external load Pmax that can be applied to the entire joint without exceeding the proof strength of the bolts. (b) Determine the maximum external load Pmax that can be applied to the entire joint without causing the members to come out of compression.

8–32 For a bolted assembly, the stiffness of each bolt is kb = 4 Mlbf/in and the stiffness of

the members is km = 12 Mlbf/in per bolt. The joint is subject to occasional disassembly for maintenance and should be preloaded accordingly. A fluctuating external load is applied to the entire joint with Pmax = 80 kips and Pmin = 20 kips. Assume the load is equally distributed to all the bolts. It has been determined to use 12 in-13 UNC grade 8 bolts with rolled threads. (a) Determine the minimum number of bolts necessary to avoid yielding of the bolts. (b) Determine the minimum number of bolts necessary to avoid joint separation.

8–33 The figure illustrates the nonpermanent connection of a steel cylinder head to a to grade 30 cast-iron pressure vessel using N bolts. A confined gasket seal has an 8–36 effective sealing diameter D. The cylinder stores gas at a maximum pressure pg.

For the specifications given in the table for the specific problem assigned, select a suitable bolt length, assuming bolts are available in increments of 14 in and 5 mm. Then determine the yielding factor of safety np, the load factor nL, and the joint separation factor n0.

F E D A B

Problems 8–33 to 8–36

C

476      Mechanical Engineering Design

Problem Number

Problem Number 8–37

8–33

8–38

8–34

8–39

8–35

8–40

8–36

Problem Number

Originating Problem Number

8–41

8–33

8–42

8–34

8–43

8–35

8–44

8–36

8–34

A

20 mm

1 2

B

20 mm

5 8

C

60 mm

D

100 mm

E F

8–35

8–36

in

20 mm

3 8

in

25 mm

1 2

2.75 in

0.7 m

2.75 in

3.5 in

0.8 m

3.25 in

200 mm

6 in

1.0 m

5.5 in

300 mm

8 in

1.1 m

7 in

N

10

10

36

8

pg

6 MPa

1500 psi

550 kPa

1200 psi

Bolt grade

ISO 9.8

SAE 5

ISO 10.9

SAE 8

M12 × 1.75

1 2

M10 × 1.5

7 16

Bolt spec.

Originating Problem Number

8–33

in-13

in in

in-14

8–37  Repeat the requirements for the problem specified in the table if the bolts and nuts to are replaced with cap screws that are threaded into tapped holes in the cast-iron 8–40 cylinder.  8–41 For the pressure vessel defined in the problem specified in the table, redesign the bolt to specifications to satisfy all of the following requirements. 8–44 ∙ Use coarse-thread bolts selecting a class from Table 8–11 for Problems 8–41 and 8–43, or a grade from Table 8–9 for Problems 8–42 and 8–44.

∙ To ensure adequate gasket sealing around the bolt circle, use enough bolts to provide a maximum center-to-center distance between bolts of four bolt diameters. ∙ Obtain a joint stiffness constant C between 0.2 and 0.3 to ensure most of the pressure load is carried by the members. ∙ The bolts may be reused, so the yielding factor of safety should be at least 1.1. ∙ The overload factor and the joint separation factor should allow for the pressure to exceed the expected pressure by 15 percent.

8–45 Bolts distributed about a bolt circle are often called upon to resist an external bending

moment as shown in the figure. The external moment is 12 kip · in and the bolt circle has a diameter of 8 in. The neutral axis for bending is a diameter of the bolt circle. What needs to be determined is the most severe external load seen by a bolt in the assembly. (a) View the effect of the bolts as placing a line load around the bolt circle whose intensity F′b, in pounds per inch, varies linearly with the distance from the neutral axis according to the relation F′b = F′b,max R sin θ. The load on any particular bolt can be viewed as the effect of the line load over the arc associated with the bolt. For example, there are 12 bolts shown in the figure. Thus each bolt load is assumed to be distributed on a 30° arc of the bolt circle. Under these conditions, what is the largest bolt load? (b) View the largest load as the intensity F′b,max multiplied by the arc length associated with each bolt and find the largest bolt load. (c) Express the load on any bolt as F = Fmax sin θ, sum the moments due to all the bolts, and estimate the largest bolt load. Compare the results of these three approaches to decide how to attack such problems in the future.

Screws, Fasteners, and the Design of Nonpermanent Joints     477

R

Problem 8–45

θ

Bolted connection subjected to bending.

M

M Neutral axis

8–46 The figure shows a cast-iron bearing block that is to be bolted to a steel ceiling joist

and is to support a gravity load of 18 kN. Bolts used are M24 ISO 8.8 with coarse threads and with 4.6-mm-thick steel washers under the bolt head and nut. The joist flanges are 20 mm in thickness, and the dimension A, shown in the figure, is 20 mm. The modulus of elasticity of the bearing block is 135 GPa.

A Problem 8–46

B d

C

(a) Find the wrench torque required if the fasteners are lubricated during assembly and the joint is to be permanent. (b) Determine the factors of safety guarding against yielding, overload, and joint separation.

8–47 The upside-down steel A frame shown in the figure is to be bolted to steel beams on

the ceiling of a machine room using ISO grade 8.8 bolts. This frame is to support the 40-kN vertical load as illustrated. The total bolt grip is 48 mm, which includes the thickness of the steel beam, the A-frame feet, and the steel washers used. The bolts are size M20 × 2.5. (a) What tightening torque should be used if the connection is permanent and the fasteners are lubricated? (b) Determine the factors of safety guarding against yielding, overload, and joint separation.

Drill 2 holes for M20 × 2.5 bolts Problem 8–47

W = 40 kN

478      Mechanical Engineering Design

8–48 For the bolted assembly in Problem 8–29, assume the external load is a repeated load. Determine the fatigue factor of safety for the bolts using the following failure criteria: (a) Goodman. (b) Gerber. (c) ASME-elliptic.

8–49 For a bolted assembly with eight bolts, the stiffness of each bolt is kb = 1.0 MN/mm

and the stiffness of the members is km = 2.6 MN/mm per bolt. The bolts are preloaded to 75 percent of proof strength. Assume the external load is equally distributed to all the bolts. The bolts are M6 × 1 class 5.8 with rolled threads. A fluctuating external load is applied to the entire joint with Pmax = 60 kN and Pmin = 20 kN. (a) Determine the yielding factor of safety. (b) Determine the overload factor of safety. (c) Determine the factor of safety based on joint separation. (d) Determine the fatigue factor of safety using the Goodman criterion.

8–50 For the bolted assembly in Problem 8–32, assume 10 bolts are used. Determine the fatigue factor of safety using the Goodman criterion.

8–51 For a bolted assembly with eight M8 × 1.25 class 9.8 bolts with rolled threads, the

stiffness of each bolt is kb = 1.5 MN/mm and the stiffness of the members is km = 3.9 MN/mm per bolt. The joint is subject to occasional disassembly for maintenance and should be preloaded accordingly. An external load of 50 kN is repeatedly applied to the entire joint. Determine the fatigue factor of safety using the following criteria: (a) Goodman.     (b) Gerber.     (c) Morrow.

Problem Number

Originating Problem Number

8–52

8–33

8–53

8–34

8–54

8–35

8–55

8–36

Problem Number

Originating Problem Number

8–56  or the pressure cylinder defined in the problem specified in the table, the gas pressure F to is cycled between pg and pg∕2. Determine the fatigue factor of safety for the bolts 8–59 using the Goodman criterion.

8–56

8–33

8–60 A 1-in-diameter hot-rolled AISI 1144 steel rod is hot-formed into an eyebolt similar

8–57

8–34

8–58

8–35

8–59

8–36

8–52  or the pressure cylinder defined in the problem specified in the table, the gas pressure F to is cycled between zero and pg. Determine the fatigue factor of safety for the bolts using 8–55 the following failure criteria: (a) Goodman. (b) Gerber. (c) ASME-elliptic.

to that shown in the figure for Problem 3–136, with an inner 3-in-diameter eye. The threads are 1 in-12 UNF and are die-cut. (a) For a repeatedly applied load collinear with the thread axis, using the Gerber criterion, is fatigue failure more likely in the thread or in the eye? (b) What can be done to strengthen the bolt at the weaker location? (c) If the factor of safety guarding against a fatigue failure is nf = 2, what repeatedly applied load can be applied to the eye?

1

3 4

in-16 UNF × 2 2 in SAE grade 5

8–61 The section of the sealed joint shown in the figure is loaded by a force cycling between 1 12

No. 40 CI

in

Problem 8–61

4 and 6 kips. The members have E = 16 Mpsi. All bolts have been carefully preloaded to Fi = 25 kip each. (a) Determine the yielding factor of safety. (b) Determine the overload factor of safety. (c) Determine the factor of safety based on joint separation. (d) Determine the fatigue factor of safety using the Goodman criterion.

Screws, Fasteners, and the Design of Nonpermanent Joints     479

8–62 Suppose the welded steel bracket shown in the figure is bolted underneath a structural-

steel ceiling beam to support a fluctuating vertical load imposed on it by a pin and yoke. The bolts are 12 -in coarse-thread SAE grade 8, tightened to recommended preload for nonpermanent assembly. The stiffnesses have already been computed and are kb  =  4 Mlbf/in and km = 16 Mlbf/in.

A C Problem 8–62 d

B

(a) Assuming that the bolts, rather than the welds, govern the strength of this design, determine the safe repeated load that can be imposed on this assembly using the Goodman criterion with the load line in Figure 8–22 and a fatigue design factor of 2. (b) Compute the static load factors based on the load found in part (a).

8–63 Using the Gerber fatigue criterion and a fatigue-design factor of 2, determine the external repeated load P that a 114 -in SAE grade 5 coarse-thread bolt can take compared with that for a fine-thread bolt. The joint constants are C = 0.30 for coarse- and 0.32 for fine-thread bolts. Assume the bolts are preloaded to 75 percent of the proof load.

8–64 An M30 × 3.5 ISO 8.8 bolt is used in a joint at recommended preload, and the joint

is subject to a repeated tensile fatigue load of P = 65 kN per bolt. The joint constant is C = 0.28. Find the static load factors and the factor of safety guarding against a fatigue failure based on the Gerber fatigue criterion.

8–65 The figure shows a fluid-pressure linear actuator (hydraulic cylinder) in which D = 4 in, t = 38 in, L = 12 in, and w = 34 in. Both brackets as well as the cylinder are of steel. The actuator has been designed for a working pressure of 2000 psi. Six 38 -in SAE grade 5 coarse-thread bolts are used, tightened to 75 percent of proof load. Assume the bolts are unthreaded within the grip. w

Problem 8–65

t

L

w

D

(a) Find the stiffnesses of the bolts and members, assuming that the entire cylinder is compressed uniformly and that the end brackets are perfectly rigid. (b) Using the Gerber fatigue criterion, find the factor of safety guarding against a fatigue failure. (c) What pressure would be required to cause total joint separation?

480      Mechanical Engineering Design

8–66 Using the Goodman fatigue criterion, repeat Problem 8–65 with the working pressure cycling between 1200 psi and 2000 psi.

8–67 The figure shows a bolted lap joint that uses SAE grade 5 bolts. The members are

made of cold-drawn AISI 1020 steel. Assume the bolt threads do not extend into the joint. Find the safe tensile shear load F that can be applied to this connection to provide a minimum factor of safety of 2 for the following failure modes: shear of bolts, bearing on bolts, bearing on members, and tension of members.

5 8

Problem 8–67

1 4

in

5 in 16

in-20 UNC

1 18 in

5 8

in 1 4

1

1 4 in

in

8–68 The bolted connection shown in the figure uses SAE grade 8 bolts. The members are hot-rolled AISI 1040 steel. A tensile shear load F = 5000 lbf is applied to the connection. Assume the bolt threads do not extend into the joint. Find the factor of safety for all possible modes of failure. 5 8

Problem 8–68

1

1 8 in

in

5 8

in

5 8

in

5 8

in 5 in-18 16

1 4

UNC

1 4

in

in

8–69 A bolted lap joint using ISO class 5.8 bolts and members made of cold-drawn SAE 1040 steel is shown in the figure. Assume the bolt threads do not extend into the joint. Find the tensile shear load F that can be applied to this connection to provide a minimum factor of safety of 2.5 for the following failure modes: shear of bolts, bearing on bolts, bearing on members, and tension of members.

40

Problem 8–69 Dimensions in millimeters.

20

M20 × 2.5

70

40 80

20

Screws, Fasteners, and the Design of Nonpermanent Joints     481

8–70 The bolted connection shown in the figure is subjected to a tensile shear load of

90 kN. The bolts are ISO class 5.8 and the material is cold-drawn AISI 1015 steel. Assume the bolt threads do not extend into the joint. Find the factor of safety of the connection for all possible modes of failure. 35

60

60

35

15 M20 × 2.5

35

Problem 8–70 Dimensions in millimeters.

35 20

8–71 The figure shows a connection that employs three SAE grade 4 bolts. The tensile shear

load on the joint is 5000 lbf. The members are cold-drawn bars of AISI 1020 steel. Assume the bolt threads do not extend into the joint. Find the factor of safety for each possible mode of failure. 5 8 5 8

1

1 8 in

in

1 4

in

5 in 16

in-20 UNC

1 in

Problem 8–71 5 8

in

3

116 in 5 in 16

2 38 in

8–72 A beam is made up by bolting together two cold-drawn bars of AISI 1018 steel as a

lap joint, as shown in the figure. The bolts used are ISO 5.8. Assume the bolt threads do not extend into the joint. Ignoring any twisting, determine the factor of safety of the connection. y A

Problem 8–72

200

50

100

3.2 kN

Dimensions in millimeters.

350

10 50

x A

M12 × 1.75

10 Section A–A

8–73 Standard design practice, as exhibited by the solutions to Problems 8–67 to 8–71, is

to assume that the bolts, or rivets, share the shear equally. For many situations, such an assumption may lead to an unsafe design. Consider the yoke bracket of Problem 8–62, for example. Suppose this bracket is bolted to a wide-flange column with the centerline through the two bolts in the vertical direction. A vertical load through the yoke-pin hole at distance B from the column flange would place a shear load on the bolts as well as a tensile load. The tensile load comes about because the bracket tends to pry itself about the bottom corner, much like a claw hammer, exerting a large tensile load on the upper bolt. In addition, it is almost certain that both the spacing of the bolt holes and their diameters will be slightly different on the column flange from what

482      Mechanical Engineering Design

they are on the yoke bracket. Thus, unless yielding occurs, only one of the bolts will take the shear load. The designer has no way of knowing which bolt this will be. In this problem the bracket is 8 in long, A = 12 in, B = 3 in, C = 6 in, and the column flange is 12 in thick. The bolts are 12 in-13 UNC × 112 in SAE grade 4. The nuts are tightened to 75 percent of proof load. The vertical yoke-pin load is 2500 lbf. If the upper bolt takes all the shear load as well as the tensile load, determine a static factor of safety for the bolt, based on the von Mises stress exceeding the proof strength.

8–74 The bearing of Problem 8–46 is bolted to a vertical surface and supports a horizontal

shaft. The bolts used have coarse threads and are M20 ISO 5.8. The joint constant is C = 0.25, and the dimensions are A = 20 mm, B = 50 mm, and C = 160 mm. The bearing base is 240 mm long. The bearing load is 14 kN. The bolts are tightened to 75 percent of proof load. Determine a static factor of safety for the bolt, based on the von Mises stress exceeding the proof strength. Use worst-case loading, as discussed in Problem 8–73.

8–75 A split-ring clamp-type shaft collar such as is described in Problem 5–80 must resist an axial load of 1000 lbf. Using a design factor of nd = 3 and a coefficient of friction of 0.12, specify an SAE Grade 5 cap screw using fine threads. What wrench torque should be used if a lubricated screw is used?

8–76 A vertical channel 152 × 76 (see Table A–7) has a cantilever beam bolted to it as shown. The channel is hot-rolled AISI 1006 steel. The bar is of hot-rolled AISI 1015 steel. The shoulder bolts are M10 × 1.5 ISO 5.8. Assume the bolt threads do not extend into the joint. For a design factor of 2.0, find the safe force F that can be applied to the cantilever.

12 F Problem 8–76 Dimensions in millimeters.

A 50

O 50

50

B 26

125

8–77 The cantilever bracket is bolted to a column with three M12 × 1.75 ISO 5.8 bolts.

The bracket is made from AISI 1020 hot-rolled steel. Assume the bolt threads do not extend into the joint. Find the factors of safety for the following failure modes: shear of bolts, bearing of bolts, bearing of bracket, and bending of bracket.

Holes for M12 × 1.75 bolts 8 mm thick 36 Problem 8–77 Dimensions in millimeters.

32

12 kN

64 36 200 Column

Screws, Fasteners, and the Design of Nonpermanent Joints     483

8–78 A 38 - × 2-in AISI 1018 cold-drawn steel bar is cantilevered to support a static load of

250 lbf as illustrated. The bar is secured to the support using two 38 in-16 UNC SAE grade 4 bolts. Assume the bolt threads do not extend into the joint. Find the factor of safety for the following modes of failure: shear of bolt, bearing on bolt, bearing on member, and strength of member.

3 8

Problem 8–78

1 in

in

12 in

3 in 1 in

250 lbf

8–79 The figure shows a welded fitting which has been tentatively designed to be bolted to

a channel so as to transfer the 2000-lbf load into the channel. The channel and the two fitting plates are of hot-rolled stock having a minimum Sy of 42 kpsi. The fitting is to be bolted using six SAE grade 4 shoulder bolts. Assume the bolt threads do not extend into the joint. Check the strength of the design by computing the factor of safety for all possible modes of failure. 6 holes for

1 2

in-13 UNC bolts

F = 2000 lbf

1 4

in

4 in 1 in

Problem 8–79

2 5 in 7 12 in

1 4

in 8 in

8 in [ 11.5

3 16

in

8–80 A plate is bolted to two walls as shown. Considering the moment and vertical force

reactions at the walls, determine the magnitude of the maximum shear stress in the bolts if they each have a cross-sectional area of 58 mm2. 10 kN 200

200

Problem 8–80 Dimensions in millimeters.

50

50

8–81 The coupled shafts shown are transmitting 250 hp at 600 rev/min. The shafts are

c­ onnected by a coupling with eight bolts. The bolt circle diameters are 5 and 10 in. (a) Determine the minimum nominal shoulder diameter of a bolt if all bolts are to have the same diameter and not exceed a shear stress of 20 kpsi. Select a preferred fraction size. (b) Determine the minimum number of bolts of the diameter determined in part (a) if the shear stress is not to exceed 20 kpsi and the bolts are to be placed only at the outer bolt circle.

484      Mechanical Engineering Design

Problem 8–81 600 rev/min

8–82 A cantilever is to be attached to the flat side of a 6-in, 13.0-lbf/in channel used as a column. The cantilever is to carry a load as shown in the figure. To a designer the choice of a bolt array is usually an a priori decision. Such decisions are made from a background of knowledge of the effectiveness of various patterns.

1 -in 2

Problem 8–82

steel plate

6 in

6 in

6 in 2000 lbf

(a) If two fasteners are used, should the array be arranged vertically, horizontally, or diagonally? How would you decide? (b) If three fasteners are used, should a linear or triangular array be used? For a triangular array, what should be the orientation of the triangle? How would you decide?

8–83 Using your experience with Problem 8–82, specify an optimal bolt pattern for two bolts for the bracket in Problem 8–82 and size the bolts.

8–84 Using your experience with Problem 8–82, specify an optimal bolt pattern for three bolts for the bracket in Problem 8–82 and size the bolts.

8–85 A gusset plate is welded to a base plate as shown. The base plate is secured to the foundation by four bolts, each with an effective cross-sectional area of 0.2 in2. Before application of the 1000 lbf force, the bolts were torqued down so that a preload of 5000 lbf tension was developed in each bolt. Determine the maximum tensile stress that develops upon the application of the 1000 lbf force. Hint: Assume that the plate is rigid and rotates about corner A.

Problem 8–85

8 in

1 in

1000 lbf A

10 in

9

Welding, Bonding, and the Design of Permanent Joints

©Ingram Publishing/SuperStock

Chapter Outline 9–1 Welding Symbols   486 9–2 Butt and Fillet Welds   488 9–3 Stresses in Welded Joints in Torsion   492 9–4 Stresses in Welded Joints in Bending   497 9–5 The Strength of Welded Joints   499 9–6 Static Loading   502 9–7 Fatigue Loading   505 9–8 Resistance Welding   507 9–9 Adhesive Bonding   508

485

486      Mechanical Engineering Design

Form can more readily pursue function with the help of joining processes such as welding, brazing, soldering, cementing, and gluing—processes that are used extensively in manufacturing today. Whenever parts have to be assembled or fabricated, there is usually good cause for considering one of these processes in preliminary design work. Particularly when sections to be joined are thin, one of these methods may lead to significant savings. The elimination of individual fasteners, with their holes and assembly costs, is an important factor. Also, some of the methods allow rapid machine assembly, furthering their attractiveness. Riveted permanent joints were common as the means of fastening rolled steel shapes to one another to form a permanent joint. The childhood fascination of seeing a cherry-red hot rivet thrown with tongs across a building skeleton to be unerringly caught by a person with a conical bucket, to be hammered pneumatically into its final shape, is all but gone. Two developments relegated riveting to lesser prominence. The first was the development of high-strength steel bolts whose preload could be controlled. The second was the improvement of welding, competing both in cost and in latitude of possible form.

9–1  Welding Symbols A weldment is fabricated by welding together a collection of metal shapes, cut to particular configurations. During welding, the several parts are held securely together, often by clamping or jigging. The welds must be precisely specified on working drawings, and this is done by using the welding symbol, shown in Figure 9–1, as standardized by the American Welding Society (AWS). The arrow of this symbol points to the joint to be welded. The body of the symbol contains as many of the following elements as are deemed necessary: ∙ Reference line ∙ Arrow

Groove angle; included angle of countersink for plug welds Length of weld

Size; size or strength for resistance welds

Tail (may be omitted when reference is not used) Basic weld symbol or detail reference

R

Other side

S

Arrow connecting reference line to arrow side of joint, to grooved member, or both L–P

Arrow side

T

Specification; process; or other reference

Pitch (center-to-center spacing) of welds

F A

Reference line sides)

The AWS standard welding symbol showing the location of the symbol elements.

Finish symbol Contour symbol Root opening; depth of filling for plug and slot welds

(Both

Figure 9–1

(N)

Field weld symbol Weld all around symbol Number of spot or projection welds

Welding, Bonding, and the Design of Permanent Joints     487

∙ ∙ ∙ ∙ ∙ ∙

Basic weld symbols as in Figure 9–2 Dimensions and other data Supplementary symbols Finish symbols Tail Specification or process

The arrow side of a joint is the line, side, area, or near member to which the arrow points. The side opposite the arrow side is the other side. Figures 9–3 to 9–6 illustrate the types of welds used most frequently by designers. For general machine elements most welds are fillet welds, though butt welds are used a great deal in designing pressure vessels. Of course, the parts to be joined must be arranged so that there is sufficient clearance for the welding operation. If unusual joints are required because of insufficient clearance or because of the section shape, the design may be a poor one and the designer should begin again and endeavor to synthesize another solution. Since heat is used in the welding operation, there are metallurgical changes in the parent metal in the vicinity of the weld. Also, residual stresses may be introduced because of clamping or holding or, sometimes, because of the order of welding. Usually these residual stresses are not severe enough to cause concern; Figure 9–2

Type of weld Bead

Fillet

Plug or slot

Arc- and gas-weld symbols.

Groove Square

V

Bevel

60

5

200

60–200 (a)

U

(b)

J

Figure 9–3 Fillet welds. (a) The number indicates the leg size; the arrow should point only to one weld when both sides are the same. (b) The symbol indicates that the welds are intermittent and staggered 60 mm along on 200-mm centers.

Figure 9–4 The circle on the weld symbol indicates that the welding is to go all around.

5

488      Mechanical Engineering Design

Figure 9–5

60°

Butt or groove welds: (a) square butt-welded on both sides; (b) single V with 60° bevel and root opening of 2 mm; (c) double V; (d) single bevel.

2 2 60° (a)

(b)

60°

45°

(d )

(c)

Figure 9–6 Special groove welds: (a) T joint for thick plates; (b) U and J welds for thick plates; (c) corner weld (may also have a bead weld on inside for greater strength but should not be used for heavy loads); (d) edge weld for sheet metal and light loads.

(a)

(b)

(c)

(d)

in some cases a light heat treatment after welding has been found helpful in relieving them. When the parts to be welded are thick, a preheating will also be of benefit. If the reliability of the component is to be quite high, a testing program should be established to learn what changes or additions to the operations are necessary to ensure the best quality.

9–2  Butt and Fillet Welds Figure 9–7a shows a single V-groove weld loaded by the tensile force F. For either tension or compression loading, the average normal stress is

σ=

F hl

(9–1)

where h is the weld throat and l is the length of the weld, as shown in the figure. Note that the value of h does not include the reinforcement. The reinforcement can

Welding, Bonding, and the Design of Permanent Joints     489 Reinforcement

Figure 9–7

Reinforcement

A

A typical butt joint.

l

l F

F

F

F

Throat h

Throat h

(a) Tensile loading

(b) Shear loading

x θ Throat D 2F

A

h

C

45°

t F

B

h

Fs

h Fn

h

F y

Figure 9–8

Figure 9–9

A transverse fillet weld.

Free body from Figure 9–8.

be desirable, but it varies somewhat and does produce stress concentration at point A in the figure. If fatigue loads exist, it is good practice to grind or machine off the reinforcement. The average stress in a butt weld due to shear loading (Figure 9–7b) is

τ=

F hl

(9–2)

Figure 9–8 illustrates a typical transverse fillet weld. In Figure 9–9 a portion of the welded joint has been isolated from Figure 9–8 as a free body. At angle θ the forces on each weldment consist of a normal force Fn and a shear force Fs. Summing forces in the x and y directions gives

Fs = F sin θ

(a)

Fn = F cos θ

(b)

Using the law of sines for the triangle in Figure 9–9 yields √2h t h h = = = sin 45° sin(180° − 45° − θ) sin(135° − θ) cos θ + sin θ

Solving for the throat thickness t gives

t=

h cos θ + sin θ

(c)

F

490      Mechanical Engineering Design

The nominal stresses at the angle θ in the weldment, τ and σ, are

τ=

Fs F sin θ(cos θ + sin θ) F = = (sin θ cos θ + sin2 θ) A hl hl

σ=

Fn F cos θ(cos θ + sin θ) F = = (cos2 θ + sin θ cos θ) (e) A hl hl

(d)

The von Mises stress σ′ at angle θ is σ′ = (σ2 + 3τ2 )1∕2 =

F [(cos2 θ + sin θ cos θ) 2 + 3(sin2 θ + sin θ cos θ) 2]1∕2 (f ) hl

The largest von Mises stress occurs at θ = 62.5° with a value of σ′ = 2.16F∕(hl). The corresponding values of τ and σ are τ = 1.196F∕(hl) and σ = 0.623F∕(hl). The maximum shear stress can be found by differentiating Equation (d) with respect to θ and equating to zero. The stationary point occurs at θ = 67.5° with a corresponding τmax = 1.207F∕(hl) and σ = 0.5F∕(hl). There are some experimental and analytical results that are helpful in evaluating Equations (d) through ( f ) and the consequences. A model of the transverse fillet weld of Figure 9–8 is easily constructed for photoelastic purposes and has the advantage of a balanced loading condition. Norris constructed such a model and reported the stress distribution along the sides AB and BC of the weld.1 An approximate graph of the results he obtained is shown as Figure 9–10a. Note that stress concentrations exist at A and B on the horizontal leg and at B on the vertical leg. Norris states that he could not determine the stresses at A and B with any certainty. Salakian2 presents data for the stress distribution across the throat of a fillet weld (Figure 9–10b). This graph is of particular interest because we have just learned that it is the throat stresses that are used in design. Again, the figure shows a stress Figure 9–10

C

Stress distribution in fillet welds: (a) stress distribution on the legs as reported by Norris; (b) distribution of principal stresses and maximum shear stress as reported by Salakian.

+ D

σ1

τ

τ m ax +

τ + A

B

σ

0 D

B σ2

(a)

1

σ

(b)

C. H. Norris, "Photoelastic Investigation of Stress Distribution in Transverse Fillet Welds," Welding J., vol. 24, 1945, p. 557s. 2 A. G. Salakian and G. E. Claussen, "Stress Distribution in Fillet Welds: A Review of the Literature," Welding J., vol. 16, May 1937, pp. 1–24.

Welding, Bonding, and the Design of Permanent Joints     491

concentration at point B. Note that Figure 9–10a applies either to the weld metal or to the parent metal, and that Figure 9–10b applies only to the weld metal. Equations (a) through ( f ) and their consequences seem familiar, and we can become comfortable with them. The net result of photoelastic and finite element analysis of transverse fillet weld geometry is more like that shown in Figure 9–10 than those given by mechanics of materials or elasticity methods. The most important concept here is that we have no analytical approach that predicts the existing stresses. The geometry of the fillet is crude by machinery standards, and even if it were ideal, the macrogeometry is too abrupt and complex for our methods. There are also subtle bending stresses due to eccentricities. Still, in the absence of robust analysis, weldments must be specified and the resulting joints must be safe. The approach has been to use a simple and conservative model, verified by testing as conservative. The approach has been to ∙ Consider the external loading to be carried by shear forces on the throat area of the weld. By ignoring the normal stress on the throat, the shearing stresses are inflated sufficiently to render the model conservative. ∙ Use distortion energy for significant stresses. ∙ Circumscribe typical cases by code.

For this model, the basis for weld analysis or design employs τ=

F 1.414F = 0.707hl hl

(9–3)

which assumes the entire force F is accounted for by a shear stress in the minimum throat area. Note that this inflates the maximum estimated shear stress by a factor of 1.414∕1.207 = 1.17. Further, consider the parallel fillet welds shown in Figure 9–11 where, as in Figure 9–8, each weld transmits a force F. However, in the case of Figure 9–11, the maximum shear stress is at the minimum throat area and corresponds to Equation (9–3). Under circumstances of combined loading we ∙ Examine primary shear stresses due to external forces. ∙ Examine secondary shear stresses due to torsional and bending moments. ∙ Estimate the strength(s) of the parent metal(s). ∙ Estimate the strength of deposited weld metal. ∙ Estimate permissible load(s) for parent metal(s). ∙ Estimate permissible load for deposited weld metal. Figure 9–11 Parallel fillet welds.

l

F h

F

2F

492      Mechanical Engineering Design

9–3  Stresses in Welded Joints in Torsion Figure 9–12 illustrates a cantilever welded to a column by two fillet welds each of length l. The reaction at the support of a cantilever always consists of a shear force V and a moment M. The shear force produces a primary shear in the welds of magnitude

τ′ =

V A

(9–4)

where A is the throat area of all the welds. The moment at the support produces secondary shear or torsion of the welds, and this stress is given by the equation

τ″ =

Mr J

(9–5)

where r is the distance from the centroid of the weld group to the point in the weld of interest and J is the second polar moment of area of the weld group about the centroid of the group. When the sizes of the welds are known, these equations can be solved and the results combined to obtain the maximum shear stress. Note that r is usually the farthest distance from the centroid of the weld group. Figure 9–13 shows two welds in a group. The rectangles represent the throat areas of the welds. Weld 1 has a throat thickness t1 = 0.707h1, and weld 2 has a throat thickness t2 = 0.707h2. Note that h1 and h2 are the respective weld sizes. The throat area of both welds together is

A = A1 + A2 = t1d + t 2 b

(a)

This is the area that is to be used in Equation (9–4).

y

F

O′

x2 r

ro

b

τ″

τ′

2 G2

O

τ

1

G1

O x1 l t1

Figure 9–12 This is a moment connection; such a connection produces torsion in the welds. The shear stresses shown are resultant stresses.

G

r2

r1

d

x

Figure 9–13

t2 y

y2 x

M

Welding, Bonding, and the Design of Permanent Joints     493

The x axis in Figure 9–13 passes through the centroid G1 of weld 1. The second moment of area about this axis is Ix =

t1d 3 12

Similarly, the second moment of area about an axis through G1 parallel to the y axis is Iy =

dt 13 12

Thus the second polar moment of area of weld 1 about its own centroid is

JG1 = Ix + Iy =

t1d 3 dt 13 + 12 12

(b)

In a similar manner, the second polar moment of area of weld 2 about its centroid is

JG2 =

bt 32 t2 b3 + 12 12

(c)

The centroid G of the weld group is located at x=

A1 y1 + A2 y2 A1x1 + A2 x2   y = A A

Using Figure 9–13 again, we see that the distances r1 and r2 from G1 and G2 to G, respectively, are r1 = [(x − x1 ) 2 + y 2]1∕2   r2 = [(y2 − y) 2 + (x2 − x) 2]1∕2 Now, using the parallel-axis theorem, we find the second polar moment of area of the weld group to be

J = (JG1 + A1r 12 ) + (JG2 + A2r 22 )

(d)

This is the quantity to be used in Equation (9–5). The distance r must be measured from G and the moment M computed about G. The reverse procedure is that in which the allowable shear stress is given and we wish to find the weld size. The usual procedure is to estimate a probable weld size and then to use iteration. Observe in Equations (b) and (c) the quantities t31 and t32, respectively, which are the cubes of the weld thicknesses. These quantities are small and can be neglected. This leaves the terms t1d3∕12 and t2b3∕12, which make JG1 and JG2 linear in the weld width. Setting the weld thicknesses t1 and t2 to unity leads to the idea of treating each fillet weld as a line. The resulting second moment of area is then a unit second polar moment of area. The advantage of treating the weld size as a line is that the value of Ju is the same regardless of the weld size. Since the throat width of a fillet weld is 0.707h, the relationship between J and the unit value is

J = 0.707hJu

(9–6)

494      Mechanical Engineering Design

in which Ju is found by conventional methods for an area having unit width. The transfer formula for Ju must be employed when the welds occur in groups, as in Figure 9–12. Table 9–1 lists the throat areas and the unit second polar moments of area for the most common fillet welds encountered. The example that follows is typical of the calculations normally made. Table 9–1  Torsional Properties of Fillet Welds* Weld Throat Area Location of G 1.

A = 0.707hd

x = 0

Unit Second Polar Moment of Area Ju = d 3∕12

G y = d∕2 d y

2.

A = 1.414hd

b

x = b∕2

Ju =

d(3b2 + d 2 ) 6

Ju =

(b + d) 4 − 6b2d 2 12(b + d)

Ju =

8b3 + 6bd 2 + d 3 b4 − 12 2b + d

Ju =

(b + d) 3 6

y = d∕2 d

G y x

3.

A = 0.707h(b + d)

b

x=

b2 2(b + d)

d2 y= 2(b + d) d G

y x

4.

A = 0.707h(2b + d)

b

x=

b2 2b + d

y = d∕2 d

G

y x

5.

A = 1.414h(b + d)

b

x = b∕2

y = d∕2 G

d

y x

6.

A = 1.414πhr r

Ju = 2πr3

G

*G is the centroid of weld group; h is weld size; plane of torque couple is in the plane of the paper; all welds are of unit width.

Welding, Bonding, and the Design of Permanent Joints     495

EXAMPLE 9–1 A 50-kN load is transferred from a welded fitting into a 200-mm steel channel as illustrated in Figure 9–14. Estimate the maximum stress in the weld. Solution3 (a) Label the ends and corners of each weld by letter. See Figure 9–15. Sometimes it is desirable to label each weld of a set by number. (b) Estimate the primary shear stress τ′. As shown in Figure 9–14, each plate is welded to the channel by means of three 6-mm fillet welds. Figure 9–15 shows that we have divided the load in half and are considering only a single plate. From case 4 of Table 9–1 we find the throat area as A = 0.707(6)[2(56) + 190] = 1280 mm2

Then the primary shear stress is

τ′ =

V 25(10) 3 = = 19.5 MPa A 1280

(c) Draw the τ′ stress, to scale, at each lettered corner or end. See Figure 9–16. (d) Locate the centroid of the weld pattern. Using case 4 of Table 9–1, we find

x=

(56) 2 = 10.4 mm 2(56) + 190

This is shown as point O on Figures 9–15 and 9–16. (e) Find the distances ri (see Figure 9–16):

rA = rB = [(190∕2) 2 + (56 − 10.4) 2]1∕2 = 105 mm

rC = rD = [(190∕2) 2 + (10.4) 2]1∕2 = 95.6 mm

These distances can also be scaled from the drawing. 25 kN 100 6

200

6

C

50 kN 6

V

56 100

6

110.4

D y

O

45.6

M B

A 95

56 200

190 6

Figure 9–14 Dimensions in millimeters. 3

x

Figure 9–15 Diagram showing the weld geometry on a single plate; all dimensions in millimeters. Note that V and M represent the reaction loads applied by the welds to the plate.

We are indebted to Professor George Piotrowski of the University of Florida for the detailed steps, presented here, of his method of weld analysis. R.G.B, J.K.N.

496      Mechanical Engineering Design

Figure 9–16

F

τ″D

τD

Free-body diagram of one of the side plates.

β

τ′C

τA

τ′D

C rC O

τ″A

α

rD rA

rB

τ′B

D

B

A′

τ″C

A

τC

τB

τ″B

( f ) Find J. Using case 4 of Table 9–1 again, with Equation (9–6), we get

J = 0.707(6) [

(56) 4 8(56) 3 + 6(56) (190) 2 + (190) 3 − 12 2(56) + 190 ]

= 7.07(10) 6 mm4

(g) Find M:

M = Fl = 25(100 + 10.4) = 2760 N · m

(h) Estimate the secondary shear stresses τ″ at each lettered end or corner:

τ″A = τ″B =

τ″C = τ″D =

Mr 2760(10) 3 (105) = = 41.0 MPa J 7.07(10) 6 2760(10) 3 (95.6) 7.07(10) 6

= 37.3 MPa

(i) Draw the τ″ stress at each corner and end. See Figure 9–16. Note that this is a free-body diagram of one of the side plates, and therefore the τ′ and τ″ stresses represent what the channel is doing to the plate (through the welds) to hold the plate in equilibrium. ( j) At each point labeled, combine the two stress components as vectors (since they apply to the same area). At point A, the angle that τA″ makes with the vertical, α, is also the angle rA makes with the horizontal, which is α = tan−1(45.6∕95) = 25.64°. This angle also applies to point B. Thus

τA = τB = √ (19.5 − 41.0 sin 25.64°) 2 + (41.0 cos 25.64°) 2 = 37.0 MPa

Similarly, for C and D, β = tan−1(10.4∕95) = 6.25°. Thus

τC = τD = √ (19.5 + 37.3 sin 6.25°) 2 + (37.3 cos 6.25°) 2 = 43.9 MPa

(k) Identify the most highly stressed point: Answer

τmax = τC = τD = 43.9 MPa

Welding, Bonding, and the Design of Permanent Joints     497

9–4  Stresses in Welded Joints in Bending Figure 9–17a shows a cantilever welded to a support by fillet welds at top and bottom. A free-body diagram of the beam would show a shear-force reaction V and a moment reaction M. The shear force produces a primary shear in the welds of magnitude

τ′ =

V A

(a)

where A is the total throat area. The moment M induces a horizontal shear stress component in the welds. Treating the two welds of Figure 9–17b as lines we find the unit second moment of area to be

Iu =

bd 2 2

(b)

The second moment of area I, based on weld throat area, is

I = 0.707hIu = 0.707h

bd 2 2

(c)

The nominal throat shear stress is now found to be

τ″ =

Md∕2 Mc 1.414M = = I bdh 0.707hbd 2∕2

(d)

The model gives the coefficient of 1.414, in contrast to the predictions of Section 9–2 of 1.197 from distortion energy, or 1.207 from maximum shear. The conservatism of the model's 1.414 is not that it is simply larger than either 1.196 or 1.207, but the tests carried out to validate the model show that it is large enough. The second moment of area in Equation (d) is based on the distance d between the two welds. If this moment is found by treating the two welds as having rectangular footprints, the distance between the weld throat centroids is approximately (d + h). This would produce a slightly larger second moment of area, and result in a smaller level of stress. This method of treating welds as a line does not interfere with the conservatism of the model. It also makes Table 9–2 possible with all the conveniences that ensue. The vertical (primary) shear of Equation (a) and the horizontal (secondary) shear of Equation (d) are then combined as vectors to give τ = (τ′ 2 + τ″ 2 ) 1∕2

(e)

For an example problem, see parts (a) and (b) of Example 9–4.

y

F

Figure 9–17

y

h

b

b x

h d

z

d h

(a)

(b) Weld pattern

A rectangular cross-section cantilever welded to a support at the top and bottom edges.

498      Mechanical Engineering Design

Table 9–2  Bending Properties of Fillet Welds* Weld

Throat Area

1.

G

Location of G

A = 0.707hd

d

x=0

y 2.

b

A = 1.414hd d

G

Iu =

d3 12

Iu =

d3 6

Iu =

bd 2 2

Iu =

d2 (6b + d) 12

Iu =

2d 3 − 2d 2 y + (b + 2d)y 2 3

Iu =

d2 (3b + d) 6

Iu =

2d 3 − 2d 2 y + (b + 2d)y 2 3

y = d∕2

x = b∕2

Unit Second Moment of Area

y = d∕2

y x

3.

b

A = 1.414hb

x = b∕2

d G

y = d∕2

y x

4.

b

A = 0.707h(2b + d)

d G y

x=

b2 2b + d

y = d∕2

x

5.

b y

A = 0.707h(b + 2d) G

d

x = b∕2 y=

d2 b + 2d

x

6.

b

A = 1.414h(b + d) d

G

y

x = b∕2 y = d∕2

x

7.

b

A = 0.707h(b + 2d)

y

G

d

x

x = b∕2 y=

d 2 b + 2d

Welding, Bonding, and the Design of Permanent Joints     499

Table 9–2  (Continued) Weld

Throat Area

8.

b

Location of G x = b∕2

A = 1.414h(b + d)

G

Unit Second Moment of Area

d

Iu =

y = d∕2

d2 (3b + d) 6

y

x

9.

A = 1.414πhr r

lu = πr3

G

*Iu, unit second moment of area, is taken about a horizontal axis through G, the centroid of the weld group, h is weld size; the plane of the bending couple is normal to the plane of the paper and parallel to the y-axis; all welds are of the same size.

9–5  The Strength of Welded Joints The matching of the electrode properties with those of the parent metal is usually not so important as speed, operator appeal, and the appearance of the completed joint. The properties of electrodes vary considerably, but Table 9–3 lists the minimum properties for some electrode classes. It is preferable, in designing welded components, to select a steel that will result in a fast, economical weld even though this may require a sacrifice of other qualities such as machinability. Under the proper conditions, all steels can be welded, but best results will be obtained if steels having a UNS specification between G10140 and G10230 are chosen. All these steels have a tensile strength in the hot-rolled condition in the range of 60 to 70 kpsi. The designer can choose factors of safety or permissible working stresses with more confidence if he or she is aware of the values of those used by others. One of Table 9–3  Minimum Weld-Metal Properties AWS Electrode Number*

Tensile Strength kpsi (MPa)

Yield Strength, kpsi (MPa)

Percent Elongation

E60xx

62 (427)

50 (345)

17–25

E70xx

70 (482)

57 (393)

22

E80xx

80 (551)

67 (462)

19

E90xx

90 (620)

77 (531)

14–17

E100xx

100 (689)

87 (600)

13–16

E120xx

120 (827)

107 (737)

14

*The American Welding Society (AWS) specification code numbering system for electrodes. This system uses an E prefixed to a four- or five-digit numbering system in which the first two or three digits designate the approximate tensile strength. The last digit includes variables in the welding ­technique, such as current supply. The next-to-last digit indicates the welding position, as, for example, flat, or vertical, or overhead. The complete set of  specifications may be obtained from the AWS upon request.

500      Mechanical Engineering Design

Table 9–4  Stresses Permitted by the AISC Code for Weld Metal Type of Loading

Type of Weld

Permissible Stress

n*

Tension Butt 0.60Sy 1.67 Bearing Butt 0.90Sy 1.11 Bending Butt 0.60–0.66Sy 1.52–1.67 Simple compression

Butt

0.60Sy 1.67

Shear

Butt or fillet

0.30S†ut

*The factor of safety n has been computed by using the distortion-energy theory. †

Shear stress on base metal should not exceed 0.40Sy of base metal.

the best standards to use is the American Institute of Steel Construction (AISC) code for building construction.4 The permissible stresses are now based on the yield strength of the material instead of the ultimate strength, and the code permits the use of a variety of ASTM structural steels having yield strengths varying from 33 to 50 kpsi. Provided the loading is the same, the code permits the same stress in the weld metal as in the parent metal. Table 9–4 lists the formulas specified by the code for calculating these permissible stresses for various loading conditions. The factors of safety implied by this code are easily calculated. For tension, n = 1∕0.60 = 1.67. For shear, n = 0.577∕0.40 = 1.44, using the distortion-energy theory as the criterion of failure. It is important to observe that the electrode material is often the strongest material present. If a bar of AISI 1010 steel is welded to one of 1018 steel, the weld metal is actually a mixture of the electrode material and the 1010 and 1018 steels. Furthermore, a welded cold-drawn bar has its cold-drawn properties replaced with the hot-rolled properties in the vicinity of the weld. Finally, remembering that the weld metal is usually the strongest, do check the stresses in the parent metals. The AISC code, as well as the AWS code, for bridges includes permissible stresses when fatigue loading is present. The designer will have no difficulty in using these codes, but their empirical nature tends to obscure the fact that they have been established by means of the same knowledge of fatigue failure already discussed in Chapter 6. Of course, for structures covered by these codes, the actual stresses cannot exceed the permissible stresses; otherwise the designer is legally liable. But in general, codes tend to conceal the actual margin of safety involved. The fatigue stress-concentration factors listed in Table 9–5 are suggested for use. These factors should be used for the parent metal as well as for the weld metal. Table 9–6 gives steady-load information and minimum fillet sizes. Table 9–5  Fatigue Stress-Concentration Factors, Kfs

4

Type of Weld

Kfs

Reinforced butt weld

1.2

Toe of transverse fillet weld

1.5

End of parallel fillet weld

2.7

T-butt joint with sharp corners

2.0

For a copy, either write the AISC, 400 N. Michigan Ave., Chicago, IL 60611, or contact on the Internet at www.aisc.org.

501

Allowable Unit Force on Fillet Weld, kip/linear in

18.0 21.0 24.0 27.0 30.0 33.0 36.0

Allowable shear stress on throat, ksi (1000 psi) of fillet weld or partial penetration groove weld

60* 70* 80 90* 100 110* 120

11.14 12.99 14.85 16.70 18.57 20.41 22.27

9.55 11.14 12.73 14.32 15.92 17.50 19.09

7.96

6.37

5.57

4.77 5.57 6.36 7.16 7.95 8.75 9.54

3.98 4.64 5.30 5.97 6.63 7.29 7.95

3.18 3.71 4.24 4.77 5.30 5.83 6.36

2.39 2.78 3.18 3.58 3.98 4.38 4.77

1.59 1.86 2.12 2.39 2.65 2.92 3.18

0.795 0.930 1.06 1.19 1.33 1.46 1.59

7∕8

3∕4

5∕8

1∕2

7∕16

3∕8

5∕16

1∕4

3∕16

1∕8

1∕16

6.50

7.42

7.42 8.35 9.28 10.21 11.14

8.48 9.54 10.61 11.67 12.73

9.28 10.61 11.93 13.27 14.58 15.91

12.73 14.85 16.97 19.09 21.21 23.33 25.45

Allowable Unit Force for Various Sizes of Fillet Welds kip/linear in

1

Leg Size h, in

f = 12.73h 14.85h 16.97h 19.09h 21.21h 23.33h 25.45h

τ =

Strength Level of Weld Metal (EXX)

Schedule A: Allowable Load for Various Sizes of Fillet Welds

For minimum fillet weld size, schedule does not go above weld for every 34 in material.

*Minimum size for bridge application does not go below

Not to exceed the thickness of the thinner part.

Over 6

3 16

1 2

Over 241

5 8

3 8

Over 121 To 241 To 6

5 16

Over 34 To 121

1 4

1 8

Over 12 To 34

incl.

3 16

1 4

5 16

in fillet

in.

Weld Size, in

Over 14 To 12

*To

Material Thickness of Thicker Part Joined, in

Schedule B: Minimum Fillet Weld Size, h

From Omer W. Blodget (ed.) Stress Allowables Affect Weldment Design, D412, The James F. Lincoln Arc Welding Foundation, Cleveland, May 1991, p. 3. Reprinted by Permission of Lincoln Electric Company.

*Fillet welds actually tested by the joint AISC-AWS Task Committee. † f = 0.707h τall.

Table 9–6  Allowable Steady Loads and Minimum Fillet Weld Sizes

502      Mechanical Engineering Design

9–6  Static Loading Some examples of statically loaded joints are useful in comparing and contrasting the conventional method of analysis and the welding code methodology. EXAMPLE 9–2 A 12 -in by 2-in rectangular-cross-section UNS G10150 HR bar carries a static load of 14 kip. It is welded to a gusset plate with a 38 -in fillet weld 2 in long on both sides with an E70XX electrode as depicted in Figure 9–18. Use the welding code method. (a) Is the weld metal strength satisfactory? (b) Is the attachment strength satisfactory? Solution (a) From Table 9–6, allowable force per unit length for a 38 -in E70 electrode metal is 5.57 kip/in of weldment; thus

F = 5.57l = 5.57(4) = 22.28 kip

Since 22.28 > 14 kip, weld metal strength is satisfactory. (b) Check shear in attachment adjacent to the welds. From Table A–20, Sy = 27.5 kpsi. Then, from Table 9–4, the allowable attachment shear stress is

τall = 0.4Sy = 0.4(27.5) = 11 kpsi

The shear stress τ on the base metal adjacent to the weld is

τ=

F 14 = = 9.3 kpsi 2hl 2(0.375)2

Since τall ≥ τ, the attachment is satisfactory near the weld beads. The tensile stress in the shank of the attachment σ is

σ=

F 14 = = 14 kpsi tl (1∕2)2

The allowable tensile stress σall, from Table 9–4, is 0.6Sy and, with welding code safety level preserved,

σall = 0.6Sy = 0.6(27.5) = 16.5 kpsi

Since σ ≤ σall, the shank tensile stress is satisfactory.

Figure 9–18

1 2

in

2 in

F = 14 kip

Welding, Bonding, and the Design of Permanent Joints     503

EXAMPLE 9–3 A specially rolled A36 structural steel section for the attachment has a cross section as shown in Figure 9–19 and has yield and ultimate tensile strengths of 36 and 58 kpsi, respectively. It is statically loaded through the attachment centroid by a load of F = 24 kip. Unsymmetrical weld tracks can compensate for eccentricity such that there is no moment to be resisted by the welds. Specify the weld track lengths l1 and l2 for a 165 -in fillet weld using an E70XX electrode. This is part of a design problem in which the design variables include weld lengths and the fillet leg size. Solution The y coordinate of the section centroid of the attachment is

y=

Σyi Ai 1(0.75)2 + 3(0.375)2 = = 1.67 in ΣAi 0.75(2) + 0.375(2)

Summing moments about point B to zero gives

∑ MB = 0 = −F1b + Fy = −F1 (4) + 24(1.67)

from which

F1 = 10 kip

It follows that

F2 = 24 − 10.0 = 14.0 kip

The weld throat areas have to be in the ratio 14∕10 = 1.4, that is, l2 = 1.4l1. The weld length design variables are coupled by this relation, so l1 is the weld length design variable. The other design variable is the fillet weld leg size h, which has been decided by the problem statement. From Table 9–4, the allowable shear stress on the throat τall is

τall = 0.3(70) = 21 kpsi

The shear stress τ on the 45° throat is

τ=

=

F F = (0.707)h(l1 + l2 ) (0.707)h(l1 + 1.4l1 ) F = τall = 21 kpsi (0.707)h(2.4l1 )

from which the weld length l1 is

l1 =

24 = 2.16 in 21(0.707)0.3125(2.4)

and

l2 = 1.4l1 = 1.4(2.16) = 3.02 in l1

F1

3 8

A

Figure 9–19

in

2 in

b 2 in

F2 B

l2

3 4

in

y

+

F = 24 kip

504      Mechanical Engineering Design

These are the weld-bead lengths required by weld metal strength. The attachment shear stress allowable in the base metal, from Table 9–4, is τall = 0.4Sy = 0.4(36) = 14.4 kpsi The shear stress τ in the base metal adjacent to the weld is from which

τ=

F F F = = = τall = 14.4 kpsi h(l1 + l2 ) h(l1 + 1.4l1 ) h(2.4l1 ) F 24 = = 2.22 in 14.4h(2.4) 14.4(0.3125)2.4

l1 =

l 2 = 1.4l1 = 1.4(2.22) = 3.11 in

These are the weld-bead lengths required by base metal (attachment) strength. The base metal controls the weld lengths. For the allowable tensile stress σall in the shank of the attachment, the AISC allowable for tension members is 0.6Sy; therefore, σall = 0.6Sy = 0.6(36) = 21.6 kpsi The nominal tensile stress σ is uniform across the attachment cross section because of the load application at the centroid. The stress σ is F 24 σ= = = 10.7 kpsi A 0.75(2) + 2(0.375) Since σ ≤ σall, the shank section is satisfactory. With l1 set to a nominal 214 in, l2 should be 1.4(2.25) = 3.15 in. Decision Set l1 = 214 in, l2 = 314 in. The small magnitude of the departure from l2∕l1 = 1.4 is not serious. The joint is essentially moment-free.

EXAMPLE 9–4 Perform an adequacy assessment of the statically loaded welded cantilever carrying 500 lbf depicted in Figure 9–20. The cantilever is made of AISI 1018 HR steel and welded with a 38 -in fillet weld as shown in the figure. An E6010 electrode was used, and the design factor was 3.0. (a) Use the conventional method for the weld metal. (b) Use the conventional method for the attachment (cantilever) metal. (c) Use a welding code for the weld metal. Figure 9–20 3 8

in

6 in 3 8

in

2 in

F = 500 lbf

Welding, Bonding, and the Design of Permanent Joints     505

Solution (a) From Table 9–3, Sy = 50 kpsi, Sut = 62 kpsi. From Table 9–2, second pattern, b = 0.375 in, d = 2 in, so

A = 1.414hd = 1.414(0.375)2 = 1.06 in2

Iu = d 3∕6 = 23∕6 = 1.33 in3

I = 0.707hIu = 0.707(0.375)1.33 = 0.353 in4

Primary shear:

τ′ =

F 500(10−3 ) = = 0.472 kpsi A 1.06

Secondary shear: Mr 500(10−3 )(6)(1) = = 8.50 kpsi I 0.353 The shear magnitude τ is from the vector addition

τ″ =

τ = (τ′ 2 + τ″ 2 ) 1∕2 = (0.4722 + 8.502 ) 1∕2 = 8.51 kpsi

The factor of safety based on a minimum strength and the distortion-energy criterion is Ssy 0.577(50) = = 3.39 τ 8.51 Since n ≥ nd, that is, 3.39 ≥ 3.0, the weld metal has satisfactory strength. (b) From Table A–20, minimum strengths are Sut = 58 kpsi and Sy = 32 kpsi. Then n=

Answer

σ=

500(10−3 )6 M M = 2 = = 12 kpsi I∕c bd ∕6 0.375(22 )∕6

Answer

n=

Sy 32 = = 2.67 σ 12

Since n < nd, that is, 2.67 < 3.0, the joint is unsatisfactory as to the attachment strength. (c) From part (a), τ = 8.51 kpsi. For an E6010 electrode Table 9–6 gives the allowable shear stress τall as 18 kpsi. Since τ < τall, the weld is satisfactory. Since the code already has a design factor of 0.577(50)∕18 = 1.6 included at the equality, the corresponding factor of safety to part (a) is Answer

n = 1.6

18 = 3.38 8.51

which is consistent.

9–7  Fatigue Loading The conventional methods for fatigue are applicable. With only shear loading, we will adapt the fluctuating-stress diagram and a fatigue failure criterion for shear stresses and shear strengths, according to the method outlined in the last subsection of Section 6–18. For the surface factor of Equation 6–18, an as-forged surface should always be assumed for weldments unless a superior finish is specified and obtained. Some examples of fatigue loading of welded joints follow.

506      Mechanical Engineering Design

EXAMPLE 9–5 The AISI 1018 HR steel strap of Figure 9–21 has a 1000 lbf, completely reversed load applied. Determine the factor of safety of the weldment for infinite life. Solution From Table A–20 for the 1018 attachment metal the strengths are Sut = 58 kpsi and Sy = 32 kpsi. For the E6010 electrode, from Table 9–3 Sut = 62 kpsi and Sy = 50 kpsi. The fatigue stress-concentration factor, from Table 9–5, is Kfs = 2.7. From Table 6–2, ka = 12.7(58)−0.758 = 0.58. For case 2 of Table 9–5, the shear area is: A = 1.414(0.375)(2) = 1.061 in2

For a uniform shear stress on the throat, kb = 1. From Equation (6–25), for torsion (shear),

k c = 0.59

kd = ke = 1

From Equations (6–10) and (6–17),

Sse = 0.58(1)0.59(1) (1)(1)0.5(58) = 9.9 kpsi

From Table 9–5, Kfs = 2.7. Only primary shear is present. So, with Fa = 1000 lbf and Fm = 0

τ′a =

Kfs Fa A

=

2.7(1000) = 2545 psi 1.061

τ′m = 0 psi

In the absence of a mean stress component, the fatigue factor of safety nf is given by nf =

Answer

Figure 9–21

Sse 9900 = = 3.9 τ′a 2545

1018

E6010

2 in

3 8

in

2 in 4- × 7.25-in channel 1 2

in 1018

1000 lbf completely reversed

Welding, Bonding, and the Design of Permanent Joints     507

EXAMPLE 9–6 The AISI 1018 HR steel strap of Figure 9–22 has a repeatedly applied load of 2000 lbf (Fa = Fm = 1000 lbf ). Determine the fatigue factor of safety of the weldment. Solution From Table 6–2, ka = 12.7(58)−0.758 = 0.58. From case 2 of Table 9–2 A = 1.414(0.375) (2) = 1.061 in2

For uniform shear stress on the throat kb = 1. Equation (6–25):

kc = 0.59

Equations (6–10) and (6–17):

Sse = (0.58) (0.59)(0.5) (58) = 9.92 kpsi

From Table 9–5, Kfs = 2. Only primary shear is present:

τ′a = τ′m =

Kfs Fa A

=

2(1000) = 1885 psi 1.061

Estimate the shear modulus of rupture with Equation (6–59),

Ssu = 0.67 Sut = 0.67 (58) = 38.9 kpsi

Choosing the Goodman fatigue failure criterion of Equation (6–41), adapted for shear, gives τ′a τ′m −1 1.885 1.885 −1 nf = ( + ) = ( + = 4.2 Sse Ssu 9.92 38.9 )

Answer

Figure 9–22

W 4- × 13-in I beam 1018

3 8

E6010 in

2 in 1018 1 2

in

2000 lbf repeatedly applied (0–2000 lbf)

9–8  Resistance Welding The heating and consequent welding that occur when an electric current is passed through several parts that are pressed together is called resistance welding. Spot welding and seam welding are forms of resistance welding most often used. The advantages of resistance welding over other forms are the speed, the accurate regulation of time and heat, the uniformity of the weld, and the mechanical properties that result. In addition the process is easy to automate, and filler metal and fluxes are not needed. The spot- and seam-welding processes are illustrated schematically in Figure 9–23. Seam welding is actually a series of overlapping spot welds, since the current is applied in pulses as the work moves between the rotating electrodes.

508      Mechanical Engineering Design

Figure 9–23 (a) Spot welding; (b) seam welding.

(a)

(b)

Failure of a resistance weld occurs either by shearing of the weld or by tearing of the metal around the weld. Because of the possibility of tearing, it is good practice to avoid loading a resistance-welded joint in tension. Thus, for the most part, design so that the spot or seam is loaded in pure shear. The shear stress is then simply the load divided by the area of the spot. Because the thinner sheet of the pair being welded may tear, the strength of spot welds is often specified by stating the load per spot based on the thickness of the thinnest sheet. Such strengths are best obtained by experiment. Somewhat larger factors of safety should be used when parts are fastened by spot welding rather than by bolts or rivets, to account for the metallurgical changes in the materials due to the welding.

9–9  Adhesive Bonding5 The use of polymeric adhesives to join components for structural, semistructural, and nonstructural applications has expanded greatly in recent years as a result of the unique advantages adhesives may offer for certain assembly processes and the development of new adhesives with improved robustness and environmental acceptability. The increasing complexity of modern assembled structures and the diverse types of materials used have led to many joining applications that would not be possible with more conventional joining techniques. Adhesives are also being used either in conjunction with or to replace mechanical fasteners and welds. Reduced weight, sealing capabilities, and reduced part count and assembly time, as well as improved fatigue and corrosion resistance, all combine to provide the designer with opportunities for customized assembly. The worldwide size of the adhesive and sealant industry is approximately 40 billion Euro dollars, and the United States market is about 12 billion US dollars.6 Figure 9–24 illustrates the numerous places where adhesives are used on a modern automobile. Indeed, the fabrication of many modern vehicles, devices, and structures is dependent on adhesives. In well-designed joints and with proper processing procedures, use of adhesives can result in significant reductions in weight. Eliminating mechanical fasteners 5

For a more extensive discussion of this topic, see J. E. Shigley and C. R. Mischke, Mechanical Engineering Design, 6th ed., McGraw-Hill, New York, 2001, Section 9–11. This section was prepared with the assistance of Professor David A. Dillard, Professor of Engineering Science and Mechanics and Director of the Center for Adhesive and Sealant Science, Virginia Polytechnic Institute and State University, Blacksburg, Virginia, and with the encouragement and technical support of the Bonding Systems Division of 3M, Saint Paul, Minnesota. 6 From E. M. Petrie, Handbook of Adhesives and Sealants, 2nd ed., McGraw-Hill, New York, 2007.

Welding, Bonding, and the Design of Permanent Joints     509 (7) Exterior Trim (14) Sound Insulation Hem Flange (1) Engine (2) Compartment (12)

Windshield/ Windows

Bumper Assembly

(4) Wheel Housing (10)

Interior Trim (11)

(1)

(8)

(2) Antiflutter

(6) Paint Shop

(5) Body-in-White

(9) Light Assemblies (13)

(15) Exterior Body Panels

(8)

(10) Brake/ Transmission

(1)

(4)

(1)

(9) (3) (13) Panel Reinforcements

(10)

Figure 9–24 Diagram of an automobile body showing at least 15 locations at which adhesives and sealants could be used or are being used. Particular note should be made of the windshield (8), which is considered a load-bearing structure in modern automobiles and is adhesively bonded. Also attention should be paid to hem flange bonding (1), in which adhesives are used to bond and seal. Adhesives are used to bond friction surfaces in brakes and clutches (10). Antiflutter adhesive bonding (2) helps control deformation of hood and trunk lids under wind shear. Thread-sealing adhesives are used in engine applications (12). Source: A. V. Pocius, Adhesion and Adhesives Technology, 2nd edition, Hanser Publishers, Munich, 2002.

eliminates the weight of the fasteners, and also may permit the use of thinner-gauge materials because stress concentrations associated with the holes are eliminated. The capability of polymeric adhesives to dissipate energy can significantly reduce noise, vibration, and harshness (NVH), crucial in modern automobile performance. Adhesives can be used to assemble heat-sensitive materials or components that might be damaged by drilling holes for mechanical fasteners. They can be used to join dissimilar materials or thin-gauge stock that cannot be joined through other means. Types of Adhesive There are numerous adhesive types for various applications. They may be classified in a variety of ways depending on their chemistry (e.g., epoxies, polyurethanes, polyimides), their form (e.g., paste, liquid, film, pellets, tape), their type (e.g., hot melt, reactive hot melt, thermosetting, pressure sensitive, contact), or their load-carrying capability (structural, semistructural, or nonstructural). Structural adhesives are relatively strong adhesives that are normally used well below their glass transition temperature; common examples include epoxies and certain acrylics. Such adhesives can carry significant stresses, and they lend themselves to structural applications. For many engineering applications, semistructural applications (where failure would be less critical) and nonstructural applications (of headliners, etc., for aesthetic purposes) are also of significant interest to the design engineer, providing cost-effective means required for assembly of finished products. These include contact adhesives, where a solution or emulsion containing an elastomeric adhesive is coated onto both adherends, the solvent is allowed to evaporate, and then the two adherends are brought into contact. Examples include rubber cement and adhesives used to bond laminates to countertops. Pressure-sensitive adhesives are very low modulus elastomers that deform easily under small pressures, permitting them to wet surfaces. When the substrate and adhesive are brought into intimate contact, van der Waals forces are sufficient to maintain the contact and provide relatively durable bonds. Pressure-sensitive adhesives are normally purchased as tapes or labels for nonstructural applications, although there are also double-sided foam tapes that can be used in semistructural applications. As the name implies, hot melts

510      Mechanical Engineering Design

Table 9–7  Mechanical Performance of Various Types of Adhesives Adhesive Chemistry or Type

Room Temperature Lap-Shear Strength, MPa (psi)

Peel Strength per Unit Width, kN/m (lbf/in)

Pressure-sensitive

0.01–0.07 (2–10)

0.18–0.88 (1–5)

Starch-based

0.07–0.7 (10–100)

0.18–0.88 (1–5)

Cellosics

0.35–3.5 (50–500)

0.18–1.8 (1–10)

Rubber-based

0.35–3.5 (50–500)

Formulated hot melt

0.35–4.8

1.8–7 (10–40)

(50–700)

0.88–3.5

(5–20)

Synthetically designed hot melt

0.7–6.9

(100–1000)

0.88–3.5

(5–20)

PVAc emulsion (white glue)

1.4–6.9

(200–1000)

0.88–1.8

(5–10)

Cyanoacrylate

6.9–13.8 (1000–2000)

0.18–3.5 (1–20)

Protein-based

6.9–13.8 (1000–2000)

0.18–1.8 (1–10)

Anaerobic acrylic

6.9–13.8

0.18–1.8

Urethane

6.9–17.2 (1000–2500)

(1000–2000)

(1–10)

1.8–8.8 (10–50)

Rubber-modified acrylic

13.8–24.1

(2000–3500)

1.8–8.8

(10–50)

Modified phenolic

13.8–27.6

(2000–4000)

3.6–7

(20–40)

Unmodified epoxy

10.3–27.6

(1500–4000)

0.35–1.8

(2–10)

Bis-maleimide

13.8–27.6 (2000–4000)

0.18–3.5 (1–20)

Polyimide

13.8–27.6 (2000–4000)

0.18–0.88 (1–5)

Rubber-modified epoxy

20.7–41.4

(3000–6000)

4.4–14

(25–80)

Source: Data from A. V. Pocius, Adhesion and Adhesives Technology, 2nd ed., Hanser Gardner Publishers, Ohio, 2002.

become liquid when heated, wetting the surfaces and then cooling into a solid polymer. These materials are increasingly applied in a wide array of engineering applications by more sophisticated versions of the glue guns in popular use. Anaerobic adhesives cure within narrow spaces deprived of oxygen; such materials have been widely used in mechanical engineering applications to lock bolts or bearings in place. Cure in other adhesives may be induced by exposure to ultraviolet light or electron beams, or it may be catalyzed by certain materials that are ubiquitous on many surfaces, such as water. Table 9–7 presents important strength properties of commonly used adhesives. Stress Distributions Good design practice normally requires that adhesive joints be constructed in such a manner that the adhesive carries the load in shear rather than tension. Bonds are typically much stronger when loaded in shear rather than in tension across the bond plate. Lap-shear joints represent an important family of joints, both for test specimens to evaluate adhesive properties and for actual incorporation into practical designs. Generic types of lap joints that commonly arise are illustrated in Figure 9–25. The simplest analysis of lap joints suggests the applied load is uniformly distributed over the bond area. Lap joint test results, such as those obtained following the ASTM D1002 for single-lap joints, report the "apparent shear strength" as the breaking load divided by the bond area. Although this simple analysis can be adequate for stiff adherends bonded with a soft adhesive over a relatively short bond length, significant peaks in shear stress occur except for the most flexible adhesives. In an effort to point out the problems associated with such practice, ASTM D4896 outlines some of the concerns associated with taking this simplistic view of stresses within lap joints.

Welding, Bonding, and the Design of Permanent Joints     511 (a)

(b)

(c)

(d)

Figure 9–25 Common types of lap joints used in mechanical design: (a) single lap; (b) double lap; (c) scarf; (d) bevel; (e) step; ( f ) butt strap; (g) double butt strap; (h) tubular lap. Source: Adapted from R. D. Adams, J. Comyn. and W. C. Wake, Structural Adhesive Joints in Engineering, 2nd ed., Chapman and Hall, New York, 1997.

(e)

(f)

(g)

(h)

In 1938, O. Volkersen presented an analysis of the lap joint, known as the shear-lag model. It provides valuable insights into the shear-stress distributions in a host of lap joints. Bending induced in the single-lap joint due to eccentricity significantly complicates the analysis, so here we will consider a symmetric double-lap joint to illustrate the principles. The shear-stress distribution for the double lap joint of Figure 9–26 is given by

τ(x) =

2Eo to − Ei ti Pω Pω cosh(ωx) + [ 4b sinh(ωl∕2) 4b cosh(ωl∕2) ( 2Eo to + Ei ti ) (αi − αo ) ΔTω + sinh(ωx) [1∕(Eo to ) + 2∕(Ei ti )] cosh(ωl∕2) ]

(9–7)

Figure 9–26

P 2 P 2

P (a) y

to

l 2

to

h

l 2

h

x

(b)

ti

Double-lap joint.

512      Mechanical Engineering Design

where ω=√

G 1 2 + h ( Eo to Ei ti )

and Eo, to, αo, and Ei, ti, αi, are the modulus, thickness, coefficient of thermal expansion for the outer and inner adherend, respectively; G, h, b, and l are the shear modulus, thickness, width, and length of the adhesive, respectively; and ΔT is a change in temperature of the joint. If the adhesive is cured at an elevated temperature such that the stress-free temperature of the joint differs from the service temperature, the mismatch in thermal expansion of the outer and inner adherends induces a thermal shear across the adhesive. EXAMPLE 9–7 The double-lap joint depicted in Figure 9–26 consists of aluminum outer adherends and an inner steel adherend. The assembly is cured at 250°F and is stress-free at 200°F. The completed bond is subjected to an axial load of 2000 lbf at a service temperature of 70°F. The width b is 1 in, the length of the bond l is 1 in. Additional information is tabulated below: G, psi

E, psi

6

α, in/(in · °F)

Thickness, in

−6

Adhesive 0.2(10 ) 55(10 ) 0.020 Outer adherend

10(106) 13.3(10−6) 0.150

Inner adherend

30(106)  6.0(10−6) 0.100

Sketch a plot of the shear stress as a function of the length of the bond due to (a) thermal stress, (b) loadinduced stress, and (c) the sum of stresses in a and b; and (d) find where the largest shear stress is m ­ aximum. Solution In Equation (9–7) the parameter ω is given by

ω=√

=√

G 1 2 + ( h Eo to Ei ti ) 0.2(106 ) 1 2 + = 3.65 in−1 6 [ 0.020 10(10 )0.15 30(106 )0.10 ]

(a) For the thermal stress component, αi − αo = 6(10−6) − 13.3(10−6) = −7.3(10−6) in/(in · °F), ΔT = 70 − 200 = −130°F,

τth (x) =

τth (x) =

(αi − αo ) ΔT ω sinh(ωx) [1∕(Eo to ) + 2∕(Ei ti )] cosh(ωl∕2) −7.3(10−6 )(−130)3.65 sinh(3.65x) 3.65(1) 1 2 [ 10(106 )0.150 + 30(106 )0.100 ] cosh[ 2 ]

= 816.4 sinh(3.65x)

The thermal stress is plotted in Figure 9–27 and tabulated at x = −0.5, 0, and 0.5 in the table below. (b) The bond is "balanced" (Eoto = Eiti∕2), so the load-induced stress is given by

τP (x) =

Pω cosh(ωx) 2000(3.65) cosh(3.65x) = = 604.1 cosh(3.65x) 4b sinh(ωl∕2) 4(1)3.0208

(1)

Welding, Bonding, and the Design of Permanent Joints     513

The load-induced stress is plotted in Figure 9–27 and tabulated at x = −0.5, 0, and 0.5 in the table below. (c) Total stress table (in psi):

τ(−0.5)

τ(0)

τ(0.5)

Thermal only

−2466

0

2466

Load-induced only

1922

604

1922

Combined

−544 604 4388

(d) The maximum shear stress predicted by the shear-lag model will always occur at the ends. See the plot in Fig­ ure 9–27. Since the residual stresses are always present, significant shear stresses may already exist prior to application of the load. The large stresses present for the combined-load case could result in local yielding of a ductile adhesive or failure of a more brittle one. The significance of the thermal stresses serves as a caution against joining dissimilar adherends when large temperature changes are involved. Note also that the average shear stress due to the load is τavg = P∕(2bl) = 1000 psi. Equation (1) produced a maximum of 1922 psi, almost double the average. Shear stress τ (psi)

Figure 9–27

4000

Plot for Example 9–7. Combined

3000

–0.4

2000

Thermal

1000

Load induced

–0.2

0.2

0.4

x (in)

–1000 –2000

Although design considerations for single-lap joints are beyond the scope of this chapter, one should note that the load eccentricity is an important aspect in the stress state of single-lap joints. Adherend bending can result in shear stresses that may be as much as double those given for the double-lap configuration (for a given total bond area). In addition, peel stresses can be quite large and often account for joint failure. Finally, plastic bending of the adherends can lead to high strains, which less ductile adhesives cannot withstand, leading to bond failure as well. Bending stresses in the adherends at the end of the overlap can be four times greater than the average stress within the adherend; thus, they must be considered in the design. Figure 9–28 shows the shear and peel stresses present in a typical single-lap joint that corresponds to the ASTM D1002 test specimen. Note that the shear stresses are significantly larger than predicted by the Volkersen analysis, a result of the increased adhesive strains associated with adherend bending. Joint Design Some basic guidelines that should be used in adhesive joint design include: ∙ Design to place bondline in shear, not peel. Beware of peel stresses focused at bond terminations. When necessary, reduce peel stresses through tapering the adherend ends, increasing bond area where peel stresses occur, or utilizing rivets at bond terminations where peel stresses can initiate failures.

514      Mechanical Engineering Design

Figure 9–28 Stresses within a single-lap joint. (a) Lap-joint tensile forces have a line of action that is not initially parallel to the adherend sides. (b) As the load increases the adherends and bond bend. (c) In the locality of the end of an adherend peel and shear stresses appear, and the peel stresses often induce joint failure. (d) The seminal Goland and Reissner stress predictions Source: J. Appl Mech., vol 77, 1944 are shown. (Note that the predicted shear-stress maximum is higher than that predicted by the Volkersen shear-lag model because of adherend bending.)

(a)

(b) Peel and shear stresses

(c) ASTM D 1002-94 l = 0.5 in (12.7 mm) t = 0.064 in (1.6 mm) Aluminum: E = 10 Msi (70 GPa) Epoxy: Ea = 500 ksi (3.5 GPa)

Stress (psi) 10000 8000 6000

σ, Goland and Reissner

Stresses shown for an applied load of P = 1000 lbf (4.4 kN) Note: For very long joints, Volkersen predicts only 50% of the G-R shear stress.

τ, Volkersen

4000

τave

2000

–0.2

0.1

–0.1

τ, Goland and Reissner

0.2

x (in)

–2000 (d )

∙ Where possible, use adhesives with adequate ductility. The ability of an adhesive to yield reduces the stress concentrations associated with the ends of joints and increases the toughness to resist debond propagation. ∙ Recognize environmental limitations of adhesives and surface preparation methods. Exposure to water, solvents, and other diluents can significantly degrade adhesive performance in some situations, through displacing the adhesive from the surface or degrading the polymer. Certain adhesives may be susceptible to environmental stress cracking in the presence of certain solvents. Exposure to ultraviolet light can also degrade adhesives. ∙ Design in a way that permits or facilitates inspections of bonds where possible. A missing rivet or bolt is often easy to detect, but debonds or unsatisfactory adhesive bonds are not readily apparent. ∙ Allow for sufficient bond area so that the joint can tolerate some debonding before going critical. This increases the likelihood that debonds can be detected. Having some regions of the overall bond at relatively low stress levels can significantly improve durability and reliability.

Welding, Bonding, and the Design of Permanent Joints     515

∙ Where possible, bond to multiple surfaces to offer support to loads in any direction. Bonding an attachment to a single surface can place peel stresses on the bond, whereas bonding to several adjacent planes tends to permit arbitrary loads to be carried predominantly in shear. ∙ Adhesives can be used in conjunction with spot welding. The process is known as weld bonding. The spot welds serve to fixture the bond until it is cured.

Figure 9–29 presents examples of improvements in adhesive bonding. Original Original

Improved Improved

Original Original

Improved Improved

Figure 9–29 Design practices that improve adhesive bonding. (a) Gray load vectors are to be avoided as resulting strength is poor. (b) Means to reduce peel stresses in lap-type joints.

(a) (a)

Peel stresses can be a problem Peel stresses be of a problem at ends of lapcan joints all types at ends of lap joints of all types

Tapered to reduce peel Tapered to reduce peel

Rivet, spot weld, or Rivet, weld, bolt to spot reduce peelor bolt to reduce peel

Mechanically reduce peel Mechanically reduce peel

Larger bond area to reduce peel Larger bond area to reduce peel (b) (b)

516      Mechanical Engineering Design

References Good references are available for analyzing and designing adhesive bonds, including the following: R. D. Adams, J. Comyn, and W. C. Wake, Structural Adhesive Joints in  Engineering, 2nd ed., Chapman and Hall, New York, 1997. G. P. Anderson, S. J. Bennett, and K. L. DeVries, Analysis and Testing of   Adhesive Bonds, Academic Press, New York, 1977. H. F. Brinson (ed.), Engineered Materials Handbook, vol. 3: Adhesives and  Sealants, ASM International, Metals Park, Ohio, 1990. A. J. Kinloch, Adhesion and Adhesives: Science and Technology, Chapman   and Hall, New York, 1987. A. J. Kinloch (ed.), Durability of Structural Adhesives, Applied Science   Publishers, New York, 1983. R. W. Messler, Jr., Joining of Materials and Structures, Elsevier Butterworth  Heinemann, Mass., 2004. E. M. Petrie, Handbook of Adhesives and Sealants, 2nd ed., McGraw-Hill,   New York, 2007. A. V. Pocius, Adhesion and Adhesives Technology: An Introduction, 2nd ed.,   Hanser Gardner, Ohio, 1997.

PROBLEMS 9–1 to 9–4 h d

F b Problems 9–1 to 9–4

9–5 to 9–8

9–9 to 9–12

The figure shows a horizontal steel bar of thickness h loaded in steady tension and welded to a vertical support. Find the load F that will cause an allowable shear stress, τallow, in the throats of the welds. Problem Number

b

d

h

τallow

9–1

50 mm

50 mm

5 mm

140 MPa

9–2

2 in

2 in

9–3

50 mm

30 mm

9–4

4 in

2 in

5 16

in

5 mm 5 16

in

25 kpsi 140 MPa 25 kpsi

For the weldments of Problems 9–1 to 9–4, the electrodes are specified in the table. For the electrode metal indicated, what is the allowable load on the weldment? Problem Number

Reference Problem

Electrode

9–5

9–1

E7010

9–6

9–2

E6010

9–7

9–3

E7010

9–8

9–4

E6010

The materials for the members being joined in Problems 9–1 to 9–4 are specified below. What load on the weldment is allowable because member metal is incorporated in the welds?

Welding, Bonding, and the Design of Permanent Joints     517

9–13 to 9–16

Problem Number

Reference Problem

Bar

Vertical Support

9–9

9–1

1018 CD

1018 HR

9–10

9–2

1020 CD

1020 CD

9–11

9–3

1035 HR

1035 CD

9–12

9–4

1035 HR

1020 CD

A steel bar of thickness h is welded to a vertical support as shown in the figure. What is the shear stress in the throat of the welds due to the force F? Problem Number

b

d

h

F

9–13

50 mm

50 mm

5 mm

100 kN

9–14

2 in

2 in

9–15

50 mm

30 mm

9–16

4 in

2 in

in

40 kip

5 mm

100 kN

5 16

5 16

h d

F b

40 kip

in

Problems 9–13 to 9–16

9–17 Prove the torsional properties for weld section 3 of Table 9–1. 9–18 Prove the torsional properties for weld section 4 of Table 9–1. steel bar of thickness h, to be used as a beam, is welded to a vertical support by A 9–19 two fillet welds as shown in the figure. to 9–22 (a) Find the safe bending force F if the allowable shear stress in the welds is τallow.

(b) In part a, you found a simple expression for F in terms of the allowable shear stress. Find the allowable load if the electrode is E7010, the bar is hot-rolled 1020, and the support is hot-rolled 1015. F

Problem Number

b

c

d

h

τallow

9–19

50 mm

150 mm

50 mm

5 mm

140 MPa

9–20

2 in

6 in

2 in

9–21

50 mm

150 mm

30 mm

9–22

4 in

6 in

2 in

5 16

in

5 mm 5 16

in

25 kpsi

h d b

140 MPa 25 kpsi

c

Problems 9–19 to 9–22

9–23 Prove the bending properties for weld section 4 of Table 9–2. 9–24 Prove the bending properties for weld section 5 of Table 9–2. 9–25 The figure shows a weldment just like that for Problems 9–19 to 9–22 except there are four welds instead of two. Find the safe bending force F if the allowable shear to 9–28 stress in the welds is τallow. F

Problem Number

b

c

d

h

τallow

9–25

50 mm

150 mm

50 mm

5 mm

140 MPa

9–26

2 in

6 in

2 in

9–27

50 mm

150 mm

30 mm

9–28

4 in

6 in

2 in

5 16

in

5 mm 5 16

in

25 kpsi 140 MPa 25 kpsi

h d b

c

Problems 9–25 to 9–28

518      Mechanical Engineering Design

A steel bar of thickness h is welded to a vertical support as shown in the figure. Find 9–29 the safe force F if the maximum allowable shear stress in the welds is τallow. to 9–32 h

Problem Number

b

c

d

h

τallow

9–29

50 mm

150 mm

50 mm

5 mm

140 MPa

9–30

2 in

6 in

2 in

9–31

50 mm

150 mm

30 mm

9–32

4 in

6 in

2 in

d b

c 45°

F

Problems 9–29 to 9–32

5 16

25 kpsi

in

5 mm 5 16

140 MPa 25 kpsi

in

9–33 The weldment shown in the figure is subjected to an alternating force F. The hot-rolled steel bar has a thickness h and is of AISI 1010 steel. The vertical support is likewise to 9–36 AISI 1010 HR steel. The electrode is given in the table below. Estimate the fatigue load F the bar will carry if three fillet welds are used.

h h d

Problem Number

b

d

h

Electrode

9–33

50 mm

50 mm

5 mm

E6010

9–34

2 in

2 in

9–35

50 mm

30 mm

9–36

4 in

2 in

F

b Problems 9–33 to 9–36

in

E6010

5 mm

E7010

5 16

5 16

E7010

in

9–37 The permissible shear stress for the weldments shown is 20 kpsi. Based on the weld-

ment of the cylinder and plate A, estimate the load, F, that will cause this stress in the weldment throat. 6 in

F

8 in

B Problem 9–37 A C

1 4

3-in dia

in 1 4

in

9–38 Repeat the analysis of Problem 9–37 for the weldment at the wall. Base your results on an analysis of points B and C,

A steel bar of thickness h is subjected to a bending force F. The vertical support is 9–39 stepped such that the horizontal welds are b1 and b2 long. Determine F if the maximum to 9–40 allowable shear stress is τallow. Problem Number

b1

b2

c

d

9–39

2 in

4 in

6 in

4 in

9–40

30 mm

50 mm

150 mm

50 mm

h 5 16

in

5 mm

τallow 25 kpsi 140 MPa

Welding, Bonding, and the Design of Permanent Joints     519 F b1 Problems 9–39 to 9–40

h

h d

b2

h

c

9–41 In the design of weldments in torsion it is helpful to have a hierarchical perception of the relative efficiency of common patterns. For example, the weld-bead patterns shown in Table 9–1 can be ranked for desirability. Assume the space available is an a × a square. Use a formal figure of merit that is directly proportional to J and inversely proportional to the volume of weld metal laid down:

fom =

0.707hJu Ju J = = 1.414 vol hl (h2∕2)l

A tactical figure of merit could omit the constant, that is, fom′ = Ju ∕(hl). Rank the six patterns of Table 9–1 from most to least efficient.

9–42 The space available for a weld-bead pattern subject to bending is a × a. Place the

patterns of Table 9–2 in hierarchical order of efficiency of weld metal placement to resist bending. A formal figure of merit can be directly proportion to I and inversely proportional to the volume of weld metal laid down:

fom =

0.707hIu Iu I = 2 = 1.414 vol hl (h ∕2)l

The tactical figure of merit can omit the constant 1.414, that is, fom′ = Iu∕(hl). Omit the patterns intended for T beams and I beams. Rank the remaining seven.

9–43 The attachment shown in the figure is made of 1018 HR steel 12 mm thick. The static

force is 100 kN. The member is 75 mm wide. Specify the weldment (give the pattern, electrode number, type of weld, length of weld, and leg size). 100 1018 HR

37.5 dia. 12

Problem 9–43

1018 HR

Dimensions in millimeters.

75 dia. F = 100 kN

225

9–44 The attachment shown carries a static bending load of 12 kN. The attachment

length, l1, is 225 mm. Specify the weldment (give the pattern, electrode number, type of weld, length of weld, and leg size). l1 150 37.5 dia.

Problem 9–44

12

Dimensions in millimeters.

75 dia.

1018 HR 1018 HR 100

F = 12 kN

520      Mechanical Engineering Design

9–45 The attachment in Problem 9–44 has not had its length determined. The static force is 12 kN. Specify the weldment (give the pattern, electrode number, type of weld, length of bead, and leg size). Specify the attachment length.

9–46 A vertical column of 1018 hot-rolled steel is 10 in wide. An attachment has been

designed to the point shown in the figure. The static load of 20 kip is applied, and the clearance a of 6.25 in has to be equaled or exceeded. The attachment is also 1018 hot-rolled steel, to be made from 12 -in plate with weld-on bosses when all dimensions are known. Specify the weldment (give the pattern, electrode number, type of weld, length of weld bead, and leg size). Specify also the length l1 for the attachment.

1018 HR

1-in dia. 6 in 2- in dia.

d Problem 9–46 1018 HR

b

a F = 20 kip l1

9–47 Write a computer program to assist with a task such as that of Problem 9–46 with a

rectangular weld-bead pattern for a torsional shear joint. In doing so solicit the force F, the clearance a, and the largest allowable shear stress. Then, as part of an iterative loop, solicit the dimensions b and d of the rectangle. These can be your design variables. Output all the parameters after the leg size has been determined by computation. In effect this will be your adequacy assessment when you stop iterating. Include the figure of merit Ju∕(hl) in the output. The fom and the leg size h with available width will give you a useful insight into the nature of this class of welds. Use your program to verify your solutions to Problem 9–46.

9–48 Fillet welds in joints resisting bending are interesting in that they can be simpler than

those resisting torsion. From Problem 9–42 you learned that your objective is to place weld metal as far away from the weld-bead centroid as you can, but distributed in an orientation parallel to the x axis. Furthermore, placement on the top and bottom of the built-in end of a cantilever with rectangular cross section results in parallel weld beads, each element of which is in the ideal position. The object of this problem is to study the full weld bead and the interrupted weld-bead pattern. Consider the case of Figure 9–17, with F = 10 kips, the beam length is 10 in, b = 8 in, and d = 8 in. For the second case, for the interrupted weld consider a centered gap of b1 = 2 in existing in the top and bottom welds. Study the two cases with τall = 12.8 kpsi. What do you notice about τ, σ, and τmax? Compare the fom′.

9–49 For a rectangular weld-bead track resisting bending, develop the necessary equations

to treat cases of vertical welds, horizontal welds, and weld-all-around patterns with depth d and width b and allowing central gaps in parallel beads of length b1 and d1. Do this by superposition of parallel tracks, vertical tracks subtracting out the gaps.

Welding, Bonding, and the Design of Permanent Joints     521

Then put the two together for a rectangular weld bead with central gaps of length b1 and d1. Show that the results are A = 1.414(b − b1 + d − d1 )h Iu =

(b − b1 )d 2 d 3 − d 31 + 2 6

I = 0.707hIu l = 2(b − b1 ) + 2(d − d1 ) fom =

Iu hl

9–50 Write a computer program based on the Problem 9–49 protocol. Solicit the largest

allowable shear stress, the force F, and the clearance a, as well as the dimensions b and d. Begin an iterative loop by soliciting b1 and d1. Either or both of these can be your design variables. Program to find the leg size corresponding to a shear-stress level at the maximum allowable at a corner. Output all your parameters including the figure of merit. Use the program to check any previous problems to which it is applicable. Play with it in a "what if " mode and learn from the trends in your parameters.

9–51 When comparing two different weldment patterns it is useful to observe the resistance

to bending or torsion and the volume of weld metal deposited. The measure of effectiveness, defined as second moment of area divided by weld-metal volume, is useful. If a 3-in by 6-in section of a cantilever carries a static 10 kip bending load 10 in from the weldment plane, with an allowable shear stress of 12 kpsi realized, compare horizontal weldments with vertical weldments by determining the measure of effectiveness for each weld pattern. The horizontal beads are to be 3 in long and the vertical beads, 6 in long.

A 2-in dia. steel bar is subjected to the loading indicated. Locate and estimate the 9–52 maximum shear stress in the weld throat. to 9–54 1 4

Problem Number

F

T

9–52

0

15 kip · in

9–53

2 kips

0

9–54

2 kips

15 kip · in

in

6 in F

T

2-in dia. Problems 9–52 to 9–54

9–55 For Problem 9–54, determine the weld size if the maximum allowable shear stress is  20 kpsi.

9–56 Find the maximum shear stress in the throat of the weld metal in the figure. F = 25 kN 150 25 Problem 9–56

50

200

Dimensions in millimeters.

25 9 25 18

25

100

25

522      Mechanical Engineering Design

9–57 The figure shows a welded steel bracket loaded by a static force F. Estimate the factor of safety if the allowable shear stress in the weld throat is 18 kpsi. F = 2 kips

5 in

1 4

Problem 9–57

in

2.5 in

5 in 45° 1 4

in

9–58 The figure shows a formed sheet-steel bracket. Instead of securing it to the support

with machine screws, welding all around the bracket support flange has been proposed. If the combined shear stress in the weld metal is limited to 1.5 kpsi, estimate the total load W the bracket will support. The dimensions of the top flange are the same as the mounting flange. 3 16

W 1 2

Problem 9–58

in 3 4

Structural support is 1030 HR steel, bracket is 1020 press cold-formed steel. The weld electrode is E6010.

3 16

- in R

in 8 in

-in dia. holes 3 4

in

0.0625 in

1 in

9–59 Without bracing, a machinist can exert only about 100 lbf on a wrench or tool handle.

The lever shown in the figure has t = 12 in and w = 2 in. We wish to specify the filletweld size to secure the lever to the tubular part at A. Both parts are of steel, and the shear stress in the weld throat should not exceed 3000 psi. Find a safe weld size. Fillet welds Rubber grip

A

t

b

B

1 2

- in ID × 1-in OD × 2 in long; 2 required

Problem 9–59 F

A

16 in

30°

3 in

w

B h

Tapered handle

9–60 Estimate the safe static load F for the weldment shown in the figure if an E6010 elec-

trode is used and the design factor is to be 2. The steel members are 1015 hot-rolled steel. Use conventional analysis.

Welding, Bonding, and the Design of Permanent Joints     523 150

Problem 9–60 Dimensions in millimeters.

100

150

9

200

6

F

9–61 Brackets, such as the one shown, are used in mooring small watercraft. Failure of such

brackets is usually caused by bearing pressure of the mooring clip against the side of the hole. Our purpose here is to get an idea of the static and dynamic margins of safety involved. We use a bracket 1∕4 in thick made of hot-rolled 1018 steel, welded with an E6010 electrode. We then assume wave action on the boat will create force F no greater than 1200 lbf. (a) Determine the moment M of the force F about the centroid of the weld G. This moment produces a shear stress on the throat resisting bending action with a 1 "tension" at A and "compression" at C. 4 in (b) Find the force component Fy that produces a shear stress at the throat resisting a "tension" throughout the weld. (c) Find the force component Fx that produces an in-line shear throughout the weld. (d) Using Table 9–2, determine A, Iu, and I for the bracket. (e) Find the shear stress τ1 at A due to Fy and M, the shear stress τ2 due to Fx, and combine to find τ. ( f ) Find the factor of safety guarding against shear yielding in the weldment. Since the weld material is comprised of a mix of the electrode material and the base material, take the conservative approach of utilizing the strength of the weaker material. (g) Find the factor of safety guarding against a static failure in the parent metal at the weld. (h) Assuming the force F alternates between zero and 1200 lbf, find the factor of safety guarding against a fatigue failure in the weld metal using a Gerber failure criterion.

y 1 -in 2 1 -in 2

dia.

Problem 9–61 Small watercraft mooring bracket.

1 4

1 -in 2

in

3

1 2

3

A

R

30° x

1 2

1 in

in

(a) y F

x 1 in

30°

dia.

2 4 in

30° x

2 4 in

y 1 -in 2

in

1 in

0.366 in

1 4

B

G

Fx

M

FG

0.732 in

1 in x

B

G

O z

1

1 4 in 1

d = 2 2 in (b) y F

C

Fy

in A

(a)

R

C

524      Mechanical Engineering Design

9–62 For the sake of perspective it is always useful to look at the matter of scale. Double

all dimensions in Problem 9–20 and find the allowable load. By what factor has it increased? First make a guess, then carry out the computation. Would you expect the same ratio if the load had been variable?

9–63 Hardware stores often sell plastic hooks that can be mounted on walls with pressure-

sensitive adhesive foam tape. Two designs are shown in (a) and (b) of the figure. Indicate which one you would buy and why. P 1 in

3.5 in

P

3.5 in

Problem 9–63

0.75 in

1 in (b)

0.2 in (a)

9–64 For a balanced double-lap joint cured at room temperature, Volkersen's equation simplifies to

τ(x) =

Pω cosh (ωx) = A1 cosh (ωx) 4b sinh (ωl∕2)

(a) Show that the average stress τ is P∕(2bl). (b) Show that the largest shear stress is Pω∕[4b tanh(ωl∕2)]. (c) Define a stress-augmentation factor K such that

τ(l∕2) = K τ and it follows that

K=

ωl∕2 Pω 2bl ωl exp (ωl∕2) + exp (−ωl∕2) = = 4b tanh (ωl∕2) P tanh (ωl∕2) 2 exp (ωl∕2) − exp (−ωl∕2)

9–65 Program the shear-lag solution for the shear-stress state into your computer using Equation (9–7). Determine the maximum shear stress for each of the following scenarios: Part

G, psi

to, in

ti, in

Eo, psi

Ei, psi

h, in

a

0.2(106)

0.125

0.250

30(106)

30(106)

0.005

b

0.2(106)

0.125

0.250

30(106)

30(106)

0.015

c

0.2(106)

0.125

0.125

30(106)

30(106)

0.005

d

0.2(106)

0.125

0.250

30(106)

10(106)

0.005

Provide plots of the actual stress distributions predicted by this analysis. You may omit thermal stresses from the calculations, assuming that the service temperature is similar to the stress-free temperature. If the allowable shear stress is 800 psi and the load to be carried is 300 lbf, estimate the respective factors of safety for each geometry. Let l = 1.25 in and b = 1 in.

10

Mechanical Springs

©Vladimir Nenezic/123RF

Chapter Outline 10–1

Stresses in Helical Springs   526

10–2

The Curvature Effect   527

10–3

Deflection of Helical Springs   528

10–4

Compression Springs   528

10–5

Stability  529

10–6

Spring Materials   531

10–7  Helical Compression Spring Design for Static Service   535 10–8  Critical Frequency of Helical Springs   542

10–9  Fatigue Loading of Helical Compression Springs  543 10–10  Helical Compression Spring Design for Fatigue Loading   547 10–11

Extension Springs   550

10–12

Helical Coil Torsion Springs   557

10–13

Belleville Springs   564

10–14

Miscellaneous Springs   565

10–15

Summary  567 525

526      Mechanical Engineering Design

When a designer wants rigidity, negligible deflection is an acceptable approximation as long as it does not compromise function. Flexibility is sometimes needed and is often provided by metal bodies with cleverly controlled geometry. These bodies can exhibit flexibility to the degree the designer seeks. Such flexibility can be linear or nonlinear in relating deflection to load. These devices allow controlled application of force or torque; the storing and release of energy can be another purpose. Flexibility allows temporary distortion for access and the immediate restoration of function. Because of machinery's value to designers, springs have been intensively studied; moreover, they are mass-­ produced (and therefore low cost), and ingenious configurations have been found for a variety of desired applications. In this chapter we will discuss the more frequently used types of springs, their necessary parametric relationships, and their design. In general, springs may be classified as wire springs, flat springs, or specialshaped springs, and there are variations within these divisions. Wire springs include helical springs of round or square wire, made to resist and deflect under tensile, compressive, or torsional loads. Flat springs include cantilever and elliptical types, wound motor- or clock-type power springs, and flat spring washers, usually called Belleville springs.

10–1  Stresses in Helical Springs F

d

Figure 10–1aF shows a round-wire helical compression spring loaded by the axial force F. We designate D as the mean coil diameter and d as the wire diameter. Isolate a section in the spring, as shown in Figure 10–1b. For equilibrium, the isolated section contains a direct shear force F and a torsional moment T = FD∕2. The maximum shear stress in the wire may be computed by superposition of the direct shear stress given by Equation (3–23), with V = F and the torsional shear stress given by Equation (3–37). The result is T = FD∕2

F (b)

τmax =

Tr F + J A

(a)

at the inside fiber of the spring. Substitution of τmax = τ, T = FD∕2, r = d∕2, J = πd 4∕32, and A = πd2∕4 gives

F D (a) F

T = FD∕2 F

τ=

8FD 4F + 2 3 πd πd

Now we define the spring index C=

(a) Axially loaded helical spring; (b) free-body diagram showing that the wire is subjected to a direct shear and a torsional shear.

D d

(10–1)

which is a measure of coil curvature. The preferred value of C ranges from 4 to 12.1 With this relation, Equation (b) can be rearranged to give

(b)

Figure 10–1

(b)

1

τ = Ks

8FD πd 3

(10–2)

Design Handbook: Engineering Guide to Spring Design, Associated Spring-Barnes Group Inc., Bristol, CT, 1987.

Mechanical Springs     527

where Ks is a shear stress-correction factor and is defined by the equation

Ks =

2C + 1 2C

(10–3)

The use of square or rectangular wire is not recommended for springs unless space limitations make it necessary. Springs of special wire shapes are not made in large quantities, unlike those of round wire; they have not had the benefit of refining development and hence may not be as strong as springs made from round wire. When space is severely limited, the use of nested round-wire springs should always be considered. They may have an economical advantage over the special-section springs, as well as a strength advantage.

10–2  The Curvature Effect Equation (10–2) is based on the wire being straight. However, the curvature of the wire causes a localized increase in stress on the inner surface of the coil, which can be accounted for with a curvature factor. This factor can be applied in the same way as a stress concentration factor. For static loading, the curvature factor is normally neglected because any localized yielding leads to localized strain strengthening. For fatigue applications, the curvature factor should be included. Unfortunately, it is necessary to find the curvature factor in a roundabout way. The reason for this is that the published equations also include the effect of the direct shear stress. Suppose Ks in Equation (10–2) is replaced by another K factor, which corrects for both curvature and direct shear. Then this factor is given by either of the equations

KW =

4C − 1 0.615 + 4C − 4 C

(10–4)

KB =

4C + 2 4C − 3

(10–5)

The first of these is called the Wahl factor, and the second, the Bergsträsser factor.2 Since the results of these two equations differ by the order of 1 percent, Equation (10–5) is preferred. The curvature correction factor can now be obtained by canceling out the effect of the direct shear. Thus, using Equation (10–5) with Equation (10–3), the curvature correction factor is found to be

Kc =

KB 2C(4C + 2) = Ks (4C − 3)(2C + 1)

(10–6)

Now, Ks, KB or KW, and Kc are simply stress-correction factors applied multiplicatively to Tr∕J at the critical location to estimate a particular stress. There is no stress-concentration factor. In this book we will use

τ = KB

8FD πd 3

(10–7)

to predict the largest shear stress. 2

Cyril Samónov, "Some Aspects of Design of Helical Compression Springs," Int. Symp. Design and Synthesis, Tokyo, 1984.

528      Mechanical Engineering Design

10–3  Deflection of Helical Springs The deflection-force relations are quite easily obtained by using Castigliano's theorem. The total strain energy for a helical spring is composed of a torsional component and a shear component. From Equations (4–18) and (4–20), the strain energy is

U=

T 2l F 2l + 2GJ 2AG

(a)

Substituting T = FD∕2, l = πDN, J = πd 4∕32, and A = πd2∕4 results in

U=

4F 2 D 3N 2F 2DN + d 4G d 2G

(b)

where N = Na = number of active coils. Then using Castigliano's theorem, Equation (4–26), to find total deflection y gives

y=

∂U 8FD3N 4FDN = + 2 ∂F d 4G d G

(c)

Since C = D∕d, Equation (c) can be rearranged to yield

y=

8FD3N 1 8FD 3N 1 + ≈ 4 2 d G ( 2C ) d 4G

(10–8)

The spring rate, also called the scale of the spring, is k = F∕y, and so

k≈

d 4G 8D 3N

(10–9)

10–4  Compression Springs The four types of ends generally used for compression springs are illustrated in Figure 10–2. The terminal end of each spring is only shown on the right-end of the spring. A spring with plain ends has a noninterrupted helicoid; the ends are the same as if a long spring had been cut into sections. A spring with plain ends that are squared or closed is obtained by deforming the ends to a zero-degree helix angle. Springs should always be both squared and ground for important applications, because a better transfer of the load is obtained. Figure 10–2 Types of ends for compression springs: (a) both ends plain; (b) both ends squared; (c) both ends squared and ground; (d) both ends plain and ground.

+

+

(a) Plain end, right hand

(c) Squared and ground end, left hand

+

(b) Squared or closed end, right hand

+

(d ) Plain end, ground, left hand

Mechanical Springs     529

Table 10–1  Formulas for the Dimensional Characteristics of Compression Springs (Na = Number of Active Coils)

Type of Spring Ends

Plain and Squared or Squared and Term Plain Ground Closed Ground End coils, Ne 0 1 2 2 Total coils, Nt

Na Na + 1

Na + 2

Na + 2

Free length, L0

pNa + d

pNa + 3d

pNa + 2d

Solid length, Ls

d(Nt + 1) dNt d(Nt + 1)

Pitch, p (L0 − d )∕Na

p(Na + 1)

L0∕(Na + 1) (L0 − 3d )∕Na

dNt (L0 − 2d )∕Na

Source: Data from Design Handbook, 1987, p. 32.

Table 10–1 shows how the type of end used affects the number of coils and the spring length.3 Note that the digits 0, 1, 2, and 3 appearing in Table 10–1 are often used without question. Some of these need closer scrutiny as they may not be integers. This depends on how a springmaker forms the ends. Forys4 pointed out that squared and ground ends give a solid length Ls of

Ls = (Nt − a)d

where a varies, with an average of 0.75, so the entry dNt in Table 10–1 may be overstated. The way to check these variations is to take springs from a particular springmaker, close them solid, and measure the solid height. Another way is to look at the spring and count the wire diameters in the solid stack. Set removal or presetting is a process used in the manufacture of compression springs to induce useful residual stresses. It is done by making the spring longer than needed and then compressing it to its solid height. This operation sets the spring to the required final free length and, since the torsional yield strength has been exceeded, induces residual stresses opposite in direction to those induced in service. Springs to be preset should be designed so that 10 to 30 percent of the initial free length is removed during the operation. If the stress at the solid height is greater than 1.3 times the torsional yield strength, distortion may occur. If this stress is much less than 1.1 times, it is difficult to control the resulting free length. Set removal increases the strength of the spring and so is especially useful when the spring is used for energy-storage purposes. However, set removal should not be used when springs are subject to fatigue.

10–5  Stability In Chapter 4 we learned that a column will buckle when the load becomes too large. Similarly, compression coil springs may buckle when the deflection becomes too large. The critical deflection is given by the equation 3

C′2 1∕2 ycr = L0C′1[ 1 − (1 − 2 ) ] λeff

(10–10)

For a thorough discussion and development of these relations, see Cyril Samónov, "Computer-Aided Design of Helical Compression Springs," ASME paper No. 80-DET-69, 1980. 4 Edward L. Forys, "Accurate Spring Heights," Machine Design, vol. 56, no. 2, January 26, 1984.

530      Mechanical Engineering Design

Table 10–2  End-Condition Constants α for Helical Compression Springs* End Condition

Constant α

Spring supported between flat parallel surfaces (fixed ends)

0.5

One end supported by flat surface perpendicular to spring axis (fixed); other end pivoted (hinged)

0.707

Both ends pivoted (hinged)

1

One end clamped; other end free

2

*Ends supported by flat surfaces must be squared and ground.

where ycr is the deflection corresponding to the onset of instability. Samónov5 states that this equation is cited by Wahl6 and verified experimentally by Haringx.7 The quantity λeff in Equation (10–10) is the effective slenderness ratio and is given by the equation

λeff =

αL0 D

(10–11)

C′1 and C′2 are dimensionless elastic constants defined by the equations C′1 =

E 2(E − G)

C′2 =

2π2 (E − G) 2G + E

Equation (10–11) contains the end-condition constant α. This depends upon how the ends of the spring are supported. Table 10–2 gives values of α for usual end conditions. Note how closely these resemble the end conditions for columns. Absolute stability occurs when, in Equation (10–10), the term C′2∕λ2eff is greater than unity. This means that the condition for absolute stability is that

L0
5.26D.

5

Cyril Samónov "Computer-Aided Design," op. cit. A. M. Wahl, Mechanical Springs, 2nd ed., McGraw-Hill, New York, 1963. 7 J. A. Haringx, "On Highly Compressible Helical Springs and Rubber Rods and Their Application for Vibration-Free Mountings," I and II, Philips Res. Rep., vol. 3, December 1948, pp. 401–449, and vol. 4, February 1949, pp. 49–80. 6

Mechanical Springs     531

10–6  Spring Materials Springs are manufactured either by hot- or cold-working processes, depending upon the size of the material, the spring index, and the properties desired. In general, prehardened wire should not be used if D∕d > 4 or if d > 14 in. Winding of the spring induces residual stresses through bending, but these are normal to the direction of the torsional working stresses in a coil spring. Quite frequently in spring manufacture, they are relieved, after winding, by a mild thermal treatment. A great variety of spring materials are available to the designer, including plain carbon steels, alloy steels, and corrosion-resisting steels, as well as nonferrous materials such as phosphor bronze, spring brass, beryllium copper, and various nickel alloys. Descriptions of the most commonly used steels will be found in Table 10–3. The UNS steels listed in Appendix A should be used in designing hot-worked, heavy-coil springs, as well as flat springs, leaf springs, and torsion bars. Spring materials may be compared by an examination of their tensile strengths; these vary so much with wire size that they cannot be specified until the wire size is known. The material and its processing also, of course, have an effect on tensile strength. It turns out that the graph of tensile strength versus wire diameter is almost Table 10–3  High-Carbon and Alloy Spring Steels Name of Material

Similar Specifications

Music wire, 0.80–0.95C

UNS G10850 AISI 1085 ASTM A228-51

This is the best, toughest, and most widely used of all spring materials for small springs. It has the highest tensile strength and can withstand higher stresses under repeated loading than any other spring material. Available in diameters 0.12 to 3 mm (0.005 to 0.125 in). Do not use above 120°C (250°F) or at subzero temperatures.

Oil-tempered wire, 0.60–0.70C

UNS G10650 AISI 1065 ASTM 229-41

This general-purpose spring steel is used for many types of coil springs where the cost of music wire is prohibitive and in sizes larger than available in music wire. Not for shock or impact loading. Available in diameters 3 to 12 mm (0.125 to 0.5000 in), but larger and smaller sizes may be obtained. Not for use above 180°C (350°F) or at subzero temperatures.

Hard-drawn wire, 0.60–0.70C

UNS G10660 AISI 1066 ASTM A227-47

This is the cheapest general-purpose spring steel and should be used only where life, accuracy, and deflection are not too important. Available in diameters 0.8 to 12 mm (0.031 to 0.500 in). Not for use above 120°C (250°F) or at subzero temperatures.

Chrome-vanadium

UNS G61500 AISI 6150 ASTM 231-41

This is the most popular alloy spring steel for conditions involving higher stresses than can be used with the high-carbon steels and for use where fatigue resistance and long endurance are needed. Also good for shock and impact loads. Widely used for aircraft-engine valve springs and for temperatures to 220°C (425°F). Available in annealed or pretempered sizes 0.8 to 12 mm (0.031 to 0.500 in) in diameter.

Chrome-silicon

UNS G92540 AISI 9254

This alloy is an excellent material for highly stressed springs that require long life and are subjected to shock loading. Rockwell hardnesses of C50 to C53 are quite common, and the material may be used up to 250°C (475°F). Available from 0.8 to 12 mm (0.031 to 0.500 in) in diameter.

Description

Source: From Harold C. R. Carlson, "Selection and Application of Spring Materials," Mechanical Engineering, vol. 78, 1956, pp. 331–334.

532      Mechanical Engineering Design

Table 10–4  Constants A and m of Sut = A∕dm for Estimating Minimum Tensile Strength of Common Spring Wires Relative ASTM Exponent Diameter, A, Diameter, A, Cost Material No. m in kpsi · inm mm MPa · mmm of Wire Music wire*

A228

0.145

0.004–0.256

201

0.10–6.5

2211

2.6

OQ&T wire

A229 0.187 0.020–0.500 147 0.5–12.7 1855

1.3

Hard-drawn wire‡

A227 0.190 0.028–0.500 140 0.7–12.7 1783

1.0

Chrome-vanadium wire§ A232 0.168 0.032–0.437 169

0.8–11.1

Chrome-silicon wire|| A401 0.108 0.063–0.375 202 1.6–9.5 #

302 Stainless wire

A313 0.146 0.013–0.10 169 0.3–2.5

0.263 0.10–0.20 128 2.5–5 5–10

3.1

1974

4.0

1867 7.6–11 2065

0.478

Phosphor-bronze wire**

0

0.004–0.022

145

0.1–0.6

0.028

0.022–0.075

121

0.6–2  913

0.064

0.075–0.30

110

2–7.5  932

B159

0.20–0.40  90

2005

2911 1000

8.0

*

Surface is smooth, free of defects, and has a bright, lustrous finish. Has a slight heat-treating scale which must be removed before plating. ‡ Surface is smooth and bright with no visible marks. § Aircraft-quality tempered wire, can also be obtained annealed. || Tempered to Rockwell C49, but may be obtained untempered. # Type 302 stainless steel. ** Temper CA510. Source: Data from Design Handbook, 1987, p. 19. †

a straight line for some materials when plotted on log-log paper. Writing the equation of this line as

Sut =

A dm

(10–14)

furnishes a good means of estimating minimum tensile strengths when the intercept A and the slope m of the line are known. Values of these constants have been worked out from recent data and are given for strengths in units of kpsi and MPa in Table 10–4. In Equation (10–14) when d is measured in millimeters, then A is in MPa · mmm and when d is measured in inches, then A is in kpsi · inm. Although the torsional yield strength is needed to design the spring and to analyze the performance, spring materials customarily are tested only for tensile strength— perhaps because it is such an easy and economical test to make. A very rough estimate of the torsional yield strength can be obtained by assuming that the tensile yield strength is between 60 and 90 percent of the tensile strength. Then the distortionenergy theory can be employed to obtain the torsional yield strength (Ssy = 0.577Sy). This approach results in the range

0.35Sut ≤ Ssy ≤ 0.52Sut

(10–15)

for steels. For wires listed in Table 10–5, the maximum allowable shear stress in a spring can be seen in column 3. Music wire and hard-drawn steel spring wire have a low

Mechanical Springs     533

Table 10–5  Mechanical Properties of Some Spring Wires Material Music wire A228

Elastic Limit, Percent of Sut Diameter

Tension 65–75

Torsion 45–60

d, in

0.125

28.0 193

11.6 80.0

HD spring A227

0.125

28.5 196.5 11.4 78.6

Oil tempered A239

85–90

45–50

28.5

196.5

11.2

77.2

Valve spring A230

85–90

50–60

29.5

203.4

11.2

77.2

Chrome-vanadium A231

88–93

65–75

29.5

203.4

11.2

77.2

        A232

88–93

29.5

203.4

11.2

77.2

Chrome-silicon A401

85–93

65–75

29.5

203.4

11.2

77.2

 A313*

65–75

45–55

28

193

10

69.0

 17-7PH

75–80

55–60

29.5

208.4

11

75.8

 414

65–70

42–55

29

200

11.2

77.2

 420

65–75

45–55

29

200

11.2

77.2

 431

72–76

50–55

30

206

11.5

79.3

Phosphor-bronze B159

75–80

45–50

15

103.4

Beryllium-copper B197

70

50

17

Inconel alloy X-750

65–70

40–45

31

Stainless steel

*Also includes 302, 304, and 316. Note: See Table 10–6 for allowable torsional stress design values.

end of range Ssy = 0.45Sut. Valve spring wire, Cr-Va, Cr-Si, and other (not shown) hardened and tempered carbon and low-alloy steel wires as a group have Ssy ≥ 0.50Sut. Many nonferrous materials (not shown) as a group have Ssy ≥ 0.35Sut. In view of this, Joerres8 uses the maximum allowable torsional stress for static application shown in Table 10–6. For specific materials for which you have torsional yield information use this table as a guide. Joerres provides set-removal information in Table 10–6, that Ssy ≥ 0.65Sut increases strength through cold work, but at the cost of an additional operation by the springmaker. Sometimes the additional operation can be done by the manufacturer during assembly. Some correlations with carbon steel springs show that the tensile yield strength of spring wire in torsion can be estimated from 0.75Sut. The corresponding estimate of the yield strength in shear based on distortion energy 8

Robert E. Joerres, "Springs," Chapter 6 in Joseph E. Shigley, Charles R. Mischke, and Thomas H. Brown, Jr. (eds.), Standard Handbook of Machine Design, 3rd ed., McGraw-Hill, New York, 2004.

6

41.4

117.2

6.5

44.8

213.7

11.2

77.2

534      Mechanical Engineering Design

Table 10–6  Maximum Allowable Torsional Stresses for Helical Compression Springs in Static Applications

Maximum Percent of Tensile Strength

Before Set Removed After Set Removed Material (includes K W or K B) (includes Ks) Music wire and cold- drawn carbon steel

45

60–70

Hardened and tempered carbon and low-alloy steel

50

65–75

Austenitic stainless steels

35

55–65

Nonferrous alloys

35

55–65

Source: Robert E. Joerres, "Springs," Chapter 6 in Joseph E. Shigley, Charles R. Mischke, and Thomas H. Brown, Jr. (eds.), Standard Handbook of Machine Design, 3rd ed., McGraw-Hill, New York, 2004.

theory is Ssy = 0.577(0.75)Sut = 0.433Sut ≈ 0.45Sut. Samónov discusses the problem of allowable stress and shows that

Ssy = τall = 0.56Sut (10–16)

for high-tensile spring steels, which is close to the value given by Joerres for hardened alloy steels. He points out that this value of allowable stress is specified by Draft Standard 2089 of the German Federal Republic when Equation (10–2) is used without stress-correction factor. EXAMPLE 10–1 A helical compression spring is made of no. 16 music wire. The outside coil diameter of the spring is 167 in. The ends are squared and there are 12 12 total turns. (a) Estimate the torsional yield strength of the wire. (b) Estimate the static load corresponding to the yield strength. (c) Estimate the scale of the spring. (d) Estimate the deflection that would be caused by the load in part (b). (e) Estimate the solid length of the spring. ( f ) What length should the spring be to ensure that when it is compressed solid and then released, there will be no permanent change in the free length? (g) Given the length found in part ( f ), is buckling a possibility? (h) What is the pitch of the body coil? Solution (a) From Table A–28, the wire diameter is d = 0.037 in. From Table 10–4, we find A = 201 kpsi · inm and m = 0.145. Therefore, from Equation (10–14) Then, from Table 10–6,

Sut =

A 201 = 324 kpsi m = d 0.0370.145

Mechanical Springs     535

Ss y = 0.45Sut = 0.45(324) = 146 kpsi

Answer

(b) The mean spring coil diameter is D = 167 − 0.037 = 0.400 in, and so the spring index is C = 0.400∕0.037 = 10.8. Then, from Equation (10–6),

KB =

4C + 2 4(10.8) + 2 = = 1.124 4C − 3 4(10.8) − 3

Now rearrange Equation (10–7) replacing τ with Ssy, and solve for F resulting in Answer

πd 3Ss y

F=

8KB D

=

π(0.0373 )146(103 ) = 6.46 lbf 8(1.124)0.400

(c) From Table 10–1, Na = 12.5 − 2 = 10.5 turns. In Table 10–5, G = 11.85 Mpsi, and the scale of the spring is found to be, from Equation (10–9), Answer

k=

0.0374 (11.85)106 d 4G = = 4.13 lbf/in 3 8D Na 8(0.4003 )10.5

Answer (d)

y=

F 6.46 = = 1.56 in k 4.13

(e) From Table 10–1, Answer Answer ( f )

Ls = (Nt + 1)d = (12.5 + 1)0.037 = 0.500 in L0 = y + Ls = 1.56 + 0.500 = 2.06 in.

(g) To avoid buckling, Equation (10–13) and Table 10–2 give

L0 < 2.63

D 0.400 = 2.63 = 2.10 in α 0.5

Mathematically, a free length of 2.06 in is less than 2.10 in, and buckling is unlikely. However, the forming of the ends will control how close α is to 0.5. This has to be investigated and an inside rod or exterior tube or hole may be needed. (h) Finally, from Table 10–1, the pitch of the body coil is Answer

p=

L0 − 3d 2.06 − 3(0.037) = = 0.186 in Na 10.5

10–7  Helical Compression Spring Design for Static Service The preferred range of the spring index is 4 ≤ C ≤ 12, with the lower indexes being more difficult to form (because of the danger of surface cracking) and springs with higher indexes tending to tangle often enough to require individual packing. This can be the first item of the design assessment. The recommended range of active turns is 3 ≤ Na ≤ 15. To maintain linearity when a spring is about to close, it is necessary to avoid the gradual touching of coils (due to nonperfect pitch). A helical coil spring forcedeflection characteristic is ideally linear. Practically, it is nearly so, but not at each end of the force-deflection curve. The spring force is not reproducible for very small deflections, and near closure, nonlinear behavior begins as the number of active turns

536      Mechanical Engineering Design

diminishes as coils begin to touch. The designer confines the spring's operating point to the central 75 percent of the curve between no load, F = 0, and closure, F = Fs. Thus, the maximum operating force should be limited to Fmax ≤ 78Fs . Defining the fractional overrun to closure as ξ, where

Fs = (1 + ξ)Fmax

(10–17)

it follows that

7 Fs = (1 + ξ)Fmax = (1 + ξ) ( )Fs 8

From the outer equality ξ = 1∕7 = 0.143 ≈ 0.15. Thus, it is recommended that ξ ≥ 0.15. In addition to the relationships and material properties for springs, we now have some recommended design conditions to follow, namely:

4 ≤ C ≤ 12

(10–18)

3 ≤ Na ≤ 15

(10–19)

ξ ≥ 0.15

ns ≥ 1.2

(10–20)

(10–21)

where ns is the factor of safety at closure (solid height). When considering designing a spring for high volume production, the figure of merit can be the cost of the wire from which the spring is wound. The fom would be proportional to the relative material cost, weight density, and volume:

fom = −(relative material cost)

γπ 2d 2 Nt D 4

(10–22)

For comparisons between steels, the specific weight γ can be omitted. Spring design is an open-ended process. There are many decisions to be made, and many possible solution paths as well as solutions. In the past, charts, nomographs, and "spring design slide rules" were used by many to simplify the spring design problem. Today, the computer enables the designer to create programs in many different formats—direct programming, spreadsheet, MATLAB, etc. Commercial programs are also available.9 There are almost as many ways to create a spring-design program as there are programmers. Here, we will suggest one possible design approach. Design Strategy Make the a priori decisions, with hard-drawn steel wire the first choice (relative material cost is 1.0). Choose a wire size d. With all decisions made, generate a column of parameters: d, D, C, OD or ID, Na, Ls, L0, (L0)cr, ns, and fom. By incrementing wire sizes available, we can scan the table of parameters and apply the design 9

For example, see Advanced Spring Design, a program developed jointly between the Spring Manufacturers Institute (SMI), www.smihq.org, and Universal Technical Systems (UTS), www.uts.com.

Mechanical Springs     537

Figure 10–3

STATIC SPRING DESIGN Choose d Over-a-rod

Free

As-wound or set

In-a-hole

As-wound Ssy = const(A) /d m †

D = d rod + d + allow C=

2α – β + 4β

Ssy α= n s

√(2α4β– β) – 3α4β 2

β=

Helical coil compression spring design flowchart for static loading.

Set removed

As-wound or set

Ssy = 0.65A /d m

D = d hole – d – allow

D=

Ssyπd 3 8ns(1 + ξ)Fmax

8(1 + ξ)Fmax πd2

D = Cd

C = D /d KB = (4C + 2) /(4C – 3) τs = 8K B(1 + ξ)FmaxD /(πd 3) ns = Ssy /τs OD = D + d ID = D – d Na = Gd 4 ymax/(8D3Fmax) Nt: Table 10 –1 Ls: Table 10 –1 L O: Table 10 –1 (LO)cr = 2.63D/α fom = –(rel. cost) γπ2d 2Nt D /4 Print or display: d, D, C, OD, ID, Na , Nt , L s , LO, (LO)cr , ns , fom Build a table, conduct design assessment by inspection Eliminate infeasible designs by showing active constraints Choose among satisfactory designs using the figure of merit †

const is found from Table 10–6.

recommendations by inspection. After wire sizes are eliminated, choose the spring design with the highest figure of merit. This will give the optimal design despite the presence of a discrete design variable d and aggregation of equality and inequality constraints. The column vector of information can be generated by using the flowchart displayed in Figure 10–3. It is general enough to accommodate to the situations of as-wound and set-removed springs, operating over a rod, or in a hole free of rod or hole. In as-wound springs the controlling equation must be solved for the spring index as follows. From Equation (10–7) with τ = Ssy∕ns, C = D∕d, KB from Equation (10–6), and Equation (10–17),

Ss y 8Fs D 4C + 2 8(1 + ξ)FmaxC = KB = ] ns 4C − 3 [ πd 3 πd 2

(a)

538      Mechanical Engineering Design

Let

α=

β=

Ss y ns 8(1 + ξ)Fmax πd 2

(b)

(c)

Substituting Equations (b) and (c) into (a) and simplifying yields a quadratic equation in C. The larger of the two solutions will yield the spring index

C=

2α − β 2 3α 2α − β + √( − 4β 4β ) 4β

(10–23)

EXAMPLE 10–2 A music wire helical compression spring is needed to support a 20-lbf load after being compressed 2 in. Because of assembly considerations the solid height cannot exceed 1 in and the free length cannot be more than 4 in. Design the spring. Solution The a priori decisions are ∙ Music wire, A228; from Table 10–4, A = 201 000 psi-inm; m = 0.145; from Table 10–5, E = 28.5 Mpsi, G = 11.75 Mpsi (expecting d > 0.064 in) ∙ Ends squared and ground ∙ Function: Fmax = 20 lbf, ymax = 2 in ∙ Safety: use design factor at solid height of (ns)d = 1.2 ∙ Robust linearity: ξ = 0.15 ∙ Use as-wound spring (cheaper), Ssy = 0.45Sut from Table 10–6 ∙ Decision variable: d = 0.080 in, music wire gauge #30, Table A–28. From Figure 10–3 and Table 10–6,

Ssy = 0.45

201 000 = 130 455 psi 0.0800.145

From Figure 10–3 or Equation (10–23) α= β= C=

Ss y 130 455 = = 108 713 psi ns 1.2 8(1 + ξ)Fmax πd

2

=

8(1 + 0.15)20 π(0.0802 )

= 9151.4 psi

2(108 713) − 9151.4 2 3(108 713) 2(108 713) − 9151.4 + √[ ] − 4(9151.4) = 10.53 4(9151.4) 4(9151.4)

Mechanical Springs     539

Continuing with Figure 10–3: D = Cd = 10.53(0.080) = 0.8424 in KB =

4(10.53) + 2 = 1.128 4(10.53) − 3

τs = 1.128 ns =

8(1 + 0.15)20(0.8424) π(0.080) 3

= 108 700 psi

130 445 = 1.2 108 700

OD = 0.843 + 0.080 = 0.923 in Na =

11.75(106 )0.0804 (2) 8(0.843) 320

= 10.05 turns

Nt = 10.05 + 2 = 12.05 total turns Ls = 0.080(12.05) = 0.964 in L0 = 0.964 + (1 + 0.15)2 = 3.264 in (L) cr = 2.63(0.843∕0.5) = 4.43 in fom = −2.6π 2 (0.080) 212.05(0.843)∕4 = −0.417 Repeat the above for other wire diameters and form a table (easily accomplished with a spreadsheet program): d

0.063 0.067 0.071 0.075 0.080 0.085 0.090 0.095

D

0.391 0.479 0.578 0.688 0.843 1.017 1.211 1.427

C

6.205 7.153 8.143 9.178 10.53 11.96 13.46 15.02

OD 0.454 0.546 0.649 0.763 0.923 1.102 1.301 1.522 Na 39.1 26.9 19.3 14.2 10.1 7.3 5.4 4.1 Ls 2.587 1.936 1.513 1.219 0.964 0.790 0.668 0.581 L0 4.887 4.236 3.813 3.519 3.264 3.090 2.968 2.881 (L0)cr 2.06 2.52 3.04 3.62 4.43 5.35 6.37 7.51 ns 1.2 1.2 1.2 1.2 1.2 1.2 1.2 1.2 fom −0.409 −0.399 −0.398 −0.404 −0.417 −0.438 −0.467 −0.505

Now examine the table and perform the adequacy assessment. The shading of the table indicates values outside the range of recommended or specified values. The spring index constraint 4 ≤ C ≤ 12 rules out diameters larger than 0.085 in. The constraint 3 ≤ Na ≤ 15 rules out wire diameters less than 0.075 in. The Ls ≤ 1 constraint rules out diameters less than 0.080 in. The L0 ≤ 4 constraint rules out diameters less than 0.071 in. The buckling criterion rules out free lengths longer than (L0)cr, which rules out diameters less than 0.075 in. The factor of safety ns is exactly 1.20 because the mathematics forced it. Had the spring been in a hole or over a rod, the helix diameter would be chosen without reference to (ns)d. The result is that there are only two springs in the feasible domain, one with a wire diameter of 0.080 in and the other with a wire diameter of 0.085. The figure of merit decides and the decision is the design with 0.080 in wire diameter (−0.417 > −0.438).

540      Mechanical Engineering Design

Having designed a spring, will we have it made to our specifications? Not necessarily. There are vendors who stock literally thousands of music wire compression springs. By browsing their catalogs, we will usually find several that are close. Maximum deflection and maximum load are listed in the display of characteristics. Check to see if this allows soliding without damage. Often it does not. Spring rates may only be close. At the very least this situation allows a small number of springs to be ordered "off the shelf" for testing. The decision often hinges on the economics of special order versus the acceptability of a close match. Spring design is not a closed-form approach and requires iteration. Example 10–2 provided an iterative approach to spring design for static service by first selecting the wire diameter. The diameter selection can be rather arbitrary. In the next example, we will first select a value for the spring index C, which is within the recommended range. EXAMPLE 10–3 Design a compression spring with plain ends using hard-drawn wire. The deflection is to be 2.25 in when the force is 18 lbf and to close solid when the force is 24 lbf. Upon closure, use a design factor of 1.2 guarding against yielding. Select the smallest gauge W&M (Washburn & Moen) wire. Solution Instead of starting with a trial wire diameter, we will start with an acceptable spring index for C after some preliminaries. From Equation (10–14) and Table 10–6 the shear strength, in kpsi, is

A Ss y = 0.45Sut = 0.45( m ) d

(1)

The shear stress given by Equation (10–7) replacing τ and F with τmax and Fmax, respectively, gives

τmax = KB

8Fmax D πd

3

= KB

8FmaxC πd 2

(2)

where the Bergsträsser factor, KB, from Equation (10–5) is

KB =

4C + 2 4C − 3

(3)

Dividing Equation (1) by the design factor ns and equating this to Equation (2), in kpsi, gives

8Fmax C −3 0.45 A = KB (10 ) ns ( d m ) πd 2

(4)

For the problem Fmax = 24 lbf and ns = 1.2. Solving for d gives

d = (0.163

KB C 1∕(2−m) A )

Try a trial spring index of C = 10. From Equation (3)

KB =

4(10) + 2 = 1.135 4(10) − 3

From Table 10–4, m = 0.190 and A = 140 kpsi · in0.190. Thus, Equation (5) gives

d = (0.163

1.135(10) 1∕(2−0.190) = 0.09160 in 140 )

(5)

Mechanical Springs     541

From Table A–28, a 12-gauge W&M wire, d = 0.105 5 in, is selected. Checking the resulting factor of safety, from Equation (4) with Fmax = 24 lbf Ad 2−m ns = 7.363 KB C 140(0.105 52−0.190 ) = 7.363 = 1.55 1.135(10)

(6)

which is pretty conservative. If we had selected the 13-gauge wire, d = 0.091 5 in, the factor of safety would be n = 1.198, which rounds to 1.2. Taking a little liberty here we will select the W&M 13-gauge wire. To continue with the design, the spring rate is

k=

F 18 = = 8 lbf/in y 2.25

From Equation (10–9) solving for the active number of coils

Na =

0.091 5(11.5)106 d 4G dG = = = 16.4 turns 3 3 8k D 8k C 8(8)103

This exceeds the recommended range of 3 ≤ Na ≤ 15. To decrease Na, increase C. Repeating the process with C = 12 gives KB = 1.111 and d = 0.100 1 in. Selecting a 12-gauge W&M wire, d = 0.105 5 in. From Equation (6), this gives n = 1.32, which is acceptable. The number of active coils is

Na =

0.105 5(11.5)106 dG = = 10.97 = 11 turns 8k C 3 8(8)123

which is acceptable. From Table 10–1, for plain ends, the total number of coils is Nt = Na = 11 turns. The deflection from free length to solid length of the spring is given by

ys =

Fmax 24 = = 3 in k 8

From Table 10–1, the solid length is Ls = d(Nt + 1) = 0.105 5(11 + 1) = 1.266 in The free length of the spring is then L 0 = Ls + ys = 1.266 + 3 = 4.266 in The mean coil diameter of the spring is

D = Cd = 12(0.105 5) = 1.266 in

and the outside coil diameter of the spring is OD = D + d = 1.266 + 0.105 5 = 1.372 in. To avoid buckling, Equation (10–13) gives D 1.266 α < 2.63 = 2.63 = 0.780 L0 4.266 From Table 10–2, the spring is stable provided it is supported between either fixed-fixed or fixed-hinged ends. The final results are: Answer W&M wire size: 12 gauge, d = 0.105 5 in Outside coil diameter: OD = 1.372 in Total number of coils: Nt = 11 turns with plain ends Free length: L0 = 4.266 in

542      Mechanical Engineering Design

10–8  Critical Frequency of Helical Springs If a wave is created by a disturbance at one end of a swimming pool, this wave will travel down the length of the pool, be reflected back at the far end, and continue in this back-and-forth motion until it is finally damped out. The same effect occurs in helical springs, and it is called spring surge. If one end of a compression spring is held against a flat surface and the other end is disturbed, a compression wave is created that travels back and forth from one end to the other exactly like the swimmingpool wave. Spring manufacturers have taken slow-motion movies of automotive valve-spring surge. These pictures show a very violent surging, with the spring actually jumping out of contact with the end plates. Figure 10–4 is a photograph of a failure caused by such surging. When helical springs are used in applications requiring a rapid reciprocating motion, the designer must be certain that the physical dimensions of the spring are not such as to create a natural vibratory frequency close to the frequency of the applied force; otherwise, resonance may occur, resulting in damaging stresses, since the internal damping of spring materials is quite low. The governing equation for the translational vibration of a spring placed between two flat and parallel plates is the wave equation

∂ 2u W ∂ 2u = ∂x 2 kgl 2 ∂t 2

where k = spring rate g = acceleration due to gravity l = length of spring between plates W = weight of spring x = coordinate along length of spring u = motion of any particle at distance x Figure 10–4 Valve-spring failure in an overrevved engine. Fracture is along the 45° line of maximum principal stress associated with pure torsional loading. (Personal photograph of Larry D. Mitchell, coauthor of Mechanical Engineering Design, 4th ed., McGraw-Hill, New York, 1983.)

(10–24)

Mechanical Springs     543

The solution to this equation is harmonic and depends on the given physical properties as well as the end conditions of the spring. The harmonic, natural, frequencies for a spring placed between two flat and parallel plates, in radians per second, are kg ω = mπ √ W

m = 1, 2, 3, . . .

where the fundamental frequency is found for m = 1, the second harmonic for m = 2, and so on. We are usually interested in the frequency in cycles per second; since ω = 2πf, we have, for the fundamental frequency in hertz, 1 kg f= √ 2 W

(10–25)

assuming the spring ends are always in contact with the plates. Wolford and Smith10 show that the frequency is

f=

1 kg 4√W

(10–26)

where the spring has one end against a flat plate and the other end free. They also point out that Equation (10–25) applies when one end is against a flat plate and the other end is driven with a sine-wave motion. The weight of the active part of a helical spring is

W = ALγ =

π 2d 2DNaγ πd 2 (πDNa )(γ) = 4 4

(10–27)

where γ is the specific weight. The fundamental critical frequency should be greater than 15 to 20 times the frequency of the force or motion of the spring in order to avoid resonance with the harmonics. If the frequency is not high enough, the spring should be redesigned to increase k or decrease W.

10–9  Fatigue Loading of Helical Compression Springs Springs are almost always subject to fatigue loading. In many instances the number of cycles of required life may be small, say, several thousand for a padlock spring or a toggle-switch spring. But the valve spring of an automotive engine must sustain millions of cycles of operation without failure; so it must be designed for infinite life. To improve the fatigue strength of dynamically loaded springs, shot peening can be used. It can increase the torsional fatigue strength by 20 percent or more. Shot size is about 641 in, so spring coil wire diameter and pitch must allow for complete coverage of the spring surface. The best data on the torsional endurance limits of spring steels are those reported by Zimmerli.11 He discovered the surprising fact that size, material, and tensile strength have no effect on the endurance limits (infinite life only) of spring steels in sizes under 38 in (10 mm). We have already observed that endurance limits tend to 10

J. C. Wolford and G. M. Smith, "Surge of Helical Springs," Mech. Eng. News, vol. 13, no. 1, February 1976, pp. 4–9. 11 F. P. Zimmerli, "Human Failures in Spring Applications," The Mainspring, no. 17, Associated Spring Corporation, Bristol, Conn., August–September 1957.

544      Mechanical Engineering Design

level out at high tensile strengths (Figure 6–22), but the reason for this is not clear. Zimmerli suggests that it may be because the original surfaces are alike or because plastic flow during testing makes them the same. Unpeened springs were tested for infinite life with a minimum torsional stress of 20 kpsi to a maximum of 90 kpsi and peened springs in the range 20 kpsi to 135 kpsi. The alternating and mean stresses correlating to these tested ranges correspond to a fluctuating-stress condition on the boundary between finite and infinite life, that is, 106 cycles for steel spring materials. According to Zimmerli, this fluctuating-stress condition is applicable for any spring steel with diameter under 38 in (10 mm). Let us define these fluctuating-stress values as the Zimmerli endurance strength components for infinite life, thus, Unpeened: Peened:

Ssa = 35 kpsi (241 MPa) Ssa = 57.5 kpsi (398 MPa)

Ssm = 55 kpsi (379 MPa)

(10–28a)

Ssm = 77.5 kpsi (534 MPa)

(10–28b)

From these endurance strength components we can estimate an equivalent completely reversed stress, and consequently an endurance limit, from one of the constant-life curves in Section 6–14. For example, selecting the Goodman criterion, we adapt Equation (6–59) to shear, and apply the endurance strength components as the alternating and mean stresses, resulting in

Sse =

Ssa Ssm 1− Ssu

(10–29a)

The Gerber criterion is one of the more commonly used for springs. Starting from the Gerber fatigue failure criterion of Equation (6–47), an equation for the endurance limit can be developed with the same constant-life approach to be

Sse =

Ssa Ssm 2 1−( ) Ssu

(10–29b)

Any of the other fatigue criterion could be used in similar fashion. As an example, given an unpeened spring with Ssu = 211.5 kpsi, the Gerber criterion of Equation (10–29b) predicts the endurance limit to be Sse =

Ssa 35 = = 37.5 kpsi 2 Ssm 55 2 1−( ) 1−( Ssu 211.5 )

For the Goodman failure criterion, Equation (10–29a) predicts the endurance limit would be 47.3 kpsi. Each possible wire size would change these numbers, since Ssu would change. An alternate approach, known as the Sines failure criterion, is also available. An extended study12 of available literature regarding torsional fatigue found that for polished, notch-free, cylindrical specimens subjected to torsional shear stress, the maximum alternating stress that may be imposed without causing failure is constant and independent of the mean stress in the cycle provided that the maximum stress range does not equal or exceed the torsional yield strength of the metal. With notches and abrupt section changes this consistency is not found. Springs are free of notches and surfaces are often very smooth. 12

Oscar J. Horger (ed.), Metals Engineering: Design Handbook, McGraw-Hill, New York, 1953, p. 84.

Mechanical Springs     545

With only shear stresses present in a helical spring, it is convenient to adapt the fluctuating-stress diagram and a fatigue failure criterion into shear stresses and shear strengths, as described at the end of Section 6–13. Common spring materials have been tested for material properties that are specifically applicable for wire that has been cold drawn to relatively small diameters.13 From this testing, a recommended relationship for estimating the torsional modulus of rupture Ssu is

Ssu = 0.67Sut

(10–30)

In the case of shafts and many other machine members, fatigue loading in the form of completely reversed stresses is quite ordinary. Helical springs, on the other hand, are never used as both compression and extension springs. In fact, they are usually assembled with a preload so that the working load is additional. Thus the stresstime diagram of Figure 6–29d expresses the usual condition for helical springs. The worst condition, then, would occur when there is no preload, that is, when τmin = 0. Now, we define Fmax − Fmin (10–31a) 2 Fmax + Fmin Fm = (10–31b) 2

Fa =

where the subscripts have the same meaning as those of Figure 6–29d when applied to the axial spring force F. Then the shear stress amplitude is

τa = KB

8Fa D πd 3

(10–32)

where KB is the Bergsträsser factor, obtained from Equation (10–5), and corrects for both direct shear and the curvature effect. As noted in Section 10–2, the Wahl factor KW can be used instead, if desired. The mean shear stress is given by the equation

τm = KB

8Fm D πd 3

(10–33)

EXAMPLE 10–4 An as-wound helical compression spring, made of music wire, has a wire size of 0.092 in, an outside coil diameter of 169 in, a free length of 438 in, 21 active coils, and both ends squared and ground. The spring is unpeened. This spring is to be assembled with a preload of 5 lbf and will operate with a maximum load of 35 lbf during use. (a) Estimate the factor of safety guarding against fatigue-failure using a torsional Gerber fatigue-failure criterion with Zimmerli data. (b) Repeat part (a) using the Sines torsional fatigue criterion (steady stress component has no effect), with Zimmerli data. (c) Repeat using a torsional Goodman failure criterion with Zimmerli data. (d) Estimate the critical frequency of the spring.

13

Associated Spring, Design Handbook: Engineering Guide to Spring Design, Associated Spring, Barnes Group Inc., Bristol, Conn., 1987.

546      Mechanical Engineering Design

Solution The mean coil diameter is D = 0.5625 − 0.092 = 0.4705 in. The spring index is C = D∕d = 0.4705∕0.092 = 5.11. Then

KB =

4C + 2 4(5.11) + 2 = = 1.287 4C − 3 4(5.11) − 3

From Equations (10–31),

Fa =

35 − 5 = 15 lbf 2

Fm =

35 + 5 = 20 lbf 2

The alternating shear-stress component is found from Equation (10–32) to be

8Fa D

τa = KB

πd

3

= (1.287)

8(15)0.4705 π(0.092) 3

(10−3 ) = 29.7 kpsi

Equation (10–33) gives the mean shear-stress component

τm = KB

8Fm D πd 3

= 1.287

8(20)0.4705 π(0.092) 3

(10−3 ) = 39.6 kpsi

From Table 10–4 we find A = 201 kpsi · inm and m = 0.145. The ultimate tensile strength is estimated from Equation (10–14) as

Sut =

A 201 = = 284.1 kpsi d m 0.0920.145

The shear modulus of rupture is estimated from Equation (10–30)

Ssu = 0.67Sut = 0.67(284.1) = 190.3 kpsi

(a) The endurance limit based on the Gerber criterion, Equation (10–29b), and the Zimmerli endurance strength components, Equation (10–28a), is

Sse =

Ssa 1 − (Ssm ∕Ssu )

2

=

35 = 38.2 kpsi 1 − (55∕190.3) 2

The Gerber fatigue criterion from Equation (6–48), adapted for shear, is Answer

2τmSse 2 1 Ssu 2 τa n f = ( ) ( )[ −1 + √ 1 + ( 2 τm Sse Ssuτa ) ] 2(39.6)(38.2) 2 1 190.3 2 29.7 = ( −1 + 1 + √ ( (190.3)(29.7) ) ] = 1.21 2 39.6 ) ( 38.2 )[

(b) The Sines failure criterion ignores Ssm so that, for the Zimmerli data of Equation (10–28a) with Ssa = 35 kpsi, nf =

Answer

Ssa 35 = = 1.18 τa 29.7

(c) The endurance limit based on the Goodman criterion, Equation (10–29a), and the Zimmerli endurance strength components, Equation (10–28a), is

Sse =

Ssa 35 = = 49.2 kpsi 1 − (Ssm ∕Ssu ) 1 − (55∕190.3)

Mechanical Springs     547

The Goodman fatigue criterion from Equation (6–41), adapted for shear, is Answer

τm −1 τa 29.7 39.6 −1 nf = ( + ) = ( + = 1.23 Sse Ssu 49.2 190.3 )

(d) Using Equation (10–9) and Table 10–5, we estimate the spring rate as

k=

0.0924[11.75(106 )] d 4G = = 48.1 lbf/in 8D 3Na 8(0.4705) 3 21

From Equation (10–27) we estimate the spring weight as

W=

π 2 (0.0922 )0.4705(21)0.284 = 0.0586 lbf 4

and from Equation (10–25) the frequency of the fundamental wave is Answer

fn =

1 48.1(386) 1∕2 = 281 Hz 2 [ 0.0586 ]

If the operating or exciting frequency is more than 281∕20 = 14.1 Hz, the spring may have to be redesigned. We used three approaches to estimate the fatigue factor of safety in Example 10–4. The results, in order of smallest to largest, were 1.18 (Sines), 1.21 (Gerber), and 1.23 (Goodman). Although the results were very close to one another, using the Zimmerli data as we have, the Sines criterion will always be the most conservative and the Goodman the least. If we perform a fatigue analysis using strength properties as was done in Chapter 6, different results would be obtained, but here the Goodman criterion would be more conservative than the Gerber criterion. Be prepared to see designers or design software using any one of ­ these techniques. This is why we cover them. Which criterion is correct? Remember, we are performing estimates and only testing will reveal the truth— statistically.

10–10  Helical Compression Spring Design for Fatigue Loading Let us begin with the statement of a problem. In order to compare a static spring to a dynamic spring, we shall design the spring in Example 10–2 for dynamic service. EXAMPLE 10–5 A music wire helical compression spring with infinite life is needed to resist a dynamic load that varies from 5 to 20 lbf at 5 Hz while the end deflection varies from 12 to 2 in. Because of assembly considerations, the solid height cannot exceed 1 in and the free length cannot be more than 4 in. The springmaker has the following wire sizes in stock: 0.069, 0.071, 0.080, 0.085, 0.090, 0.095, 0.105, and 0.112 in.

548      Mechanical Engineering Design

Solution The a priori decisions are: ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙

Material and condition: for music wire, A = 201 kpsi · inm, m = 0.145, G = 11.75(106) psi; relative cost is 2.6 Surface treatment: unpeened End treatment: squared and ground Robust linearity: ξ = 0.15 Set: use in as-wound condition Fatigue-safe: nf = 1.5 using the Sines-Zimmerli fatigue-failure criterion Function: Fmin = 5 lbf, Fmax = 20 lbf, ymin = 0.5 in, ymax = 2 in, spring operates free (no rod or hole) Decision variable: wire size d

The figure of merit will be the cost of wire to wind the spring, Equation (10–22) without density. The design strategy will be to set wire size d, build a table, inspect the table, and choose the satisfactory spring with the highest figure of merit. Set d = 0.112 in. Then

20 − 5 20 + 5 = 7.5 lbf Fm = = 12.5 lbf 2 2 Fmax 20 k= = = 10 lbf/in ymax 2 201 Sut = = 276.1 kpsi 0.1120.145

Fa =

Ssu = 0.67(276.1) = 185.0 kpsi

Ssy = 0.45(276.1) = 124.2 kpsi

From Equation (10–28a), with the Sines criterion, Sse = Ssa = 35 kpsi. Equation (10–23) can be used to determine C with Sse, nf, and Fa in place of Ssy, ns, and (1 + ξ)Fmax, respectively. Thus, Sse 35 000 = = 23 333 psi nf 1.5

α=

β=

C=

D = Cd = 14.005(0.112) = 1.569 in

Fs = (1 + ξ)Fmax = (1 + 0.15)20 = 23 lbf

Na =

Nt = Na + 2 = 5.98 + 2 = 7.98 turns

L s = dNt = 0.112(7.98) = 0.894 in

L0 = Ls +

8Fa πd

2

=

8(7.5)

π(0.1122 )

= 1522.5 psi

2(23 333) − 1522.5 2 3(23 333) 2(23 333) − 1522.5 + √[ ] − 4(1522.5) = 14.005 4(1522.5) 4(1522.5)

0.1124 (11.75) (106 ) d 4G = = 5.98 turns 8D 3k 8(1.569) 310

Fs 23 = 0.894 + = 3.194 in k 10

Mechanical Springs     549

ID = 1.569 − 0.112 = 1.457 in

OD = 1.569 + 0.112 = 1.681 in

ys = L 0 − L s = 3.194 − 0.894 = 2.30 in (1.569) 2.63D = 2.63 = 8.253 in α 0.5

(L 0 ) cr
L0

We see that none of the diameters satisfy the given constraints. The 0.105-in-diameter wire is the closest to satisfying all requirements. The value of C = 12.14 is not a serious deviation and can be tolerated. However, the tight constraint on Ls needs to be addressed. If the assembly conditions can be relaxed to accept a solid height of 1.116 in, we have a solution. If not, the only other possibility is to use the 0.112-in diameter and accept a value C = 14, individually package the springs, and possibly reconsider supporting the spring in service.

10–11  Extension Springs Extension springs differ from compression springs in that they carry tensile loading, they require some means of transferring the load from the support to the body of the spring, and the spring body is wound with an initial tension. The load transfer can be done with a threaded plug or a swivel hook; both of these add to the cost of the finished product, and so one of the methods shown in Figure 10–5 is usually employed. Stresses in the body of the extension spring are handled the same as compression springs. In designing a spring with a hook end, bending and torsion in the hook must be included in the analysis. In Figure 10–6a and b a commonly used method of designing the end is shown. The maximum tensile stress at A, due to bending and axial loading, is given by 16D 4 σA = F [ (K) A 3 + 2 ] πd πd

(10–34)

Figure 10–5 +

+ +

+

Types of ends used on extension springs. (Courtesy of Associated Spring.)

(a) Machine half loop–open

(b) Raised hook

(c) Short twisted loop

(d) Full twisted loop

B

(a) F

F

d

A r1

F

F

d d

(b) Mechanical Springs     551

A r1

r2

r2 B

B

(c)

(b)

(a)

(d )

Note: Radius r1 is in the plane of the end coil for curved beam bending stress. Radius r2 is at a right angle to the end coil for torsional shear stress. Note: Radius r1 is in the plane of the end coil for curved beam Figure bending stress. Radius r2 is at a right angle to theview end of part a; stress is Ends for extension springs. (a) Usual design; stress at A is due to combined axial force and bending moment. (b) Side coil for torsional shear stress. F

10–6F

mostly torsion at B. (c) Improved design; stress at A is due to combined axial force and bending moment. (d) Side view of part c; stress at B is mostly torsion. d A r2

r1

B

where (K)A is a bending stress-correction factor for curvature, given by

(K) A = (c)

4C 21 − C1 − 1 4C1 (C1 − 1)

C1 =

(d )

2r1 d

(10–35)

The maximum torsional stress at point B is given by

Note: Radius r1 is in the plane of the end coil for curved beam bending stress. Radius rτ2 is = B at a right angle to the end coil for torsional shear stress.

(K) B

8FD πd 3

(10–36)

where the stress-correction factor for curvature, (K)B, is

(K) B =

4C2 − 1 4C2 − 4

C2 =

2r2 d

(10–37)

Figures 10–6c and d show an improved design due to a reduced coil diameter. When extension springs are made with coils in contact with one another, they are said to be close-wound. Spring manufacturers prefer some initial tension in closewound springs in order to hold the free length more accurately. The corresponding load-deflection curve is shown in Figure 10–7a, where y is the extension beyond the free length L0 and Fi is the initial tension in the spring that must be exceeded before the spring deflects. The load-deflection relation is then

F = Fi + ky

(10–38)

where k is the spring rate. The free length L0 of a spring measured inside the end loops or hooks as shown in Figure 10–7b can be expressed as

L0 = 2(D − d) + (Nb + 1)d = (2C − 1 + Nb )d

(10–39)

where D is the mean coil diameter, Nb is the number of body coils, and C is the spring index. With ordinary twisted end loops as shown in Figure 10–7b, to account for the

552      Mechanical Engineering Design

Free length F Wire diameter

Fi y y

Outside diameter

Length of body

Gap

Inside diameter

-

+

Hook length

Loop length

Mean diameter

(b)

(a) 300 Difficult to attain

275 250

40 35

225 30

Available upon special request from springmaker

200 175

25

150 125 100 75

15 10

Difficult to control

50 25

20

Preferred range

Torsional stress (uncorrected) caused by initial tension (10 3 psi)

(a) Geometry of the force F and extension y curve of an extension spring; (b) geometry of the extension spring; and (c) torsional stresses due to initial tension as a function of spring index C in helical extension springs.

F

Torsional stress (uncorrected) caused by initial tension MPa

Figure 10–7

5 4

6

8

10

12

14

16

Index (c)

deflection of the loops in determining the spring rate k, the equivalent number of active helical turns Na for use in Equation (10–9) is

Na = Nb +

G E

(10–40)

where G and E are the shear and tensile moduli of elasticity, respectively (see Problem 10–38). The initial tension in an extension spring is created in the winding process by twisting the wire as it is wound onto the mandrel. When the spring is completed and removed from the mandrel, the initial tension is locked in because the spring cannot get any shorter. The amount of initial tension that a springmaker can routinely incorporate is as shown in Figure 10–7c. The preferred range can be expressed in terms of the initial torsional stress τi (uncorrected for curvature) as

τi =

33 500 C−3 ± 1000(4 − psi exp(0.105C) 6.5 )

where C is the spring index.

(10–41)

Mechanical Springs     553

Table 10–7  Maximum Allowable Stresses (KW or KB corrected) for Helical Extension Springs in Static Applications

Percent of Tensile Strength

In Torsion

In Bending

Materials

Body End

End

Patented, cold-drawn or hardened and tempered carbon and low-alloy steels

45–50

40

75

Austenitic stainless steel and nonferrous alloys

35

30

55

This information is based on the following conditions: set not removed and low temperature heat treatment ­applied. For springs that require high initial tension, use the same percent of tensile strength as for  end. Source: Data from Design Handbook, 1987, p. 52.

Guidelines for the maximum allowable corrected stresses for static applications of extension springs are given in Table 10–7.

EXAMPLE 10–6 A hard-drawn steel wire extension spring has a wire diameter of 0.035 in, an outside coil diameter of 0.248 in, hook radii of r1 = 0.106 in and r2 = 0.089 in, and an initial tension of 1.19 lbf. The number of body turns is 12.17. From the given information: (a) Determine the physical parameters of the spring. (b) Check the initial preload stress conditions. (c) Find the factors of safety under a static 5.25-lbf load. Solution (a)

D = OD − d = 0.248 − 0.035 = 0.213 in

C=

KB =

D 0.213 = = 6.086 d 0.035 4C + 2 4(6.086) + 2 = = 1.234 4C − 3 4(6.086) − 3

Equation (10–40) and Table 10–5: Equation (10–9): Equation (10–39):

Na = Nb + G∕E = 12.17 + 11.6∕28.7 = 12.57 turns k=

0.0354 (11.6)106 d 4G = = 17.91 lbf/in 8D 3Na 8(0.2133 )12.57

L 0 = (2C − 1 + Nb )d = [2(6.086) − 1 + 12.17]0.035 = 0.817 in

The deflection under the service load is

ymax =

Fmax − Fi 5.25 − 1.19 = = 0.227 in k 17.91

where the spring length becomes L = L0 + y = 0.817 + 0.227 = 1.044 in.

554      Mechanical Engineering Design

(b) The uncorrected initial stress is given by Equation (10–2) without the correction factor. That is,

(τi ) uncorr =

8Fi D πd

3

=

8(1.19)0.213(10−3 ) π(0.0353 )

= 15.1 kpsi

The preferred range is given by Equation (10–41) and for this case is (τi ) pref = =

33 500 C−3 ± 1000(4 − exp(0.105C) 6.5 )

33 500 6.086 − 3 ± 1000(4 − exp[0.105(6.086)] 6.5 )

= 17 681 ± 3525 = 21 206, 14 156 psi = 21.2, 14.2 kpsi Answer Thus, the initial tension of 15.1 kpsi is in the preferred range. (c) For hard-drawn wire, Table 10–4 gives m = 0.190 and A = 140 kpsi · inm. From Equation (10–14)

Sut =

A 140 = 264.7 kpsi m = d 0.0350.190

For torsional shear in the main body of the spring, from Table 10–7, Ssy = 0.45 Sut = 0.45(264.7) = 119.1 kpsi

The shear stress under the service load is

τmax =

8KB Fmax D πd 3

=

8(1.234)5.25(0.213) π(0.0353 )

(10−3 ) = 82.0 kpsi

Thus, the factor of safety is n=

Answer

Ssy 119.1 = = 1.45 τmax 82.0

For the end-hook bending at A, C1 = 2r1∕d = 2(0.106)∕0.035 = 6.057

From Equation (10–35) From Equation (10–34)

(K) A =

4C 12 − C1 − 1 4(6.0572 ) − 6.057 − 1 = = 1.14 4C1 (C1 − 1) 4(6.057)(6.057 − 1)

σA = Fmax [ (K) A

16D 4 + 2] 3 πd πd

16(0.213) 4 = 5.25 [ 1.14 + (10−3 ) = 156.9 kpsi 3 π(0.035 ) π(0.0352 ) ]

The yield strength, from Table 10–7, is given by

Sy = 0.75Sut = 0.75(264.7) = 198.5 kpsi

The factor of safety for end-hook bending at A is then Answer

nA =

Sy 198.5 = = 1.27 σA 156.9

Mechanical Springs     555

For the end-hook in torsion at B, from Equation (10–37) C2 = 2r2 ∕d = 2(0.089)∕0.035 = 5.086

(K) B =

4C2 − 1 4(5.086) − 1 = = 1.18 4C2 − 4 4(5.086) − 4

and the corresponding stress, given by Equation (10–36), is

τB = (K) B

8FmaxD πd

3

= 1.18

8(5.25)0.213 π(0.0353 )

(10−3 ) = 78.4 kpsi

Using Table 10–7 for yield strength, the factor of safety for end-hook torsion at B is nB =

Answer

(Ssy ) B 0.4(264.7) = = 1.35 τB 78.4

Yield due to bending of the end hook will occur first.

Next, let us consider a fatigue problem. EXAMPLE 10–7 The helical coil extension spring of Example 10–6 is subjected to a dynamic loading from 1.5 to 5 lbf. Estimate the factors of safety using the Gerber failure criterion for (a) coil fatigue, (b) coil yielding, (c) end-hook bending fatigue at point A of Figure 10–6a, and (d) end-hook torsional fatigue at point B of Figure 10–6b. Solution A number of quantities are the same as in Example 10–6: d = 0.035 in, Sut = 264.7 kpsi, D = 0.213 in, r1 = 0.106 in, C = 6.086, KB = 1.234, (K)A = 1.14, (K)B = 1.18, Nb = 12.17 turns, L0 = 0.817 in, k = 17.91 lbf/in, Fi = 1.19 lbf, and (τi)uncorr = 15.1 kpsi. Then

Fa = (Fmax − Fmin )∕2 = (5 − 1.5)∕2 = 1.75 lbf

Fm = (Fmax + Fmin )∕2 = (5 + 1.5)∕2 = 3.25 lbf

The strengths from Example 10–6 include Sut = 264.7 kpsi, Sy = 198.5 kpsi, and Ssy = 119.1 kpsi. The ultimate shear strength is estimated from Equation (10–30) as

Ssu = 0.67Sut = 0.67(264.7) = 177.3 kpsi

(a) Body-coil fatigue:

τa =

τm =

8KB Fa D πd 3

=

8(1.234)1.75(0.213) π(0.0353 )

(10−3 ) = 27.3 kpsi

Fm 3.25 τa = 27.3 = 50.7 kpsi Fa 1.75

Using the Zimmerli data of Equation (10–28a) with the Gerber criterion of Equation (10-29b) gives

Sse =

Ssa 35 = = 38.7 kpsi Ssm 2 55 2 1−( ) 1−( Ssu 177.3 )

556      Mechanical Engineering Design

From Equation (6–48), the Gerber fatigue criterion for shear is Answer

τm Sse 2 1 Ssu 2 τa (n f ) body = ( ) −1 + 1 + 2 √ ( Ssu τa ) ] 2 τm Sse [ 1 177.3 2 27.3 50.7 38.7 2 = ( −1 + 1 + 2 √ ( 177.3 27.3 ) ] = 1.24 2 50.7 ) 38.7 [

(b) The load-line for the coil body begins at Ssm = τi and has a slope r = τa∕(τm − τi). It can be shown that the intersection with the yield line is given by (Ssa)y = [r∕(r + 1)] (Ssy − τi). Consequently, τi = (Fi∕Fa)τa = (1.19∕1.75)27.3 = 18.6 kpsi, r = 27.3∕(50.7 − 18.6) = 0.850, and

(Ssa ) y =

0.850 (119.1 − 18.6) = 46.2 kpsi 0.850 + 1

Thus, (ny ) body =

Answer

(Ssa ) y 46.2 = = 1.69 τa 27.3

(c) End-hook bending fatigue: using Equations (10–34) and (10–35) gives

16D 4 σa = Fa [ (K) A 3 + 2 ] πd πd

16(0.213) 4 = 1.75 [ 1.14 + (10−3 ) = 52.3 kpsi π(0.0353 ) π(0.0352 ) ] σm =

Fm 3.25 σa = 52.3 = 97.1 kpsi Fa 1.75

To estimate the tensile endurance limit using the distortion-energy theory, Se = Sse∕0.577 = 38.7∕0.577 = 67.1 kpsi

Using the Gerber criterion for tension from Equation (6–48) gives Answer

σm Se 2 1 Sut 2 σa (n f ) A = ( ) −1 + 1 + 2 √ ( Sut σa ) ] 2 σm Se [ 1 264.7 2 52.3 97.1 67.1 2 = ( −1 + 1 + 2 √ ( 264.7 52.3 ) ] = 1.08 2 97.1 ) 67.1 [

(d) End-hook torsional fatigue: from Equation (10–36)

(τa ) B = (K) B

(τm ) B =

8Fa D πd

3

= 1.18

8(1.75)0.213 π(0.0353 )

(10−3 ) = 26.1 kpsi

Fm 3.25 (τa ) B = 26.1 = 48.5 kpsi Fa 1.75

Then, again using the Gerber criterion, we obtain Answer

τm Sse 2 1 Ssu 2 τa (n f ) B = ( ) −1 + 1 + 2 √ ( Ssu τa ) ] 2 τm Sse [ 1 177.3 2 26.1 48.5 38.7 2 = ( −1 + 1 + 2 √ ( 177.3 26.1 ) ] = 1.30 2 48.5 ) 38.7 [

Mechanical Springs     557

Table 10–8  Maximum Allowable Stresses for ASTM A228 and Type 302 Stainless Steel Helical Extension Springs in Cyclic Applications

Percent of Tensile Strength

Number

In Torsion

of Cycles

Body

In Bending

End

End

5

36 34

51

6

10

33 30

47

107

30 28

45

10

This information is based on the following conditions: not shot-peened, no surging and ambient environment with a low temperature heat treatment applied. Stress ratio = 0. Source: Data from Design Handbook, 1987, p. 52.

The analyses in Examples 10–6 and 10–7 show how extension springs differ from compression springs. The end hooks are usually the weakest part, with bending usually controlling. We should also appreciate that a fatigue failure separates the extension spring under load. Flying fragments, lost load, and machine shutdown are threats to personal safety as well as machine function. For these reasons higher design factors are used in extension-spring design than in the design of compression springs. In Example 10–7 we estimated the endurance limit for the hook in bending using the Zimmerli data, which are based on torsion in compression springs and the distortion theory. An alternative method is to use Table 10–8, which is based on a stressratio of R = τmin∕τmax = 0. For this case, τa = τm = τmax∕2. Label the strength values of Table 10–8 as Sr for bending or Ssr for torsion. Then for torsion, for example, Ssa = Ssm = Ssr∕2. From the Gerber criterion, Equation (10–29b), the endurance limit is

Sse =

Ssa 1 − (Ssm ∕Ssu )

2

=

Ssr∕2 Ssr∕2 2 1−( Ssu )

(10–42)

So in Example 10–7 an estimate for the bending endurance limit from Table 10–8 would be Sr = 0.45Sut = 0.45(264.7) = 119.1 kpsi and from Equation (10–42)

Se =

Sr ∕2 2

1 − [Sr∕(2Sut )]

=

119.1∕2 = 62.7 kpsi 119.1∕2 2 1−( 264.7 )

Using this in place of 67.1 kpsi in Example 10–7 results in (nf)A = 1.03, a reduction of 5 percent.

10–12  Helical Coil Torsion Springs When a helical coil spring is subjected to end torsion, it is called a torsion spring. It is usually close-wound, as is a helical coil extension spring, but with negligible initial tension. There are single-bodied and double-bodied types as depicted in

558      Mechanical Engineering Design

Figure 10–8 Torsion springs. (Courtesy of Associated Spring.)

Special ends

Short hook ends

Hinge ends

Double torsion

Straight offset

Straight torsion

Figure 10–8. As shown in the figure, torsion springs have ends configured to apply torsion to the coil body in a convenient manner, with short hook, hinged straight offset, straight torsion, and special ends. The ends ultimately connect a force at a distance from the coil axis to apply a torque. The most frequently encountered (and least expensive) end is the straight torsion end. If intercoil friction is to be avoided completely, the spring can be wound with a pitch that just separates the body coils. Helical coil torsion springs are usually used with a rod or arbor for reactive support when ends cannot be built in, to maintain alignment, and to provide buckling resistance if necessary. The wire in a torsion spring is in bending, in contrast to the torsion encountered in helical coil compression and extension springs. The springs are designed to wind tighter in service. As the applied torque increases, the inside diameter of the coil decreases. Care must be taken so that the coils do not interfere with the pin, rod, or arbor. The bending mode in the coil might seem to invite square- or rectangular-crosssection wire, but cost, range of materials, and availability discourage its use. Torsion springs are familiar in clothespins, window shades, and animal traps, where they may be seen around the house, and out of sight in counterbalance mechanisms, ratchets, and a variety of other machine components. There are many stock springs that can be purchased off-the-shelf from a vendor. This selection can add economy of scale to small projects, avoiding the cost of custom design and small-run manufacture. Describing the End Location In specifying a torsion spring, the ends must be located relative to each other. Commercial tolerances on these relative positions are listed in Table 10–9. The simplest scheme for expressing the initial unloaded location of one end with respect to the other is in terms of an angle β defining the partial turn present in the coil body

Mechanical Springs     559

Table 10–9  End Position Tolerances for Helical Coil Torsion Springs (for D∕d Ratios up to and Including 16)

θ F l

α

Tolerance: ± Degrees*

Total Coils

Up to 3  8 Over 3–10

10

Over 10–20

15

Over 20–30

20

Over 30

25

β

Figure 10–9 The free-end location angle is β. The rotational coordinate θ is proportional to the product Fl. Its back angle is α. For all positions of the moving end θ + α = Σ = constant.

*Closer tolerances available on request. Source: Data from Design Handbook, 1987, p. 52.

as Np = β∕360°, as shown in Figure 10–9. For analysis purposes the nomenclature of Figure 10–9 can be used. Communication with a springmaker is often in terms of the back-angle α. The number of body turns Nb is the number of turns in the free spring body by count. The body-turn count is related to the initial position angle β by Nb = integer +

β = integer + Np 360°

where Np is the number of partial turns. The above equation means that Nb takes on noninteger, discrete values such as 5.3, 6.3, 7.3, . . . , with successive differences of 1 as possibilities in designing a specific spring. This consideration will be discussed later. Bending Stress A torsion spring has bending induced in the coils, rather than torsion. This means that residual stresses built in during winding are in the same direction but of opposite sign to the working stresses that occur during use. The strain-strengthening locks in residual stresses opposing working stresses provided the load is always applied in the winding sense. Torsion springs can operate at bending stresses exceeding the yield strength of the wire from which it was wound. The bending stress can be obtained from curved-beam theory expressed in the form σ=K

Mc I

where K is a stress-correction factor. The value of K depends on the shape of the wire cross section and whether the stress sought is at the inner or outer fiber. Wahl analytically determined the values of K to be, for round wire,

Ki =

4C 2 − C − 1 4C(C − 1)

Ko =

4C 2 + C − 1 4C(C + 1)

(10–43)

where C is the spring index and the subscripts i and o refer to the inner and outer fibers, respectively. In view of the fact that Ko is always less than unity, we shall use

560      Mechanical Engineering Design

Ki to estimate the stresses. When the bending moment is M = Fr and the section modulus I∕c = d3∕32, we express the bending equation as σ = Ki

32Fr πd 3

(10–44)

which gives the bending stress for a round-wire torsion spring. Deflection and Spring Rate For torsion springs, angular deflection can be expressed in radians or revolutions (turns). If a term contains revolution units the term will be expressed with a prime sign. The spring rate k′ is expressed in units of torque/revolution (lbf · in/rev or N · mm/rev) and moment is proportional to angle θ′ expressed in turns rather than radians. The spring rate, if linear, can be expressed as k′ =

M1 M2 M2 − M1 = = θ′1 θ′2 θ′2 − θ′1

(10–45)

where the moment M can be expressed as Fl or Fr. The angle subtended by the end deflection of a cantilever, when viewed from the built-in ends, is y∕l rad. From Table A–9–1,

θe =

y Fl2 Fl2 64Ml = = = 4 l 3EI 3E(πd ∕64) 3πd 4E

(10–46)

For a straight torsion end spring, end corrections such as Equation (10–46) must be added to the body-coil deflection. The strain energy in bending is, from Equation (4–23), U=

2

∫ M2EIdx

For a torsion spring, M = Fl = Fr, and integration must be accomplished over the length of the body-coil wire. The force F will deflect through a distance rθ where θ is the angular deflection of the coil body, in radians. Applying Castigliano's theorem gives rθ =

∂U = ∂F

πDNb

0

∂ F 2r 2 dx = ∂F ( 2EI )

πDNb

0

Fr 2 dx EI

Substituting I = πd 4∕64 for round wire and solving for θ gives θ=

64FrDNb d 4E

=

64MDNb d 4E

The total angular deflection in radians is obtained by adding Equation (10–46) for each end of lengths l1, l2:

θt =

64MDNb d 4E

+

64Ml1

3πd 4E

+

64Ml2

3πd 4E

=

l1 + l2 64MD Nb + 3πD ) d 4E (

(10–47)

The equivalent number of active turns Na is expressed as

Na = Nb +

l1 + l2 3πD

(10–48)

The spring rate k in torque per radian is

k=

Fr M d 4E = = θt θt 64DNa

(10–49)

Mechanical Springs     561

The spring rate may also be expressed as torque per turn. The expression for this is obtained by multiplying Equation (10–49) by 2π rad/turn. Thus spring rate k′ (units torque/turn) is

k′ =

2πd 4E d 4E = 64DNa 10.2DNa

(10–50)

Tests show that the effect of friction between the coils and arbor is such that the constant 10.2 should be increased to 10.8. The equation above becomes

k′ =

d 4E 10.8DNa

(10–51)

(units torque per turn). Equation (10–51) gives better results. Also Equation (10–47) becomes

θ′t =

l1 + l2 10.8MD Nb + 4 ( 3πD ) d E

(10–52)

Torsion springs are frequently used over a round bar or pin. When the load is applied to a torsion spring, the spring winds up, causing a decrease in the inside diameter of the coil body. It is necessary to ensure that the inside diameter of the coil never becomes equal to or less than the diameter of the pin, in which case loss of spring function would ensue. The helix diameter of the coil D′ becomes

D′ =

Nb D Nb + θ′c

(10–53)

where θ′c is the angular deflection of the body of the coil in number of turns, given by

θ′c =

10.8MDNb d 4E

(10–54)

The new inside diameter D′i = D′ − d makes the diametral clearance Δ between the body coil and the pin of diameter Dp equal to

Δ = D′ − d − Dp =

Nb D − d − Dp Nb + θ′c

(10–55)

Equation (10–55) solved for Nb is

Nb =

θ′c ( Δ + d + Dp ) D − Δ − d − Dp

(10–56)

which gives the number of body turns corresponding to a specified diametral clearance of the arbor. This angle may not be in agreement with the necessary partial-turn remainder. Thus the diametral clearance may be exceeded but not equaled. Static Strength First column entries in Table 10–6 can be divided by 0.577 (from distortion-energy theory) to give

 0.78 Sut  Sy =  0.87 Sut   0.61 Sut

Music wire and cold-drawn carbon steels OQ&T carbon and low-alloy steels (10–57) Austenitic stainless steel and nonferrous alloys

562      Mechanical Engineering Design

Table 10–10  Maximum Recommended Bending Stresses (KB Corrected) for Helical Torsion Springs in Cyclic Applications as Percent of Sut Fatigue

ASTM A228 and Type 302 Stainless Steel

ASTM A230 and A232

Life, Not Shot- Not Shot Cycles Peened Shot-Peened* Peened Shot-Peened* 105 53 62 55 64 106 50 60 53 62 This information is based on the following conditions: no surging, springs are in the "as-stress-relieved" condition. *Not always possible. Source: Data from Associated Spring.

Fatigue Strength Since the spring wire is in bending, the Sines equation is not applicable. The Sines model is in the presence of pure torsion. Since Zimmerli's results were for compression springs (wire in pure torsion), we will use the repeated bending stress (R = 0) values provided by Associated Spring in Table 10–10. As in Equation (10–40) we will use the Gerber fatigue-failure criterion incorporating the Associated Spring R = 0 fatigue strength Sr:

Se =

Sr∕2 Sr∕2 2 1−( Sut )

(10–58)

The value of Sr (and Se) has been corrected for size, surface condition, and type of loading, but not for temperature or miscellaneous effects. The Gerber fatigue failure criterion from Equation (6–48) is applicable, and is repeated here:

nf =

σm Se 2 1 σa Sut 2 −1 + 1 + 2 √ ( Sut σa ) ] 2 Se ( σm ) [

(10–59)

EXAMPLE 10–8 A stock spring is shown in Figure 10–10. It is made from 0.072-in-diameter music wire and has 414 body turns with straight torsion ends. It works over a pin of 0.400 in diameter. The coil outside diameter is 19 32 in. (a) Find the maximum operating torque and corresponding rotation for static loading. (b) Estimate the inside coil diameter and pin diametral clearance when the spring is subjected to the torque in part (a). (c) Estimate the fatigue factor of safety nf if the applied moment varies between Mmin = 1 to Mmax = 5 lbf · in. Solution (a) For music wire, from Table 10–4 we find that A = 201 kpsi · inm and m = 0.145. Therefore, Sut =

A 201 = = 294.4 kpsi d m (0.072) 0.145

Using Equation (10–57) gives

Sy = 0.78Sut = 0.78(294.4) = 229.6 kpsi

Mechanical Springs     563

Figure 10–10 θ

F

β

Angles α, β, and θ are measured between the straight-end centerline translated to the coil axis. Coil OD is 19 32 in.

1 in

α

1 in

F 2 in

The mean coil diameter is D = 19∕32 − 0.072 = 0.5218 in. The spring index C = D∕d = 0.5218∕0.072 = 7.247. The bending stress-correction factor Ki from Equation (10–43), is

Ki =

4(7.247) 2 − 7.247 − 1 = 1.115 4(7.247)(7.247 − 1)

Now rearrange Equation (10–44), substitute Sy for σ, and solve for the maximum torque Fr to obtain

Mmax = (Fr) max =

πd 3Sy 32 Ki

=

π(0.072) 3229 600 = 7.546 lbf · in 32(1.115)

Note that no factor of safety has been used. Next, from Equation (10–54) and Table 10–5, the number of turns of the coil body θ′c is

θ′c =

10.8MDNb 4

d E

=

10.8(7.546)0.5218(4.25) 0.0724 (28.5)106

= 0.236 turn

(θ′c ) deg = 0.236(360°) = 85.0°

Answer

The active number of turns Na, from Equation (10–48), is

Na = Nb +

l1 + l2 1+1 = 4.25 + = 4.657 turns 3πD 3π(0.5218)

The spring rate of the complete spring, from Equation (10–51), is

k′ =

0.0724 (28.5)106 = 29.18 lbf · in/turn 10.8(0.5218)4.657

The number of turns of the complete spring θ′ is Answer

θ′ =

M 7.546 = = 0.259 turn k′ 29.18

(θ′s ) deg = 0.259(360°) = 93.24°

564      Mechanical Engineering Design

(b) With no load, the mean coil diameter of the spring is 0.5218 in. From Equation (10–53),

D′ =

Nb D 4.25(0.5218) = = 0.494 in Nb + θ′c 4.25 + 0.236

The diametral clearance between the inside of the spring coil and the pin at load is Answer

Δ = D′ − d − Dp = 0.494 − 0.072 − 0.400 = 0.022 in

(c) Fatigue:

Ma = (Mmax − Mmin )∕2 = (5 − 1)∕2 = 2 lbf · in

Mm = (Mmax + Mmin )∕2 = (5 + 1)∕2 = 3 lbf · in

r=

σa = Ki

σm =

32Ma πd

3

Ma 2 = Mm 3

= 1.115

32(2)

π 0.0723

= 60 857 psi

Mm 3 σa = (60 857) = 91 286 psi Ma 2

From Table 10–10, Sr = 0.50Sut = 0.50(294.4) = 147.2 kpsi. Then Se =

147.2∕2 = 78.51 kpsi 147.2∕2 2 1−( 294.4 )

Applying the Gerber criterion with Equation (10–59), Answer

2σmSe 2 1 Sut 2 σa n f = ( ) ( )[ −1 + √ 1 + ( 2 σm Se Sutσa ) ] 2(91.29) (78.51) 2 1 294.4 2 60.86 = ( −1 + 1 + √ ( (294.4) (60.86) ) ] = 1.13 2 91.29 ) ( 78.51 )[

10–13  Belleville Springs The inset of Figure 10–11 shows the cross-section of a coned-disk spring, commonly called a Belleville spring. Although the mathematical treatment is beyond the scope of this book, you should at least become familiar with the remarkable characteristics of these springs. Aside from the obvious advantage that a Belleville spring occupies only a small space, variation in the h∕t ratio will produce a wide variety of load-deflection curve shapes, as illustrated in Figure 10–11. For example, using an h∕t ratio of 2.83 or larger gives an S curve that might be useful for snap-acting mechanisms. A reduction of the ratio to a value between 1.41 and 2.1 causes the central portion of the curve to become horizontal, which means that the load is constant over a considerable deflection range. A higher load for a given deflection may be obtained by nesting, that is, by stacking the springs in parallel. On the other hand, stacking in series provides a larger deflection for the same load, but in this case there is danger of instability.

Mechanical Springs     565

Figure 10–11

Load

Load-deflection curves for Belleville springs. (Courtesy of Associated Spring.)

t 500

3

h ⁄t =

200

3 .5 0

= 2.8

h ⁄t

h ⁄t

= 2.1

1

= 1 .4

0. 7

h ⁄t =

300

h ⁄t

Load, lbf

400

h ⁄t =

0

h

100 0 –100

0

0.04

0.08

0.12

0.16

0.20

0.24

0.28

0.32

Deflection, in

10–14  Miscellaneous Springs The extension spring shown in Figure 10–12 is made of slightly curved strip steel, not flat, so that the force required to uncoil it remains constant; thus it is called a constant-force spring. This is equivalent to a zero spring rate. Such springs can also be manufactured having either a positive or a negative spring rate. A volute spring, shown in Figure 10–13a, is a wide, thin strip, or "flat," of material wound on the flat so that the coils fit inside one another. Since the coils do not stack, the solid height of the spring is the width of the strip. A variable-spring scale, in a compression volute spring, is obtained by permitting the coils to contact the support. Thus, as the deflection increases, the number of active coils decreases. The volute spring has another important advantage that cannot be obtained with round-wire springs: if the coils are wound so as to contact or slide on one another during action, the sliding friction will serve to damp out vibrations or other unwanted transient disturbances.

F

Initial deflection

h

F

l

l

F

h

x

Rated load b bo ID

bo

b

t F

(a) (a)

Figure 10–12

Figure 10–13

Constant-force spring. (Courtesy of Vulcan Spring & Mfg. Co. Telford, PA. www.vulcanspring.com.)

(a) A volute spring; (b) a flat triangular spring.

(b) (b)

b

x

F

566      Mechanical Engineering Design

A conical spring, as the name implies, is a coil spring wound in the shape of a cone (see Problem 10–29). Most conical springs are compression springs and are wound with round wire. But a volute spring is a conical spring too. Probably the principal advantage of this type of spring is that it can be wound so that the solid height is only a single wire diameter. Flat stock is used for a great variety of springs, such as clock springs, power springs, torsion springs, cantilever springs, and hair springs; frequently it is specially shaped to create certain spring actions for fuse clips, relay springs, spring washers, snap rings, and retainers. In designing many springs of flat stock or strip material, it is often economical and of value to proportion the material so as to obtain a constant stress throughout the spring material. A uniform-section cantilever spring has a stress

σ=

M Fx = I∕c I∕c

(a)

which is proportional to the distance x if I∕c is a constant. But there is no reason why I∕c need be a constant. For example, one might design such a spring as that shown in Figure 10–13b, in which the thickness h is constant but the width b is permitted to vary. Since, for a rectangular section, I∕c = bh2∕6, we have, from Equation (a), bh2 Fx = σ 6

or

b=

6Fx h2σ

(b)

Since b is linearly related to x, the width bo at the base of the spring is

bo =

6Fl h2σ

(10–60)

Good approximations for deflections can be found easily by using Castigliano's theorem. To demonstrate this, assume that deflection of the triangular flat spring is primarily due to bending and we can neglect the transverse shear force.14 The bending moment as a function of x is M = −Fx and the beam width at x can be expressed as b = box∕l. Thus, the deflection of F is given by Equation (4–31), as

y=

l

l

0

1 3 0 12 (bo x∕l)h

1 dx = ∫ ∫ M(∂M∕∂F) EI E l

12 Fl 6Fl3 = x dx = bo h3E 0 bo h3E

−Fx(−x)

dx

(10–61)

Thus the spring constant, k = F∕y, is estimated as

k=

bo h3E 6l3

(10–62)

The methods of stress and deflection analysis illustrated in previous sections of this chapter have served to illustrate that springs may be analyzed and designed by using the fundamentals discussed in the earlier chapters of this book. This is also true 14

Note that, because of shear, the width of the beam cannot be zero at x = 0. So, there is already some simplification in the design model. All of this can be accounted for in a more sophisticated model.

Mechanical Springs     567

for most of the miscellaneous springs mentioned in this section, and you should now experience no difficulty in reading and understanding the literature of such springs.

10–15  Summary In this chapter we have considered helical coil springs in considerable detail in order to show the importance of viewpoint in approaching engineering problems, their analysis, and design. For compression springs undergoing static and fatigue loads, the complete design process was presented. This was not done for extension and torsion springs, as the process is the same, although the governing conditions are not. The governing conditions, however, were provided and extension to the design process from what was provided for the compression spring should be straightforward. Problems are provided at the end of the chapter, and it is hoped that the reader will develop additional, similar, problems to tackle. As spring problems become more computationally involved, programmable calculators and computers must be used. Spreadsheet programming is very popular for repetitive calculations. As mentioned earlier, commercial programs are available. With these programs, backsolving can be performed; that is, when the final objective criteria are entered, the program determines the input values.

PROBLEMS 10–1

Within the range of recommended values of the spring index, C, determine the maximum and minimum percentage difference between the Bergsträsser factor, KB, and the Wahl factor, KW.

10–2 It is instructive to examine the question of the units of the parameter A of Equation

(10–14). Show that for U.S. customary units the units for Auscu are kpsi · inm and for SI units are MPa · mmm for ASI, which make the dimensions of both Auscu and ASI different for every material to which Equation (10–14) applies. Also show that the conversion from Auscu to ASI is given by

ASI = 6.895(25.40) m Auscu

10–3 A helical compression spring is wound using 2.5-mm-diameter music wire. The spring has an outside diameter of 31 mm with plain ground ends, and 14 total coils. (a) Estimate the spring rate. (b) What force is needed to compress this spring to closure? (c) What should the free length be to ensure that when the spring is compressed solid the torsional stress does not exceed the yield strength? (d) Is there a possibility that the spring might buckle in service?

10–4 The spring in Problem 10–3 is to be used with a static load of 130 N. Perform a design

assessment represented by Equations (10–13) and (10–18) through (10–21) if the spring is closed to solid height.

10–5 A helical compression spring is made with oil-tempered wire with wire diameter of 0.2 in,

mean coil diameter of 2 in, a total of 12 coils, a free length of 5 in, with squared ends. (a) Find the solid length. (b) Find the force necessary to deflect the spring to its solid length. (c) Find the factor of safety guarding against yielding when the spring is compressed to its solid length.

568      Mechanical Engineering Design

10–6 A helical compression spring is to be made of oil-tempered wire of 4-mm diameter with a spring index of C = 10. The spring is to operate inside a hole, so buckling is not a problem and the ends can be left plain. The free length of the spring should be 80 mm. A force of 50 N should deflect the spring 15 mm. (a) Determine the spring rate. (b) Determine the minimum hole diameter for the spring to operate in. (c) Determine the total number of coils needed. (d) Determine the solid length. (e) Determine a static factor of safety based on the yielding of the spring if it is compressed to its solid length.

10–7 A helical compression spring is made of hard-drawn spring steel wire 0.080-in in diameter and has an outside diameter of 0.880 in. The ends are plain and ground, and there are 8 total coils. (a) The spring is wound to a free length, which is the largest possible with a solid-safe property. Find this free length. (b) What is the pitch of this spring? (c) What force is needed to compress the spring to its solid length? (d) Estimate the spring rate. (e) Will the spring buckle in service?

10–8 The spring of Problem 10–7 is to be used with a static load of 16.5 lbf. Perform a

design assessment represented by Equations (10–13) and (10–18) through (10–21) if the spring closed to solid height.

10–9 Listed in the tables are six springs described in customary units and five springs to described in SI units. Investigate these squared-and-ground-ended helical compression 10–19 springs to see if they are solid-safe. If not, what is the largest free length to which they can be wound using ns = 1.2?

Problem Number d, in 10–9

0.007

OD, in

L0, in

0.038

0.58

Nt Material 38

A228 music wire

10–10 0.014 0.128 0.50 16 B159 phosphor-bronze 10–11

0.050

0.250

0.68

10–12

0.148

2.12

2.5

10–13

0.138

0.92

2.86

10–14 0.185 2.75

d, mm

OD, mm

11.2 5.75 12

A313 stainless steel A227 hard-drawn steel A229 OQ&T steel

7.5 8 A232 chrome-vanadium

L0, mm

Nt Material

10–15 0.25 0.95 12.1 38 A313 stainless steel 10–16

1.2

6.5

15.7

10.2

10–17

3.5

50.6

75.5

5.5

10–18 3.8 31.4 10–19

4.5

69.2

A228 music wire A229 OQ&T spring steel

71.4 12.8 B159 phosphor-bronze 215.6

10–20 Consider the steel spring in the illustration.

8.2

A232 chrome-vanadium

(a) Find the pitch, solid height, and number of active turns. (b) Find the spring rate. Assume the material is A227 HD steel.

Mechanical Springs     569

(c) Find the force Fs required to close the spring solid. (d) Find the shear stress in the spring due to the force Fs. 3

4 4 in

Problem 10–20

2 in

0.135 in

10–21 Consider a plain end hard-drawn steel compression helical spring. The spring is made

of 12-gauge Washburn & Moen wire, has an outside coil diameter of 0.75 in, 20 total coils, and a free length of 3.75 in. Determine the (a) spring index, C. (b) pitch of the coil, p. (c) spring deflection when closed solid, ys. (d) force to close the spring solid, Fs. (e) shear stress at closure, τs. ( f ) factor of safety in yield at closure, ns. (g) spring rate, k, using the exact and the approximate formulation, and compare the two results.

10–22 Consider a squared and ground oil-tempered compression helical spring. The spring

is made of 3-mm wire, has an outer diameter of 30 mm, 32 total coils, and a free length of 240 mm. Determine the (a) spring index, C. (b) pitch of the coil, p. (c) spring deflection when closed solid, ys. (d) force to close the spring solid, Fs. (e) shear stress at closure, τs. ( f ) factor of safety in yield at closure, ns. (g) spring rate, k, using the exact and the approximate formulation, and compare the two results.

10–23 Starting with a spring index of C = 10, design a compression coil spring of 302 stainless

wire with squared ends. The spring is to deflect y = 50 mm, for an applied force of F = 90 N and to close solid when ys = 60 mm. At closure, use a design factor of 1.2 guarding against yielding. Select the smallest diameter wire where wire diameters are available in 0.2-mm increments between 0.2 to 3.2 mm. For the design, specify the wire diameter, the inside and outside diameters of the coil, the spring rate, the total number of coils, the free length of the spring, the final factor of safety, and stability conditions.

10–24 Starting with a spring index of C = 10, design a compression coil spring of phosphor

bronze with closed ends. The spring is to deflect y = 2 in, for an applied force of F = 15 lbf and to close solid when ys = 3 in. At closure, use a design factor of 1.2 guarding against yielding. Determine the smallest-diameter wire based on preferred sizes. For the design, specify the wire diameter, the inside and outside diameters of the coil, the spring rate, the total number of coils, the free length of the spring, the final factor of safety, and stability conditions.

570      Mechanical Engineering Design

10–25 After determining the wire diameter in Problem 10–24, determine what the spring index should be to maintain a design factor of 1.2 at closure. For this design, specify the wire diameter, the inside and outside diameters of the coil, the spring rate, the total number of coils, the free length of the spring, the final factor of safety, and stability conditions.

10–26 A static service music wire helical compression spring is needed to support a 20-lbf

load after being compressed 2 in. The solid height of the spring cannot exceed 112 in. The free length must not exceed 4 in. The static factor of safety must equal or exceed 1.2. For robust linearity use a fractional overrun to closure ξ of 0.15. There are two springs to be designed. Start with a wire diameter of 0.075 in. (a) The spring must operate over a 34 -in rod. A 0.050-in diametral clearance allowance should be adequate to avoid interference between the rod and the spring due to out-of-round coils. Design the spring. (b) The spring must operate in a 1-in-diameter hole. A 0.050-in diametral clearance allowance should be adequate to avoid interference between the spring and the hole due to swelling of the spring diameter as the spring is compressed and outof-round coils. Design the spring.

10–27 Solve Problem 10–26 by iterating with an initial value of C = 10. If you have already solved Problem 10–26, compare the steps and the results.

10–28 A holding fixture for a workpiece 37.5 mm thick at the clamp locations is being

designed. The detail of one of the clamps is shown in the figure. A spring is required to drive the clamp upward when removing the workpiece with a starting force of 45 N. The clamp screw has an M10 × 1.25 thread. Allow a diametral clearance of 1.25 mm between it and the uncompressed spring. It is further specified that the free length of the spring should be L0 ≤ 48 mm, the solid height Ls ≤ 31.5 mm, and the safety factor when closed solid should be ns ≥ 1.2. Starting with d = 2 mm, design a suitable helical coil compression spring for this fixture. For A227 HD steel, wire diameters are available in 0.2-mm increments between 0.2 to 3.2 mm. Clamp screw

Spherical washer Slot

Clamp

Problem 10–28 Clamping fixture.

Groove Workpiece Pin

10–29 Solve Problem 10–28 by iterating with an initial value of C = 8. If you have already solved Problem 10–28, compare the steps and the results.

Mechanical Springs     571

10–30 Your instructor will provide you with a stock spring supplier's catalog, or pages repro-

duced from it. Accomplish the task of Problem 10–28 by selecting an available stock spring. (This is design by selection.)

10–31 A compression spring is needed to fit over a 0.5-in diameter rod. To allow for some

clearance, the inside diameter of the spring is to be 0.6 in. To ensure a reasonable coil, use a spring index of 10. The spring is to be used in a machine by compressing it from a free length of 5 in through a stroke of 3 in to its solid length. The spring should have squared and ground ends, unpeened, and is to be made from cold-drawn wire. (a) Determine a suitable wire diameter. (b) Determine a suitable total number of coils. (c) Determine the spring constant. (d) Determine the static factor of safety when compressed to solid length. (e) Determine the fatigue factor of safety when repeatedly cycled from free length to solid length. Use the Gerber-Zimmerli fatigue-failure criterion.

10–32 A compression spring is needed to fit within a 1-in diameter hole. To allow for some

clearance, the outside diameter of the spring is to be no larger than 0.9 in. To ensure a reasonable coil, use a spring index of 8. The spring is to be used in a machine by compressing it from a free length of 3 in to a solid length of 1 in. The spring should have squared ends, and is unpeened, and is to be made from music wire. (a) Determine a suitable wire diameter. (b) Determine a suitable total number of coils. (c) Determine the spring constant. (d) Determine the static factor of safety when compressed to solid length. (e) Determine the fatigue factor of safety when repeatedly cycled from free length to solid length. Use the Gerber-Zimmerli fatigue-failure criterion.

10–33 A helical compression spring is to be cycled between 150 lbf and 300 lbf with a 1-in

stroke. The number of cycles is low, so fatigue is not an issue. The coil must fit in a 2.1-in diameter hole with a 0.1-in clearance all the way around the spring. Use unpeened oil tempered wire with squared and ground ends. (a) Determine a suitable wire diameter, using a spring index of C = 7. (b) Determine a suitable mean coil diameter. (c) Determine the necessary spring constant. (d) Determine a suitable total number of coils. (e) Determine the necessary free length so that if the spring were compressed to its solid length, there would be no yielding.

10–34 The figure shows a conical compression helical coil spring where R1 and R2 are

F R1

the initial and final coil radii, respectively, d is the diameter of the wire, and Na is the total number of active coils. The wire cross section primarily transmits a torsional moment, which changes with the coil radius. Let the coil radius be given by R = R1 +

d

R 2 − R1 θ 2πNa

where θ is in radians. Use Castigliano's method to estimate the spring rate as

k=

d 4G 16Na (R2 + R1 ) (R22 + R12 )

R2 Problem 10–34

572      Mechanical Engineering Design

10–35 A helical coil compression spring is needed for food service machinery. The load­ varies from a minimum of 4 lbf to a maximum of 18 lbf. The spring rate k is to be 9.5 lbf/in. The outside diameter of the spring cannot exceed 212 in. The springmaker has available suitable dies for drawing 0.080-, 0.0915-, 0.1055-, and 0.1205-in-diameter wire. Use squared and ground ends, and an unpeened material. Using a fatigue design factor nf of 1.5, and the Gerber-Zimmerli fatigue-failure criterion, design a suitable spring.

10–36 Solve Problem 10–35 using the Goodman-Zimmerli fatigue-failure criterion. 10–37 Solve Problem 10–35 using the Sines-Zimmerli fatigue-failure criterion. 10–38 Design the spring of Example 10–5 using the Gerber-Zimmerli fatigue-failure criterion. 10–39 Solve Problem 10–38 using the Goodman-Zimmerli fatigue-failure criterion. 10–40 A hard-drawn spring steel extension spring is to be designed to carry a static load of

18 lbf with an extension of 12 in using a design factor of ny = 1.5 in bending. Use full-coil end hooks with the fullest bend radius of r = D∕2 and r2 = 2d. The free length must be less than 3 in, and the body turns must be fewer than 30. (Note: Integer and half-integer body turns allow end hooks to be placed in the same plane. However, this adds extra cost and is done only when necessary.)

10–41 The extension spring shown in the figure has full-twisted loop ends. The material is AISI 1065 OQ&T wire. The spring has 84 coils and is close-wound with a preload of 16 lbf. (a) Find the closed length of the spring. (b) Find the torsional stress in the spring corresponding to the preload. (c) Estimate the spring rate. (d) What load would cause permanent deformation? (e) What is the spring deflection corresponding to the load found in part d?

0.162 in 1 12

Problem 10–41 1 - in 4

R.

in 1 - in 2

R.

10–42 Design an infinite-life helical coil extension spring with full end loops and generous

loop-bend radii for a minimum load of 9 lbf and a maximum load of 18 lbf, with an accompanying stretch of 14 in. The spring is for food-service equipment and must be stainless steel. The outside diameter of the coil cannot exceed 1 in, and the free length cannot exceed 212 in. Using a fatigue design factor of nf = 2, complete the design. Use the Gerber criterion with Table 10–8.

10–43 Prove Equation (10–40). Hint: Using Castigliano's theorem, determine the deflection

due to bending of an end hook alone as if the hook were fixed at the end connecting it to the body of the spring. Consider the wire diameter d small as compared to the mean radius of the hook, R = D∕2. Add the deflections of the end hooks to the deflection of the main body to determine the final spring constant, then equate it to Equation (10–9).

Mechanical Springs     573

10–44 The figure shows a finger exerciser used by law-enforcement officers and athletes to

strengthen their grip. It is formed by winding A227 hard-drawn steel wire around a mandrel to obtain 212 turns when the grip is in the closed position. After winding, the wire is cut to leave the two legs as handles. The plastic handles are then molded on, the grip is squeezed together, and a wire clip is placed around the legs to obtain initial "tension" and to space the handles for the best initial gripping position. The clip is formed like a figure 8 to prevent it from coming off. When the grip is in the closed position, the stress in the spring should not exceed the permissible stress. (a) Determine the configuration of the spring before the grip is assembled. (b) Find the force necessary to close the grip. 16 R.

4 dia

Wire clip

Problem 10–44 Dimensions in millimeters.

+

Molded plastic handle

87.5

112.5

75

10–45 The rat trap shown in the figure uses two opposite-image torsion springs. The wire

has a diameter of 0.081 in, and the outside diameter of the spring in the position shown is 12 in. Each spring has 11 turns. Use of a fish scale revealed a force of about 8 lbf is needed to set the trap. (a) Find the probable configuration of the spring prior to assembly. (b) Find the maximum stress in the spring when the trap is set.

Problem 10–45 A

1

1 2 in

VICTOR

5

3 16 in

10–46 Wire form springs can be made in a variety of shapes. The clip shown operates by

applying a force F. The wire diameter is d, the length of the straight section is l, and Young's modulus is E. Consider the effects of bending only, with d ≪ R. (a) Use Castigliano's theorem to determine the spring rate k. (b) Determine the spring rate if the clip is made from 2-mm diameter A227 harddrawn steel wire with R = 6 mm and l = 25 mm. (c) For part (b), estimate the value of the load F, which will cause the wire to yield.

574      Mechanical Engineering Design l

R Problem 10–46 R F

F

10–47 For the wire form shown, the wire diameter is d, the length of the straight section is l,

and Young's modulus is E. Consider the effects of bending only, with d ≪ R. (a) Use Castigliano's method to determine the spring rate k. (b) Determine the spring rate if the form is made from 0.063-in diameter A313 stainless wire with R = 58 in and l = 12 in. (c) For part (b), estimate the value of the load F, which will cause the wire to yield. C

Problem 10–47

F

R

A B l

10–48 Figure 10–13b shows a spring of constant thickness and constant stress. A constant

stress spring can be designed where the width b is constant as shown. (a) Determine how h varies as a function of x. (b) Given Young's modulus E, determine the spring rate k in terms of E, l, b, and ho. Verify the units of k. F

l

ho Problem 10–48

h x

b

10–49 Using the experience gained with Problem 10–35, write a computer program that would help in the design of helical coil compression springs.

10–50 Using the experience gained with Problem 10–42, write a computer program that would help in the design of a helical coil extension spring.

11

Rolling-Contact Bearings

©Андрей Радченко/123RF

Chapter Outline

11–7

11–1

Bearing Types   576

11–2

Bearing Life   579

11–8  Selection of Ball and Cylindrical Roller Bearings  593

11–3

Bearing Load Life at Rated Reliability   580

11–4  Reliability versus Life—The Weibull Distribution  582 11–5

Relating Load, Life, and Reliability   583

11–6

Combined Radial and Thrust Loading   585

11–9

Variable Loading   590

Selection of Tapered Roller Bearings   596

11–10  Design Assessment for Selected RollingContact Bearings   604 11–11

Lubrication  608

11–12

Mounting and Enclosure   609 575

576      Mechanical Engineering Design

The terms rolling-contact bearing, antifriction bearing, and rolling bearing are all used to describe that class of bearing in which the main load is transferred through elements in rolling contact rather than in sliding contact. In a rolling bearing the starting friction is about twice the running friction, but still it is negligible in comparison with the starting friction of a sleeve bearing. Load, speed, and the operating viscosity of the lubricant do affect the frictional characteristics of a rolling bearing. It is probably a mistake to describe a rolling bearing as "antifriction," but the term is used generally throughout the industry. From the mechanical designer's standpoint, the study of antifriction bearings differs in several respects when compared with the study of other topics because the bearings they specify have already been designed. The specialist in antifrictionbearing design is confronted with the problem of designing a group of elements that compose a rolling bearing: these elements must be designed to fit into a space whose dimensions are specified; they must be designed to receive a load having certain characteristics; and finally, these elements must be designed to have a satisfactory life when operated under the specified conditions. Bearing specialists must therefore consider such matters as fatigue loading, friction, heat, corrosion resistance, kinematic problems, material properties, lubrication, machining tolerances, assembly, use, and cost. From a consideration of all these factors, bearing specialists arrive at a compromise that, in their judgment, is a good solution to the problem as stated. We begin with an overview of bearing types; then we note that bearing life cannot be described in deterministic form. We introduce the invariant, the statistical distribution of bearing life, which is described by the Weibull distribution. There are some useful deterministic equations addressing load versus life at constant reliability, and the catalog rating as rating life is introduced. The load-life-reliability relationship combines statistical and deterministic relationships, which gives the designer a way to move from the desired load and life to the catalog rating in one equation. Ball bearings also resist thrust, and a unit of thrust does different damage per revolution than a unit of radial load, so we must find the equivalent pure radial load that does the same damage as the existing radial and thrust loads. Next, variable loading, stepwise and continuous, is approached, and the equivalent pure radial load doing the same damage is quantified. Oscillatory loading is mentioned. With this preparation we have the tools to consider the selection of ball and cylindrical roller bearings. The question of misalignment is quantitatively approached. Tapered roller bearings have some complications, and our experience so far contributes to understanding them. Having the tools to find the proper catalog ratings, we make decisions (selections), we perform a design assessment, and the bearing reliability is quantified. Lubrication and mounting conclude our introduction. Vendors' manuals should be consulted for specific details relating to bearings of their manufacture.

11–1  Bearing Types Bearings are manufactured to take pure radial loads, pure thrust loads, or a combination of the two kinds of loads. The nomenclature of a ball bearing is illustrated in Figure 11–1, which also shows the four essential parts of a bearing. These are the outer ring, the inner ring, the balls or rolling elements, and the separator. In low-priced bearings, the separator is sometimes omitted, but it has the important function of separating the elements so that rubbing contact will not occur.

Rolling-Contact Bearings     577

Figure 11–1

Outside diameter Inner ring

Bore

Face

Nomenclature of a ball bearing. (Source: Based on General Motors Corp., GM Media Archives)

Outer ring

Width Outer ring ball race

Separator (retainer)

Inner ring ball race

Corner radius

Shoulders Corner radius

Figure 11–2 +

+

+

+

(a) Deep groove

(b) Filling notch

(c) Angular contact

(d) Shielded

+

+

+

(e) Sealed

+ + +

(f) External self-aligning

(g) Double row

(h) Self-aligning

(i) Thrust

( j) Self-aligning thrust

In this section we include a selection from the many types of standardized bearings that are manufactured. Most bearing manufacturers provide engineering manuals and brochures containing lavish descriptions of the various types available. In the small space available here, only a meager outline of some of the most common types can be given. So you should include a survey of bearing manufacturers' literature in your studies of this section. Some of the various types of standardized bearings that are manufactured are shown in Figure 11–2. The single-row deep-groove bearing will take radial load as well as some thrust load. The balls are inserted into the grooves by moving the inner ring to an eccentric position. The balls are separated after loading, and the separator is then inserted. The use of a filling notch (Figure 11–2b) in the inner and outer rings enables a greater number of balls to be inserted, thus increasing the load capacity. The thrust capacity is decreased, however, because of the bumping of the balls against the edge of the notch when thrust loads are present. The angular-contact bearing (Figure 11–2c) provides a greater thrust capacity. All these bearings may be obtained with shields on one or both sides. The shields are not a complete closure but do offer a measure of protection against dirt. A variety

Various types of ball bearings.

578      Mechanical Engineering Design

of bearings are manufactured with seals on one or both sides. When the seals are on both sides, the bearings are lubricated at the factory. Although a sealed bearing is supposed to be lubricated for life, a method of relubrication is sometimes provided. Single-row bearings will withstand a small amount of shaft misalignment of deflection, but where this is severe, self-aligning bearings may be used. Double-row bearings are made in a variety of types and sizes to carry heavier radial and thrust loads. Sometimes two single-row bearings are used together for the same reason, although a double-row bearing will generally require fewer parts and occupy less space. The one-way ball thrust bearings (Figure 11–2i) are made in many types and sizes. Some of the large variety of standard roller bearings available are illustrated in Figure 11–3. Straight roller bearings (Figure 11–3a) will carry a greater radial load than ball bearings of the same size because of the greater contact area. However, they have the disadvantage of requiring almost perfect geometry of the raceways and rollers. A slight misalignment will cause the rollers to skew and get out of line. For this reason, the retainer must be heavy. Straight roller bearings will not, of course, take thrust loads. Helical rollers are made by winding rectangular material into rollers, after which they are hardened and ground. Because of the inherent flexibility, they will take considerable misalignment. If necessary, the shaft and housing can be used for raceways instead of separate inner and outer races. This is especially important if radial space is limited. The spherical-roller thrust bearing (Figure 11–3b) is useful where heavy loads and misalignment occur. The spherical elements have the advantage of increasing their contact area as the load is increased. Needle bearings (Figure 11–3d) are very useful where radial space is limited. They have a high load capacity when separators are used, but may be obtained without separators. They are furnished both with and without races.

(a)

(b)

(c)

(a)

(b)

(c)

(a)

(b)

(c)

(d )

(e)

(f)

(d )

(e)

(f)

Figure 11–3 Types of roller bearings: (a) straight roller; (b) spherical roller, thrust; (c) tapered roller, thrust; (d) needle; (e) tapered roller; ( f ) steep-angle tapered roller. (Source: Redrawn from material Furnished by The Timken Company.)

Rolling-Contact Bearings     579

Tapered roller bearings (Figure 11–3e, f ) combine the advantages of ball and straight roller bearings, since they can take either radial or thrust loads or any combination of the two, and in addition, they have the high load-carrying capacity of straight roller bearings. The tapered roller bearing is designed so that all elements in the roller surface and the raceways intersect at a common point on the bearing axis. The bearings described here represent only a small portion of the many available for selection. Many special-purpose bearings are manufactured, and bearings are also made for particular classes of machinery. Typical of these are: ∙ Instrument bearings, which are high-precision and are available in stainless steel and high-temperature materials ∙ Nonprecision bearings, usually made with no separator and sometimes having split or stamped sheet-metal races ∙ Ball bushings, which permit either rotation or sliding motion or both ∙ Bearings with flexible rollers

11–2  Bearing Life When the ball or roller of rolling-contact bearings rolls, contact stresses occur on the inner ring, the rolling element, and on the outer ring. Because the curvature of the contacting elements in the axial direction is different from that in the radial direction, the equations for these stresses are more involved than in the Hertz equations presented in Chapter 3. If a bearing is clean and properly lubricated, is mounted and sealed against the entrance of dust and dirt, is maintained in this condition, and is operated at reasonable temperatures, then metal fatigue will be the only cause of failure. Inasmuch as metal fatigue implies many millions of stress applications successfully endured, we need a quantitative life measure. Common life measures are ∙ Number of revolutions of the inner ring (outer ring stationary) until the first tangible evidence of fatigue ∙ Number of hours of use at a standard angular speed until the first tangible evidence of fatigue The commonly used term is bearing life, which is applied to either of the measures just mentioned. It is important to realize, as in all fatigue, life as defined above is a stochastic variable and, as such, has both a distribution and associated statistical parameters. The life measure of an individual bearing is defined as the total number of revolutions (or hours at a constant speed) of bearing operation until the failure criterion is developed. Under ideal conditions, the fatigue failure consists of spalling of the load-carrying surfaces. The American Bearing Manufacturers Association (ABMA) standard states that the failure criterion is the first evidence of fatigue. The fatigue criterion used by the Timken Company laboratories is the spalling or pitting of an area of 0.01 in2. Timken also observes that the useful life of the bearing may extend considerably beyond this point. This is an operational definition of fatigue failure in rolling bearings. The rating life is a term sanctioned by the ABMA and used by most manufacturers. The rating life of a group of nominally identical ball or roller bearings is defined as the number of revolutions (or hours at a constant speed) that 90 percent of a group of bearings will achieve or exceed before the failure criterion develops.

580      Mechanical Engineering Design

The terms minimum life, L10 life, and B10 life are also used as synonyms for rating life. The rating life is the 10th percentile location of the bearing group's revolutionsto-failure distribution. Median life is the 50th percentile life of a group of bearings. The term average life has been used as a synonym for median life, contributing to confusion. When many groups of bearings are tested, the median life is between 4 and 5 times the L10 life. Each bearing manufacturer will choose a specific rating life for which load ratings of its bearings are reported. The most commonly used rating life is 106 revolutions. The Timken Company is a well-known exception, rating its bearings at 3000 hours at 500 rev/min, which is 90(106) revolutions. These levels of rating life are actually quite low for today's bearings, but since rating life is an arbitrary reference point, the traditional values have generally been maintained.

11–3  Bearing Load Life at Rated Reliability When nominally identical groups are tested to the life-failure criterion at different loads, the data are plotted on a graph as depicted in Figure 11–4 using a log-log transformation. To establish a single point, load F1 and the rating life of group one (L10)1 are the coordinates that are logarithmically transformed. The reliability associated with this point, and all other points, is 0.90. Thus we gain a glimpse of the load-life function at 0.90 reliability. Using a regression equation of the form FL1∕a = constant

(11–1)

the result of many tests for various kinds of bearings result in ∙ a = 3 for ball bearings ∙ a = 10∕3 for roller bearings (cylindrical and tapered roller) A catalog load rating is defined as the radial load that causes 10 percent of a group of bearings to fail at the bearing manufacturer's rating life. We shall denote the catalog load rating as C10. The catalog load rating is often referred to as a Basic Dynamic Load Rating, or sometimes just Basic Load Rating, if the manufacturer's rating life is 106 revolutions. The radial load that would be necessary to cause failure at such a low life would be unrealistically high. Consequently, the Basic Load Rating should be viewed as a reference value, and not as an actual load to be achieved by a bearing. Figure 11–4

log F

Typical bearing load-life log-log curve.

0

log L

Rolling-Contact Bearings     581

In selecting a bearing for a given application, it is necessary to relate the desired load and life requirements to the published catalog load rating corresponding to the catalog rating life. From Equation (11–1) we can write 1∕a F1 L1∕a 1 = F2 L 2 (11–2)

where the subscripts 1 and 2 can refer to any set of load and life conditions. Letting F1 and L1 correlate with the catalog load rating and rating life, and F2 and L2 correlate with desired load and life for the application, we can express Equation (11–2) as 1∕a FR L1∕a R = FD LD

(a)

where the units of LR and LD are revolutions, and the subscripts R and D stand for Rated and Desired. It is sometimes convenient to express the life in hours at a given speed. Accordingly, any life L in revolutions can be expressed as

L = 60 ℒn

(b)

where ℒ is in hours, n is in rev/min, and 60 min/h is the appropriate conversion factor. Incorporating Equation (b) into Equation (a), FR (ℒR n R 60) 1∕a = FD (ℒD nD 60) 1∕a

(c)

catalog rating, lbf or kN             desired speed, rev/min     rating life in hours          desired life, hours    rating speed, rev/min          desired radial load, lbf or kN Solving Equation (c) for FR, and noting that it is simply an alternate notation for the catalog load rating C10, we obtain an expression for a catalog load rating as a function of the desired load, desired life, and catalog rating life. L D 1∕a ℒD nD 60 1∕a C10 = FR = FD ( ) = FD ( LR ℒR n R 60 )

(11–3)

It is sometimes convenient to define xD = LD∕LR as a dimensionless multiple of rating life.

EXAMPLE 11–1 Consider SKF, which rates its bearings for 1 million revolutions. If you desire a life of 5000 h at 1725 rev/min with a load of 400 lbf with a reliability of 90 percent, for which catalog rating would you search in an SKF catalog? Solution The rating life is L10 = LR = ℒRnR60 = 106 revolutions. From Equation (11–3), Answer

ℒD n D 60 1∕a 5000(1725)60 1∕3 C10 = FD ( = 400 [ ] = 3211 lbf = 14.3 kN ℒR nR 60 ) 106

582      Mechanical Engineering Design

11–4  Reliability versus Life—The Weibull Distribution At constant load, the life measure distribution of rolling-contact bearings is right skewed. Because of its robust ability to adjust to varying amounts of skewness, the three-parameter Weibull distribution is used exclusively for expressing the reliability of rolling-contact bearings. Unlike the development of the normal distribution in Section 1–12, we will begin with the definition of the reliability, R, for a Weibull distribution of the life measure, x, as

x − x0 b R = exp [ − ( θ − x0 ) ]

(11–4)

where the three parameters are1 x0 = guaranteed, or "minimum," value of x θ = c haracteristic parameter. For rolling-contact bearings, this corresponds to the 63.2121 percentile value of x b = shape parameter that controls the skewness. For rolling-contact bearings, b ≈ 1.5 The life measure is expressed in dimensionless form as x = L∕L10. From Equation (1–8), R = 1 − p, where p is the probability of a value of x occurring between −∞ and x, and is the integral of the probability distribution, f (x), between those limits. Accordingly, f (x) = −dR∕dx. Thus, from the derivative of Equation (11–4), the Weibull probability density function, f (x), is given by

 b x − x0 b−1 x − x0 b exp −  θ − x (θ − x ) [ ( θ − x0 ) ] 0 0 f(x) =   0 

x ≥ x0 ≥ 0 x < x0

(11–5)

The mean and standard deviation of f (x) are

μx = x0 + (θ − x 0 ) Γ(1 + 1∕b)

(11–6)

σˆx = (θ − x 0 ) √Γ(1 + 2∕b) − Γ 2 (1 + 1∕b)

(11–7)

where Γ is the gamma function, and is found tabulated in Table A–34. Given a specific required reliability, solving Equation (11–4) for x yields

1 1∕b x = x0 + (θ − x0 ) (ln ) R

(11–8)

EXAMPLE 11–2 Construct the distributional properties of a 02–30 mm deep-groove ball bearing if the Weibull parameters are x0 = 0.020, θ = 4.459, and b = 1.483. Find the mean, median, 10th percentile life, standard deviation, and coefficient of variation.

1

To estimate the Weibull parameters from data, see J. E. Shigley and C. R. Mischke, Mechanical Engineering Design, 5th ed., McGraw-Hill, New York, 1989, Sec. 4–12, Ex. 4–10.

Rolling-Contact Bearings     583

Solution From Equation (11–6) and interpolating Table A–34, the mean dimensionless life is Answer

μx = x0 + (θ − x0 )Γ(1 + 1∕b)

= 0.020 + (4.459 − 0.020)Γ(1 + 1∕1.483)

= 0.020 + 4.439Γ(1.67431) = 0.020 + 4.439(0.9040) = 4.033

This says that the average bearing life is 4.033 L10. The median dimensionless life corresponds to R = 0.50, or L50, and from Equation (11–8) is Answer

1 1∕b x0.50 = x0 + (θ − x0 ) (ln 0.50 ) 1 1∕1.483 = 0.020 + (4.459 − 0.020) (ln = 3.487 0.50 )

or, L = 3.487 L10. The 10th percentile value of the dimensionless life x is Answer

x 0.10 = 0.020 + (4.459 − 0.020) (ln

1 1∕1.483 ≈ 1   (as it should be) 0.90 )

The standard deviation of the dimensionless life, given by Equation (11–7), is Answer

σˆx = (θ − x 0 ) √Γ(1 + 2∕b) − Γ 2 (1 + 1∕b)

= (4.459 − 0.020) √Γ(1 + 2∕1.483) − Γ 2 (1 + 1∕1.483)

= 4.439 √Γ(2.349) − Γ 2 (1.674) = 4.439 √1.2023 − 0.90402

= 2.755

The coefficient of variation of the dimensionless life is Cx =

Answer

σˆx 2.755 = = 0.683 μx 4.033

11–5  Relating Load, Life, and Reliability This is the designer's problem. The desired load is not the manufacturer's test load or catalog entry. The desired speed is different from the vendor's test speed, and the reliability expectation is typically much higher than the 0.90 accompanying the catalog entry. Figure 11–5 shows the situation. The catalog information is plotted as point A, whose coordinates are (the logs of) C10 and x10 = L10∕L10 = 1, a point on the 0.90 reliability contour. The design point is at D, with the coordinates (the logs of) FD and xD, a point that is on the R = RD reliability contour. The designer must move from point D to point A via point B as follows. Along a constant reliability contour (BD), Equation (11–2) applies: FB xB1∕a = FD xD1∕a from which

FB = FD(

xD 1∕a xB )

(a)

584      Mechanical Engineering Design

Figure 11–5

log F

Constant reliability contours. Point A represents the catalog rating C10 at x = L∕L10 = 1. Point B is on the target reliability design line RD, with a load of C10. Point D is a point on the desired reliability contour exhibiting the design life xD = LD∕L10 at the design load FD.

Rated line

C10

B

A

R=

0.90

D

FD

R=

RD Design line log x

x10 xD Dimensionless life measure x

Along a constant load line (AB), Equation (11–4) applies: RD = exp [ − (

xB − x 0 b θ − x0 ) ]

Solving for xB gives xB = x0 + (θ − x0 ) (ln

1 1∕b RD )

Now substitute this in Equation (a) to obtain 1∕a xD 1∕a xD FB = FD ( ) = FD [ xB x 0 + (θ − x 0 )[ln(1∕RD )]1∕b ]

Noting that FB = C10, and including an application factor af with the design load, C10 = af FD [

x 0 + (θ − x 0 )[ln (1∕RD )]1∕b ]

1∕a

xD

(11–9)

The application factor serves as a factor of safety to increase the design load to take into account overload, dynamic loading, and uncertainty. Typical load application factors for certain types of applications will be discussed shortly. Equation (11–9) can be simplified slightly for calculator entry by noting that ln

1 1 = ln = ln(1 + pf + … ) ≈ pf = 1 − RD RD 1 − pf

where pf is the probability for failure. Equation (11–9) can be written as

C10 ≈ af FD [

x 0 + (θ − x 0 )(1 − RD ) 1∕b ] xD

1∕a

R ≥ 0.90

(11–10)

Either Equation (11–9) or Equation (11–10) may be used to convert from a design situation with a desired load, life, and reliability to a catalog load rating based on a rating life at 90 percent reliability. Note that when RD = 0.90, the denominator is equal to one, and the equation reduces to Equation (11–3). The Weibull parameters are usually provided in the manufacturer's catalog. Typical values are given at the beginning of the end-of-chapter problems.

Rolling-Contact Bearings     585

EXAMPLE 11–3 The design load on a ball bearing is 413 lbf and an application factor of 1.2 is appropriate. The speed of the shaft is to be 300 rev/min, the life to be 30 kh with a reliability of 0.99. What is the C10 catalog entry to be sought (or exceeded) when searching for a deep-groove bearing in a manufacturer's catalog on the basis of 106 revolutions for rating life? The Weibull parameters are x0 = 0.02, (θ − x0) = 4.439, and b = 1.483. Solution xD =

LD 60ℒD nD 60(30 000)300 = = = 540 LR L 10 106

Thus, the design life is 540 times the L10 life. For a ball bearing, a = 3. Then, from Equation (11–10), Answer

1∕3 540 C10 = (1.2)(413) [ = 6696 lbf 1∕1.483 ] 0.02 + 4.439(1 − 0.99)

Shafts generally have two bearings. Often these bearings are different. If the bearing reliability of the shaft with its pair of bearings is to be R, then R is related to the individual bearing reliabilities RA and RB, using Equation (1–9), as R = RARB First, we observe that if the product RARB equals R, then, in general, RA and RB are both greater than R. Since the failure of either or both of the bearings results in the shutdown of the shaft, then A or B or both can create a failure. Second, in sizing bearings one can begin by making RA and RB equal to the square root of the reliability goal, √R. In Example 11–3, if the bearing was one of a pair, the reliability goal would be √0.99, or 0.995. The bearings selected are discrete in their reliability property in your problem, so the selection procedure "rounds up," and the overall reliability exceeds the goal R. Third, it may be possible, if RA > √R, to round down on B yet have the product RARB still exceed the goal R.

11–6  Combined Radial and Thrust Loading A ball bearing is capable of resisting radial loading and a thrust loading. Furthermore, these can be combined. Consider Fa and Fr to be the axial thrust and radial loads, respectively, and Fe to be the equivalent radial load that does the same damage as the combined radial and thrust loads together. A rotation factor V is defined such that V = 1 when the inner ring rotates and V = 1.2 when the outer ring rotates. Two dimensionless groups can now be formed: Fe∕(VFr) and Fa∕(VFr). When these two dimensionless groups are plotted as in Figure 11–6, the data fall in a gentle curve that is well approximated by two straight-line segments. The abscissa e is defined by the intersection of the two lines. The equations for the two lines shown in Figure 11–6 are

Fe =1 VFr

Fa Fe =X+Y VFr VFr

when

Fa ≤e VFr

(11–11a)

when

Fa > e VFr

(11–11b)

586      Mechanical Engineering Design

Table 11–1  Equivalent Radial Load Factors for Ball Bearings

Fe VFr

Fa∕(VFr) ≤ e

1 Slope Y X

Fa VFr 0

e

e

X1

Y1

X2

Y2

0.014*

0.19

1.00

0

0.56

2.30

0.021

0.21

1.00

0

0.56

2.15

0.028

0.22

1.00

0

0.56

1.99

0.042

0.24

1.00

0

0.56

1.85

0.056

0.26

1.00

0

0.56

1.71

0.070

0.27

1.00

0

0.56

1.63

0.084

0.28

1.00

0

0.56

1.55

0.110

0.30

1.00

0

0.56

1.45

0.17

0.34

1.00

0

0.56

1.31

0.28

0.38

1.00

0

0.56

1.15

0.42

0.42

1.00

0

0.56

1.04

0.56

0.44

1.00

0

0.56

1.00

*Use 0.014 if Fa∕C0 < 0.014.

Figure 11–6 The relationship of dimensionless group Fe∕(VFr) and Fa∕(VFr) and the straight-line segments representing the data.

Fa∕(VFr) > e

Fa∕C0

where, as shown, X is the ordinate intercept and Y is the slope of the line for Fa∕(VFr) >  e. It is common to express Equations (11–11a) and (11–11b) as a single equation,

Fe = Xi VFr + Yi Fa

(11–12)

where i = 1 when Fa∕(VFr) ≤ e and i = 2 when Fa∕(VFr) > e. The X and Y factors depend upon the geometry and construction of the specific bearing. Table 11–1 lists representative values of X1, Y1, X2, and Y2 as a function of e, which in turn is a function of Fa∕C0, where C0 is the basic static load rating. The basic static load rating is the load that will produce a total permanent deformation in the raceway and rolling element at any contact point of 0.0001 times the diameter of the rolling element. The basic static load rating is typically tabulated, along with the basic dynamic load rating C10, in bearing manufacturers' publications. See Table 11–2, for example. In these equations, the rotation factor V is intended to correct for the rotating-ring conditions. The factor of 1.2 for outer-ring rotation is simply an acknowledgment that the fatigue life is reduced under these conditions. Self-aligning bearings are an exception: they have V = 1 for rotation of either ring. Since straight or cylindrical roller bearings will take no axial load, or very little, the Y factor is always zero. The ABMA has established standard boundary dimensions for bearings, which define the bearing bore, the outside diameter (OD), the width, and the fillet sizes on the shaft and housing shoulders. The basic plan covers all ball and straight roller bearings in the metric sizes. The plan is quite flexible in that, for a given bore, there is an assortment of widths and outside diameters. Furthermore, the outside diameters selected are such that, for a particular outside diameter, one can usually find a variety of bearings having different bores and widths.

Rolling-Contact Bearings     587

Table 11–2  Dimensions and Load Ratings for Single-Row 02-Series Deep-Groove and Angular-Contact Ball Bearings Bore, mm

OD, mm

Width, mm

Fillet Radius, mm

Shoulder

Load Ratings, kN

Diameter, mm dS

Deep Groove

dH

C0

C10

Angular Contact C10

C0

10 30 9 0.6 12.5 27 5.07 2.24 4.94 2.12 12 32 10 0.6 14.5 28 6.89 3.10 7.02 3.05 15 35 11 0.6 17.5 31 7.80 3.55 8.06 3.65 17 40 12 0.6 19.5 34 9.56 4.50 9.95 4.75 20 47 14 1.0 25 41 12.7 6.20 13.3 6.55 25 52 15 1.0 30 47 14.0 6.95 14.8 7.65 30 62 16 1.0 35 55 19.5 10.0 20.3 11.0 35 72 17 1.0 41 65 25.5 13.7 27.0 15.0 40 80 18 1.0 46 72 30.7 16.6 31.9 18.6 45 85 19 1.0 52 77 33.2 18.6 35.8 21.2 50 90 20 1.0 56 82 35.1 19.6 37.7 22.8 55 100 21 1.5 63 90 43.6 25.0 46.2 28.5 60 110 22 1.5 70 99 47.5 28.0 55.9 35.5 65 120 23 1.5 74 109 55.9 34.0 63.7 41.5 70 125 24 1.5 79 114 61.8 37.5 68.9 45.5 75 130 25 1.5 86 119 66.3 40.5 71.5 49.0 80 140 26 2.0 93 127 70.2 45.0 80.6 55.0 85 150 28 2.0 99 136 83.2 53.0 90.4 63.0 90 160 30 2.0 104 146 95.6 62.0 106

73.5

95 170 32 2.0 110 156 108 69.5 121 85.0

0

1

2

r

33

32

31

23

22

20

13

10 12

Dimension series

r

00 02 03 04

Diameter 3 series 2 1 0

Figure 11–7

3

4

30

Width series

OD

Bore

This basic ABMA plan is illustrated in Figure 11–7. The bearings are identified by a two-digit number called the dimension-series code. The first number in the code is from the width series, 0, 1, 2, 3, 4, 5, and 6. The second number is from the diameter series (outside), 8, 9, 0, 1, 2, 3, and 4. Figure 11–7 shows the variety of bearings that may be obtained with a particular bore. Since the

The basic ABMA plan for boundary dimensions. These apply to ball bearings, straight roller bearings, and spherical roller bearings, but not to inchseries ball bearings or tapered roller bearings. The contour of the corner is not specified. It may be rounded or chamfered, but it must be small enough to clear the fillet radius specified in the standards.

588      Mechanical Engineering Design

dS

dH

Figure 11–8 Shaft and housing shoulder diameters dS and dH should be adequate to ensure good bearing support.

dimension-series code does not reveal the dimensions directly, it is necessary to resort to tabulations. The 02 series is used here as an example of what is available. See Table 11–2. The housing and shaft shoulder diameters listed in the tables should be used whenever possible to secure adequate support for the bearing and to resist the maximum thrust loads (Figure 11–8). Table 11–3 lists the dimensions and load ratings of some straight roller bearings. To assist the designer in the selection of bearings, most of the manufacturers' handbooks contain data on bearing life for many classes of machinery, as well as information on load-application factors. Such information has been accumulated the hard way, that is, by experience, and the beginner designer should utilize this information until he or she gains enough experience to know when deviations are possible. Table 11–4 contains recommendations on bearing life for some classes of machinery. The load-application factors in Table 11–5 serve the same purpose as factors of safety; use them to increase the equivalent load before selecting a bearing.

Table 11–3  Dimensions and Basic Load Ratings for Cylindrical Roller Bearings 02-Series Bore, mm

OD, mm

Width, mm 15

03-Series

Load Rating, kN C10

Load Rating, kN OD, Width, C0 mm mm C10 C0

16.8

8.8

25

52

62

17

28.6

15.0

30

62 16

22.4 12.0 72 19

36.9 20.0

35

72 17

31.9 17.6 80 21

44.6 27.1

40

80 18

41.8 24.0 90 23

56.1 32.5

45

85 19

44.0 25.5 100 25

72.1 45.4

50

90 20

45.7 27.5 110 27

88.0 52.0

55 100 21 56.1 34.0 120 29 102 67.2 60 110 22 64.4 43.1 130 31 123 76.5 65 120 23 76.5 51.2 140 33 138 85.0 70

125

24

79.2

51.2

150

35

151

102

75

130

25

93.1

63.2

160

37

183

125

80 140 26 106 69.4 170 39 190 125 85 150 28 119 78.3 180 41 212 149 90 160 30 142 100 190 43 242 160 95 170 32 165 112 200 45 264 189 100 180 34 183 125 215 47 303 220 110 200 38 229 167 240 50 391 304 120 215 40 260 183 260 55 457 340 130 230 40 270 193 280 58 539 408 140 250 42 319 240 300 62 682 454 150 270 45 446 260 320 65 781 502

Rolling-Contact Bearings     589

Table 11–4  Bearing-Life Recommendations for Various Classes of Machinery Type of Application

Life, kh

Instruments and apparatus for infrequent use

Up to 0.5

Aircraft engines

0.5–2

Machines for short or intermittent operation where service interruption is of minor importance

4–8

Machines for intermittent service where reliable operation is of great importance

8–14

Machines for 8-h service that are not always fully utilized

14–20

Machines for 8-h service that are fully utilized

20–30

Machines for continuous 24-h service

50–60

Machines for continuous 24-h service where reliability is of extreme importance

100–200

Table 11–5  Load-Application Factors Type of Application

Load Factor

Precision gearing

1.0–1.1

Commercial gearing

1.1–1.3

Applications with poor bearing seals

1.2

Machinery with no impact

1.0–1.2

Machinery with light impact

1.2–1.5

Machinery with moderate impact

1.5–3.0

EXAMPLE 11–4 An SKF 6210 angular-contact ball bearing has an axial load Fa of 400 lbf and a radial load Fr of 500 lbf applied with the outer ring stationary. The basic static load rating C0 is 4450 lbf and the basic load rating C10 is 7900 lbf. Estimate the ℒ10 life at a speed of 720 rev/min. Solution V = 1 and Fa∕C0 = 400∕4450 = 0.090. Interpolate for e in Table 11–1: Fa∕C0

e

0.084 0.28 0.090

e

from which e = 0.285

0.110 0.30

Fa∕(VFr) = 400∕[(1)500] = 0.8 > 0.285. Thus, interpolate for Y2: Fa∕C0

Y2

0.084 1.55 0.090

Y 2

0.110 1.45

from which Y2 = 1.527

590      Mechanical Engineering Design

From Equation (11–12), Fe = X2VFr + Y2Fa = 0.56(1)500 + 1.527(400) = 890.8 lbf With ℒD = ℒ10 and FD = Fe, solving Equation (11–3) for ℒ10 gives ℒ10 =

Answer

60ℒR nR C10 a 106 7900 3 = = 16 150 h 60nD ( Fe ) 60(720) ( 890.8 )

We now know how to combine a steady radial load and a steady thrust load into an equivalent steady radial load Fe that inflicts the same damage per revolution as the radial-thrust combination.

11–7  Variable Loading Bearing loads are frequently variable and occur in some identifiable patterns: ∙ Piecewise constant loading in a cyclic pattern ∙ Continuously variable loading in a repeatable cyclic pattern ∙ Random variation Equation (11–1) can be written as F aL = constant = K (a) Note that F may already be an equivalent steady radial load for a radial–thrust load combination. Figure 11–9 is a plot of F a as ordinate and L as abscissa for Equation (a). If a load level of F1 is selected and run to the failure criterion, then the area under the F 1a-L1 trace is numerically equal to K. The same is true for a load level F2; that is, the area under the F a2-L2 trace is numerically equal to K. The linear damage theory says that in the case of load level F1, the area from L = 0 to L = LA does damage measured by F 1aLA = D. Consider the piecewise continuous cycle depicted in Figure 11–10. The loads Fei are equivalent steady radial loads for combined radial–thrust loads. The damage done by loads Fe1, Fe2, and Fe3 is a a a D = Fe1 l1 + Fe2 l2 + Fe3 l3 (b) Fa

Fa

Fe2a

F 1a

A

F aeq

B Fe1a

Fe3a

F 2a

l1 0

L1

L2

L

l2

l3 l

Figure 11–9

Figure 11–10

Plot of F a as ordinate and L as abscissa for F aL = constant. The linear damage hypothesis says that in the case of load F1, the area under the curve from L = 0 to L = L A is a measure of the damage D = F 1aL A. The complete damage to failure is measured by C10a LB.

A three-part piecewise-continuous periodic loading cycle involving loads Fe1, Fe2, and Fe3. Feq is the equivalent steady load inflicting the same damage when run for l1 + l2 + l3 revolutions, doing the same damage D per period.

Rolling-Contact Bearings     591

where li is the number of revolutions at life Li. The equivalent steady load Feq when run for l1 + l2 + l3 revolutions does the same damage D. Thus a D = Feq (l1 + l2 + l3 )

(c)

Equating Equations (b) and (c), and solving for Feq, we get Feq = [

a a a 1∕a F e1 l1 + F e2 l2 + F e3 l3 1∕a a ∑ = f F ei i ] [ ] l1 + l2 + l3

(11–13)

where fi is the fraction of revolution run up under load Fei. Since li can be expressed as niti, where ni is the rotational speed at load Fei and ti is the duration of that speed, then it follows that

∑ ni ti Feia 1∕a ⌉ Feq = ⌈ ⌊ ∑ ni ti ⌋

(11–14)

The character of the individual loads can change, so an application factor (af) can be prefixed to each Fei as (afiFei)a; then Equation (11–13) can be written Feq = [ ∑ fi (afi Fei ) a ]

1∕a

L eq =

K a F eq

(11–15)

EXAMPLE 11–5 A ball bearing is run at four piecewise continuous steady loads as shown in the following table. Columns (1), (2), and (5) to (8) are given. (1)

(2)

Time Speed, Fraction rev/min

(3)

(4)

(5) (6) (7) (8) (9)

Product, Turns Column Fraction, Fri, Fai, Fei, afi Fei, (1) × (2) (3)∕Σ(3) lbf lbf lbf afi lbf

0.1

2000

200

0.077

600 300 794 1.10 873

0.1

3000

300

0.115

300 300 626 1.25 795

0.3

3000

900

0.346

750 300 878 1.10 966

0.5

2400

1200

0.462

375 300 668 1.25 835

2600

1.000

Columns 1 and 2 are multiplied to obtain column 3. The column 3 entry is divided by the sum of column 3, 2600, to give column 4. Columns 5, 6, and 7 are the radial, axial, and equivalent loads, respectively. Column 8 is the appropriate application factor. Column 9 is the product of columns 7 and 8. Solution From Equation (11–13), with a = 3, the equivalent radial load Fe is Answer

Fe = [0.077(873) 3 + 0.115(795) 3 + 0.346(966) 3 + 0.462(835) 3]1∕3 = 884 lbf

592      Mechanical Engineering Design

Sometimes the question after several levels of loading is: How much life is left if the next level of stress is held until failure? Failure occurs under the linear damage hypothesis when the damage D equals the constant K = F aL. Taking the first form of Equation (11–13), we write a a a a Feq Leq = Fe1 l1 + Fe2 l2 + Fe3 l3

and note that a a a K = Fe1 L1 = Fe2 L2 = Fe3 L3

and K also equals a a K = Fe1a l1 + Fe2 l2 + Fe3 l3 =

li K K K l1 + l2 + l3 = K ∑ L1 L2 L3 Li

From the outer parts of the preceding equation we obtain

li = 1 Li

(11–16)

This equation was advanced by Palmgren in 1924, and again by Miner in 1945. See Equation (6–69). The second kind of load variation mentioned is continuous, periodic variation, depicted by Figure 11–11. The differential damage done by F a during rotation through the angle dθ is dD = F a dθ An example of this would be a cam whose bearings rotate with the cam through the angle dθ. The total damage during a complete cam rotation is given by ϕ

D = dD =

∫ F dθ = F ϕ a

0

a eq

from which, solving for the equivalent load, we obtain 1 Feq = [ ϕ

ϕ

0

F a dθ ]

1∕a

L eq =

K a Feq

(11–17)

The value of ϕ is often 2π, although other values occur. Numerical integration is often useful to carry out the indicated integration, particularly when a is not an integer and trigonometric functions are involved. We have now learned how to find the steady equivalent load that does the same damage as a continuously varying cyclic load. Figure 11–11

Fa

A continuous load variation of a cyclic nature whose period is ϕ. Fa

dθ θ 0

ϕ

Rolling-Contact Bearings     593

EXAMPLE 11–6 The operation of a particular rotary pump involves a power demand of P = P + A′ sin θ where P is the average power. The bearings feel the same variation as F = F + A sin θ. Develop an application factor af for this application of ball bearings. Solution From Equation (11–17), with a = 3, 1 Feq = ( 2π =[

0

1 2π (

+ A3

F a dθ)

1∕a

3

2

0

(F + A sin θ) 3 dθ)

0

1∕3

2

sin θ dθ + 3FA

0

2

sin θ dθ

sin3 θ dθ)]

0

1 2π

F dθ + 3F A

0

=(

1∕3

1∕3 1 3 A 2 1∕3 Feq = [ (2πF 3 + 0 + 3πFA2 + 0) ] = F [ 1 + ( ) ] 2π 2 F

In terms of F , the application factor is 3 A 2 1∕3 af = [1 + ( ) ] 2 F

Answer We can present the result in tabular form: A∕F

af

0

1

0.2

1.02

0.4

1.07

0.6

1.15

0.8

1.25

1.0

1.36

11–8  Selection of Ball and Cylindrical Roller Bearings We have enough information concerning the loading of rolling-contact ball and roller bearings to develop the steady equivalent radial load that will do as much damage to the bearing as the existing loading. Now let's put it to work. EXAMPLE 11–7 The second shaft on a parallel-shaft 25-hp foundry crane speed reducer contains a helical gear with a pitch diameter of 8.08 in. Helical gears transmit components of force in the tangential, radial, and axial directions (see Chapter 13). The components of the gear force transmitted to the second shaft are shown in Figure 11–12, at point A. The bearing reactions at C and D, assuming simple-supports, are also shown.

594      Mechanical Engineering Design

Figure 11–12

356.6

Forces in pounds applied to the second shaft of the helical gear speed reducer of Example 11–7.

C

x

344 297

y

n 3i

.5

B

n

3i D

297 .

z

5

4.04 in

106.6

595

34

4

A 250

A ball bearing is to be selected for location C to accept the thrust, and a cylindrical roller bearing is to be utilized at location D. The life goal of the speed reducer is 10 kh, with a reliability factor for the ensemble of all four bearings (both shafts) to equal or exceed 0.96 for the Weibull parameters of Example 11–3. The application factor is to be 1.2. (a) Select the roller bearing for location D. (b) Select the ball bearing (angular contact) for location C, assuming the inner ring rotates. Solution The torque transmitted is T = 595(4.04) = 2404 lbf · in. The speed at the rated horsepower, given by Equation (3–42), is

nD =

63 025H 63 025(25) = = 655.4 rev/min T 2404

The radial load at D is √106.62 + 297.52 = 316.0 lbf, and the radial load at C is √356.62 + 297.52 = 464.4 4 lbf. The individual bearing reliabilities, if equal, must be at least √ 0.96 = 0.98985 ≈ 0.99. The dimensionless design life for both bearings is

xD =

LD 60ℒD nD 60(10 000)655.4 = = = 393.2 L10 L10 106

(a) From Equation (11–10), the Weibull parameters of Example 11–3, an application factor of 1.2, and a  = 10∕3 for the roller bearing at D, the catalog rating should be equal to or greater than C10 = af FD [

x0 + (θ − x0 )(1 − RD ) 1∕b ] xD

1∕a

3∕10 393.2 = 1.2(316.0) [ = 3591 lbf = 16.0 kN 0.02 + 4.439(1 − 0.99) 1∕1.483 ]

Rolling-Contact Bearings     595

Answer The absence of a thrust component makes the selection procedure simple. Choose a 02-25 mm series, or a 03-25 mm series cylindrical roller bearing from Table 11–3. (b) The ball bearing at C involves a thrust component. This selection procedure requires an iterative procedure. Assuming Fa∕(VFr) > e, 1 2 3 4 5 6 7

Choose Y2 from Table 11–1. Find C10. Tentatively identify a suitable bearing from Table 11–2, note C0. Using Fa∕C0 enter Table 11–1 to obtain a new value of Y2. Find C10. If the same bearing is obtained, stop. If not, take next bearing and go to step 4.

As a first approximation, take the middle entry from Table 11–1:

X2 = 0.56

Y2 = 1.63.

From Equation (11–12), with V = 1,

Fe = XVFr + YFa = 0.56(1)(464.4) + 1.63(344) = 821 lbf = 3.65 kN

From Equation (11–10), with a = 3,

1∕3 393.2 C10 = 1.2(3.65) [ = 53.2 kN 0.02 + 4.439(1 − 0.99) 1∕1.483 ]

From Table 11–2, angular-contact bearing 02-60 mm has C10 = 55.9 kN. C0 is 35.5 kN. Step 4 becomes, with Fa in kN, Fa 344(4.45)10−3 = = 0.0431 C0 35.5 which makes e from Table 11–1 approximately 0.24. Now Fa∕(VFr) = 344∕[(1) 464.4] = 0.74, which is greater than 0.24, so we find Y2 by interpolation: Fa∕C0

Y2

0.042

1.85

0.043

Y2

0.056

1.71

from which Y2 = 1.84

From Equation (11–12),

Fe = 0.56(1) (464.4) + 1.84(344) = 893 lbf = 3.97 kN

The prior calculation for C10 changes only in Fe, so

C10 =

3.97 53.2 = 57.9 kN 3.65

From Table 11–2 an angular contact bearing 02-65 mm has C10 = 63.7 kN and C0 of 41.5 kN. Again,

Fa 344(4.45)10−3 = = 0.0369 C0 41.5

596      Mechanical Engineering Design

making e approximately 0.23. Now from before, Fa∕(VFr) = 0.74, which is greater than 0.23. We find Y2 again by interpolation: Fa∕C0

Y2

0.028

1.99

0.0369

Y2

0.042

1.85

from which Y2 = 1.90

From Equation (11–12),

Fe = 0.56(1)(464.4) + 1.90(344) = 914 lbf = 4.07 kN

The prior calculation for C10 changes only in Fe, so

C10 =

4.07 53.2 = 59.3 kN 3.65

Answer From Table 11–2 an angular-contact 02-65 mm is still selected, so the iteration is complete.

11–9  Selection of Tapered Roller Bearings Tapered roller bearings have a number of features that make them complicated. As we address the differences between tapered roller and ball and cylindrical roller bearings, note that the underlying fundamentals are the same, but that there are differences in detail. Moreover, bearing and cup combinations are not necessarily priced in proportion to capacity. Any catalog displays a mix of high-production, low-production, and successful special-order designs. Bearing suppliers have computer programs that will take your problem descriptions, give intermediate design assessment information, and list a number of satisfactory cup-and-cone combinations in order of decreasing cost. Company sales offices provide access to comprehensive engineering services to help designers select and apply their bearings. At a large original equipment manufacturer's plant, there may be a resident bearing company representative. Bearing suppliers provide a wealth of engineering information and detail in their catalogs and engineering guides, both in print and online. It is strongly recommended that the designer become familiar with the specifics of the supplier. It will usually utilize a similar approach as presented here, but may include various modifying factors for such things as temperature and lubrication. Many of the suppliers will provide online software tools to aid in bearing selection. The engineer will always benefit from a general understanding of the theory utilized in such software tools. Our goal here is to introduce the vocabulary, show congruence to fundamentals that were learned earlier, offer examples, and develop confidence. Finally, problems should reinforce the learning experience. The four components of a tapered roller bearing assembly are the 1 Cone (inner ring) 2 Cup (outer ring)

3 Tapered rollers 4 Cage (spacer-retainer)

The assembled bearing consists of two separable parts: (1) the cone assembly: the cone, the rollers, and the cage; and (2) the cup. Bearings can be made as single-row, two-row, four-row, and thrust-bearing assemblies. Additionally, auxiliary components such as spacers and closures can be used. Figure 11–13 shows the nomenclature of

Rolling-Contact Bearings     597 Bearing width T Cup front face radius

Cup length C

Cup front face

Cup back face radius r

Direct mounting ag

Cup back face

Cone back face rib

ae

G a Cone front face radius

Cup outside diameter (OD) D

Cone bore d

(a)

Cone front face rib Cone length B

Cone back face radius R

Bc

Ac

Cage Cone back face

Bo

Ao

Bearing A

Bearing B 90° (b)

Cone front face

Cone Roller

Ac Cup

Bc

ag a

Standout F

ae

a

Indirect mounting

Figure 11–13

Figure 11–14

Nomenclature of a tapered roller bearing. Point G is the location of the effective load center; use this point to estimate the radial bearing load. (Source: Redrawn from material Furnished by The Timken Company.)

Comparison of mounting stability between indirect and direct mountings. (Source: Redrawn from material Furnished by The Timken Company.)

a tapered roller bearing, and the point G through which radial and axial components of load act. A tapered roller bearing can carry both radial and thrust (axial) loads, or any combination of the two. However, even when an external thrust load is not present, the radial load will induce a thrust reaction within the bearing because of the taper. To avoid the separation of the races and the rollers, this thrust must be resisted by an equal and opposite force. One way of generating this force is to always use at least two tapered roller bearings on a shaft. Two bearings can be mounted with the cone backs facing each other, in a configuration called direct mounting, or with the cone fronts facing each other, in what is called indirect mounting. Figure 11–14 shows a pair of tapered roller bearings mounted directly (a) and indirectly (b) with the bearing reaction locations A0 and B0 shown for the shaft. For the shaft as a beam, the span is ae, the effective spread. It is through points A0 and B0 that the radial loads act perpendicular to the shaft axis, and the thrust loads act along the shaft axis. The geometric spread ag for the direct mounting is greater than for the indirect mounting. With indirect mounting the bearings are closer together compared to the direct mounting; however, the system stability is the same (ae is the same in both cases). Thus direct and indirect mounting involve space and compactness needed or desired, but with the same system stability. In addition to the usual ratings and geometry information, catalog data for tapered roller bearings will include the location of the effective force center. Two sample pages from a Timken catalog are shown in Figure 11–15.

598      Mechanical Engineering Design

SINGLE-ROW STRAIGHT BORE D Db da

T C

r a

B

R

d db Da

cone rating at 500 rpm for 3000 hours L 10 onerow thrust radial

part numbers

cup backing shoulder diameters

backing shoulder diameters

max shaft fillet radius

width

R1

B

db

da

r1

C

Db

Da

30205

1.0 0.04

15.000 0.5906

30.5 1.20

29.0 1.14

1.0 0.04

13.000 0.5118

46.0 1.81

48.5 1.91

32205-B

32205-B

1.0 0.04

18.000 0.7087

34.0 1.34

31.0 1.22

1.0 0.04

15.000 0.5906

43.5 1.71

49.5 1.95

–7.6 –0.30

33205

33205

1.0 0.04

22.000 0.8661

34.0 1.34

30.5 1.20

1.0 0.04

18.000 0.7087

44.5 1.75

49.0 1.93

1.95

–5.1 –0.20

30305

30305

1.5 0.06

17.000 0.6693

32.5 1.28

30.0 1.18

1.5 0.06

15.000 0.5906

55.0 2.17

57.0 2.24

8930 2010

1.95

–9.7 –0.38

32305

32305

1.5 0.06

24.000 0.9449

35.0 1.38

31.5 1.24

1.5 0.06

20.000 0.7874

54.0 2.13

57.0 2.24

6990 1570

4810 1080

1.45

–2.8 –0.11

07096

07196

1.5 0.06

14.260 0.5614

31.5 1.24

29.5 1.16

1.0 0.04

9.525 0.3750

44.5 1.75

47.0 1.85

13.495 0.5313

6990 1570

4810 1080

1.45

–2.8 –0.11

07100

07196

1.0 0.04

14.260 0.5614

30.5 1.20

29.5 1.16

1.0 0.04

9.525 0.3750

44.5 1.75

47.0 1.85

50.005 1.9687

13.495 0.5313

6990 1570

4810 1080

1.45

–2.8 –0.11

07100-S

07196

1.5 0.06

14.260 0.5614

31.5 1.24

29.5 1.16

1.0 0.04

9.525 0.3750

44.5 1.75

47.0 1.85

25.400 1.0000

50.292 1.9800

14.224 0.5600

7210 1620

4620 1040

1.56

–3.3 –0.13

L44642

L44610

14.732 0.5800

36.0 1.42

29.5 1.16

1.3 0.05

10.668 0.4200

44.5 1.75

47.0 1.85

25.400 1.0000

50.292 1.9800

14.224 0.5600

7210 1620

4620 1040

1.56

–3.3 –0.13

L44643

. L44610

3.5 0.14 1.3 0.05

14.732 0.5800

31.5 1.24

29.5 1.16

1.3 0.05

10.668 0.4200

44.5 1.75

47.0 1.85

25.400 1.0000

51.994 2.0470

15.011 0.5910

6990 1570

4810 1080

1.45

–2.8 –0.11

07100

07204

1.0 0.04

14.260 0.5614

30.5 1.20

29.5 1.16

1.3 0.05

12.700 0.5000

45.0 1.77

48.0 1.89

25.400 1.0000

56.896 2.2400

19.368 0.7625

10900 2450

5740 1290

1.90

–6.9 –0.27

1780

1729

0.8 0.03

19.837 0.7810

30.5 1.20

30.0 1.18

1.3 0.05

15.875 0.6250

49.0 1.93

51.0 2.01

25.400 1.0000

57.150 2.2500

19.431 0.7650

11700 2620

10900 2450

1.07

–3.0 –0.12

M84548

M84510

1.5 0.06

19.431 0.7650

36.0 1.42

33.0 1.30

1.5 0.06

14.732 0.5800

48.5 1.91

54.0 2.13

25.400 1.0000

58.738 2.3125

19.050 0.7500

11600 2610

6560 1470

1.77

–5.8 –0.23

1986

1932

1.3 0.05

19.355 0.7620

32.5 1.28

30.5 1.20

1.3 0.05

15.080 0.5937

52.0 2.05

54.0 2.13

25.400 1.0000

59.530 2.3437

23.368 0.9200

13900 3140

13000 2930

1.07

–5.1 –0.20

M84249

M84210

0.8 0.03

23.114 0.9100

36.0 1.42

32.5 1.27

1.5 0.06

18.288 0.7200

49.5 1.95

56.0 2.20

25.400 1.0000

60.325 2.3750

19.842 0.7812

11000 2480

6550 1470

1.69

–5.1 –0.20

15578

15523

1.3 0.05

17.462 0.6875

32.5 1.28

30.5 1.20

1.5 0.06

15.875 0.6250

51.0 2.01

54.0 2.13

25.400 1.0000

61.912 2.4375

19.050 0.7500

12100 2730

7280 1640

1.67

–5.8 –0.23

15101

15243

0.8 0.03

20.638 0.8125

32.5 1.28

31.5 1.24

2.0 0.08

14.288 0.5625

54.0 2.13

58.0 2.28

25.400 1.0000

62.000 2.4409

19.050 0.7500

12100 2730

7280 1640

1.67

–5.8 –0.23

15100

15245

3.5 0.14

20.638 0.8125

38.0 1.50

31.5 1.24

1.3 0.05

14.288 0.5625

55.0 2.17

58.0 2.28

25.400 1.0000

62.000 2.4409

19.050 0.7500

12100 2730

7280 1640

1.67

–5.8 –0.23

15101

15245

0.8 0.03

20.638 0.8125

32.5 1.28

31.5 1.24

1.3 0.05

14.288 0.5625

55.0 2.17

58.0 2.28

bore

outside diameter

width

d

D

T

N lbf

25.000 0.9843

52.000 2.0472

16.250 0.6398

25.000 0.9843

52.000 2.0472

25.000 0.9843

factor

eff. load center

N lbf

K

a2

8190 1840

5260 1180

1.56

–3.6 –0.14

30205

19.250 0.7579

9520 2140

9510 2140

1.00

–3.0 –0.12

52.000 2.0472

22.000 0.8661

13200 2980

7960 1790

1.66

25.000 0.9843

62.000 2.4409

18.250 0.7185

13000 2930

6680 1500

25.000 0.9843

62.000 2.4409

25.250 0.9941

17400 3910

25.159 0.9905

50.005 1.9687

13.495 0.5313

25.400 1.0000

50.005 1.9687

25.400 1.0000

cone

cup

max houswidth ing fillet radius

Figure 11–15  (Continued on next page) Catalog entry of single-row straight-bore Timken roller bearings, in part. (Courtesy of The Timken Company.)

Rolling-Contact Bearings     599

SINGLE-ROW STRAIGHT BORE cone rating at 500 rpm for 3000 hours L 10 onerow thrust radial

part numbers

cup backing shoulder diameters

backing shoulder diameters

max shaft fillet radius

width

R1

B

db

da

r1

C

Db

Da

15245

1.5 0.06

20.638 0.8125

34.0 1.34

31.5 1.24

1.3 0.05

14.288 0.5625

55.0 2.17

58.0 2.28

15101

15244

0.8 0.03

20.638 0.8125

32.5 1.28

31.5 1.24

1.3 0.05

15.875 0.6250

55.0 2.17

58.0 2.28

–5.8 –0.23

15101

15250

0.8 0.03

20.638 0.8125

32.5 1.28

31.5 1.24

1.3 0.05

15.875 0.6250

56.0 2.20

59.0 2.32

1.67

–5.8 –0.23

15101

15250X

0.8 0.03

20.638 0.8125

32.5 1.28

31.5 1.24

1.5 0.06

15.875 0.6250

55.0 2.17

59.0 2.32

13500 3040

1.07

–3.3 –0.13

M86643

M86610

1.5 0.06

21.433 0.8438

38.0 1.50

36.5 1.44

1.5 0.06

16.670 0.6563

54.0 2.13

61.0 2.40

13100 2950

16400 3690

0.80

–2.3 –0.09

23100

23256

1.5 0.06

21.463 0.8450

39.0 1.54

34.5 1.36

1.5 0.06

15.875 0.6250

53.0 2.09

63.0 2.48

23.812 0.9375

18400 4140

8000 1800

2.30

–9.4 –0.37

2687

2631

1.3 0.05

25.433 1.0013

33.5 1.32

31.5 1.24

1.3 0.05

19.050 0.7500

58.0 2.28

60.0 2.36

68.262 2.6875

22.225 0.8750

15300 3440

10900 2450

1.40

–5.1 –0.20

02473

02420

0.8 0.03

22.225 0.8750

34.5 1.36

33.5 1.32

1.5 0.06

17.462 0.6875

59.0 2.32

63.0 2.48

25.400 1.0000

72.233 2.8438

25.400 1.0000

18400 4140

17200 3870

1.07

–4.6 –0.18

HM88630

HM88610

0.8 0.03

25.400 1.0000

39.5 1.56

39.5 1.56

2.3 0.09

19.842 0.7812

60.0 2.36

69.0 2.72

25.400 1.0000

72.626 2.8593

30.162 1.1875

22700 5110

13000 2910

1.76

–10.2 –0.40

3189

3120

0.8 0.03

29.997 1.1810

35.5 1.40

35.0 1.38

3.3 0.13

23.812 0.9375

61.0 2.40

67.0 2.64

26.157 1.0298

62.000 2.4409

19.050 0.7500

12100 2730

7280 1640

1.67

–5.8 –0.23

15103

15245

0.8 0.03

20.638 0.8125

33.0 1.30

32.5 1.28

1.3 0.05

14.288 0.5625

55.0 2.17

58.0 2.28

26.162 1.0300

63.100 2.4843

23.812 0.9375

18400 4140

8000 1800

2.30

–9.4 –0.37

2682

2630

1.5 0.06

25.433 1.0013

34.5 1.36

32.0 1.26

0.8 0.03

19.050 0.7500

57.0 2.24

59.0 2.32

26.162 1.0300

66.421 2.6150

23.812 0.9375

18400 4140

8000 1800

2.30

–9.4 –0.37

2682

2631

1.5 0.06

25.433 1.0013

34.5 1.36

32.0 1.26

1.3 0.05

19.050 0.7500

58.0 2.28

60.0 2.36

26.975 1.0620

58.738 2.3125

19.050 0.7500

11600 2610

6560 1470

1.77

–5.8 –0.23

1987

1932

0.8 0.03

19.355 0.7620

32.5 1.28

31.5 1.24

1.3 0.05

15.080 0.5937

52.0 2.05

54.0 2.13

† 26.988 † 1.0625

50.292 1.9800

14.224 0.5600

7210 1620

4620 1040

1.56

–3.3 –0.13

L44649

L44610

3.5 0.14

14.732 0.5800

37.5 1.48

31.0 1.22

1.3 0.05

10.668 0.4200

44.5 1.75

47.0 1.85

† 26.988 † 1.0625

60.325 2.3750

19.842 0.7812

11000 2480

6550 1470

1.69

–5.1 –0.20

15580

15523

3.5 0.14

17.462 0.6875

38.5 1.52

32.0 1.26

1.5 0.06

15.875 0.6250

51.0 2.01

54.0 2.13

† 26.988 † 1.0625

62.000 2.4409

19.050 0.7500

12100 2730

7280 1640

1.67

–5.8 –0.23

15106

15245

0.8 0.03

20.638 0.8125

33.5 1.32

33.0 1.30

1.3 0.05

14.288 0.5625

55.0 2.17

58.0 2.28

† 26.988 † 1.0625

66.421 2.6150

23.812 0.9375

18400 4140

8000 1800

2.30

–9.4 –0.37

2688

2631

1.5 0.06

25.433 1.0013

35.0 1.38

33.0 1.30

1.3 0.05

19.050 0.7500

58.0 2.28

60.0 2.36

28.575 1.1250

56.896 2.2400

19.845 0.7813

11600 2610

6560 1470

1.77

–5.8 –0.23

1985

1930

0.8 0.03

19.355 0.7620

34.0 1.34

33.5 1.32

0.8 0.03

15.875 0.6250

51.0 2.01

54.0 2.11

28.575 1.1250

57.150 2.2500

17.462 0.6875

11000 2480

6550 1470

1.69

–5.1 –0.20

15590

15520

3.5 0.14

17.462 0.6875

39.5 1.56

33.5 1.32

1.5 0.06

13.495 0.5313

51.0 2.01

53.0 2.09

28.575 1.1250

58.738 2.3125

19.050 0.7500

11600 2610

6560 1470

1.77

–5.8 –0.23

1985

1932

0.8 0.03

19.355 0.7620

34.0 1.34

33.5 1.32

1.3 0.05

15.080 0.5937

52.0 2.05

54.0 2.13

28.575 1.1250

58.738 2.3125

19.050 0.7500

11600 2610

6560 1470

1.77

–5.8 –0.23

1988

1932

3.5 0.14

19.355 0.7620

39.5 1.56

33.5 1.32

1.3 0.05

15.080 0.5937

52.0 2.05

54.0 2.13

28.575 1.1250

60.325 2.3750

19.842 0.7812

11000 2480

6550 1470

1.69

–5.1 –0.20

15590

15523

3.5 0.14

17.462 0.6875

39.5 1.56

33.5 1.32

1.5 0.06

15.875 0.6250

51.0 2.01

54.0 2.13

28.575 1.1250

60.325 2.3750

19.845 0.7813

11600 2610

6560 1470

1.77

–5.8 –0.23

1985

1931

0.5 0.03

19.355 0.7620

34.0 1.34

33.5 1.32

1.3 0.05

15.875 0.6250

52.0 2.05

55.0 2.17

bore

outside diameter

width

d

D

T

N lbf

25.400 1.0000

62.000 2.4409

19.050 0.7500

25.400 1.0000

62.000 2.4409

25.400 1.0000

factor

eff. load center

N lbf

K

a2

12100 2730

7280 1640

1.67

–5.8 –0.23

15102

20.638 0.8125

12100 2730

7280 1640

1.67

–5.8 –0.23

63.500 2.5000

20.638 0.8125

12100 2730

7280 1640

1.67

25.400 1.0000

63.500 2.5000

20.638 0.8125

12100 2730

7280 1640

25.400 1.0000

64.292 2.5312

21.433 0.8438

14500 3250

25.400 1.0000

65.088 2.5625

22.225 0.8750

25.400 1.0000

66.421 2.6150

25.400 1.0000

cone

cup

These maximum fillet radii will be cleared by the bearing corners. Minus value indicates center is inside cone backface. † For standard class ONLY, the maximum metric size is a whole mm value. For "J" part tolerances—see metric tolerances, page 73, and fitting practice, page 65. ISO cone and cup combinations are designated with a common part number and should be purchased as an assembly. For ISO bearing tolerances—see metric tolerances, page 73, and fitting practice, page 65. 1 2

Figure 11–15  (Continued)

max houswidth ing fillet radius

600      Mechanical Engineering Design

A radial load on a tapered roller bearing will induce a thrust reaction. The load zone includes about half the rollers and subtends an angle of approximately 180°. Using the symbol Fi for the induced thrust load from a radial load with a 180° load zone, Timken provides the equation

Fi =

0.47Fr K

(11–18)

where the K factor is geometry-specific, and is the ratio of the radial load rating to the thrust load rating. The K factor can be first approximated with 1.5 for a radial bearing and 0.75 for a steep angle bearing in the preliminary selection process. After a possible bearing is identified, the exact value of K for each bearing can be found in the bearing catalog. A shaft supported by a pair of direct-mounted tapered roller bearings is shown in Figure 11–16. Force vectors are shown as applied to the shaft. FrA and FrB are the radial loads carried by the bearings, applied at the effective force centers GA and GB. The induced loads FiA and FiB due to the effect of the radial loads on the tapered bearings are also shown. Additionally, there may be an externally applied thrust load Fae on the shaft from some other source, such as the axial load on a helical gear. Since the bearings experience both radial and thrust loads, it is necessary to determine equivalent radial loads. Following the form of Equation (11–12), where Fe = XVFr + YFa, Timken recommends using X = 0.4 and V = 1 for all cases, and using the K factor for the specific bearing for Y. This gives an equation of the form

Fe = 0.4Fr + KFa

(a)

The axial load Fa is the net axial load carried by the bearing due to the combination of the induced axial load from the other bearing and the external axial load. However, only one of the bearings will carry the net axial load, and which one it is depends on the direction the bearings are mounted, the relative magnitudes of the induced loads, the direction of the external load, and whether the shaft or the housing is the moving part. Timken handles it with a table containing each of the configurations and a sign convention on the external loads. It further requires the application to be oriented horizontally with left and right bearings that must match the left and right sign conventions. Here, we will present a method that gives Figure 11–16 Direct-mounted tapered roller bearings, showing radial, induced thrust, and external thrust loads.

FrA

FiA GA

A

FrB

Fae

GB

FiB

B

Rolling-Contact Bearings     601

equivalent results, but that is perhaps more conducive to visualizing and understanding the logic behind it. First, determine visually which bearing is being "squeezed" by the external thrust load, and label it as bearing A. Label the other bearing as bearing B. For example, in Figure 11–16, the external thrust Fae causes the shaft to push to the left against the cone of the left bearing, squeezing it against the rollers and the cup. On the other hand, it tends to pull apart the cup from the right bearing. The left bearing is therefore labeled as bearing A. If the direction of Fae were reversed, then the right bearing would be labeled as bearing A. This approach to labeling the bearing being squeezed by the external thrust is applied similarly regardless of whether the bearings are mounted directly or indirectly, regardless of whether the shaft or the housing carries the external thrust, and regardless of the orientation of the assembly. To clarify by example, consider the vertical shaft and cylinder in Figure 11–17 with direct-mounted bearings. In Figure 11–17a, an external load is applied in the upward direction to a rotating shaft, compressing the top bearing, which should be labeled as bearing A. On the other hand, in Figure 11–17b, an upward external load is applied to a rotating outer cylinder with a stationary shaft. In this case, the lower bearing is being squeezed and should be labeled as bearing A. If there is no external thrust, then either bearing can arbitrarily be labeled as bearing A. Second, determine which bearing actually carries the net axial load. Generally, it would be expected that bearing A would carry the axial load, since the external thrust Fae is directed toward A, along with the induced thrust FiB from bearing B. However, if the induced thrust FiA from bearing A happens to be larger than the combination of the external thrust and the thrust induced by bearing B, then bearing B will carry the net thrust load. We will use Equation (a) for the bearing carrying the thrust load. Timken recommends leaving the other bearing at its original radial load, rather than reducing it due to the negative net thrust load. The results are presented in equation form below, where the induced thrusts are defined by Equation (11–18). Figure 11–17

Bearing A

Bearing B

Fae Fae

Bearing B

(a)

Bearing A

(b)

Examples of determining which bearing carries the external thrust load. In each case, the compressed bearing is labeled as bearing A. (a) External thrust applied to rotating shaft; (b) External thrust applied to rotating cylinder.

602      Mechanical Engineering Design

If

FiA ≤ (FiB + Fae )

FeA = 0.4FrA + KA (FiB + Fae ) {FeB = FrB

If

FiA > (FiB + Fae )

FeB = 0.4FrB + KB (FiA − Fae ) {FeA = FrA

(11–19a) (11–19b) (11–20a) (11–20b)

In any case, if the equivalent radial load is ever less than the original radial load, then the original radial load should be used. Once the equivalent radial loads are determined, they should be used to find the catalog rating load using any of Equations (11–3), (11–9), or (11–10) as before. Timken uses a Weibell model with x0 = 0, θ = 4.48, and b = 3∕2. Note that since KA and KB are dependent on the specific bearing chosen, it may be necessary to iterate the process. EXAMPLE 11–8 The shaft depicted in Figure 11–18a carries a helical gear with a tangential force of 3980 N, a radial force of 1770 N, and a thrust force of 1690 N at the pitch cylinder with directions shown. The pitch diameter of the gear is 200 mm. The shaft runs at a speed of 800 rev/min, and the span (effective spread) between the directmount bearings is 150 mm. The design life is to be 5000 h and an application factor of 1 is appropriate. If the reliability of the bearing set is to be 0.99, select suitable single-row tapered-roller Timken bearings. Solution The reactions in the xz plane from Figure 11–18b are

Rz A =

3980(50) = 1327 N 150

Rz B =

3980(100) = 2653 N 150

Figure 11–18

1770

1690

3980

Essential geometry of helical gear and shaft. Length dimensions in mm, loads in N, couple in N · mm. (a) Sketch (not to scale) showing thrust, radial, and tangential forces. (b) Forces in xz plane. (c) Forces in xy plane.

x

20

0

B

50 y 150

100

A z

(a) 3980 A

B RzB

Rz A

1770

y x

1690

A RyA

169 000

z (b)

(c)

B RyB

x

Rolling-Contact Bearings     603

The reactions in the xy plane from Figure 11–18c are

RyA =

1770(50) 169 000 + = 1716.7 = 1717 N 150 150

RyB =

1770(100) 169 000 − = 53.3 N 150 150

The radial loads FrA and FrB are the vector additions of RyA and RzA, and RyB and RzB, respectively:

2 1∕2 Fr A = (R 2z A + R yA ) = (13272 + 17172 ) 1∕2 = 2170 N

FrB = (R 2zB + R 2yB ) 1∕2 = (26532 + 53.32 ) 1∕2 = 2654 N

Trial 1: With direct mounting of the bearings and application of the external thrust to the shaft, the squeezed bearing is bearing A as labeled in Figure 11–18a. Using K of 1.5 as the initial guess for each bearing, the induced loads from the bearings are

FiA =

0.47FrA 0.47(2170) = = 680 N KA 1.5

FiB =

0.47FrB 0.47(2654) = = 832 N KB 1.5

Since FiA is clearly less than FiB + Fae, bearing A carries the net thrust load, and Equation (11–19) is applicable. Therefore, the dynamic equivalent loads are

FeA = 0.4FrA + KA (FiB + Fae ) = 0.4(2170) + 1.5(832 + 1690) = 4651 N

FeB = FrB = 2654 N

The multiple of rating life is xD =

LD ℒD nD 60 (5000)(800) (60) = = = 2.67 LR LR 90(106 )

Estimate RD as √0.99 = 0.995 for each bearing. For bearing A, from Equation (11–10) the catalog entry C10 should equal or exceed

3∕10 2.67 C10 = (1)(4651) [ = 11 486 N (4.48)(1 − 0.995) 2∕3 ]

From Figure 11–15, tentatively select type TS 15100 cone and 15245 cup, which will work: KA = 1.67, C10  =  12 100 N. For bearing B, from Equation (11–10), the catalog entry C10 should equal or exceed

3∕10 2.67 C10 = (1)2654 [ = 6554 N 2∕3 ] (4.48)(1 − 0.995)

Tentatively select the bearing identical to bearing A, which will work: KB = 1.67, C10 = 12 100 N. Trial 2: Repeat the process with KA = KB = 1.67 from tentative bearing selection.

FiA =

0.47FrA 0.47(2170) = = 611 N KA 1.67

FiB =

0.47FrB 0.47(2654) = = 747 N KB 1.67

604      Mechanical Engineering Design

Since FiA is still less than FiB + Fae, Equation (11–19) is still applicable.

FeA = 0.4FrA + K A (FiB + Fae ) = 0.4(2170) + 1.67(747 + 1690) = 4938 N

FeB = FrB = 2654 N

For bearing A, from Equation (11–10) the corrected catalog entry C10 should equal or exceed

3∕10 2.67 C10 = (1)(4938) [ = 12 195 N (4.48)(1 − 0.995) 2∕3 ]

Although this catalog entry exceeds slightly the tentative selection for bearing A, we will keep it since the reliability of bearing B exceeds 0.995. In the next section we will quantitatively show that the combined reliability of bearing A and B will exceed the reliability goal of 0.99. For bearing B, FeB = FrB = 2654 N. From Equation (11–10),

3∕10 2.67 C10 = (1)2654 [ = 6554 N 2∕3 ] (4.48)(1 − 0.995)

Select cone and cup 15100 and 15245, respectively, for both bearing A and B. Note from Figure 11–14 the effective load center is located at a = −5.8 mm, that is, 5.8 mm into the cup from the back. Thus the shoulderto-shoulder dimension should be 150 − 2(5.8) = 138.4 mm. Note that in each iteration of Equation (11–10) to find the catalog load rating, the bracketed portion of the equation is identical and need not be re-entered on a calculator each time.

11–10  Design Assessment for Selected Rolling-Contact Bearings In textbooks, machine elements typically are treated singly. This can lead the reader to the presumption that a design assessment involves only that element, in this case a rolling-contact bearing. The immediately adjacent elements (the shaft journal and the housing bore) have immediate influence on the performance. Other elements, further removed (gears producing the bearing load), also have influence. Just as some say, "If you pull on something in the environment, you find that it is attached to everything else." This should be intuitively obvious to those involved with machinery. How, then, can one check shaft attributes that aren't mentioned in a problem statement? Possibly, because the bearing hasn't been designed yet (in fine detail). All this points out the necessary iterative nature of designing, say, a speed reducer. If power, speed, and reduction are stipulated, then gear sets can be roughed in, their sizes, geometry, and location estimated, shaft forces and moments identified, bearings tentatively selected, seals identified; the bulk is beginning to make itself evident, the housing and lubricating scheme as well as the cooling considerations become clearer, shaft overhangs and coupling accommodations appear. It is time to iterate, now addressing each element again, knowing much more about all of the others. When you have completed the necessary iterations, you will know what you need for the design assessment for the bearings. In the meantime you do as much of the design assessment as you can, avoiding bad selections, even if tentative. Always keep in mind that you eventually have to do it all in order to pronounce your completed design satisfactory.

Rolling-Contact Bearings     605

An outline of a design assessment for a rolling contact bearing includes, at a minimum, ∙ ∙ ∙ ∙ ∙

Bearing reliability for the load imposed and life expected Shouldering on shaft and housing satisfactory Journal finish, diameter and tolerance compatible Housing finish, diameter and tolerance compatible Lubricant type according to manufacturer's recommendations; lubricant paths and volume supplied to keep operating temperature satisfactory ∙ Preloads, if required, are supplied Since we are focusing on rolling-contact bearings, we can address bearing reliability quantitatively, as well as shouldering. Other quantitative treatment will have to wait until the materials for shaft and housing, surface quality, and diameters and tolerances are known. Bearing Reliability Equation (11–9) can be solved for the reliability RD in terms of C10, the basic load rating of the selected bearing:

  R = exp  −  

a f FD a b   x − x 0    D( C ) 10    θ − x0      

(11–21)

Equation (11–10) can likewise be solved for RD:

a f FD a b   xD ( C ) − x 0  10  R≈1−  θ − x0    

R ≥ 0.90

(11–22)

EXAMPLE 11–9 In Example 11–3, the minimum required load rating for 99 percent reliability, at xD = LD∕L10 = 540, is C10 = 6696 lbf = 29.8 kN. From Table 11–2 a 02-40 mm deep-groove ball bearing would satisfy the requirement. If the bore in the application had to be 70 mm or larger (selecting a 02-70 mm deep-groove ball bearing), what is the resulting reliability? Solution From Table 11–2, for a 02-70 mm deep-groove ball bearing, C10 = 61.8 kN = 13 888 lbf. Using Equation (11–22), recalling from Example 11–3 that af = 1.2, FD = 413 lbf, x0 = 0.02, (θ − x0) = 4.439, and b = 1.483, we can write

Answer

 1.483  1.2(413) 3  [ 540[ 13 888 ] − 0.02 ]   R≈1− = 0.999 963 4.439    

which, as expected, is much higher than 0.99 from Example 11–3.

606      Mechanical Engineering Design

In tapered roller bearings, or other bearings for a two-parameter Weibull distribution, Equation (11–21) becomes, for x0 = 0, θ = 4.48, b = 32 , b xD R = exp {− [ a] } θ[C10∕(a f FD )]

3∕2 xD = exp {− [ 4.48[C10∕(a f FD )]10∕3 ] }

(11–23)

and Equation (11–22) becomes R≈1−{

b 3∕2 xD xD a} = 1 − { θ[C10∕(a f FD )] 4.48[C10∕(a f FD )]10∕3}

(11–24)

EXAMPLE 11–10 In Example 11–8 bearings A and B (cone 15100 and cup 15245) have C10 = 12 100 N. What is the reliability of the pair of bearings A and B? Solution The desired life xD was 5000(800)60∕[90(106)] = 2.67 rating lives. Using Equation (11–24) for bearing A, where from Example 11–8, FD = FeA = 4938 N, and af = 1, gives

RA ≈ 1 − {

3∕2 2.67 = 0.994 791 4.48[12 100∕(1 × 4938)]10∕3}

which is less than 0.995, as expected. Using Equation (11–24) for bearing B with FD = FeB = 2654 N gives

RB ≈ 1 − {

3∕2 2.67 = 0.999 766 4.48[12 100∕(1 × 2654)]10∕3}

Answer The reliability of the bearing pair is

R = RA RB = 0.994 791(0.999 766) = 0.994 558

which is greater than the overall reliability goal of 0.99. When two bearings are made identical for simplicity, or reducing the number of spares, or other stipulation, and the loading is not the same, both can be made smaller and still meet a reliability goal. If the loading is disparate, then the more heavily loaded bearing can be chosen for a reliability goal just slightly larger than the overall goal. An additional example is useful to show what happens in cases of pure thrust loading. EXAMPLE 11–11 Consider a constrained housing as depicted in Figure 11–19 with two direct-mount tapered roller bearings resisting an external thrust Fae of 8000 N. The shaft speed is 950 rev/min, the desired life is 10 000 h, the expected shaft diameter is approximately 1 in. The reliability goal is 0.95. The application factor is appropriately af = 1. (a) Choose a suitable tapered roller bearing for A. (b) Choose a suitable tapered roller bearing for B. (c) Find the reliabilities RA, RB, and R.

Rolling-Contact Bearings     607

Solution (a) By inspection, note that the left bearing carries the axial load and is properly labeled as bearing A. The bearing reactions at A are

FrA = FrB = 0

FaA = Fae = 8000 N

Since bearing B is unloaded, we will start with R = RA = 0.95. With no radial loads, there are no induced thrust loads. Equation (11–19) is applicable.

FeA = 0.4FrA + KA (FiB + Fae ) = K A Fae

If we set KA = 1, we can find C10 in the thrust column and avoid iteration:

FeA = (1)8000 = 8000 N

FeB = FrB = 0

The multiple of rating life is

xD =

LD ℒD nD 60 (10 000)(950)(60) = = = 6.333 LR LR 90(106 )

Then, from Equation (11–10), for bearing A C10 = a f FeA [

4.48(1 − RD ) 2∕3 ]

= (1) 8000 [

xD

3∕10

3∕10 6.33 = 16 159 N 4.48(1 − 0.95) 2∕3 ]

Answer Figure 11–15 presents one possibility in the 1-in bore (25.4-mm) size: cone, HM88630, cup HM88610 with a thrust rating (C10)a = 17 200 N. Answer (b) Bearing B experiences no load, and the cheapest bearing of this bore size will do, including a ball or roller bearing. (c) The actual reliability of bearing A, from Equation (11–24), is Bearing A

Figure 11–19

Bearing B

The constrained housing of Example 11–11.

Fae = 8000 N

608      Mechanical Engineering Design

Answer

RA ≈ 1 − { ≈1−{

4.48[C10∕(a f FD )]10∕3} xD

3∕2

3∕2 6.333 = 0.963 4.48[17 200∕(1 × 8000)]10∕3}

which is greater than 0.95, as one would expect. For bearing B, FD = FeB = 0

Answer

RB ≈ 1 − [

3∕2 6.333 =1−0=1 0.85(17 200∕0) 10∕3 ]

as one would expect. The combined reliability of bearings A and B as a pair is Answer

R = RARB = 0.963(1) = 0.963

which is greater than the reliability goal of 0.95, as one would expect.

Matters of Fit Table 11–2 (and Figure 11–8), which shows the rating of single-row, 02-series, deepgroove and angular-contact ball bearings, includes shoulder diameters recommended for the shaft seat of the inner ring and the shoulder diameter of the outer ring, denoted dS and dH, respectively. The shaft shoulder can be greater than dS but not enough to obstruct the annulus. It is important to maintain concentricity and perpendicularity with the shaft centerline, and to that end the shoulder diameter should equal or exceed dS. The housing shoulder diameter dH is to be equal to or less than dH to maintain concentricity and perpendicularity with the housing bore axis. Neither the shaft shoulder nor the housing shoulder features should allow interference with the free movement of lubricant through the bearing annulus. In a tapered roller bearing (Figure 11–15), the cup housing shoulder diameter should be equal to or less than Db. The shaft shoulder for the cone should be equal to or greater than db. Additionally, free lubricant flow is not to be impeded by obstructing any of the annulus. In splash lubrication, common in speed reducers, the lubricant is thrown to the housing cover (ceiling) and is directed in its draining by ribs to a bearing. In direct mounting, a tapered roller bearing pumps oil from outboard to inboard. An oil passageway to the outboard side of the bearing needs to be provided. The oil returns to the sump as a consequence of bearing pump action. With an indirect mount, the oil is directed to the inboard annulus, the bearing pumping it to the outboard side. An oil passage from the outboard side to the sump has to be provided.

11–11  Lubrication The contacting surfaces in rolling bearings have a relative motion that is both rolling and sliding, and so it is difficult to understand exactly what happens. If the relative velocity of the sliding surfaces is high enough, then the lubricant action is hydrodynamic (see Chapter 12). Elastohydrodynamic lubrication (EHD) is the phenomenon that occurs when a lubricant is introduced between surfaces that are in pure rolling contact. The contact of gear teeth and that found in rolling bearings and in cam-andfollower surfaces are typical examples. When a lubricant is trapped between two

Rolling-Contact Bearings     609

surfaces in rolling contact, a tremendous increase in the pressure within the lubricant film occurs. But viscosity is exponentially related to pressure, and so a very large increase in viscosity occurs in the lubricant that is trapped between the surfaces. Leibensperger2 observes that the change in viscosity in and out of contact pressure is equivalent to the difference between cold asphalt and light sewing machine oil. The purposes of an antifriction-bearing lubricant may be summarized as follows: ∙ ∙ ∙ ∙

To To To To

provide a film of lubricant between the sliding and rolling surfaces help distribute and dissipate heat prevent corrosion of the bearing surfaces protect the parts from the entrance of foreign matter

Either oil or grease may be employed as a lubricant. The following rules may help in deciding between them. Use Grease When

Use Oil When

∙ The temperature is not over 200°F.

∙ Speeds are high.

∙ The speed is low.

∙ Temperatures are high.

∙U  nusual protection is required from the entrance of foreign matter.

∙ Oiltight seals are readily employed.

∙ Simple bearing enclosures are desired. ∙O  peration for long periods without attention is desired.

∙B  earing type is not suitable for grease lubrication. ∙T  he bearing is lubricated from a central supply which is also used for other machine parts.

11–12  Mounting and Enclosure There are so many methods of mounting antifriction bearings that each new design is a real challenge to the ingenuity of the designer. The housing bore and shaft outside diameter must be held to very close limits, which of course is expensive. There are usually one or more counterboring operations, several facing operations and drilling, tapping, and threading operations, all of which must be performed on the shaft, housing, or cover plate. Each of these operations contributes to the cost of production, so that the designer, in ferreting out a trouble-free and low-cost mounting, is faced with a difficult and important problem. The various bearing manufacturers' handbooks give many mounting details in almost every design area. In a text of this nature, however, it is possible to give only the barest details. The most frequently encountered mounting problem is that which requires one bearing at each end of a shaft. Such a design might use one ball bearing at each end, one tapered roller bearing at each end, or a ball bearing at one end and a straight roller bearing at the other. One of the bearings usually has the added function of positioning or axially locating the shaft. Figure 11–20 shows a very common solution to this problem. The inner rings are backed up against the shaft shoulders and are held in position by round nuts threaded onto the shaft. The outer ring of the left-hand bearing is backed up against a housing shoulder and is held in position by a device that is not shown. The outer ring of the right-hand bearing floats in the housing. 2

R. L. Leibensperger, "When Selecting a Bearing," Machine Design, vol. 47, no. 8, April 3, 1975, pp. 142–147.

610      Mechanical Engineering Design

Figure 11–20

Figure 11–21

A common bearing mounting.

An alternative bearing mounting to that in Figure 11–20.

There are many variations possible on the method shown in Figure 11–20. For example, the function of the shaft shoulder may be performed by retaining rings, by the hub of a gear or pulley, or by spacing tubes or rings. The round nuts may be replaced by retaining rings or by washers locked in position by screws, cotters, or taper pins. The housing shoulder may be replaced by a retaining ring; the outer ring of the bearing may be grooved for a retaining ring, or a flanged outer ring may be used. The force against the outer ring of the left-hand bearing is usually applied by the cover plate, but if no thrust is present, the ring may be held in place by retaining rings. Figure 11–21 shows an alternative method of mounting in which the inner races are backed up against the shaft shoulders as before but no retaining devices are required. With this method the outer races are completely retained. This eliminates the grooves or threads, which cause stress concentration on the overhanging end, but it requires accurate dimensions in an axial direction or the employment of adjusting means. This method has the disadvantage that if the distance between the bearings is great, the temperature rise during operation may expand the shaft enough to destroy the bearings. It is frequently necessary to use two or more bearings at one end of a shaft. For example, two bearings could be used to obtain additional rigidity or increased load capacity or to cantilever a shaft. Several two-bearing mountings are shown in Figure 11–22. These may be used with tapered roller bearings, as shown, or with ball bearings. In either case it should be noted that the effect of the mounting is to preload the bearings in an axial direction. Figure 11–22 Two-bearing mountings. (Source: Redrawn from material Furnished by The Timken Company.)

(a)

(b)

Rolling-Contact Bearings     611

Figure 11–23 Mounting for a washing-machine spindle. (Source: Redrawn from material Furnished by The Timken Company.)

Figure 11–24

(a)

(b)

(c)

Figure 11–23 shows another two-bearing mounting. Note the use of washers against the cone backs. When maximum stiffness and resistance to shaft misalignment is desired, pairs of angular-contact ball bearings (Figure 11–2) are often used in an arrangement called duplexing. Bearings manufactured for duplex mounting have their rings ground with an offset, so that when a pair of bearings is tightly clamped together, a preload is automatically established. As shown in Figure 11–24, three mounting arrangements are used. The face-to-face mounting, called DF, will take heavy radial loads and thrust loads from either direction. The DB mounting (back to back) has the greatest aligning stiffness and is also good for heavy radial loads and thrust loads from either direction. The tandem arrangement, called the DT mounting, is used where the thrust is always in the same direction; since the two bearings have their thrust functions in the same direction, a preload, if required, must be obtained in some other manner. Bearings are usually mounted with the rotating ring a press fit, whether it be the inner or outer ring. The stationary ring is then mounted with a push fit. This permits the stationary ring to creep in its mounting slightly, bringing new portions of the ring into the load-bearing zone to equalize wear. Preloading The object of preloading is to: ∙ Remove the internal clearance usually found in bearings ∙ Increase the fatigue life ∙ Decrease the shaft slope at the bearing

Arrangements of angular ball bearings. (a) DF mounting; (b) DB mounting; (c) DT mounting. (Source: Redrawn from material Furnished by The Timken Company.)

612      Mechanical Engineering Design Clearance

Figure 11–25 shows a typical bearing in which the clearance is exaggerated for clarity. Preloading of straight roller bearings may be obtained by: ∙ Mounting the bearing on a tapered shaft or sleeve to expand the inner ring ∙ Using an interference fit for the outer ring ∙ Purchasing a bearing with the outer ring preshrunk over the rollers

Figure 11–25 Clearance in an off-the-shelf bearing, exaggerated for clarity.

Ball bearings are usually preloaded by the axial load built in during assembly. However, the bearings of Figure 11–24a and b are preloaded in assembly because of the differences in widths of the inner and outer rings. It is always good practice to follow manufacturers' recommendations in determining preload, since too much will lead to early failure. Alignment The permissible misalignment in bearings depends on the type of bearing and the geometric and material properties of the specific bearing. Manufacturers' catalogs should be referenced for detailed specifications on a given bearing. In general, cylindrical and tapered roller bearings require alignments that are closer than deep-groove ball bearings. Spherical ball bearings and self-aligning bearings are the most forgiving. Table 7–2 gives typical maximum ranges for each type of bearing. The life of the bearing decreases significantly when the misalignment exceeds the allowable limits. Additional protection against misalignment is obtained by providing the full shoulders (see Figure 11–8) recommended by the manufacturer. Also, if there is any misalignment at all, it is good practice to provide a safety factor of around 2 to account for possible increases during assembly. Enclosures To exclude dirt and foreign matter and to retain the lubricant, the bearing mountings must include a seal. The three principal methods of sealings are the felt seal, the commercial seal, and the labyrinth seal (Figure 11–26). 1 Felt seals may be used with grease lubrication when the speeds are low. The rubbing surfaces should have a high polish. Felt seals should be protected from dirt by placing them in machined grooves or by using metal stampings as shields. 2 The commercial seal is an assembly consisting of the rubbing element and, generally, a spring backing, which are retained in a sheet-metal jacket. These seals are usually made by press fitting them into a counterbored hole in the bearing cover. Since they obtain the sealing action by rubbing, they should not be used for high speeds.

Figure 11–26 Typical sealing methods. (Source: Based on General Motors Corp., GM Media Archives)

(a) Felt seal

(b) Commercial seal

(c) Labyrinth seal

Rolling-Contact Bearings     613

3 The labyrinth seal is especially effective for high-speed installations and may be used with either oil or grease. It is sometimes used with flingers. At least three grooves should be used, and they may be cut on either the bore or the outside diameter. The clearance may vary from 0.010 to 0.040 in, depending upon the speed and temperature.

PROBLEMS Problems marked with an asterisk (*) are linked to problems in other chapters, as summarized in Table 1–2 of Section 1–17. Since each bearing manufacturer makes individual decisions with respect to materials, treatments, and manufacturing processes, manufacturers' experiences with bearing life distribution differ. In solving the following problems, we will use the experience of two manufacturers, tabulated in Table 11–6.

Table 11–6  Typical Weibull Parameters for Two Manufacturers Weibull Parameters Rating Lives Manufacturer 1 2

Rating Life, Revolutions 6 6

90(10 )  1(10 )

x0

θ

b

0

4.48

1.5

0.02

4.459

1.483

Tables 11–2 and 11–3 are based on manufacturer 2.

11–1

Timken rates its bearings for 3000 hours at 500 rev/min. Determine the catalog rating for a ball bearing running for 10 000 hours at 1800 rev/min with a load of 2.75 kN with a reliability of 90 percent.

11–2

A certain application requires a ball bearing with the inner ring rotating, with a design life of 25 kh at a speed of 350 rev/min. The radial load is 2.5 kN and an application factor of 1.2 is appropriate. The reliability goal is 0.90. Find the multiple of rating life required, xD, and the catalog rating C10 with which to enter a bearing table. Choose a 02-series deep-groove ball bearing from Table 11–2, and estimate the ­reliability in use.

11–3

An angular-contact, inner ring rotating, 02-series ball bearing is required for an application in which the life requirement is 40 kh at 520 rev/min. The design radial load is 725 lbf. The application factor is 1.4. The reliability goal is 0.90. Find the multiple of rating life xD required and the catalog rating C10 with which to enter Table 11–2. Choose a bearing and estimate the existing reliability in service.

11–4

The other bearing on the shaft of Problem 11–3 is to be a 03-series cylindrical roller bearing with inner ring rotating. For a 2235-lbf radial load, find the catalog rating C10 with which to enter Table 11–3. The reliability goal is 0.90. Choose a bearing and estimate its reliability in use.

614      Mechanical Engineering Design

11–5

Problems 11–3 and 11–4 raise the question of the reliability of the bearing pair on the shaft. Since the combined reliabilities R is R1R2, what is the reliability of the two bearings (probability that either or both will not fail) as a result of your decisions in Problems 11–3 and 11–4? What does this mean in setting reliability goals for each of the bearings of the pair on the shaft?

11–6

Combine Problems 11–3 and 11–4 for an overall reliability of R = 0.90. Reconsider your selections, and meet this overall reliability goal.

11–7

For Problem 11–1, determine the catalog rating for a ball bearing with a reliability of 96 percent with an application factor of 1.2. The Weibull parameters for Timken bearings are x0 = 0, θ = 4.48, and b = 1.5.

11–8

A straight (cylindrical) roller bearing is subjected to a radial load of 20 kN. The life is to be 8000 h at a speed of 950 rev/min and exhibit a reliability of 0.95. What basic load rating should be used in selecting the bearing from a catalog of manufacturer 2 in Table 11–6?

11–9

Two ball bearings from different manufacturers are being considered for a certain application. Bearing A has a catalog rating of 2.0 kN based on a catalog rating system of 3000 hours at 500 rev/min. Bearing B has a catalog rating of 7.0 kN based on a catalog that rates at 106 cycles. For a given application, determine which bearing can carry the larger load.

11–10 F  or the bearing application specifications given in the table for the assigned to problem, determine the Basic Load Rating for a ball bearing with which to enter 11–15 a bearing catalog of manufacturer 2 in Table 11–6. Assume an application factor of one.

Problem Number

Radial Load

Design Life

Desired Reliability

11–10

2 kN

109 rev

90%

11–11

800 lbf

12 kh, 350 rev/min

90%

11–12

4 kN

8 kh, 500 rev/min

90%

11–13

650 lbf

5 yrs, 40 h/week, 400 rev/min

95%

8

11–14

9 kN

10 rev

99%

11–15

11 kips

20 kh, 200 rev/min

99%

11–16* For the problem specified in the table, build upon the results of the original probto lem to obtain a Basic Load Rating for a ball bearing at C with a 95 percent reli11–19* ability, assuming distribution data from manufacturer 2 in Table 11–6. The shaft rotates at 1200 rev/min, and the desired bearing life is 15 kh. Use an application factor of 1.2. Problem Number

Original Problem Number

11–16*

3–79

11–17*

3–80

11–18*

3–81

11–19*

3–82

Rolling-Contact Bearings     615

11–20* For the shaft application defined in Problem 3–88, the input shaft EG is driven at a

constant speed of 191 rev/min. Obtain a Basic Load Rating for a ball bearing at A for a life of 12 kh with a 95 percent reliability, assuming distribution data from manufacturer 2 in Table 11–6.

11–21* For the shaft application defined in Problem 3–90, the input shaft EG is driven at a

constant speed of 280 rev/min. Obtain a Basic Load Rating for a cylindrical roller bearing at A for a life of 14 kh with a 98 percent reliability, assuming distribution data from manufacturer 2 in Table 11–6.

11–22 An 02-series single-row deep-groove ball bearing with a 65-mm bore (see Tables 11–1

and 11–2 for specifications) is loaded with a 3-kN axial load and a 7-kN radial load. The outer ring rotates at 500 rev/min. (a) Determine the equivalent radial load that will be experienced by this particular bearing. (b) Determine whether this bearing should be expected to carry this load with a 95  percent reliability for 10 kh.

11–23 An 02-series single-row deep-groove ball bearing with a 30-mm bore (see Tables 11–1 and 11–2 for specifications) is loaded with a 2-kN axial load and a 5-kN radial load. The inner ring rotates at 400 rev/min. (a) Determine the equivalent radial load that will be experienced by this particular bearing. (b) Determine the predicted life (in revolutions) that this bearing could be expected to give in this application with a 99 percent reliability.

11–24  An 02-series single-row deep-groove ball bearing is to be selected from Table 11–2 to for the application conditions specified in the table. Assume Table 11–1 is applicable 11–28 if needed. Specify the smallest bore size from Table 11–2 that can satisfy these conditions. Problem Number

Radial Load

Axial Load

Design Life

Ring Rotating

Desired Reliability

11–24

  8 kN

0 kN

109 rev

Inner

90%

11–25

  8 kN

2 kN

10 kh, 400 rev/min

Inner

99%

11–26

  8 kN

3 kN

108 rev

Outer

90%

11–27

10 kN

5 kN

12 kh, 300 rev/min

Inner

95%

11–28

  9 kN

3 kN

108 rev

Outer

99%

11–29* The shaft shown in the figure is proposed as a preliminary design for the appli-

cation defined in Problem 3–83. The effective centers of the gears for force transmission are shown. The dimensions for the bearing surfaces (indicated with cross markings) have been estimated. The shaft rotates at 1200 rev/min, and the desired bearing life is 15 kh with a 95 percent reliability in each bearing, assuming distribution data from manufacturer 2 in Table 11–6. Use an application factor of 1.2. (a) Obtain a Basic Load Rating for a ball bearing at the right end. (b) Use an online bearing catalog to find a specific bearing that satisfies the needed Basic Load Rating and the geometry requirements. If necessary, indicate appropriate adjustments to the dimensions of the bearing surface.

616      Mechanical Engineering Design Gear center

Gear center

16

0.5

1.75

1.3

1.00

14 2.5

9 1.75

1.3

1.00

Problem 11–29* All fillets

1 16

in. Dimensions in inches.

2

1 15

10 11

17 41

11–30* Repeat the requirements of Problem 11–29 for the bearing at the left end of the  shaft.

11–31* The shaft shown in the figure is proposed as a preliminary design for the application

defined in Problem 3–84. The effective centers of the gears for force transmission are shown. The dimensions for the bearing surfaces (indicated with cross markings) have been estimated. The shaft rotates at 900 rev/min, and the desired bearing life is 12 kh with a 98 percent reliability in each bearing, assuming distribution data from manufacturer 2 in Table 11–6. Use an application factor of 1.2. (a) Obtain a Basic Load Rating for a ball bearing at the right end. (b) Use an online bearing catalog to find a specific bearing that satisfies the needed Basic Load Rating and the geometry requirements. If necessary, indicate appropriate adjustments to the dimensions of the bearing surface.

Gear center 15

Gear center

400 50

40

30

350 75

300 50

42

30

Problem 11–31* All fillets 2 mm. Dimensions in millimeters.

30 285

385

30

325

425 1080

11–32* Repeat the requirements of Problem 11–31 for the bearing at the left end of the shaft. 11–33 Shown in the figure is a gear-driven squeeze roll that mates with an idler roll. The

roll is designed to exert a normal force of 35 lbf/in of roll length and a pull of 28 lbf/in on the material being processed. The roll speed is 350 rev/min, and a design life of 35 kh is desired. Use an application factor of 1.2, and select a pair of angular-contact 02-series ball bearings from Table 11–2 to be mounted at 0 and A. Use the same size bearings at both locations and a combined reliability of at least 0.92, assuming distribution data from manufacturer 2 in Table 11–6.

Rolling-Contact Bearings     617 y

O

4 dia. F

Problem 11–33 Dimensions in inches.

z

A

3

14

20°

3

B

8 3

14

3

24

2

Gear 4 3 dia.

x

11–34 The figure shown is a geared countershaft with an overhanging pinion at C. Select an

angular-contact ball bearing from Table 11–2 for mounting at O and an 02-series cylindrical roller bearing from Table 11–3 for mounting at B. The force on gear A is FA = 600 lbf, and the shaft is to run at a speed of 420 rev/min. Solution of the statics problem gives force of bearings against the shaft at O as RO = −387j + 467k lbf, and at B as RB = 316j − 1615k lbf. Specify the bearings required, using an application factor of 1.2, a desired life of 40 kh, and a combined reliability goal of 0.95, assuming distribution data from manufacturer 2 in Table 11–6.

y 20 16

O

FC 10

Problem 11–34 Dimensions in inches.

20°

z Gear 3 24 dia.

B A

A

Gear 4 10 dia.

FA

CC

2

x

20°

11–35 The figure is a schematic drawing of a countershaft that supports two V-belt pulleys.

The countershaft runs at 1500 rev/min and the bearings are to have a life of 60 kh at  a combined reliability of 0.98, assuming distribution data from manufacturer 2 in Table 11–6. The belt tension on the loose side of pulley A is 15 percent of the tension on the tight side. Select deep-groove bearings from Table 11–2 for use at O and E, using an application factor of unity.

618      Mechanical Engineering Design y 300 45° O

400

P2 Problem 11–35 z Dimensions in millimeters.

P1 A

150

2

B

250 dia.

300 dia.

C E

50 N

3 4

D

x

270 N

11–36 A gear-reduction unit uses the countershaft depicted in the figure. Find the two bear-

ing reactions. The bearings are to be angular-contact ball bearings, having a desired life of 50 kh when used at 300 rev/min. Use 1.2 for the application factor and a reliability goal for the bearing pair of 0.96, assuming distribution data from manufacturer 2 in Table 11–6. Select the bearings from Table 11–2.

y 16 F

14 O Problem 11–36

25°

12

240 lbf 20°

Dimensions in inches. z

A Gear 3, 24 dia. B Gear 4, 12 dia.

C 2 x

11–37 The worm shaft shown in part a of the figure transmits 1.2 hp at 500 rev/min. A

static force analysis gave the results shown in part b of the figure. Bearing A is to be an angular-contact ball bearing selected from Table 11–2, mounted to take the 555-lbf thrust load. The bearing at B is to take only the radial load, so an 02-series cylindrical roller bearing from Table 11–3 will be employed. Use an application factor of 1.2, a desired life of 30 kh, and a combined reliability goal of 0.99, assuming distribution data from manufacturer 2 in Table 11–6. Specify each bearing.

Rolling-Contact Bearings     619 y B Worm pitch cylinder A

Gear pitch cylinder

y 36 B

Problem 11–37 (a) Worm and worm gear; (b) force analysis of worm shaft, forces in pounds.

z

x 67 212

36

555

x A (a)

z

555

72

T 145 (b)

11–38 In bearings tested at 2000 rev/min with a steady radial load of 18 kN, a set of bearings showed an L10 life of 115 h and an L80 life of 600 h. The basic load rating of this bearing is 39.6 kN. Estimate the Weibull shape factor b and the characteristic life θ for a two-parameter model. This manufacturer rates ball bearings at 1 million revolutions.

11–39 A 16-tooth pinion drives the double-reduction spur-gear train in the figure. All gears

have 25° pressure angles. The pinion rotates ccw at 1200 rev/min and transmits power to the gear train. The shaft has not yet been designed, but the free bodies have been generated. The shaft speeds are 1200 rev/min, 240 rev/min, and 80 rev/min. A bearing study is commencing with a 10-kh life and a gearbox bearing ensemble reliability of 0.99, assuming distribution data from manufacturer 2 in Table 11–6. An application factor of 1.2 is appropriate. For each shaft, specify a matched pair of 02-series cylindrical roller bearings from Table 11–3. 2385

3

8

E

8 657

3 1530 3280

F

C

80 T

16 T

2 393

1530

6

613

3

874

20 T

2274 9

B A 9

a

B

1075

a

12

1314 2

b

A

613

1314

D

C

b

D

3280

c 16

c

F

1113

60 T E

417

895

2

Developed view (a) Problem 11–39 (a) Drive detail; (b) force analysis on shafts. Forces in pounds; linear dimensions in inches.

239

111 (b) Developed view

502

620      Mechanical Engineering Design

11–40 A gear reduction unit has input shaft AB and output shaft CD, with an input

torque of Ti = 200 lbf · in at constant speed ωi = 60 rev/min driving an output load torque To at output speed ωo. Shaft AB (shown separately with dimensions) is supported by deep-groove ball bearings at A and B, which can be treated as simple supports. The pitch radii of the gears are r 1 = 1.0 in and r2 = 2.5 in. The pressure angle for the spur gears is 20°, as shown. The targeted combined reliability for the entire set of four bearings is 92 percent, for a life of 30 000 hours of operation. (a) Determine the target reliability for each individual bearing. (b) Determine the radial force to be carried by the bearing at A. (c) Determine the load rating with which to select a bearing from a catalog that rates bearings for an L10 life of 1 million cycles.

F

To

B

Problem 11–40

y

E

A ωo

D G2

2 in

1.5 in

ωi, Ti = 200 lbf ⋅in

G1

5 in

z B

G1

r1

200 A

C H

20° FG

x

I 4 in

11–41 The gearbox of Problem 11–40 contains four bearings at locations A, B, C, and D.

Gear 1 has 30 teeth and gear 2 has 60 teeth. The gearbox is rated for a maximum input speed of 200 rpm. The targeted combined reliability for the entire set of bearings is 90 percent for a life cycle of 8 hours per day, 5 days per week, for 6 years. The bearing catalog uses a rating life of 1 million revolutions, and Weibull parameters for reliability ratings are xo = 0.02, θ = 4.459, and b = 1.483. Three of the bearings (A, B, and C) have already been specified, and the individual reliability for each of these bearings is determined to be 97 percent, 96 percent, and 98 percent, respectively. Bearing D is to be a 02-series single-row deep groove ball bearing (specifications in Tables 11–1 and 11–2). A free body diagram determines that bearing D is loaded with a 5-kN axial load and a 9-kN radial load. (a) What should be the minimum reliability goal for bearing D, to achieve the combined reliability goal? (b) Determine the smallest bore size from Table 11–2 that satisfies the conditions.

Rolling-Contact Bearings     621

11–42 Given a 03-series 30-mm cylindrical roller bearing, which was subjected to 300 000

revolutions with a radial load of 5 kips. Estimate the remaining life of the bearing if subsequently subjected to 8 kips.

11–43 A shaft cycles such that each bearing undergoes the following radial loads: 30 percent

of the time at 20 kN, 50 percent of the time at 25 kN, and 20 percent of the time at 30 kN. If the design life of a bearing is to be at least 5 (106) revolutions, select the smallest 03-series cylindrical roller bearing that will accomplish this.

11–44 Estimate the remaining life in revolutions of an 02-30 mm angular-contact ball bearing already subjected to 200 000 revolutions with a radial load of 18 kN, if it is now to be subjected to a change in load to 30 kN.

11–45 The same 02-30 angular-contact ball bearing as in Problem 11–44 is to be subjected

to a two-step loading cycle of 4 min with a loading of 18 kN, and one of 6 min with a loading of 30 kN. This cycle is to be repeated until failure. Estimate the total life in revolutions, hours, and loading cycles.

11–46 A countershaft is supported by two tapered roller bearings using an indirect mounting.

The radial bearing loads are 560 lbf for the left-hand bearing and 1095 for the righthand bearing. An axial load of 200 lbf is carried by the left bearing. The shaft rotates at 400 rev/min and is to have a desired life of 40 kh. Use an application factor of 1.4 and a combined reliability goal of 0.90, assuming distribution data from manufacturer 1 in Table 11–6. Using an initial K = 1.5, find the required radial rating for each bearing. Select the bearings from Figure 11–15.

11–47* For the shaft application defined in Problem 3–85, perform a preliminary specification

for tapered roller bearings at C and D. A bearing life of 108 revolutions is desired with a 90 percent combined reliability for the bearing set, assuming distribution data from manufacturer 1 in Table 11–6. Should the bearings be oriented with direct mounting or indirect mounting for the axial thrust to be carried by the bearing at C? Assuming bearings are available with K = 1.5, find the required radial rating for each bearing. For this preliminary design, assume an application factor of one.

11–48* For the shaft application defined in Problem 3–87, perform a preliminary specification

for tapered roller bearings at A and B. A bearing life of 500 million revolutions is desired with a 90 percent combined reliability for the bearing set, assuming distribution data from manufacturer 1 in Table 11–6. Should the bearings be oriented with direct mounting or indirect mounting for the axial thrust to be carried by the bearing at A? Assuming bearings are available with K = 1.5, find the required radial rating for each bearing. For this preliminary design, assume an application factor of one.

11–49 An outer hub rotates around a stationary shaft, supported by two tapered roller bearings as shown in Figure 11–23. The device is to operate at 250 rev/min, 8 hours per day, 5 days per week, for 5 years, before bearing replacement is necessary. A reliability of 90 percent on each bearing is acceptable. A free body analysis determines the radial force carried by the upper bearing to be 12 kN and the radial force at the lower bearing to be 25 kN. In addition, the outer hub applies a downward force of 5 kN. Assuming bearings are available from manufacturer 1 in Table 11–6 with K = 1.5, find the required radial rating for each bearing. Assume an application factor of 1.2.

622      Mechanical Engineering Design

11–50 The gear-reduction unit shown has a gear that is press fit onto a cylindrical sleeve

that rotates around a stationary shaft. The helical gear transmits an axial thrust load T of 250 lbf as shown in the figure. Tangential and radial loads (not shown) are also transmitted through the gear, producing radial ground reaction forces at the bearings of 875 lbf for bearing A and 625 lbf for bearing B. The desired life for each bearing is 90 kh at a speed of 150 rev/min with a 90 percent reliability. The first iteration of the shaft design indicates approximate diameters of 118 in at A and 1 in at B. Assuming distribution data from manufacturer 1 in Table 11–6, select suitable tapered roller bearings from Figure 11–15.

T B

Problem 11–50 (Source: Redrawn from material Furnished by The Timken Company.)

A

12

Lubrication and Journal Bearings

©Chase Somero/Shutterstock

Chapter Outline 12–1

Types of Lubrication   624

12–9  Steady-State Conditions in Self-Contained Bearings  649

12–2

Viscosity  625

12–3

Petroff's Equation   627

12–4

Stable Lubrication   632

12–5

Thick-Film Lubrication   633

12–6

Hydrodynamic Theory   634

12–7

Design Variables   639

12–14  Dynamically Loaded Journal Bearings  663

12–8

The Relations of the Variables   640

12–15

12–10

Clearance  653

12–11

Pressure-Fed Bearings   655

12–12

Loads and Materials   661

12–13

Bearing Types   662

Boundary-Lubricated Bearings   670

623

624      Mechanical Engineering Design

The object of lubrication is to reduce friction, wear, and heating of machine parts that move relative to each other. A lubricant is any substance that, when inserted between the moving surfaces, accomplishes these purposes. In a sleeve bearing, a shaft, or journal, rotates or oscillates within a sleeve, or bushing, and the relative motion is sliding. In an antifriction bearing, the main relative motion is rolling. A follower may either roll or slide on the cam. Gear teeth mate with each other by a combination of rolling and sliding. Pistons slide within their cylinders. All these applications require lubrication to reduce friction, wear, and heating. The field of application for journal bearings is immense. The crankshaft and connecting-rod bearings of an automotive engine must operate for thousands of miles at high temperatures and under varying load conditions. The journal bearings used in the steam turbines of power-generating stations are said to have reliabilities approaching 100 percent. At the other extreme there are thousands of applications in which the loads are light and the service relatively unimportant; a simple, easily installed bearing is required, using little or no lubrication. In such cases an antifriction bearing might be a poor answer because of the cost, the elaborate enclosures, the close tolerances, the radial space required, the high speeds, or the increased inertial effects. Instead, a nylon bearing requiring no lubrication, a powder-metallurgy bearing with the lubrication "built in," or a bronze bearing with ring oiling, wick feeding, or solid-lubricant film or grease lubrication might be a very satisfactory solution. Recent metallurgy developments in bearing materials, combined with increased knowledge of the lubrication process, now make it possible to design journal bearings with satisfactory lives and very good reliabilities. Much of the material we have studied thus far in this book has been based on fundamental engineering studies, such as statics, dynamics, the mechanics of solids, metal processing, mathematics, and metallurgy. In the study of lubrication and journal bearings, additional fundamental studies, such as chemistry, fluid mechanics, thermodynamics, and heat transfer, must be utilized in developing the material. While we shall not utilize all of them in the material to be included here, you can now begin to appreciate better how the study of mechanical engineering design is really an integration of most of your previous studies and a directing of this total background toward the resolution of a single objective.

12–1  Types of Lubrication Five distinct forms of lubrication may be identified:

1 Hydrodynamic 2 Hydrostatic 3 Elastohydrodynamic 4 Boundary 5 Solid film

Hydrodynamic lubrication means that the load-carrying surfaces of the bearing are separated by a relatively thick film of lubricant, so as to prevent metal-to-metal contact. Hydrodynamic lubrication does not depend upon the introduction of the lubricant under pressure, though that may occur; but it does require the existence of an adequate supply at all times. The film pressure is created by the moving surface itself pulling the lubricant into a wedge-shaped zone at a velocity sufficiently high to create the pressure necessary to separate the surfaces against the load on the bearing. Hydrodynamic lubrication is also called full-film, or fluid lubrication.

Lubrication and Journal Bearings     625

Hydrostatic lubrication is obtained by introducing the lubricant, which is sometimes air or water, into the load-bearing area at a pressure high enough to separate the surfaces with a relatively thick film of lubricant. So, unlike hydrodynamic lubrication, this kind of lubrication does not require motion of one surface relative to another. We shall not deal with hydrostatic lubrication in this book, but the subject should be considered in designing bearings where the velocities are small or zero and where the frictional resistance is to be an absolute minimum. Elastohydrodynamic lubrication (EHL) is the phenomenon where surface deformations play a key role in the development of a lubricant film. Hard EHL occurs when a lubricant is introduced between surfaces that are in rolling contact, such as mating gears, cam-tappet contacts, or roller bearings. The size of lubricant film region between the surfaces is of the order of that predicted by Hertzian contact mechanics and is thus much smaller than the dimensions of the bodies themselves. On the other hand, soft EHL occurs where the size of the contact region can extend over a substantial portion of the mating journal and sleeve surfaces, for example, in big-end connecting rod bearings. Insufficient surface area, a drop in the velocity of the moving surface, a lessening in the quantity of lubricant delivered to a bearing, an increase in the bearing load, or an increase in lubricant temperature resulting in a decrease in viscosity—any one of these—may prevent the buildup of a film thick enough for full-film lubrication. When this happens, the highest asperities may be separated by lubricant films only several molecular dimensions in thickness. This is called boundary lubrication. The change from hydrodynamic to boundary lubrication is not at all a sudden or abrupt one. It is probable that a mixed hydrodynamic- and boundary-type lubrication occurs first, and as the surfaces move closer together, the boundary-type lubrication becomes predominant. The viscosity of the lubricant is not of as much importance with boundary lubrication as is the chemical composition. When bearings must be operated at extreme temperatures, a solid-film lubricant such as graphite or molybdenum disulfide must be used because the ordinary mineral oils are not satisfactory. Much research is currently being carried out in an effort to find composite bearing materials with low wear rates as well as small frictional coefficients.

12–2  Viscosity In Figure 12–1 consider a plate with surface area A moving with a velocity U on a film of lubricant of thickness h. We imagine the film as composed of a series of horizontal layers and the force F causing these layers to deform or slide on one another just like a deck of cards. The layers in contact with the moving plate are Figure 12–1

U Plate

F

u

h y

626      Mechanical Engineering Design

assumed to have a velocity U; those in contact with the stationary surface are assumed to have a zero velocity. Intermediate layers have velocities that depend upon their distances y from the stationary surface. The fluid is said to be Newtonian when the shear stress in the fluid is proportional to the rate of change of velocity with respect to y. Thus

τ=

F du =μ A dy

(12–1)

where μ is the constant of proportionality and defines dynamic viscosity, also called absolute viscosity. The derivative du∕dy is the rate of change of velocity with distance and may be called the rate of shear, or the velocity gradient. The viscosity μ is thus a measure of the internal frictional resistance of the fluid. For most lubricating fluids, the rate of shear is constant, and du∕dy = U∕h. Thus, from Equation (12–1),

τ=

F U =μ A h

(12–2)

The unit of dynamic viscosity in the ips system is seen to be the pound-force-second per square inch; this is the same as stress or pressure multiplied by time. The ips unit is called the reyn, in honor of Sir Osborne Reynolds. When using ips units, the microreyn (μreyn) is often more convenient. The symbol μ′ will be used to designate viscosity in μreyn such that μ = μ′∕(106). The dynamic viscosity in SI units is measured by the pascal-second (Pa · s); this is the same as a Newton-second per square meter. The conversion from ips units to SI is the same as for stress. For example, multiply the dynamic viscosity in reyns by 6890 to convert to units of Pa · s. The American Society of Mechanical Engineers (ASME) has published a list of cgs units that are not to be used in ASME documents.1 This list results from a recommendation by the International Committee of Weights and Measures (CIPM) that the use of cgs units with special names be discouraged. Included in this list is a unit of force called the dyne (dyn), a unit of dynamic viscosity called the poise (P), and a unit of kinematic viscosity called the stoke (St). All of these units have been, and still are, used extensively in lubrication studies. The poise is the cgs unit of dynamic or absolute viscosity, and its unit is the dyne-second per square centimeter (dyn · s/cm2). It has been customary to use the centipoise (cP) in analysis, because its value is more convenient. When the viscosity is expressed in centipoises, it is often designated by Z. The conversion from cgs units to SI and ips units is as follows: 1

μ(Pa · s) = (10)−3Z (cP) μ(reyn) =

Z (cP) 6.89(10) 6

μ(mPa · s) = 6.89 μ′ (μreyn)

ASME Orientation and Guide for Use of Metric Units, 2nd ed., American Society of Mechanical Engineers, 1972, p. 13.

Lubrication and Journal Bearings     627

The ASTM standard method for determining viscosity uses an instrument called the Saybolt Universal Viscosimeter. The method consists of measuring the time in seconds for 60 mL of lubricant at a specified temperature to run through a tube 1.76 mm in diameter and 12.25 mm long. The result is called the kinematic viscosity, and in the past the unit of the square centimeter per second has been used. One square centimeter per second is defined as a stoke. By the use of the Hagen-Poiseuille law, the kinematic viscosity based upon seconds Saybolt, also called Saybolt Universal viscosity (SUV) in seconds, is 180 Zk = (0.22t − t )

(12–3)

where Zk is in centistokes (cSt) and t is the number of seconds Saybolt. In SI, the kinematic viscosity ν has the unit of the square meter per second (m2/s), and the conversion is ν(m2/s) = 10−6Zk (cSt) Thus, Equation (12–3) becomes

ν = (0.22t −

180 (10−6 ) t )

(12–4)

To convert to dynamic viscosity, we multiply ν by the density in SI units. Designating the density as ρ with the unit of the kilogram per cubic meter, we have

180 μ = ρ (0.22t − (10−6 ) t )

(12–5)

where μ is in pascal-seconds. Viscosity can vary considerably with temperature in a nonlinear fashion. Figures 12–2 to 12–4 show the dynamic viscosity in ips and SI units for common grades of lubricating oils employed in machinery. The ordinates in Figures 12–2 to 12–4 are not logarithmic, as the decades are of differing vertical length. Figure 12–5 shows the dynamic viscosity in the ips system for a number of other fluids often used for lubrication purposes and their variation in temperature. The viscosity-temperature trends are provided here only in graphical form. However, various curve fits have been developed. See Table 12–1 for one particular example. Booker2 provides general viscosity-temperature functional forms based on sets of two or three measured temperature-viscosity data points.

12–3  Petroff's Equation The phenomenon of bearing friction was first explained by Petroff on the assumption that the shaft is concentric with its bushing. Though we shall seldom make use of Petroff's method of analysis in the material to follow, it is important

2

J. F. Booker, "Dynamically-Loaded Journal Bearings: Numerical Application of the Mobility Method," ASME Journal of Lubrication Technology, vol. 93, 1971, pp. 168–176, 315.

628      Mechanical Engineering Design

Figure 12–2

10 4

Viscosity–temperature chart in U.S. customary units. (Source: Raimondi and Boyd.)

5 3 2 10 3 5 3 2 10 2 5 4 3 2

SA E

Dynamic viscosity (μreyn)

10

5 4 10

3

20

30

40

50

60

70

2

1

0.5 0.4

0.3

0.2 30

50

100

150

200

250

300

Temperature (°F)

because it defines groups of dimensionless parameters and because the coefficient of friction predicted by this law turns out to be quite good even when the shaft is not concentric. Let us now consider a vertical shaft rotating in a guide bearing. It is assumed that the bearing carries a very small load, that the clearance space is uniform and completely filled with oil, and that leakage is negligible (Figure 12–6). We denote the radius of the shaft by r, the radial clearance by c, and the length of the bearing by l, all dimensions being in inches. Note, in tribology literature, the clearance may be

Lubrication and Journal Bearings     629

Figure 12–3

10 4 5

Viscosity–temperature chart in SI units. (Source: Adapted from Figure 12–2.)

3 2 103 5 3 2

Dynamic viscosity (mPa·s)

102

SA E

5 4

50

3 2

10

20

30

60

70

40

10

5 4

3

2 10

20

30

40

50

60

70

80

90

100

110

120

130

140

Temperature (°C)

denoted as C. If the shaft rotates at N rev/s, then its surface velocity is U = 2πrN in/s. Since the shearing stress in the lubricant is equal to the velocity gradient times the viscosity, from Equation (12–2) we have

τ=μ

U 2πrμ N = c h

(a)

where the radial clearance c has been substituted for the distance h. The force required to shear the film is the stress times the area. The torque is the force times the lever arm r. Thus

T = (τA)(r) = (

4π 2r 3lμ N 2πrμ N (2πrl) (r) = c ) c

(12–6)

630      Mechanical Engineering Design

Figure 12–4

103

Chart for multiviscosity lubricants. This chart was derived from known viscosities at two points, 100° and 210°F, and the results are believed to be correct for other temperatures.

5 4 3 2 102 5 4 3 2

Dynamic viscosity (μreyn)

10 20

W

5 4

–5

0

20

W

–4

0

3

10

W

2

–3

0

1

5W

–3

0

10

W

20

W

0.5 0.4

0.3

0.2 50

100

150

200

250

300

Temperature (°F)

If we now designate a small force on the bearing by W, in pounds-force, then the bearing pressure P acting upon the projected area of the bearing is

P=

W 2rl

(12–7)

The frictional force is f W, where f is the coefficient of friction, and so the frictional torque is

T = f Wr = ( f )(2rlP)(r) = 2r 2 flP (12–8)

Lubrication and Journal Bearings     631

Table 12–1  Curve Fits* to Approximate the Viscosity versus Temperature Functions for SAE Grades 10 to 60

10 –3

Dynamic viscosity, reyn

10 –4

Ca sto ro il SA E3 0o il

10 –5

Oil Grade, SAE

Viscosity μ0, reyn

Constant b, °F

10

0.0158(10−6)

1157.5

20

10 –6

10

0.0141(10 )

1360.0

40

0.0121(10−6)

1474.4

60

Gasoline

50

1509.6

−6

1564.0

0.0170(10 ) 0.0187(10 )

Source: A. S. Seireg and S. Dandage, "Empirical Design Procedure for the Thermodynamic Behavior of Journal Bearings," J. Lubrication Technology, vol. 104, April 1982, pp. 135–148.

Air

0

−6

*μ = μ0 exp [b∕(T + 95)], T in °F.

10 –8

10 –9

1271.6

−6

0.0136(10 )

30 50 Water

–7

−6

100

150

200

Temperature, °F

Figure 12–5 A comparison of the viscosities of various fluids.

Substituting the value of the torque from Equation (12–8) in Equation (12–6) and solving for the coefficient of friction, we find f = 2π 2

μN r (12–9) P c

Equation (12–9) is called Petroff's equation and was first published in 1883. The two quantities μN∕P and r∕c are very important parameters in lubrication. Substitution of the appropriate dimensions in each parameter will show that they are dimensionless. Figure 12–6

"Keyway" sump Oilfill hole A

Bushing (bearing) Journal (shaft)

W

N

W r

U c

W A

W l Section A-A

Side leakage negligible

Petroff's lightly loaded journal bearing consisting of a shaft journal and a bushing with an axial-groove internal lubricant reservoir. The linear velocity gradient is shown in the end view. The radial clearance c is several thousandths of an inch and is grossly exaggerated for presentation purposes.

632      Mechanical Engineering Design

The bearing characteristic number, or the Sommerfeld number, is defined by the equation r 2 μN S=( ) c P

(12–10)

The Sommerfeld number is very important in lubrication analysis because it contains many of the parameters that are specified by the designer. Note that it is also dimensionless. The quantity r∕c is called the radial clearance ratio. If we multiply both sides of Equation (12–9) by this ratio, we obtain the interesting relation

f

μN r 2 r = 2π 2 = 2π 2 S c P (c)

(12–11)

12–4  Stable Lubrication The difference between boundary and hydrodynamic lubrication can be explained by reference to Figure 12–7. This plot of the change in the coefficient of friction versus the bearing modulus μN∕P was obtained by the McKee brothers in an actual test of friction.3 The plot is important because it defines stability of lubrication and helps us to understand hydrodynamic and boundary, or thin-film, lubrication. Recall Petroff's bearing model in the form of Equation (12–9) predicts that f is proportional to μN∕P, that is, a straight line from the origin in the first quadrant. On the coordinates of Figure 12–7 the locus to the right of point C is an example. Petroff's model presumes thick-film lubrication, that is, no metal-to-metal contact, the surfaces being completely separated by a lubricant film. The McKee abscissa was ZN∕P (centipoise × rev/min/psi) and the value of abscissa B in Figure 12–7 was 30. The corresponding μN∕P (reyn × rev/s/psi) is 0.33(10−6). Designers keep μN∕P ≥ 1.7(10−6), which corresponds to ZN∕P ≥ 150. A design constraint to keep thick-film lubrication is to be sure that μN ≥ 1.7(10−6 ) P

(a)

Suppose we are operating to the right of line BA and something happens, say, an increase in lubricant temperature. This results in a lower viscosity and hence a smaller value of μN∕P. The coefficient of friction decreases and not as much heat is generated in shearing the lubricant. Consequently, the lubricant temperature drops, and the Figure 12–7 Coefficient of friction f

The variation of the coefficient of friction f with μN∕P.

A

Thick film (stable)

Thin film (unstable)

C

B Bearing modulus, μN/P

3

S. A. McKee and T. R. McKee, "Journal Bearing Friction in the Region of Thin Film Lubrication," SAE J., vol. 31, 1932, pp. (T)371–377.

Lubrication and Journal Bearings     633

viscosity increases. Thus the region to the right of line BA defines stable lubrication because variations are self-correcting. To the left of line BA, a decrease in viscosity would increase the friction. A temperature rise would ensue, and the viscosity would be reduced still more. The result would be compounded. Thus the region to the left of line BA represents unstable lubrication. It is also helpful to see that a small viscosity, and hence a small μN∕P, means that the lubricant film is very thin and that there will be a greater possibility of some metal-to-metal contact, and hence of more friction. Thus, point C is a reasonable estimation of the beginning of metal-to-metal contact as μN∕P becomes smaller.

12–5  Thick-Film Lubrication The nomenclature of a complete journal bearing is shown in Figure 12–8. The bearing comprises a journal and bushing (or sleeve) separated by a thin lubricant film. Both journal and bushing surfaces are assumed perfectly cylindrical, and their respective axes of rotation are assumed parallel. The dimension c is the radial clearance and is the difference in the radii of the journal and bushing. The radial clearance as drawn in Figure 12–8 is exaggerated for clarity. The radial clearance is typically very small compared to the journal radius (typically c∕r ≈ 0.001, where r is journal radius). The displacement of the journal center O relative to the sleeve center O′ in the clearance space is called the eccentricity and is denoted by e. The film thickness at any point on the bearing is denoted as h and to a high approximation can be written as

h = c + e cos θ

(12–12)

where bearing angle θ is measured from the line of action of the journal and sleeve centers as shown. For design purposes, it is convenient to define the eccentricity ratio, ε, as

e ε= c

(12–13)

where the kinematic limit ε = 1 denotes journal to bushing contact. Suppose a steady load W is applied to a journal rotating clockwise at a steady speed N. The attitude angle ϕ of the bearing is defined as the included angle between the eccentricity and load vector. When a lubricant is introduced into the top of the bearing, the action of the rotating journal is to pump the lubricant around the bearing in a clockwise direction. The lubricant is drawn into a convergent wedge-shaped space and forces the journal center to a steady position over to the side. The minimum film thickness h0 occurs, not along the load line, but at a point displaced clockwise from the load. The magnitude of h0 is c (1 − ε) and is found by setting θ = 180° in Equation (12–12). A complete lubricant film is established from the inlet region to a point just beyond the minimum film thickness value. It is in this fluid-converging region that film pressure is generated in the lubricant film to support the applied load. As the fluid continues to flow clockwise, the journal and bushing surfaces diverge, which results in a significant pressure drop. The film pressure in this region drops to subambient levels, whereupon the fluid film ruptures or cavitates into liquid streamers separated by a liquid–gas mixture. The cavitation region typically extends over approximately half of the bearing surface.

634      Mechanical Engineering Design Journal

Q (flow) O'

Cavitating film region

O

e

Bushing

Journal θ O' e

h0

N

W

O ϕ

N

β/2

h

β/2

r Complete film region

h0

Bushing

Figure 12–9 Nomenclature of a partial journal bearing.

W

Figure 12–8 Nomenclature of a complete journal bearing.

A partial journal bearing is shown in Figure 12–9. If the radius of the bushing is the same as that of the journal (c = 0), it is known as a fitted bearing. The angle β defines the angular extent of the partial bearing, in this case measured from the load line. If the angular extent is small and if h0 is near the edge, cavitation is unlikely to occur in the lubricant film.

12–6  Hydrodynamic Theory The present theory of hydrodynamic lubrication originated in the laboratory of Beauchamp Tower in the early 1880s in England. Tower had been employed to study the friction in railroad journal bearings and learn the best methods of lubricating them. It was an accident or error, during the course of this investigation, that prompted Tower to look at the problem in more detail and that resulted in a discovery that eventually led to the development of the theory. Figure 12–10 is a schematic drawing of the journal bearing that Tower investigated. It is a partial bearing, having a diameter of 4 in, a length of 6 in, a bearing arc of 157°, and bath-type lubrication, as shown. The coefficients of friction obtained by Tower in his investigations on this bearing were quite low, which is now not surprising. After testing this bearing, Tower later drilled a 12 -in-diameter lubricator hole Figure 12–10 Schematic representation of the partial bearing used by Tower.

Lubricator hole

W

N

Journal

Partial bronze bearing

Lubricant level

Lubrication and Journal Bearings     635

Figure 12–11

pmax

Approximate pressure-distribution curves obtained by Tower.

p=0

N l = 6 in d = 4 in

through the top. But when the apparatus was set in motion, oil flowed out of this hole. In an effort to prevent this, a cork stopper was used, but this popped out, and so it was necessary to drive a wooden plug into the hole. When the wooden plug was pushed out too, Tower, at this point, undoubtedly realized that he was on the verge of discovery. A pressure gauge connected to the hole indicated a pressure of more than twice the unit bearing load. He further investigated the bearing film pressures in detail throughout the bearing width and length and reported a distribution similar to that of Figure 12–11.4 The results obtained by Tower had such regularity that Osborne Reynolds concluded that there must be a definite equation relating the friction, the pressure, and the surface velocity. The present mathematical theory of lubrication is based upon Reynolds's work following the experiment by Tower.5 The solution is a challenging problem that has interested many investigators ever since then, and it is still the starting point for lubrication studies. Reynolds pictured the lubricant as adhering to both surfaces and being pulled by the moving surface into a narrowing, wedge-shaped space so as to create a fluid or film pressure of sufficient intensity to support the bearing load. One of the important simplifying assumptions resulted from Reynolds's realization that the fluid films were so thin in comparison with the bearing radius that the curvature could be neglected. This enabled him to replace the cylindrical bearing with a flat "unwrapped" bearing, called a plane slider bearing. Other assumptions made were:

1 2 3 4 5

The The The The The

lubricant is Newtonian, Equation (12–1). forces due to the inertia of the lubricant are neglected. lubricant is assumed to be incompressible. viscosity is assumed to be constant throughout the film. pressure does not vary in the axial direction.

Figure 12–12a shows a journal rotating in the clockwise direction at a constant angular velocity ωj, supported by a film of lubricant of variable thickness h on a 4

Beauchamp Tower, "First Report on Friction Experiments," Proc. Inst. Mech. Eng., November 1883, pp. 632–666; "Second Report," ibid., 1885, pp. 58–70; "Third Report," ibid., 1888, pp. 173–205; "Fourth Report," ibid., 1891, pp. 111–140. 5 Osborne Reynolds, "Theory of Lubrication, Part I," Phil. Trans. Roy. Soc. London, 1886.

636      Mechanical Engineering Design

θ Rotating journal

u=U Journal

Flow of lubricant

(τ + ∂τ dy) dx dz ∂y (p +

y

h

dp dx) dy dz dx

dy

U

p dy dz

dx z τ dx dz

dx

x dy

h Stationary partial bushing Partial bushing (a)

(b)

Figure 12–12 partial bearing, which is fixed. We specify that the journal has a constant surface velocity, U = rωj. Using Reynolds's assumption that curvature can be neglected, we fix a right-handed xyz reference system to the stationary bushing, where x = rθ. We now make the following additional assumptions: 6 The bushing and journal extend infinitely in the z direction; this means there can be no lubricant flow in the z direction. 7 The film pressure is constant in the y direction. Thus the pressure depends only on the coordinate x. 8 The velocity of any particle of lubricant in the film depends only on the coordinates x and y. We now select an element of lubricant in the film (Figure 12–12a) of dimensions dx, dy, and dz, and compute the forces that act on the sides of this element. As shown in Figure 12–12b, normal forces, due to the pressure, act upon the right and left sides of the element, and shear forces, due to the viscosity and to  the velocity, act upon the top and bottom sides. Summing the forces in the x  direction gives

∑ Fx = p dy dz − (p +

dp ∂τ dx)dy dz − τ dx dz + (τ + dy dx dz = 0 dx ∂y )

(a)

This reduces to

dp ∂τ = dx ∂y

(b)

∂u ∂y

(c)

From Equation (12–1), we have

τ=μ

Lubrication and Journal Bearings     637

where the partial derivative is used because the velocity u depends upon both x and y. Substituting Equation (c) in Equation (b), we obtain dp ∂ 2u = μ 2 dx ∂y

(d)

Holding x constant, we now integrate this expression twice with respect to y. This gives ∂u 1 dp = y + C1 ∂y μ dx

u=

1 dp 2 y + C1y + C2 2μ dx

(e)

Note that the act of holding x constant means that C1 and C2 can be functions of x. We now assume that there is no slip between the lubricant and the boundary surfaces. This gives two sets of boundary conditions for evaluating C1 and C2:

At  y = 0, u = 0

At  y = h, u = U

(f )

Notice, in the second condition, that h is a function of x. Substituting these conditions in Equation (e) and solving for C1 and C2 gives

C1 =

U h dp − h 2μ dx

u=

1 dp 2 U (y − hy) + y 2μ dx h

or

C2 = 0

(12–14)

This equation gives the velocity distribution of the lubricant in the film as a function of the coordinate y and the pressure gradient dp∕dx. The equation shows that the velocity distribution across the film (from y = 0 to y = h) is obtained by superposing a parabolic distribution onto a linear distribution. Figure 12–13 shows the superposition of these distributions to obtain the velocity for particular values of x and dp∕dx. In general, the parabolic term may be additive or subtractive to the linear term, depending upon the sign of the pressure gradient. When the pressure is maximum, dp∕dx = 0 and the velocity is

u=

U y h

(g)

which is a linear relation. Figure 12–13

Rotating journal y

Velocity of the lubricant.

U

u h y

x

Stationary bushing

Flow of lubricant

dp >0 dx dp =0 dx dp 1 μm, so the bearing specifications are acceptable for this application.

12–15  Boundary-Lubricated Bearings When two surfaces slide relative to each other with only a partial lubricant film between them, boundary lubrication is said to exist. Boundary- or thin-film lubrication occurs in hydrodynamically lubricated bearings when they are starting or stopping, when the load increases, when the supply of lubricant decreases, or whenever other operating changes happen to occur. There are, of course, a very large number of cases in design in which boundary-lubricated bearings must be used because of the type of application or the competitive situation. The coefficient of friction for boundary-lubricated surfaces may be greatly decreased by the use of animal or vegetable oils mixed with the mineral oil or grease. Fatty acids, such as stearic acid, palmitic acid, or oleic acid, or several of these, which occur in animal and vegetable fats, are called oiliness agents. These acids appear to reduce friction, either because of their strong affinity for certain metallic surfaces or because they form a soap film that binds itself to the metallic surfaces by a chemical reaction. Thus the fatty-acid molecules bind themselves to the journal and bearing surfaces with such great strength that the metallic asperities of the rubbing metals do not weld or shear. Fatty acids will break down at temperatures of 250°F or more, causing increased friction and wear in thin-film-lubricated bearings. In such cases the extreme-pressure,

Lubrication and Journal Bearings     671

or EP, lubricants may be mixed with the fatty-acid lubricant. These are composed of chemicals such as chlorinated esters or tricresyl phosphate, which form an organic film between the rubbing surfaces. Though the EP lubricants make it possible to operate at higher temperatures, there is the added possibility of excessive chemical corrosion of the sliding surfaces. When a bearing operates partly under hydrodynamic conditions and partly under dry or thin-film conditions, a mixed-film lubrication exists. If the lubricant is supplied by hand oiling, by drop or mechanical feed, or by wick feed, for example, the bearing is operating under mixed-film conditions. In addition to occurring with a scarcity of lubricant, mixed-film conditions may be present when ∙ ∙ ∙ ∙ ∙

The viscosity is too low. The bearing speed is too low. The bearing is overloaded. The clearance is too tight. Journal and bearing are not properly aligned.

Relative motion between surfaces in contact in the presence of a lubricant is called boundary lubrication. This condition is present in hydrodynamic film bearings during starting, stopping, overloading, or lubricant deficiency. Some bearings are boundary lubricated (or dry) at all times. To signal this an adjective is placed before the word "bearing." Commonly applied adjectives (to name a few) are thin-film, boundary friction, Oilite, Oiles, and bushed-pin. The applications include situations in which thick film will not develop and there are low journal speed, oscillating journal, padded slides, light loads, and lifetime lubrication. The characteristics include considerable friction, ability to tolerate expected wear without losing function, and light loading. Such bearings are limited by lubricant temperature, speed, pressure, galling, and cumulative wear. Table 12–8 gives some properties of a range of bushing materials. Table 12–8  Some Materials for Boundary-Lubricated Bearings and Their Operating Limits Maximum Load, psi

Maximum Temperature, °F

Maximum Speed, fpm

Maximum PV Value*

Cast bronze

4 500

325

1 500

50 000

Porous bronze

4 500

150

1 500

50 000

Porous iron

8 000

150

800

50 000

Phenolics

6 000

200

2 500

15 000

Nylon

1 000

200

1 000

3 000

Material

Teflon

500

500

100

1 000

2 500

500

1 000

10 000

60 000

500

50

25 000

1 000

180

1 000

3 000

600

750

2 500

15 000

50

150

4 000

2 000

150

2 000

Reinforced Teflon Teflon fabric Delrin Carbon-graphite Rubber Wood *P = load, psi; V = speed, fpm.

15 000

672      Mechanical Engineering Design

Figure 12–39

F

F

Sliding block subjected to wear. fs PA τA

fs PA A

w

τA

S

PA

PA

Linear Sliding Wear Consider the sliding block depicted in Figure 12–39, moving along a plate with contact pressure P′ acting over area A, in the presence of a coefficient of sliding friction fs. The linear measure of wear w is expressed in inches or millimeters. The work done by force fs PA during displacement S is fs PAS or fs PAVt, where V is the sliding velocity and t is time. The material volume removed due to wear is wA and is proportional to the work done, that is, wA ∝ fs PAVt, or wA = KPAVt where K is the proportionality factor, which includes fs, and is determined from laboratory testing. The linear wear is then expressed as

w = KPVt

(12–38)

In US customary units, P is expressed in psi, V in fpm (i.e., ft/min), and t in hours. This makes the units of K in3 · min/(lbf · ft · h). SI units commonly used for K are cm3 · min/(kgf · m · h), where 1 kgf = 9.806 N. Tables 12–9 and 12–10 give some wear factors and coefficients of friction from one manufacturer. Some care in the selection of the wear coefficient K is warranted here, as K can be dependent on a variety of factors, such as surface finish, application of coatings, and surface speeds.

Table 12–9  Wear Factors in U.S. Customary Units* Bushing Material

Wear Factor K

Table 12–10  Coefficients of Friction Limiting PV

Type

Bearing

fs

Placetic

Oiles 80

0.05

Composite

Drymet ST

0.03

Toughmet

0.05

Cermet M

0.05

Oiles 2000

0.03

Oiles 300

0.03

Oiles 500SP

0.03

Oiles 800

3(10−10)

18 000

Oiles 500

0.6(10−10)

46 700

Polyactal copolymer

50(10−10)

5 000

Polyactal homopolymer

60(10−10)

3 000

200(10−10)

2 000

66 nylon + 15% PTFE

13(10−10)

7 000

     + 15% PTFE + 30% glass

16(10−10)

10 000

200(10−10)

2 000

200(10−10)

2 000

75(10−10)

7 000

66 nylon

    + 2.5% MoS2 6 nylon Polycarbonate + 15% PTFE Sintered bronze Phenol + 25% glass fiber

−10

102(10

)

8 500

8(10−10)

11 500

*dim[K] = in3 · min/(lbf · ft · h), dim [PV] = psi · ft/min. Source: Data from Oiles America Corp., Plymouth, MI 48170.

Met

Source: Data from Oiles America Corp., Plymouth, MI 48170.

Lubrication and Journal Bearings     673

Bushing Wear Consider a pin of diameter D, rotating at speed N, in a bushing of length L, and supporting a stationary radial load F. The nominal pressure P is given by

F DL

P=

(12–39)

and if N is in rev/min and D is in inches, velocity in ft/min is given by

V=

Thus PV, in psi · ft/min, is

πDN 12

(12–40)

F πDN π FN = (12–41) DL 12 12 L Note the independence of PV from the journal diameter D. A time-wear equation similar to Equation (12–38) can be written. However, before doing so, it is important to note that Equation (12–39) provides the nominal value of P. Figure 12–40 provides a more accurate representation of the pressure distribution, which can be written as π π p = Pmax cos θ − ≤θ≤ 2 2 The vertical component of p dA is p dA cos θ = [pL(D∕2) dθ]cos θ = Pmax(DL∕2) cos2 θ dθ. Integrating this from θ = −π∕2 to π∕2 yields F. Thus,

or

PV =

π∕2

−π∕2

Pmax (

DL π cos2 θ dθ = Pmax DL = F 2 ) 4 Pmax =

4 F π DL

(12–42)

Substituting V from Equation (12–40) and Pmax for P from Equation (12–42) into Equation (12–37) gives 4 F πDNt KFNt w=K = (12–43) π DL 12 3L Figure 12–40

F

Pressure distribution on a boundary-lubricated bushing.

D/2

θ

P Pmax

674      Mechanical Engineering Design

In designing a bushing, because of various trade-offs it is recommended that the length/diameter ratio be in the range

0.5 ≤ L∕D ≤ 2

(12–44)

EXAMPLE 12–9 An Oiles SP 500 alloy brass bushing is 1 in long with a 1-in bore and operates in a clean environment at 70°F. The allowable wear without loss of function is 0.005 in. The radial load is 700 lbf. The peripheral velocity is 33 ft/min. Estimate the number of revolutions for radial wear to be 0.005 in. See Figure 12–41 and Table 12–11 from the manufacturer. Solution From Table 12–9, K = 0.6(10−10) in3 · min/(lbf · ft · h); Table 12–11, PV = 46 700 psi · ft/min, Pmax = 3560 psi, Vmax = 100 ft/min. From Equations (12–42), (12–40), and (12–41), 4 F 4 700 = = 891 psi < 3560 psi  (OK) π DL π (1)(1)

Pmax =

P=

V = 33 ft/min < 100 ft/min

F 700 = = 700 psi DL (1)(1) (OK)

PV = 700(33) = 23 100 psi · ft/min < 46 700 psi · ft/min  (OK)

Figure 12–41 Journal/bushing for Example 12–9.

700 lbf 1 in

1 in

Table 12–11  Oiles 500 SP (SPBN · SPWN) Service Range and Properties Service Range

Units

Allowable

Characteristic pressure Pmax

psi

1.6

Specific gravity Source: Data from Oiles America Corp., Plymouth, MI 48170.

8.2

Lubrication and Journal Bearings     675

Equation (12–43) with Equation (12–40) is

w=K

4 F πDNt 4 F =K Vt π DL 12 π DL

Solving for t gives

t=

The rotational speed is

π(1) (1)0.005 πDLw = = 2830 h = 169 800 min 4 KVF 4(0.6)10 −10 (33)700

N=

12V 12(33) = = 126 r/min πD π(1)

Cycles = Nt = 126(169 800) = 21.4(106 ) rev

Answer

Temperature Rise At steady state, the rate at which work is done against bearing friction equals the rate at which heat is transferred from the bearing housing to the surroundings by convection and radiation. The rate of heat generation in Btu/h is given by fsFV∕J, or

Hgen =

fs F(πD)(60N) 5π fs FDN = 12J J

(12–45)

where N is journal speed in rev/min and J = 778 ft · lbf/Btu. The rate at which heat is transferred to the surroundings, in Btu/h, is

Hloss = h CR AΔ T = h CR A(Tb − T∞ ) =

h CR A (Tf − T∞ ) 2

(12–46)

where  A = housing surface area, ft2 hCR = overall combined coefficient of heat transfer, Btu/(h · ft2 · °F)   Tb = housing metal temperature, °F   Tf = lubricant temperature, °F The empirical observation that Tb is about midway between Tf and T∞ has been incorporated in Equation (12–46). Equating Equations (12–45) and (12–46) gives

Tf = T ∞ +

10πfs FDN J h CR A

(12–47)

Although this equation seems to indicate the temperature rise Tf − T∞ is independent of length L, the housing surface area generally is a function of L. The housing surface area can be initially estimated, and as tuning of the design proceeds, improved results will converge. If the bushing is to be housed in a pillow block, the surface area can be roughly estimated from

A≈

2πDL 144

(12–48)

Substituting Equation (12–48) into Equation (12–47) gives

Tf ≈ T ∞ +

10πfs FDN 720 fs FN = T∞ + J h CR (2πDL∕144) J h CR L

(12–49)

676      Mechanical Engineering Design

EXAMPLE 12–10 Choose an Oiles 500 bushing to give a maximum wear of 0.001 in for 800 h of use with a 300 rev/min journal and 50 lbf radial load. Use h CR = 2.7 Btu/(h · ft2 · °F), Tmax = 300°F, fs = 0.03, and a design factor nd = 2. Table 12–12 lists the available bushing sizes from the manufacturer. Solution Using Equation (12–49) with nd F for F, fs = 0.03 from Table 12–10, and h CR = 2.7 Btu∕(h · ft2 · °F), gives

L≥

720 fs nd FN 720(0.03)2(50)300 = = 1.34 in J h CR (Tf − T∞ ) 778(2.7)(300 − 70)

From Table 12–12, the smallest available bushing has an ID = 58 in, OD = 78 in, and L = 112 in. However, for this case L∕D = 1.5∕0.625 = 2.4, and is outside of the recommendations of Equation (12–44). Thus, for the first trial, try the bushing with ID = 34 in, OD = 118 in, and L = 112 in (L∕D = 1.5∕0.75 = 2). Thus, Equation (12–42):   Pmax =

4 nd F 4 2(50) = = 113 psi < 3560 psi  (OK) π DL π 0.75(1.5)

nd F 2(50) = = 88.9 psi DL 0.75(1.5)

P=

Table 12–12  Available Bushing Sizes (in inches) of One Manufacturer* L ID OD 1 2 5 8 3 4 7 8

1 1 114 112 134

3 4 7 8 118 114 138 112 158

2 214

1 2

5 8

3 4

7 8

1

141

121

143

2

221

3

321

4

∙ ∙

212 234

3

412

338 358 418 434 538

5

6

2 214 212 234 3 312 4

5

∙ ∙

*In a display such as this a manufacturer is likely to show catolog numbers where the ∙ appears.

Lubrication and Journal Bearings     677

Equation (12–40):   V =

πDN π(0.75)300 = = 58.9 ft/min < 100 ft/min  (OK) 12 12

PV = 88.9(58.9) = 5240 psi · ft/min < 46 700 psi · ft/min  (OK)

Equation (12–43), with Table 12–9:

w=

KndFNt 6(10−11 )2(50)300(800) = = 0.000320 in < 0.001 in  (OK) 3L 3(1.5)

Answer Select ID = 34 in, OD = 118 in, and L = 112 in.

PROBLEMS 12–1

A full journal bearing has a journal diameter of 25 mm, with a unilateral tolerance of −0.03 mm. The bushing bore has a diameter of 25.03 mm and a unilateral tolerance of 0.04 mm. The l∕d ratio is 1∕2. The load is 1.2 kN and the journal runs at 1100 rev/min. If the average viscosity is 55 mPa · s, find the minimum film thickness, the power loss, and the side flow for the minimum clearance assembly.

12–2 A full journal bearing has a journal diameter of 32 mm, with a unilateral tolerance of −0.012 mm. The bushing bore has a diameter of 32.05 mm and a unilateral tolerance of 0.032 mm. The bearing is 64 mm long. The journal load is 1.75 kN and it runs at a speed of 900 rev/ min. Using an average viscosity of 55 mPa · s find the minimum film thickness, the maximum film pressure, and the total oil-flow rate for the minimum clearance assembly.

12–3 A journal bearing has a journal diameter of 3.000 in, with a unilateral tolerance of −0.001

in. The bushing bore has a diameter of 3.005 in and a unilateral tolerance of 0.004 in. The bushing is 1.5 in long. The journal speed is 600 rev/min and the load is 800 lbf. For both SAE 10 and SAE 40, lubricants, find the minimum film thickness and the maximum film pressure for an operating temperature of 150°F for the minimum clearance assembly.

12–4 A journal bearing has a journal diameter of 3.250 in with a unilateral tolerance of

−0.003 in. The bushing bore has a diameter of 3.256 in and a unilateral tolerance of 0.004 in. The bushing is 3 in long and supports a 800-lbf load. The journal speed is 1000 rev/min. Find the minimum oil film thickness and the maximum film pressure for both SAE 20 and SAE 20W-40 lubricants, for the tightest assembly if the operating film temperature is 150°F.

12–5 A full journal bearing has a journal with a diameter of 2.000 in and a unilateral toler-

ance of −0.0012 in. The bushing has a bore with a diameter of 2.0024 and a unilateral tolerance of 0.002 in. The bushing is 1 in long and supports a load of 600 lbf at a speed of 800 rev/min. Find the minimum film thickness, the power loss, and the total lubricant flow if the average film temperature is 130°F and SAE 20 lubricant is used. The tightest assembly is to be analyzed.

12–6 A full journal bearing has a shaft journal diameter of 25 mm with a unilateral tolerance

of −0.01 mm. The bushing bore has a diameter of 25.04 mm with a unilateral tolerance of 0.03 mm. The l∕d ratio is unity. The bushing load is 1.25 kN, and the journal rotates at 1200 rev/min. Analyze the minimum clearance assembly if the average viscosity is 50 mPa · s to find the minimum oil film thickness, the power loss, and the percentage of side flow.

678      Mechanical Engineering Design

12–7 A full journal bearing has a shaft journal with a diameter of 1.25 in and a unilateral

tolerance of −0.0006 in. The bushing bore has a diameter of 1.252 in with a unilateral tolerance of 0.0014 in. The bushing bore is 2 in in length. The bearing load is 620 lbf and the journal rotates at 1120 rev/min. Analyze the minimum clearance assembly and find the minimum film thickness, the coefficient of friction, and the total oil flow if the average viscosity is 8.5 μreyn.

12–8 A journal bearing has a shaft diameter of 75.00 mm with a unilateral tolerance of

−0.02 mm. The bushing bore has a diameter of 75.10 mm with a unilateral tolerance of 0.06 mm. The bushing is 36 mm long and supports a load of 2 kN. The journal speed is 720 rev/min. For the minimum clearance assembly find the minimum film thickness, the heat loss rate, and the maximum lubricant pressure for SAE 20 and SAE 40 lubricants operating at an average film temperature of 60°C.

12–9 A full journal bearing is 28 mm long. The shaft journal has a diameter of 56 mm with a unilateral tolerance of −0.012 mm. The bushing bore has a diameter of 56.05 mm with a unilateral tolerance of 0.012 mm. The load is 2.4 kN and the journal speed is 900 rev/min. For the minimum clearance assembly find the minimum oil-film thickness, the power loss, and the side flow if the operating temperature is 65°C and SAE 40 lubricating oil is used.

12–10 A full journal bearing has a shaft diameter of 3.000 in with a unilateral tolerance of −0.0004 in. The l∕d ratio is unity. The bushing has a bore diameter of 3.003 in with a unilateral tolerance of 0.0012 in. The SAE 40 oil supply is in an axial-groove sump with a steady-state temperature of 140°F. The radial load is 675 lbf. Estimate the average film temperature, the minimum film thickness, the heat loss rate, and the lubricant side-flow rate for the minimum clearance assembly, if the journal speed is 10 rev/s.

12–11 A 212 × 221-in sleeve bearing uses grade 20 lubricant. The axial-groove sump has a steady-

state temperature of 110°F. The shaft journal has a diameter of 2.500 in with a unilateral tolerance of −0.001 in. The bushing bore has a diameter of 2.504 in with a unilateral tolerance of 0.001 in. The journal speed is 1120 rev/min and the radial load is 1200 lbf. Estimate (a) The magnitude and location of the minimum oil-film thickness. (b) The eccentricity. (c) The coefficient of friction. (d) The power loss rate. (e) Both the total and side oil-flow rates. ( f ) The maximum oil-film pressure and its angular location. (g) The terminating position of the oil film. (h) The average temperature of the side flow. (i) The oil temperature at the terminating position of the oil film.

12–12 A set of sleeve bearings has a specification of shaft journal diameter of 1.250 in with a

unilateral tolerance of −0.001 in. The bushing bore has a diameter of 1.252  in with a unilateral tolerance of 0.003 in. The bushing is 141 in long. The radial load is 250 lbf and the shaft rotational speed is 1750 rev/min. The lubricant is SAE 10 oil and the axial-groove sump temperature at steady state Ts is 120°F. For the cmin, cmedian, and cmax assemblies analyze the bearings and observe the changes in S, ε, f, Q, Qs, ΔT, Tmax, T f , and h0.

12–13 An interpolation equation was given by Raimondi and Boyd, and it is displayed as

Equation (12–20). This equation is a good candidate for a computer program. Write such a program for interactive use. Once ready for service it can save time and reduce errors. Another version of this program can be used with a subprogram that contains curve fits to Raimondi and Boyd charts for computer use.

Lubrication and Journal Bearings     679

12–14 A natural-circulation pillow-block bearing with l∕d = 1 has a journal diameter D of

2.500 in with a unilateral tolerance of −0.001 in. The bushing bore diameter B is 2.504 in with a unilateral tolerance of 0.004 in. The shaft runs at an angular speed of 1120 rev/min; the bearing uses SAE grade 20 oil and carries a steady load of 300 lbf in shaft-stirred air at 70°F with α = 1. The lateral area of the pillow-block housing is 60 in2. Perform a design assessment using minimum radial clearance for a load of 600 lbf and 300 lbf. Use Trumpler's criteria.

12–15 An eight-cylinder diesel engine has a front main bearing with a journal diameter of

3.500 in and a unilateral tolerance of −0.003 in. The bushing bore diameter is 3.505 in with a unilateral tolerance of +0.005 in. The bushing length is 2 in. The pressure-fed bearing has a central annular groove 0.250 in wide. The SAE 30 oil comes from a sump at 120°F using a supply pressure of 50 psig. The sump's heat-dissipation capacity is 5000 Btu/h per bearing. For a minimum radial clearance, a speed of 2000 rev/min, and a radial load of 4600 lbf, find the average film temperature and apply Trumpler's criteria in your design assessment.

12–16 A pressure-fed bearing has a journal diameter of 50.00 mm with a unilateral tolerance of −0.05 mm. The bushing bore diameter is 50.084 mm with a unilateral tolerance of 0.10 mm. The length of the bushing is 55 mm. Its central annular groove is 5 mm wide and is fed by SAE 30 oil is 55°C at 200 kPa supply gauge pressure. The journal speed is 2880 rev/ min carrying a load of 10 kN. The sump can dissipate 300 watts per bearing if necessary. For minimum radial clearances, perform a design assessment using Trumpler's criteria.

12–17 Design a central annular-groove pressure-fed bearing with an l′∕d ratio of 0.5, using

SAE grade 20 oil, the lubricant supplied at 30 psig. The exterior oil cooler can ­maintain the sump temperature at 120°F for heat dissipation rates up to 1500 Btu/h. The load to be carried is 900 lbf at 3000 rev/min. The groove width is 14 in. Use nominal journal diameter d as one design variable and c as the other. Use Trumpler's criteria for your adequacy assessment.

12–18 Repeat design problem Problem 12–17 using the nominal bushing bore B as one decision variable and the radial clearance c as the other. Again, Trumpler's criteria to be used.

12–19 Table 12–1 gives the Seireg and Dandage curve fit approximation for the absolute viscosity in customary U.S. engineering units. Show that in SI units of mPa · s and a temperature of C degrees Celsius, the viscosity can be expressed as

μ = 6.89(106 )μ0 exp[(b∕(1.8C + 127))] where μ0 and b are from Table 12–1. If the viscosity μ′0 is expressed in μreyn, then μ = 6.89μ′0 exp[(b∕(1.8C + 127))] What is the viscosity of a grade 50 oil at 70°C? Compare your results with Figure 12–3.

12–20 For Problem 12–17 a satisfactory design is

+0 +0.003 d = 2.000−0.001 in  b = 2.005−0 in

Double the size of the bearing dimensions and quadruple the load to 3600 lbf. (a) Analyze the scaled-up bearing for median assembly. (b) Compare the results of a similar analysis for the 2-in bearing, median assembly.

12–21 Consider a journal bearing of length 25 mm, diameter 50 mm, and radial clearance

25 μm. The dynamic viscosity of the lubricant is 7 mPa · s. In the computing reference frame of Figure 12–34, the journal is rotating at a constant angular velocity of 200 rad/s, and the sleeve is fixed. At time t = 0, the journal is initially positioned with journal

680      Mechanical Engineering Design

eccentricity ratio components εx = −0.8, εy = −0.3. At time t = 0, a steady load with components Fx = 3000 N, Fy = 0 N is applied to the journal. (a) Calculate and plot the journal orbit in the clearance space from t = 0 to the time it takes the journal to rotate one revolution (360°). Use Euler integration and a time step equal to 0.5° of journal rotation. Indicate the location of the journal eccentricity ratio at 90° intervals on the plot. (b) In part (a), the journal has essentially reached an equilibrium position. Compare the minimum film thickness and attitude angle obtained at journal rotation of 360° with that predicted by Figures 12–15 and 12–16.

12–22 With the same bearing dimensional specifications and fluid viscosity used in Problem

12–21, the load now rotates with the journal; i.e. Fx = 3000 cos ωt, Fy = 3000 sin ωt, where ω = 200 rad/s is the steady journal angular velocity. (a) Starting the journal at the same initial position as in Problem 12–21, calculate and plot the journal orbit in the clearance space from t = 0 to the time it takes the journal to rotate one revolution (360°). Use Euler integration and a time step equal to 0.5° of journal rotation. Indicate the location of the journal eccentricity ratio at 90° intervals on the plot. (b) Describe the resulting periodic orbit shape and estimate the cyclic minimum film thickness. Compare your answer with that obtained from Problem 12–21.

12–23 With the same bearing dimensional specifications and fluid viscosity used in Problem

12–21, the load now rotates at the half speed of the journal; i.e. Fx = 3000 cos ωt∕2, Fy = 3000 sin ωt∕2, where ω = 200 rad/s is the steady journal angular velocity. (a) Starting the journal at the same initial position as in Problem 12–21, calculate and plot the journal orbit in the clearance space from t = 0 to the time it takes the journal to rotate four revolutions (1440°). Use Euler integration and a time step equal to 0.5° of journal rotation. Indicate the location of the journal eccentricity ratio at 360° intervals on the plot. (b) Describe and explain the resulting journal orbit.

12–24 In Problem 12–21, assume the journal has reached its equilibrium position under the specified steady load and journal speed. If the load is suddenly removed entirely, describe and explain the resulting journal orbit.

12–25 Using the bearing specifications in Example 12–8, prepare a plot of cyclic minimum

film thickness in a big-end connecting bearing as a function of engine speed over the range 500–8000 rev/min. Assume the peak cylinder pressure is constant over the specified speed range. Indicate on the plot the speed range where the predicted film thicknesses from the chart are applicable, and indicate the range of speeds where the bearing performance is acceptable for this application.

12–26 An Oiles SP 500 alloy brass bushing is 0.75 in long with a 0.75-in dia bore and oper-

ates in a clean environment at 70°F. The allowable wear without loss of function is 0.004 in. The radial load is 400 lbf. The shaft speed is 250 rev/min. Estimate the number of revolutions for radial wear to be 0.004 in.

12–27 Choose an Oiles SP 500 alloy brass bushing to give a maximum wear of 0.002 in

for 1000 h of use with a 200 rev/min journal and 100 lbf radial load. Use h CR = 2.7 Btu/(h · ft 2 · °F) , Tmax = 300°F, fs = 0.03, and a design factor nd = 2. The bearing is to operate in a clean environment at 70°F. Table 12–12 lists the bushing sizes available from the manufacturer.

13

Gears—General

©iStockphoto/Getty Images

Chapter Outline 13–1

Types of Gears   682

13–10

Parallel Helical Gears   696

13–2

Nomenclature  683

13–11

Worm Gears   700

13–3

Conjugate Action   684

13–12

Tooth Systems   701

13–4

Involute Properties   685

13–13

Gear Trains   703

13–5

Fundamentals  686

13–14

Force Analysis—Spur Gearing   710

13–6

Contact Ratio   689

13–15

Force Analysis—Bevel Gearing   713

13–7

Interference  690

13–8

The Forming of Gear Teeth   693

13–16  Force Analysis—Helical Gearing  716

13–9

Straight Bevel Gears   695

13–17

Force Analysis—Worm Gearing   719 681

682      Mechanical Engineering Design

This chapter addresses gear geometry, the kinematic relations, and the forces transmitted by the four principal types of gears: spur, helical, bevel, and worm gears. The forces transmitted between meshing gears supply torsional moments to shafts for motion and power transmission and create forces and moments that affect the shaft and its bearings. The next two chapters will address stress, strength, safety, and reliability of the four types of gears.

13–1  Types of Gears Spur gears, illustrated in Figure 13–1, have teeth parallel to the axis of rotation and are used to transmit motion from one shaft to another, parallel, shaft. Of all types, the spur gear is the simplest and, for this reason, will be used to develop the primary kinematic relationships of the tooth form. Helical gears, shown in Figure 13–2, have teeth inclined to the axis of rotation. Helical gears can be used for the same applications as spur gears and, when so used, are not as noisy, because of the more gradual engagement of the teeth during meshing. The inclined tooth also develops thrust loads and bending couples, which are not present with spur gearing. Sometimes helical gears are used to transmit motion between nonparallel shafts. Bevel gears, shown in Figure 13–3, have teeth formed on conical surfaces and are used mostly for transmitting motion between intersecting shafts. The figure actually illustrates straight-tooth bevel gears. Spiral bevel gears are cut so the tooth is no longer straight, but forms a circular arc. Hypoid gears are quite similar to spiral bevel gears except that the shafts are offset and nonintersecting. Worms and worm gears, shown in Figure 13–4, represent the fourth basic gear type. As shown, the worm resembles a screw. The direction of rotation of the worm gear, also called the worm wheel, depends upon the direction of rotation of the worm and upon whether the worm teeth are cut right-hand or left-hand. Worm gearsets are also made so that the teeth of one or both wrap partly around the other. Such sets are called single-enveloping and double-enveloping worm gearsets. Worm gearsets are mostly used when the speed ratios of the two shafts are quite high, say, 3 or more.

Figure 13–1

Figure 13–2

Spur gears are used to transmit rotary motion between parallel shafts.

Helical gears are used to transmit motion between parallel or nonparallel shafts.

Gears—General     683

Figure 13–4

Figure 13–3

Worm gearsets are used to transmit rotary motion between nonparallel and nonintersecting shafts.

Bevel gears are used to transmit rotary motion between intersecting shafts.

13–2  Nomenclature The terminology of spur-gear teeth is illustrated in Figure 13–5. ∙ Pitch circle is a theoretical circle upon which all calculations are usually based. The pitch circles of a pair of mating gears are tangent to each other. ∙ Pitch diameter is the diameter of the pitch circle. ∙ Pinion is the smaller of two mating gears. The larger is often called the gear. Figure 13–5

Top land

Fa

ce

w

id

th

Nomenclature of spur-gear teeth.

Addendum circle

Flank Pitch c

ircle

Tooth thickness

m

lan

d

Width of space

tto

Dedendum

Circular pitch

Clearance

Fillet radius

Bo

Addendum

Face

Dedendum circle

Clearance circle

684      Mechanical Engineering Design

∙ Circular pitch p is the distance, measured on the pitch circle, from a point on one tooth to a corresponding point on an adjacent tooth. Thus the circular pitch is equal to the sum of the tooth thickness and the width of space. ∙ Module m is the ratio of the pitch diameter to the number of teeth. The customary unit of length used is the millimeter. The module is the index of tooth size in SI. ∙ Diametral pitch P is the ratio of the number of teeth on the gear to the pitch. diameter. Thus, it is the reciprocal of the module. Since diametral pitch is used only with U.S. units, it is expressed as teeth per inch. ∙ Addendum a is the radial distance between the top land and the pitch circle. ∙ Dedendum b is the radial distance from the bottom land to the pitch circle. ∙ Whole depth ht is the sum of the addendum and the dedendum. ∙ Clearance circle is a circle that is tangent to the addendum circle of the mating gear. ∙ Clearance c is the amount by which the dedendum in a given gear exceeds the addendum of its mating gear. ∙ Backlash is the amount by which the width of a tooth space exceeds the thickness of the engaging tooth measured on the pitch circles.

You should prove for yourself the validity of the following useful relations:

P=

N d

(13–1)

m=

d N

(13–2)

p=

πd = πm N

(13–3)

pP = π

where P N d m p

= = = = =

(13–4)

diametral pitch, teeth per inch number of teeth pitch diameter, in or mm module, mm circular pitch, in or mm

13–3  Conjugate Action The following discussion assumes the teeth to be perfectly formed, perfectly smooth, and absolutely rigid. Such an assumption is, of course, unrealistic, because the application of forces will cause deflections. Mating gear teeth acting against each other to produce rotary motion are similar to cams. When the tooth profiles, or cams, are designed so as to produce a constant angular-velocity ratio during meshing, these are said to have conjugate action. In theory, at least, it is possible arbitrarily to select any profile for one tooth and then to find a profile for the meshing tooth that will give conjugate action. One of these solutions is the involute profile, which, with few exceptions, is in universal use for gear teeth and is the only one with which we will be concerned. When one curved surface pushes against another (Figure 13–6), the point of contact occurs where the two surfaces are tangent to each other (point c), and the

Gears—General     685

Figure 13–6 A

O rA

a c P

b

rB

B O

forces at any instant are directed along the common normal ab to the two curves. The line ab, representing the direction of action of the forces, is called the line of action. The line of action will intersect the line of centers O-O at some point P. The angular-velocity ratio between the two arms is inversely proportional to their radii to the point P. Circles drawn through point P from each center are called pitch circles, and the radius of each circle is called the pitch radius. Point P is called the pitch point. Figure 13–6 is useful in making another observation. A pair of gears is really a pair of cams that act through a small arc and, before running off the involute contour, are replaced by another identical pair of cams. The cams can run in either direction and are configured to transmit a constant angular-velocity ratio. If involute curves are used, the gears tolerate changes in center-to-center distance with no variation in constant angular-velocity ratio. Furthermore, the rack profiles are straight-flanked, making primary tooling simpler. To transmit motion at a constant angular-velocity ratio, the pitch point must remain fixed; that is, all the lines of action for every instantaneous point of contact must pass through the same point P. In the case of the involute profile, it will be shown that all points of contact occur on the same straight line ab, that all normals to the tooth profiles at the point of contact coincide with the line ab, and, thus, that these profiles transmit uniform rotary motion.

13–4  Involute Properties An involute curve may be generated as shown in Figure 13–7a. Cord def, held tight, is wrapped around cylinder A. Point b on the cord represents the tracing point, and as the cord is wrapped and unwrapped about the cylinder, point b will trace out the involute curve ac. The radius of the curvature of the involute varies continuously, being zero at point a and a maximum at point c. At point b the radius is equal to the distance be, since point b is instantaneously rotating about point e. Thus the generating line de is normal to the involute at all points of intersection and, at the same time, is always tangent to the cylinder A. The circle on which the involute is generated is called the base circle.

Cam A and follower B in contact. When the contacting surfaces are involute profiles, the ensuing conjugate action produces a constant angularvelocity ratio.

686      Mechanical Engineering Design

Figure 13–7

c

d

(a) Generation of an involute; (b) involute action.

b

Base circle

ω1

a

Pitch circle

e

Gear 1 + Pressure line

+

r1

rb1 ϕ

ϕ

O

O1

c

f

a g A

f

d e

ω2

b

Pitch circle

rb2 ϕ

r2 Gear 2 +

Base circle O2

Let us now examine the involute profile to see how it satisfies the requirement for the transmission of uniform motion. In Figure 13–7b, two gear blanks with fixed centers at O1 and O2 are shown having base circles whose respective radii are O1a and O2b. We now imagine that a cord is wound counterclockwise around the base circle of gear 1, pulled tight between points a and b, and wound clockwise around the base circle of gear 2. If, now, the base circles are rotated in different directions so as to keep the cord tight, a point g on the cord will trace out the involutes cd on gear 1 and ef on gear 2. The involutes are thus generated simultaneously by the tracing point. The tracing point, therefore, represents the point of contact, while the portion of the cord ab is the generating line. The point of contact moves along the generating line; the generating line does not change position, because it is always tangent to the base circles; and since the generating line is always normal to the involutes at the point of contact, the requirement for uniform motion is satisfied.

13–5  Fundamentals When two gears are in mesh, their pitch circles roll on one another without slipping. As shown in Figure 13–7b, designate the pitch radii as r1 and r2 and the angular velocities as ω1 and ω2, respectively. Then the pitch-line velocity is V = ∣ r1ω1∣ = ∣ r2 ω2∣ Thus the relation between the radii on the angular velocities is

ω1

r2

= ⎹ ω⎹ r1 2

(13–5)

Suppose now we wish to design a speed reducer such that the input speed is 1800 rev/min and the output speed is 1200 rev/min. This is a ratio of 3:2; the gear pitch diameters would be in the same ratio, for example, a 4-in pinion driving a 6-in gear. The various dimensions found in gearing are always based on the pitch circles.

Gears—General     687 Dedendum circle Base circle Pitch circle Addendum circle

Pinion (driver) O1 Angle of recess

Angle of approach

Pressure line

a b

P

Angle of recess

Angle of approach

Addendum circle Pitch circle Base circle Dedendum circle

Gear (driven)

O2

The line ab in Figure 13–7b has three names, all of which are in general use. It is called the pressure line, the generating line, and the line of action. The angle between the pressure line and the perpendicular to the line of the gear centers, O1O2, is called the pressure angle, ϕ. The pressure angle usually has values of 20° or 25°, although 14 12° was once used. The pressure line is tangent to each gear at their respective base circle, where the base radius of gear i is given by

rbi = ri cos ϕ  i = 1, 2

(13–6)

where ri is the pitch radius of gear i. Figure 13–8 contains further definitions of the gears and depicts the meshing process. The initial contact will take place when the flank of the driver comes into contact with the tip of the driven tooth. This occurs at point a where the addendum circle of the driven gear crosses the pressure line. This defines the angle of approach for each gear. As the mesh proceeds, the point of contact will slide up the side of the driving tooth so that the tip of the driver will be in contact just before contact ends. The final point of contact will therefore be where the addendum circle of the driver crosses the pressure line. This is point b in Figure 13–8. This defines the angle of recess for each gear in a manner similar to that of the angle of approach. The sum of the angle of approach and the angle of recess for either gear is called the angle of action. We may imagine a rack as a spur gear having an infinitely large pitch diameter. Therefore, the rack has an infinite number of teeth and a base circle that is an infinite distance from the pitch point. The sides of involute teeth on a rack are straight lines making an angle to the line of centers equal to the pressure angle. Figure 13–9 shows an involute rack in mesh with a pinion. Corresponding sides on involute teeth are parallel curves; the base pitch is the constant and fundamental distance between them

Figure 13–8 Tooth action.

688      Mechanical Engineering Design

Figure 13–9

Base pitch

Involute-toothed pinion and rack.

pb ϕ pc Circular pitch

Figure 13–10 Internal gear and pinion.

Pressure line Pitch circle

Dedendum circle

ω3 3 Base circle

2

Pitch circle Base circle

Addendum circle

ω2 O2

along a common normal as shown in Figure 13–9. The base pitch is related to the circular pitch by the equation

pb = pc cos ϕ

(13–7)

where pb is the base pitch. Figure 13–10 shows a pinion in mesh with an internal, or ring, gear. Note that both of the gears now have their centers of rotation on the same side of the pitch point. Thus the positions of the addendum and dedendum circles with respect to the pitch circle are reversed; the addendum circle of the internal gear lies inside the pitch circle. Note, too, from Figure 13–10, that the base circle of the internal gear lies inside the pitch circle near the addendum circle. Another interesting observation concerns the fact that the operating diameters of the pitch circles of a pair of meshing gears need not be the same as the respective design pitch diameters of the gears. If we increase the center distance, we create two new operating pitch circles having larger diameters because they must be tangent to each other at the pitch point. Thus the pitch circles of gears really do not come into existence until a pair of gears are brought into mesh. Changing the center distance has no effect on the base circles, because these were used to generate the tooth profiles. Thus the base circle is basic to a gear. Increasing the center distance increases the pressure angle and decreases the length of the line of action, but the teeth are still conjugate, the requirement for uniform motion transmission is still satisfied, and the angular-velocity ratio has not changed.

Gears—General     689

EXAMPLE 13–1 A gearset consists of a 16-tooth pinion driving a 40-tooth gear. The diametral pitch is 2, and the addendum and dedendum are 1∕P and 1.25∕P, respectively. The gears are cut using a pressure angle of 20°. (a) Compute the circular pitch, the center distance, and the radii of the base circles. (b) In mounting these gears, the center distance was incorrectly made 14 in larger. Compute the new values of the pressure angle and the pitch-circle diameters. Solution Answer  (a)

p=

π π = = 1.571 in P 2

The pitch diameters of the pinion and gear are, respectively,

dP =

NP 16 = = 8 in P 2

dG =

NG 40 = = 20 in P 2

Therefore the center distance is Answer

dP + d G 8 + 20 = = 14 in 2 2

From Equation (13–6), with a 20° pressure angle, the base radii are Answer

(rb ) pinion =

Answer

(rb ) gear =

8 cos 20° = 3.759 in 2

20 cos 20° = 9.397 in 2

(b) Designating d′P and d′G as the new pitch-circle diameters, the 14 -in increase in the center distance requires that d′P + d′G = 14.250 2

(1)

Also, the velocity ratio does not change, and hence d′P 16 = d′G 40

(2)

Solving Equations (1) and (2) simultaneously yields Answer

d′P = 8.143 in

d′G = 20.357 in

Since rb = r cos ϕ, using either the pinion or gear, the new pressure angle is Answer

ϕ′ = cos−1

(rb ) pinion d′P∕2

= cos−1

3.759 = 22.59° 8.143∕2

13–6  Contact Ratio The zone of action of meshing gear teeth is shown in Figure 13–11. We recall that tooth contact begins and ends at the intersections of the two addendum circles with the pressure line. In Figure 13–11 initial contact occurs at a and final contact at b. Tooth profiles drawn through these points intersect the pitch circle at A and B, respectively. As shown, the distance AP is called the arc of approach qa, and the distance PB, the arc of recess qr. The sum of these is the arc of action qt.

690      Mechanical Engineering Design

Figure 13–11

Arc of recess qr

Arc of approach qa

Definition of contact ratio.

sure

Pres

line

b

ϕ

P A

B Addendum circle

a Addendum circle

Motion

Pitch circle

Lab

Now, consider a situation in which the arc of action is exactly equal to the circular pitch, that is, qt = p. This means that one tooth and its space will occupy the entire arc AB. In other words, when a tooth is just beginning contact at a, the previous tooth is simultaneously ending its contact at b. Therefore, during the tooth action from a to b, there will be exactly one pair of teeth in contact. Next, consider a situation in which the arc of action is greater than the circular pitch, but not very much greater, say, qt ≈ 1.2p. This means that when one pair of teeth is just entering contact at a, another pair, already in contact, will not yet have reached b. Thus, for a short period of time, there will be two teeth in contact, one in the vicinity of A and another near B. As the meshing proceeds, the pair near B must cease contact, leaving only a single pair of contacting teeth, until the procedure repeats itself. Because of the nature of this tooth action, either one or two pairs of teeth in contact, it is convenient to define the term contact ratio mc as

mc =

qt p

(13–8)

a number that indicates the average number of pairs of teeth in contact. Note that this ratio is also equal to the length of the path of contact divided by the base pitch. Gears should not generally be designed having contact ratios less than about 1.20, because inaccuracies in mounting might reduce the contact ratio even more, increasing the possibility of impact between the teeth as well as an increase in the noise level. An easier way to obtain the contact ratio is to measure the line of action ab instead of the arc distance AB. Since ab in Figure 13–11 is tangent to the base circle when extended, the base pitch pb must be used to calculate mc instead of the circular pitch as in Equation (13–8). If the length of the line of action is Lab, the contact ratio is

mc =

L ab p cos ϕ

(13–9)

in which Equation (13–7) was used for the base pitch.

13–7  Interference The contact of portions of tooth profiles that are not conjugate is called interference. Consider Figure 13–12. Illustrated are two 16-tooth gears that have been cut to the now obsolete 14 12° pressure angle. The driver, gear 2, turns clockwise. The initial and final points of contact are designated A and B, respectively, and are located on the pressure line. Now notice that the points of tangency of the pressure line with the base circles C and D are located inside of points A and B. Interference is present.

Gears—General     691

Figure 13–12 ω3

Interference in the action of gear teeth.

O3 Driven gear 3

Base circle This portion of profile is not an involute

D Pressure line

B Addendum circles

C A

Interference is on flank of driver during approach

Base circle

This portion of profile is not an involute

Driving gear 2 ω2

O2

The interference is explained as follows. Contact begins when the tip of the driven tooth contacts the flank of the driving tooth. In this case the flank of the driving tooth first makes contact with the driven tooth at point A, and this occurs before the involute portion of the driving tooth comes within range. In other words, contact is occurring below the base circle of gear 2 on the noninvolute portion of the flank. The actual effect is that the involute tip or face of the driven gear tends to dig out the noninvolute flank of the driver. In this example the same effect occurs again as the teeth leave contact. Contact should end at point D or before. Since it does not end until point B, the effect is for the tip of the driving tooth to dig out, or interfere with, the flank of the driven tooth. When gear teeth are produced by a generation process, interference is automatically eliminated because the cutting tool removes the interfering portion of the flank. This effect is called undercutting; if undercutting is at all pronounced, the undercut tooth is considerably weakened. Thus the effect of eliminating interference by a generation process is merely to substitute another problem for the original one. The smallest number of teeth on a spur pinion and gear,1 one-to-one gear ratio, which can exist without interference is NP. This number of teeth for spur gears is given by 2k NP = (1 + √1 + 3 sin2 ϕ) (13–10) 3 sin2 ϕ where k = 1 for full-depth teeth, 0.8 for stub teeth and ϕ = pressure angle. 1

Robert Lipp, "Avoiding Tooth Interference in Gears," Machine Design, vol. 54, no. 1, 1982, pp. 122–124.

692      Mechanical Engineering Design

For a 20° pressure angle, with k = 1, NP =

2(1)

3 sin2 20°

(1 + √1 + 3 sin2 20°) = 12.3 = 13 teeth

Thus 13 teeth on pinion and gear are interference-free. Realize that 12.3 teeth is possible in meshing arcs, but for fully rotating gears, 13 teeth represents the least number. For a 14 12° pressure angle, NP = 23 teeth, so one can appreciate why few 14 12° -tooth systems are used, as the higher pressure angles can produce a smaller pinion with accompanying smaller center-to-center distances. If the mating gear has more teeth than the pinion, that is, mG = NG∕NP = m is more than one, then the smallest number of teeth on the pinion without interference is given by

NP =

2k (m + √m2 + (1 + 2m) sin2 ϕ) (1 + 2m) sin2 ϕ

(13–11)

For example, if m = 4, ϕ = 20°, NP =

2(1)

[1 + 2(4)] sin2 20°

[4 + √42 + [1 + 2(4)] sin2 20°] = 15.4 = 16 teeth

Thus a 16-tooth pinion will mesh with a 64-tooth gear without interference. The largest gear with a specified pinion that is interference-free is

NG =

NP2 sin2 ϕ − 4k 2 4k − 2NP sin2 ϕ

(13–12)

For example, for a 13-tooth pinion with a pressure angle ϕ of 20°, NG =

132 sin2 20° − 4(1) 2

4(1) − 2(13) sin2 20°

= 16.45 = 16 teeth

For a 13-tooth spur pinion, the maximum number of gear teeth possible without interference is 16. The smallest spur pinion that will operate with a rack without interference is

NP =

2(k)

sin2 ϕ

(13–13)

For a 20° pressure angle full-depth tooth the smallest number of pinion teeth to mesh with a rack is NP =

2(1)

sin2 20°

= 17.1 = 18 teeth

Since gear-shaping tools amount to contact with a rack, and the gear-hobbing process is similar, the minimum number of teeth to prevent interference to prevent undercutting by the hobbing process is equal to the value of NP when N G is infinite. The importance of the problem of teeth that have been weakened by undercutting cannot be overemphasized. Of course, interference can be eliminated by using more teeth on the pinion. However, if the pinion is to transmit a given amount of power, more teeth can be used only by increasing the pitch diameter.

Gears—General     693

Interference can also be reduced by using a larger pressure angle. This results in a smaller base circle, so that more of the tooth profile becomes involute. The demand for smaller pinions with fewer teeth thus favors the use of a 25° pressure angle even though the frictional forces and bearing loads are increased and the contact ratio decreased.

13–8  The Forming of Gear Teeth There are a large number of ways of forming the teeth of gears, such as sand casting, shell molding, investment casting, permanent-mold casting, die casting, and centrifugal casting. Teeth can also be formed by using the powder-metallurgy process; or, by using extrusion, a single bar of aluminum may be formed and then sliced into gears. Gears that carry large loads in comparison with their size are usually made of steel and are cut with either form cutters or generating cutters. In form cutting, the tooth space takes the exact form of the cutter. In generating, a tool having a shape different from the tooth profile is moved relative to the gear blank so as to obtain the proper tooth shape. One of the newest and most promising of the methods of forming teeth is called cold forming, or cold rolling, in which dies are rolled against steel blanks to form the teeth. The mechanical properties of the metal are greatly improved by the rolling process, and a high-quality generated profile is obtained at the same time. Gear teeth may be machined by milling, shaping, or hobbing. They may be finished by shaving, burnishing, grinding, or lapping. Gears made of thermoplastics such as nylon, polycarbonate, and acetal are quite popular and are easily manufactured by injection molding. These gears are of low to moderate precision, low in cost for high production quantities, and capable of light loads, and can run without lubrication. Milling Gear teeth may be cut with a form milling cutter shaped to conform to the tooth space. With this method it is theoretically necessary to use a different cutter for each gear, because a gear having 25 teeth, for example, will have a different-shaped tooth space from one having, say, 24 teeth. Actually, the change in space is not too great, and it has been found that eight cutters may be used to cut with reasonable accuracy any gear in the range of 12 teeth to a rack. A separate set of cutters is, of course, required for each pitch. Shaping Teeth may be generated with either a pinion cutter or a rack cutter. The pinion cutter (Figure 13–13) reciprocates along the vertical axis and is slowly fed into the gear blank to the required depth. When the pitch circles are tangent, both the cutter and the blank rotate slightly after each cutting stroke. Since each tooth of the cutter is a cutting tool, the teeth are all cut after the blank has completed one rotation. The sides of an involute rack tooth are straight. For this reason, a rack-generating tool provides an accurate method of cutting gear teeth. This is also a shaping operation and is illustrated by the drawing of Figure 13–14. In operation, the cutter reciprocates and is first fed into the gear blank until the pitch circles are tangent. Then, after each cutting stroke, the gear blank and cutter roll slightly on their pitch circles. When the blank and cutter have rolled a distance equal to the circular pitch, the cutter is returned to the starting point, and the process is continued until all the teeth have been cut.

694      Mechanical Engineering Design

Figure 13–13 Generating a spur gear with a pinion cutter. (Courtesy of Boston Gear­—Altra Industrial Motion)

Figure 13–14 Shaping teeth with a rack.

Gear blank rotates in this direction

Rack cutter reciprocates in a direction perpendicular to this page

Hobbing The hobbing process is illustrated in Figure 13–15. The hob is simply a cutting tool that is shaped like a worm. The teeth have straight sides, as in a rack, but the hob axis must be turned through the lead angle in order to cut spur-gear teeth. For this reason, the teeth generated by a hob have a slightly different shape from those generated by a rack cutter. Both the hob and the blank must be rotated at the proper angular-velocity ratio. The hob is then fed slowly across the face of the blank until all the teeth have been cut.

Gears—General     695

Figure 13–15 Hobbing a spur gear. (©Dmitry Kalinovsky/Shutterstock)

Finishing Gears that run at high speeds and transmit large forces may be subjected to additional dynamic forces if there are errors in tooth profiles. Errors may be diminished somewhat by finishing the tooth profiles. The teeth may be finished, after cutting, by either shaving or burnishing. Several shaving machines are available that cut off a minute amount of metal, bringing the accuracy of the tooth profile within the limits of 250 μin. Burnishing, like shaving, is used with gears that have been cut but not heattreated. In burnishing, hardened gears with slightly oversize teeth are run in mesh with the gear until the surfaces become smooth. Grinding and lapping are used for hardened gear teeth after heat treatment. The grinding operation employs the generating principle and produces very accurate teeth. In lapping, the teeth of the gear and lap slide axially so that the whole surface of the teeth is abraded equally.

13–9  Straight Bevel Gears When gears are used to transmit motion between intersecting shafts, some form of bevel gear is required. A bevel gearset is shown in Figure 13–16. Although bevel gears are usually made for a shaft angle of 90°, they may be produced for almost any angle. The teeth may be cast, milled, or generated. Only the generated teeth may be classed as accurate. The terminology of bevel gears is illustrated in Figure 13–16. The pitch of bevel gears is measured at the large end of the tooth, and both the circular pitch and the pitch diameter are calculated in the same manner as for spur gears. It should be noted that the clearance is uniform. The pitch angles are defined by the pitch cones meeting at the apex, as shown in the figure. They are related to the tooth numbers as follows:

tan γ =

NP NG

tan Γ =

NG NP

(13–14)

where the subscripts P and G refer to the pinion and gear, respectively, and where γ and Γ are, respectively, the pitch angles of the pinion and gear.

696      Mechanical Engineering Design

Figure 13–16 Pitch angle

Terminology of bevel gears.

γ

e Ao

ne

Co

F

nc ista

d

Face

Γ

Uniform clearance

Pitch angle

Pitch diameter DG Back cone

Back-cone radius, r b

Figure 13–16 shows that the shape of the teeth, when projected on the back cone, is the same as in a spur gear having a radius equal to the back-cone distance rb. This is called Tredgold's approximation. The number of teeth in this imaginary gear is

N′ =

2π rb p

(13–15)

where N′ is the virtual number of teeth and p is the circular pitch measured at the large end of the teeth. Standard straight-tooth bevel gears are cut by using a 20° pressure angle, unequal addenda and dedenda, and full-depth teeth. This increases the contact ratio, avoids undercut, and increases the strength of the pinion.

13–10  Parallel Helical Gears Helical gears, used to transmit motion between parallel shafts, are shown in Figure 13–2. The helix angle is the same on each gear, but one gear must have a right-hand helix and the other a left-hand helix. The shape of the tooth is an involute helicoid and is illustrated in Figure 13–17. If a piece of paper cut in the shape of a parallelogram is wrapped around a cylinder, the angular edge of the paper becomes a helix. If we unwind this paper, each point on the angular edge generates an involute curve. This surface obtained when every point on the edge generates an involute is called an involute helicoid. The initial contact of spur-gear teeth is a line extending all the way across the face of the tooth. The initial contact of helical-gear teeth is a point that extends into a line as the teeth come into more engagement. In spur gears the line of contact is parallel to the axis of rotation; in helical gears the line is diagonal across the face of the tooth. It is this gradual engagement of the teeth and the smooth transfer of load from one tooth to another that gives helical gears the ability to transmit heavy loads

Gears—General     697

(a) ϕn Section B–B

b

pn

d ψ

A

px

Involute

e Edge of paper

a

ψ pt

Base helix angle Section A–A

Figure 13–17

Figure 13–18

An involute helicoid.

Nomenclature of helical gears.

at high speeds. Because of the nature of contact between helical gears, the contact ratio is of only minor importance, and it is the contact area, which is proportional to the face width of the gear, that becomes significant. Helical gears subject the shaft bearings to both radial and thrust loads. When the thrust loads become high or are objectionable for other reasons, it may be desirable to use double helical gears. A double helical gear (herringbone) is equivalent to two helical gears of opposite hand, mounted side by side on the same shaft. They develop opposite thrust reactions and thus cancel out the thrust load. When two or more single helical gears are mounted on the same shaft, the hand of the gears should be selected so as to produce the minimum thrust load. Figure 13–18 represents a portion of the top view of a helical rack. Lines ab and cd are the centerlines of two adjacent helical teeth taken on the same pitch plane. The angle ψ is the helix angle. The distance ac is the transverse circular pitch pt in the plane of rotation (usually called the circular pitch). The distance ae is the normal circular pitch pn and is related to the transverse circular pitch as follows: pn = pt cos ψ

(13–16)

The distance ad is called the axial pitch px and is related by the expression px =

pt tan ψ

(13–17)

Since pn Pn = π, the normal diametral pitch is

Pn =

Pt cos ψ

c

(c)

Base cylinder

(b)

ϕt

B

A B

(13–18)

R

698      Mechanical Engineering Design

Figure 13–19 ψ

A cylinder cut by an oblique plane.

(a)

b

a +

D R

(b)

(a)

The pressure angle ϕn in the normal direction is different from the pressure angle ϕt in the direction of rotation, because of the angularity of the teeth. These angles are related by the equation D

+

cos ψ =

tan ϕn tan ϕt

(13–19)

Figure 13–19 illustrates a cylinder cut by an oblique plane ab at an angle ψ to a (b) right section. The oblique plane cuts out an arc having a radius of curvature of R. For the condition that ψ = 0, the radius of curvature is R = D∕2. If we imagine the angle ψ to be slowly increased from zero to 90°, we see that R begins at a value of D∕2 and increases until, when ψ = 90°, R = ∞. The radius R is the apparent pitch radius of a helical-gear tooth when viewed in the direction of the tooth elements. A gear of the same pitch and with the radius R will have a greater number of teeth, because of the increased radius. In helical-gear terminology this is called the virtual number of teeth. It can be shown by analytical geometry that the virtual number of teeth is related to the actual number by the equation

N′ =

N cos3 ψ

(13–20)

where N′ is the virtual number of teeth and N is the actual number of teeth. It is necessary to know the virtual number of teeth in design for strength and also, sometimes, in cutting helical teeth. This apparently larger radius of curvature means that few teeth may be used on helical gears, because there will be less undercutting. EXAMPLE 13–2 A stock helical gear has a normal pressure angle of 20°, a helix angle of 25°, and a transverse diametral pitch of 6 teeth/in, and has 18 teeth. Find: (a) The pitch diameter (b) The transverse, the normal, and the axial pitches (c) The normal diametral pitch (d) The transverse pressure angle Solution Answer  (a)

d=

N 18 = = 3 in Pt 6

Gears—General     699

Answer  (b)

pt =

π π = = 0.5236 in Pt 6

pn = pt cos ψ = 0.5236 cos 25° = 0.4745 in

Answer

px =

Answer Answer  (c)

Pn =

pt 0.5236 = = 1.123 in tan ψ tan 45°

Pt 6 = = 6.620 teeth/in cos ψ cos 25°

tan ϕn tan 20° ϕt = tan−1 ( = tan−1 ( = 21.88° ) cos ψ cos 25° )

Answer  (d)

Just like teeth on spur gears, helical-gear teeth can interfere. Equation (13–19) can be solved for the pressure angle ϕt in the tangential (rotation) direction to give ϕt = tan−1 (

tan ϕn cos ψ )

The smallest tooth number NP of a helical-spur pinion that will run without interference2 with a gear with the same number of teeth is

NP =

2k cos ψ 3 sin2 ϕt

(1 + √1 + 3 sin2 ϕt )

(13–21)

For example, if the normal pressure angle ϕn is 20°, the helix angle ψ is 30°, then ϕt is tan 20° ϕt = tan−1( = 22.80° cos 30° )

NP =

2(1) cos 30°

3 sin2 22.80°

(1 + √1 + 3 sin2 22.80°) = 8.48 = 9 teeth

For a given gear ratio mG = NG∕NP = m, the smallest pinion tooth count is NP =

2k cos ψ (1 + 2m) sin2 ϕt

[m + √m2 + (1 + 2m) sin2 ϕt ]

The largest gear with a specified pinion is given by

NG =

NP2 sin2 ϕt − 4k 2 cos2 ψ 4k cos ψ − 2NP sin2 ϕt

(13–22)

(13–23)

For example, for a nine-tooth pinion with a pressure angle ϕn of 20°, a helix angle ψ of 30°, and recalling that the tangential pressure angle ϕt is 22.80°, NG =

92 sin2 22.80° − 4(1) 2 cos2 30°

4(1) cos 30° − 2(9) sin2 22.80°

= 12.02 = 12

The smallest pinion that can be run with a rack is 2

NP =

2k cos ψ sin2 ϕt

Op. cit., Robert Lipp, Machine Design, pp. 122–124.

(13–24)

700      Mechanical Engineering Design

For a normal pressure angle ϕn of 20° and a helix angle ψ of 30°, and ϕt = 22.80°, NP =

2(1) cos 30° sin2 22.80°

= 11.5 = 12 teeth

For helical-gear teeth the number of teeth in mesh across the width of the gear will be greater than unity and a term called face-contact ratio is used to describe it. This increase of contact ratio, and the gradual sliding engagement of each tooth, results in quieter gears.

13–11  Worm Gears The nomenclature of a worm gearset is shown in Figure 13–20. The worm and worm gear of a set have the same hand of helix as for crossed helical gears, but the helix angles are usually quite different. The helix angle on the worm is generally quite large, and that on the gear very small. Because of this, it is usual to specify the lead angle λ on the worm and helix angle ψG on the gear; the two angles are equal for a 90° shaft angle. The worm lead angle is the complement of the worm helix angle, as shown in Figure 13–20. In specifying the pitch of worm gearsets, it is customary to state the axial pitch px of the worm and the transverse circular pitch pt, often simply called the circular pitch, of the mating gear. These are equal if the shaft angle is 90°. The pitch diameter of the gear is the diameter measured on a plane containing the worm axis, as shown in Figure 13–20; it is the same as for spur gears and is

dG =

NG pt π

(13–25)

Since it is not related to the number of teeth, the worm may have any pitch diameter; this diameter should, however, be the same as the pitch diameter of the hob used Figure 13–20

Root diameter

Pitch cylinder

Helix

ψW, helix angle Worm

Axial pitch px

Lead angle λ Lead L Pitch diameter dG

Nomenclature of a singleenveloping worm gearset.

Pitch diameter dw

Worm gear

Gears—General     701

to cut the worm-gear teeth. Generally, the pitch diameter of the worm should be selected so as to fall into the range C 0.875 C 0.875 ≤ dW ≤ 3.0 1.7

(13–26)

where C is the center distance. These proportions appear to result in optimum horsepower capacity of the gearset. The lead L and the lead angle λ of the worm have the following relations:

L = px NW tan λ =

(13–27)

L π dW

(13–28)

13–12  Tooth Systems3 A tooth system is a standard that specifies the relationships involving addendum, dedendum, working depth, tooth thickness, and pressure angle. The standards were originally planned to attain interchangeability of gears of all tooth numbers, but of the same pressure angle and pitch. Table 13–1 contains the standards most used for spur gears. A 1412° pressure angle was once used for these but is now obsolete; the resulting gears had to be comparatively larger to avoid interference problems. Table 13–2 is particularly useful in selecting the pitch or module of a gear. Cutters are generally available for the sizes shown in this table. Table 13–1  Standard and Commonly Used Tooth Systems for Spur Gears Tooth System

Pressure Angle ϕ, deg

Addendum a Dedendum b

20

1∕P or m 1.25∕P or 1.25m

Full depth

1.35∕P or 1.35m 2212 1∕P or m 1.25∕P or 1.25m

1.35∕P or 1.35m

25 1∕P or m 1.25∕P or 1.25m

1.35∕P or 1.35m Stub

20 0.8∕P or 0.8m 1∕P    or m

Table 13–2  Tooth Sizes in General Uses Diametral Pitch P (teeth/in) Coarse 2, 214 , 212 , 3, 4, 6, 8, 10, 12, 16 Fine

20, 24, 32, 40, 48, 64, 80, 96, 120, 150, 200

Module m (mm/tooth) Preferred

1, 1.25, 1.5, 2, 2.5, 3, 4, 5, 6, 8, 10, 12, 16, 20, 25, 32, 40, 50

Next Choice 1.125, 1.375, 1.75, 2.25, 2.75, 3.5, 4.5, 5.5, 7, 9, 11, 14, 18, 22, 28, 36, 45 3

Standardized by the American Gear Manufacturers Association (AGMA). Write AGMA for a complete list of standards, because changes are made from time to time. The address is: 1001 N. Fairfax Street, Suite 500, Alexandria, VA 22314-1587; or, www.agma.org.

702      Mechanical Engineering Design

Table 13–3 lists the standard tooth proportions for straight bevel gears. These sizes apply to the large end of the teeth. The nomenclature is defined in Figure 13–16. Standard tooth proportions for helical gears are listed in Table 13–4. Tooth proportions are based on the normal pressure angle; these angles are standardized Table 13–3  Tooth Proportions for 20° Straight Bevel-Gear Teeth Item Formula Working depth

hk = 2.0∕P

Clearance

c = (0.188∕P) + 0.002 in

Addendum of gear

aG =

Gear ratio

mG = NG∕NP

Equivalent 90° ratio

m90 = mG when Γ = 90°

m 90 = √ mG

Face width

F = 0.3A0 or F =

Minimum number of teeth

Pinion  16  15  14  13

Gear     16  17  20  30

0.54 0.460 + P P(m90 ) 2

cos γ when Γ ≠ 90° cos Γ 10 , whichever is smaller P

Table 13–4  Standard Tooth Proportions for Helical Gears Quantity*

Formula

Quantity* Formula

Addendum

1.00 Pn

Dedendum

1.25 D+d   Standard center distance Pn 2

Pinion pitch diameter

NP   Gear outside diameter Pn cos ψ

D + 2a

Gear pitch diameter

NG   Pinion outside diameter Pn cos ψ

d + 2a

Normal arc tooth thickness†

Bn π −   Gear root diameter 2Pn 2

D − 2b

Pinion base diameter

d cos ϕt   Pinion root diameter

d − 2b

External gears:

Internal gears:

Gear base diameter

D cos ϕt   Center distance

D−d 2

Base helix angle

tan−1 (tan ψ cos ϕt)   Inside diameter

D − 2a

  Root diameter *All dimensions are in inches, and angles are in degrees. † Bn is the normal backlash.

D + 2b

Gears—General     703

Table 13–5  Recommended Pressure Angles and Tooth Depths for Worm Gearing Lead Angle λ, deg

Pressure Angle ϕn, deg

Addendum a

Dedendum bG

 0–15

1412 0.3683px 0.3683px

15–30

20

0.3683px 0.3683px

30–35

25

0.2865px 0.3314px

35–40

25

0.2546px 0.2947px

40–45

30

0.2228px 0.2578px

the same as for spur gears. Though there will be exceptions, the face width of helical gears should be at least 2 times the axial pitch to obtain good helical-gear action. Tooth forms for worm gearing have not been highly standardized, perhaps because there has been less need for it. The pressure angles used depend upon the lead angles and must be large enough to avoid undercutting of the worm-gear tooth on the side at which contact ends. A satisfactory tooth depth, which remains in about the right proportion to the lead angle, may be obtained by making the depth a proportion of the axial circular pitch. Table 13–5 summarizes what may be regarded as good practice for pressure angle and tooth depth. The face width FG of the worm gear should be made equal to the length of a tangent to the worm pitch circle between its points of intersection with the addendum circle, as shown in Figure 13–21.

13–13  Gear Trains Consider a pinion 2 driving a gear 3. The speed of the driven gear is

n3 =

∣ ∣ ∣ ∣

N2 d2 n2 = n2 N3 d3

(13–29)

where n = revolutions or rev/min N = number of teeth d = pitch diameter Equation (13–29) applies to any gearset no matter whether the gears are spur, helical, bevel, or worm. The absolute-value signs are used to permit complete freedom in choosing positive and negative directions. In the case of spur and parallel helical gears, the directions in the viewing plane ordinarily correspond to the right-hand rule—positive for counterclockwise rotation and negative for clockwise rotation. Rotational directions are somewhat more difficult to deduce for worm and crossed helical gearsets. Figure 13–22 will be of help in these situations. The gear train shown in Figure 13–23 is made up of five gears. Considering gear 2 to be the primary driving gear, the speed of gear 6 is

n6 = −

N2 N3 N5 n2 N3 N4 N6

(a)

FG

Figure 13–21 A graphical depiction of the face width of the worm of a worm gearset.

704      Mechanical Engineering Design

Figure 13–22 Thrust, rotation, and hand relations for crossed helical gears. Note that each pair of drawings refers to a single gearset. These relations also apply to worm gearsets. (Reproduced by permission, Boston Gear Division, Colfax Corp.)

Driver

Thrust bearing

Driver

(a)

Driver

Thrust bearing

Driver

(c)

Figure 13–23

(d)

Left hand

N4

A gear train.

(b)

Right hand

N5

N6

n2

2

+

3

+

4

+ 5

N2

6

+ n6

N3

Hence we notice that gear 3 is an idler, that its tooth numbers cancel in Equation (a), and hence that it affects only the direction of rotation of gear 6. We notice, furthermore, that gears 2, 3, and 5 are drivers, while 3, 4, and 6 are driven members. We define the train value e as

e=±

product of driving tooth numbers product of driven tooth numbers

(13–30)

Note that pitch diameters can be used in Equation (13–30) as well. When Equation (13–30) is used for spur gears, e is positive if the last gear rotates in the same sense as the first, and negative if the last rotates in the opposite sense. Another way of establishing the sign is to count the number of meshes. If the number is odd, then e is negative. If the number is even, then e is positive. Do not count the mesh when a mesh involves an internal gear such as the ring gear in Figure 13–26. Now we can write

n L = enF

(13–31)

where nL is the speed of the last gear in the train and nF is the speed of the first. As a rough guideline, a train value of up to 10 to 1 can be obtained with one pair of gears. Greater ratios can be obtained in less space and with fewer dynamic

Gears—General     705

Figure 13–24 A two-stage compound gear train.

problems by compounding additional pairs of gears. A two-stage compound gear train, such as shown in Figure 13–24, can obtain a train value of up to 100 to 1. The design of gear trains to accomplish a specific train value is straightforward. Since numbers of teeth on gears must be integers, it is better to determine them first, and then obtain pitch diameters second. Determine the number of stages necessary to obtain the overall ratio, then divide the overall ratio into portions to be accomplished in each stage. To minimize package size, keep the portions as evenly divided between the stages as possible. In cases where the overall train value need only be approximated, each stage can be identical. For example, in a two-stage compound gear train, assign the square root of the overall train value to each stage. If an exact train value is needed, attempt to factor the overall train value into integer components for each stage. Then assign the smallest gear(s) to the minimum number of teeth allowed for the specific ratio of each stage, in order to avoid interference (see Section 13–7). Finally, applying the ratio for each stage, determine the necessary number of teeth for the mating gears. Round to the nearest integer and check that the resulting overall ratio is within acceptable tolerance.

EXAMPLE 13–3 A gearbox is needed to provide a 30:1 (±1 percent) increase in speed, while minimizing the overall gearbox size. Specify appropriate tooth numbers. Solution Since the ratio is greater than 10:1, but less than 100:1, a two-stage compound gear train, such as in Figure 13–24, is needed. The portion to be accomplished in each stage is √30 = 5.4772. For this ratio, assuming a typical 20°  pressure angle, the minimum number of teeth to avoid interference is 16, according to Equation (13–11). The number of teeth necessary for the mating gears is Answer

16 √30 = 87.64 ≈ 88

From Equation (13–30), the overall train value is e = (88∕16)(88∕16) = 30.25 This is within the 1 percent tolerance. If a closer tolerance is desired, then increase the pinion size to the next integer and try again.

706      Mechanical Engineering Design

EXAMPLE 13–4 A gearbox is needed to provide an exact 30:1 increase in speed, while minimizing the overall gearbox size. Specify appropriate teeth numbers. Solution The previous example demonstrated the difficulty with finding integer numbers of teeth to provide an exact ratio. In order to obtain integers, factor the overall ratio into two integer stages. e = 30 = (6) (5) N2∕N3 = 6

and

N4∕N5 = 5

With two equations and four unknown numbers of teeth, two free choices are available. Choose N3 and N5 to be as small as possible without interference. Assuming a 20° pressure angle, Equation (13–11) gives the minimum as 16. Then N2 = 6N3 = 6(16) = 96 N4 = 5N5 = 5(16) = 80 The overall train value is then exact. e = (96∕16)(80∕16) = (6) (5) = 30 It is sometimes desirable for the input shaft and the output shaft of a two-stage compound gear train to be in-line, as shown in Figure 13–25. This configuration is called a compound reverted gear train. This requires the distances between the shafts to be the same for both stages of the train, which adds to the complexity of the design task. The distance constraint is d2∕2 + d3∕2 = d4∕2 + d5∕2 The diametral pitch relates the diameters and the numbers of teeth, P = N∕d. Replacing all the diameters gives N2∕(2P) + N3∕(2P) = N4∕(2P) + N5∕(2P) Assuming a constant diametral pitch in both stages, we have the geometry condition stated in terms of numbers of teeth: N2 + N3 = N4 + N5 This condition must be exactly satisfied, in addition to the previous ratio equations, to provide for the in-line condition on the input and output shafts. Figure 13–25

2

2

A compound reverted gear train.

5

5

3

3 4

4

Gears—General     707

EXAMPLE 13–5 A gearbox is needed to provide an exact 30:1 increase in speed, while minimizing the overall gearbox size. The input and output shafts should be in-line. Specify appropriate teeth numbers. Solution The governing equations are N2∕N3 = 6 N4∕N5 = 5 N2 + N3 = N4 + N5 With three equations and four unknown numbers of teeth, only one free choice is available. Of the two smaller gears, N3 and N5, the free choice should be used to minimize N3 since a greater gear ratio is to be achieved in this stage. To avoid interference, the minimum for N3 is 16. Applying the governing equations yields N2 = 6N3 = 6(16) = 96 N2 + N3 = 96 + 16 = 112 = N4 + N5 Substituting N4 = 5N5 gives 112 = 5N5 + N5 = 6N5 N5 = 112∕6 = 18.67 If the train value need only be approximated, then this can be rounded to the nearest integer. But for an exact solution, it is necessary to choose the initial free choice for N3 such that solution of the rest of the teeth numbers results exactly in integers. This can be done by trial and error, letting N3 = 17, then 18, etc., until it works. Or, the problem can be normalized to quickly determine the minimum free choice. Beginning again, let the free choice be N3 = 1. Applying the governing equations gives N2 = 6N3 = 6(1) = 6 N2 + N3 = 6 + 1 = 7 = N4 + N5 Substituting N4 = 5N5, we find 7 = 5N5 + N5 = 6N5 N5 = 7∕6 This fraction could be eliminated if it were multiplied by a multiple of 6. The free choice for the smallest gear N3 should be selected as a multiple of 6 that is greater than the minimum allowed to avoid interference. This would indicate that N3 = 18. Repeating the application of the governing equations for the final time yields N2 = 6N3 = 6(18) = 108 N2 + N3 = 108 + 18 = 126 = N4 + N5 126 = 5N5 + N5 = 6N5 N5 = 126∕6 = 21 N4 = 5N5 = 5(21) = 105

708      Mechanical Engineering Design

Thus, Answer

N2 = 108

N3 = 18

N4 = 105

N5 = 21

Checking, we calculate e = (108∕18)(105∕21) = (6)(5) = 30. And checking the geometry constraint for the in-line requirement, we calculate

N2 + N3 = N4 + N5

108 + 18 = 105 + 21

126 = 126 Unusual effects can be obtained in a gear train by permitting some of the gear axes to rotate about others. Such trains are called planetary, or epicyclic, gear trains. Planetary trains always include a sun gear, a planet carrier or arm, and one or more planet gears, as shown in Figure 13–26. Planetary gear trains are unusual mechanisms because they have two degrees of freedom; that is, for constrained motion, a planetary train must have two inputs. For example, in Figure 13–26 these two inputs could be the motion of any two of the elements of the train. We might, say, specify that the sun gear rotates at 100 rev/min clockwise and that the ring gear rotates at 50 rev/min counterclockwise; these are the inputs. The output would be the motion of the arm. In most planetary trains one of the elements is attached to the frame and has no motion. Figure 13–27 shows a planetary train composed of a sun gear 2, an arm or carrier 3, and planet gears 4 and 5. The angular velocity of gear 2 relative to the arm in rev/min is

n23 = n2 − n3

(b)

80T 5 Sun gear

2

Arm

30T

4 3

Arm 2 5

3

20T

4

Planet gear

Ring gear

Figure 13–26 A planetary gear train.

Figure 13–27 A gear train on the arm of a planetary gear train.

Gears—General     709

Also, the velocity of gear 5 relative to the arm is

n53 = n5 − n3

(c)

Dividing Equation (c) by Equation (b) gives n53 n5 − n3 = n23 n2 − n3

(d)

Equation (d) expresses the ratio of gear 5 to that of gear 2, and both velocities are taken relative to the arm. Now this ratio is the same and is proportional to the tooth numbers, whether the arm is rotating or not. It is the train value. Therefore, we may write

e=

n5 − n3 n2 − n3

(e)

This equation can be used to solve for the output motion of any planetary train. It is more conveniently written in the form

e=

nL − nA nF − nA

(13–32)

where nF = rev/min of first gear in planetary train nL = rev/min of last gear in planetary train nA = rev/min of arm EXAMPLE 13–6 In Figure 13–26 the sun gear is the input, and it is driven clockwise at 100 rev/min. The ring gear is held stationary by being fastened to the frame. Find the rev/min and direction of rotation of the arm and gear 4. Solution Let nF = n2 = −100 rev/min, and nL = n5 = 0. For e, unlock gear 5 and fix the arm. Then, planet gear 4 and ring gear 5 rotate in the same direction, opposite of sun gear 2. Thus, e is negative. Alternatively, the number of meshes, not counting the one internal mesh is one, again making e negative and N2 N4 20 30 e = −( )( ) = −( )( ) = −0.25 N4 N5 30 80 Substituting this value in Equation (13–32) gives −0.25 =

0 − nA (−100) − nA

or Answer

nA = −20 rev/min = 20 rev/min clockwise

To obtain the speed of gear 4, we follow the procedure outlined by Equations (b), (c), and (d). Thus

n43 = n4 − n3

n23 = n2 − n3

and so

n43 n4 − n3 = n23 n2 − n3

(1)

710      Mechanical Engineering Design

But n43 20 2 =− =− n23 30 3

(2)

Substituting the known values in Equation (1) gives n4 − (−20) 2 − = 3 (−100) − (−20)

Solving gives

n 4 = +33 31 rev/min = 33 31 rev/min counterclockwise

Answer

13–14  Force Analysis—Spur Gearing Before beginning the force analysis of gear trains, let us agree on the notation to be used. Beginning with the numeral 1 for the frame of the machine, we shall designate the input gear as gear 2, and then number the gears successively 3, 4, etc., until we arrive at the last gear in the train. Next, there may be several shafts involved, and usually one or two gears are mounted on each shaft as well as other elements. We shall designate the shafts, using lowercase letters of the alphabet, a, b, c, etc. With this notation we can now speak of the force exerted by gear 2 against gear 3 as F23. The force of gear 2 against shaft a is F2a. We can also write Fa2 to mean the force of shaft a against gear 2. Unfortunately, it is also necessary to use superscripts to indicate directions. The coordinate directions will usually be indicated by the x, y, and z coordinates, and the radial and tangential directions by superscripts r and t. With this notation, F t43 is the tangential component of the force of gear 4 acting against gear 3. Figure 13–28a shows a pinion mounted on shaft a rotating clockwise at n2 rev/min and driving a gear on shaft b at n3 rev/min. The reactions between the mating teeth occur along the pressure line. In Figure 13–28b the pinion has been separated from Figure 13–28 Free-body diagrams of the forces and moments acting upon two gears of a simple gear train.

Gear

Tb3 3

n3

Fb3 ϕ b

3 b ϕ F23

ϕ

(c)

F32

2

ϕ Ta 2

a n2

a Pinion (a)

2

Fa 2 (b)

Gears—General     711 F32

F r32

Figure 13–29 Resolution of gear forces.

F t32

n2 Ta2 F ta2

a F ra2

2

Fa 2 d2

the gear and the shaft, and their effects have been replaced by forces. Fa2 and Ta2 are the force and torque, respectively, exerted by shaft a against pinion 2. F32 is the force exerted by gear 3 against the pinion. Using a similar approach, we obtain the freebody diagram of the gear shown in Figure 13–28c. In Figure 13–29, the free-body diagram of the pinion has been redrawn and the forces have been resolved into tangential and radial components. We now define

t Wt = F 32

(a)

as the transmitted load. This tangential load is really the useful component, because r the radial component F32 serves no useful purpose. It does not transmit power. The applied torque and the transmitted load are seen to be related by the equation

T=

d Wt 2

(b)

where we have used T = Ta2 and d = d2 to obtain a general relation. The power H transmitted through a rotating gear can be obtained from the standard relationship of the product of torque T and angular velocity ω.

H = Tω = (Wt d∕2) ω

(13–33)

While any units can be used in this equation, the units of the resulting power will obviously be dependent on the units of the other parameters. It will often be desirable to work with the power in either horsepower or kilowatts, and appropriate conversion factors should be used. Since meshed gears are reasonably efficient, with losses of less than 2 percent, the power is generally treated as constant through the mesh. Consequently, with a pair of meshed gears, Equation (13–33) will give the same power regardless of which gear is used for d and ω. Gear data is often tabulated using pitch-line velocity, which is the linear velocity of a point on the gear at the radius of the pitch circle; thus V = (d∕2)ω. Converting this to customary units gives

V = πdn∕12

where V = pitch-line velocity, ft/min d = gear diameter, in n = gear speed, rev/min

(13–34)

712      Mechanical Engineering Design

Many gear design problems will specify the power and speed, so it is convenient to solve Equation (13–33) for Wt. With the pitch-line velocity and appropriate conversion factors incorporated, Equation (13–33) can be rearranged and expressed in U.S. customary units as

H V

(13–35)

60 000H πdn

(13–36)

Wt = 33 000

where Wt = transmitted load, lbf H = power, hp V = pitch-line velocity, ft/min

The corresponding equation in SI units is

Wt =

where Wt = transmitted load, kN H = power, kW d = gear diameter, mm n = speed, rev/min EXAMPLE 13–7 Pinion 2 in Figure 13–30a runs at 1750 rev/min and transmits 2.5 kW to idler gear 3. The teeth are cut on the 20° full-depth system and have a module of m = 2.5 mm. Draw a free-body diagram of gear 3 and show all the forces that act upon it. Solution The pitch diameters of gears 2 and 3 are

d 2 = N2 m = 20(2.5) = 50 mm

d3 = N3 m = 50(2.5) = 125 mm

From Equation (13–36) we find the transmitted load to be

Wt =

60 000(2.5) 60 000H = = 0.546 kN πd2n π(50) (1750)

t Thus, the tangential force of gear 2 on gear 3 is F 23 = 0.546 kN, as shown in Figure 13–30b. Therefore

t F r23 = F 23 tan 20° = (0.546) tan 20° = 0.199 kN

and so

F23 =

t F 23 0.546 = = 0.581 kN cos 20° cos 20°

Since gear 3 is an idler, it transmits no power (torque) to its shaft, and so the tangential reaction of gear 4 on gear 3 is also equal to Wt. Therefore

t F 43 = 0.546 kN

and the directions are shown in Figure 13–30b.

r F 43 = 0.199 kN

F43 = 0.581 kN

Gears—General     713

Figure 13–30

y y

3

50T

20°

4

F xb3

c b 30T

3

x

F

b x

r 43

20T 2

A gear train containing an idler gear. (a) The gear train. (b) Free-body of the idler gear.

F t43

F43

a

F

y b3

Fb3 F t23 F r23

(a)

(b)

20° F23

The shaft reactions in the x and y directions are

x t r F b3 = −(F 23 + F 43 ) = −(−0.546 + 0.199) = 0.347 kN

r t F yb3 = −(F 23 + F 43 ) = −(0.199 − 0.546) = 0.347 kN

The resultant shaft reaction is Fb3 = √ (0.347) 2 + (0.347) 2 = 0.491 kN

These are shown on the figure.

13–15  Force Analysis—Bevel Gearing In determining shaft and bearing loads for bevel-gear applications, the usual practice is to use the tangential or transmitted load that would occur if all the forces were concentrated at the midpoint of the tooth. While the actual resultant occurs somewhere between the midpoint and the large end of the tooth, there is only a small error in making this assumption. For the transmitted load, this gives

Wt =

T rav

(13–37)

where T is the torque and rav is the pitch radius at the midpoint of the tooth for the gear under consideration. The forces acting at the center of the tooth are shown in Figure 13–31. The resultant force W has three components: a tangential force Wt, a radial force Wr, and an axial force Wa. From the trigonometry of the figure,

Wr = Wt tan ϕ cos γ

Wa = Wt tan ϕ sin γ

(13–38)

The three forces Wt, Wr, and Wa are at right angles to each other and can be used to determine the bearing loads by using the methods of statics.

714      Mechanical Engineering Design

Figure 13–31

y

Bevel-gear tooth forces.

x Wt

W ϕ

rav

z Wa

Wr γ

EXAMPLE 13–8 The bevel pinion in Figure 13–32a rotates at 600 rev/min in the direction shown and transmits 5 hp to the gear. The mounting distances, the location of all bearings, and the average pitch radii of the pinion and gear are shown in the figure. For simplicity, the teeth have been replaced by pitch cones. Bearings A and C should take the thrust loads. Find the bearing forces on the gearshaft. Solution The pitch angles are 3 γ = tan−1( ) = 18.4° 9

9 Γ = tan−1( ) = 71.6° 3

The pitch-line velocity corresponding to the average pitch radius is

V=

2πrP n 2π(1.293) (600) = = 406 ft/min 12 12

Therefore the transmitted load is

Wt =

33 000H (33 000) (5) = = 406 lbf V 406

and from Equation (13–38), with Γ replacing γ, we have

Wr = Wt tan ϕ cos Γ = 406 tan 20° cos 71.6° = 46.6 lbf

Wa = Wt tan ϕ sin Γ = 406 tan 20° sin 71.6° = 140 lbf

Gears—General     715

Figure 13–32

y 1

62

(a) Bevel gearset of Example 13–8. (b) Free-body diagram of shaft CD. Dimensions in inches.

3

3.88 5

1 16 D 1 22

A

3 (a)

B x

γ

1.293 5

38

15-tooth pinion P = 5 teeth / in

Γ 45-tooth gear C

9 z

x

(b)

y

F zD Wa F xD 2

2

Wr

G

D 1

8

3.8

1.2

93

F zC 3

8

Wt

5

C F

x C

F yC T

where Wt acts in the positive z direction, Wr in the −x direction, and Wa in the −y direction, as illustrated in the isometric sketch of Figure 13–32b. In preparing to take a sum of the moments about bearing D, define the position vector from D to G as

R G = 3.88i − (2.5 + 1.293) j = 3.88i − 3.793j

We shall also require a vector from D to C:

R C = −(2.5 + 3.625)j = −6.125j

716      Mechanical Engineering Design

Then, summing moments about D gives

R G × W + R C × FC + T = 0

(1)

When we place the details in Equation (1), we get

(3.88i − 3.793j) × (−46.6i − 140j + 406k) + (−6.125j) × (F Cx i + F Cy j + F Cz k) + T j = 0

(2)

After the two cross products are taken, the equation becomes (−1540i − 1575j − 720k) + (−6.125F Cz i + 6.125F Cx k) + T j = 0

from which

FCx = 118 lbf

T = 1575j lbf · in

F Cz = −251 lbf

(3)

Now sum the forces to zero. Thus

FD + FC + W = 0

(4)

When the details are inserted, Equation (4) becomes

(FDx i + FDz k) + (118i + F Cy j − 251k) + (−46.6i − 140j + 406k) = 0

(5)

First we see that F Cy = 140 lbf, and so FC = 118i + 140j − 251k lbf

Answer Then, from Equation (5),

FD = −71.4i − 155k lbf

Answer

These are all shown in Figure 13–32b in the proper directions. The analysis for the pinion shaft is quite similar.

13–16  Force Analysis—Helical Gearing Figure 13–33 is a three-dimensional view of the forces acting against a helical-gear tooth. The point of application of the forces is in the pitch plane and in the center of the gear face. From the geometry of the figure, the three components of the total (normal) tooth force W are

Wr = W sin ϕn

Wt = W cos ϕn cos ψ

Wa = W cos ϕn sin ψ

where W Wr Wt Wa

= = = =

(13–39)

total force radial component tangential component, also called the transmitted load axial component, also called the thrust load

Usually Wt is given and the other forces are desired. In this case, it is not difficult to discover that

Wr = Wt tan ϕt

Wa = Wt tan ψ

Wt W= cos ϕn cos ψ

(13–40)

Gears—General     717

Figure 13–33 Tooth forces acting on a right-hand helical gear.

W

ϕn Wr ϕt

Wt

Wa

ψ Tooth element ψ

y

x z

Pitch cylinder

EXAMPLE 13–9 In Figure 13–34 an electric motor transmits 1-hp at 1800 rev/min in the clockwise direction, as viewed from the positive x axis. Keyed to the motor shaft is an 18-tooth helical pinion having a normal pressure angle of 20°, a helix angle of 30°, and a normal diametral pitch of 12 teeth/in. The hand of the helix is shown in the figure. Make a three-dimensional sketch of the motor shaft and pinion, and show the forces acting on the pinion and the bearing reactions at A and B. The thrust should be taken out at A. Solution From Equation (13–19) we find

ϕt = tan−1

tan ϕn tan 20° = tan−1 = 22.8° cos ψ cos 30°

Also, Pt = Pn cos ψ = 12 cos 30° = 10.39 teeth/in. Therefore the pitch diameter of the pinion is dp = 18∕10.39 = 1.732 in. The pitch-line velocity is

V=

πdn π(1.732)(1800) = = 816 ft/min 12 12 Figure 13–34 The motor and gear train of Example 13–9.

y 36T A

B x 18T

10 in

3 in

718      Mechanical Engineering Design

Figure 13–35

y

Free-body diagram of motor shaft of Example 13–9. Forces in lbf.

F yA

Wr C F zA

F xA

10 in A

F zB T

The transmitted load is Wt =

3 in

Wa dp∕2 = 1.732∕2 = 0.866 in

B

x

F yB

z

Wt

33 000H (33 000) (1) = = 40.4 lbf V 816

From Equation (13–40) we find Wr = Wt tan ϕt = (40.4) tan 22.8° = 17.0 lbf

Wa = Wt tan ψ = (40.4) tan 30° = 23.3 lbf

W=

Wt 40.4 = = 49.6 lbf cos ϕn cos ψ cos 20° cos 30°

These three forces, Wr = 17.0 lbf in the −y direction, Wa = 23.3 lbf in the −x direction, and Wt = 40.4 lbf in the +z direction, are shown acting at point C in Figure 13–35. We assume bearing reactions at A and B as shown. Then F Ax = Wa = 23.3 lbf. Taking moments about the z axis, −(17.0)(13) + (23.3) (0.866) + 10F By = 0

or FBy = 20.1 lbf. Summing forces in the y direction then gives FAy = 3.1 lbf. Taking moments about the y axis, next 10FBz − (40.4) (13) = 0

or F Bz = 52.5 lbf. Summing forces in the z direction and solving gives F zA = 12.1 lbf. Also, the torque is T = Wt dp∕2 = (40.4)(1.732∕2) = 35 lbf · in. For comparison, solve the problem again using vectors. The force at C is

W = −23.3i − 17.0j + 40.4k lbf

Position vectors to B and C from origin A are

RB = 10i

RC = 13i + 0.866j

Taking moments about A, we have

RB × FB + T + RC × W = 0

Using the directions assumed in Figure 13–35 and substituting values gives

10i × (FBy j − FBz k) − T i + (13i + 0.866j) × (−23.3i − 17.0j + 40.4k) = 0

When the cross products are evaluated we get

(10F By k + 10FBz j) − T i + (35i − 525j − 201k) = 0

obtaining T = 35 lbf · in, F By = 20.1 lbf, and FBz = 52.5 lbf. Next, FA = −FB − W, and so FA = 23.3i − 3.1j + 12.1k lbf.

Gears—General     719

13–17  Force Analysis—Worm Gearing If friction is neglected, then the only force exerted by the gear will be the force W, shown in Figure 13–36, having the three orthogonal components W x, W y, and W z. From the geometry of the figure, we see that

W x = W cos ϕn sin λ

W y = W sin ϕn

(13–41)

z

W = W cos ϕn cos λ

We now use the subscripts W and G to indicate forces acting against the worm and gear, respectively. We note that W y is the separating, or radial, force for both the worm and the gear. The tangential force on the worm is W x and is W z on the gear, assuming a 90° shaft angle. The axial force on the worm is W z, and on the gear, W x. Since the gear forces are opposite to the worm forces, we can summarize these relations by writing

WWt = −WGa = W x

WWr = −WGr = W y

WWa = −WGt = W

(13–42)

z

It is helpful in using Equation (13–41) and also Equation (13–42) to observe that the gear axis is parallel to the x direction and the worm axis is parallel to the z direction and that we are employing a right-handed coordinate system. In our study of spur-gear teeth we have learned that the motion of one tooth relative to the mating tooth is primarily a rolling motion; in fact, when contact occurs at the pitch point, the motion is pure rolling. In contrast, the relative motion between worm and worm-gear teeth is pure sliding, and so we must expect that friction plays an important role in the performance of worm gearing. By introducing a coefficient of friction f, we can develop another set of relations similar to those of Equation (13–41). In Figure 13–36 we see that the force W acting normal to the worm-tooth profile produces a frictional force Wf = f W, having a component f W cos λ in the negative x

y

Figure 13–36

Wy

Drawing of the pitch cylinder of a worm, showing the forces exerted upon it by the worm gear. f W sin λ

ϕt

W λ

Wx f W cos λ

ϕn x Wz nW z

Pitch helix Pitch cylinder

λ Wf = f W

720      Mechanical Engineering Design

direction and another component f W sin λ in the positive z direction. Equation (13–41) therefore becomes

W x = W(cos ϕn sin λ + f cos λ)

W y = W sin ϕn

W z = W(cos ϕn cos λ − f sin λ)

(13–43)

Equation (13–42), of course, still applies. Inserting −WGt from Equation (13–42) for W z in Equation (13–43) and multiplying both sides by f, we find the frictional force Wf to be

Wf = f W =

f WGt f sin λ − cos ϕn cos λ

(13–44)

A useful relation between the two tangential forces, WWt and WGt, can be obtained by equating the first and third parts of Equations (13–42) and (13–43) and eliminating W. The result is

WWt = WGt

cos ϕn sin λ + f cos λ f sin λ − cos ϕn cos λ

(13–45)

Efficiency η can be defined by using the equation

η=

WWt (without friction) WWt (with friction)

(a)

Substitute Equation (13–45) with f = 0 in the numerator of Equation (a) and the same equation in the denominator. After some rearranging, you will find the efficiency to be

η=

cos ϕn − f tan λ cos ϕn + f cot λ

(13–46)

Selecting a typical value of the coefficient of friction, say f = 0.05, and the pressure angles shown in Table 13–5, we can use Equation (13–46) to get some useful design information. Solving this equation for lead angles from 1° to 30° gives the interesting results shown in Table 13–6. Table 13–6  Efficiency of Worm Gearsets for f = 0.05 Lead Angle λ, Efficiency η, deg % 1.0 25.2 2.5 45.7 5.0 62.6 7.5 71.3 10.0 76.6 15.0 82.7 20.0 85.6 30.0 88.7

Gears—General     721 0.10

Coefficient of friction, f

Gear

Worm above VW

+ λ

VG

Gear axis

VS

0.08 0.06 A 0.04 B 0.02 0

Worm axis

0

400

800

1200

1600

2000

Sliding velocity VS , ft /min

Figure 13–38

Figure 13–37

Representative values of the coefficient of friction for worm gearing. These values are based on good lubrication. Use curve B for high-quality materials, such as a casehardened steel worm mating with a phosphor-bronze gear. Use curve A when more friction is expected, as with a cast-iron worm mating with a cast-iron worm gear.

Velocity components in worm gearing.

Many experiments have shown that the coefficient of friction is dependent on the relative or sliding velocity. In Figure 13–37, VG is the pitch-line velocity of the gear and VW the pitch-line velocity of the worm. Vectorially, VW = VG + VS; consequently, the sliding velocity is

VS =

VW cos λ

(13–47)

Published values of the coefficient of friction vary as much as 20 percent, undoubtedly because of the differences in surface finish, materials, and lubrication. The values on the chart of Figure 13–38 are representative and indicate the general trend.

EXAMPLE 13–10 A 2-tooth right-hand worm transmits 1 hp at 1200 rev/min to a 30-tooth worm gear. The gear has a transverse diametral pitch of 6 teeth/in and a face width of 1 in. The worm has a pitch diameter of 2 in and a face width of 2 21 in. The normal pressure angle is 1412°. The materials and quality of the gearing to be used are such that curve B of Figure 13–38 should be used to obtain the coefficient of friction. (a) Find the axial pitch, the center distance, the lead, and the lead angle. (b) Figure 13–39 is a drawing of the worm gear oriented with respect to the coordinate system described earlier in this section; the gear is supported by bearings A and B. Find the forces exerted by the bearings against the worm-gear shaft, and the output torque. Solution (a) The axial pitch is the same as the transverse circular pitch of the gear, which is Answer

px = pt =

π π = = 0.5236 in P 6

722      Mechanical Engineering Design

Figure 13–39

y Worm pitch cylinder

The pitch cylinders of the worm gear train of Example 13–10.

1200 rev/min A

Gear pitch cylinder

B 1

1 2 in

z

1

2 2 in x

The pitch diameter of the gear is dG = NG∕P = 30∕6 = 5 in. Therefore, the center distance is C=

Answer

dW + dG 2 + 5 = = 3.5 in 2 2

From Equation (13–27), the lead is

L = px N W = (0.5236)(2) = 1.0472 in

Answer Also using Equation (13–28), we find Answer

λ = tan−1

L 1.0472 = tan−1 = 9.46° πdW π(2)

(b) Using the right-hand rule for the rotation of the worm, you will see that your thumb points in the positive z direction. Now use the bolt-and-nut analogy (the worm is right-handed, as is the screw thread of a bolt), and turn the bolt clockwise with the right hand while preventing nut rotation with the left. The nut will move axially along the bolt toward your right hand. Therefore the surface of the gear (Figure 13–39) in contact with the worm will move in the negative z direction. Thus, viewing the gear in the negative x direction, the gear rotates clockwise about the x axis The pitch-line velocity of the worm is The speed of the gear is nG =

VW =

πd W nW π(2) (1200) = = 628 ft/min 12 12

( 302 ) (1200) = 80 rev/min. Therefore the pitch-line velocity of the gear is VG =

πdG nG π(5)(80) = = 105 ft/min 12 12

Then, from Equation (13–47), the sliding velocity VS is found to be

VS =

VW 628 = = 637 ft/min cos λ cos 9.46°

Gears—General     723

Getting to the forces now, we begin with the horsepower formula WWt =

33 000H (33 000) (1) = = 52.5 lbf VW 628

This force acts in the negative x direction, the same as in Figure 13–36. Using Figure 13–38, we find f = 0.03. Then, the first equation of Equation (13–43) gives

W=

=

Wx cos ϕn sin λ + f cos λ 52.5 = 278 lbf cos 14.5° sin 9.46° + 0.03 cos 9.46°

Also, from Equation (13–43),

W y = W sin ϕn = 278 sin 14.5° = 69.6 lbf

W z = W(cos ϕn cos λ − f sin λ)

= 278(cos 14.5° cos 9.46° − 0.03 sin 9.46°) = 264 lbf

We now identify the components acting on the gear as

WG a = −W x = 52.5 lbf

WGr = −W y = −69.6 lbf

WGt = −W z = −264 lbf

A free-body diagram showing the forces and torsion acting on the gearshaft is shown in Figure 13–40. We shall make B a thrust bearing in order to place the gearshaft in compression. Thus, summing forces in the x direction gives FBx = −52.5 lbf

Answer

Figure 13–40 69. 6

Free-body diagram for Example 13–10. Forces are given in lbf.

52.5 y

G 69.6 2 12 in

A

F zA

F yA z

1 12 in 2 12 in F zB B F xB F yB T x

724      Mechanical Engineering Design

Taking moments about the z axis, we have Answer

−(52.5)(2.5) − (69.6) (1.5) + 4F By = 0

F By = 58.9 lbf

Taking moments about the y axis, (264) (1.5) − 4F Bz = 0

Answer

F Bz = 99 lbf

Summing forces in the y direction, −69.6 + 58.9 + F Ay = 0

Answer

F Ay = 10.7 lbf

Similarly, summing forces in the z direction, −264 + 99 + F Az = 0

Answer

F Az = 165 lbf

We still have one more equation to write. Summing moments about x, −(264)(2.5) + T = 0

Answer

T = 660 lbf · in

It is because of the frictional loss that this output torque is less than the product of the gear ratio and the input torque.

PROBLEMS Problems marked with an asterisk (*) are linked to problems in other chapters, as summarized in Table 1–2 of Section 1–17.

13–1

A 17-tooth spur pinion has a diametral pitch of 8 teeth/in, runs at 1120 rev/min, and drives a gear at a speed of 544 rev/min. Find the number of teeth on the gear and the theoretical center-to-center distance.

13–2 A 15-tooth spur pinion has a module of 3 mm and runs at a speed of 1600 rev/min.

The driven gear has 60 teeth. Find the speed of the driven gear, the circular pitch, and the theoretical center-to-center distance.

13–3 A spur gearset has a module of 6 mm and a velocity ratio of 4. The pinion has 16 teeth. Find the number of teeth on the driven gear, the pitch diameters, and the theoretical center-to-center distance.

13–4 A 21-tooth spur pinion mates with a 28-tooth gear. The diametral pitch is 3 teeth/in

and the pressure angle is 20°. Make a drawing of the gears showing one tooth on each gear. Find and tabulate the following results: the addendum, dedendum, clearance, circular pitch, tooth thickness, and base-circle diameters; the lengths of the arc of approach, recess, and action; and the base pitch and contact ratio.

13–5 A 20° straight-tooth bevel pinion having 14 teeth and a diametral pitch of 6 teeth/in drives a 32-tooth gear. The two shafts are at right angles and in the same plane. Find: (a) The cone distance (b) The pitch angles (c) The pitch diameters (d) The face width

Gears—General     725

13–6 A parallel helical gearset uses a 20-tooth pinion driving a 36-tooth gear. The pinion has a right-hand helix angle of 30°, a normal pressure angle of 25°, and a normal diametral pitch of 4 teeth/in. Find: (a) The normal, transverse, and axial circular pitches (b) The normal base circular pitch (c) The transverse diametral pitch and the transverse pressure angle (d) The addendum, dedendum, and pitch diameter of each gear

13–7 A parallel helical gearset consists of a 19-tooth pinion driving a 57-tooth gear. The

pinion has a left-hand helix angle of 30°, a normal pressure angle of 20°, and a normal module of 2.5 mm. Find: (a) The normal, transverse, and axial circular pitches (b) The transverse diametral pitch and the transverse pressure angle (c) The addendum, dedendum, and pitch diameter of each gear

13–8 To avoid the problem of interference in a pair of spur gears using a 20° pressure angle, specify the minimum number of teeth allowed on the pinion for each of the following gear ratios. (a) 2 to 1 (b) 3 to 1 (c) 4 to 1 (d) 5 to 1

13–9 Repeat Problem 13–8 with a 25° pressure angle. 13–10 For a spur gearset with ϕ = 20°, while avoiding interference, find:

(a) The smallest pinion tooth count that will run with itself (b) The smallest pinion tooth count at a ratio mG = 2.5, and the largest gear tooth count possible with this pinion (c) The smallest pinion that will run with a rack

13–11 Repeat Problem 13–10 for a helical gearset with ϕn = 20° and ψ = 30°. 13–12 The decision has been made to use ϕn = 20°, Pt = 6 teeth/in, and ψ = 30° for a 2:1

reduction. Choose the smallest acceptable full-depth pinion and gear tooth count to avoid interference.

13–13 Repeat Problem 13–12 with ψ = 45°. 13–14 By employing a pressure angle larger than standard, it is possible to use fewer pinion teeth, and hence obtain smaller gears without undercutting during machining. If the gears are full-depth spur gears, what is the smallest possible pressure angle ϕ that can be obtained without undercutting for a 9-tooth pinion to mesh with a rack?

13–15 A parallel-shaft gearset consists of an 18-tooth helical pinion driving a 32-tooth gear. The pinion has a left-hand helix angle of 25°, a normal pressure angle of 20°, and a normal module of 3 mm. Find: (a) The normal, transverse, and axial circular pitches (b) The transverse module and the transverse pressure angle (c) The pitch diameters of the two gears

726      Mechanical Engineering Design

13–16 The double-reduction helical gearset shown in the figure is driven through shaft a at

a speed of 700 rev/min. Gears 2 and 3 have a normal diametral pitch of 12 teeth/in, a 30° helix angle, and a normal pressure angle of 20°. The second pair of gears in the train, gears 4 and 5, have a normal diametral pitch of 8 teeth/in, a 25° helix angle, and a normal pressure angle of 20°. The tooth numbers are: N2 = 12, N3 = 48, N4 = 16, N5 = 36. Find: (a) The directions of the thrust force exerted by each gear upon its shaft (b) The speed and direction of shaft c (c) The center distance between shafts

y 1

14

y

1

1

22

14

5

c E

F 5

Problem 13–16

3

Dimensions in inches.

a b

2

4

x

z C 3 4

4

3

A 2

3 4

2 1

D

B 3 4

13–17 Shaft a in the figure rotates at 600 rev/min in the direction shown. Find the speed and direction of rotation of shaft d. 20T, ψ = 30° RH a

Problem 13–17

8T, ψ = 60° RH

2 b 3

7

6

5

40T

c

17T, ψ = 30° RH

20T

d

60T

13–18 The mechanism train shown consists of an assortment of gears and pulleys to drive

gear 9. Pulley 2 rotates at 1200 rev/min in the direction shown. Determine the speed and direction of rotation of gear 9.

Gears—General     727 2

6-in dia.

3

10-in dia. 18T 4 38T 5

Problem 13–18

48T

7

20T

6

36T 9

Worm 3T · R.H.

8

13–19 The figure shows a gear train consisting of a pair of helical gears and a pair of miter gears. The helical gears have a 1712° normal pressure angle and a helix angle as shown. Find: (a) The speed of shaft c (b) The distance between shafts a and b (c) The pitch diameter of the miter gears x

4P, 32T

B

3

34

45°

a 540 rev/min

A

a

4

Problem 13–19 Dimensions in inches.

8 normal DP, 12T, 23° ψ

5 8

2

1

12

F

x

5

28

c

b

y 1

54

3

5 E

32T

2

3

18

z

b

C

D

1

14

40T 1

14

13–20 A compound reverted gear train is to be designed as a speed increaser to provide a total increase of speed of exactly 45 to 1. With a 20° pressure angle, specify appropriate numbers of teeth to minimize the gearbox size while avoiding the interference problem in the teeth. Assume all gears will have the same diametral pitch.

13–21 Repeat Problem 13–20 with a 25° pressure angle. 13–22 Repeat Problem 13–20 for a gear ratio of exactly 30 to 1. 13–23 Repeat Problem 13–20 for a gear ratio of approximately 45 to 1. 13–24 A gearbox is to be designed with a compound reverted gear train that transmits 25

horsepower with an input speed of 2500 rev/min. The output should deliver the power at a rotational speed in the range of 280 to 300 rev/min. Spur gears with 20° pressure

728      Mechanical Engineering Design

angle are to be used. Determine suitable numbers of teeth for each gear, to minimize the gearbox size while providing an output speed within the specified range. Be sure to avoid an interference problem in the teeth.

13–25 The tooth numbers for the automotive differential shown in the figure are N2 = 16,

N3 = 48, N4 = 14, N5 = N6 = 20. The drive shaft turns at 900 rev/min. (a) What are the wheel speeds if the car is traveling in a straight line on a good road surface? (b) Suppose the right wheel is jacked up and the left wheel resting on a good road surface. What is the speed of the right wheel? (c) Suppose, with a rear-wheel drive vehicle, the auto is parked with the right wheel resting on a wet icy surface. Does the answer to part (b) give you any hint as to what would happen if you started the car and attempted to drive on?

Drive shaft

2

3 Ring gear 4

5

Problem 13–25 To rear wheel

To rear wheel

6

Planet gears

13–26 The figure illustrates an all-wheel drive concept using three differentials, one for the

front axle, another for the rear, and the third connected to the drive shaft. (a) Explain why this concept may allow greater acceleration. (b) Suppose either the center or the rear differential, or both, can be locked for certain road conditions. Would either or both of these actions provide greater traction? Why?

Front differential Problem 13–26 The Audi "Quattro concept," showing the three differentials that provide permanent all-wheel drive. (Reprinted by permission of Audi of America, Inc.)

Center differential

Driveshaft Rear differential

Gears—General     729

13–27 In the reverted planetary train illustrated, find the speed and direction of rotation of

the arm if gear 2 is unable to rotate and gear 6 is driven at 12 rev/min in the clockwise direction as viewed from the bottom of the figure. 3 4

20T

30T

2 Problem 13–27

16T 5

6

13–28 In the gear train of Problem 13–27, let gear 6 be driven at 85 rev/min counterclockwise (as viewed from the bottom of the figure) while gear 2 is held stationary. What is the speed and direction of rotation of the arm?

13–29 A compound reverted gear train is to be designed to provide the appropriate ratio between the minute hand and the hour hand of a clock. The hour hand is attached to the shaft of gear 5, such that the rotational speed of the hour hand will equal the rotational speed of gear 5. The shaft of gear 2 passes through the center of the hollow shaft of gear 5, and has the minute hand attached to it. Use spur gears with a 20° pressure angle. Specify appropriate numbers of teeth, while minimizing the gearbox size and avoiding the interference problem in the teeth. Minute hand

Hour hand 5 Problem 13–29

4 3 2

13–30 Tooth numbers for the gear train shown in the figure are N2 = 12, N3 = 16, and N4 = 12.

How many teeth must internal gear 5 have? Suppose gear 5 is fixed. What is the speed of the arm if shaft a rotates at 320 rev/min counterclockwise as viewed from the left side of the figure? 5 4

3

Problem 13–30

6 a

2

b

730      Mechanical Engineering Design

13–31 The tooth numbers for the gear train illustrated are N2 = 20, N3 = 16, N4 = 30, N6 = 36, and N7 = 46. Gear 7 is fixed. If shaft a is turned through 10 revolutions, how many turns will shaft b make?

4

6 5

3

Problem 13–31

a

7

b

13–32 Shaft a in the figure has a power input of 75 kW at a speed of 1000 rev/min in the

y

counterclockwise direction. The gears have a module of 5 mm and a 20° pressure angle. Gear 3 is an idler. (a) Find the force F3b that gear 3 exerts against shaft b. (b) Find the torque T4c that gear 4 exerts on shaft c.

51T

c 4

34T

13–33 The 24T 6-pitch 20° pinion 2 shown in the figure rotates clockwise at 1000 rev/min

and is driven at a power of 25 hp. Gears 4, 5, and 6 have 24, 36, and 144 teeth, respectively. What torque can arm 3 deliver to its output shaft? Draw free-body diagrams of the arm and of each gear and show all forces that act upon them.

b 3

2

17T a

x

2

y

Problem 13–32

6

4

2 Problem 13–33

+

+ 3

+

x

5

Fixed

Gears—General     731

13–34 A compound reverted gear train is to be designed as a speed increaser to provide a

total increase of speed of exactly 40 to 1. Use spur gears with a 20° pressure angle and a diametral pitch of 6 teeth/in. (a) Specify appropriate numbers of teeth to minimize the gearbox size while avoiding the interference problem in the teeth. (b) Determine the gearbox outside dimension Y, assuming the wall thickness of the box is 0.75 in, and allowing 0.5 in clearances between the tips of the gear teeth and the walls.

2

2 5

5

Problem 13–34

Y

3

3 4

4

13–35 Repeat Problem 13–34, using a 25° pressure angle. Take note of the effect of this variation on the overall size of the gearbox.

13–36 Repeat Problem 13–34, using helical gears with a 20° pressure angle and a 45° helix angle. Take note of the effect of this variation on the overall size of the gearbox.

13–37 Repeat Problem 13–34, except instead of an exact speed ratio of 40 to 1, the speed

ratio may be anywhere between 38 and 42. Take note of the effect of this variation on the overall size of the gearbox.

13–38 In the gearbox shown, gears 4 and 5 are compounded to the same shaft. The gearbox receives an input power of 4 hp at a speed of 300 rpm. The gears have a diametral pitch of 6 teeth/in, a 20° pressure angle, and the following numbers of teeth: N7 = 80 teeth, N5 = N3 = 20 teeth, N4 = 60 teeth, N2 = 30 teeth. Assume all shafts lie in the same plane. Assume the gearbox is reasonably efficient, so losses can be neglected for this analysis. Determine (a) the number of teeth needed on gear 6 to make the input and output shafts be in-line. (b) the minimum inside dimension of the gearbox Y before adding any clearance. (c) the pitch line velocity for gear 2, in units of ft/min. (d) the tangential force transmitted between gears 2 and 3. (e) the radial force transmitted between gears 2 and 3. ( f ) the input torque, in units of lbf-ft. (g) the output torque, in units of lbf-ft.

732      Mechanical Engineering Design

(h) the output speed, in units of rev/min. (i) the output power, in units of hp. Input

4

3

2

Problem 13–38

5

6

7

Output Y

13–39 The gears shown in the figure have a module of 12 mm and a 20° pressure angle. The

pinion rotates at 1800 rev/min clockwise and transmits 150 kW through the idler pair to gear 5 on shaft c. What forces do gears 3 and 4 transmit to the idler shaft? y

5

18T Problem 13–39

3

b

4

48T

c

32T y

a

x

2

18T

3 D

C

13–40 The figure shows a pair of shaft-mounted spur gears having a diametral pitch of

b

B a

2

A

Tin x

3 in 3 in Problem 13–40

5 teeth/in with an 18-tooth 20° pinion driving a 45-tooth gear. The power input is 32-hp at 1800 rev/min. Find the direction and magnitude of the forces acting on ­bearings A, B, C, and D.

13–41 The figure shows the electric-motor frame dimensions for a 30-hp 900 rev/min motor.

The frame is bolted to its support using four 34 -in bolts spaced 1114 in apart in the view shown and 14 in apart when viewed from the end of the motor. A 4 diametral pitch 20° spur pinion having 20 teeth and a face width of 2 in is keyed to and flush with the end of the motor shaft. This pinion drives another gear whose axis is in the same

Gears—General     733

xz plane and directly behind the motor shaft. Determine the maximum shear and tensile forces on the mounting bolts based on 200 percent overload torque. Does the direction of rotation matter?

y Key

5 8

×

5 8

× 4 14 7

18

Problem 13–41 NEMA No. 364 frame; dimensions in inches. The z axis is directed out of the paper.

x

z 5

58

9

3 4 5

58

5

58 1

15 4

1

11 2

13–42 Continue Problem 13–24 by finding the following information, assuming a diametral

pitch of 6 teeth/in. (a) Determine pitch diameters for each of the gears. (b) Determine the pitch line velocities (in ft/min) for each set of gears. (c) Determine the magnitudes of the tangential, radial, and total forces transmitted between each set of gears. (d) Determine the input torque. (e) Determine the output torque, neglecting frictional losses.

13–43 A speed-reducer gearbox containing a compound reverted gear train transmits 35 horsepower with an input speed of 1200 rev/min. Spur gears with 20° pressure angle are used, with 16 teeth on each of the small gears and 48 teeth on each of the larger gears. A diametral pitch of 10 teeth/in is proposed. (a) Determine the speeds of the intermediate and output shafts. (b) Determine the pitch line velocities (in ft/min) for each set of gears. (c) Determine the magnitudes of the tangential, radial, and total forces transmitted between each set of gears. (d) Determine the input torque. (e) Determine the output torque, neglecting frictional losses.

13–44* For the countershaft in Problem 3–83 assume the gear ratio from gear B to its mating

gear is 2 to 1. (a) Determine the minimum number of teeth that can be used on gear B without an interference problem in the teeth. (b) Using the number of teeth from part (a), what diametral pitch is required to also achieve the given 8-in pitch diameter?

734      Mechanical Engineering Design

(c) Suppose the 20° pressure angle gears are exchanged for gears with 25° pressure angle, while maintaining the same pitch diameters and diametral pitch. Determine the new forces FA and FB if the same power is to be transmitted.

13–45* For the countershaft in Problem 3–84, assume the gear ratio from gear B to its mating

gear is 5 to 1. (a) Determine the minimum number of teeth that can be used on gear B without an interference problem in the teeth. (b) Using the number of teeth from part (a), what module is required to also achieve the given 300-mm pitch diameter? (c) Suppose the 20° pressure angle for gear A is exchanged for a gear with 25° pressure angle, while maintaining the same pitch diameters and module. Determine the new forces FA and FB if the same power is to be transmitted.

13–46* For the gear and sprocket assembly analyzed in Problem 3–88, information for the

gear sizes and the forces transmitted through the gears was provided in the problem statement. In this problem, we will perform the preceding design steps necessary to acquire the information for the analysis. A motor providing 2 kW is to operate at 191 rev/min. A gear unit is needed to reduce the motor speed by half to drive a chain sprocket. (a) Specify appropriate numbers of teeth on gears F and C to minimize the size while avoiding the interference problem in the teeth. (b) Assuming an initial guess of 125-mm pitch diameter for gear F, what is the module that should be used for the stress analysis of the gear teeth? (c) Calculate the input torque applied to shaft EFG. (d) Calculate the magnitudes of the radial, tangential, and total forces transmitted between gears F and C.

13–47* For the gear and sprocket assembly analyzed in Problem 3–90, information for the

gear sizes and the forces transmitted through the gears was provided in the problem statement. In this problem, we will perform the preceding design steps necessary to acquire the information for the analysis. A motor providing 1 hp is to operate at 70 rev/min. A gear unit is needed to double the motor speed to drive a chain sprocket. (a) Specify appropriate numbers of teeth on gears F and C to minimize the size while avoiding the interference problem in the teeth. (b) Assuming an initial guess of 10-in pitch diameter for gear F, what is the diametral pitch that should be used for the stress analysis of the gear teeth? (c) Calculate the input torque applied to shaft EFG. (d) Calculate the magnitudes of the radial, tangential, and total forces transmitted between gears F and C.

13–48* For the bevel gearset in Problems 3–85 and 3–87, shaft AB is rotating at 600 rev/min

and transmits 10 hp. The gears have a 20° pressure angle. (a) Determine the bevel angle γ for the gear on shaft AB. (b) Determine the pitch-line velocity. (c) Determine the tangential, radial, and axial forces acting on the pinion. Were the forces given in Problem 3–85 correct?

13–49 The figure shows a 16T 20° straight bevel pinion driving a 32T gear, and the location of the bearing centerlines. Pinion shaft a receives 2.5 hp at 240 rev/min. Determine the bearing reactions at A and B if A is to take both radial and thrust loads.

Gears—General     735 y 1

22

1

2

32

1 12

C

D b

x

O Problem 13–49

4

a

3

Dimensions in inches.

2 2

B

1

22 A

13–50 The figure shows a 10 diametral pitch 18-tooth 20° straight bevel pinion driving a 30-tooth gear. The transmitted load is 25 lbf. Find the bearing reactions at C and D on the output shaft if D is to take both radial and thrust loads. 5 8

y

5 8

1 2

2 B

A a

x

Problem 13–50 Dimensions in inches.

3 D

C

9 16

5 8

y

2

b

3 x

13–51 The gears shown in the figure have a normal diametral pitch of 5 teeth/in, a normal

pressure angle of 20°, and a 30° helix angle. The transmitted load is 800 lbf. The pinion rotates counterclockwise about the y axis, as viewed from the positive y axis. Find the force exerted by each gear on its shaft.

a

18T, LH

b

32T, RH

Problem 13−51 y

2

3

4

13–52 The gears shown in the figure have a normal diametral pitch of 5 teeth/in, a normal

pressure angle of 20°, and a 30° helix angle. The transmitted load is 800 lbf. Gear 2 rotates clockwise about the y axis, as viewed from the positive y axis. Gear 3 is an idler. Find the forces exerted by gears 2 and 3 on their shafts.

x 16T

a

24T

b

Problem 13−52

c

18T

736      Mechanical Engineering Design

13–53 A gear train is composed of four helical gears with the three shaft axes in a single plane, as shown in the figure. The gears have a normal pressure angle of 20° and a 30° helix angle. Gear 2 is the driver, and is rotating counterclockwise as viewed from the top. Shaft b is an idler and the transmitted load from gear 2 to gear 3 is 500 lbf. The gears on shaft b both have a normal diametral pitch of 7 teeth/in and have 54 and 14 teeth, respectively. Find the forces exerted by gears 3 and 4 on shaft b.

5

LH 4 Problem 13–53

2

RH

3

RH

LH a

b

c

13–54 In the figure for Problem 13–35, pinion 2 is to be a right-hand helical gear having a helix angle of 30°, a normal pressure angle of 20°, 16 teeth, and a normal diametral pitch of 6 teeth/in. A motor delivers 25 hp to shaft a at a speed of 1720 rev/min clockwise about the x axis. Gear 3 has 42 teeth. Find the reaction exerted by bearings C and D on shaft b. One of these bearings is to take both radial and thrust loads. This bearing should be selected so as to place the shaft in compression.

13–55 Gear 2, in the figure, has 16 teeth, a 20° transverse pressure angle, a 15° helix angle,

and a module of 4 mm. Gear 2 drives the idler on shaft b, which has 36 teeth. The driven gear on shaft c has 28 teeth. If the driver rotates at 1600 rev/min and transmits 6 kW, find the radial and thrust load on each shaft. LH 3

4 90° b

c

Problem 13–55 RH

2 a RH

13–56 The figure shows a double-reduction helical gearset. Pinion 2 is the driver, and it receives a torque of 1200 lbf · in from its shaft in the direction shown. Pinion 2 has a normal diametral pitch of 8 teeth/in, 14 teeth, and a normal pressure angle of 20° and is cut right-handed with a helix angle of 30°. The mating gear 3 on shaft b has 36 teeth. Gear 4, which is the driver for the second pair of gears in the train, has a normal diametral pitch of 5 teeth/in, 15 teeth, and a normal pressure angle of 20° and is cut left-handed with a helix angle of 15°. Mating gear 5 has 45 teeth. Find the magnitude and direction of the force exerted by the bearings C and D on shaft b if bearing C can take only a radial load while bearing D is mounted to take both radial and thrust loads.

Gears—General     737 y 1

1

34

y

34

5 c

c E

Problem 13–56

F 5

Dimensions in inches.

3

3

b

b

4

C

a T2

2

D

4

a

z

x

A 3

B

2

T2

1

12

2

2

13–57 A right-hand single-tooth hardened-steel (hardness not specified) worm has a catalog

rating of 2000 W at 600 rev/min when meshed with a 48-tooth cast-iron gear. The axial pitch of the worm is 25 mm, the normal pressure angle is 14 12°, the pitch diameter of the worm is 100 mm, and the face widths of the worm and gear are, respectively, 100 mm and 50 mm. Bearings are centered at locations A and B on the worm shaft. Determine which should be the thrust bearing (so that the axial load in the shaft is in compression), and find the magnitudes and directions of the forces exerted by both bearings. 50 50

100 y B Worm pitch cylinder

A

Gear pitch cylinder

Problem 13–57 Dimensions in millimeters. z

x

13–58 The hub diameter and projection for the gear of Problem 13–57 are 100 and 37.5 mm,

respectively. The face width of the gear is 50 mm. Locate bearings C and D on opposite sides, spacing C 10 mm from the gear on the hidden face (see figure) and D 10 mm from the hub face. Choose one as the thrust bearing, so that the axial load in the shaft is in compression. Find the output torque and the magnitudes and directions of the forces exerted by the bearings on the gearshaft.

738      Mechanical Engineering Design

13–59 A 2-tooth left-hand worm transmits

3 4

hp at 600 rev/min to a 36-tooth gear having a transverse diametral pitch of 8 teeth/in. The worm has a normal pressure angle of 20°, a pitch diameter of 112 in, and a face width of 112 in. Use a coefficient of friction of 0.05 and find the force exerted by the gear on the worm and the torque input. For the same geometry as shown for Problem 13–57, the worm velocity is clockwise as viewed from the positive z axis.

13–60 Write a computer program that will analyze a spur gear or helical-mesh gear, accept-

ing ϕn, ψ, Pt, NP, and NG; compute mG, dP, dG, pt, pn, px, and ϕt; and give advice as to the smallest tooth count that will allow a pinion to run with itself without interference, run with its gear, and run with a rack. Also have it give the largest tooth count possible with the intended pinion.

14

Spur and Helical Gears

©koi88/Alamy Stock Photo

Chapter Outline 14–1

The Lewis Bending Equation   740

14–10

Size Factor Ks  765

14–2

Surface Durability   749

14–11

Load-Distribution Factor Km (KH)  765

14–3

AGMA Stress Equations   751

14–12

Hardness-Ratio Factor CH (ZW)  767

14–4

AGMA Strength Equations   752

14–13

Stress-Cycle Factors YN and ZN  768

14–5

 eometry Factors I and J G (ZI and YJ)  757

14–14

Reliability Factor KR (YZ)  769

14–6

The Elastic Coefficient Cp (ZE)  761

14–15

Temperature Factor KT (Yθ)  770

14–7

Dynamic Factor Kv  763

14–16

Rim-Thickness Factor KB  770

14–8

Overload Factor Ko  764

14–17

Safety Factors SF and SH  771

14–9

Surface Condition Factor Cf (ZR)  764

14–18

Analysis  771

14–19

Design of a Gear Mesh   781

739

740      Mechanical Engineering Design

This chapter is devoted primarily to analysis and design of spur and helical gears to resist bending failure of the teeth as well as pitting failure of tooth surfaces. Failure by bending will occur when the significant tooth stress equals or exceeds either the yield strength or the bending endurance strength. A surface failure occurs when the significant contact stress equals or exceeds the surface endurance strength. The first two sections present a little of the history of the analyses from which current methodology developed. The American Gear Manufacturers Association1 (AGMA) has for many years been the responsible authority for the dissemination of knowledge pertaining to the design and analysis of gearing. The methods this organization presents are in general use in the United States when strength and wear are primary considerations. In view of this fact it is important that the AGMA approach to the subject be presented here. The general AGMA approach requires a great many charts and graphs—too many for a single chapter in this book. We have omitted many of these here by choosing a single pressure angle and by using only full-depth teeth. This simplification reduces the complexity but does not prevent the development of a basic understanding of the approach. Furthermore, the simplification makes possible a better development of the fundamentals and hence should constitute an ideal introduction to the use of the general AGMA method.2 Sections 14–1 and 14–2 are elementary and serve as an examination of the foundations of the AGMA method. Table 14–1 is largely AGMA nomenclature.

14–1  The Lewis Bending Equation Wilfred Lewis introduced an equation for estimating the bending stress in gear teeth in which the tooth form entered into the formulation. The equation, announced in 1892, still remains the basis for most gear design today. To derive the basic Lewis equation, refer to Figure 14–1a, which shows a rectangular cantilever beam of cross-sectional dimensions F and t, having a length l and a load W t, uniformly distributed across the face width F. The section modulus I∕c is Ft2∕6, and therefore the bending stress is

σ=

M 6W tl = I∕c Ft 2

(a)

Gear designers denote the components of gear-tooth forces as Wt, Wr, Wa or W t, W r, W a interchangeably. The latter notation leaves room for post-subscripts essential to t free-body diagrams. For instance, for gears 2 and 3 in mesh, W 23 is the transmitted 1

1001 N. Fairfax Street, Suite 500, Alexandria, VA 22314-1587. The standards ANSI/AGMA 2001-D04 (revised AGMA 2001-C95) and ANSI/AGMA 2101-D04 (metric edition of ANSI/AGMA 2001-D04), Fundamental Rating Factors and Calculation Methods for Involute Spur and Helical Gear Teeth, are used in this chapter. The use of American National Standards is completely voluntary; their existence does not in any respect preclude people, whether they have approved the standards or not, from manufacturing, marketing, purchasing, or using products, processes, or procedures not conforming to the standards. The American National Standards Institute does not develop standards and will in no circumstances give an interpretation of any American National Standard. Requests for interpretation of these standards should be addressed to the American Gear Manufacturers Association. [Tables or other self-supporting sections may be quoted or extracted in their entirety. Credit line should read: "Extracted from ANSI/AGMA Standard 2001-D04 or 2101-D04 Fundamental Rating Factors and Calculation Methods for Involute Spur and Helical Gear Teeth" with the permission of the publisher, American Gear Manufacturers Association, 1001 N. Fairfax Street, Suite 500, Alexandria, Virginia 22314-1587.] The foregoing is adapted in part from the ANSI foreword to these standards. 2

Spur and Helical Gears     741

Table 14–1  Symbols, Their Names, and Locations Symbol*

Name

Where Found

Ce

Mesh alignment correction factor

Eq. (14–35)

Cf (ZR )

Surface condition factor

Eq. (14–16)

CH (ZW)

Hardness-ratio factor

Eq. (14–18)

Cma

Mesh alignment factor

Eq. (14–34)

Cmc

Load correction factor

Eq. (14–31)

Cmf

Face load-distribution factor

Eq. (14–30)

Cp (ZE)

Elastic coefficient

Eq. (14–13)

Cpf

Pinion proportion factor

Eq. (14–32)

Cpm

Pinion proportion modifier

Eq. (14–33)

d

Pitch diameter

Ex. (14–1)

dP

Pitch diameter, pinion

Eq. (14–22)

dG

Pitch diameter, gear

Eq. (14–22)

F (b)

Net face width of narrowest member

Eq. (14–15)

fP

Pinion surface finish

Fig. 14–13

H

Power

Fig. 14–17

HB

Brinell hardness

Ex. 14–3

HBG

Brinell hardness of gear

Sec. 14–12

HBP

Brinell hardness of pinion

Sec. 14–12

hp

Horsepower

Ex. 14–1

ht

Gear-tooth whole depth

Sec. 14–16

I(ZI )

Geometry factor of pitting resistance

Eq. (14–16)

J(YJ)

Geometry factor for bending strength

Eq. (14–15)

KB

Rim-thickness factor

Eq. (14–40)

Kf

Fatigue stress-concentration factor

Eq. (14–9)

Km (KH)

Load-distribution factor

Eq. (14–30)

Ko

Overload factor

Eq. (14–15)

KR (YZ )

Reliability factor

Eq. (14–17)

Ks

Size factor

Sec. 14–10

KT (Yθ)

Temperature factor

Eq. (14–17)

Kv

Dynamic factor

Eq. (14–27)

m

Module

Eq. (14–15)

mB

Backup ratio

Eq. (14–39)

mF

Face-contact ratio

Eq. (14–19)

mG

Gear ratio (never less than 1)

Eq. (14–22)

mN

Load-sharing ratio

Eq. (14–21)

mt

Transverse module

Eq. (14–15)

N

Number of stress cycles

Fig. 14–14

NG

Number of teeth on gear

Eq. (14–22)

NP

Number of teeth on pinion

Eq. (14–22)

n

Speed, in rev/min

Eq. (13–34) (Continued)

742      Mechanical Engineering Design

Table 14–1  Symbols, Their Names, and Locations  (Continued) Symbol*

Name

Where Found

nP

Pinion speed, in rev/min

Ex. 14–4

P

Diametral pitch

Eq. (14–2)

Pd

Transverse diametral pitch

Eq. (14–15)

pN

Normal base pitch

Eq. (14–24)

pn

Normal circular pitch

Eq. (14–24)

px

Axial pitch

Eq. (14–19)

Qv

Quality number

Eq. (14–29)

R

Reliability

Eq. (14–38)

Ra

Root-mean-squared roughness

Fig. 14–13

rf

Tooth fillet radius

Fig. 14–1

rG

Pitch-circle radius, gear

In standard

rP

Pitch-circle radius, pinion

In standard

rbP

Pinion base-circle radius

Eq. (14–25)

rbG

Gear base-circle radius

Eq. (14–25)

SC

Buckingham surface endurance strength

Ex. 14–3

Sc

AGMA surface endurance strength

Eq. (14–18)

St

AGMA bending strength

Eq. (14–17)

S

Bearing span

Fig. 14–10

S1

Pinion offset from center span

Fig. 14–10

SF

Safety factor—bending

Eq. (14–41)

SH

Safety factor—pitting

Eq. (14–42)

W t or Wt

Transmitted load

Fig. 14–1

YN

Stress-cycle factor for bending strength

Fig. 14–14

ZN

Stress-cycle factor for pitting resistance

Fig. 14–15

β

Exponent

Eq. (14–44)

σ

Bending stress, AGMA

Eq. (14–15)

σC

Contact stress from Hertzian relationships

Eq. (14–14)

σc

Contact stress from AGMA relationships

Eq. (14–16)

σall

Allowable bending stress, AGMA

Eq. (14–17)

σc,all

Allowable contact stress, AGMA

Eq. (14–18)

ϕ

Pressure angle

Eq. (14–12)

ϕn

Normal pressure angle

Eq. (14–24)

ϕt

Transverse pressure angle

Eq. (14–23)

ψ

Helix angle

Ex. 14–5

*Where applicable, the alternate symbol for the metric standard is shown in parenthesis. t force of body 2 on body 3, and W 32 is the transmitted force of body 3 on body 2. When working with double- or triple-reduction speed reducers, this notation is compact and essential to clear thinking. Since gear-force components rarely take exponents, this causes no complication. Pythagorean combinations, if necessary, can be treated with parentheses or avoided by expressing the relations trigonometrically.

Spur and Helical Gears     743 Wr

W

Wt l Wt F

rf a t

x t

l (a)

(b)

Referring now to Figure 14–1b, we assume that the maximum stress in a gear tooth occurs at point a. By similar triangles, you can write

t∕2 l = x t∕2

or

x=

t2 4l

or

l=

t2 4x

(b)

By rearranging Equation (a),

σ=

6W tl W t 1 Wt 1 1 = = F t 2∕6l F t 2∕4l 46 Ft 2

(c)

If we now substitute the value of l from Equation (b) in Equation (c) and multiply the numerator and denominator by the circular pitch p, we find

σ=

W tp F( 23 ) xp

(d)

Letting y = 2x∕(3p), we have

σ=

Wt Fp y

(14–1)

This completes the development of the original Lewis equation. The factor y is called the Lewis form factor, and it may be obtained by a graphical layout of the gear tooth or by digital computation. In using this equation, most engineers prefer to employ the diametral pitch in determining the stresses. This is done by substituting p = π∕P and y = Y∕π in Equation (14–1). This gives

σ=

W tP FY

(14–2)

Y=

2x P 3

(14–3)

where

The use of this equation for Y means that only the bending of the tooth is considered and that the compression due to the radial component of the force is neglected. Values of Y obtained from this equation are tabulated in Table 14–2.

Figure 14–1

744      Mechanical Engineering Design

Table 14–2  Values of the Lewis Form Factor Y (These Values Are for a Normal Pressure Angle of 20°, Full-Depth Teeth, and a Diametral Pitch of Unity in the Plane of Rotation) Number of Teeth

Y

Number of Teeth

12

0.245

28

0.353

13

0.261

30

0.359

14

0.277

34

0.371

15

0.290

38

0.384

16

0.296

43

0.397

17

0.303

50

0.409

18

0.309

60

0.422

19

0.314

75

0.435

20

0.322

100

0.447

21

0.328

150

0.460

22

0.331

300

0.472

24

0.337

400

0.480

26

0.346

Rack

0.485

Y

The use of Equation (14–3) also implies that the teeth do not share the load and that the greatest force is exerted at the tip of the tooth. But we have already learned that the contact ratio should be somewhat greater than unity, say about 1.5, to achieve a quality gearset. If, in fact, the gears are cut with sufficient accuracy, the tip-load condition is not the worst, because another pair of teeth will be in contact when this condition occurs. Examination of run-in teeth will show that the heaviest loads occur near the middle of the tooth. Therefore the maximum stress probably occurs while a single pair of teeth is carrying the full load, at a point where another pair of teeth is just on the verge of coming into contact. Dynamic Effects When a pair of gears is driven at moderate or high speed and noise is generated, it is certain that dynamic effects are present. One of the earliest efforts to account for an increase in the load due to velocity employed a number of gears of the same size, material, and strength. Several of these gears were tested to destruction by meshing and loading them at zero velocity. The remaining gears were tested to destruction at various pitch-line velocities. For example, if a pair of gears failed at 500 lbf tangential load at zero velocity and at 250 lbf at velocity V1, then a velocity factor, designated Kv, of 2 was specified for the gears at velocity V1. Then another, identical, pair of gears running at a pitch-line velocity V1 could be assumed to have a load equal to twice the tangential or transmitted load. Note that the definition of dynamic factor Kv has been altered. AGMA standards ANSI/AGMA 2001-D04 and 2101-D04 contain this caution: Dynamic factor Kv has been redefined as the reciprocal of that used in previous AGMA standards. It is now greater than 1.0. In earlier AGMA standards it was less than 1.0.

Care must be taken in referring to work done prior to this change in the standards.

Spur and Helical Gears     745

In the nineteenth century, Carl G. Barth first expressed the velocity factor, and in terms of the current AGMA standards, they are represented as

Kv =

600 + V 600

(cast iron, cast profile)

(14–4a)

Kv =

1200 + V 1200

(cut or milled profile)

(14–4b)

where V is the pitch-line velocity in feet per minute. It is also quite probable, because of the date that the tests were made, that the tests were conducted on teeth having a cycloidal profile instead of an involute profile. Cycloidal teeth were in general use in the nineteenth century because they were easier to cast than involute teeth. Equation (14–4a) is called the Barth equation. The Barth equation is often modified into Equation (14–4b), for cut or milled teeth. Later, AGMA added 50 + √V 50

Kv =

Kv = √

(hobbed or shaped profile)

(14–5a)

(shaved or ground profile)

(14–5b)

78 + √V 78

In SI units, Equations (14–4a) through (14–5b) become

Kv =

3.05 + V 3.05

Kv =

6.1 + V 6.1

Kv =

3.56 + √V 3.56

Kv = √

(cast iron, cast profile)

(14–6a)

(14–6b)

(hobbed or shaped profile)

(14–6c)

(shaved or ground profile)

(14–6d)

(cut or milled profile)

5.56 + √V 5.56

where V is in meters per second (m/s). Introducing the velocity factor into Equation (14–2) gives

σ=

Kv W t P FY

(14–7)

Kv W t FmY

(14–8)

The metric version of this equation is

σ=

where the face width F and the module m are both in millimeters (mm). Expressing the tangential component of load W t in newtons (N) then results in stress units of megapascals (MPa). As a general rule, spur gears should have a face width F from three to five times the circular pitch p. Equations (14–7) and (14–8) are important because they form the basis for the AGMA approach to the bending strength of gear teeth. They are in general use for estimating the capacity of gear drives when life and reliability are not important considerations. The equations can be useful in obtaining a preliminary estimate of gear sizes needed for various applications.

746      Mechanical Engineering Design

EXAMPLE 14–1 A stock spur gear is available having a diametral pitch of 8 teeth/in, a 112 -in face, 16 teeth, and a pressure angle of 20° with full-depth teeth. The material is AISI 1020 steel in as-rolled condition. Use a design factor of nd = 3 to rate the horsepower output of the gear corresponding to a speed of 1200 rev/m and moderate applications. Solution The term moderate applications seems to imply that the gear can be rated by using the yield strength as a criterion of failure. From Table A–20, we find Sut = 55 kpsi and Sy = 30 kpsi. A design factor of 3 means that the allowable bending stress is 30∕3 = 10 kpsi. The pitch diameter is d = N∕P = 16∕8 = 2 in, so the pitch-line velocity is

πdn π(2)1200 = = 628 ft/min 12 12

V=

The velocity factor from Equation (14–4b) is found to be

Kv =

1200 + V 1200 + 628 = = 1.52 1200 1200

Table 14–2 gives the form factor as Y = 0.296 for 16 teeth. We now arrange and substitute in Equation (14–7) as follows:

Wt =

FYσall 1.5(0.296)10 000 = = 365 lbf Kv P 1.52(8)

The horsepower that can be transmitted is Answer

hp =

365(628) W tV = = 6.95 hp 33 000 33 000

It is important to emphasize that this is a rough estimate, and that this approach must not be used for important applications. The example is intended to help you understand some of the fundamentals that will be involved in the AGMA approach.

EXAMPLE 14–2 Estimate the horsepower rating of the gear in the previous example based on obtaining an infinite life in bending. Solution The rotating-beam endurance limit is estimated from Eq. (6–10),

S′e = 0.5Sut = 0.5(55) = 27.5 kpsi

To obtain the surface finish Marin factor ka we refer to Table 6–3 for machined surface, finding a = 2.00 and b = −0.217. Then Eq. (6–18) gives the surface finish Marin factor ka as

ka = aSutb = 2.00(55) −0.217 = 0.838

The next step is to estimate the size factor kb. From Table 13–1, the sum of the addendum and dedendum is

l=

1 1.25 1 1.25 + = + = 0.281 in P P 8 8

Spur and Helical Gears     747

The tooth thickness t in Fig. 14–1b is given in Sec. 14–1 [Eq. (b)] as t = (4lx)1∕2 when x = 3Y∕(2P) from Eq. (14–3). Therefore, since from Ex. 14–1 Y = 0.296 and P = 8,

x=

3Y 3(0.296) = = 0.0555 in 2P 2(8)

then t = (4lx) 1∕2 = [4(0.281)0.0555]1∕2 = 0.250 in

We have recognized the tooth as a cantilever beam of rectangular cross section, so the equivalent rotating-beam diameter must be obtained from Eq. (6–24):

de = 0.808(hb) 1∕2 = 0.808(Ft) 1∕2 = 0.808[1.5(0.250)]1∕2 = 0.495 in

Then, Eq. (6–19) gives kb as

de −0.107 0.495 −0.107 kb = ( = = 0.948 ( 0.30 ) 0.30 )

The load factor kc from Eq. (6–25) is unity. With no information given concerning temperature and reliability we will set kd = ke = 1. In general, a gear tooth is subjected only to one-way bending. Exceptions include idler gears and gears used in reversing mechanisms. We will account for one-way bending by establishing a miscellaneous-effects Marin factor kf . For one-way bending the steady and alternating stress components are σa = σm = σ∕2 where σ is the largest repeatedly applied bending stress as given in Eq. (14–7). If a material exhibited a Goodman failure locus, σa σm + =1 S′e Sut Since σa and σm are equal for one-way bending, we substitute σa for σm and solve the preceding equation for σa, giving

S′e Sut S′e + Sut Now replace σa with σ∕2, and in the denominator replace S′e with 0.5Sut to obtain

σa =

2S′e Sut 2S′e = = 1.33S′e 0.5Sut + Sut 0.5 + 1 Now defining a miscellaneous Marin factor kf = σ∕S′e = 1.33S′e∕S′e = 1.33. Similarly, if we were to use a Gerber fatigue locus,

σ=

σa σm 2 +( ) =1 S′e Sut

Setting σa = σm and solving the quadratic in σa gives

σa =

S 2ut 4S′e2 −1 + 1 + √ 2S′e ( S2ut )

Setting σa = σ∕2, Sut = S′e∕0.5 gives S′e

[−1 + √1 + 4(0.5) 2 ] = 1.66S′e 0.52 and kf = σ∕S′e = 1.66. Since a Gerber locus runs in and among fatigue data and Goodman does not, we will use kf = 1.66. The Marin equation for the fully corrected endurance strength is

σ=

Se = ka kb k c k d k e k f S′e = 0.838(0.948)(1)(1)(1)1.66(27.5) = 36.3 kpsi

748      Mechanical Engineering Design

For stress, we will first determine the fatigue stress-concentration factor Kf. For a 20° full-depth tooth the radius of the root fillet is denoted rf, with a typically proportioned value of

rf =

From Fig. A–15–6

0.300 0.300 = = 0.0375 in P 8

r rf 0.0375 = = = 0.15 d t 0.250 Since D∕d = ∞, we approximate with D∕d = 3, giving Kt = 1.68. From Fig. 6–26, q = 0.62. From Eq. (6–32),

Kf = 1 + (0.62)(1.68 − 1) = 1.42

For a design factor of nd = 3, as used in Ex. 14–1, applied to the load or strength, the maximum bending stress is Se σmax = K f σall = nd

σall =

Se 36.3 = = 8.52 kpsi Kf nd 1.42(3)

The transmitted load W t is Wt =

FYσall 1.5(0.296)8520 = = 311 lbf Kv P 1.52(8)

and the power is, with V = 628 ft/min from Ex. 14–1, hp =

311(628) W tV = = 5.9 hp 33 000 33 000

Again, it should be emphasized that these results should be accepted only as preliminary estimates to alert you to the nature of bending in gear teeth. In Example 14–2 our resources (Figure A–15–6) did not directly address stress concentration in gear teeth. A photoelastic investigation by Dolan and Broghamer reported in 1942 constitutes a primary source of information on stress concentration.3 Mitchiner and Mabie4 interpret the results in term of fatigue stress-concentration factor Kf as

t L t M Kf = H + ( ) ( ) r l

(14–9)

where  H = 0.34 − 0.458 366 2ϕ L = 0.316 − 0.458 366 2ϕ M = 0.290 + 0.458 366 2ϕ (b − rf ) 2 r= (d∕2) + b − rf 3

T. J. Dolan and E. I. Broghamer, A Photoelastic Study of the Stresses in Gear Tooth Fillets, Bulletin 335, Univ. Ill. Exp. Sta., March 1942, See also W. D. Pilkey and D. F. Pilkey, Peterson's Stress-Concentration Factors, 3rd ed., John Wiley & Sons, Hoboken, NJ, 2008, pp. 407–409, 434–437. 4 R. G. Mitchiner and H. H. Mabie, "Determination of the Lewis Form Factor and the AGMA Geometry Factor J of External Spur Gear Teeth," J. Mech. Des., Vol. 104, No. 1, Jan. 1982, pp. 148–158.

Spur and Helical Gears     749

In these equations l and t are from the layout in Figure 14–1, ϕ is the pressure angle, rf is the fillet radius, b is the dedendum, and d is the pitch diameter. It is left as an exercise for the reader to compare Kf from Equation (14–9) with the results of using the approximation of Figure A–15–6 in Example 14–2.

14–2  Surface Durability In this section we are interested in the failure of the surfaces of gear teeth, which is generally called wear. Pitting, as explained in Section 6–18, is a surface fatigue failure due to many repetitions of high contact stresses. Other surface failures are scoring, which is a lubrication failure, and abrasion, which is wear due to the presence of foreign material. To obtain an expression for the surface-contact stress, we shall employ the Hertz theory. In Equation (3–74), it was shown that the contact stress between two cylinders may be computed from the equation

pmax =

2F πbl

(a)

where  pmax = largest surface pressure F = force pressing the two cylinders together l = length of cylinders and half-width b is obtained from Equation (3–73), given by

b=[

2 2 2F (1 − ν 1 )∕E1 + (1 − ν 2 )∕E 2 1∕2 ] πl 1∕d1 + 1∕d 2

(14–10)

where ν1, ν2, E1, and E2 are the elastic constants and d1 and d2 are the diameters, respectively, of the two contacting cylinders. To adapt these relations to the notation used in gearing, we replace F by W t∕cos ϕ, d by 2r, and l by the face width F. With these changes, we can substitute the value of b as given by Equation (14–10) in Equation (a). Replacing pmax by σC, the surface compressive stress (Hertzian stress) is found from the equation

1∕2 1∕r1 + 1∕r2 Wt σC = −[ 2 2 πF cos ϕ (1 − ν1 )∕E1 + (1 − ν2 )∕E2 ]

(14–11)

where r1 and r2 are the instantaneous values of the radii of curvature on the pinion- and gear-tooth profiles, respectively, at the point of contact. By accounting for load sharing in the value of W t used, Equation (14–11) can be solved for the Hertzian stress for any or all points from the beginning to the end of tooth contact. Of course, pure rolling exists only at the pitch point. Elsewhere the motion is a mixture of rolling and sliding. Equation (14–11) does not account for any sliding action in the evaluation of stress. We note that AGMA uses μ for Poisson's ratio instead of ν as is used here. We have already noted that the first evidence of wear occurs near the pitch line. The radii of curvature of the tooth profiles at the pitch point are

r1 =

dP sin ϕ 2

r2 =

dG sin ϕ 2

(14–12)

where ϕ is the pressure angle and dP and dG are the pitch diameters of the pinion and gear, respectively.

750      Mechanical Engineering Design

Note, in Equation (14–11), that the denominator of the second group of terms contains four elastic constants, two for the pinion and two for the gear. As a simple means of combining and tabulating the results for various combinations of pinion and gear materials, AGMA defines an elastic coefficient Cp by the equation Cp =

[

1

1 − ν2P 1 − ν2G π( + EP EG )

]

1∕2

(14–13)

With this simplification, and the addition of a velocity factor Kv, Equation (14–11) can be written as KvW t 1 1 1∕2 σC = −Cp [ + (14–14) F cos ϕ ( r1 r2 )] where the sign is negative because σC is a compressive stress. EXAMPLE 14–3 The pinion of Exs. 14–1 and 14–2 is to be mated with a 50-tooth gear manufactured of ASTM No. 50 cast iron. Using the tangential load of 382 lbf, estimate the factor of safety of the drive based on the possibility of a surface fatigue failure. The surface endurance strength of cast iron can be estimated from Sc = 0.32 HB kpsi for 108 cycles. Solution From Table A–5 we find the elastic constants to be EP = 30 Mpsi, νP = 0.292, EG = 14.5 Mpsi, νG = 0.211. We substitute these in Eq. (14–13) to get the elastic coefficient as Cp = {π [

= 1817 (psi) 1∕2 30(106 ) 14.5(106 ) ]} From Ex. 14–1, the pinion pitch diameter is dP = 2 in. The value for the gear is dG = 50∕8 = 6.25 in. Then Eq. (14–12) is used to obtain the radii of curvature at the pitch points. Thus

1 − (0.292) 2

+

1 − (0.211) 2

−(1∕2)

2 sin 20° 6.25 sin 20° = 0.342 in r2 = = 1.069 in 2 2 The face width is given as F = 1.5 in. Use Kv = 1.52 from Ex. 14–1. Substituting all these values in Eq. (14–14) with ϕ = 20° gives the contact stress as

r1 =

1∕2 1.52(380) 1 1 σC = −1817 [ + = −72 400 psi 1.5 cos 20° ( 0.342 1.069 )] Table A–24 gives HB = 262 for ASTM No. 50 cast iron. Therefore SC = 0.32(262) = 83.8 kpsi. Contact stress is not linear with respect to the transmitted load [see Eq. (14–14)]. If the factor of safety is defined as the loss-offunction load divided by the imposed load, then the ratio of loads is the ratio of stresses squared. In other words,

loss-of-function load S 2C 83.8 2 = 2 =( = 1.34 imposed load 72.4 ) σC One is free to define the factor of safety as SC∕σC. Awkwardness comes when one compares the factor of safety in bending fatigue with the factor of safety in surface fatigue for a particular gear. Suppose the factor of safety of this gear in bending fatigue is 1.20 and the factor of safety in surface fatigue is 1.34 as above. The threat, since 1.34 is greater than 1.20, is in bending fatigue since both numbers are based on load ratios. If the factor of safety in surface fatigue is based on SC∕σC = √1.34 = 1.16, then 1.20 is greater than 1.16, but the threat is not from surface fatigue. The surface fatigue factor of safety can be defined either way. One way has the burden of requiring a squared number before numbers that instinctively seem comparable can be compared.

n=

Spur and Helical Gears     751

In addition to the dynamic factor Kv already introduced, there are transmitted load excursions, nonuniform distribution of the transmitted load over the tooth contact, and the influence of rim thickness on bending stress. Tabulated strength values can be means, ASTM minimums, or of unknown heritage. In surface fatigue there are no endurance limits. Endurance strengths have to be qualified as to corresponding cycle count, and the slope of the S-N curve needs to be known. In bending fatigue there is a definite change in slope of the S-N curve near 106 cycles, but some evidence indicates that an endurance limit does not exist. Gearing experience leads to cycle counts of 1011 or more. Evidence of diminishing endurance strengths in bending have been included in AGMA methodology.

14–3  AGMA Stress Equations Two fundamental stress equations are used in the AGMA methodology, one for bending stress and another for pitting resistance (contact stress). In AGMA terminology, these are called stress numbers, as contrasted with actual applied stresses, and are designated by a lowercase letter s instead of the Greek lowercase σ we have used in this book (and shall continue to use). The fundamental equations for bending are

 Pd Km KB  W tKo Kv Ks  F J σ=  t 1 KH KB  W Ko Kv Ks  bmt YJ

(U.S. customary units)

(14–15)

(SI units)

where for U.S. customary units (SI units), W t is the tangential transmitted load, lbf (N) Ko is the overload factor Kv is the dynamic factor Ks is the size factor Pd is the transverse diametral pitch F (b) is the face width of the narrower member, in (mm) Km (KH) is the load-distribution factor KB is the rim-thickness factor J (YJ) is the geometry factor for bending strength (which includes root fillet stress-concentration factor Kf) (mt) is the transverse metric module Before you try to digest the meaning of all these terms in Equation (14–15), view them as advice concerning items the designer should consider whether he or she follows the voluntary standard or not. These items include issues such as ∙ ∙ ∙ ∙ ∙ ∙

Transmitted load magnitude Overload Dynamic augmentation of transmitted load Size Geometry: pitch and face width Distribution of load across the teeth ∙ Rim support of the tooth ∙ Lewis form factor and root fillet stress concentration

752      Mechanical Engineering Design

The fundamental equations for pitting resistance (contact stress) are  K  Cp W t Ko Kv Ks m √  dP F σc =   KH  ZE W tKo Kv Ks √ dw1b 

Cf I ZR ZI

(U.S. customary units)

(14–16)

(SI units)

where W t, Ko, Kv, Ks, Km, F, and b are the same terms as defined for Equation (14–15). For U.S. customary units (SI units), the additional terms are Cp (ZE) is an elastic coefficient, √lbf/in2 ( √N/mm2 ) Cf (ZR) is the surface condition factor dP (dw1) is the pitch diameter of the pinion, in (mm) I (ZI) is the geometry factor for pitting resistance The evaluation of all these factors is explained in the sections that follow. The development of Equation (14–16) is clarified in the second part of Section 14–5.

14–4  AGMA Strength Equations Instead of using the term strength, AGMA uses data termed allowable stress numbers and designates these by the symbols sat and sac. It will be less confusing here if we continue the practice in this book of using the uppercase letter S to designate strength and the lowercase Greek letters σ and τ for stress. To make it perfectly clear we shall use the term gear strength as a replacement for the phrase allowable stress numbers as used by AGMA. Following this convention, values for gear bending strength, designated here as St, are to be found in Figures 14–2, 14–3, and 14–4, and in Tables 14–3 and 14–4. Since gear strengths are not identified with other strengths such as Sut, Se, or Sy as used elsewhere in this book, their use should be restricted to gear problems. In this approach the strengths are modified by various factors that produce limiting values of the bending stress and the contact stress. Figure 14–2

Metallurgical and quality control procedure required Gear bending strength, St kpsi

Gear bending strength for through-hardened steels, St. The SI equations are: St = 0.533HB + 88.3 MPa, grade 1, and St = 0.703HB + 113 MPa, grade 2. (Source: ANSI/AGMA 2001-D04 and 2101-D04.)

50

Grade 2 St = 102 HB + 16 400 psi

40

30

Grade 1 St = 77.3 HB + 12 800 psi

20

10 150

200

250

300 Brinell hardness, HB

350

400

450

Spur and Helical Gears     753

Figure 14–3

80 Metallurgical and quality control procedures required

Gear bending strength for nitrided through-hardened steel gears (i.e., AISI 4140, 4340), St. The SI equations are: St = 0.568HB + 83.8 MPa, grade 1, and St = 0.749HB + 110 MPa, grade 2. (Source: ANSI/AGMA 2001-D04 and 2101-D04.)

Gear bending strength, St kpsi

70

60

Grade 2 St = 108.6HB + 15 890 psi

50

40 Grade 1 St = 82.3HB + 12 150 psi

30

20 250

275

300

325

350

Core hardness, HB

70

Figure 14–4

Metallurgical and quality control procedures required

Gear bending strength, St kpsi

Grade 3 – 2.5% Chrome St = 105.2HB + 29 280 psi 60

Grade 2 – 2.5% Chrome St = 105.2HB + 22 280 psi

Grade 2 – Nitralloy St = 113.8HB + 16 650 psi

50

Grade 1 – 2.5% Chrome St = 105.2HB + 9280 psi 40 Grade 1 – Nitralloy St = 86.2HB + 12 730 psi 30 250

275

300

325

350

Core hardness, HB

The equation for the allowable bending stress is

σall

   =   

St SF St SF

YN KT KR YN Yθ YZ

(U.S. customary units) (SI units)

where for U.S. customary units (SI units), St is the gear bending strength, lbf/in2 (N/mm2) YN is the stress-cycle factor for bending stress KT (Yθ) are the temperature factors KR (YZ) are the reliability factors SF is the AGMA factor of safety, a stress ratio

(14–17)

Gear bending strength for nitriding steel gears, St. The SI equations are: St = 0.594HB + 87.76 MPa Nitralloy grade 1 St = 0.784HB + 114.81 MPa Nitralloy grade 2 St = 0.7255HB + 63.89 MPa 2.5% chrome, grade 1 St = 0.7255HB + 153.63 MPa 2.5% chrome, grade 2 St = 0.7255HB + 201.91 MPa 2.5% chrome, grade 3 (Source: ANSI/AGMA 2001-D04, 2101-D04.)

754      Mechanical Engineering Design

Table 14–3  Repeatedly Applied Gear Bending Strength St at 107 Cycles and 0.99 Reliability for Steel Gears Material Designation Steel3

Nitralloy 135M, Nitralloy N, and 2.5% chrome (no aluminum)

Gear Bending Strength St,2 psi

Minimum Surface Hardness1

Grade 1

Grade 2

Grade 3

Through-hardened Flame4 or induction hardened4 with type A pattern5

See Fig. 14–2 See Table 8*

See Fig. 14–2 45 000

See Fig. 14–2 55 000

— —

Flame4 or induction hardened4 with type B pattern5

See Table 8*

22 000

22 000

Carburized and hardened

See Table 9*

55 000

65 000 or 70 0006

75 000

Nitrided4,7 (throughhardened steels)

83.5 HR15N

See Fig. 14–3

See Fig. 14–3

Nitrided4,7

87.5 HR15N

See Fig. 14–4

See Fig. 14–4

See Fig. 14–4

Heat Treatment

Notes: See ANSI/AGMA 2001-D04 for references cited in notes 1–7. 1 Hardness to be equivalent to that at the root diameter in the center of the tooth space and face width. 2 See tables 7 through 10 for major metallurgical factors for each stress grade of steel gears. 3 The steel selected must be compatible with the heat treatment process selected and hardness required. 4 The gear bending strength indicated may be used with the case depths prescribed in 16.1. 5 See figure 12 for type A and type B hardness patterns. 6 If bainite and microcracks are limited to grade 3 levels, 70 000 psi may be used. 7 The overload capacity of nitrided gears is low. Since the shape of the effective S-N curve is flat, the sensitivity to shock should be investigated before proceeding with the design. [7] *Tables 8 and 9 of ANSI/AGMA 2001-D04 are comprehensive tabulations of the major metallurgical factors affecting St and Sc of flamehardened and induction-hardened (Table 8) and carburized and hardened (Table 9) steel gears. Source: ANSI/AGMA 2001-D04.

The equation for the allowable contact stress σc,all is

σc,all

   =   

Sc SH Sc SH

Z N CH K T KR Z N ZW Yθ YZ

(U.S. customary units)

(14–18)

(SI units)

where the upper equation is in U.S. customary units and the lower equation is in SI units. Also, Sc is the gear contact strength, lbf/in2 (N/mm2) ZN is the stress-cycle factor CH (ZW) are the hardness ratio factors for pitting resistance KT (Yθ) are the temperature factors KR (YZ) are the reliability factors SH is the AGMA factor of safety, a stress ratio The values for the gear contact strength, designated here as Sc, are to be found in Figure 14–5 and Tables 14–5, 14–6, and 14–7.

Spur and Helical Gears     755

Table 14–4  Repeatedly Applied Gear Bending Strength St for Iron and Bronze Gears at 107 Cycles and 0.99 Reliability Material Designation1

Material ASTM A48 gray cast iron

ASTM A536 ductile (nodular) Iron

Heat Treatment

Typical Minimum Surface Hardness2

Gear Bending Strength, St,3 psi

Class 20

As cast

5000

Class 30

As cast

174 HB

8500

Class 40

As cast

201 HB

13 000

Grade 60–40–18

Annealed

140 HB

22 000–33 000

Grade 80–55–06

Quenched and tempered

179 HB

22 000–33 000

Grade 100–70–03

Quenched and tempered

229 HB

27 000–40 000

Grade 120–90–02

Quenched and tempered

269 HB

31 000–44 000

Sand cast

Minimum tensile strength 40 000 psi

5700

Heat treated

Minimum tensile strength 90 000 psi

23 600

Bronze ASTM B–148 Alloy 954

Notes: 1 See ANSI/AGMA 2004-B89, Gear Materials and Heat Treatment Manual. 2 Measured hardness to be equivalent to that which would be measured at the root diameter in the center of the tooth space and face width. 3 The lower values should be used for general design purposes. The upper values may be used when:   High quality material is used.   Section size and design allow maximum response to heat treatment.   Proper quality control is effected by adequate inspection.   Operating experience justifies their use.

1000 lb/in2

Source: ANSI/AGMA 2001-D04.

Figure 14–5 Gear contact strength Sc at 107 cycles and 0.99 reliability for through-hardened steel gears. The SI equations are: Sc = 2.22HB + 200 MPa, grade 1, and Sc = 2.41HB + 237 MPa, grade 2. (Source: ANSI/AGMA 2001-D04 and 2101-D04.)

Metallurgical and quality control procedures required

Gear contact strength, Sc

175 Grade 2 Sc = 349 HB + 34 300 psi

150 125

Grade 1 Sc = 322 HB + 29 100psi

100 75 150

200

250

300 Brinell hardness, HB

350

400

450

756      Mechanical Engineering Design

Table 14–5  Nominal Temperature Used in Nitriding and Hardnesses Obtained Temperature Before Nitriding, °F

Nitriding, °F

Nitralloy 135*

1150

Nitralloy 135M

Steel

Hardness, Rockwell C Scale Case

Core

975

62–65

30–35

1150

975

62–65

32–36

Nitralloy N

1000

975

62–65

40–44

AISI 4340

1100

975

48–53

27–35

AISI 4140

1100

975

49–54

27–35

31 Cr Mo V 9

1100

975

58–62

27–33

*Nitralloy is a trademark of the Nitralloy Corp., New York. Source: Darle W. Dudley, Handbook of Practical Gear Design, rev. ed., McGraw-Hill, New York, 1984.

Table 14–6  Repeatedly Applied Gear Contact Strength Sc at 107 Cycles and 0.99 Reliability for Steel Gears Material Designation 3

Steel

Heat Treatment 4

Through hardened 5

Minimum Surface Hardness1

Grade 1

Grade 2

Grade 3

See Fig. 14–5

See Fig. 14–5

See Fig. 14–5

Gear Contact Strength,2 Sc, psi

Flame or induction hardened5

50 HRC

170 000

190 000

54 HRC

175 000

195 000

Carburized and hardened5

See Table 9*

180 000

225 000

275 000

Nitrided5 (through hardened steels)

83.5 HR15N

150 000

163 000

175 000

84.5 HR15N

155 000

168 000

180 000

2.5% chrome (no aluminum)

5

Nitrided

87.5 HR15N

155 000

172 000

189 000

Nitralloy 135M

Nitrided5

90.0 HR15N

170 000

183 000

195 000

Nitralloy N

5

Nitrided

90.0 HR15N

172 000

188 000

205 000

2.5% chrome (no aluminum)

Nitrided5

90.0 HR15N

176 000

196 000

216 000

Notes: See ANSI/AGMA 2001-D04 for references cited in notes 1–5. 1 Hardness to be equivalent to that at the start of active profile in the center of the face width. 2 See Tables 7 through 10 for major metallurgical factors for each stress grade of steel gears. 3 The steel selected must be compatible with the heat treatment process selected and hardness required. 4 These materials must be annealed or normalized as a minimum. 5 The gear contact strengths indicated may be used with the case depths prescribed in 16.1. *Table 9 of ANSI/AGMA 2001-D04 is a comprehensive tabulation of the major metallurgical factors affecting St and Sc of carburized and hardened steel gears. Source: ANSI/AGMA 2001-D04.

Spur and Helical Gears     757

Table 14–7  Repeatedly Applied Gear Contact Strength Sc 107 Cycles and 0.99 Reliability for Iron and Bronze Gears Material ASTM A48 gray cast iron ASTM A536 ductile (nodular) iron

Bronze

Material Designation1

Heat Treatment

Typical Minimum Surface Hardness2

Gear Contact Strength,3 Sc, psi

Class 20

As cast

50 000–60 000

Class 30

As cast

174 HB

65 000–75 000

Class 40

As cast

201 HB

75 000–85 000

Grade 60–40–18

Annealed

140 HB

77 000–92 000

Grade 80–55–06

Quenched and tempered

179 HB

77 000–92 000

Grade 100–70–03

Quenched and tempered

229 HB

92 000–112 000

Grade 120–90–02

Quenched and tempered

269 HB

103 000–126 000

Sand cast

Minimum tensile strength 40 000 psi

30 000

ASTM B-148 Alloy 954

Heat treated

Minimum tensile strength 90 000 psi

65 000

Notes: 1 See ANSI/AGMA 2004-B89, Gear Materials and Heat Treatment Manual. 2 Hardness to be equivalent to that at the start of active profile in the center of the face width. 3 The lower values should be used for general design purposes. The upper values may be used when:   High-quality material is used.   Section size and design allow maximum response to heat treatment.   Proper quality control is effected by adequate inspection.   Operating experience justifies their use. Source: ANSI/AGMA 2001-D04.

AGMA allowable stress numbers (strengths) for bending and contact stress are for

∙ Unidirectional loading ∙ 10 million stress cycles ∙ 99 percent reliability The factors in this section, too, will be evaluated in subsequent sections. When two-way (reversed) loading occurs, as with idler gears, AGMA recommends using 70 percent of St values. This is equivalent to 1∕0.70 = 1.43 as a value of kf in Example 14–2. The recommendation falls between the value of kf = 1.33 for a Goodman failure locus and kf = 1.66 for a Gerber failure locus.

14–5  Geometry Factors I and J (ZI and YJ) We have seen how the factor Y is used in the Lewis equation to introduce the effect of tooth form into the stress equation. The AGMA factors5 I and J are intended to accomplish the same purpose in a more involved manner. 5

A useful reference is AGMA 908-B89, Geometry Factors for Determining Pitting Resistance and Bending Strength of Spur, Helical and Herringbone Gear Teeth.

758      Mechanical Engineering Design

The determination of I and J depends upon the face-contact ratio mF. This is defined as

mF =

F px

(14–19)

where px is the axial pitch and F is the face width. For spur gears, mF = 0. Low-contact-ratio (LCR) helical gears having a small helix angle or a thin face width, or both, have face-contact ratios less than unity (mF ≤ 1), and will not be considered here. Such gears have a noise level not too different from that for spur gears. Consequently we shall consider here only spur gears with mF = 0 and conventional helical gears with mF > 1. Bending-Strength Geometry Factor J (YJ) The AGMA factor J employs a modified value of the Lewis form factor, also denoted by Y; a fatigue stress-concentration factor Kf; and a tooth load-sharing ratio mN. The resulting equation for J for spur and helical gears is

J=

Y Kf m N

(14–20)

It is important to note that the form factor Y in Equation (14–20) is not the Lewis factor at all. The value of Y here is obtained from calculations within AGMA 908-B89, and is often based on the highest point of single-tooth contact. The factor Kf in Equation (14–20) is called a stress-correction factor by AGMA. It is based on a formula deduced from a photoelastic investigation of stress concentration in gear teeth over 50 years ago. The load-sharing ratio mN is equal to the face width divided by the minimum total length of the lines of contact. This factor depends on the transverse contact ratio mp, the face-contact ratio mF, the effects of any profile modifications, and the tooth deflection. For spur gears, mN = 1.0. For helical gears having a face-contact ratio mF > 2.0, a conservative approximation is given by the equation

mN =

pN 0.95Z

(14–21)

where pN is the normal base pitch and Z is the length of the line of action in the transverse plane (distance Lab in Figure 13–11). Use Figure 14–6 to obtain the geometry factor J for spur gears having a 20° pressure angle and full-depth teeth. Use Figures 14–7 and 14–8 for helical gears having a 20° normal pressure angle and face-contact ratios of mF = 2 or greater. For other gears, consult the AGMA standard. Surface-Strength Geometry Factor I (ZI) The factor I is also called the pitting-resistance geometry factor by AGMA. We will develop an expression for I by noting that the sum of the reciprocals of Equation (14–14), from Equation (14–12), can be expressed as

1 1 2 1 1 + = + r1 r2 sin ϕt ( dP d G )

(a)

Spur and Helical Gears     759

Pinion addendum 1.000

2.400 Whole depth

0.55

Geometry factor J

0.50

0.45

0.35 rT

1000 170 85 50 35 25 17

20°

Generating rack 1 pitch

Load applied at highest point of single-tooth contact

Gear addendum 1.000

Addendum 1.000

0.60

Number of teeth in mating gear

0.40

0.50

0.45

0.40

0.35

0.30

0.30 Load applied at tip of tooth

0.25

0.20

12

0.20

15

17

20

24

30

35

40 45 50

60

80

125

275

Number of teeth for which geometry factor is desired

Figure 14–6 Spur-gear geometry factors J. Source: The graph is from AGMA 218.01, which is consistent with tabular data from the current AGMA 908-B89. The graph is convenient for design purposes.

where we have replaced ϕ by ϕt, the transverse pressure angle, so that the relation will apply to helical gears too. Now define speed ratio mG as

mG =

NG dG = NP dP

(14–22)

Equation (a) can now be written mG + 1 1 1 2 + = r1 r2 dP sin ϕt mG

(b)

Now substitute Equation (b) for the sum of the reciprocals in Equation (14–14). The result is found to be

0.55

0.35

0.25

0.60

σc = ∣σC∣ = Cp

[

Kv W t 1 dP F cos ϕt sin ϕt m G 2 mG + 1

]

1∕2

(c)

760      Mechanical Engineering Design

Add. 1.0 Pnd 2.355 Pnd

Tooth height

Generating rack

20°

rT = 0.4276 Pnd (a) mN =

pN 0.95Z

Value for Z is for an element of indicated numbers of teeth and a 75-tooth mate Normal tooth thickness of pinion and gear tooth each reduced 0.024 in to provide 0.048 in total backlash for one normal diametral pitch 0.70

500 150 60

0.50

30

Number of teeth

Geometry factor J '

0.60 Factors are for teeth cut with a full fillet hob

20 0.40

0.30 0°

10°

15°

20°

25°

30°

35°

Helix angle ψ (b)

Figure 14–7 Helical-gear geometry factors J′. Source: The graph is from AGMA 218.01, which is consistent with tabular data from the current AGMA 908-B89. The graph is convenient for design purposes.

The geometry factor I for external spur and helical gears is the denominator of the second term in the brackets in Equation (c). By adding the load-sharing ratio mN, we obtain a factor valid for both spur and helical gears. The equation is then written as

 cos ϕt sin ϕt mG   2mN mG + 1 I=  cos ϕt sin ϕt mG  2mN mG − 1 

external gears internal gears

(14–23)

Spur and Helical Gears     761

Figure 14–8

1.05

500 150 75 50

Modifying factor

1.00

30 20

0.95

Number of teeth in mating element

The modifying factor can be applied to the J factor when other than 75 teeth are used in the mating element

0.90

0.85 0°

10°

15°

20°

25°

30°

35°

Helix angle ψ

where mN = 1 for spur gears. In solving Equation (14–21) for mN, note that

pN = pn cos ϕn

(14–24)

where pn is the normal circular pitch. The quantity Z, for use in Equation (14–21), can be obtained from the equation

Z = [(rP + a) 2 − rb2P]1∕2 + [(rG + a) 2 − rb2G]1∕2 − (rP + rG ) sin ϕt

(14–25)

where rP and rG are the pitch radii and rbP and rbG the base-circle radii of the pinion and gear, respectively.6 Recall from Equation (13–6), the radius of the base circle is

rb = r cos ϕt

(14–26)

Certain precautions must be taken in using Equation (14–25). The tooth profiles are not conjugate below the base circle, and consequently, if either one or the other of the first two terms in brackets is larger than the third term, then it should be replaced by the third term. In addition, the effective outside radius is sometimes less than r + a, owing to removal of burrs or rounding of the tips of the teeth. When this is the case, always use the effective outside radius instead of r + a.

14–6  The Elastic Coefficient Cp (ZE) Values of Cp may be computed directly from Equation (14–13) or obtained from Table 14–8.

6

For a development, see Joseph E. Shigley and John J. Uicker Jr., Theory of Machines and Mechanisms, McGraw-Hill, New York, 1980, p. 262.

J′-factor multipliers for use with Figure 14–7 to find J. Source: The graph is from AGMA 218.01, which is consistent with tabular data from the current AGMA 908-B89. The graph is convenient for design purposes.

762 √psi (√MPa)

Gear Material and Modulus of Elasticity EG, lbf/in2 (MPa)*

25 × 106 2180 2090 2070 2020 1900 1850 (1.7 × 105) (181) (174) (172) (168) (158) (154) 24 × 106 2160 2070 2050 2000 1880 1830 (1.7 × 105) (179) (172) (170) (166) (156) (152) 22 × 106 2100 2020 2000 1960 1850 1800 (1.5 × 105) (174) (168) (166) (163) (154) (149) 17.5 × 106 1950 1900 1880 1850 1750 1700 (1.2 × 105) (162) (158) (156) (154) (145) (141) 16 × 106 1900 1850 1830 1800 1700 1650 (1.1 × 105) (158) (154) (152) (149) (141) (137)

Malleable iron

Nodular iron

Cast iron

Aluminum bronze

Tin bronze

Source: AGMA 218.01

*When more exact values for modulus of elasticity are obtained from roller contact tests, they may be used.

Poisson's ratio = 0.30.

30 × 106 2300 2180 2160 2100 1950 1900 (2 × 105) (191) (181) (179) (174) (162) (158)

Steel

Malleable Nodular Cast Aluminum Tin Pinion Modulus of Steel Iron Iron Iron Bronze Bronze Pinion Elasticity Ep 30 × 106 25 × 106 24 × 106 22 × 106 17.5 × 106 16 × 106 Material psi (MPa)* (2 × 105) (1.7 × 105) (1.7 × 105) (1.5 × 105) (1.2 × 105) (1.1 × 105)

Table 14–8  Elastic Coefficient Cp (ZE),

Spur and Helical Gears     763

14–7  Dynamic Factor Kv As noted earlier, dynamic factors are used to account for inaccuracies in the manufacture and meshing of gear teeth in action. Transmission error is defined as the departure from uniform angular velocity of the gear pair. Some of the effects that produce transmission error are: ∙ Inaccuracies produced in the generation of the tooth profile; these include errors in tooth spacing, profile lead, and runout ∙ Vibration of the tooth during meshing due to the tooth stiffness ∙ Magnitude of the pitch-line velocity ∙ Dynamic unbalance of the rotating members ∙ Wear and permanent deformation of contacting portions of the teeth ∙ Gearshaft misalignment and the linear and angular deflection of the shaft ∙ Tooth friction In an attempt to account for these effects, AGMA has defined a set of quality numbers, Qv.7 These numbers define the tolerances for gears of various sizes manufactured to a specified accuracy. Quality numbers 3 to 7 will include most commercialquality gears. Quality numbers 8 to 12 are of precision quality. The following equations for the dynamic factor are based on these Qv numbers:

 A + √V B  ) ( A Kv =   A + √200V B  ) A (

V in ft/min

(14–27)

V in m/s

where

A = 50 + 56(1 − B)

B = 0.25(12 − Qv ) 2∕3

(14–28)

Figure 14–9 graphically represents Equation (14–27). The maximum recommended pitch-line velocity for a given quality number is represented by the end point of each Qv curve, and is given by

(Vt ) max

 [A + (Q − 3)]2 v  =  2  [A + (Qv − 3)] 

200

7

ft/min

(14–29)

m/s

AGMA 2000-A88. ANSI/AGMA 2001-D04, adopted in 2004, replaced the quality number Qv with the transmission accuracy level number Av and incorporated ANSI/AGMA 2015-1-A01. Av ranges from 6 to 12, with lower numbers representing greater accuracy. The Qv approach was maintained as an alternate approach, and resulting Kv values are comparable.

764      Mechanical Engineering Design

Figure 14–9 Qv = 5

1.8

Qv = 6

1.7

Dynamic factor, Kv

Dynamic factor Kv. The equations to these curves are given by Equation (14–27) and the end points by Equation (14–29). (ANSI/AGMA 2001-D04, Annex A)

Qv = 7

1.6 Qv = 8

1.5

Qv = 9

1.4 1.3

Qv = 10

1.2

Qv = 11

1.1 1.0

"Very Accurate Gearing" 0

2000

4000

6000

8000

10 000

Pitch-line velocity, Vt , ft /min

14–8  Overload Factor Ko The overload factor Ko is intended to make allowance for all externally applied loads in excess of the nominal tangential load W t in a particular application (see Figures 14–17 and 14–18 for tables). Examples include variations in torque from the mean value due to firing of cylinders in an internal combustion engine or reaction to torque variations in a piston pump drive. There are other similar factors such as application factor or service factor. These factors are established after considerable field experience in a particular application.8

14–9  Surface Condition Factor Cf (ZR) The surface condition factor Cf or ZR is used only in the pitting resistance equation, Equation (14–16). It depends on ∙ Surface finish as affected by, but not limited to, cutting, shaving, lapping, grinding, shotpeening ∙ Residual stress ∙ Plastic effects (work hardening) Standard surface conditions for gear teeth have not yet been established. When a detrimental surface finish effect is known to exist, AGMA specifies a value of Cf greater than unity.

8

An extensive list of service factors appears in Howard B. Schwerdlin, "Couplings," Chap. 16 in Joseph E. Shigley, Charles R. Mischke, and Thomas H. Brown, Jr. (eds.), Standard Handbook of Machine Design, 3rd ed., McGraw-Hill, New York, 2004.

Spur and Helical Gears     765

14–10  Size Factor Ks The size factor reflects nonuniformity of material properties due to size. It depends upon ∙ ∙ ∙ ∙ ∙ ∙ ∙

Tooth size Diameter of part Ratio of tooth size to diameter of part Face width Area of stress pattern Ratio of case depth to tooth size Hardenability and heat treatment

Standard size factors for gear teeth have not yet been established for cases where there is a detrimental size effect. In such cases AGMA recommends a size factor greater than unity. If there is no detrimental size effect, use unity. AGMA has identified and provided a symbol for size factor. Also, AGMA suggests Ks = 1, which makes Ks a placeholder in Equations (14–15) and (14–16) until more information is gathered. Following the standard in this manner is a failure to apply all of your knowledge. From Table 13–1, l = a + b = 2.25∕P. The tooth thickness t in Figure 14–6 is given in Section 14–1, Equation (b), as t = √4lx where x = 3Y∕(2P) from Equation (14–3). From Equation (6–24), the equivalent diameter de of a rectangular section in bending is de = 0.808 √Ft. From Equation (6–19), kb = (de∕0.3)−0.107. Noting that Ks is the reciprocal of kb, we find the result of all the algebraic substitution is

Ks =

1 F √Y 0.0535 = 1.192 ( kb P )

(a)

Ks can be viewed as Lewis's geometry incorporated into the Marin size factor in fatigue. You may set Ks = 1, or you may elect to use the preceding Equation (a). This is a point to discuss with your instructor. We will use Equation (a) to remind you that you have a choice. If Ks in Equation (a) is less than 1, use Ks = 1.

14–11  Load-Distribution Factor Km (KH) The load-distribution factor modified the stress equations to reflect nonuniform distribution of load across the line of contact. The ideal is to locate the gear "midspan" between two bearings at the zero slope place when the load is applied. However, this is not always possible. The following procedure is applicable to ∙ ∙ ∙ ∙

Net face width to pinion pitch diameter ratio F∕dP ≤ 2 Gear elements mounted between the bearings Face widths up to 40 in Contact, when loaded, across the full width of the narrowest member

The load-distribution factor under these conditions is currently given by the face load distribution factor, Cmf, where

Km = Cmf = 1 + Cmc (Cp f Cpm + Cma Ce )

(14–30)

766      Mechanical Engineering Design

where

1 Cmc = { 0.8

for uncrowned teeth for crowned teeth

 F − 0.025   10dP  F Cpf =  − 0.0375 + 0.0125F  10dP  F  − 0.1109 + 0.0207F − 0.000 228F 2  10dP

(14–31)

F ≤ 1 in 1 < F ≤ 17 in

(14–32)

17 < F ≤ 40 in

Note that for values of F∕(10dP) < 0.05, F∕(10dP) = 0.05 is used. 1 Cpm = { 1.1

for straddle-mounted pinion with S1∕S < 0.175 for straddle-mounted pinion with S1∕S ≥ 0.175

(14–33)

   Cma = A + BF + CF 2    (see Table 14–9 for values of A, B, and C) (14–34)  0.8  Ce =   1

for gearing adjusted at assembly, or compatibility is improved by lapping, or both

(14–35)

for all other conditions

See Figure 14–10 for definitions of S and S1 for use with Equation (14–33), and see Figure 14–11 for graph of Cma. Table 14–9  Empirical Constants A, B, and C for Equation (14–34), Face Width F in Inches* Condition

A

C

Open gearing

0.247

0.0167

−0.765(10−4)

Commercial, enclosed units

0.127

0.0158

−0.930(10−4)

Precision, enclosed units

0.0675

0.0128

−0.926(10−4)

Extraprecision enclosed gear units

0.00360

0.0102

−0.822(10−4)

*See ANSI/AGMA 2101-D04, pp. 20–22, for SI formulation. Source: ANSI/AGMA 2001-D04.

Figure 14–10 Definition of distances S and S1 used in evaluating Cpm, Equation (14–33). (ANSI/AGMA 2001-D04.)

B

Centerline of gear face Centerline of bearing

Centerline of bearing

S 2

S1 S

Spur and Helical Gears     767

0.90 Open gearing

Mesh alignment factor, Cma

0.80 0.70 0.60

Commercial enclosed gear units Curve 1

0.50 0.40

Curve 2

0.30

Curve 3

Precision enclosed gear units Extra precision enclosed gear units

0.20 Curve 4

0.10 0.0

For determination of Cma , see Eq. (14–34) 0

5

10

15

20

25

30

Face width, F (in)

Figure 14–11 Mesh alignment factor Cma. Curve-fit equations in Table 14–9. (ANSI/AGMA 2001-D04.)

14–12  Hardness-Ratio Factor CH (ZW) The pinion generally has a smaller number of teeth than the gear and consequently is subjected to more cycles of contact stress. If both the pinion and the gear are throughhardened, then a uniform surface strength can be obtained by making the pinion harder than the gear. A similar effect can be obtained when a surface-hardened pinion is mated with a through-hardened gear. The hardness-ratio factor CH is used only for the gear. Its purpose is to adjust the surface strengths for this effect. For the pinion, CH  = 1. For the gear, CH is obtained from the equation where

CH = 1.0 + A′(m G − 1.0) A′ = 8.98(10−3 ) (

HBP − 8.29(10−3 ) HBG )

1.2 ≤

(14–36)

HBP ≤ 1.7 HBG

The terms HBP and HBG are the Brinell hardness (10-mm ball at 3000-kg load) of the pinion and gear, respectively. The term mG is the speed ratio and is given by Equation (14–22). See Figure 14–12 for a graph of Equation (14–36). For

HBP < 1.2, HBG

A′ = 0

HBP > 1.7, HBG

A′ = 0.006 98

When surface-hardened pinions with hardnesses of 48 Rockwell C scale (Rockwell C48) or harder are run with through-hardened gears (180–400 Brinell), a work hardening occurs. The CH factor is a function of pinion surface finish fP and the mating gear hardness. Figure 14–13 displays the relationships:

CH = 1 + B′(450 − HBG )

(14–37)

35

768      Mechanical Engineering Design

1.12

1.7 1.6

1.10

1.5

1.08

1.4 1.3

1.06

1.2

1.04 When HBP < 1.2, HBG Use CH = 1

1.02 1.00

HBP HBG

Hardness-ratio factor CH (through-hardened steel). (ANSI/AGMA 2001-D04.)

Calculated hardness-ratio,

1.14

Hardness-ratio factor, CH

Figure 14–12

0

2

4

6

8

10

12

14

16

18

20

Single reduction gear ratio mG

Figure 14–13

Surface Finish of Pinion, fP, microinches, Ra

1.16

Hardness-ratio factor CH (surface-hardened steel pinion). (ANSI/AGMA 2001-D04.) Hardness-ratio factor, CH

1.14 fP = 16

1.12

fP = 32

1.10 1.08

fP = 64

1.06 1.04 1.02 1.00 180

When fP > 64 use CH = 1.0 200

250

300

350

400

Brinell hardness of the gear, HBG

where B′ = 0.000 75 exp[−0.0112fP] and fP is the surface finish of the pinion expressed as root-mean-square roughness Ra in μ in.

14–13  Stress-Cycle Factors YN and ZN The AGMA strengths as given in Figures 14–2 through 14–4, in Tables 14–3 and 14–4 for bending fatigue, and in Figure 14–5 and Tables 14–5 and 14–6 for contact-stress fatigue are based on 107 load cycles applied. The purpose of the stress-cycle factors YN and ZN is to modify the gear strength for lives other than 107 cycles. Values for these factors are given in Figures 14–14 and 14–15. Note that for 107 cycles YN = ZN = 1 on each graph. Note also that the equations for YN and ZN change on either side of 107 cycles. For life goals slightly higher than 107 cycles, the mating gear may be experiencing fewer than 107 cycles and the equations for (YN)P and (YN)G can be different. The same comment applies to (ZN)P and (ZN)G.

Spur and Helical Gears     769 5.0

Stress-cycle factor, YN

4.0

YN = 9.4518 N

400 HB

YN = 6.1514 N

3.0 Case carb. 250 HB Nitrided 2.0 160 HB

Repeatedly applied bending strength stress-cycle factor YN. (ANSI/AGMA 2001-D04.)

Pitchline velocity Gear material cleanliness Residual stress Material ductility and fracture toughness

– 0.1192

YN = 4.9404 N – 0.1045

YN = 3.517 N – 0.0817 YN = 1.3558 N – 0.0178

YN = 2.3194 N – 0.0538 1.0 0.9 0.8 0.7 0.6 0.5 10 2

Figure 14–14

NOTE: The choice of YN in the shaded area is influenced by:

– 0.148

YN = 1.6831 N – 0.0323

10 3

10 4

10 5

10 6

10 7

10 8

10 9

1.0 0.9 0.8 0.7 0.6 0.5 10 10

Number of load cycles, N

5.0

NOTE: The choice of Z N in the shaded zone is influenced by:

4.0

Lubrication regime Failure criteria Smoothness of operation required Pitchline velocity Gear material cleanliness Material ductility and fracture toughness Residual stress

Stress-cycle factor, Z N

3.0 2.0 ZN = 2.466 N – 0.056

ZN = 1.4488 N – 0.023 1.1 1.0 0.9 0.8 0.7 0.6 0.5 102

Nitrided ZN = 1.249 N – 0.0138

103

104

105

106

107

108

109

1010

Number of load cycles, N

14–14  Reliability Factor KR (YZ) The reliability factor accounts for the effect of the statistical distributions of material fatigue failures. Load variation is not addressed here. The gear strengths St and Sc are based on a reliability of 99 percent. Table 14–10 is based on data developed by the U.S. Navy for bending and contact-stress fatigue failures. The functional relationship between KR and reliability is highly nonlinear. When interpolation is required, linear interpolation is too crude. A log transformation to each quantity produces a linear string. A least-squares regression fit is

0.658 − 0.0759 ln(1 − R) KR = { 0.50 − 0.109 ln(1 − R)

0.5 < R < 0.99 0.99 ≤ R ≤ 0.9999

(14–38)

For cardinal values of R, take KR from the table. Otherwise use the logarithmic interpolation afforded by Equations (14–38).

Figure 14–15 Pitting resistance stress-cycle factor ZN. (ANSI/AGMA 2001-D04.)

770      Mechanical Engineering Design

Table 14–10  Reliability Factors KR (YZ) Reliability

KR (YZ)

0.9999

1.50

0.999

1.25

0.99

1.00

0.90

0.85

0.50

0.70

Source: ANSI/AGMA 2001-D04.

14–15  Temperature Factor KT (Yθ) For oil or gear-blank temperatures up to 250°F (120°C), use KT = Yθ = 1.0. For higher temperatures, the factor should be greater than unity. Heat exchangers may be used to ensure that operating temperatures are considerably below this value, as is desirable for the lubricant.

14–16  Rim-Thickness Factor KB When the rim thickness is not sufficient to provide full support for the tooth root, the location of bending fatigue failure may be through the gear rim rather than at the tooth fillet. In such cases, the use of a stress-modifying factor KB is recommended. This factor, the rim-thickness factor KB, adjusts the estimated bending stress for the thin-rimmed gear. It is a function of the backup ratio mB,

mB =

tR ht

(14–39)

where tR = rim thickness below the tooth, and ht = the tooth height. The geometry is depicted in Figure 14–16. The rim-thickness factor KB is given by 2.242   1.6 ln m B KB =  1 

Figure 14–16

2.4 2.2 Rim-thickness factor, KB

Rim-thickness factor KB. (ANSI/AGMA 2001-D04.)

2.0

For mB < 1.2 KB = 1.6 ln 2.242 mB

( (

1.2

tR

For mB ≥ 1.2 KB = 1.0

1.4

mB ≥ 1.2

ht

1.8 1.6

mB < 1.2

mB =

tR ht

1.0

0 0.5 0.6

0.8

1.0

1.2

2

3

Backup ratio, mB

4

5

6

7 8 9 10

(14–40)

Spur and Helical Gears     771

Figure 14–16 also gives the value of KB graphically. The rim-thickness factor KB is applied in addition to the 0.70 reverse-loading factor when applicable.

14–17  Safety Factors SF and SH The ANSI/AGMA standards 2001-D04 and 2101-D04 contain a safety factor SF guarding against bending fatigue failure and safety factor SH guarding against pitting failure. The definition of SF, from Equation (14–17), for U.S. customary units, is

SF =

StYN ∕(KT KR ) fully corrected bending strength = (14–41) σ bending stress

where σ is estimated from Equation (14–15), for U.S. customary units. It is a strengthover-stress definition in a case where the stress is linear with the transmitted load. The definition of SH, from Equation (14–18), is

SH =

Sc ZN CH ∕(KT K R ) fully corrected contact strength = (14–42) σc contact stress

when σc is estimated from Equation (14–16). This, too, is a strength-over-stress definition but in a case where the stress is not linear with the transmitted load W t. While the definition of SH does not interfere with its intended function, a caution is required when comparing SF with SH in an analysis in order to ascertain the nature and severity of the threat to loss of function. To render SH linear with the transmitted load, W t it could have been defined as

SH = (

fully corrected contact strength 2 ) contact stress imposed

(14–43)

with the exponent 2 for linear or helical contact, or an exponent of 3 for crowned teeth (spherical contact). With the definition, Equation (14–42), compare SF with SH2 (or SH3 for crowned teeth) when trying to identify the threat to loss of function with confidence. The role of the overload factor Ko is to include predictable excursions of load beyond W t based on experience. A safety factor is intended to account for unquantifiable elements in addition to Ko. When designing a gear mesh, the quantity SF becomes a design factor (SF)d within the meanings used in this book. The quantity SF evaluated as part of a design assessment is a factor of safety. This applies equally well to the quantity SH.

14–18  Analysis Description of the procedure based on the AGMA standard is highly detailed. The best review is a "road map" for bending fatigue and contact-stress fatigue. Figure 14–17 identifies the bending stress equation, the endurance strength in bending equation, and the factor of safety SF. Figure 14–18 displays the contactstress equation, the contact fatigue endurance strength equation, and the factor of safety SH. The equations in these figures are in terms of U.S. customary units. Similar road maps can readily be generated in terms of SI units. The following example of a gear mesh analysis is intended to make all the details presented concerning the AGMA method more familiar.

772      Mechanical Engineering Design

SPUR GEAR BENDING Based on ANSI/AGMA 2001-D04 (U.S. customary units) NP Pd

dP =

V = πdn 12 W = 33 000 H V

1 [or Eq. (a), Sec. 14 –10]

t

Gear bending stress equation Eq. (14 –15)

Eq. (14 –30)

P K K σ = W KoKυKs d m B J F

Eq. (14 – 40)

t

Fig. 14 – 6 Eq. (14 –27) Table below 0.99 (St )107

Gear bending endurance strength equation Eq. (14–17) Bending factor of safety Eq. (14–41)

Tables 14 –3, 14 – 4 St YN SF KT KR

σall =

Fig. 14 –14 Table 14 –10, Eq. (14 –38)

1 if T < 250°F

SF =

St YN /(KT KR) σ

Remember to compare SF with S 2H when deciding whether bending or wear is the threat to function. For crowned gears compare SF with S 3H . Table of Overload Factors, Ko Driven Machine Power source Uniform Light shock Medium shock

Uniform Moderate shock Heavy shock 1.00 1.25 1.50

1.25 1.50 1.75

1.75 2.00 2.25

Figure 14–17 Road map of gear bending equations based on AGMA standards. (ANSI/AGMA 2001-D04.)

Spur and Helical Gears     773

SPUR GEAR WEAR Based on ANSI/AGMA 2001-D04 (U.S. customary units) NP Pd

dP =

V = πdn 12

1 [or Eq. (a), Sec. 14 –10] Eq. (14 –30) 1

W = 33 000 H V t

Gear contact stress equation Eq. (14 –16)

(

K C σσc = Cp W tKoKυKs m f dP F I

Gear contact endurance strength Eq. (14–18) Wear factor of safety Eq. (14–42)

1/2

Eq. (14 –23)

Eq. (14 –27)

Eq. (14 –13), Table 14 – 8 0.99 (Sc )107

)

Table below Tables 14 –6, 14 –7 Fig. 14 –15

σσcc,all =

Sc Z N CH SH KT KR

Section 14 –12, gear only

Table 14 –10, Eq. (14 –38) 1 if T < 250°F Gear only SH =

Sc Z N CH /(KT KR) σc

Remember to compare SF with S 2H when deciding whether bending or wear is the threat to function. For crowned gears compare SF with S 3H . Table of Overload Factors, Ko Driven Machine Power source Uniform Light shock Medium shock

Uniform Moderate shock Heavy shock 1.00 1.25 1.50

1.25 1.50 1.75

1.75 2.00 2.25

Figure 14–18 Road map of gear wear equations based on AGMA standards. (ANSI/AGMA 2001-D04.)

774      Mechanical Engineering Design

EXAMPLE 14–4 A 17-tooth 20° pressure angle spur pinion rotates at 1800 rev/min and transmits 4 hp to a 52-tooth disk gear. The diametral pitch is 10 teeth/in, the face width 1.5 in, and the quality standard is No. 6. The gears are straddle-mounted with bearings immediately adjacent. The pinion is a grade 1 steel with a hardness of 240 Brinell tooth surface and through-hardened core. The gear is steel, through-hardened also, grade 1 material, with a Brinell hardness of 200, tooth surface and core. Poisson's ratio is 0.30, JP = 0.30, JG = 0.40, and Young's modulus is 30(106) psi. The loading is smooth because of motor and load. Assume a pinion life of 108 cycles and a reliability of 0.90, and use YN = 1.3558N−0.0178, ZN = 1.4488N−0.023. The tooth profile is uncrowned. This is a commercial enclosed gear unit. (a) Find the factor of safety of the gears in bending. (b) Find the factor of safety of the gears in wear. (c) By examining the factors of safety, identify the threat to each gear and to the mesh. Solution There will be many terms to obtain so use Figs. 14–17 and 14–18 as guides to what is needed.

dP = NP∕Pd = 17∕10 = 1.7 in

V=

πdP nP π(1.7)1800 = = 801.1 ft/min 12 12

Wt =

33 000 H 33 000(4) = = 164.8 lbf V 801.1

dG = 52∕10 = 5.2 in

Assuming uniform loading, Ko = 1. To evaluate Kv, from Eq. (14–28) with a quality number Qv = 6,

B = 0.25(12 − 6) 2∕3 = 0.8255

A = 50 + 56(1 − 0.8255) = 59.77

Then from Eq. (14–27) the dynamic factor is

Kv = (

59.77 + √801.1 0.8255 = 1.377 ) 59.77

To determine the size factor, Ks, the Lewis form factor is needed. From Table 14–2, with NP = 17 teeth, YP = 0.303. Interpolation for the gear with NG = 52 teeth yields YG = 0.412. Thus from Eq. (a) of Sec. 14–10, with F = 1.5 in,

(Ks ) P = 1.192(

1.5 √0.303 0.0535 = 1.043 ) 10

(Ks ) G = 1.192(

1.5 √0.412 0.0535 = 1.052 ) 10

The load distribution factor Km is determined from Eq. (14–30), where five terms are needed. They are, where F = 1.5 in when needed: Uncrowned, Eq. (14–30): Cmc = 1, Eq. (14–32): Cpf = 1.5∕[10(1.7)] − 0.0375 + 0.0125(1.5) = 0.0695 Bearings immediately adjacent, Eq. (14–33): Cpm = 1 Commercial enclosed gear units (Fig. 14–11): Cma = 0.15 Eq. (14–35): Ce = 1

Spur and Helical Gears     775

Thus,

Km = 1 + Cmc (Cpf Cpm + Cma Ce ) = 1 + (1)[0.0695(1) + 0.15(1)] = 1.22

Assuming constant thickness gears, the rim-thickness factor KB = 1. The speed ratio is mG = NG∕NP = 52∕17 = 3.059. The load cycle factors given in the problem statement, with N(pinion) = 108 cycles and N(gear) = 108∕mG = 108∕3.059 cycles, are

(YN ) P = 1.3558(108 ) −0.0178 = 0.977

(YN ) G = 1.3558(108∕3.059) −0.0178 = 0.996

From Table 14–10, with a reliability of 0.9, KR = 0.85. From Fig. 14–18, the temperature and surface condition factors are KT = 1 and Cf = 1. From Eq. (14–23), with mN = 1 for spur gears,

I=

cos 20° sin 20° 3.059 = 0.121 2 3.059 + 1

From Table 14–8, Cp = 2300 √psi. Next, we need the terms for the gear endurance strength equations. From Table 14–3, for grade 1 steel with HBP = 240 and HBG = 200, we use Fig. 14–2, which gives

(St ) P = 77.3(240) + 12 800 = 31 350 psi

(St ) G = 77.3(200) + 12 800 = 28 260 psi

Similarly, from Table 14–6, we use Fig. 14–5, which gives

(Sc ) P = 322(240) + 29 100 = 106 400 psi

(Sc ) G = 322(200) + 29 100 = 93 500 psi

From Fig. 14–15,

(ZN ) P = 1.4488(108 ) −0.023 = 0.948

(ZN ) G = 1.4488(108∕3.059) −0.023 = 0.973

For the hardness ratio factor CH, the hardness ratio is HBP∕HBG = 240∕200 = 1.2. Then, from Sec. 14–12, A′ = 8.98(10−3 ) (HBP∕HBG ) − 8.29(10−3 )

= 8.98(10−3 ) (1.2) − 8.29(10−3 ) = 0.002 49

Thus, from Eq. (14–36),

CH = 1 + 0.002 49(3.059 − 1) = 1.005

(a) Pinion tooth bending. Substituting the appropriate terms for the pinion into Eq. (14–15) gives

(σ) P = (W tKo Kv Ks

= 6417 psi

Pd Km KB 10 1.22(1) = 164.8(1)1.377(1.043) F J )P 1.5 0.30

Substituting the appropriate terms for the pinion into Eq. (14–41) gives Answer

(SF ) P = (

St YN∕(KT KR ) 31 350(0.977)∕[1(0.85)] = = 5.62 )P σ 6417

776      Mechanical Engineering Design

Gear tooth bending. Substituting the appropriate terms for the gear into Eq. (14–15) gives

(σ) G = 164.8(1)1.377(1.052)

10 1.22(1) = 4854 psi 1.5 0.40

Substituting the appropriate terms for the gear into Eq. (14–41) gives (SF ) G =

Answer

28 260(0.996)∕[1(0.85)] = 6.82 4854

(b) Pinion tooth wear. Substituting the appropriate terms for the pinion into Eq. (14–16) gives

Km Cf 1∕2 (σc ) P = CP(W t Ko Kv Ks dP F I )P 1.22 1 1∕2 = 2300 [ 164.8(1)1.377(1.043) = 70 360 psi 1.7(1.5) 0.121 ]

Substituting the appropriate terms for the pinion into Eq. (14–42) gives Answer

(SH ) P = [

Sc ZN∕(KT KR ) 106 400(0.948)∕[1(0.85)] = 1.69 ]P = σc 70 360

Gear tooth wear. The only term in Eq. (14–16) that changes for the gear is Ks. Thus,

(σc ) G = [

(Ks ) G 1∕2 1.052 1∕2 (σ ) = c P ( 1.043 ) 70 360 = 70 660 psi (Ks ) P ]

Substituting the appropriate terms for the gear into Eq. (14–42) with CH = 1.005 gives Answer

(SH ) G =

93 500(0.973)1.005∕[1(0.85)] = 1.52 70 660

(c) For the pinion, we compare (SF)P with (SH)P2, or 5.73 with 1.692 = 2.86, so the threat in the pinion is from wear. For the gear, we compare (SF)G with (SH)G2 , or 6.96 with 1.522 = 2.31, so the threat in the gear is also from wear.

There are perspectives to be gained from Example 14–4. First, the pinion is overly strong in bending compared to wear. The performance in wear can be improved by surface-hardening techniques, such as flame or induction hardening, nitriding, or carburizing and case hardening, as well as shot peening. This in turn permits the gearset to be made smaller. Second, in bending, the gear is stronger than the pinion, indicating that both the gear core hardness and tooth size could be reduced; that is, we may increase P and reduce the diameters of the gears, or perhaps allow a cheaper material. Third, in wear, surface strength equations have the ratio (ZN)∕KR. The values of (ZN)P and (ZN)G are affected by gear ratio mG. The designer can control strength by specifying surface hardness. This point will be elaborated later. Having followed a spur-gear analysis in detail in Example 14–4, it is timely to analyze a helical gearset under similar circumstances to observe similarities and differences.

Spur and Helical Gears     777

EXAMPLE 14–5 A 17-tooth 20° normal pitch-angle helical pinion with a right-hand helix angle of 30° rotates at 1800 rev/min when transmitting 4 hp to a 52-tooth helical gear. The normal diametral pitch is 10 teeth/in, the face width is 1.5 in, and the set has a quality number of 6. The gears are straddle-mounted with bearings immediately adjacent. The pinion and gear are made from a through-hardened steel with surface and core hardnesses of 240 Brinell on the pinion and surface and core hardnesses of 200 Brinell on the gear. The transmission is smooth, connecting an electric motor and a centrifugal pump. Assume a pinion life of 108 cycles and a reliability of 0.9 and use the upper curves in Figs. 14–14 and 14–15. (a) Find the factors of safety of the gears in bending. (b) Find the factors of safety of the gears in wear. (c) By examining the factors of safety identify the threat to each gear and to the mesh. Solution All of the parameters in this example are the same as in Ex. 14–4 with the exception that we are using helical gears. Thus, several terms will be the same as Ex. 14–4. The reader should verify that the following terms remain unchanged: Ko = 1, YP = 0.303, YG = 0.412, mG = 3.059, (Ks)P = 1.043, (Ks)G = 1.052, (YN)P = 0.977, (YN)G = 0.996, KR = 0.85, KT = 1, Cf = 1, Cp = 2300 √psi, (St)P = 31 350 psi, (St)G = 28 260 psi, (Sc)P = 106 380 psi, (Sc)G = 93 500 psi, (ZN)P = 0.948, (ZN)G = 0.973, and CH = 1.005. For helical gears, the transverse diametral pitch, given by Eq. (13–18), is

Pt = Pn cos ψ = 10 cos 30° = 8.660 teeth/in

Thus, the pitch diameters are dP = NP∕Pt = 17∕8.660 = 1.963 in and dG = 52∕8.660 = 6.005 in. The pitchline velocity and transmitted force are πdP nP π(1.963)1800 = = 925 ft/min 12 12 33 000H 33 000(4) Wt = = = 142.7 lbf V 925 As in Ex. 14–4, for the dynamic factor, B = 0.8255 and A = 59.77. Thus, Eq. (14–27) gives

V=

Kv = (

59.77 + √925 0.8255 = 1.404 ) 59.77

The geometry factor I for helical gears requires a little work. First, the transverse pressure angle is given by Eq. (13–19),

ϕt = tan−1 (

tan ϕn tan 20° = tan−1 ( = 22.80° ) cos ψ cos 30° )

The radii of the pinion and gear are rP = 1.963∕2 = 0.9815 in and rG = 6.004∕2 = 3.002 in, respectively. The addendum is a = 1∕Pn = 1∕10 = 0.1, and the base-circle radii of the pinion and gear are given by Eq. (13–6), with ϕ = ϕt: (rb ) P = rP cos ϕt = 0.9815 cos 22.80° = 0.9048 in

(rb ) G = 3.002 cos 22.80° = 2.767 in

From Eq. (14–25), the surface strength geometry factor

Z = √ (0.9815 + 0.1) 2 − 0.90482 + √ (3.004 + 0.1) 2 − 2.7692 −(0.9815 + 3.004) sin 22.80° = 0.5924 + 1.4027 − 1.544 4 = 0.4507 in

778      Mechanical Engineering Design

Since the first two terms are less than 1.544 4, the equation for Z stands. From Eq. (14–24) the normal circular pitch pN is

pN = pn cos ϕn =

π π cos 20° = cos 20° = 0.2952 in Pn 10

From Eq. (14–21), the load sharing ratio

mN =

pN 0.2952 = = 0.6895 0.95Z 0.95(0.4507)

Substituting in Eq. (14–23), the geometry factor I is

I=

sin 22.80° cos 22.80° 3.06 = 0.195 2(0.6895) 3.06 + 1

From Fig. 14–7, geometry factors J′P = 0.45 and J′G = 0.54. Also from Fig. 14–8 the J-factor multipliers are 0.94 and 0.98, correcting J′P and J′G to

JP = 0.45(0.94) = 0.423

JG = 0.54(0.98) = 0.529

The load-distribution factor Km is estimated from Eq. (14–32):

Cp f =

1.5 − 0.0375 + 0.0125(1.5) = 0.0577 10(1.963)

with Cmc = 1, Cpm = 1, Cma = 0.15 from Fig. 14–11, and Ce = 1. Therefore, from Eq. (14–30),

Km = 1 + (1)[0.0577(1) + 0.15(1)] = 1.208

(a) Pinion tooth bending. Substituting the appropriate terms into Eq. (14–15) using Pt gives

Pt Km KB 8.66 1.208(1) (σ) P = (W t K o K v Ks = 142.7(1)1.404(1.043) ) F J 1.5 0.423 P = 3445 psi

Substituting the appropriate terms for the pinion into Eq. (14–41) gives Answer

(SF ) P = (

StYN∕(KT KR ) 31 350(0.977)∕[1(0.85)] = = 10.5 ) σ 3445 P

Gear tooth bending. Substituting the appropriate terms for the gear into Eq. (14–15) gives

(σ) G = 142.7(1)1.404(1.052)

8.66 1.208(1) = 2779 psi 1.5 0.529

Substituting the appropriate terms for the gear into Eq. (14–41) gives Answer

(SF ) G =

28 260(0.996)∕[1(0.85)] = 11.9 2779

Spur and Helical Gears     779

(b) Pinion tooth wear. Substituting the appropriate terms for the pinion into Eq. (14–16) gives Km Cf 1∕2 (σc ) P = Cp (W tKo K v Ks dP F I )P

1.208 1 1∕2 = 2300 [ 142.7(1)1.404(1.043) = 48 230 psi 1.963(1.5) 0.195 ]

Substituting the appropriate terms for the pinion into Eq. (14–42) gives Answer

(SH ) P = (

Sc ZN∕(KT KR ) 106 400(0.948)∕[1(0.85)] = 2.46 )P = σc 48 230

Gear tooth wear. The only term in Eq. (14–16) that changes for the gear is Ks. Thus,

(σc ) G =

(Ks ) G 1∕2 1.052 1∕2 (σc ) P = 48 230 = 48 440 psi [ (Ks ) P ] ( 1.043 )

Substituting the appropriate terms for the gear into Eq. (14–42) with CH = 1.005 gives (SH ) G =

Answer

93 500(0.973)1.005∕[1(0.85)] = 2.22 48 440

(c) For the pinion we compare SF with S2H, or 10.5 with 2.462 = 6.05, so the threat in the pinion is from wear. For the gear we compare SF with S2H, or 11.9 with 2.222 = 4.93, so the threat is also from wear in the gear. For the meshing gearset wear controls.

It is worthwhile to compare Example 14–4 with Example 14–5. The spur and helical gearsets were placed in nearly identical circumstances. The helical gear teeth are of greater length because of the helix and identical face widths. The pitch diameters of the helical gears are larger. The J factors and the I factor are larger, thereby reducing stresses. The result is larger factors of safety. In the design phase the gearsets in Example 14–4 and Example 14–5 can be made smaller with control of materials and relative hardnesses. Now that examples have given the AGMA parameters substance, it is time to examine some desirable (and necessary) relationships between material properties of spur gears in mesh. In bending, the AGMA equations are displayed side by side:

Pd K m KB σP = (W tKo K v Ks F J )P

(SF ) P = (

StYN∕(KT KR ) )P σ

Pd Km KB σG = (W tKo Kv Ks F J )G (SF ) G = (

St YN∕(KT KR ) )G σ

Equating the factors of safety, substituting for stress and strength, canceling identical terms (Ks virtually equal or exactly equal), and solving for (St)G gives

(St ) G = (St ) P

(YN ) P JP (YN ) G JG

(a)

780      Mechanical Engineering Design

The stress-cycle factor YN comes from Figure 14–14, where for a particular hardness, YN = αNβ. For the pinion, (YN)P = αNPβ, and for the gear, (YN)G = α(NP∕mG)β. Substituting these into Equation (a) and simplifying gives (St ) G = (St ) P m βG

JP JG

(14–44)

Normally, mG > 1 and JG > JP, so Equation (14–44) shows that the gear can be less strong (lower Brinell hardness) than the pinion for the same safety factor.

EXAMPLE 14–6 In a set of spur gears, a 300-Brinell 18-tooth 16-pitch 20° full-depth pinion meshes with a 64-tooth gear. Both gear and pinion are of grade 1 through-hardened steel. Using β = −0.110, what hardness can the gear have for the same factor of safety? Solution For through-hardened grade 1 steel the pinion strength (St)P is given in Figure 14–2:

(St ) P = 77.3(300) + 12 800 = 35 990 psi

From Figure 14–6 the form factors are JP = 0.32 and JG = 0.41. Equation (14–44) gives (St ) G = 35 990(

64 −0.110 0.32 = 24 430 psi 18 ) 0.41

Use the equation in Figure 14–2 again. (HB ) G =

Answer

24 430 − 12 800 = 150 Brinell 77.3

The AGMA contact-stress equations also are displayed side by side:

Km Cf 1∕2 (σc ) P = Cp(W t K o Kv Ks dP F I )P

(SH ) P = (

Sc Z N∕(KT KR ) )P σc

(σc ) G = Cp(W t K o K v Ks

(SH ) G = (

Km Cf 1∕2 dP F I )G

Sc Z N CH∕(K T K R ) )G σc

Equating the factors of safety, substituting the stress relations, and canceling identical terms including Ks gives, after solving for (Sc)G,

(Sc ) G = (Sc ) P

(Z N ) P 1 1 = (SC ) P m βG( ) (Z N ) G ( CH )G CH G

where, as in the development of Equation (14–44), (ZN)P∕(ZN)G = mGβ and the value of β for wear comes from Figure 14–15. Since CH is so close to unity, it is usually neglected; therefore

(Sc ) G = (Sc ) P m Gβ

(14–45)

Spur and Helical Gears     781

EXAMPLE 14–7 For β = −0.056 for a through-hardened steel, grade 1, continue Example 14–6 for wear. Solution From Figure 14–5,

(Sc ) P = 322(300) + 29 100 = 125 700 psi

From Equation (14–45), (Sc ) G = (Sc ) P (

Answer

64 −0.056 64 −0.056 = 125 700 = 117 100 psi ( 18 ) 18 )

(HB ) G =

117 100 − 29 200 = 273 Brinell 322

which is slightly less than the pinion hardness of 300 Brinell.

Equations (14–44) and (14–45) apply as well to helical gears.

14–19  Design of a Gear Mesh A useful decision set for spur and helical gears includes ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙

 Function: load, speed, reliability, life, Ko  Unquantifiable risk: design factor nd  Tooth system: ϕ, ψ, addendum, dedendum, root fillet radius  a priori decisions  Gear ratio mG, Np, NG   Quality number Qv  Diametral pitch Pd   Face width F  design decisions  Pinion material, core hardness, case hardness   Gear material, core hardness, case hardness

The first item to notice is the dimensionality of the decision set. There are four design decision categories, eight different decisions if you count them separately. This is a larger number than we have encountered before. It is important to use a design strategy that is convenient in either longhand execution or computer implementation. The design decisions have been placed in order of importance (impact on the amount of work to be redone in iterations). The steps, after the a priori decisions have been made are 1 Choose a diametral pitch. 2 Examine implications on face width, pitch diameters, and material properties. If not satisfactory, return to pitch decision for change. 3 Choose a pinion material and examine core and case hardness requirements. If not satisfactory, return to pitch decision and iterate until no decisions are changed. 4 Choose a gear material and examine core and case hardness requirements. If not satisfactory, return to pitch decision and iterate until no decisions are changed.

782      Mechanical Engineering Design

With these plan steps in mind, we can consider them in more detail. First select a trial diametral pitch. Pinion bending: ∙ Select a median face width for this pitch, 4π∕P ∙ Find the range of necessary ultimate strengths ∙ Choose a material and a core hardness ∙ Find face width to meet factor of safety in bending ∙ Choose face width ∙ Check factor of safety in bending Gear bending: ∙ Find necessary companion core hardness ∙ Choose a material and core hardness ∙ Check factor of safety in bending Pinion wear: ∙ Find necessary Sc and attendant case hardness ∙ Choose a case hardness ∙ Check factor of safety in wear Gear wear: ∙ Find companion case hardness ∙ Choose a case hardness ∙ Check factor of safety in wear Completing this set of steps will yield a satisfactory design. Additional designs with diametral pitches adjacent to the first satisfactory design will produce several among which to choose. A figure of merit is necessary in order to choose the best. Unfortunately, a figure of merit in gear design is complex in an academic environment because material and processing costs vary. The possibility of using a process depends on the manufacturing facility if gears are made in house. After examining Example 14–4 and Example 14–5 and seeing the wide range of factors of safety, one might entertain the notion of setting all factors of safety equal.9 In steel gears, wear is usually controlling and (SH)P and (SH)G can be brought close to equality. The use of softer cores can bring down (SF)P and (SF)G, but there is value in keeping them higher. A tooth broken by bending fatigue not only can destroy the gearset, but can bend shafts, damage bearings, and produce inertial stresses up- and downstream in the power train, causing damage elsewhere if the gear box locks. 9

In designing gears it makes sense to define the factor of safety in wear as (S)H2 for uncrowned teeth, so that there is no mix-up. ANSI, in the preface to ANSI/AGMA 2001-D04 and 2101-D04, states "the use is completely voluntary . . . does not preclude anyone from using . . . procedures . . . not conforming to the standards."

Spur and Helical Gears     783

EXAMPLE 14–8 Design a 4:1 spur-gear reduction for a 100-hp, three-phase squirrel-cage induction motor running at 1120 rev/min. The load is smooth, providing a reliability of 0.95 at 109 revolutions of the pinion. Gearing space is meager. Use Nitralloy 135M, grade 1 material to keep the gear size small. The gears are heat-treated first then nitrided. Solution Make the a priori decisions: ∙ ∙ ∙ ∙ ∙ ∙

Function: 100 hp, 1120 rev/min, R = 0.95, N = 109 cycles, Ko = 1 Design factor for unquantifiable exingencies: nd = 2 Tooth system: ϕn = 20° Tooth count: NP = 18 teeth, NG = 72 teeth (no interference, Sec. 13–7) Quality number: Qv = 6, use grade 1 material Assume mB ≥ 1.2 in Eq. (14–40), KB = 1

Pitch: Select a trial diametral pitch of Pd = 4 teeth/in. Thus, dP = 18∕4 = 4.5 in and dG = 72∕4 = 18 in. From Table 14–2, YP = 0.309, YG = 0.4324 (interpolated). From Fig. 14–6, JP = 0.32, JG = 0.415.

V=

πdP nP π(4.5)1120 = = 1319 ft/min 12 12

Wt =

33 000H 33 000(100) = = 2502 lbf V 1319

From Eqs. (14–28) and (14–27),

B = 0.25(12 − Qv ) 2∕3 = 0.25(12 − 6) 2∕3 = 0.8255

A = 50 + 56(1 − 0.8255) = 59.77

Kv = (

59.77 + √1319 0.8255 = 1.480 ) 59.77

From Eq. (14–38), KR = 0.658 − 0.0759ln(1 − 0.95) = 0.885. From Fig. 14–14,

(YN ) P = 1.3558(109 ) −0.0178 = 0.938

(YN ) G = 1.3558(109∕4) −0.0178 = 0.961

From Fig. 14–15,

(ZN ) P = 1.4488(109 )−0.023 = 0.900

(ZN ) G = 1.4488(109∕4) −0.023 = 0.929

From the recommendation after Eq. (14–8), 3p ≤ F ≤ 5p. Try F = 4p = 4π∕P = 4π∕4 = 3.14 in. From Eq.  (a), Sec. 14–10,

F √Y 0.0535 3.14 √0.309 0.0535 Ks = 1.192( = 1.192 = 1.140 ( ) P ) 4

From Eqs. (14–31), (14–33) and (14–35), Cmc = Cpm = Ce = 1. From Fig. 14–11, Cma = 0.175 for commercial enclosed gear units. From Eq. (14–32), F∕(10dP) =3.14∕[10(4.5)] = 0.0698. Thus,

Cpf = 0.0698 − 0.0375 + 0.0125(3.14) = 0.0715

784      Mechanical Engineering Design

From Eq. (14–30),

Km = 1 + (1)[0.0715(1) + 0.175(1)] = 1.247

From Table 14–8, for steel gears, Cp = 2300 √psi. From Eq. (14–23), with mG = 4 and mN = 1,

I=

cos 20° sin 20° 4 = 0.1286 2 4+1

Pinion tooth bending. With the above estimates of Ks and Km from the trial diametral pitch, we check to see if the mesh width F is controlled by bending or wear considerations. Equating Eqs. (14–15) and (14–17), substituting ndW t for W t, and solving for the face width (F)bend necessary to resist bending fatigue, we obtain (F) bend = nd W tKo Kv Ks Pd

Km KB KT KR JP St YN

(1)

Equating Eqs. (14–16) and (14–18), substituting ndW t for W t, and solving for the face width (F)wear necessary to resist wear fatigue, we obtain (F) wear = (

Cp KT KR Sc Z N

Km Cf t ) nd W Ko Kv Ks dP I 2

(2)

From Table 14–5 the hardness range of Nitralloy 135M is Rockwell C32–36 (302–335 Brinell). Choosing a midrange hardness as attainable, using 320 Brinell. From Fig. 14–4, St = 86.2(320) + 12 730 = 40 310 psi

Inserting the numerical value of St in Eq. (1) to estimate the face width gives

(F) bend = 2(2502)(1)1.48(1.14)4

1.247(1)(1)0.885 = 3.08 in 0.32(40 310)0.938

From Table 14–6 for Nitralloy 135M, Sc = 170 000 psi. Inserting this in Eq. (2), we find

(F) wear = (

1.247(1) 2300(1)(0.885) 2 2(2502)1(1.48)1.14 = 3.22 in 170 000(0.900) ) 4.5(0.1286)

Decision Make face width 3.50 in. Correct Ks and Km: Ks = 1.192(

3.50 √0.309 0.0535 = 1.147 ) 4

F 3.50 = = 0.0778 10dP 10(4.5) C p f = 0.0778 − 0.0375 + 0.0125(3.50) = 0.0841

Km = 1 + (1)[0.0841(1) + 0.175(1)] = 1.259 t

The bending stress induced by W  in bending, from Eq. (14–15), is

(σ) P = 2502(1)1.48(1.147)

4 1.259(1) = 19 100 psi 3.50 0.32

The AGMA factor of safety in bending of the pinion, from Eq. (14–41), is

(SF ) P =

40 310(0.938)∕[1(0.885)] = 2.24 19 100

Spur and Helical Gears     785

Decision Gear tooth bending. Use cast gear blank because of the 18-in pitch diameter. Use the same material, heat treatment, and nitriding. The load-induced bending stress is in the ratio of JP∕JG. Then

(σ) G = 19 100

0.32 = 14 730 psi 0.415

The factor of safety of the gear in bending is

(SF ) G =

40 310(0.961)∕[1(0.885)] = 2.97 14 730

Pinion tooth wear. The contact stress, given by Eq. (14–16), is

1.259 1 1∕2 (σc ) P = 2300[ 2502(1)1.48(1.147) = 118 000 psi 4.5(3.5) 0.129 ]

The factor of safety from Eq. (14–42), is

(SH ) P =

170 000(0.900)∕[1(0.885)] = 1.465 118 000

By our definition of factor of safety, pinion bending is (SF)P = 2.24, and wear is (SH)P2 = (1.465)2 = 2.15. Gear tooth wear. The hardness of the gear and pinion are the same. Thus, from Fig. 14–12, CH = 1, the contact stress on the gear is the same as the pinion, (σc)G = 118 000 psi. The wear strength is also the same, Sc = 170 000 psi. The factor of safety of the gear in wear is

(SH ) G =

170 000(0.929)∕[1(0.885)] = 1.51 118 000

So, for the gear in bending, (SF)G = 2.97, and wear (SH)G2 = (1.51)2 = 2.29. Rim. Keep mB ≥ 1.2. The whole depth is ht = addendum + dedendum = 1∕Pd + 1.25∕Pd = 2.25∕Pd = 2.25∕4 = 0.5625 in. The rim thickness tR is

tR ≥ mB ht = 1.2(0.5625) = 0.675 in

In the design of the gear blank, be sure the rim thickness exceeds 0.675 in; if it does not, review and modify this mesh design.

This design example showed a satisfactory design for a four-pitch spur-gear mesh. Material could be changed, as could pitch. There are a number of other satisfactory designs, thus a figure of merit is needed to identify the best. One can appreciate that gear design was one of the early applications of the digital computer to mechanical engineering. A design program should be interactive, presenting results of calculations, pausing for a decision by the designer, and showing the consequences of the decision, with a loop back to change a decision for the better. The program can be structured in totem-pole fashion, with the most influential decision at the top, then tumbling down, decision after decision, ending with the ability to change the current decision or to begin again. Such a program would make a fine class project. Troubleshooting the coding will reinforce your knowledge, adding flexibility as well as bells and whistles in subsequent terms.

786      Mechanical Engineering Design

Standard gears may not be the most economical design that meets the functional requirements, because no application is standard in all respects.10 Methods of designing custom gears are well understood and frequently used in mobile equipment to provide good weight-to-performance index. The required calculations including optimizations are within the capability of a personal computer.

PROBLEMS Problems marked with an asterisk (*) are linked to problems in other chapters, as summarized in Table 1–2 of Section 1–17. Because the results will vary depending on the method used, the problems are presented by section.

Section 14–1 14–1

A steel spur pinion has a pitch of 6 teeth/in, 22 full-depth teeth, and a 20° pressure angle. The pinion runs at a speed of 1200 rev/min and transmits 15 hp to a 60-tooth gear. If the face width is 2 in, estimate the bending stress.

14–2 A steel spur pinion has a diametral pitch of 10 teeth/in, 18 teeth cut full-depth with a 20° pressure angle, and a face width of 1 in. This pinion is expected to transmit 2 hp at a speed of 600 rev/min. Determine the bending stress.

14–3 A steel spur pinion has a module of 1.25 mm, 18 teeth cut on the 20° full-depth system, and a face width of 12 mm. At a speed of 1800 rev/min, this pinion is expected to carry a steady load of 0.5 kW. Determine the bending stress.

14–4 A steel spur pinion has 16 teeth cut on the 20° full-depth system with a module of

8  mm and a face width of 90 mm. The pinion rotates at 150 rev/min and transmits 6  kW to the mating steel gear. What is the bending stress?

14–5 A steel spur pinion has a module of 1 mm and 16 teeth cut on the 20° full-depth system and is to carry 0.15 kW at 400 rev/min. Determine a suitable face width based on an allowable bending stress of 150 MPa.

14–6 A 20° full-depth steel spur pinion has 20 teeth and a module of 2 mm and is to trans-

mit 0.5 kW at a speed of 200 rev/min. Find an appropriate face width if the bending stress is not to exceed 75 MPa.

14–7 A 20° full-depth steel spur pinion has a diametral pitch of 5 teeth/in and 24 teeth and transmits 6 hp at a speed of 50 rev/min. Find an appropriate face width if the allowable bending stress is 20 kpsi.

14–8 A steel spur pinion is to transmit 20 hp at a speed of 400 rev/min. The pinion is cut

on the 20° full-depth system and has a diametral pitch of 4 teeth/in and 16 teeth. Find a suitable face width based on an allowable stress of 12 kpsi.

14–9 A 20° full-depth steel spur pinion with 18 teeth is to transmit 2.5 hp at a speed of 600 rev/min. Determine appropriate values for the face width and diametral pitch based on an allowable bending stress of 10 kpsi.

14–10 A 20° full-depth steel spur pinion is to transmit 1.5 kW hp at a speed of 900 rev/min.

If the pinion has 18 teeth, determine suitable values for the module and face width. The bending stress should not exceed 75 MPa.

10

See H. W. Van Gerpen, C. K. Reece, and J. K. Jensen, Computer Aided Design of Custom Gears, Van Gerpen–Reece Engineering, Cedar Falls, Iowa, 1996.

Spur and Helical Gears     787

Section 14–2 14–11 The steel pinion of Problem 14–4 is to mesh with a steel gear with a gear ratio of 4:1.

The Brinell hardness of the teeth is 200, and the tangential load transmitted by the gears is 6 kN. If the contact fatigue strength of the steel can be estimated from the AGMA formula of Sc = 2.22 HB + 200 MPa, estimate the factor of safety of the drive based on a surface fatigue failure.

14–12 A speed reducer has 20° full-depth teeth and consists of a 20-tooth steel spur pinion

driving a 50-tooth cast-iron gear. The horsepower transmitted is 12 at a pinion speed of 1200 rev/min. For a diametral pitch of 8 teeth/in and a face width of 1.5 in, find the contact stress.

14–13 A gear drive consists of a 16-tooth 20° steel spur pinion and a 48-tooth cast-iron

gear having a pitch of 12 teeth/in. For a power input of 1.5 hp at a pinion speed of 700 rev/min, select a face width based on an allowable contact stress of 100 kpsi.

14–14 A gearset has a module of 5 mm, a 20° pressure angle, and a 24-tooth cast-iron spur

pinion driving a 48-tooth cast-iron gear. The pinion is to rotate at 50 rev/min. What horsepower input can be used with this gearset if the contact stress is limited to 690 MPa and F = 60 mm?

14–15 A 20° 20-tooth cast-iron spur pinion having a module of 4 mm drives a 32-tooth castiron gear. Find the contact stress if the pinion speed is 1000 rev/min, the face width is 50 mm, and 10 kW of power is transmitted.

14–16 A steel spur pinion and gear have a diametral pitch of 12 teeth/in, milled teeth, 17 and

30 teeth, respectively, a 20° pressure angle, a face width of 78 in, and a pinion speed of 525 rev/min. The tooth properties are Sut = 76 kpsi, Sy = 42 kpsi and the Brinell hardness is 149. Use the Gerber criteria to compensate for one-way bending. For a design factor of 2.25, what is the power rating of the gearset?

14–17 A milled-teeth steel pinion and gear pair have Sut = 113 kpsi, Sy = 86 kpsi and a hard-

ness at the involute surface of 262 Brinell. The diametral pitch is 3 teeth/in, the face width is 2.5 in, and the pinion speed is 870 rev/min. The tooth counts are 20 and 100. Use the Gerber criteria to compensate for one-way bending. For a design factor of 1.5, rate the gearset for power considering both bending and wear.

14–18 A 20° full-depth steel spur pinion rotates at 1145 rev/min. It has a module of 6 mm,

a face width of 75 mm, and 16 milled teeth. The ultimate tensile strength at the involute is 900 MPa exhibiting a Brinell hardness of 260. The gear is steel with 30 teeth and has identical material strengths. Use the Gerber criteria to compensate for one-way bending. For a design factor of 1.3 find the power rating of the gearset based on the pinion and the gear resisting bending and wear fatigue.

14–19 A steel spur pinion has a pitch of 6 teeth/in, 17 full-depth milled teeth, and a pres-

sure angle of 20°. The pinion has an ultimate tensile strength at the involute surface of 116 kpsi, a Brinell hardness of 232, and a yield strength of 90 kpsi. Its shaft speed is 1120 rev/min, its face width is 2 in, and its mating gear has 51 teeth. Use a design factor of 2. (a) Pinion bending fatigue imposes what power limitation? Use the Gerber criteria to compensate for one-way bending. (b) Pinion surface fatigue imposes what power limitation? The gear has identical strengths to the pinion with regard to material properties. (c) Determine power limitations due to gear bending and wear. (d) Specify the power rating for the gearset.

788      Mechanical Engineering Design

Section 14–3 to 14–19 14–20 A commercial enclosed gear drive is to be designed. Initially, the following is specified.

The drive is to transmit 5 hp with an input pinion speed of 300 rev/min. The drive is to consist of two spur gears, with a 20° pressure angle, and the pinion is to have 16 teeth driving a 48-tooth gear. The gears are to be grade 1 steel, through-hardened at 200 Brinell, made to No. 6 quality standards. The gears will be uncrowned, centered on their shafts between bearings. A pinion life of 108 cycles is desired, with a 90 percent reliability. (a) Use a trial value of diametral pitch of 6 teeth/in and a face width of 2 in. For the pinion, (i) determine the bending stress, allowable bending stress, and bending factor of safety. (ii)  determine the contact stress, allowable contact stress, and contact factor of safety. (b) Assume gears are readily available in face-width increments of 0.5 in, with face widths in the range of three to five times the circular pitch. Specify a diametral pitch and face width such that the minimum factor of safety for the pinion is equal to 2.

14–21 A 20° spur pinion with 20 teeth and a module of 2.5 mm transmits 120 W to a 36-tooth

gear. The pinion speed is 100 rev/min, and the gears are grade 1, 18-mm face width, through-hardened steel at 200 Brinell, uncrowned, manufactured to a No. 6 quality standard, and considered to be of open gearing quality installation. Find the AGMA bending and contact stresses and the corresponding factors of safety for a pinion life of 108 cycles and a reliability of 0.95.

14–22 Repeat Problem 14–20, part (a), using helical gears each with a 20° normal pitch angle and a helix angle of 30° and a normal diametral pitch of 6 teeth/in.

14–23 A spur gearset has 17 teeth on the pinion and 51 teeth on the gear. The pressure angle

is 20° and the overload factor Ko = 1. The diametral pitch is 6 teeth/in and the face width is 2 in. The pinion speed is 1120 rev/min and its cycle life is to be 108 revolutions at a reliability R = 0.99. The quality number is 5. The material is a throughhardened steel, grade 1, with Brinell hardnesses of 232 core and case of both gears. For a design factor of 2, rate the gearset for these conditions using the AGMA method.

14–24 In Section 14–10, Equation (a) is given for Ks based on the procedure in Example 14–2. Derive this equation.

14–25 A speed-reducer has 20° full-depth teeth, and the single-reduction spur-gear gearset has 22 and 60 teeth. The diametral pitch is 4 teeth/in and the face width is 314 in. The pinion shaft speed is 1145 rev/min. The life goal of 5-year 24-hour-per-day service is about 3(109) pinion revolutions. The absolute value of the pitch variation is such that the quality number is 6. The materials are 4340 through-hardened grade 1 steels, heattreated to 250 Brinell, core and case, both gears. The load is moderate shock and the power is smooth. For a reliability of 0.99, rate the speed reducer for power.

14–26 The speed reducer of Problem 14–25 is to be used for an application requiring 40 hp

at 1145 rev/min. For the gear and the pinion, estimate the AGMA factors of safety for bending and wear, that is, (SF)P, (SF)G, (SH)P, and (SH)G. By examining the factors of safety, identify the threat to each gear and to the mesh.

14–27 The gearset of Problem 14–25 needs improvement of wear capacity. Toward this end

the gears are nitrided so that the grade 1 materials have hardnesses as follows: The pinion core is 250 and the pinion case hardness is 390 Brinell, and the gear core hardness is 250 core and 390 case. Estimate the power rating for the new gearset.

Spur and Helical Gears     789

14–28 The gearset of Problem 14–25 has had its gear specification changed to 9310 for carburizing and surface hardening with the result that the pinion Brinell hardnesses are 285 core and 580–600 case, and the gear hardnesses are 285 core and 580–600 case. Estimate the power rating for the new gearset.

14–29 The gearset of Problem 14–28 is going to be upgraded in material to a quality of grade 2 (9310) steel. Estimate the power rating for the new gearset.

14–30 Matters of scale always improve insight and perspective. Reduce the physical size of

the gearset in Problem 14–25 by one-half and note the result on the estimates of transmitted load W t and power.

14–31 AGMA procedures with cast-iron gear pairs differ from those with steels because life

predictions are difficult; consequently (YN)P, (YN)G, (ZN)P, and (ZN)G are set to unity. The consequence of this is that the fatigue strengths of the pinion and gear materials are the same. The reliability is 0.99 and the life is 107 revolution of the pinion (KR = 1). For longer lives the reducer is derated in power. For the pinion and gearset of Problem 14–25, use grade 40 cast iron for both gears (HB = 201 Brinell). Rate the reducer for power with SF and SH equal to unity.

14–32 Spur-gear teeth have rolling and slipping contact (often about 8 percent slip). Spur gears tested to wear failure are reported at 108 cycles as Buckingham's surface fatigue load-stress factor K. This factor is related to Hertzian contact strength SC by

1.4K SC = √ (1∕E1 + 1∕E2 ) sin ϕ where ϕ is the normal pressure angle. Cast iron grade 20 gears with ϕ = 14 12° and 20° pressure angle exhibit a minimum K of 81 and 112 psi, respectively. How does this compare with SC = 0.32HB kpsi?

14–33 You've probably noticed that although the AGMA method is based on two equa-

tions, the details of assembling all the factors is computationally intensive. To reduce error and omissions, a computer program would be useful. Write a program to perform a power rating of an existing gearset, then use Problem 14–25, 14–27, 14–28, 14–29, and 14–30 to test your program by comparing the results to your longhand solutions.

14–34 In Example 14–5 use nitrided grade 1 steel (4140) which produces Brinell hardnesses of 250 core and 500 at the surface (case). Use the upper fatigue curves on Figures 14–14 and 14–15. Estimate the power capacity of the mesh with factors of safety of SF = SH = 1.

14–35 In Example 14–5 use carburized and case-hardened gears of grade 1. Carburizing and

case-hardening can produce a 550 Brinell case. The core hardnesses are 200 Brinell. Estimate the power capacity of the mesh with factors of safety of SF = SH = 1, using the lower fatigue curves in Figures 14–14 and 14–15.

14–36 In Example 14–5, use carburized and case-hardened gears of grade 2 steel. The core hardnesses are 200, and surface hardnesses are 600 Brinell. Use the lower fatigue curves of Figures 14–14 and 14–15. Estimate the power capacity of the mesh using SF = SH = 1. Compare the power capacity with the results of Problem 14–35.

14–37* The countershaft in Problem 3–83 is part of a speed reducing compound gear train using 20° spur gears. A gear on the input shaft drives gear A. Gear B drives a gear

790      Mechanical Engineering Design

on the output shaft. The input shaft runs at 2400 rev/min. Each gear reduces the speed (and thus increases the torque) by a 2 to 1 ratio. All gears are to be of the same material. Since gear B is the smallest gear, transmitting the largest load, it will likely be critical, so a preliminary analysis is to be performed on it. Use a diametral pitch of 2 teeth/in, a face width of 4 times the circular pitch, a Grade 2 steel through-hardened to a Brinell hardness of 300, and a desired life of 15 kh with a 95 percent reliability. Determine factors of safety for bending and wear.

14–38* The countershaft in Problem 3–84 is part of a speed reducing compound gear train

using 20° spur gears. A gear on the input shaft drives gear A with a 2 to 1 speed reduction. Gear B drives a gear on the output shaft with a 5 to 1 speed reduction. The input shaft runs at 1800 rev/min. All gears are to be of the same material. Since gear B is the smallest gear, transmitting the largest load, it will likely be critical, so a preliminary analysis is to be performed on it. Use a module of 18.75 mm/tooth, a face width of 4 times the circular pitch, a Grade 2 steel through-hardened to a Brinell hardness of 300, and a desired life of 12 kh with a 98 percent reliability. Determine factors of safety for bending and wear.

14–39* Build on the results of Problem 13–46 to find factors of safety for bending and wear

for gear F. Both gears are made from Grade 2 carburized and hardened steel. Use a face width of 4 times the circular pitch. The desired life is 12 kh with a 95 percent reliability.

14–40* Build on the results of Problem 13–47 to find factors of safety for bending and wear

for gear C. Both gears are made from Grade 2 carburized and hardened steel. Use a face width of 4 times the circular pitch. The desired life is 14 kh with a 98 percent reliability.

15

Bevel and Worm Gears

©Dmitry Kalinovsky/123 RF

Chapter Outline 15–1 Bevel Gearing—General   792

15–6

15–2

15–7 Worm-Gear Analysis   818

Bevel-Gear Stresses and Strengths   794

Worm Gearing—AGMA Equation   814

15–3 AGMA Equation Factors   797

15–8

Designing a Worm-Gear Mesh   822

15–4

Straight-Bevel Gear Analysis   808

15–9

Buckingham Wear Load   825

15–5

Design of a Straight-Bevel Gear Mesh   811 791

792      Mechanical Engineering Design

The American Gear Manufacturers Association (AGMA) has established standards for the analysis and design of the various kinds of bevel and worm gears. Chapter 14 was an introduction to the AGMA methods for spur and helical gears and contains many of the definitions of terms used in this chapter. AGMA has established similar methods for other types of gearing, which all follow the same general approach.

15–1  Bevel Gearing—General Bevel gears may be classified as follows: ∙ ∙ ∙ ∙ ∙

Straight bevel gears Spiral bevel gears Zerol bevel gears Hypoid gears Spiroid gears

A straight bevel gear was illustrated in Figure 13–31. These gears are usually used for pitch-line velocities up to 1000 ft/min (5 m/s) when the noise level is not an important consideration. They are available in many stock sizes and are less expensive to produce than other bevel gears, especially in small quantities. A spiral bevel gear is shown in Figure 15–1; the definition of the spiral angle is illustrated in Figure 15–2. These gears are recommended for higher speeds and where the noise level is an important consideration. Spiral bevel gears are the bevel counterpart of the helical gear; it can be seen in Figure 15–1 that the pitch surfaces and the nature of contact are the same as for straight bevel gears except for the differences brought about by the spiral-shaped teeth. The Zerol bevel gear is a patented gear having curved teeth but with a zero spiral angle. The axial thrust loads permissible for Zerol bevel gears are not as large as those for the spiral bevel gear, and so they are often used instead of straight bevel gears. The Zerol bevel gear is generated by the same tool used for regular spiral bevel Figure 15–1 Spiral bevel gears. (©Jim Francis/Shutterstock)

Bevel and Worm Gears     793

Figure 15–2

Circular pitch

Cutting spiral-gear teeth on the basic crown rack.

Face advance

Mean radius of crown rack

Spiral angle

ψ

Cutter radius

Basic crown rack

gears. For design purposes, use the same procedure as for straight bevel gears and then simply substitute a Zerol bevel gear. It is frequently desirable, as in the case of automotive differential applications, to have gearing similar to bevel gears but with the shafts offset. Such gears are called hypoid gears, because their pitch surfaces are hyperboloids of revolution. The tooth action between such gears is a combination of rolling and sliding along a straight line and has much in common with that of worm gears. Figure 15–3 shows a pair of hypoid gears in mesh. Figure 15–3 Hypoid gears. (Courtesy of Gleason Works, Rochester, N.Y.)

794      Mechanical Engineering Design

Figure 15–4 Comparison of intersecting- and offset-shaft bevel-type gearings. (From Gear Handbook by Darle W. Dudley, 1962, pp. 2–24.)

Worm

Spiroid Ring gear

Hypoid

Spiral bevel

Figure 15–4 is included to assist in the classification of spiral bevel gearing. It is seen that the hypoid gear has a relatively small shaft offset. For larger offsets, the pinion begins to resemble a tapered worm and the set is then called spiroid gearing.

15–2  Bevel-Gear Stresses and Strengths In a typical bevel-gear mounting, Figure 13–32, for example, one of the gears is often mounted outboard of the bearings. This means that the shaft deflections can be more pronounced and can have a greater effect on the nature of the tooth contact. Another difficulty that occurs in predicting the stress in bevel-gear teeth is the fact that the teeth are tapered. Thus, to achieve perfect line contact passing through the cone center, the teeth should bend more at the large end than at the small end. To obtain this condition requires that the load be proportionately greater at the large end. Because of this varying load across the face of the tooth, it is desirable to have a fairly short face width. Because of the complexity of bevel, spiral bevel, Zerol bevel, hypoid, and spiroid gears, as well as the limitations of space, only a portion of the applicable standards that refer to straight-bevel gears is presented here.1 Table 15–1 gives the symbols used in ANSI/AGMA 2003-B97. Fundamental Contact Stress Equation 1

1∕2 Wt sc = σc = Cp ( Ko Kv Km Cs Cxc) FdP I 1∕2 1000W t σ H = ZE ( KA Kv KHβ Zx Zxc) bd Z1

(U.S. customary units)

(15–1)

(SI units)

Figures 15–5 to 15–13 and Tables 15–1 to 15–7 have been extracted from ANSI/AGMA 2003-B97, Rating the Pitting Resistance and Bending Strength of Generated Straight Bevel, Zerol Bevel and Spiral Bevel Gear Teeth with the permission of the publisher, the American Gear Manufacturers Association, 1001 N. Fairfax Street, Suite 500, Alexandria, VA, 22314-1587.

Bevel and Worm Gears     795

Table 15–1  Symbols Used in Bevel Gear Rating Equations, ANSI/AGMA 2003-B97 Standard AGMA ISO Symbol Symbol Description Am Rm A0 Re CH ZW Ci Zi CL ZNT

Mean cone distance Outer cone distance Hardness ratio factor for pitting resistance Inertia factor for pitting resistance Stress cycle factor for pitting resistance

Cp ZE Elastic coefficient CR ZZ Reliability factor for pitting CSF Service factor for pitting resistance CS Zx Size factor for pitting resistance Cxc Zxc Crowning factor for pitting resistance D, d de2, de1 Outer pitch diameters of gear and pinion, respectively EG, EP E2, E1 Young's modulus of elasticity for materials of gear and pinion, respectively e e Base of natural (Napierian) logarithms F b Net face width FeG, FeP b′2, b′1 Effective face widths of gear and pinion, respectively fP Ra1 Pinion surface roughness HBG HB2 Minimum Brinell hardness number for gear material HBP HB1 Minimum Brinell hardness number for pinion material hc Eht min Minimum total case depth at tooth middepth he h′c Minimum effective case depth he lim h′c lim Suggested maximum effective case depth limit at tooth middepth I ZI Geometry factor for pitting resistance J YJ Geometry factor for bending strength JG, JP YJ2, YJ1 Geometry factor for bending strength for gear and pinion, respectively KF YF Stress correction and concentration factor Ki Yi Inertia factor for bending strength KL YNT Stress cycle factor for bending strength Km KHβ Load distribution factor Ko KA Overload factor KR Yz Reliability factor for bending strength KS YX Size factor for bending strength KSF Service factor for bending strength KT Kθ Temperature factor Kv Kv Dynamic factor Kx Y β Lengthwise curvature factor for bending strength met Outer transverse module mmt Mean transverse module mmn Mean normal module mNI εNI Load sharing ratio, pitting mNJ εNJ Load sharing ratio, bending N z2 Number of gear teeth NL nL Number of load cycles n z1 Number of pinion teeth nP n1 Pinion speed

Units in (mm) in (mm)

[lbf/in2]0.5 ([N/mm2]0.5)

in (mm) lbf/in2 (N/mm2) in (mm) in (mm) μin (μm) HB HB in (mm) in (mm) in (mm)

(mm) (mm) (mm)

rev/min

(Continued)

796      Mechanical Engineering Design

Table 15–1  Symbols Used in Bevel Gear Rating Equations, ANSI/AGMA 2003-B97 Standard (Continued) AGMA ISO Symbol Symbol Description P P Design power through gear pair Pa Pa Allowable transmitted power Pac Paz Allowable transmitted power for pitting resistance Pacu Pazu Allowable transmitted power for pitting resistance at unity service factor Pat Pay Allowable transmitted power for bending strength Patu Payu Allowable transmitted power for bending strength at unity service factor Pd Outer transverse diametral pitch Pm Mean transverse diametral pitch Pmn Mean normal diametral pitch Qv Qv Transmission accuracy number q q Exponent used in formula for lengthwise curvature factor R, r rmpt 2, rmpt1 Mean transverse pitch radii for gear and pinion, respectively Rt, rt rmyo2, rmyo1 Mean transverse radii to point of load application for gear and pinion, respectively rc rc 0 Cutter radius used for producing Zerol bevel and spiral bevel gears s gc Length of the instantaneous line of contact between mating tooth surfaces sac σH lim Allowable contact stress number sat σF lim Bending stress number (allowable) sc σH Calculated contact stress number SF SF Bending safety factor SH SH Contact safety factor st σF Calculated bending stress number swc σHP Permissible contact stress number swt σFP Permissible bending stress number TP T1 Operating pinion torque TT θT Operating gear blank temperature t0 sai Normal tooth top land thickness at narrowest point Uc Uc Core hardness coefficient for nitrided gear UH UH Hardening process factor for steel vt vet Pitch-line velocity at outer pitch circle YKG, YKP YK2, YK1 Tooth form factors including stress-concentration factor for gear and pinion, respectively μG, μp ν2, ν1 Poisson's ratio for materials of gear and pinion, respectively ρ0 ρyo Relative radius of profile curvature at point of maximum contact stress between mating tooth surfaces ϕ αn Normal pressure angle at pitch surface ϕt αwt Transverse pressure angle at pitch point ψ βm Mean spiral angle at pitch surface ψb βmb Mean base spiral angle Source: ANSI/AGMA 2003-B97.

Units hp (kW) hp (kW) hp (kW) hp (kW) hp (kW) hp (kW) teeth/in teeth/in teeth/in

in (mm) in (mm) in (mm) in (mm) lbf/in2 (N/mm2) lbf/in2 (N/mm2) lbf/in2 (N/mm2)

lbf/in2 (N/mm2) lbf/in2 (N/mm2) lbf/in2 (N/mm2) lbf in (Nm) °F(°C) in (mm) lbf/in2 (N/mm2) lbf/in2 (N/mm2) ft/min (m/s)

in (mm)

Bevel and Worm Gears     797

The first term in each equation is the AGMA symbol, whereas σc, our normal notation, is directly equivalent. Permissible Contact Stress Number (Strength) Equation sacCLCH SH KT CR

swc = (σc ) all =

σH lim ZNT ZW σHP = SH Kθ ZZ

(U.S. customary units)

(15–2)

(SI units)

Bending Stress

st =

σF =

Ks Km Wt Pd KoKv F Kx J

(U.S. customary units)

1000W t KA Kv Yx KHβ met Yβ YJ b

(SI units)

(15–3)

Permissible Bending Stress Equation

swt =

sat KL SF KT KR

(U.S. customary units)

σF lim YNT σFP = SF Kθ Yz

(15–4)

(SI units)

15–3  AGMA Equation Factors Overload Factor Ko (KA ) The overload factor makes allowance for any externally applied loads in excess of the nominal transmitted load. Table 15–2, from Appendix A of 2003-B97, is included for your guidance. Safety Factors SH and SF The factors of safety SH and SF as defined in 2003-B97 are adjustments to strength, not load, and consequently cannot be used as is to assess (by comparison) whether the threat is from wear fatigue or bending fatigue. Since W t is the same for the pinion and gear, the comparison of √SH to SF allows direct comparison. Table 15–2  Overload Factors Ko (KA) Character of Prime Mover

Character of Load on Driven Machine Uniform

Light Shock

Medium Shock

Heavy Shock

Uniform

1.00

1.25

1.50

1.75 or higher

Light shock

1.10

1.35

1.60

1.85 or higher

Medium shock

1.25

1.50

1.75

2.00 or higher

Heavy shock

1.50

1.75

2.00

2.25 or higher

Note: This table is for speed-decreasing drives. For speed-increasing drives, add 0.01(N/n)2 or 0.01(z2 /z1)2 to the above factors. Source: ANSI/AGMA 2003-B97.

798      Mechanical Engineering Design

Figure 15–5 Dynamic factor Kv. (Source: ANSI/AGMA 2003-B97.)

2.0

0

Pitch-line velocity, υet (m/s) 20 30

10

1.9

50

Qυ = 6

1.8

Qυ = 7

1.7 Dynamic factor, Kv

40

Qυ = 5

1.6

Qυ = 8

1.5

Qυ = 9

1.4

Qυ = 10

1.3

Qυ = 11

1.2 1.1 1.0

0

2000

4000 6000 Pitch-line velocity, υt (ft/min)

8000

10 000

Dynamic Factor Kv In 2003-C87 AGMA changed the definition of Kv to its reciprocal but used the same symbol. Other standards have yet to follow this move. The dynamic factor Kv makes allowance for the effect of gear-tooth quality related to speed and load, and the increase in stress that follows. AGMA uses a transmission accuracy number Qv to describe the precision with which tooth profiles are spaced along the pitch circle. Figure 15–5 shows graphically how pitch-line velocity and transmission accuracy number are related to the dynamic factor Kv. Curve fits are

Kv = (

A + √vt B ) A

Kv = (

A+

√200vet

A

(U.S. customary units) )

B

(15–5)

(SI units)

where

A = 50 + 56(1 − B)

B = 0.25(12 − Qv ) 2∕3

(15–6)

and vt(vet) is the pitch-line velocity at outside pitch diameter, expressed in ft/min (m/s):

vt = π dP nP∕12

(U.S. customary units)

vet = 5.236(10−5 )d1 n1

(SI units)

(15–7)

The maximum recommended pitch-line velocity is associated with the abscissa of the terminal points of the curve in Figure 15–5:

vt max = [A + (Qv − 3)]2 vet max =

where vt max and vet

max

2

[A + (Qv − 3)] 200

(U.S. customary units) (SI units)

are in ft/min and m/s, respectively.

(15–8)

Bevel and Worm Gears     799

Size Factor for Pitting Resistance Cs (Zx)  0.5  Cs =  0.125F + 0.4375   1

F < 0.5 in 0.5 ≤ F ≤ 4.5 in      (U.S. customary units) F > 4.5 in

 0.5  Zx =  0.004 92b + 0.4375   1

b < 12.7 mm 12.7 ≤ b ≤ 114.3 mm   (SI units) b > 114.3 mm

(15–9)

Size Factor for Bending Ks (Yx )

 0.4867 + 0.2132∕Pd 0.5 ≤ Pd ≤ 16 teeth/in       (U.S. customary units)  0.5 Pd > 16 teeth/in

KS = 

 0.5

Yx = 

 0.4867 + 0.008 339m et

   m et < 1.6 mm    (SI units) 1.6 ≤ m et ≤ 50 mm

(15–10)

Load-Distribution Factor Km (KHβ)

Km = Kmb + 0.0036 F 2    (U.S. customary units) KHβ = Kmb + 5.6(10−6 )b2  (SI units)

where Kmb

 1.00  =  1.10   1.25

(15–11)

both members straddle-mounted one member straddle-mounted neither member straddle-mounted

Crowning Factor for Pitting Cxc (Zxc) The teeth of most bevel gears are crowned in the lengthwise direction during manufacture to accommodate the deflection of the mountings.

 1.5  2.0

Cxc = Zxc = 

properly crowned teeth or larger uncrowned teeth

(15–12)

Lengthwise Curvature Factor for Bending Strength Kx (Yβ ) For straight-bevel gears,

Kx = Yβ = 1

(15–13)

Pitting Resistance Geometry Factor I (ZI) Figure 15–6 shows the geometry factor I (ZI) for straight-bevel gears with a 20° pressure angle and 90° shaft angle. Enter the figure ordinate with the number of pinion teeth, move to the number of gear-teeth contour, and read from the abscissa. Bending Strength Geometry Factor J (YJ) Figure 15–7 shows the geometry factor J for straight-bevel gears with a 20° pressure angle and 90° shaft angle.

800      Mechanical Engineering Design

Figure 15–6

Number of gear teeth

Contact geometry factor I (ZI) for coniflex straight-bevel gears with a 20° normal pressure angle and a 90° shaft angle. (Source: ANSI/AGMA 2003-B97.)

50

50

60

70

80

90

100

45 40

Number of pinion teeth

40

35 30

30 25 20

20 15

10 0.05

0.06

Figure 15–7

0.08 0.09 Geometry factor, I (Z I )

0.10

0.11

Number of teeth in mate 13 15

100

Number of teeth on gear for which geometry factor is desired

Bending factor J (YJ) for coniflex straight-bevel gears with a 20° normal pressure angle and 90° shaft angle. (Source: ANSI/AGMA 2003-B97.)

0.07

20

25

30 35 40 45 50

100 90

90 80

80 70

70 60

60 50 40 30 20 10 0.16

0.18

0.20

0.22

0.24

0.26

0.28

0.30

Geometry factor, J (YJ)

0.32

0.34

0.36

0.38

0.40

Bevel and Worm Gears     801

Figure 15–8

5.0

Contact stress-cycle factor for pitting resistance CL (ZNT) for carburized case-hardened steel bevel gears. (Source: ANSI/AGMA 2003-B97.)

4.0

Stress-cycle factor, CL (Z N T )

3.0

2.0 CL = 3.4822 NL–0.0602 ZNT = 3.4822 nL–0.0602 1.0 0.9 0.8 0.7 0.6 0.5 103

104

105

106 107 Number of load cycles, NL (nL )

108

109

1010

Stress-Cycle Factor for Pitting Resistance CL (ZNT) 

2 CL =   3.4822NL−0.0602

103 ≤ NL < 104 104 ≤ NL ≤ 1010

ZNT

103 ≤ nL < 104 104 ≤ nL ≤ 1010

2 =  −0.0602  3.4822nL

(15–14)

See Figure 15–8 for a graphical presentation of Equations (15–14). Stress-Cycle Factor for Bending Strength KL (YNT)

   KL =    

YNT

   =   

2.7 6.1514NL−0.1192 1.683NL−0.0323 1.3558NL−0.0178

102 ≤ NL < 103 103 ≤ NL < 3(106 ) 3(106 ) ≤ NL ≤ 1010 3(106 ) ≤ NL ≤ 1010

critical general

2.7 6.1514nL−0.1192 1.683nL−0.0323 1.3558n−0.0178 L

102 ≤ nL < 103 103 ≤ nL < 3(106 ) 3(106 ) ≤ nL ≤ 1010 3(106 ) ≤ nL ≤ 1010

critical general

(15–15)

See Figure 15–9 for a plot of Equations (15–15). Hardness-Ratio Factor CH (ZW)

CH = 1 + B1 (N∕n − 1)

Z W = 1 + B1 (z2∕z1 − 1)

B1 = 0.008 98(HBP∕HBG ) − 0.008 29 (15–16) B1 = 0.008 98(HB1∕HB2 ) − 0.008 29

The preceding equations are valid when 1.2 ≤ HBP∕HBG ≤ 1.7 (1.2 ≤ HB1∕HB2 ≤ 1.7). Figure 15–10 graphically displays Equations (15–16). When a surface-hardened pinion (48 HRC or harder) is run with a through-hardened gear (180 ≤ HB ≤ 400),

802      Mechanical Engineering Design 3.5

Stress-cycle factor for bending strength KL (YNT) for carburized case-hardened steel bevel gears. (Source: ANSI/AGMA 2003-B97.)

3.0

2.0

NOTE: The choice of KL (YNT) is influenced by: Pitch-line velocity Gear material cleanliness Residual stress Material ductility and fracture toughness

KL = 6.1514 NL–0.1192 YNT = 6.1514 nL–0.1192

1.5

KL = 1.3558 NL–0.0178 YNT = 1.3558 nL–0.0178

1.0 0.9

1.0 0.9

0.8

0.8

KL = 1.683 NL–0.0323 YNT = 1.683 nL–0.0323

0.7 0.6 0.5 102

103

Figure 15–10

1.14

Hardness-ratio factor CH (ZW) for through-hardened pinion and gear. (Source: ANSI/AGMA 2003-B97.)

1.12

104

105 106 107 Number of load cycles, NL (nL)

108

0.7 0.6 0.5 1010

109

1.4

1.08

1.3

1.06

1.2 1.04

HB1

HB2

1.5

1.10

HBP

1.6

HBG

Hardness-ratio factor, CH (Z W )

1.7

Calculated hardness ratio,

Stress-cycle factor, KL (YN T )

Figure 15–9

When

1.02

HBP

HB1

HBG

HB2

< 1.2

use CH (ZW) = 1 1.00

0

2

4

6

8 10 12 14 Reduction gear ratio, N/n (z2 /z1)

16

18

20

a work-hardening effect occurs. The CH (ZW) factor varies with pinion surface roughness fP(Ra1) and the mating-gear hardness:

CH = 1 + B2 (450 − HBG )

B2 = 0.000 75 exp(−0.0122 fP )

ZW = 1 + B2 (450 − HB2 )

B2 = 0.000 75 exp(−0.52 R a1 )

(15–17)

where fP (Ra1) = pinion surface hardness μin (μm) HBG (HB2) = minimum Brinell hardness of the gear See Figure 15–11 for carburized steel gear pairs of approximately equal hardness CH = ZW = 1.

Bevel and Worm Gears     803

Figure 15–11

Hardness-ratio factor CH (Z W )

1.20

16 μin (0.4 μm)

1.15

Hardness-ratio factor CH (ZW) for surface-hardened pinions. (Source: ANSI/AGMA 2003-B97.)

Surface roughness of pinion, fP (Ra1)

32 μin (0.8 μm)

1.10

63 μin (1.6 μm) 1.05 125 μin (3.2 μm) 1.00 180

200

250

300 Brinell hardness of the gear HB

350

400

Temperature Factor KT (Kθ)

1 KT = { (460 + t)∕710

32°F ≤ t ≤ 250°F t > 250°F

1 Kθ = { (273 + θ)∕393

0°C ≤ θ ≤ 120°C θ > 120°C

Reliability Factors CR (ZZ) and KR (YZ) Table 15–3 displays the reliability factors. Note that CR = Logarithmic interpolation equations are 0.50 − 0.25 log(1 − R) YZ = KR = { 0.70 − 0.15 log(1 − R)

(15–18)

√KR

and ZZ =

0.99 ≤ R ≤ 0.999 0.90 ≤ R < 0.99

Table 15–3  Reliability Factors Requirements of Application

Reliability Factors for Steel* CR (ZZ)

KR (YZ)†

Fewer than one failure in 10 000

1.22

1.50

Fewer than one failure in 1000

1.12

1.25

Fewer than one failure in 100

1.00

1.00

Fewer than one failure in 10

0.92

0.85‡

Fewer than one failure in 2

0.84

0.70§

*At the present time there are insufficient data concerning the reliability of bevel gears made from other materials. † Tooth breakage is sometimes considered a greater hazard than pitting. In such cases a greater value of KR (YZ ) is selected for bending. ‡ At this value plastic flow might occur rather than ­pitting. § From test data extrapolation. Source: ANSI/AGMA 2003-B97.

√YZ.

(15–19) (15–20)

804      Mechanical Engineering Design

Table 15–4  Allowable Contact Stress Number for Steel Gears, sac (σH  lim) Minimum Material Heat Surface* Designation

Treatment

Allowable Contact Stress Number, sac (σH lim) lbf/in2 (N/mm2)

Hardness ‡

Steel Through-hardened

Fig. 15–12

Grade 1†

Grade 2†

Fig. 15–12

Fig. 15–12

Flame or induction 50 HRC hardened§

175 000 190 000 (1210) (1310)

200 000 (1380)

Carburized and case hardened§

2003-B97 Table 8

225 000 (1550)

Grade 3†

250 000 (1720)

AISI 4140 Nitrided§ 84.5 HR15N 145 000 (1000) Nitralloy 135M Nitrided§ 90.0 HR15N

160 000 (1100)

*Hardness to be equivalent to that at the tooth middepth in the center of the face width. † See ANSI/AGMA 2003-B97, Tables 8 through 11, for metallurgical factors for each stress grade of steel gears. ‡ These materials must be annealed or normalized as a minumum. § The allowable stress numbers indicated may be used with the case depths prescribed in 21.1, ANSI/AGMA 2003-B97. Source: ANSI/AGMA 2003-B97.

Table 15–5  Allowable Contact Stress Number for Iron Gears, sac (σH

lim)

Typical Minimum Allowable Contact Material Designation Heat Surface Stress Number, sac Material ASTM ISO Treatment Hardness (σH lim) lbf/in2 (N/mm2) Cast iron

ASTM A48

ISO/DR 185

Class 30

Grade 200

As cast

175 HB   50 000 (345)

Class 40

Grade 300

As cast

200 HB   65 000 (450)

Ductile (nodular) iron

ASTM A536

ISO/DIS 1083

Grade 80-55-06

Grade 600-370-03

Grade 120-90-02

Grade 800-480-02

Quenched and tempered

300 HB

180 HB   94 000 (650) 135 000 (930)

Source: ANSI/AGMA 2003-B97.

The reliability of the stress (fatigue) numbers allowable in Tables 15–4, 15–5, 15–6, and 15–7 is 0.99. Elastic Coefficient for Pitting Resistance Cp (ZE)

1 Cp = √ π[(1 − ν2P )∕EP + (1 − ν2G )∕EG]

ZE = √

where Cp = ZE = EP and EG = E1 and E2 =

π[(1 −

ν21 )∕E1

1 + (1 − ν22 )∕E2]

(15–21)

elastic coefficient, 2290 √psi for steel elastic coefficient, 190 √N/mm2 for steel Young's moduli for pinion and gear respectively, psi Young's moduli for pinion and gear respectively, N/mm2

Bevel and Worm Gears     805

Table 15–6  Allowable Bending Stress Numbers for Steel Gears, sat (σF lim) Minimum Material Heat Surface Designation

Treatment

Steel

Bending Stress Number (Allowable), sat (σF lim) lbf/in2 (N/mm2)

Hardness

Through-hardened

Grade 1*

Fig. 15–13

Grade 2*

Fig. 15–13

Fig. 15–13

Flame or induction hardened    Unhardened roots 50 HRC    Hardened roots

15 000 (85) 22 500 (154)

13 500 (95)

Carburized and case   hardened†

30 000 (205)

2003-B97 Table 8

Grade 3*

35 000 (240)

AISI 4140

Nitrided†,‡

84.5 HR15N

22 000 (150)

Nitralloy 135M

Nitrided†,‡

90.0 HR15N

24 000 (165)

40 000 (275)

*See ANSI/AGMA 2003-B97, Tables 8–11, for metallurgical factors for each stress grade of steel gears. † The allowable stress numbers indicated may be used with the case depths prescribed in 21.1, ANSI/AGMA 2003-B97. ‡ The overload capacity of nitrided gears is low. Since the shape of the effective S-N curve is flat, the sensitivity to shock should be investigated before proceeding with the design. Source: ANSI/AGMA 2003-B97.

Table 15–7  Allowable Bending Stress Number for Iron Gears, sat (σF lim) Typical Minimum Bending Stress Number Material Designation Heat Surface (Allowable), sat Material ASTM ISO Treatment Hardness (SF lim) lbf/in2 (N/mm2) Cast iron

ASTM A48

ISO/DR 185

Class 30

Grade 200

As cast

175 HB

4500 (30)

Class 40

Grade 300

As cast

200 HB

6500 (45)

Ductile (nodular) iron

ASTM A536

ISO/DIS 1083

Grade 80-55-06

Grade 600-370-03

10 000 (70)

Grade 800-480-02

Quenched and tempered

180 HB

Grade 120-90-02

300 HB

13 500 (95)

Source: ANSI/AGMA 2003-B97.

Allowable contact stress number sac , kpsi

200 1300 175

1200

Maximum for grade 2 sac = 363.6 HB + 29 560 (σH lim = 2.51 HB + 203.86)

150

1100 1000

125

900

Maximum for grade 1 sac = 341 HB + 23 620 (σH lim = 2.35 HB + 162.89)

100

800 700 600

75 150

200

250

300 Brinell hardness HB

350

400

450

Allowable contact stress number σH lim , M Pa

Allowable Contact Stress Tables 15–4 and 15–5 provide values of sac(σH) for steel gears and for iron gears, respectively. Figure 15–12 graphically displays allowable stress for grade 1 and 2 materials. Figure 15–12 Allowable contact stress number for through-hardened steel gears, sac(σH lim). (Source: ANSI/AGMA 2003-B97.)

Allowable bending stress number for through-hardened steel gears, sat(σF lim). (Source: ANSI/AGMA 2003-B97.)

60 350

50

300

40 Maximum for grade 2 sat = 48 HB + 5980 (σF lim = 0.33 HB + 41.24)

30

Maximum for grade 1 sat = 44 HB + 2100 (σF lim = 0.30 HB + 14.48)

250 200 150

20

100 10 150

200

250

300

350

400

450

Bending stress number (allowable) σF lim (MPa)

Figure 15–13

Bending stress number (allowable) sat (kpsi)

806      Mechanical Engineering Design

Brinell hardness HB

The equations are sac = 341HB + 23 620 psi σH lim = 2.35HB + 162.89 MPa sac = 363.6HB + 29 560 psi σH lim = 2.51HB + 203.86 MPa

grade 1 grade 1 grade 2 grade 2

(15–22)

Allowable Bending Stress Numbers Tables 15–6 and 15–7 provide sat (σF lim) for steel gears and for iron gears, respectively. Figure 15–13 shows graphically allowable bending stress sat (σH lim) for throughhardened steels. The equations are

sat = 44HB + 2100 psi σF lim = 0.30HB + 14.48 MPa sat = 48HB + 5980 psi σH lim = 0.33HB + 41.24 MPa

grade 1 grade 1 grade 2 grade 2

(15–23)

Reversed Loading AGMA recommends use of 70 percent of allowable strength in cases where tooth load is completely reversed, as in idler gears and reversing mechanisms. Summary Figure 15–14 is a "road map" for straight-bevel gear wear relations using 2003-B97. Figure 15–15 is a similar guide for straight-bevel gear bending using 2003-B97. The standard does not mention specific steel but mentions the hardness attainable by heat treatments such as through-hardening, carburizing and case-hardening, flame-hardening, and nitriding. Through-hardening results depend on size (diametral pitch). Through-hardened materials and the corresponding Rockwell C-scale hardness at the 90 percent martensite shown in parentheses following include 1045 (50), 1060 (54), 1335 (46), 2340 (49), 3140 (49), 4047 (52), 4130 (44), 4140 (49), 4340 (49), 5145 (51), E52100 (60), 6150 (53), 8640 (50), and 9840 (49). For carburized case-hard materials the approximate core hardnesses are 1015 (22), 1025 (37), 1118 (33), 1320 (35), 2317 (30), 4320 (35), 4620 (35), 4820 (35), 6120

Bevel and Worm Gears     807

Figure 15–14 STRAIGHT-BEVEL GEAR WEAR BASED ON ANSI /AGMA 2003-B97 (U.S. customary units)

Geometry

Force Analysis

Strength Analysis

N dP = P Pd

W = 2T d av

W t = 2T dP

W r = W t tanϕ cosγ

W r = W t tanϕ cosγ

W a = W t tanϕ sinγ

W a = W t tanϕ sinγ

γ = tan–1

t

NP NG

–1 NG

Γ = tan

NP

At large end of tooth Table 15-2 Eqs. (15-5) to (15-8) Eq. (15-11)

d av = d P – F cos Γ

Gear contact stress

"Road map" summary of principal straight-bevel gear wear equations and their parameters.

Sc = σc = Cp

( FdW I K K K t

P

o

υ

)

m Cs Cxc

Fig. 15-6 Eq. (15-21)

1/2

Eq. (15-12) Eq. (15-9)

Tables 15-4, 15-5, Fig. 15-12, Eq. (15-22) Fig. 15-8, Eq. (15-14) Eqs. (15-16), (15-17), gear only Gear wear strength

Swc = (σc )all =

sac CL CH SH KT CR Eqs. (15-19), (15-20), Table 15-3 Eq. (15-18)

Wear factor of safety

(σ ) SH = σc all , based on strength c

( )

nw =

(σc )all σc

2

, based on W t ; can be compared directly with SF

(35), 8620 (35), and E9310 (30). The conversion from HRC to HB (300-kg load, 10-mm ball) is HRC 42 40 38 36 34 32 30 28 26 24 22 20 18 16 14 12 10 HB    388 375 352 331 321 301 285 269 259 248 235 223 217 207 199 192 187

Most bevel-gear sets are made from carburized case-hardened steel, and the factors incorporated in 2003-B97 largely address these high-performance gears. For through-hardened gears, 2003-B97 is silent on KL and CL, and Figures 15–8 and 15–9 should prudently be considered as approximate.

808      Mechanical Engineering Design

Figure 15–15 "Road map" summary of principal straight-bevel gear bending equations and their parameters.

STRAIGHT-BEVEL GEAR BENDING BASED ON ANSI /AGMA 2003-B97 (U.S. customary units)

Geometry

Force Analysis

Strength Analysis

N dP = P Pd

W = 2T d av

W t = 2T dP

W r = W t tanϕ cosγ

W r = W t tanϕ cosγ

W a = W t tanϕ sinγ

W a = W t tanϕ sinγ

γ = tan–1

t

NP NG

–1 NG

Γ = tan

NP

Table 15-2 Eqs. (15-5) to (15-8)

d av = d P – F cos Γ

Eq. (15-10) Eq. (15-11)

At large end of tooth Gear bending stress

t KK St = σ = W Pd Ko Kυ s m F Kx J

Fig. 15-7 Eq. (15-13)

Table 15-6 or 15-7 Fig. 15-9, Eq. (15-15) Gear bending strength

Swt = σall =

sa t KL SF KT KR Eqs. (15-19), (15-20), Table 15-3 Eq. (15-18)

Bending factor of safety

σ SF = σall , based on strength σ n B = σall , based on W t , same as SF

15–4  Straight-Bevel Gear Analysis EXAMPLE 15–1 A pair of identical straight-tooth miter gears listed in a catalog has a diametral pitch of 5 at the large end, 25 teeth, a 1.10-in face width, and a 20° normal pressure angle; the gears are grade 1 steel through-hardened with a core and case hardness of 180 Brinell. The gears are uncrowned and intended for general industrial use. They have a quality number of Qv = 7. It is likely that the application intended will require outboard mounting of the gears. Use a safety factor of 1, a 107 cycle life, and a 0.99 reliability. (a) For a speed of 600 rev/min find the power rating of this gearset based on AGMA bending strength. (b) For the same conditions as in part (a) find the power rating of this gearset based on AGMA wear strength.

Bevel and Worm Gears     809

(c) For a reliability of 0.995, a gear life of 109 revolutions, and a safety factor of SF = SH = 1.5, find the power rating for this gearset using AGMA strengths. Solution From Figs. 15–14 and 15–15, dP = NP∕Pd = 25∕5 = 5.000 in vt = πdP nP∕12 = π(5) 600∕12 = 785.4 ft/min Overload factor: uniform-uniform loading, Table 15–2, Ko = 1.00. Safety factor: SF = 1, SH = 1. Dynamic factor Kv: from Eq. (15–6),

B = 0.25(12 − 7) 2∕3 = 0.731

A = 50 + 56(1 − 0.731) = 65.06 Kv = (

65.06 + √785.4 0.731 = 1.299 ) 65.06

From Eq. (15–8), vt max = [65.06 + (7 − 3)]2 = 4769 ft/min

vt < vt

max,

that is, 785.4 < 4769 ft/min, therefore Kv is valid. From Eq. (15–10),

Ks = 0.4867 + 0.2132∕5 = 0.529

From Eq. (15–11),

Kmb = 1.25

and

Km = 1.25 + 0.0036(1.10) 2 = 1.254

From Eq. (15–13), Kx = 1. From Fig. 15–6, I = 0.065; from Fig. 15–7, JP = 0.216, JG = 0.216. From Eq. (15–15), KL = 1.683(107 ) −0.0323 = 0.999 96 ≈ 1

From Eq. (15–14),

CL = 3.4822(107 ) −0.0602 = 1.32

Since HBP∕HBG = 1, then from Fig. 15–10, CH = 1. From Eqs. (15–13) and (15–18), Kx = 1 and KT = 1, respectively. From Eq. (15–20),

KR = 0.70 − 0.15 log(1 − 0.99) = 1,

CR =

√KR

=

√1

=1

(a) Bending: From Eq. (15–23), sat = 44(180) + 2100 = 10 020 psi

From Eq. (15–3),

st = σ =

Ks K m 0.529(1.254) Wt Wt Pd Ko Kv = (5)(1)1.299 F Kx J 1.10 (1)0.216

= 18.13 W t

From Eq. (15–4),

swt =

sat KL 10 020(1) = = 10 020 psi SF KT KR (1)(1) (1)

810      Mechanical Engineering Design

Equating st and swt,

18.13W t = 10 020

Answer

H=

W t = 552.6 lbf

W tvt 552.6(785.4) = = 13.2 hp 33 000 33 000

(b) Wear: From Fig. 15–12 or Eq. (15–22), sac = 341(180) + 23 620 = 85 000 psi From Eq. (15–2),

σc, all =

sacCLCH 85 000(1.32) (1) = = 112 200 psi SH KT CR (1) (1)(1)

Now Cp = 2290 √psi from definitions following Eq. (15–21). From Eq. (15–9), Cs = 0.125(1.1) + 0.4375 = 0.575

From Eq. (15–12), Cxc = 2. Substituting in Eq. (15–1) gives 1∕2 Wt σc = Cp( Ko Kv KmCsCxc) FdP I

1∕2 Wt = 2290 [ (1)1.299(1.254)0.575(2) ] = 5242 √W t 1.10(5)0.065

Equating σc and σc,all gives

5242 √W t = 112 200,

H=

W t = 458.1 lbf

458.1(785.4) = 10.9 hp 33 000

Rated power for the gearset is H = min(12.9, 10.9) = 10.9 hp

Answer

(c) Life goal 109 cycles, R = 0.995, SF = SH = 1.5, and from Eq. (15–15), KL = 1.683(109 ) −0.0323 = 0.8618

From Eq. (15–19),

KR = 0.50 − 0.25 log(1 − 0.995) = 1.075,

CR =

√KR

=

√1.075

From Eq. (15–14), CL = 3.4822(109 ) −0.0602 = 1

Bending: From Eq. (15–23) and part (a), sat = 10 020 psi. From Eq. (15–3),

st = σ =

0.529(1.254) Wt 5(1)1.299 = 18.13W t 1.10 (1)0.216

= 1.037

Bevel and Worm Gears     811

From Eq. (15–4),

swt =

sat KL 10 020(0.8618) = = 5355 psi SF KT KR 1.5(1)1.075

Equating st to swt gives

18.13W t = 5355

H=

W t = 295.4 lbf

295.4(785.4) = 7.0 hp 33 000

Wear: From Eq. (15–22), and part (b), sac = 85 000 psi. Substituting into Eq. (15–2) gives

σc,all =

sacCLCH 85 000(1)(1) = = 54 640 psi SH KT CR 1.5(1)1.037

From part (b), σc = 5242 √W t and equating σc to σc,all gives

σc = σc,all = 54 640 = 5242 √W t

W t = 108.6 lbf

The wear power is H=

108.6(785.4) = 2.58 hp 33 000

Answer The mesh rated power is H = min(7.0, 2.58) = 2.6 hp.

15–5  Design of a Straight-Bevel Gear Mesh A useful decision set for straight-bevel gear design is ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙

Function: power, speed, mG, R Design factor: nd Tooth system Tooth count: NP, NG Pitch and face width: Pd, F Quality number: Qv Gear material, core and case hardness Pinion material, core and case hardness

    A priori decisions        Design decisions   

In bevel gears the quality number is linked to the wear strength. The J factor for the gear can be smaller than for the pinion. Bending strength is not linear with face width, because added material is placed at the small end of the teeth. Consequently, face width is roughly prescribed as

F = min(0.3A 0, 10∕Pd )

(15–24)

where A0 is the cone distance (see Figure 13–16), given by

A0 =

dG dP = 2 sin γ 2 sin Γ

(15–25)

812      Mechanical Engineering Design

EXAMPLE 15–2 Design a straight-bevel gear mesh for shaft centerlines that intersect perpendicularly, to deliver 6.85 hp at 900 rev/min with a gear ratio of 3:1, temperature of 300°F, normal pressure angle of 20°, using a design f­actor of 2. The load is uniform-uniform. Use a pinion of 20 teeth. The material is to be AGMA grade 1 and the teeth are to be crowned. The reliability goal is 0.995 with a pinion life of 109 revolutions. Solution First we list the a priori decisions and their immediate consequences. Function: 6.85 hp at 900 rev/min, gear ratio mG = 3, 300°F environment, neither gear straddle-mounted, Kmb = 1.25 [Eq. (15–11)], R = 0.995 at 109 revolutions of the pinion, Eq. (15–14):

(CL ) G = 3.4822(109∕3) −0.0602 = 1.068

(CL ) P = 3.4822(10 9 ) −0.0602 = 1

Eq. (15–15):

(KL ) G = 1.683(109∕3) −0.0323 = 0.8929

(KL ) P = 1.683(109 ) −0.0323 = 0.8618

Eq. (15–19):

KR = 0.50 − 0.25 log(1 − 0.995) = 1.075

CR =

Eq. (15–18):

KT = CT = (460 + 300)∕710 = 1.070

Design factor: nd = 2, SF = 2, SH =

√2

√KR

=

√1.075

= 1.037

= 1.414.

Tooth system: crowned, straight-bevel gears, normal pressure angle 20°, Eq. (15–13):

Kx = 1

Eq. (15–12):

Cxc = 1.5.

With NP = 20 teeth, NG = (3)20 = 60 teeth and from Fig. 15–14, γ = tan−1 (NP∕NG ) = tan−1 (20∕60) = 18.43°

Γ = tan−1 (60∕20) = 71.57°

From Figs. 15–6 and 15–7, I = 0.0825, JP = 0.248, and JG = 0.202. Note that JP > JG. Decision 1: Trial diametral pitch, Pd = 8 teeth/in. Eq. (15–10):

Ks = 0.4867 + 0.2132∕8 = 0.5134

dP = NP∕Pd = 20∕8 = 2.5 in

dG = 2.5(3) = 7.5 in

vt = πdP nP∕12 = π(2.5)900∕12 = 589.0 ft/min

W t = 33 000 hp/vt = 33 000(6.85)∕589.0 = 383.8 lbf

Eq. (15–25):

A 0 = dP∕(2 sin γ) = 2.5∕(2 sin 18.43°) = 3.954 in

Eq. (15–24):

F = min(0.3A 0, 10∕Pd ) = min[0.3(3.954), 10∕8] = min(1.186, 1.25) = 1.186 in

Decision 2: Let F = 1.25 in. Then, Eq. (15–9):

Cs = 0.125(1.25) + 0.4375 = 0.5937

Eq. (15–11):

Km = 1.25 + 0.0036(1.25) 2 = 1.256

Bevel and Worm Gears     813

Decision 3: Let the transmission accuracy number be 6. Then, from Eq. (15–6),

B = 0.25(12 − 6) 2∕3 = 0.8255

A = 50 + 56(1 − 0.8255) = 59.77 Kv = (

Eq. (15–5):

59.77 + √589.0 0.8255 = 1.325 ) 59.77

Decision 4: Pinion and gear material and treatment. Carburize and case-harden grade ASTM 1320 to

Core 21 HRC (HB is 229 Brinell) Case 55-64 HRC (HB is 515 Brinell)

From Table 15–4, sac = 200 000 psi and from Table 15–6, sat = 30 000 psi. Gear bending: From Eq. (15–3), the bending stress is

(st ) G =

Ks Km 383.8 0.5134(1.256) Wt Pd Ko Kv = 8(1)1.325 F Kx JG 1.25 (1)0.202

= 10 390 psi

The bending strength, from Eq. (15–4), is given by

(swt ) G = (

sat KL 30 000(0.8929) = = 11 640 psi ) SF KT KR G 2(1.070)1.075

The strength exceeds the stress by a factor of 11640∕10390 = 1.12, giving an actual factor of safety of (SF)G = 2(1.12) = 2.24. Pinion bending: The bending stress can be found from

(st ) P = (st ) G

JG 0.202 = 10 390 = 8463 psi JP 0.248

The bending strength, again from Eq. (15–4), is given by

(swt ) P = (

sat KL 30 000(0.8618) = = 11 240 psi SF KT KR )P 2(1.070)1.075

The strength exceeds the stress by a factor of 11 240∕8463 = 1.33, giving an actual factor of safety of (SF)P = 2(1.33) = 2.66. Gear wear: The load-induced contact stress for the pinion and gear, from Eq. (15–1), is

1∕2 Wt sc = Cp( Ko Kv KmCsCxc) FdP I

1∕2 383.8 = 2290 [ (1)1.325(1.256) 0.5937(1.5) ] 1.25(2.5)0.0825

= 107 560 psi

From Eq. (15–2) the contact strength of the gear is

(swc ) G = (

sacCLCH 200 000(1.068)(1) = = 136 120 psi SH KT CR )G √2(1.070)1.037

814      Mechanical Engineering Design

The strength exceeds the stress by a factor of 136 120∕107 560 = 1.266, giving an actual factor of safety of (SH ) 2G = 1.2662 (2) = 3.21. Pinion wear: From Eq. (15–2) the contact strength of the pinion is (swc ) P = (

sacCLCH 200 000(1) (1) = = 127 450 psi SH KT CR )P √2(1.070)1.037

The strength exceeds the stress by a factor of 127 450∕107 560 = 1.185, giving an actual factor of safety of (SH ) 2P = 1.1852 (2) = 2.81. The actual factors of safety are 2.24, 2.66, 3.21, and 2.81. Making a direct comparison of the factors, we note that the primary threat is from gear bending. We also note that the other three factors of safety are considerably higher than the target design factor. If optimization is desired, our goal would be to make changes in the design decisions that drive the factors closer to 2.

15–6  Worm Gearing—AGMA Equation Sections 13–11 and 13–17 introduced worm gearing and its force analysis and efficiency. Here, we will present a condensed version of the AGMA recommendations for cylindrical (single-enveloping) worm gearing.2 For brevity, the equations will be shown for U.S. customary units only. Similar equations for SI units are available in the AGMA standards. Since they are essentially nonenveloping worm gears, the crossed helical gears, shown in Figure 15–16, can be considered with other worm gearing. Crossed helical gears, and worm gears too, usually have a 90° shaft angle, though this need not be so. The relation between the shaft and helix angles is

∑ = ψP ± ψG

Figure 15–16

(15–26)

Pitch cylinder of B

View of the pitch cylinders of a pair of crossed helical gears.

Axis of B

Axis of A

2

Pitch cylinder of A

ANSI/AGMA 6034-B92, February 1992, Practice for Enclosed Cylindrical Wormgear Speed-Reducers and Gear Motors; and ANSI/AGMA 6022-C93, Dec. 1993, Design Manual for Cylindrical Wormgearing. Note: Equations (15–32) to (15–38) are contained in Annex C of 6034-B92 for informational purposes only. To comply with ANSI/AGMA 6034-B92, use the tabulations of these rating factors provided in the standard.

Bevel and Worm Gears     815

Table 15–8  Cylindrical Worm Dimensions Common to Both Worm and Gear* ϕn

14.5° 20° 25° Quantity Symbol NW ≤ 2 NW ≤ 2 NW > 2 Addendum

a 0.3183px 0.3183px 0.286px

Dedendum

b 0.3683px 0.3683px 0.349px

Whole depth

ht 0.6866px 0.6866px 0.635px

*The table entries are for a tangential diametral pitch of the gear of Pt = 1.

where Σ is the shaft angle. The plus sign is used when both helix angles are of the same hand, and the minus sign when they are of opposite hand. The subscript P in Equation (15–26) refers to the pinion (worm); the subscript W is used for this same purpose. The subscript G refers to the gear, also called gear wheel, worm wheel, or simply the wheel. Table 15–8 gives cylindrical worm dimensions common to worm and gear. Good proportions indicate the worm pitch diameter d falls in the range

C 0.875 C 0.875 ≤d≤ 3 1.6

(15–27)

where C is the center-to-center distance. AGMA relates the allowable tangential force on the worm-gear tooth (W t)all to other parameters by

(W t ) all = Cs D 0.8 m FeCmCv

(15–28)

where Cs = materials factor Dm = mean gear diameter, in Fe = effective face width of the gear (actual face width, but not to exceed 0.67dm, the mean worm diameter), in Cm = ratio correction factor Cv = velocity factor The friction force Wf is given by

Wf =

f Wt cos λ cos ϕn

(15–29)

where f = coefficient of friction λ = lead angle at mean worm diameter ϕn = normal pressure angle The sliding velocity Vs at the mean worm diameter, in feet per minute, is

Vs =

πnW d m 12 cos λ

(15–30)

where nW = rotative speed of the worm and dm = mean worm diameter. The torque at the worm gear is W t Dm TG = (15–31) 2 where Dm is the mean gear diameter.

816      Mechanical Engineering Design

The parameters in Equation (15–28) are, quantitatively, Cs = 720 + 10.37C 3

C ≤ 3 in

(15–32)

For sand-cast gears,

1000 Cs = { 1190 − 477 log Dm

C > 3 C > 3

Dm ≤ 2.5 in Dm > 2.5 in

(15–33)

Dm ≤ 8 in Dm > 8 in

(15–34)

Dm ≤ 25 in Dm > 25 in

(15–35)

For chilled-cast gears,

1000 Cs = { 1412 − 456 log Dm

C > 3 C > 3

For centrifugally cast gears,

1000 Cs = { 1251 − 180 log Dm

C > 3 C > 3

The ratio correction factor Cm for gear ratio mG is given by

 2  0.02 √−mG + 40mG − 76 + 0.46  Cm =  0.0107 √−m2G + 56mG + 5145    1.1483 − 0.006 58mG

3 < mG ≤ 20 20 < mG ≤ 76

(15–36)

mG > 76

The velocity factor Cv is given by

 0.659 exp (−0.0011Vs )  Cv =  13.31 V s−0.571   65.52 V −0.774 s

Vs < 700 ft /min 700 ≤ Vs < 3000 ft /min Vs > 3000 ft /min

(15–37)

AGMA reports the coefficient of friction f as

 0.15  f =  0.124 exp (−0.074V s0.645 )   0.103 exp (−0.110V 0.450 ) + 0.012 s

Vs = 0 0 < Vs ≤ 10 ft /min Vs > 10 ft /min

(15–38)

Now we examine some worm-gear mesh geometry. The addendum a and dedendum b are

a=

px = 0.3183px π

(15–39)

b=

1.157px = 0.3683px π

(15–40)

The full depth ht is

 2.157px  = 0.6866px  π ht =   2.200px  + 0.002 = 0.7003px + 0.002  π

px ≥ 0.16 in

(15–41)

px < 0.16 in

The worm outside diameter do is

do = d + 2a

(15–42)

dr = d − 2b

(15–43)

The worm root diameter dr is

Bevel and Worm Gears     817

The worm-gear throat diameter Dt is

Dt = D + 2a

(15–44)

where D is the worm-gear pitch diameter. The worm-gear root diameter Dr is

Dr = D − 2b

(15–45)

c = b − a

(15–46)

The clearance c is

The worm face width (maximum) (FW)max is 2 Dt 2 D (FW ) max = 2 √ ( ) − ( − a) = 2 √2Da 2 2

(15–47)

which was simplified using Equation (15–44). The worm-gear face width FG is 2d m ∕3 FG = { 1.125 √ (do + 2c) 2 − (do − 4a) 2

px > 0.16 in px ≤ 0.16 in

(15–48)

The heat loss rate Hloss from the worm-gear case in ft · lbf/min is

Hloss = 33 000(1 − e)Hin

(15–49)

where e is efficiency, given by Equation (13–46), and Hin is the input horsepower from the worm. The overall coefficient hCR for combined convective and radiative heat transfer from the worm-gear case in ft · lbf/(min · in2 · °F) is

hCR =

      

nW + 0.13 6494 nW + 0.13 3939

no fan on worm shaft fan on worm shaft

(15–50)

The temperature of the oil sump ts is given by

ts = ta +

Hloss 33 000(1 − e)(H ) in = + ta hCR A hCR A

(15–51)

where A is the case lateral area in in2, and ta is the ambient temperature in °F. Bypassing Equations (15–49), (15–50), and (15–51) one can apply the AGMA recommendation for minimum lateral area Amin in in2 using Amin = 43.20C1.7

(15–52)

Because worm teeth are inherently much stronger than worm-gear teeth, they are not considered. The teeth in worm gears are short and thick on the edges of the face; midplane they are thinner as well as curved. Buckingham3 adapted the Lewis equation for this case: WGt σa = (15–53) pn Fe y where pn = px cos λ and y is the Lewis form factor related to circular pitch. For ϕn = 14.5°, y = 0.100; ϕn = 20°, y = 0.125; ϕn = 25°, y = 0.150; ϕn = 30°, y = 0.175. 3

Earle Buckingham, Analytical Mechanics of Gears, McGraw-Hill, New York, 1949, p. 495.

818      Mechanical Engineering Design

15–7  Worm-Gear Analysis Compared to other gearing systems worm-gear meshes have a much lower mechanical efficiency. Cooling, for the benefit of the lubricant, becomes a design constraint sometimes resulting in what appears to be an oversize gear case in light of its contents. If the heat can be dissipated by natural cooling, or simply with a fan on the wormshaft, simplicity persists. Water coils within the gear case or lubricant outpumping to an external cooler is the next level of complexity. For this reason, gear-case area is a design decision. To reduce cooling load, use multiple-thread worms. Also keep the worm pitch diameter as small as possible. Multiple-thread worms can remove the self-locking feature of many worm-gear drives. When the worm drives the gearset, the mechanical efficiency eW is given by

cos ϕn − f tan λ cos ϕn + f cot λ

eW =

(15–54)

With the gear driving the gearset, the mechanical efficiency eG is given by

eG =

cos ϕn − f cot λ cos ϕn + f tan λ

(15–55)

To ensure that the worm gear will drive the worm,

f stat < cos ϕn tan λ

(15–56)

where values of fstat can be found in ANSI/AGMA 6034-B92. To prevent the worm gear from driving the worm, refer to clause 9 of 6034-B92 for a discussion of selflocking in the static condition. It is important to have a way to relate the tangential component of the gear force t WGt to the tangential component of the worm force W W , which includes the role of t friction and the angularities of ϕn and λ. Refer to Equation (13–45), solved for W W :

WWt = WGt

cos ϕn sin λ + f cos λ cos ϕn cos λ − f sin λ

(15–57)

In the absence of friction WWt = WGt tan λ The mechanical efficiency of most gearing is very high, which allows power in and power out to be used almost interchangeably. Worm gearsets have such poor efficiencies that we work with, and speak of, output power. The magnitude of the gear transmitted force WGt can be related to the output horsepower H0, the application factor Ka, the efficiency e, and design factor nd by

WGt =

33 000nd H0 Ka VG e

(15–58)

t We use Equation (15–57) to obtain the corresponding worm force W W . It follows that the worm and gear transmitted powers in hp are

HW =

WWt VW πdW nW WWt = 33 000 12(33 000)

(15–59)

HG =

WGt VG π dG nG WGt = 33 000 12(33 000)

(15–60)

Bevel and Worm Gears     819

(15–61)

Table 15–9  Largest Lead Angle Associated with a Normal Pressure Angle ϕn for Worm Gearing

(15–62)

Maximum Lead ϕn Angle λmax

From Equation (13–44),

Wf =

f WGt f sin λ − cos ϕn cos λ

The sliding velocity of the worm at the pitch cylinder Vs is Vs =

π dnW 12 cos λ

and the friction power Hf is given by Hf =

∣Wf ∣ Vs 33 000

hp

(15–63)

Table 15–9 gives the largest lead angle λmax associated with normal pressure angle ϕn.

14.5°

16°

20°

25°

25°

35°

30°

45°

EXAMPLE 15–3 A single-thread steel worm rotates at 1800 rev/min, meshing with a 24-tooth worm gear transmitting 3 hp to the output shaft. The worm pitch diameter is 3 in and the tangential diametral pitch of the gear is 4 teeth/in. The normal pressure angle is 14.5°. The ambient temperature is 70°F. The application factor is 1.25 and the design factor is 1; gear face width is 2 in, lateral case area 600 in2, and the gear is chill-cast bronze. (a) Find the gear geometry. (b) Find the transmitted gear forces and the mesh efficiency. (c) Is the mesh sufficient to handle the loading? (d) Estimate the lubricant sump temperature. Solution (a) mG = NG∕NW = 24∕1 = 24, gear: D = NG ∕Pt = 24∕4 = 6.000 in, worm: d = 3.000 in. The axial circular pitch px is px = π∕Pt = π∕4 = 0.7854 in. C = (3 + 6)∕2 = 4.5 in. Eq. (15–39):

a = px∕π = 0.7854∕π = 0.250 in

Eq. (15–40):

b = 0.3683px = 0.3683(0.7854) = 0.289 in

Eq. (15–41):

h t = 0.6866px = 0.6866(0.7854) = 0.539 in

Eq. (15–42):

do = 3 + 2(0.250) = 3.500 in

Eq. (15–43):

d r = 3 − 2(0.289) = 2.422 in

Eq. (15–44):

D t = 6 + 2(0.250) = 6.500 in

Eq. (15–45):

Dr = 6 − 2(0.289) = 5.422 in

Eq. (15–46): Eq. (15–47):

c = 0.289 − 0.250 = 0.039 in (FW ) max = 2 √2(6)0.250 = 3.464 in

The tangential speeds of the worm, VW, and gear, VG, are, respectively,

VW = π(3)1800∕12 = 1414 ft/min

VG =

π(6)1800∕24 = 117.8 ft/min 12

820      Mechanical Engineering Design

The lead of the worm, from Eq. (13–27), is L = px NW = 0.7854(1) = 0.7854 in. The lead angle λ, from Eq. (13–28), is λ = tan−1

L 0.7854 = tan−1 = 4.764° πd π(3)

The normal diametral pitch for a worm gear is the same as for a helical gear, which from Eq. (13–18), with ψ = λ is

Pn =

Pt 4 = = 4.014 cos λ cos 4.764°

pn =

π π = = 0.7827 in Pn 4.014

The sliding velocity, from Eq. (15–62), is

Vs =

π dnW π(3)1800 = = 1419 ft/min 12 cos λ 12 cos 4.764°

(b) The coefficient of friction, from Eq. (15–38), is f = 0.103 exp[−0.110(1419) 0.450] + 0.012 = 0.0178

The efficiency e, from Eq. (13–46), is Answer

e=

cos ϕn − f tan λ cos 14.5° − 0.0178 tan 4.764° = = 0.818 cos ϕn + f cot λ cos 14.5° + 0.0178 cot 4.764°

The designer used nd = 1, Ka = 1.25 and an output horsepower of H0 = 3 hp. The gear tangential force component W Gt , from Eq. (15–58), is Answer

WGt =

33 000nd H0 Ka 33 000(1)3(1.25) = = 1284 lbf VG e 117.8(0.818)

Answer The tangential force on the worm is given by Eq. (15–57):

WWt = WGt

cos ϕn sin λ + f cos λ cos ϕn cos λ − f sin λ

= 1284

cos 14.5° sin 4.764° + 0.0178 cos 4.764° = 131 lbf cos 14.5° cos 4.764° − 0.0178 sin 4.764°

(c) Eq. (15–34):

Cs = 1000

Eq. (15–36):

Cm = 0.0107 √−242 + 56(24) + 5145 = 0.823

Eq. (15–37):

Cv = 13.31(1419) −0.571 = 0.211

Eq. (15–28):

(W t ) all = Cs D0.8 (Fe ) G Cm Cv = 1000(6) 0.8 (2)0.823(0.211) = 1456 lbf

Bevel and Worm Gears     821

The friction force Wf is given by Eq. (15–61):

Wf =

f WGt 0.0178(1284) = f sin λ − cos ϕn cos λ 0.0178 sin 4.764° − cos 14.5° cos 4.764°

= −23.7 lbf

The power dissipated in frictional work Hf is given by Eq. (15–63): Hf =

∣Wf ∣Vs 33 000

=

∣−23.7∣1419 = 1.02 hp 33 000

The worm and gear transmitted powers, HW and HG, are given by

HW =

WWt VW 131(1414) = = 5.61 hp 33 000 33 000

HG =

WGt VG 1284(117.8) = = 4.58 hp 33 000 33 000

Answer Gear power is satisfactory. Now,

Pn = Pt∕cos λ = 4∕cos 4.764° = 4.014

pn = π∕Pn = π∕4.014 = 0.7827 in

The bending stress in a gear tooth is given by Buckingham's adaptation of the Lewis equation, Eq. (15–53), as

(σ) G =

WGt 1284 = = 8200 psi pn FG y 0.7827(2) (0.1)

Answer Stress in gear is satisfactory. (d) Amin = 43.2C1.7 = 43.2(4.5) 1.7 = 557 in2

Eq. (15–52):

The gear case has a lateral area of 600 in2. Eq. (15–49): Eq. (15–50): Answer  Eq. (15–51):

Hloss = 33 000(1 − e)Hin = 33 000(1 − 0.818)5.61 = 33 690 ft · lbf/min hCR =

nW 1800 + 0.13 = + 0.13 = 0.587 ft · lbf/(min · in2 · °F) 3939 3939

ts = ta +

Hloss 33 690 = 70 + = 166°F hCR A 0.587(600)

822      Mechanical Engineering Design

15–8  Designing a Worm-Gear Mesh A usable decision set for a worm-gear mesh includes ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙

Function: power, speed, mG, Ka Design factor: nd Tooth system Materials and processes Number of threads on the worm: NW Axial pitch of worm: px Pitch diameter of the worm: dW Face width of gear: FG Lateral area of case: A

     A priori decisions         Design decisions   

Reliability information for worm gearing is not well developed at this time. The use of Equation (15–28) together with the factors Cs, Cm, and Cv, with an alloy steel casehardened worm together with customary nonferrous worm-wheel materials, will result in lives in excess of 25 000 h. The worm-gear materials in the experience base are principally bronzes: ∙ Tin- and nickel-bronzes (chilled-casting produces hardest surfaces) ∙ Lead-bronze (high-speed applications) ∙ Aluminum- and silicon-bronze (heavy load, slow-speed application)

Table 15–10  Minimum Number of Gear Teeth for Normal Pressure Angle ϕn   ϕn (NG)min 14.5 40 17.5 27 20 21 22.5 17 25 14 27.5 12 30 10

The factor Cs for bronze in the spectrum sand-cast, chilled-cast, and centrifugally cast increases in the same order. Standardization of tooth systems is not as far along as it is in other types of gearing. For the designer this represents freedom of action, but acquisition of tooling for tooth-forming is more of a problem for in-house manufacturing. When using a subcontractor the designer must be aware of what the supplier is capable of providing with on-hand tooling. Axial pitches for the worm are usually integers, and quotients of integers are common. Typical pitches are 14, 165 , 38, 12, 43, 1, 54, 64, 74 , and 2, but others are possible. Table 15–8 shows dimensions common to both worm gear and cylindrical worm for proportions often used. Teeth frequently are stubbed when lead angles are 30° or larger. Worm-gear design is constrained by available tooling, space restrictions, shaft center-to-center distances, gear ratios needed, and the designer's experience. ANSI/AGMA 6022-C93, Design Manual for Cylindrical Wormgearing offers the following guidance. Normal pressure angles are chosen from 14.5°, 17.5°, 20°, 22.5°, 25°, 27.5°, and 30°. The recommended minimum number of gear teeth is given in Table 15–10. The normal range of the number of threads on the worm is 1 through 10. Mean worm pitch diameter is usually chosen in the range given by Equation (15–27). A design decision is the axial pitch of the worm. Since acceptable proportions are couched in terms of the center-to-center distance, which is not yet known, one chooses a trial axial pitch px. Having NW and a trial worm diameter d, NG = mG NW

Pt =

π px

D=

NG Pt

Bevel and Worm Gears     823

Then (d) lo = C 0.87 5∕3

(d) hi = C 0.875∕1.6

Examine (d)lo ≤ d ≤ (d)hi, and refine the selection of mean worm-pitch diameter to d1 if necessary. Recompute the center-to-center distance as C = (d1 + D)∕2. There is even an opportunity to make C a round number. Choose C and set d2 = 2C − D Equations (15–39) through (15–48) apply to one usual set of proportions.

EXAMPLE 15–4 Design a 10-hp 11:1 worm-gear speed-reducer mesh for a lumber mill planer feed drive for 3- to 10-h daily use. A 1720-rev/min squirrel-cage induction motor drives the planer feed (Ka = 1.25), and the ambient temperature is 70°F. Solution Function: H0 = 10 hp, mG = 11, nW = 1720 rev/min. Design factor: nd = 1.2. Materials and processes: case-hardened alloy steel worm, sand-cast bronze gear. Worm threads: double, NW = 2, NG = mG NW = 11(2) = 22 gear teeth acceptable for ϕn = 20°, according to Table 15–10. Decision 1: Choose an axial pitch of worm px = 1.5 in. Then,

Pt = π∕px = π∕1.5 = 2.0944

D = NG∕Pt = 22∕2.0944 = 10.504 in

Eq. (15–39):

a = 0.3183px = 0.3183(1.5) = 0.4775 in (addendum)

Eq. (15–40):

b = 0.3683(1.5) = 0.5525 in (dedendum)

Eq. (15–41):

h t = 0.6866(1.5) = 1.030 in

Decision 2: Choose a mean worm diameter d = 2.000 in. Then

C = (d + D)∕2 = (2.000 + 10.504)∕2 = 6.252 in

(d) lo = 6.2520.875∕3 = 1.657 in

(d) hi = 6.2520.875∕1.6 = 3.107 in

The range, given by Eq. (15–27), is 1.657 ≤ d ≤ 3.107 in, which is satisfactory. Try d = 2.500 in. Recompute C:

C = (2.5 + 10.504)∕2 = 6.502 in

The range is now 1.715 ≤ d ≤ 3.216 in, which is still satisfactory. Decision: d = 2.500 in. Then Eq. (13–27):

L = px NW = 1.5(2) = 3.000 in

Eq. (13–28):

λ = tan−1[L∕(πd)] = tan−1[3∕(π 2.5)] = 20.905°  (from Table 15–9 lead angle OK)

824      Mechanical Engineering Design

Eq. (15–62):

Vs =

π d nW π(2.5)1720 = = 1205.1 ft/min 12 cos λ 12 cos 20.905°

VW =

π d nW π(2.5)1720 = = 1125.7 ft/min 12 12

VG =

π D nG π(10.504)1720∕11 = = 430.0 ft/min 12 12

Eq. (15–33):

Cs = 1190 − 477 log 10.504 = 702.8

Eq. (15–36):

Cm = 0.02 √−112 + 40(11) − 76 + 0.46 = 0.772

Eq. (15–37):

Cv = 13.31(1205.1)−0.571 = 0.232 f = 0.103 exp[−0.11(1205.1) 0.45] + 0.012 = 0.0191

Eq. (15–38): Eq. (15–54):

eW =

cos 20° − 0.0191 tan 20.905° = 0.942 cos 20° + 0.0191 cot 20.905°

(If the worm gear drives, eG = 0.939.) To ensure nominal 10-hp output, with adjustments for Ka, nd, and e, cos 20° sin 20.905° + 0.0191 cos 20.905° = 495.4 lbf cos 20° cos 20.905° − 0.0191 sin 20.905°

Eq. (15–57):

WWt = 1222

Eq. (15–58):

WGt =

33 000(1.2)10(1.25) = 1222 lbf 430(0.942)

Eq. (15–59):

HW =

π(2.5)1720(495.4) = 16.9 hp 12(33 000)

Eq. (15–60):

HG =

π(10.504)1720∕11(1222) = 15.92 hp 12(33 000)

Eq. (15–61):

Wf =

0.0191(1222) = −26.8 lbf 0.0191 sin 20.905° − cos 20° cos 20.905°

Eq. (15–63):

Hf =

∣−26.8 ∣1205.1 = 0.979 hp 33 000

From Eq. (15–28), with Cs = 702.8, Cm = 0.772, and Cv = 0.232,

(Fe ) req =

WGt 0.8

Cs D Cm Cv

=

1222 = 1.479 in 702.8(10.504) 0.80.772(0.232)

Decision 3: The available range of (Fe)G is 1.479 ≤ (Fe)G ≤ 2d∕3 or 1.479 ≤ (Fe)G ≤ 1.667 in. Set (Fe)G = 1.5 in. Wallt = 702.8(10.504) 0.81.5(0.772)0.232 = 1239 lbf

Eq. (15–28):

This is greater than 1222 lbf. There is a little excess capacity. The force analysis stands. Decision 4: Eq. (15–50): Eq. (15–49):

hCR =

nW 1720 + 0.13 = + 0.13 = 0.395 ft · lbf/(min · in2 · °F) 6494 6494

Hloss = 33 000(1 − e)HW = 33 000(1 − 0.942)16.9 = 32 347 ft · lbf/min

Bevel and Worm Gears     825

The AGMA area, from Eq. (15–52), is Amin = 43.2C1.7 = 43.2(6.502)1.7 = 1041.5 in2. A rough estimate of the lateral area for 6-in clearances: Vertical:  d + D + 6 = 2.5 + 10.5 + 6 = 19 in Width:   D + 6 = 10.5 + 6 = 16.5 in Thickness:  d + 6 = 2.5 + 6 = 8.5 in Area:    2(19)16.5 + 2(8.5)19 + 16.5(8.5) ≈ 1090 in2 Expect an area of 1100 in2. Choose: Air-cooled, no fan on worm, with an ambient temperature of 70°F.

ts = t a +

Hloss 32 350 = 70 + = 70 + 74.5 = 144.5°F hCR A 0.395(1100)

Lubricant is safe with some margin for smaller area. Eq. (13–18):

Pn =

Pt 2.094 = = 2.242 cos λ cos 20.905°

pn =

π π = = 1.401 in Pn 2.242

Gear bending stress, for reference, is Eq. (15–53):

σ=

WGt 1222 = = 4652 psi pn Fe y 1.401(1.5)0.125

The risk is from wear, which is addressed by the AGMA method that provides (W Gt )all.

15–9  Buckingham Wear Load A precursor to the AGMA method was the method of Buckingham, which identified an allowable wear load in worm gearing. Buckingham showed that the allowable geartooth loading for wear can be estimated from (WGt ) all = Kw dG Fe

(15–64)

where Kw = worm-gear load factor dG = gear-pitch diameter Fe = worm-gear effective face width Table 15–11 gives values for Kw for worm gearsets as a function of the material pairing and the normal pressure angle. EXAMPLE 15–5 Estimate the allowable gear wear load (WGt )all for the gearset of Example 15–4 using Buckingham's wear equation. Solution From Table 15–11 for a hardened steel worm and a bronze bear, Kw is given as 80 for ϕn = 20°. Equation (15–64) gives (WGt ) all = 80(10.504)1.5 = 1260 lbf which is larger than the 1239 lbf of the AGMA method. The method of Buckingham does not have refinements of the AGMA method. [Is (WGt )all linear with gear diameter?]

826      Mechanical Engineering Design

Table 15–11  Wear Factor Kw for Worm Gearing

Material

Thread Angle ϕn

Worm Gear 14 21 ° 20° 25° 30° Hardened steel*

Chilled bronze

90

125

150

180

Hardened steel*

Bronze

60

80

100

120

Steel, 250 BHN (min.)

Bronze

36

50

60

72

115

140

165

High-test cast iron

Bronze

80

Gray iron†

Aluminum

10 12 15 18

High-test cast iron

Gray iron

90

125

150

180

High-test cast iron

Cast steel

22

31

37

45

High-test cast iron

High-test cast iron

135

185

225

270

Steel 250 BHN (min.)

Laminated phenolic

47

64

80

95

Gray iron

Laminated phenolic

70

96

120

140

*Over 500 BHN surface. † For steel worms, multiply given values by 0.6. Source: Earle Buckingham, Design of Worm and Spiral Gears, Industrial Press, New York, 1981.

For material combinations not addressed by AGMA, Buckingham's method allows quantitative treatment.

PROBLEMS 15–1

An uncrowned straight-bevel pinion has 20 teeth, a diametral pitch of 6 teeth/in, and a transmission accuracy number of 6. Both the pinion and gear are made of throughhardened steel with a Brinell hardness of 300. The driven gear has 60 teeth. The gearset has a life goal of 109 revolutions of the pinion with a reliability of 0.999. The shaft angle is 90°, and the pinion speed is 900 rev/min. The face width is 1.25 in, and the normal pressure angle is 20°. The pinion is mounted outboard of its bearings, and the gear is straddle-mounted. Based on the AGMA bending strength, what is the power rating of the gearset? Use K0 = 1 and SF = SH = 1.

15–2 For the gearset and conditions of Problem 15–1, find the power rating based on the AGMA surface durability.

15–3 An uncrowned straight-bevel pinion has 30 teeth, a diametral pitch of 6, and a trans-

mission accuracy number of 6. The driven gear has 60 teeth. Both are made of No.  30 cast iron. The shaft angle is 90°. The face width is 1.25 in, the pinion speed is 900 rev/min, and the normal pressure angle is 20°. The pinion is mounted outboard of its bearings and the bearings of the gear straddle it. What is the power rating based on AGMA bending strength? Note: For cast iron gearsets reliability information has not yet been developed. We say that if the life is greater than 107 revolutions, then set KL = 1, CL = 1, CR = 1, KR = 1, and apply a factor of safety. Use SF = 2 and SH = √2.

15–4 For the gearset and conditions of Problem 15–3, find the power rating based on AGMA surface durability.

15–5 An uncrowned straight-bevel pinion has 22 teeth, a module of 4 mm, and a transmis-

sion accuracy number of 5. The pinion and the gear are made of through-hardened steel, both having core and case hardnesses of 180 Brinell. The pinion drives the

Bevel and Worm Gears     827

24-tooth bevel gear. The shaft angle is 90°, the pinion speed is 1800 rev/min, the face width is 25 mm, and the normal pressure angle is 20°. Both gears have an outboard mounting. Find the power rating based on AGMA pitting resistance if the life goal is 109 revolutions of the pinion at 0.999 reliability.

15–6 For the gearset and conditions of Problem 15–5, find the power rating for AGMA bending strength.

15–7 In straight-bevel gearing, there are some analogs to Equations (14–44) and (14–45), respectively. If we have a pinion core with a hardness of (HB)11 and we try equal power ratings, the transmitted load W t can be made equal in all four cases. It is possible to find these relations: Core Case Pinion (HB)11

(HB)12

Gear (HB)21 (HB)22

(a) For carburized case-hardened gear steel with core AGMA bending strength (sat)G and pinion core strength (sat)P, show that the relationship is

(sat ) G = (sat ) P

Jp JG

m−0.0323 G

This allows (HB)21 to be related to (HB)11. (b) Show that the AGMA contact strength of the gear case (sac)G can be related to the AGMA core bending strength of the pinion core (sat)P by (sac ) G =

S2H (sat ) P (KL ) P Kx JP KT Cs Cxc (CL ) G CH √ SF NP I Ks Cp

If factors of safety are applied to the transmitted load Wt, then SH = is unity. The result allows (HB)22 to be related to (HB)11.

√SF

and SH2∕SF

(c) Show that the AGMA contact strength of the gear (sac)G is related to the contact strength of the pinion (sac)P by (sac ) P = (sac ) G m0.0602 CH G

15–8 Refer to your solution to Problems 15–1 and 15–2. If the pinion core hardness is 300  Brinell, use the relations from Problem 15–7 to determine the required hardness of the gear core and the case hardnesses of both gears to ensure equal power ratings.

15–9 Repeat Problems 15–1 and 15–2 with the hardness protocol Core Case

Pinion 300

373

Gear 339

345

which can be established by the relations in Problem 15–7, and see if the result matches transmitted loads W t in all four cases.

15–10 A catalog of stock bevel gears lists a power rating of 5.2 hp at 1200 rev/min pinion speed

for a straight-bevel gearset consisting of a 20-tooth pinion driving a 40-tooth gear. This gear pair has a 20° normal pressure angle, a face width of 0.71 in, a diametral pitch of 10 teeth/in, and is through-hardened to 300 BHN. Assume the gears are for general industrial use, are generated to a transmission accuracy number of 5, and are uncrowned.

828      Mechanical Engineering Design

Also assume the gears are rated for a life of 3 × 106 revolutions with a 99 percent reliability. Given these data, what do you think about the stated catalog power rating?

15–11 Apply the relations of Problem 15–7 to Example 15–1 and find the Brinell case hard-

ness of the gears for equal allowable load W t in bending and wear. Check your work by reworking Example 15–1 to see if you are correct. How would you go about the heat treatment of the gears?

15–12 Your experience with Example 15–1 and problems based on it will enable you to write

an interactive computer program for power rating of through-hardened steel gears. Test your understanding of bevel-gear analysis by noting the ease with which the coding develops. The hardness protocol developed in Problem 15–7 can be incorporated at the end of your code, first to display it, then as an option to loop back and see the consequences of it.

15–13 Use your experience with Problem 15–11 and Example 15–2 to design an interactive

computer-aided design program for straight-steel bevel gears, implementing the ANSI/ AGMA 2003-B97 standard. It will be helpful to follow the decision set in Section 15–5, allowing the return to earlier decisions for revision as the consequences of earlier decisions develop.

15–14 A single-threaded steel worm rotates at 1725 rev/min, meshing with a 56-tooth worm gear

transmitting 1 hp to the output shaft. The pitch diameter of the worm is 1.50. The tangential diametral pitch of the gear is 8 teeth per inch and the normal pressure angle is 20°. The ambient temperature is 70°F, the application factor is 1.25, the design factor is 1, the gear face is 0.5 in, the lateral case area is 850 in2, and the gear is sand-cast bronze. (a) Determine and evaluate the geometric properties of the gears. (b) Determine the transmitted gear forces and the mesh efficiency. (c) Is the mesh sufficient to handle the loading? (d) Estimate the lubricant sump temperature, assuming fan-stirred air.

As in Example 15–4, design a cylindrical worm-gear mesh to connect a squirrel-cage 15–15 induction motor to a liquid agitator. The motor speed is 1125 rev/min, and the velocity to 15–22 ratio is to be 10:1. The output power requirement is 25 hp. The shaft axes are 90° to each other. An overload factor Ko (see Table 15–2) makes allowance for external dynamic excursions of load from the nominal or average load W t. For this service Ko = 1.25 is appropriate. Additionally, a design factor nd of 1.1 is to be included to address other unquantifiable risks. For Problems 15–15 to 15–17 use the AGMA method for (W Gt )all whereas for Problems 15–18 to 15–22, use the Buckingham method. See Table 15–12.

Table 15–12  Table Supporting Problems 15–15 to 15–22 Problem No. Method

Materials Worm

Gear

15–15

AGMA

Steel, HRC 58

Sand-cast bronze

15–16

AGMA

Steel, HRC 58

Chilled-cast bronze

15–17

AGMA

Steel, HRC 58

Centrifugal-cast bronze

15–18

Buckingham

Steel, 500 Bhn

Chilled-cast bronze

15–19

Buckingham

Steel, 500 Bhn

Cast bronze

15–20

Buckingham

Steel, 250 Bhn

Cast bronze

15–21

Buckingham

High-test cast iron

Cast bronze

15–22

Buckingham

High-test cast iron

High-test cast iron

16

Clutches, Brakes, Couplings, and Flywheels

©pheeraphol suthongsa/123RF

Chapter Outline 16–1  Static Analysis of Clutches and Brakes  831 16–2  Internal Expanding Rim Clutches and Brakes  836 16–3  External Contracting Rim Clutches and Brakes  844 16–4

Band-Type Clutches and Brakes   847

16–5  Frictional-Contact Axial Clutches  849

16–6

Disk Brakes   852

16–7

Cone Clutches and Brakes   856

16–8

Energy Considerations   858

16–9

Temperature Rise   860

16–10

Friction Materials   863

16–11  Miscellaneous Clutches and Couplings  866 16–12

Flywheels  868

829

830      Mechanical Engineering Design

This chapter is concerned with a group of elements usually associated with rotation that have in common the function of storing and/or transferring rotational energy. Because of this similarity of function, clutches, brakes, couplings, and flywheels are treated together in this chapter. A simplified dynamic representation of a friction clutch or brake is shown in Figure 16–1a. Two inertias, I1 and I2, traveling at the respective angular velocities ω1 and ω2, one of which may be zero in the case of brakes, are to be brought to the same speed by engaging the clutch or brake. Slippage occurs because the two elements are running at different speeds and energy is dissipated during actuation, resulting in a temperature rise. In analyzing the performance of these devices we shall be interested in: ∙ ∙ ∙ ∙

The The The The

actuating force torque transmitted energy loss temperature rise

The torque transmitted is related to the actuating force, the coefficient of friction, and the geometry of the clutch or brake. This is a problem in statics, which will have to be studied separately for each geometric configuration. However, temperature rise is related to energy loss and can be studied without regard to the type of brake or clutch, because the geometry of interest is that of the heat-dissipating surfaces. The various types of devices to be studied may be classified as follows: ∙ ∙ ∙ ∙ ∙ ∙

Rim types with internal expanding shoes Rim types with external contracting shoes Band types Disk or axial types Cone types Miscellaneous types

A flywheel is an inertial energy-storage device. It absorbs mechanical energy by increasing its angular velocity and delivers energy by decreasing its velocity. Figure 16–1b is a mathematical representation of a flywheel. An input torque Ti, corresponding to a coordinate θi, will cause the flywheel speed to increase. And a load or output torque To, with coordinate θo, will absorb energy from the flywheel and cause it to slow down. We shall be interested in designing flywheels so as to obtain a specified amount of speed regulation. Figure 16–1

Clutch or brake

(a) Dynamic representation of a clutch or brake; (b) mathematical representation of a flywheel. ω1

I2

I1

(a) Ti , θi

To, θo I, θ (b)

ω2

Clutches, Brakes, Couplings, and Flywheels     831

16–1  Static Analysis of Clutches and Brakes Many types of clutches and brakes can be analyzed by following a general procedure. The procedure entails the following tasks: ∙ Estimate, model, or measure the pressure distribution on the friction surfaces. ∙ Find a relationship between the largest pressure and the pressure at any point. ∙ Use the conditions of static equilibrium to find the braking force or torque and the support reactions. Let us apply these tasks to the doorstop depicted in Figure 16–2a. The stop is hinged at pin A. A normal pressure distribution p(u) is shown under the friction pad as a function of position u, taken from the right edge of the pad. A similar distribution of shearing frictional traction is on the surface, of intensity fp(u), in the direction of the motion of the floor relative to the pad, where f is the coefficient of friction. The width of the pad into the page is w2. The net force in the y direction and moment about C from the pressure are respectively,

N = w2 w2

w1

0

w1

0

p(u) du = pav w1w2

p(u)u du = u w2

w1

0

(a)

p(u) du = pav w1w2 u

(b)

We sum the forces in the x-direction to obtain

∑ Fx = Rx ∓ w2

w1

0

f p(u) du = 0

where − or + is for rightward or leftward relative motion of the floor, respectively. Assuming f constant, solving for Rx gives

Rx = ±w2

w1

0

f p(u) du = ± f w1w2 pav

(c)

Summing the forces in the y direction gives

∑ Fy = −F + w2

w1

0

p(u) du + Ry = 0

from which

Ry = F − w2

w1

0

p(u) du = F − pav w1w2

(d)

for either direction. Summing moments about the pin located at A we have

∑ MA = Fb − w2

w1

0

p(u)(c + u) du ∓ af w2

w1

0

p(u) du = 0

A brake shoe is self-energizing if its moment sense helps set the brake, self-deenergizing if the moment resists setting the brake. Continuing,

F=

w2 b [

w1

0

p(u)(c + u) du ± af

w1

0

p(u) du ]

(e)

832      Mechanical Engineering Design y

Figure 16–2 A common doorstop. (a) Free body of the doorstop. (b) Trapezoidal pressure distribution on the foot pad based on linear deformation of pad. (c) Free-body diagram for leftward movement of the floor, uniform pressure, Example 16–1. (d) Free-body diagram for rightward movement of the floor, uniform pressure, Example 16–1. (e) Free-body diagram for leftward movement of the floor, trapezoidal pressure, Example 16–1. For (c), (d), and (e), units of forces and dimensions are lbf and in, respectively.

Ry Rx

A

A

x

Plan view of pad

α β

w2 w1

a r1

r2

a

F b w1 Friction pad

β B

C

w1 α y2

Relative motion p(u)

c Δϕ

y1

B

r1 Δϕ

c

u

C

r2 Δϕ

u

υ

υ υ Center of pressure

(a)

(b)

4.595

30 2.162

10

16

2.139

10

2.162 2.1 5.405

10

16 2.1 40

(c)

4.652

2.139 2.14 5.348

(d)

(e)

Clutches, Brakes, Couplings, and Flywheels     833

Can F be equal to or less than zero? Only during rightward motion of the floor when the expression in brackets in Equation (e) is equal to or less than zero. We set the brackets to zero or less:

w1

0

p(u)(c + u) du − af

w1

0

p(u) du ≤ 0

from which

fcr ≥

∫ 1

w1

0

a

p(u)(c + u) du

w1

p(u) du

1 = a

c

w1

0

p(u) du +

0

fcr ≥

w1

w1

p(u)u du

0

(f )

p(u) du

0

c+u a

where u is the distance of the center of pressure from the right edge of the pad. The conclusion that a self-acting or self-locking phenomenon is present is independent of our knowledge of the normal pressure distribution p(u). Our ability to find the critical value of the coefficient of friction fcr is dependent on our knowledge of p(u), from which we derive u. EXAMPLE 16–1 The doorstop depicted in Figure 16–2a has the following dimensions: a = 4 in, b = 2 in, c = 1.6 in, w1 = 1 in, w2 = 0.75 in, where w2 is the depth of the pad into the plane of the paper. (a) For a leftward relative movement of the floor, an actuating force F of 10 lbf, a coefficient of friction of 0.4, use a uniform pressure distribution pav, find Rx, Ry, pav, and the largest pressure pa. (b) Repeat part a for rightward relative movement of the floor. (c) Model the normal pressure to be the "crush" of the pad, much as if it were composed of many small helical coil springs. Find Rx, Ry, pav, and pa for leftward relative movement of the floor and other conditions as in part a. (d) For rightward relative movement of the floor, is the doorstop a self-acting brake? Solution  (a)

Eq. (c):  Rx = f pav w1w2 = 0.4(1)(0.75)pav = 0.3pav

Eq. (d):  Ry = F − pav w1w2 = 10 − pav (1)(0.75) = 10 − 0.75pav

Eq. (e):

F=

w2 b [

1

∫p 0

av (c

+ u) du + af

1

∫p 0

av

du ]

=

w2 pav c b (

=

w2 pav 0.75 (c + 0.5 + af ) = [1.6 + 0.5 + 4(0.4)] pav b 2

= 1.3875 pav

1

0

du + pav

1

0

u du + af pav

1

0

du)

834      Mechanical Engineering Design

Solving for pav gives

pav =

F 10 = = 7.207 psi 1.3875 1.3875

We evaluate Rx and Ry as Answer

Rx = 0.3(7.207) = 2.162 lbf

Answer

Ry = 10 − 0.75(7.207) = 4.595 lbf

The normal force N on the pad is F − Ry = 10 − 4.595 = 5.405 lbf, upward. The line of action is through the center of pressure, which is at the center of the pad. The friction force is f N = 0.4(5.405) = 2.162 lbf directed to the left. A check of the moments about A gives

∑ MA = Fb − f Na − N(w1∕2 + c)

= 10(2) − 0.4(5.405)4 − 5.405(1∕2 + 1.6) ≈ 0

Answer The maximum pressure pa = pav = 7.207 psi. (b) Eq. (c):

Rx = −f pavw1w2 = −0.4(1) (0.75)pav = −0.3pav

Eq. (d):

Ry = F − pavw1w2 = 10 − pav (1) (0.75) = 10 − 0.75pav

Eq. (e):

F=

w2 b [

1

∫p 0

av (c

+ u) du + af

1

∫p 0

av

du ]

=

w2 pav c b (

=

0.75 pav [1.6 + 0.5 − 4(0.4)] = 0.1875pav 2

1

0

du + pav

1

0

u du + a f pav

1

0

du)

from which

pav =

F 10 = = 53.33 psi 0.1875 0.1875

which makes Answer

Rx = −0.3(53.33) = −16 lbf

Answer

Ry = 10 − 0.75(53.33) = −30 lbf

The normal force N on the pad is 10 + 30 = 40 lbf upward. The friction shearing force is f N = 0.4(40) = 16 lbf to the right. We now check the moments about A:

MA = f Na + Fb − N(c + 0.5) = 16(4) + 10(2) − 40(1.6 + 0.5) = 0

Note the change in average pressure from 7.207 psi in part a to 53.3 psi. Also note how directions of forces have changed. The maximum pressure pa is the same as pav, which has changed from 7.207 psi to 53.3 psi. (c) We will model the deformation of the pad as follows. If the doorstop rotates Δϕ counterclockwise, the right and left edges of the pad will deform down y1 and y2, respectively (Figure 16–2b). From similar triangles,

Clutches, Brakes, Couplings, and Flywheels     835

y1∕(r1 Δϕ) = c∕r1 and y2∕(r2 Δϕ) = (c + w1)∕r2. Thus, y1 = c Δϕ and y2 = (c + w1) Δϕ. This means that y is directly proportional to the horizontal distance from the pivot point A. That is, y = C1v, where C1 is a constant (see Figure 16–2b). Assuming the pressure is directly proportional to deformation, then p(v) = C2v, where C2 is a constant. In terms of u, the pressure is p(u) = C2(c + u) = C2(1.6 + u). Eq. (e):

F=

w2 b [

0.75 2 [

w1

0

1

p(u)c du +

w1

0

p(u)u du + af

1

w1

0

p(u) du ]

∫ C (1.6 + u) du]

=

= 0.375C2[(1.6 + 0.5)1.6 + (0.8 + 0.3333) + 4(0.4)(1.6 + 0.5)] = 2.945C2

0

C2 (1.6 + u) u du + a f

1

0

C2 (1.6 + u)1.6du +

0

2

Since F = 10 lbf, then C2 = 10∕2.945 = 3.396 psi/in, and p(u) = 3.396(1.6 + u). The average pressure is given by Answer

pav =

1 w1

w1 0

p(u) du =

1 1

1

0

3.396(1.6 + u) du = 3.396(1.6 + 0.5) = 7.132 psi

The maximum pressure occurs at u = 1 in, and is pa = 3.396(1.6 + 1) = 8.83 psi

Answer

Equations (c) and (d) of Sec. 16–1 are still valid. Thus, Answer

Rx = 0.3pav = 0.3(7.131) = 2.139 lbf

Ry = 10 − 0.75pav = 10 − 0.75(7.131) = 4.652 lbf

From statics, the floor normal force is 5.348 lbf at a horizontal distance from A of 2.14 in. The average pressure is pav = 7.13 psi and the maximum pressure is pa = 8.83 psi, which is approximately 24 percent higher than the average pressure. The presumption that the pressure was uniform in part a (because the pad was small, or because the arithmetic would be easier?) underestimated the peak pressure. Modeling the pad as a onedimensional springset is better, but the pad is really a three-dimensional continuum. A theory of elasticity approach or a finite element modeling may be overkill, given uncertainties inherent in this problem, but it still represents better modeling. (d) To evaluate u we need to evaluate two integrations

c

0

p(u)u du =

1

∫ 3.396(1.6 + u)u du = 3.396(0.8 + 0.3333) = 3.849 lbf 0

c

1

0

0

∫ p(u) du = ∫ 3.396(1.6 + u)du = 3.396(1.6 + 0.5) = 7.132 lbf/in

Thus u = 3.849∕7.132 = 0.5397 in. Then, from Eq. ( f ) of Sec. 16–1, the critical coefficient of friction is Answer

fcr ≥

c + u 1.6 + 0.5397 = = 0.535 a 4

The doorstop friction pad does not have a high enough coefficient of friction to make the doorstop a selfacting brake. The configuration must change and/or the pad material specification must be changed to sustain the function of a doorstop.

836      Mechanical Engineering Design

16–2  Internal Expanding Rim Clutches and Brakes The internal-shoe rim clutch shown in Figure 16–3 consists essentially of three elements: 1 the mating frictional surface 2 the means of transmitting the torque to and from the surfaces 3 the actuating mechanism Depending upon the operating mechanism, such clutches are further classified as ∙ ∙ ∙ ∙ ∙

expanding-ring centrifugal magnetic hydraulic pneumatic

The expanding-ring clutch is often used in textile machinery, excavators, and machine tools where the clutch may be located within the driving pulley. Expandingring clutches benefit from centrifugal effects; transmit high torque, even at low speeds; and require both positive engagement and ample release force. The centrifugal clutch is used mostly for automatic operation. If no spring is used, the torque transmitted is proportional to the square of the speed. This is particularly useful for electric-motor drives where, during starting, the driven machine comes up to speed without shock. Springs can also be used to prevent engagement until a certain motor speed is reached, but some shock may occur. Magnetic clutches are particularly useful for automatic and remote-control systems. Such clutches are also useful in drives subject to complex load cycles (see Section 11–7). Hydraulic and pneumatic clutches are also useful in drives having complex loading cycles and in automatic machinery, or in robots. Here the fluid flow can be controlled remotely using solenoid valves. These clutches are also available as disk, cone, and multiple-plate clutches. In braking systems, the internal-shoe or drum brake is used mostly for automotive applications.

Figure 16–3 An internal expanding centrifugalacting rim clutch.

Clutches, Brakes, Couplings, and Flywheels     837

Rim rotation

h Δϕ cos θ 2 θ 2 B

y

h ∆ϕ

r

r dθ

h

θ 2

F dθ

θ 2

θ2 θ A

O

x

r

A

a r

Figure 16–4

Figure 16–5

Internal friction shoe geometry.

The geometry associated with an arbitrary point on the shoe.

To analyze an internal-shoe device, refer to Figure 16–4, which shows a shoe pivoted at point A, with the actuating force acting at the other end of the shoe. Since the shoe is long, we cannot make the assumption that the distribution of normal forces is uniform. The mechanical arrangement permits no pressure to be applied at the heel, and we will therefore assume the pressure at this point to be zero. It is the usual practice to omit the friction material for a short distance away from the heel (point A). This eliminates interference, and the material would contribute little to the performance anyway, as will be shown. In some designs the hinge pin is made movable to provide additional heel pressure. This gives the effect of a floating shoe. (Floating shoes will not be treated in this book, although their design follows the same general principles.) Let us consider the pressure p acting upon an element of area of the frictional material located at an angle θ from the hinge pin (Figure 16–4). We designate the maximum pressure pa located at an angle θa from the hinge pin. To find the pressure distribution on the periphery of the internal shoe, consider point B on the shoe (Figure 16–5). As in Example 16–1, if the shoe deforms by an infinitesimal rotation Δϕ about the pivot point A, deformation perpendicular to AB is h Δϕ. From the isosceles triangle AOB, h = 2r sin(θ∕2), so

h Δϕ = 2r Δϕ sin(θ∕2)

The deformation perpendicular to the rim is h Δϕ cos(θ∕2), which is

h Δϕ cos (θ∕2) = 2r Δϕ sin(θ∕2) cos (θ∕2) = r Δϕ sin θ

Thus, the deformation, and consequently the pressure, is proportional to sin θ. In terms of the pressure at B and where the pressure is a maximum, this means

p pa = sin θ sin θa

(a)

Rearranging gives

p=

pa sin θ sin θa

(16–1)

838      Mechanical Engineering Design p

This pressure distribution has interesting and useful characteristics: θ θ1

π

θ2 θa (a)

p

θ θ1

θa

θ2

π

(b)

Figure 16–6 Defining the angle θa at which the maximum pressure pa occurs when (a) shoe exists in zone θ1 ≤ θ2 ≤ π∕2 and (b) shoe exists in zone θ1 ≤ π∕2 ≤ θ2.

∙ The pressure distribution is sinusoidal with respect to the angle θ. ∙ If the shoe is short, as shown in Figure 16–6a, the largest pressure on the shoe is pa occurring at the end of the shoe, θ2. ∙ If the shoe is long, as shown in Figure 16–6b, the largest pressure on the shoe is pa occurring at θa = 90°. Since limitations on friction materials are expressed in terms of the largest allowable pressure on the lining, the designer wants to think in terms of pa and not about the amplitude of the sinusoidal distribution that addresses locations off the shoe. When θ = 0, Equation (16–1) shows that the pressure is zero. The frictional material located at the heel therefore contributes very little to the braking action and might as well be omitted. A good design would concentrate as much frictional material as possible in the neighborhood of the point of maximum pressure. Such a design is shown in Figure 16–7. In this figure the frictional material begins at an angle θ1, measured from the hinge pin A, and ends at an angle θ2. Any arrangement such as this will give a good distribution of the frictional material. Proceeding now (Figure 16–7), the hinge-pin reactions are Rx and Ry. The actuating force F has components Fx and Fy and operates at distance c from the hinge pin. At any angle θ from the hinge pin there acts a differential normal force d N whose magnitude is

d N = pbr d θ

(b)

where b is the face width (perpendicular to the page) of the friction material. Substituting the value of the pressure from Equation (16–1), the normal force is

dN =

pa br sin θ dθ sin θa

(c)

The normal force d N has horizontal and vertical components d N cos θ and d N sin θ, as shown in the figure. The frictional force f d N has horizontal and vertical components Figure 16–7 Forces on the shoe.

as

in

y dN

f dN cos θ

dN sin θ

f dN u

θ

Fx

dN cos θ f dN sin θ

F Fy

θ2

ac

os θ

θ

r–

A

θ1 Rx

Ry

c a r

Rotation

θ

x

Clutches, Brakes, Couplings, and Flywheels     839

whose magnitudes are f d N sin θ and f d N cos θ, respectively. By applying the conditions of static equilibrium, we may find the actuating force F, the torque T, and the pin reactions Rx and Ry. We shall find the actuating force F, using the condition that the summation of the moments about the hinge pin is zero. The frictional forces have a moment arm about the pin of r − a cos θ. The moment Mf of these frictional forces is

Mf =

f p br ∫ f dN(r − a cos θ) = sin θ ∫ a

a

θ2

sin θ (r − a cos θ) dθ

(16–2)

θ1

which is obtained by substituting the value of d N from Equation (c). It is convenient to integrate Equation (16–2) for each problem, and we shall therefore retain it in this form. The moment arm of the normal force d N about the pin is a sin θ. Designating the moment of the normal forces by MN and summing these about the hinge pin give

MN =

∫ dN(a sin θ) = psinbraθ ∫ a

a

θ2

sin2 θ dθ

(16–3)

θ1

The actuating force F must balance these moments. Thus

F=

MN − Mf c

(16–4)

We see here that a condition for zero actuating force exists. In other words, if we make MN = Mf , self-locking is obtained, and no actuating force is required. This furnishes us with a method for obtaining the dimensions for some self-energizing action. So to avoid self-locking, the dimension a in Figure 16–7 must be such that

MN > Mf

(16–5)

The torque T applied to the drum by the brake shoe is the sum of the frictional forces f d N times the radius of the drum:

2

∫ fr dN = fsinp brθ ∫ a

T=

a

θ2

sin θ dθ

θ1

f pa br 2 (cos θ1 − cos θ2 ) sin θa

=

(16–6)

The hinge-pin reactions are found by taking a summation of the horizontal and vertical forces. Thus, for Rx, we have

Rx =

∫ dN cos θ − ∫ f dN sin θ − F

x

pa br = sin θa (

θ2

sin θ dθ) − Fx

(d)

sin θ cos θ dθ) − Fy

(e)

sin θ cos θ dθ − f

θ1

θ2

θ1

2

The vertical reaction is found in the same way:

Ry = dN sin θ +

pa br = sin θa (

θ2

θ1

∫ f dN cos θ − F

y

2

sin θ dθ + f

θ2

θ1

840      Mechanical Engineering Design

The direction of the frictional forces is reversed if the rotation is reversed. Thus, for counterclockwise rotation the actuating force is

F=

MN + Mf c

(16–7)

and since both moments have the same sense, the self-energizing effect is lost. Also, for counterclockwise rotation the signs of the frictional terms in the equations for the pin reactions change, and Equations (d) and (e) become

Rx = Ry =

pa br sin θa ( pa br sin θa (

θ2

θ2

θ1

(f  )

sin θ cos θ dθ) − Fy

(g)

θ2

θ1

sin2 θ dθ − f

θ1

θ2

θ1

Equations (d ), (e), ( f ), and (g) can be simplified to ease computations. Thus, let A=

θ2

θ2

θ1

sin2 θ dθ) − Fx

sin θ cos θ dθ + f

B=

θ1

θ2 1 sin θ cos θ dθ = ( sin2 θ) 2 θ1 θ2 θ 1 sin2 θ dθ = ( − sin 2θ) 2 4 θ1

(16–8)

Then, for clockwise rotation as shown in Figure 16–7, the hinge-pin reactions are pa br (A − f B) − Fx sin θa

Rx =

pa br Ry = (B + f A) − Fy sin θa

(16–9)

For counterclockwise rotation, Equations ( f ) and (g) become pa br (A + f B) − Fx sin θa

Rx =

pa br Ry = (B − f A) − Fy sin θa

(16–10)

In using these equations, the reference system always has its origin at the center of the drum. The positive x axis is taken through the hinge pin. The positive y axis is always in the direction of the shoe, even if this should result in a left-handed system. The following assumptions are implied by the preceding analysis: 1 The pressure at any point on the shoe is assumed to be proportional to the distance from the hinge pin, being zero at the heel. This should be considered from the standpoint that pressures specified by manufacturers are averages rather than maxima. 2 The effect of centrifugal force has been neglected. In the case of brakes, the shoes are not rotating, and no centrifugal force exists. In clutch design, the effect of this force must be considered in writing the equations of static equilibrium.

Clutches, Brakes, Couplings, and Flywheels     841

3 The shoe is assumed to be rigid. Since this cannot be true, some deflection will occur, depending upon the load, pressure, and stiffness of the shoe. The resulting pressure distribution may be different from that which has been assumed. 4 The entire analysis has been based upon a coefficient of friction that does not vary with pressure. Actually, the coefficient may vary with a number of conditions, including temperature, wear, and environment. EXAMPLE 16–2 The brake shown in Figure 16–8 is 300 mm in diameter and is actuated by a mechanism that exerts the same force F on each shoe. The shoes are identical and have a face width of 32 mm. The lining is a molded asbestos having a coefficient of friction of 0.32 and a pressure limitation of 1000 kPa. Estimate the maximum (a) Actuating force F.   (b) Braking capacity.   (c) Hinge-pin reactions. Solution (a) The right-hand shoe is self-energizing, and so the force F is found on the basis that the maximum pressure will occur on this shoe. Here θ1 = 0°, θ2 = 126°, θa = 90°, and sin θa = 1. Also, a = √ (112) 2 + (50) 2 = 122.7 mm

Integrating Equation (16–2) from 0 to θ2 yields

Mf =

=

θ2 f pa br 1 2 θ2 −r cos θ − a )0 ( 2 sin θ)0 ] sin θa [(

f pa br a r − r cos θ2 − sin2 θ2) ( sin θa 2

Changing all lengths to meters, we have Mf = (0.32)[1000(10) 3](0.032) (0.150) × [ 0.150 − 0.150 cos 126° − ( = 304 N · m

62

62

F

F

100

0.1227 sin2 126° ] 2 ) Figure 16–8

30°

Brake with internal expanding shoes; dimensions in millimeters.

150 126°

112

50 Rotation

50 24°

842      Mechanical Engineering Design

The moment of the normal forces is obtained from Equation (16–3). Integrating from 0 to θ2 gives θ2 pa bra θ 1 − sin 2θ )0 sin θa ( 2 4

MN =

=

= [1000(10) 3](0.032)(0.150)(0.1227) {

= 788 N · m

pa bra θ2 1 − sin 2θ2) sin θa ( 2 4 π 126 1 − sin[(2) (126°)]} 2 180 4

From Equation (16–4), the actuating force is F=

Answer

MN − Mf 788 − 304 = = 2.28 kN c 100 + 112

(b) From Equation (16–6), the torque applied by the right-hand shoe is

TR =

=

f pa br 2 (cos θ1 − cos θ2 ) sin θa 0.32[1000(10) 3](0.032) (0.150) 2 (cos 0° − cos 126°) = 366 N · m sin 90°

The torque contributed by the left-hand shoe cannot be obtained until we learn its maximum operating pressure. Equations (16–2) and (16–3) indicate that the frictional and normal moments are proportional to this pressure. Thus, for the left-hand shoe,

MN =

788pa 1000

Mf =

304pa 1000

Then, from Equation (16–7),

F=

MN + Mf c

or

2.28 =

(788∕1000)pa + (304∕1000)pa 100 + 112

Solving gives pa = 443 kPa. Then, from Equation (16–6), the torque on the left-hand shoe is

TL =

f pa br 2 (cos θ1 − cos θ2 ) sin θa

Since sin θa = sin 90° = 1, we have

TL = 0.32[443(10) 3](0.032)(0.150) 2 (cos 0° − cos 126°) = 162 N · m

The braking capacity is the total torque: Answer

T = TR + TL = 366 + 162 = 528 N · m

Clutches, Brakes, Couplings, and Flywheels     843

(c) In order to find the hinge-pin reactions, we note that sin θa = 1 and θ1 = 0. Then Equation (16–8) gives

A=

1 2 1 sin θ2 = sin2 126° = 0.3273 2 2

B=

θ2 1 π(126) 1 − sin 2θ2 = − sin[(2)(126°)] = 1.3373 2 4 2(180) 4

Also, let

D=

pa br 1000(0.032) (0.150) = = 4.8 kN sin θa 1

where pa = 1000 kPa for the right-hand shoe. Then, using Equation (16–9), we have

Rx = D(A − f B) − Fx = 4.8[0.3273 − 0.32(1.3373)] − 2.28 sin 24° = −1.410 kN

Ry = D(B + f A) − Fy = 4.8[1.3373 + 0.32(0.3273)] − 2.28 cos 24°

= 4.839 kN

The resultant on this hinge pin is Answer

R = √ (−1.410) 2 + (4.839) 2 = 5.04 kN

The reactions at the hinge pin of the left-hand shoe are found using Equations (16–10) for a pressure of 443 kPa. They are found to be Rx = 0.678 kN and Ry = 0.538 kN. The resultant is Answer

R = √ (0.678) 2 + (0.538) 2 = 0.866 kN

The reactions for both hinge pins, together with their directions, are shown in Figure 16–9. Figure 16–9 Fx

Fx F F Fy

Fy 24°

24°

y

y

Rx R Ry

Ry Rx

R x

x

844      Mechanical Engineering Design

This example dramatically shows the benefit to be gained by arranging the shoes to be self-energizing. If the left-hand shoe were turned over so as to place the hinge pin at the top, it could apply the same torque as the right-hand shoe. This would make the capacity of the brake (2)(366) = 732 N · m instead of the present 528 N · m, a 30 percent improvement. In addition, some of the friction material at the heel could be eliminated without seriously affecting the capacity, because of the low pressure in this area. This change might actually improve the overall design because the additional rim exposure would improve the heat-dissipation capacity.

16–3  External Contracting Rim Clutches and Brakes The patented clutch-brake of Figure 16–10 has external contracting friction elements, but the actuating mechanism is pneumatic. Here we shall study only pivoted external shoe brakes and clutches, though the methods presented can easily be adapted to the clutch-brake of Figure 16–10. Operating mechanisms can be classified as: ∙ ∙ ∙ ∙

Solenoids Levers, linkages, or toggle devices Linkages with spring loading Hydraulic and pneumatic devices

The static analysis required for these devices has already been covered in Section 3–1. The methods there apply to any mechanism system, including all those used in brakes and clutches. It is not necessary to repeat the material in Chapter 3 that applies directly to such mechanisms. Omitting the operating mechanisms from consideration allows us to concentrate on brake and clutch performance without the extraneous influences introduced by the need to analyze the statics of the control mechanisms. The notation for external contracting shoes is shown in Figure 16–11. The moments of the frictional and normal forces about the hinge pin are the same as for the internal expanding shoes. Equations (16–2) and (16–3) apply and are repeated here for convenience:

Mf = MN =

f pa br sin θa

θ2

pa bra sin θa

θ2

sin θ(r − a cos θ) dθ

(16–2)

sin2 θ dθ

(16–3)

θ1

θ1

Both these equations give positive values for clockwise moments (Figure 16–11) when used for external contracting shoes. The actuating force must be large enough to balance both moments: MN + Mf F= (16–11) c Figure 16–10 An external contracting clutch-brake that is engaged by expanding the flexible tube with compressed air. (Courtesy of Twin Disc Clutch Company.)

The horizontal and vertical reactions at the hinge pin are found in the same manner as for internal expanding shoes. They are

∫ f d N sin θ − F

(a)

∫ dN sin θ + F

(b)

Rx = d N cos θ +

Ry = f dN cos θ −

x

y

Clutches, Brakes, Couplings, and Flywheels     845

Figure 16–11 Notation of external contracting shoes.

Fx Fy

F

c f dN sin θ

y

θ f dN

f dN cos θ

dN θ

θ2 θ1

dN sin θ dN cos θ

θ

Rx A

x

Ry

r a

Rotation

By using Equation (16–8) and Equation (c) from Section 16–2, we have pa br (A + f B) − Fx sin θa

Rx =

pa br Ry = ( f A − B) + Fy sin θa

(16–12)

If the rotation is counterclockwise, the sign of the frictional term in each equation is reversed. Thus Equation (16–11) for the actuating force becomes

F=

MN − Mf c

(16–13)

and self-energization exists for counterclockwise rotation. The horizontal and vertical reactions are found, in the same manner as before, to be pa br (A − f B) − Fx sin θa

Rx =

pa br Ry = (−f A − B) + Fy sin θa

(16–14)

It should be noted that, when external contracting designs are used as clutches, the effect of centrifugal force is to decrease the normal force. Thus, as the speed increases, a larger value of the actuating force F is required. A special case arises when the pivot is symmetrically located and also placed so that the moment of the friction forces about the pivot is zero. The geometry of such a brake will be similar to that of Figure 16–12a. To get a pressure-distribution relation, we note that lining wear is such as to retain the cylindrical shape, much as a

846      Mechanical Engineering Design

Figure 16–12

y y

(a) Brake with symmetrical pivoted shoe; (b) wear of brake lining.

f dN sin θ

w(θ)

f dN cos θ

f dN

Rotation

w0 a cos θ – r θ dN sin θ

dN

r

wo

θ

(b)

dN cos θ

θ2

x

A Rx

θ1

x

r cos θ Ry

a (a)

milling machine cutter feeding in the x direction would do to the shoe held in a vise. See Figure 16–12b. This means the abscissa component of wear is w0 for all positions θ. If wear in the radial direction is expressed as w(θ), then

w(θ) = w0 cos θ

Using Equation (12–26), to express radial wear w(θ) as

w(θ) = KPVt

where K is a material constant, P is pressure, V is rim velocity, and t is time. Then, denoting P as p(θ) above and solving for p(θ) gives

p(θ) =

w(θ) w0 cos θ = KVt KV t

Since all elemental surface areas of the friction material see the same rubbing speed for the same duration, w0∕(KVt) is a constant and

p(θ) = (constant) cos θ = pa cos θ

(c)

where pa is the maximum value of p(θ). Proceeding to the force analysis, we observe from Figure 16–12a that

dN = pbr dθ

(d)

dN = pa br cos θ dθ

(e)

or

The distance a to the pivot is chosen by finding where the moment of the frictional forces Mf is zero. First, this ensures that reaction Ry is at the correct location to establish symmetrical wear. Second, a cosinusoidal pressure distribution is sustained, preserving our predictive ability. Symmetry means θ1 = θ2, so

Mf = 2

θ2

0

( f dN)(a cos θ − r) = 0

Clutches, Brakes, Couplings, and Flywheels     847

Substituting Equation (e) gives

2 f pa br

θ2

0

(a cos2 θ − r cos θ) dθ = 0

from which

a=

4r sin θ2 2θ2 + sin 2θ2

(16–15)

The distance a depends on the pressure distribution. Mislocating the pivot makes Mf zero about a different location, so the brake lining adjusts its local contact pressure, through wear, to compensate. The result is unsymmetrical wear, retiring the shoe lining, hence the shoe, sooner. With the pivot located according to Equation (16–15), the moment about the pin is zero, and the horizontal and vertical reactions are

Rx = 2

θ2

0

dN cos θ = 2

θ2

0

( pabr cos θ dθ) cos θ =

pa br (2θ2 + sin 2θ2 ) 2

(16–16)

where, because of symmetry,

∫ f dN sin θ = 0

Also, Ry = 2

θ2

0

f dN cos θ = 2

θ2

0

f ( pabr cos θ dθ) cos θ =

pa br f (2θ2 + sin 2θ2 ) (16–17) 2

where

∫ dN sin θ = 0

also because of symmetry. Note, too, that Rx = N and Ry = f N, as might be expected for the particular choice of the dimension a. Also, it can be shown that the torque is

T = a f N

(16–18)

16–4  Band-Type Clutches and Brakes Flexible clutch and brake bands are used in power excavators and in hoisting and other machinery. The analysis follows the notation of Figure 16–13. Because of friction and the rotation of the drum, the actuating force P2 is less than the pin reaction P1. Any element of the band, of angular length dθ, will be in equilibrium under the action of the forces shown in the figure. Summing these forces in the vertical direction, we have

dN − P sin

dθ dθ − (P + dP) sin = 0 2 2

For small (infinitesimal) dθ and dP, sin  dθ∕2 ≈ dθ∕2 and dP  sin  dθ∕2 ≈ 0. Thus, Equation (a) reduces to

dN = P dθ

(b)

848      Mechanical Engineering Design

Figure 16–13

r dθ dθ

Forces on a brake band.

P + dP

P

ϕ dN

θ

O

D

fdN Drum rotation

P2

P1

r

O

(b)

(a)

Summing the forces in the horizontal direction gives

f dN + P cos

dθ dθ − (P + dP) cos = 0 2 2

(c)

For small (infinitesimal) dθ, cos  (dθ∕2) ≈ 1, and Equation (c) simplifies to

dP = f dN

(d)

Substituting the value of dN from Equation (b) in (d) and integrating gives

P1

P2

dP =f P

ϕ

or

0

ln

P1 = fϕ P2

and thus

P1 = e fϕ P2

(16–19)

The torque may be obtained from the equation

T = (P1 − P2 )

D 2

(16–20)

The normal force dN acting on an element of area of width b and length r dθ is

dN = pbr dθ

(e)

where p is the pressure. Substitution of the value of dN from Equation (b) gives

P dθ = pbr dθ

Therefore

p=

P 2P = br bD

(16–21)

The pressure is therefore proportional to the tension in the band. The maximum pressure pa will occur at the toe and has the value

pa =

2P1 bD

(16–22)

Clutches, Brakes, Couplings, and Flywheels     849

16–5  Frictional-Contact Axial Clutches An axial clutch is one in which the mating frictional members are moved in a direction parallel to the shaft. One of the earliest of these is the cone clutch, which is simple in construction and quite powerful. However, except for relatively simple installations, it has been largely displaced by the disk clutch employing one or more disks as the operating members. Advantages of the disk clutch include the freedom from centrifugal effects, the large frictional area that can be installed in a small space, the more effective heat-dissipation surfaces, and the favorable pressure distribution. Figure 16–14 shows a single-plate disk clutch; a multiple-disk clutch-brake is shown in Figure 16–15. Let us now determine the capacity of such a clutch or brake in terms of the material and geometry. Figure 16–16 shows a friction disk having an outside diameter D and an inside diameter d. We are interested in obtaining the axial force F necessary to produce a certain torque T and pressure p. Two methods of solving the problem, depending upon the construction of the clutch, are in general use. If the disks are rigid, then the greatest amount of wear will at first occur in the outer areas, since the work of friction is greater in those areas. After a certain amount of wear has taken place, the pressure distribution will change so as to permit the wear to be uniform. This is the basis of the first method of solution. Another method of construction employs springs to obtain a uniform pressure over the area. It is this assumption of uniform pressure that is used in the second method of solution. Uniform Wear After initial wear has taken place and the disks have worn down to a point where uniform wear is established, the axial wear can be expressed by Equation (12–38) as

w = KPVt

A

B

C

Figure 16–14

Figure 16–15

Cross-sectional view of a single-plate clutch; A, driver; B, driven plate (keyed to driven shaft); C, actuator.

An oil-actuated multiple-disk clutch-brake for operation in an oil bath or spray. It is especially useful for rapid cycling. (Courtesy of Twin Disc Clutch Company.)

850      Mechanical Engineering Design

Figure 16–16

dr

Disk friction member.

r

F

d

D

where the pressure P and velocity V can vary in the wear area. For uniform wear, w is constant, therefore PV is constant. Setting p = P, and V = rω, where ω is the angular velocity of the rotating member, we find that in the wear area, pr is constant. The maximum pressure pa occurs where r is minimum, r = d∕2, and thus d pr = pa (a) 2 We can take an expression from Equation (a), which is the condition for having the same amount of work done at radius r as is done at radius d∕2. Referring to Figure 16–16, we have an element of area of radius r and thickness dr. The area of this element is 2πr dr, so that the normal force acting upon this element is dF = 2πpr dr. We can find the total normal force by letting r vary from d∕2 to D∕2 and integrating. Thus, with Equation (a),

F=

D∕2

2πpr dr = πpa d

d∕2

D∕2

dr =

d∕2

πpa d (D − d) 2

(16–23)

The torque is found by integrating the product of the frictional force and the radius:

T=

D∕2

2πf pr 2 dr = πf pa d

d∕2

D∕2

d∕2

r dr =

πf pa d 2 (D − d 2 ) 8

(16–24)

By substituting the value of F from Equation (16–23) we may obtain a more convenient expression for the torque. Thus Ff (D + d) (16–25) 4 In use, Equation (16–23) gives the actuating force for the selected maximum pressure pa. This equation holds for any number of friction pairs or surfaces. Equation (16–25), however, gives the torque capacity for only a single friction surface.

T=

Uniform Pressure When uniform pressure pa can be assumed over the area of the disk, the actuating force F is simply the product of the pressure and the area. This gives πpa (D 2 − d 2 ) (16–26) 4 As before, the torque is found by integrating the product of the frictional force and the radius:

F=

T = 2πf pa

D∕2

d∕2

r 2 dr =

πf pa 3 (D − d 3 ) 12

(16–27)

Clutches, Brakes, Couplings, and Flywheels     851

Figure 16–17

T f FD

Dimensionless plot of Equations (b) and (c).

1

Uniform pressure (new clutch)

0.5

Uniform wear (old clutch) 0

0

0.5

d D 1

From Equation (16–26) we can rewrite Equation (16–27) as

T=

F f D3 − d3 3 D2 − d 2

(16–28)

It should be noted for both equations that the torque is for a single pair of mating surfaces. This value must therefore be multiplied by the number of pairs of surfaces in contact. Let us express Equation (16–25) for torque during uniform wear (old clutch) as 1 + d∕D T = f FD 4

(b)

and Equation (16–28) for torque during uniform pressure (new clutch) as 3 T 1 1 − (d∕D) = f FD 3 1 − (d∕D) 2

(c)

and plot these in Figure 16–17. What we see is a dimensionless presentation of Equations (b) and (c) that reduces the number of variables from five (T, f, F, D, and d) to three (T∕FD, f, and d∕D), which are dimensionless. This is the method of Buckingham. The dimensionless groups (called pi terms) are

π1 =

T FD

π2 = f

π3 =

d D

This allows a five-dimensional space to be reduced to a three-dimensional space. Further, because of the "multiplicative" relation between f and T in Equations (b) and (c), it is possible to plot π1∕π2 versus π3 in a two-dimensional space (the plane of a sheet of paper) to view all cases over the domain of existence of Equations (b) and (c) and to compare, without risk of oversight! By examining Figure 16–17 we can conclude that a new clutch, Equation (c), always transmits more torque than an old clutch, Equation (b). Furthermore, since clutches of this type are typically proportioned to make the diameter ratio d∕D fall in the range 0.6 ≤ d∕D ≤ 1, the largest discrepancy between Equation (b) and Equation (c) will be

T 1 + 0.6 = = 0.400 f FD 4

T 1 1 − 0.63 = = 0.4083   uniform pressure (new clutch) f FD 3 1 − 0.62

   uniform wear (old clutch)

852      Mechanical Engineering Design

so the proportional error is (0.4083 − 0.400)∕0.400 = 0.021, or about 2 percent. Given the uncertainties in the actual coefficient of friction and the certainty that new clutches get old, there is little reason to use anything but Equations (16–23), (16–24), and (16–25).

16–6  Disk Brakes As indicated in Figure 16–16, there is no fundamental difference between a disk clutch and a disk brake. The analysis of the preceding section applies to disk brakes too. We have seen that rim or drum brakes can be designed for self-energization. While this feature is important in reducing the braking effort required, it also has a disadvantage. When drum brakes are used as vehicle brakes, only a slight change in the coefficient of friction will cause a large change in the pedal force required for braking. A not unusual 30 percent reduction in the coefficient of friction due to a temperature change or moisture, for example, can result in a 50 percent change in the pedal force required to obtain the same braking torque obtainable prior to the change. The disk brake has no self-energization, and hence is not so susceptible to changes in the coefficient of friction. Another type of disk brake is the floating caliper brake, shown in Figure 16–18. The caliper supports a single floating piston actuated by hydraulic pressure. The Figure 16–18 An automotive disk brake. (Courtesy DaimlerChrysler Corporation.)

Caliper

Wheel

Boot

Seal Piston

Brake fluid

Shoe and lining

Wheel stud

Inner bearing

Seal

Spindle

Adapter

Mounting bolt Outer bearing

Steering knuckle

Braking disk

Splash shield

Clutches, Brakes, Couplings, and Flywheels     853 y

F

Figure 16–19 Geometry of contact area of an annular-pad segment of a caliper brake.

F r

θ2

ri

ro

x

θ1

action is much like that of a screw clamp, with the piston replacing the function of the screw. The floating action also compensates for wear and ensures a fairly constant pressure over the area of the friction pads. The seal and boot of Figure 16–18 are designed to obtain clearance by backing off from the piston when the piston is released. Caliper brakes (named for the nature of the actuating linkage) and disk brakes (named for the shape of the unlined surface) press friction material against the face(s) of a rotating disk. Depicted in Figure 16–19 is the geometry of an annular-pad brake contact area. The governing axial wear equation is Equation (12–38)

w = KPVt

The coordinate r locates the line of action of force F that intersects the y axis. Of interest also is the effective radius re, which is the radius of an equivalent shoe of infinitesimal radial thickness. If p is the local contact pressure, the actuating force F and the friction torque T are given by

F= T=

θ2

∫ ∫

ro

θ1

ri

θ2

ro

θ1

ri

pr dr dθ = (θ2 − θ1 )

∫ ∫

ro

(16–29)

pr 2 dr

(16–30)

pr dr

ri

f pr 2 dr dθ = (θ2 − θ1 ) f

ro

ri

The equivalent radius re can be found from f Fre = T, or

re =

T = fF

ro

pr 2 dr

ri ro

(16–31)

pr dr

ri

The locating coordinate r of the activating force is found by taking moments about the x axis:

Mx = F r =

θ2

ro

θ1

ri

∫ ∫

pr (r sin θ)dr dθ = (cos θ1 − cos θ2 )

Mx ( cos θ1 − cos θ2 ) r= = re F θ2 − θ1

ro

pr 2 dr (16–32)

ri

854      Mechanical Engineering Design

Uniform Wear It is clear from Equation (12–38) that for the axial wear to be the same everywhere, the product PV must be a constant. From Equation (a), Section 16–5, the pressure p can be expressed in terms of the largest allowable pressure pa (which occurs at the inner radius ri) as p = pari∕r. Equation (16–29) becomes ro

F = (θ2 − θ1 )pari

∫ dr = (θ

− θ1 )pari (ro − ri )

(16–33)

1 (θ2 − θ1 ) f pa ri (ro2 − ri2 ) 2

(16–34)

2

ri

Equation (16–30) becomes

ro

T = (θ2 − θ1 ) f pa ri

r dr =

ri

Equation (16–31) becomes

pa ri

ro

r dr

ri

re = pa ri

=

ro

dr

ro2 − ri2 1 ro + ri = ro − ri 2 2

(16–35)

ri

Equation (16–32) becomes

r=

cos θ1 − cos θ2 ro + ri 2 θ2 − θ1

(16–36)

Uniform Pressure In this situation, approximated by a new brake, p = pa. Equation (16–29) becomes

F = (θ2 − θ1 )pa

ro

r dr =

1 (θ2 − θ1 )pa (ro2 − ri2 ) 2

(16–37)

r 2 dr =

1 (θ2 − θ1 ) f pa (ro3 − ri3 ) 3

(16–38)

ri

Equation (16–30) becomes

T = (θ2 − θ1 ) f pa

ro

ri

Equation (16–31) becomes pa

ro

r 2 dr

ri ro

re = pa

= r dr

ro3 − ri3 2 2 ro3 − ri3 = 3 ro2 − ri2 3 ro2 − ri2

(16–39)

ri

Equation (16–32) becomes

r=

cos θ1 − cos θ2 2 ro3 − ri3 2 ro3 − ri3 cos θ1 − cos θ2 = θ2 − θ1 3 ro2 − ri2 3 ro2 − ri2 θ2 − θ1

(16–40)

Clutches, Brakes, Couplings, and Flywheels     855

EXAMPLE 16–3 Two annular pads, ri = 3.875 in, ro = 5.50 in, subtend an angle of 108°, have a coefficient of friction of 0.37, and are actuated by a pair of hydraulic cylinders 1.5 in in diameter. The torque requirement is 13 000 lbf · in. For uniform wear (a) Find the largest normal pressure pa. (b) Estimate the actuating force F. (c) Find the equivalent radius re and force location r. (d) Estimate the required hydraulic pressure. Solution (a) From Eq. (16–34), with T = 13 000∕2 = 6500 lbf · in for each pad, pa =

Answer

=

2T (θ2 − θ1 ) f ri (ro2 − ri2 ) 2(6500)

(144° − 36°)(π∕180)0.37(3.875) (5.52 − 3.8752 )

= 315.8 psi

(b) From Eq. (16–33), Answer

F = (θ2 − θ1 )pa ri (ro − ri ) = (144° − 36°) (π∕180)315.8(3.875)(5.5 − 3.875) = 3748 lbf

(c) From Eq. (16–35), re =

Answer From Eq. (16–36), Answer

r=

ro + ri 5.50 + 3.875 = = 4.688 in 2 2

cos θ1 − cos θ2 ro + ri cos 36° − cos 144° 5.50 + 3.875 = θ2 − θ1 2 (144° − 36°)(π∕180) 2

= 4.024 in

(d) Each cylinder supplies the actuating force, 3748 lbf. Answer

phydraulic =

F 3748 = = 2121 psi AP π(1.52∕4) R

Circular (Button or Puck) Pad Caliper Brake Figure 16–20 displays the pad geometry. Numerical integration is necessary to analyze this brake since the boundaries are difficult to handle in closed form. Table 16–1 gives the parameters for this brake as determined by Fazekas. The effective radius is given by

re = δe

(16–41)

F = π R 2 pav

(16–42)

T = f Fre

(16–43)

e

The actuating force is given by and the torque is given by

Figure 16–20 Geometry of circular pad of a caliper brake.

856      Mechanical Engineering Design

Table 16–1  Parameters for a Circular-Pad Caliper Brake re e

pmax pav

R e

δ=

0.0

1.000

1.000

0.1

0.983

1.093

0.2

0.969

1.212

0.3

0.957

1.367

0.4

0.947

1.578

0.5

0.938

1.875

Source: G. A. Fazekas, "On Circular Spot Brakes," Trans. ASME, J. Engineering for Industry, vol. 94, Series B, No. 3, August 1972, pp. 859–863.

EXAMPLE 16–4 A button-pad disk brake uses dry sintered metal pads. The pad radius is 12 in, and its center is 2 in from the axis of rotation of the 312 -in-diameter disk. Using half of the largest allowable pressure, pmax = 350 psi, find the actuating force and the brake torque. The coefficient of friction is 0.31. Solution Since the pad radius R = 0.5 in and eccentricity e = 2 in, R 0.5 = = 0.25 e 2 From Table 16–1, by interpolation, δ = 0.963 and pmax∕pav = 1.290. It follows that the effective radius e is found from Equation (16–41):

re = δe = 0.963(2) = 1.926 in

and the average pressure is pmax∕2 350∕2 = = 135.7 psi 1.290 1.290 The actuating force F is found from Equation (16–42) to be

Answer

pav =

F = πR 2 pav = π(0.5) 2135.7 = 106.6 lbf   (one side)

The brake torque T is Answer

T = f Fre = 0.31(106.6)1.926 = 63.65 lbf · in  (one side)

16–7  Cone Clutches and Brakes The drawing of a cone clutch in Figure 16–21 shows that it consists of a cup keyed or splined to one of the shafts, a cone that must slide axially on splines or keys on the mating shaft, and a helical spring to hold the clutch in engagement. The clutch is disengaged by means of a fork that fits into the shifting groove on the friction cone. The important geometric design parameters are the cone angle α and the diameter and face width of the cone. If the cone angle is too small, say, less than about 8°, then the force required to disengage the clutch may be quite large. The wedging effect lessens rapidly when larger

Clutches, Brakes, Couplings, and Flywheels     857 α p dA

α Cone angle

dr sin α

dr

Cone α

r

D F

d Spring Shifting groove

Cup

(a)

(b)

Figure 16–21

Figure 16–22

Cross section of a cone clutch.

Contact area of a cone clutch.

cone angles are used. Depending upon the characteristics of the friction materials, a good compromise can usually be found using cone angles between 10° and 15°. To find a relation between the operating force F and the torque transmitted, designate the dimensions of the friction cone as shown in Figure 16–22. As in the case of the axial clutch, we can obtain one set of relations for a uniform-wear and another set for a uniform-pressure assumption. Uniform Wear The pressure relation is the same as for the axial clutch:

p = pa

d 2r

(a)

Next, referring to Figure 16–22, we see that we have an element of area dA of radius r and width dr∕sin α. Thus dA = (2πrdr)∕sin α. As shown in Figure 16–22, the operating force will be the integral of the axial component of the differential force pdA. Thus

F=

pdA sin α =

D∕2

d∕2

= πpa d

D∕2

d∕2

d 2πr dr (pa 2r ) ( sin α ) (sin α)

(16–44)

πpa d dr = (D − d) 2

which is the same result as in Equation (16–23). The differential friction force is f pdA, and the torque is the integral of the product of this force with the radius. Thus,

T=

r f pdA =

π f pa d = sin α

D∕2

d∕2

D∕2

d∕2

(r f ) (pa

d 2πr dr ) ( 2r sin α )

π f pa d r dr = (D 2 − d 2 ) 8 sin α

(16–45)

858      Mechanical Engineering Design

Note that Equation (16–24) is a special case of Equation (16–45), with α = 90°. Using Equation (16–44), we find that the torque can also be written

T=

Ff (D + d) 4 sin α

(16–46)

Uniform Pressure Using p = pa, the actuating force is found to be

F=

∫ p dA sin α = ∫

D∕2

a

d∕2

(pa ) (

πpa 2πr dr (sin α) = (D 2 − d 2 ) ) sin α 4

(16–47)

The torque is

T=

r fpa dA =

D∕2

d∕2

(r fpa ) (

π fpa 2πr dr = (D3 − d 3 ) ) sin α 12 sin α

(16–48)

Using Equation (16–47) in Equation (16–48) gives

T=

F f D3 − d 3 3 sin α D2 − d 2

(16–49)

As in the case of the axial clutch, we can write Equation (16–46) dimensionlessly as

T sin α 1 + d∕D = f FD 4

(b)

3 T sin α 1 1 − (d∕D) = f FD 3 1 − (d∕D) 2

(c)

and write Equation (16–49) as

This time there are six (T, α, f, F, D, and d) parameters and four pi terms:

π1 =

T FD

π2 = f

π3 = sin α

π4 =

d D

As in Figure 16–17, we plot T sin α∕( f FD) as ordinate and d∕D as abscissa. The plots and conclusions are the same. There is little reason for using equations other than Equations (16–44), (16–45), and (16–46).

16–8  Energy Considerations When the rotating members of a machine are caused to stop by means of a brake, the kinetic energy of rotation must be absorbed by the brake. This energy appears in the brake in the form of heat. In the same way, when the members of a machine that are initially at rest are brought up to speed, slipping must occur in the clutch until the driven members have the same speed as the driver. Kinetic energy is absorbed during slippage of either a clutch or a brake, and this energy appears as heat. We have seen how the torque capacity of a clutch or brake depends upon the coefficient of friction of the material and upon a safe normal pressure. However, the character of the load may be such that, if this torque value is permitted, the clutch or brake may be destroyed by its own generated heat. The capacity of a clutch is therefore limited by two factors, the characteristics of the material and

Clutches, Brakes, Couplings, and Flywheels     859

the ability of the clutch to dissipate heat. In this section we shall consider the amount of heat generated by a clutching or braking operation. If the heat is generated faster than it is dissipated, we have a temperature-rise problem; that is the subject of the next section. To get a clear picture of what happens during a simple clutching or braking operation, refer to Figure 16–1a, which is a mathematical model of a two-inertia system connected by a clutch. As shown, inertias I1 and I2 have initial angular velocities of ω1 and ω2, respectively. During the clutch operation both angular velocities change and eventually become equal. We assume that the two shafts are rigid and that the clutch torque is constant. Writing the equation of motion for inertia 1 gives I1θ1̈ = −T

(a)

where θ1̈ is the angular acceleration of I1 and T is the clutch torque. A similar equation for I2 is I2θ ̈2 = T

(b)

We can determine the instantaneous angular velocities θ1̇ and θ 2̇ of I1 and I2 after any period of time t has elapsed by integrating Equations (a) and (b). The results are

θ1̇ = −

θ ̇2 =

T t + ω1 I1

T t + ω2 I2

(c) (d)

where θ1̇ = ω1 and θ2̇ = ω2 at t = 0. The difference in the velocities, sometimes called the relative velocity, is

θ ̇ = θ 1̇ − θ2̇ = −

I1 + I2 T T t + ω1 − ( t + ω2) = ω1 − ω2 − T( t I1 I2 I1I2 )

(e)

It is convenient to manipulate the inertial terms in Equation (e) into a form that mathematically works well for clutches (where both inertias are needed) and for brakes (where the inertia of the grounded body, I2, can be set to infinity). This gives

θ ̇ = ω1 − ω2 − (

I1∕I2 + 1 )Tt I1

(16–50)

The clutching operation is completed at the instant in which the two angular velocities θ1̇ and θ 2̇ become equal. Let the time required for the entire operation be t1. Then θ ̇ = 0 when θ1̇ = θ 2̇ , and so Equation (16–50) gives the time as

I1 ω1 − ω2 t1 = ( I1∕I2 + 1 ) T

(16–51)

This equation shows that the time required for the engagement operation is directly proportional to the velocity difference and inversely proportional to the torque. We have assumed the clutch torque to be constant. Therefore, using Equation (16–50), we find the rate of energy-dissipation during the clutching operation to be

u = Tθ ̇ = T[ ω1 − ω2 − (

I1∕I2 + 1 )Tt ] I1

(f )

860      Mechanical Engineering Design

This equation shows that the energy-dissipation rate is greatest at the start, when t = 0. The total energy dissipated during the clutching operation or braking cycle is obtained by integrating Equation (e) from t = 0 to t = t1. The result is found to be

E=

∫ u dt = T ∫ [ω t1

t1

0

0

1

− ω2 − (

I1 (ω1 − ω2 ) 2 =( I1∕I2 + 1 ) 2

I1∕I2 + 1 )Tt ]dt I1

(16–52)

where Equation (16–51) was employed. Note that the energy dissipated is proportional to the velocity difference squared and is independent of the clutch torque. Note that E in Equation (16–52) is the energy lost or dissipated and is the energy that is absorbed by the clutch or brake. If the inertias are expressed in U.S. customary units (lbf · in · s2), then the energy absorbed by the clutch assembly is in in · lbf. Using these units, the heat generated in Btu is

E 9336

H=

(16–53)

In SI, the inertias are expressed in kilogram-meter2 units, and the energy dissipated is expressed in joules.

16–9  Temperature Rise The temperature rise of the clutch or brake assembly can be approximated by the classic expression

ΔT =

H Cp W

(16–54)

where ΔT = temperature rise, °F Cp = specific heat capacity, Btu/(lbm · °F); use 0.12 for steel or cast iron W = mass of clutch or brake parts, lbm A similar equation can be written for SI units. It is

ΔT =

E Cp m

(16–55)

where ΔT = temperature rise, °C Cp = specific heat capacity; use 500 J/kg · °C for steel or cast iron m = mass of clutch or brake parts, kg The temperature-rise equations above can be used to explain what happens when a clutch or brake is operated. However, there are so many variables involved that it would be most unlikely that such an analysis would even approximate experimental results. For this reason such analyses are most useful, for repetitive cycling, in pinpointing those design parameters that have the greatest effect on performance. If an object is at initial temperature T1 in an environment of temperature T∞, then Newton's cooling model is expressed as

h CR A T − T∞ = exp (− t T1 − T ∞ WCp )

(16–56)

Clutches, Brakes, Couplings, and Flywheels     861

Figure 16–23 The effect of clutching or braking operations on temperature. T∞ is the ambient temperature. Note that the temperature rise ΔT may be different for each operation.

Instantaneous temperature Ti

T2

A

T1

ΔT

ΔT B C T∞

tA

tB

tC

Time t

where T = temperature at time t, °F T1 = initial temperature, °F T∞ = environmental temperature, °F h CR = overall coefficient of heat transfer, Btu/(in2 · s · °F) A = lateral surface area, in2 W = mass of the object, lbm Cp = specific heat capacity of the object, Btu/(lbm · °F) Figure 16–23 shows an application of Equation (16–56). The curve ABC is the exponential decline of temperature given by Equation (16–56). At time tB a second application of the brake occurs. The temperature quickly rises to temperature T2, and a new cooling curve is started. For repetitive brake applications, subsequent temperature peaks T3, T4, . . . , may be higher than the previous peaks if insufficient cooling has occurred between applications. If this is a production situation with brake applications every t1 seconds, then a steady state develops in which all the peaks Tmax and all the valleys Tmin are repetitive. The heat-dissipation capacity of disk brakes has to be planned to avoid reaching disk and pad temperatures that are detrimental to the parts. When a disk brake has a rhythm such as discussed above, then the rate of heat transfer is described by another Newtonian equation: where Hloss h CR hr hc fv T T∞

Hloss = h CR A(T − T∞ ) = (hr + fv hc )A(T − T∞ ) = = = = = = =

(16–57)

rate of energy loss, Btu/s overall coefficient of heat transfer, Btu/(in2 · s · °F) radiation component of h CR, Btu/(in2 · s · °F), Figure 16–24a convective component of h CR, Btu/(in2 · s · °F), Figure 16–24a ventilation factor, Figure 16–24b disk temperature, °F ambient temperature, °F

hc

2

862      Mechanical Engineering Design

0

0

100

200

12

400

500

600

700

(a)

hr

10

8 8

Multiplying factor fv

Heat-transfer coefficient (hc or hr) (10−6 Btu /s · in2 · °F)

300

Temperature rise T – T∞ (°F)

6

4 hc

2

0

0

100

200

300

400

500

600

6

4

2

0

700

0

20

40

60

Temperature rise T – T∞ (°F)

Forced ventilation velocity (ft/s)

(a)

(b)

80

Figure 16–24

Multiplying factor fv

(a) Heat-transfer coefficient in still air. (b) Ventilation factors. (Courtesy of Tolo-o-matic.) 8

The energy E absorbed by the brake stopping a rotary inertia I in terms of original and final angular velocities ωo and ωf is the change in kinetic energy, I(ω2o − ω2f )∕2. Expressing this in Btu,

6

1 I (ω2o − ω2f ) 2 9336 The temperature rise ΔT due to a single stop is

4

2

0

(16–58)

E (16–59) WC 20 80 enough to transfer E Btu in t seconds. For steady state, rearrange T40max has 60 to be high 1 Forced ventilation velocity (ft/s) Equation (16–56) as (b) Tmin − T∞ = exp (−βt1 ) Tmax − T∞ where β = h CR A∕(WCp ). Cross-multiply, multiply the equation by −1, add Tmax to both sides, set Tmax − Tmin = ΔT, and rearrange, obtaining

0

E=

ΔT =

Tmax = T∞ +

ΔT 1 − exp (−βt1 )

(16–60)

EXAMPLE 16–5 A caliper brake is used 24 times per hour to arrest a machine shaft from a speed of 250 rev/min to rest. The ventilation of the brake provides a mean air speed of 25 ft/s. The equivalent rotary inertia of the machine as seen from the brake shaft is 289 lbm · in · s. The disk is steel with a density γ = 0.282 lbm/in3, a specific heat capacity of 0.108 Btu/(lbm · °F), a diameter of 6 in, a thickness of 14 in. The pads are dry sintered metal. The lateral area of the brake surface is 50 in2. Find Tmax and Tmin for the steady-state operation if T∞ = 70°F.

Clutches, Brakes, Couplings, and Flywheels     863

t1 = 602∕24 = 150 s

Solution

Assuming a temperature rise of Tmax − T∞ = 200°F, from Fig. 16–24a,

hr = 3.0(10−6 ) Btu/(in2 · s · °F)

hc = 2.0(10−6 ) Btu/(in2 · s · °F)

Fig. 16–24b

fv = 4.8 −6

h CR = hr + fv hc = 3.0(10 ) + 4.8(2.0)10−6 = 12.6(10−6 ) Btu/(in2 · s · °F)

The mass of the disk is

W=

πγD 2h π(0.282)62 (0.25) = = 1.99 lbm 4 4

Eq. (16–58):

E=

2 1 I 289 2π (ω2o − ω2f ) = 250 ) = 10.6 Btu 2 9336 2(9336) ( 60

β=

h CR A 12.6(10−6 )50 = = 2.93(10−3 ) s−1 WCp 1.99(0.108)

Eq. (16–59): Answer  Eq. (16–60):

ΔT = Tmax = 70 +

Answer

E 10.6 = = 49.3°F WCp 1.99(0.108) 49.3 = 209°F 1 − exp[−2.93(10−3 )150]

Tmin = 209 − 49.3 = 160°F

The predicted temperature rise here is Tmax − T∞ = 209 − 70 = 139°F. Iterating with revised values of hr and hc from Fig. 16–24a, we can make the solution converge to Tmax = 220°F and Tmin = 171°F. Table 16–3 for dry sintered metal pads gives a continuous operating maximum temperature of 570–660°F. There is no danger of overheating.

16–10  Friction Materials A brake or friction clutch should have the following lining material characteristics to a degree that is dependent on the severity of service: ∙ High and reproducible coefficient of friction ∙ Imperviousness to environmental conditions, such as moisture ∙ The ability to withstand high temperatures, together with good thermal conductivity and diffusivity, as well as high specific heat capacity ∙ Good resiliency ∙ High resistance to wear, scoring, and galling ∙ Compatible with the environment ∙ Flexibility

864      Mechanical Engineering Design

Table 16–2  Area of Friction Material Required for a Given Average Braking Power Ratio of Area to Average Braking Power, in2/(Btu/s) Band and Drum Brakes

Plate Disk Brakes

Caliper Disk Brakes

Emergency brakes

0.85

2.8

0.28

Intermittent

Elevators, cranes, and winches

2.8

7.1

0.70

Heavy-duty

Excavators, presses

5.6–6.9

13.6

1.41

Duty Cycle

Typical Applications

Infrequent

Sources: M. J. Neale, The Tribology Handbook, Butterworth, London, 1973; Friction Materials for Engineers, Ferodo Ltd., Chapel-en-le-frith, England, 1968.

Table 16–2 gives area of friction surface required for several braking powers. Table 16–3 gives important characteristics of some friction materials for brakes and clutches. The manufacture of friction materials is a highly specialized process, and it is advisable to consult manufacturers' catalogs and handbooks, as well as manufacturers directly, in selecting friction materials for specific applications. Selection involves a consideration of the many characteristics as well as the standard sizes available. The woven-cotton lining is produced as a fabric belt that is impregnated with resins and polymerized. It is used mostly in heavy machinery and is usually supplied in rolls up to 50 ft in length. Thicknesses available range from 18 to 1 in, in widths up to about 12 in. A woven-asbestos lining is made in a similar manner to the cotton lining and may also contain metal particles. It is not quite as flexible as the cotton lining and comes in a smaller range of sizes. Along with the cotton lining, the asbestos lining was widely used as a brake material in heavy machinery. Molded-asbestos linings contain asbestos fiber and friction modifiers; a thermoset polymer is used, with heat, to form a rigid or semirigid molding. The principal use was in drum brakes. Molded-asbestos pads are similar to molded linings but have no flexibility; they were used for both clutches and brakes. Sintered-metal pads are made of a mixture of copper and/or iron particles with friction modifiers, molded under high pressure and then heated to a high temperature to fuse the material. These pads are used in both brakes and clutches for heavy-duty applications. Cermet pads are similar to the sintered-metal pads and have a substantial ceramic content. Table 16–4 lists properties of typical brake linings. The linings may consist of a mixture of fibers to provide strength and ability to withstand high temperatures, various friction particles to obtain a degree of wear resistance as well as a higher coefficient of friction, and bonding materials. Table 16–5 includes a wider variety of clutch friction materials, together with some of their properties. Some of these materials may be run wet by allowing them to dip in oil or to be sprayed by oil. This reduces the coefficient of friction somewhat but carries away more heat and permits higher pressures to be used.

865

0.38

Flexible molded asbestos

Wound asbestos yarn and wire

100

100

100 660

660–750

660

100–150

350

300

300–350

300

500–750

440–660

3600

3600

3600

4800–7500

4800

3600

Vehicle clutches

Clutches and brakes

Clutches and brakes

Clutches and brakes

Disk brakes

Industrial clutches

Drum brakes and clutches

PV < 500 000 Clutches and transmission psi · ft/min bands

Sources: Ferodo Ltd., Chapel-en-le-frith, England; Scan-pac, Mequon, Wisc.; Raybestos, New York, N.Y. and Stratford, Conn.; Gatke Corp., Chicago, Ill.; General Metals Powder Co., Akron, Ohio; D. A. B. Industries, Troy, Mich.; Friction Products Co., Medina, Ohio.

Resilient paper (wet) 0.09–0.15 400 300

Industrial clutches and brakes

0.39–0.45

Semirigid molded asbestos

660 930–1380

3600

Woven cotton 0.47 100 230 170 3600

0.37–0.41

Rigid molded nonasbestos

750

300

350

3600

Industrial clutches and brakes

0.33–0.63

Rigid molded asbestos pads

660–750

570

Woven asbestos yarn and 0.38 100 500 260 3600 wire

0.06 0.31–0.49

Rigid molded asbestos (wet)

100

930

0.35–0.41

500

Rigid molded asbestos (dry)

Brakes and clutches

Clutches

750

0.06–0.08

1500

Sintered metal (wet)

150

Applications Clutches and caliper disk brakes

0.32

Friction Maximum Maximum Temperature Maximum Coefficient Pressure Instantaneous, Continuous, Velocity Vmax , f pmax , psi °F °F ft/min

Sintered metal (dry) 0.29–0.33 300–400 930–1020 570–660 3600

Cermet

Material

Table 16–3  Characteristics of Friction Materials for Brakes and Clutches

866      Mechanical Engineering Design

Table 16–4  Some Properties of Brake Linings Woven Lining

Molded Lining

Rigid Block

Compressive strength, kpsi

10–15

10–18

10–15

Compressive strength, MPa

70–100

70–125

70–100

Tensile strength, kpsi

2.5–3

4–5

3–4

Tensile strength, MPa

17–21

27–35

21–27

Max. temperature, °F

400–500

500

750

Max. temperature, °C

200–260

260

400

7500

5000

7500

Max. speed, ft/min Max. speed, m/s

38

25

38

Max. pressure, psi

 50–100

100

150

Max. pressure, kPa

340–690

690

1000

0.45

0.47

0.40–45

Frictional coefficient, mean

Table 16–5  Friction Materials for Clutches Material Cast iron on cast iron

Friction Coefficient

Max. Temperature

Wet

Dry

°F

°C

psi

Max. Pressure kPa

0.05

0.15–0.20

600

320

150–250

1000–1750

Powdered metal* on cast iron

0.05–0.1

0.1–0.4

1000

540

150

1000

Powdered metal* on hard steel

0.05–0.1

0.1–0.3

1000

540

300

2100

Wood on steel or cast iron

0.16

0.2–0.35

300

150

60–90

400–620

Leather on steel or cast iron

0.12

0.3–0.5

200

100

10–40

70–280

Cork on steel or cast iron

0.15–0.25

0.3–0.5

200

100

8–14

50–100

Felt on steel or cast iron

0.18

0.22

280

140

5–10

35–70

Woven asbestos* on steel or cast iron

0.1–0.2

0.3–0.6

350–500

175–260

50–100

350–700

Molded asbestos* on steel or cast iron

0.08–0.12

0.2–0.5

500

260

50–150

350–1000

Impregnated asbestos* on steel or cast iron

0.12

0.32

500–750

260–400

150

1000

Carbon graphite on steel

0.05–0.1

0.25

700–1000

370–540

300

2100

*The friction coefficient can be maintained with ±5 percent for specific materials in this group.

16–11  Miscellaneous Clutches and Couplings The square-jaw clutch shown in Figure 16–25a is one form of positive-contact clutch. These clutches have the following characteristics: ∙ ∙ ∙ ∙ ∙

They do not slip. No heat is generated. They cannot be engaged at high speeds. Sometimes they cannot be engaged when both shafts are at rest. Engagement at any speed is accompanied by shock.

Clutches, Brakes, Couplings, and Flywheels     867

Figure 16–25 (a) Square-jaw clutch; (b) overload release clutch using a detent.

Shift lever

(a)

(b)

The greatest differences among the various types of positive clutches are concerned with the design of the jaws. To provide a longer period of time for shift action during engagement, the jaws may be ratchet-shaped, spiral-shaped, or gear-tooth-shaped. Sometimes a great many teeth or jaws are used, and they may be cut either circumferentially, so that they engage by cylindrical mating, or on the faces of the mating elements. Although positive clutches are not used to the extent of the frictional-contact types, they do have important applications where synchronous operation is required, as, for example, in power presses or rolling-mill screw-downs. Devices such as linear drives or motor-operated screwdrivers must run to a definite limit and then come to a stop. An overload-release type of clutch is required for these applications. Figure 16–25b is a schematic drawing illustrating the principle of operation of such a clutch. These clutches are usually spring-loaded so as to release at a predetermined torque. The clicking sound that is heard when the overload point is reached is considered to be a desirable signal. Both fatigue and shock loads must be considered in obtaining the stresses and deflections of the various portions of positive clutches. In addition, wear must generally be considered. The application of the fundamentals discussed in Parts 1 and 2 of this book is usually sufficient for the complete design of these devices. An overrunning clutch or coupling permits the driven member of a machine to "freewheel" or "overrun" because the driver is stopped or because another source of power increases the speed of the driven mechanism. The construction uses rollers or balls mounted between an outer sleeve and an inner member having cam flats machined around the periphery. Driving action is obtained by wedging the rollers between the sleeve and the cam flats. This clutch is therefore equivalent to a pawl and ratchet with an infinite number of teeth. There are many varieties of overrunning clutches available, and they are built in capacities up to hundreds of horsepower. Since no slippage is involved, the only power loss is that due to bearing friction and windage. The shaft couplings shown in Figure 16–26 are representative of the selection available in catalogs.

(a)

(b)

(c)

Figure 16–26 Shaft couplings. (a) Plain. (b) Light-duty toothed coupling. (c) Three-jaw coupling available with bronze, rubber, or polyurethane insert to minimize vibration. (Reproduced by permission, Boston Gear Division, Colfax Corp.)

868      Mechanical Engineering Design

16–12  Flywheels The equation of motion for the flywheel represented in Figure 16–1b is ∑ M = Ti (θi, θ ̇i ) − To (θo, θ ̇o ) − Iθ ̈ = 0 or Iθ ̈ = Ti (θi, ωi ) − To (θo, ωo )

(a)

where Ti and To are the input and output (load) torques, respectively; and where θ ̇ and θ ̈ are the first and second time derivatives of θ, respectively. Note that both Ti and To may depend for their values on the angular displacements θi and θo as well as their angular velocities ωi and ωo. In many cases the torque characteristic depends upon only one of these. Thus, the torque delivered by an induction motor depends upon the speed of the motor. In fact, motor manufacturers publish charts detailing the torque-speed characteristics of their various motors. When the input and output torque functions are given, Equation (a) can be solved for the motion of the flywheel using well-known techniques for solving linear and nonlinear differential equations. We can dispense with this here by assuming a rigid shaft, giving θi = θo = θ and ωi = ωo = ω. Thus, Equation (a) becomes Iθ ̈ = Ti (θ, ω) − To (θ, ω) (b) When the two torque functions are known and the starting values of the displacement θ and velocity ω are given, Equation (b) can be solved for θ, ω, and θ ̈ as functions of time. However, we are not really interested in the instantaneous values of these terms at all. Primarily we want to know the overall performance of the flywheel. What should its moment of inertia be? How do we match the power source to the load? And what are the resulting performance characteristics of the system that we have selected? To gain insight into the problem, a hypothetical situation is diagrammed in Figure 16–27. An input power source subjects a flywheel to a constant torque Ti while the shaft rotates from θ1 to θ2. This is a positive torque and is plotted upward. Equation (b) indicates that a positive acceleration θ ̈ will be the result, and so the shaft velocity increases from ω1 to ω2. As shown, the shaft now rotates from θ2 to θ3 with zero torque and hence, from Equation (b), with zero acceleration. Therefore ω3 = ω2. From θ3 to θ4 a load, or output torque, of constant magnitude is applied, causing the shaft to slow down from ω3 to ω4. Note that the output torque is plotted in the negative direction in accordance with Equation (b). The work input to the flywheel is the area of the rectangle between θ1 and θ2, or Figure 16–27

Ui = Ti (θ2 − θ1 )

T, ω Ti

ω2

ω1

ω3

ω ω4

Ui θ3 θ1

θ2

θ4 Uo

To 1 cycle

θ

(c)

Clutches, Brakes, Couplings, and Flywheels     869

The work output of the flywheel is the area of the rectangle from θ3 to θ4, or Uo = To (θ4 − θ3 )

(d)

If Uo is greater than Ui, the load uses more energy than has been delivered to the flywheel and so ω4 will be less than ω1. If Uo = Ui, assuming no friction losses, ω4 will be equal to ω1 because the gains and losses are equal. And finally, ω4 will be greater than ω1 if Ui > Uo. We can also write these relations in terms of kinetic energy. At θ = θ1 the flywheel has a velocity of ω1 rad/s, and so its kinetic energy is

E1 =

1 2 Iω1 2

(e)

E2 =

1 2 Iω2 2

(f )

At θ = θ2 the velocity is ω2, and so Thus the change in kinetic energy is

E2 − E1 =

1 I(ω22 − ω21 ) 2

(16–61)

Many of the torque displacement functions encountered in practical engineering situations are so complicated that they must be integrated by numerical methods. Figure 16–28, for example, is a typical plot of the engine torque for one cycle of motion of a single-cylinder internal combustion engine. Since a part of the torque curve is negative, the flywheel must return part of the energy back to the engine. Integrating this curve from θ = 0 to 4π and dividing the result by 4π yields the mean torque Tm available to drive a load during the cycle. It is convenient to define a coefficient of speed fluctuation as

Cs =

ω 2 − ω1 ω

(16–62)

Figure 16–28

Crank torque T

Relation between torque and crank angle for a one-cylinder, four-stroke–cycle internal combustion engine.

Tm 180°

360° Crank angle θ

540°

720°

870      Mechanical Engineering Design

where ω is the nominal angular velocity, given by

ω=

ω2 + ω1 2

(16–63)

Equation (16–61) can be factored to give

E2 − E1 =

I (ω2 − ω1 )(ω2 + ω1 ) 2

Since ω2 − ω1 = Csω and ω2 + ω1 = 2ω, we have E2 − E1 = Cs Iω2

(16–64)

Equation (16–64) can be used to obtain an appropriate flywheel inertia corresponding to the energy change E2 − E1. EXAMPLE 16–6 Table 16–6 lists values of the torque used to plot Figure 16–28. The nominal speed of the engine is to be 250 rad/s. (a) Integrate the torque-displacement function for one cycle and find the energy that can be delivered to a load during the cycle. (b) Determine the mean torque Tm (see Figure 16–28). (c) The greatest energy fluctuation is approximately between θ = 15° and θ = 150° on the torque diagram; see Figure 16–28 and note that To = −Tm. Using a coefficient of speed fluctuation Cs = 0.1, find a suitable value for the flywheel inertia. (d) Find ω2 and ω1. Solution (a) Using n = 48 intervals of Δθ = 4π∕48, numerical integration of the data of Table 16–6 yields E = 3368 in · lbf

Answer

Table 16–6  Plotting Data for Figure 16–28 θ, deg

T, lbf · in

θ, deg

T, lbf · in

θ, deg

T, lbf · in

θ, deg

T, lbf · in

0

0

195

−107

375

−85

555

−107

15

2800

210

−206

390

−125

570

−206

30

2090

225

−260

405

−89

585

−292

45

2430

240

−323

420

8

600

−355

60

2160

255

−310

435

126

615

−371

75

1840

270

−242

450

242

630

−362

90

1590

285

−126

465

310

645

−312

105

1210

300

−8

480

323

660

−272

120

1066

315

89

495

280

675

−274

135

803

330

125

510

206

690

−548

150

532

345

85

525

107

705

−760

165

184

360

0

540

0

720

0

180

0

Clutches, Brakes, Couplings, and Flywheels     871

This is the energy that can be delivered to the load. Answer  (b)

Tm =

3368 = 268 lbf · in 4π

(c) The largest positive loop on the torque-displacement diagram occurs between θ = 0° and θ = 180°. We select this loop as yielding the largest speed change. Subtracting 268 lbf · in from the values in Table 16–6 for this loop gives, respectively, −268, 2532, 1822, 2162, 1892, 1572, 1322, 942, 798, 535, 264, −84, and −268 lbf · in. Numerically integrating T − Tm with respect to θ yields E2 − E1 = 3531 lbf · in. We now solve Equation (16–64) for I. This gives I=

Answer

E2 − E1 Cs ω2

=

3531 = 0.565 lbf · s2 in 0.1(250) 2

(d) Equations (16–62) and (16–63) can be solved simultaneously for ω2 and ω1. Substituting appropriate values in these two equations yields ω 250 (2 + Cs ) = (2 + 0.1) = 262.5 rad/s 2 2

Answer

ω2 =

Answer

ω1 = 2ω − ω2 = 2(250) − 262.5 = 237.5 rad/s

These two speeds occur at θ = 180° and θ = 0°, respectively.

Punch-press torque demand often takes the form of a severe impulse and the running friction of the drive train. The motor overcomes the minor task of overcoming friction while attending to the major task of restoring the flywheel's angular speed. The situation can be idealized as shown in Figure 16–29. Neglecting the running friction, Euler's equation can be written as T(θ1 − 0) =

1 I(ω21 − ω22 ) = E2 − E1 2

where the only significant inertia is that of the flywheel. Punch presses can have the motor and flywheel on one shaft, then, through a gear reduction, drive a slider-crank mechanism that carries the punching tool. The motor can be connected to the punch Torque T

Figure 16–29

Torque TM

(a) Punch-press torque demand during punching. (b) Squirrelcage electric motor torque-speed characteristic. Tr

0

0

0 θ1

0

ωr ωs

Rotation θ

Angular velocity ω

(a)

(b)

872      Mechanical Engineering Design

continuously, creating a punching rhythm, or it can be connected on command through a clutch that allows one punch and a disconnect. The motor and flywheel must be sized for the most demanding service, which is steady punching. The work done is given by

W=

θ2

[T (θ) − T ]dθ =

θ1

1 I(ω2max − ω2min ) 2

This equation can be arranged to include the coefficient of speed fluctuation Cs as follows: 1 I W = I(ω2max − ω2min ) = (ω max − ω min )(ω max + ω min ) 2 2

=

I (Csω) (2ω0 ) = ICsωω0 2

When the speed fluctuation is low, ω 0 ≈ ω, and

I=

W Cs ω 2

An induction motor has a linear torque characteristic TM = aω + b in the range of operation. The constants a and b can be found from the nameplate speed ωr and the synchronous speed ωs: T r − Ts Tr Tr = =− ωr − ωs ωr − ωs ωs − ωr

a=

Tr ωs − Ts ωr Tr ωs b= = ωs − ωr ωs − ωr

(16–65)

For example, a 3-hp three-phase squirrel-cage ac motor with a synchronous speed of 1200 rev/min is rated at 1125 rev/min and has a torque of Tr = 63 025(3)∕1125 = 168.1 lbf · in. The rated angular velocity is ωr = 2πnr∕60 = 2π(1125)∕60 = 117.81 rad/s, and the synchronous angular velocity ωs = 2π(1200)∕60 = 125.66 rad/s. Thus a = −168.1∕(125.66 − 117.81) = −21.41 lbf · in · s/rad, and b = −125.66(−21.41) = 2690.9 lbf · in, and we can express T(ω) as aω + b. During the interval from t1 to t2 the motor accelerates the flywheel according to Iθ ̈ = TM (i.e., Idω∕dt = TM). Separating the equation TM = Idω∕dt we have

t2

dt =

t1

ω2

ωr

Idω =I TM

ω2

dω I aω2 + b I T2 = ln = ln aω + b a aωr + b a Tr

ωr

or

t2 − t1 =

I T2 ln a Tr

(16–66)

For the deceleration interval when the motor and flywheel feel the punch (load) torque on the shaft as TL, (TM − TL) = Idω∕dt, or

t1

0

dt = I

ωr

ω2

dω =I T M − TL

ωr

ω2

or

t1 =

dω I aωr + b − TL = ln aω + b − TL a aω2 + b − TL

I T r − TL ln a T 2 − TL

(16–67)

Clutches, Brakes, Couplings, and Flywheels     873

We can divide Equation (16–66) by Equation (16–67) to obtain T2 TL − Tr (t2 −t1)∕t1 =( Tr T L − T2 )

(16–68)

Equation (16–68) can be solved for T2 numerically. Having T2 the flywheel inertia is, from Equation (16–66),

I=

a(t2 − t1 ) ln (T2 ∕Tr )

(16–69)

It is important that a be in units of lbf · in · s/rad so that I has proper units. The constant a should not be in lbf · in per rev/min or lbf · in per rev/s.

PROBLEMS 16–1

The figure shows an internal rim-type brake having an inside rim diameter of 300 mm and a dimension R = 125 mm. The shoes have a face width of 40 mm and are both actuated by a force of 2.2 kN. The drum rotates clockwise. The mean coefficient of friction is 0.28. (a) Find the maximum pressure and indicate the shoe on which it occurs. (b) Estimate the braking torque effected by each shoe, and find the total braking torque. (c) Estimate the resulting hinge-pin reactions. 30°

30°

F

F R

Problem 16–1

120°

120° Pin

30°

Pin

30°

16–2 For the brake in Problem 16–1, consider the pin and actuator locations to be the same. However, instead of 120°, let the friction surface of the brake shoes be 90° and centrally located. Find the maximum pressure and the total braking torque.

16–3 In the figure for Problem 16–1, the inside rim diameter is 11 in and the dimension R

is 3.5 in. The shoes have a face width of 1.25 in. Find the braking torque and the maximum pressure for each shoe if the actuating force is 225 lbf, the drum rotation is counterclockwise, and f = 0.30.

16–4 The figure shows a 400-mm-diameter brake drum with four internally expanding

shoes. Each of the hinge pins A and B supports a pair of shoes. The actuating mechanism is to be arranged to produce the same force F on each shoe. The face width of the shoes is 75 mm. The material used permits a coefficient of friction of 0.24 and a maximum pressure of 1000 kPa.

874      Mechanical Engineering Design 15°

15°

F F d Problem 16–4 The dimensions in millimeters are a = 150, c = 165, R = 200, and d = 50.

c

d

a

a

10°

10° A

10°

B R

d

d

10°

c

F F

15°

15°

(a) Determine the maximum actuating force. (b) Estimate the brake capacity. (c) Noting that rotation may be in either direction, estimate the hinge-pin reactions.

16–5 The block-type hand brake shown in the figure has a face width of 1.25 in and a mean coefficient of friction of 0.25. For an estimated actuating force of 90 lbf, find the maximum pressure on the shoe and find the braking torque. 12

8

F

Problem 16–5 Dimensions in inches.

90°

6

45°

6R

Rotation

16–6 Suppose the standard deviation of the coefficient of friction in Problem 16–5 is σˆf = 0.025, where the deviation from the mean is due entirely to environmental conditions. Find the brake torques corresponding to ±3σˆf .

16–7 The brake shown in the figure has a coefficient of friction of 0.30, a face width of 2 in, and a limiting shoe lining pressure of 150 psi. Find the limiting actuating force F and the torque capacity.

16–8 Refer to the symmetrical pivoted external brake shoe of Figure 16–12 and Equation

(16–15). Suppose the pressure distribution was uniform, that is, the pressure p is independent of θ. What would the pivot distance a′ be? If θ1 = θ2 = 60°, compare a with a′.

16–9 The shoes on the brake depicted in the figure subtend a 90° arc on the drum of this external pivoted-shoe brake. The actuation force P is applied to the lever. The rotation direction of the drum is counterclockwise, and the coefficient of friction is 0.30. (a) What should the dimension e be, in order to eliminate frictional moments on each shoe? (b) Draw the free-body diagrams of the handle lever and both shoe levers, with forces expressed in terms of the actuation force P. (c) Does the direction of rotation of the drum affect the braking torque?

Clutches, Brakes, Couplings, and Flywheels     875 5

5

F

16

4 30°

30°

ion

12

Ro tat

Problem 16–7 Dimensions in inches.

130°

130° 10 R

12

20° 20°

A

B 3

3

P 3 Shoe 3

68

Problem 16–9 Dimensions in inches.

15.28 13.5

7.78

e

16–10 Problem 16–9 is preliminary to analyzing the brake. A rigid molded non-asbestos

lining is used dry in the brake of Problem 16–9 on a cast-iron drum. The shoes are 6  in wide and subtend a 90° arc. Conservatively estimate the maximum allowable actuation force and the braking torque.

16–11 The maximum band interface pressure on the brake shown in the figure is 620 kPa. Use

a 350 mm-diameter drum, a band width of 25 mm, a coefficient of friction of 0.30, and an angle-of-wrap of 270°. Find the maximum band tensions and the torque capacity. Rotation

ϕ

Problem 16–11

P1

P2

876      Mechanical Engineering Design

16–12 The drum for the band brake in Problem 16–11 is 12 in in diameter. The band selected

has a mean coefficient of friction of 0.28 and a width of 3.25 in. It can safely support a tension of 1.8 kip. If the angle of wrap is 270°, find the maximum lining pressure and the corresponding torque capacity.

16–13 The brake shown in the figure has a coefficient of friction of 0.30 and is to operate using a maximum force F of 400 N. If the band width is 50 mm, find the maximum band tensions and the braking torque. 200

Problem 16–13 Dimensions in millimeters.

160

F

100

225

16–14 The figure depicts a band brake whose drum rotates counterclockwise at 200 rev/min. The drum diameter is 16 in and the band lining 3 in wide. The coefficient of friction is 0.20. The maximum lining interface pressure is 70 psi. (a) Find the maximum brake torque, necessary force P, and steady-state power. (b) Complete the free-body diagram of the drum. Find the bearing radial load that a pair of straddle-mounted bearings would have to carry. (c) What is the lining pressure p at both ends of the contact arc?

P

Problem 16–14

3 in

10 in

16–15 The figure shows a band brake designed to prevent "backward" rotation of the shaft.

The angle of wrap is 270°, the band width is 281 in, and the coefficient of friction is 0.20. The torque to be resisted by the brake is 150 lbf · ft. The diameter of the pulley is 841 in. (a) What dimension c1 will just prevent backward motion? (b) If the rocker was designed with c1 = 1 in, what is the maximum pressure between the band and drum at 150 lbf · ft back torque? (c) If the back-torque demand is 100 lbf · in, what is the largest pressure between the band and drum?

Clutches, Brakes, Couplings, and Flywheels     877

Problem 16–15

c1 1

2 4 in P2

P1 Rocker detail

16–16 A plate clutch has a single pair of mating friction surfaces 250-mm OD by 175-mm ID.

The mean value of the coefficient of friction is 0.30, and the actuating force is 4 kN. (a) Find the maximum pressure and the torque capacity using the uniform-wear model. (b) Find the maximum pressure and the torque capacity using the uniform-pressure model.

16–17 A hydraulically operated multidisk plate clutch has an effective disk outer diameter of

6.5 in and an inner diameter of 4 in. The coefficient of friction is 0.24, and the limiting pressure is 120 psi. There are six planes of sliding present. (a) Using the uniform wear model, estimate the limiting axial force F and the torque T. (b) Let the inner diameter of the friction pairs d be a variable. Complete the following table: d, in

2 3 4 5 6

T, lbf · in   (c) What does the table show?

16–18 Look again at Problem 16–17.

(a) Show how the optimal diameter d* is related to the outside diameter D. (b) What is the optimal inner diameter? (c) What does the tabulation show about maxima? (d) Common proportions for such plate clutches lie in the range 0.45 ≤ d∕D ≤ 0.80. Is the result in part a useful?

16–19 A cone clutch has D = 12 in, d = 11 in, a cone length of 2.25 in, and a coefficient

of friction of 0.28. A torque of 1.8 kip · in is to be transmitted. For this requirement, estimate the actuating force and maximum pressure by both models.

16–20 Show that for the caliper brake the T∕( f F D) versus d∕D plots are the same as Equations (b) and (c) of Section 16–5.

16–21 A two-jaw clutch has the dimensions shown in the figure and is made of ductile steel.

The clutch has been designed to transmit 2 kW at 500 rev/min. Find the bearing and shear stresses in the key and the jaws.

878      Mechanical Engineering Design 45°

Dimensions in millimeters. 3

10

45 dia.

6

24 dia.

Problem 16–21

26 dia.

1.5 (typ.)

50

16–22 A brake has a normal braking torque of 2.8 kip · in and heat-dissipating cast-iron

surfaces whose mass is 40 lbm. Suppose a load is brought to rest in 8.0 s from an initial angular speed of 1600 rev/min using the normal braking torque; estimate the temperature rise of the heat-dissipating surfaces.

16–23 A cast-iron flywheel has a rim whose OD is 1.5 m and whose ID is 1.4 m. The flywheel

weight is to be such that an energy fluctuation of 6.75 J will cause the angular speed to vary no more than 240 to 260 rev/min. Estimate the coefficient of speed fluctuation. If the weight of the spokes is neglected, what should be the width of the rim?

16–24 A single-geared blanking press has a stroke of 200 mm and a rated capacity of 320 kN.

A cam-driven ram is assumed to be capable of delivering the full press load at constant force during the last 15 percent of a constant-velocity stroke. The camshaft has an average speed of 90 rev/min and is geared to the flywheel shaft at a 6:1 ratio. The total work done is to include an allowance of 20 percent for friction. (a) Estimate the maximum energy fluctuation. (b) Find the rim weight for an effective diameter of 1.2 m and a coefficient of speed fluctuation of 0.10.

16–25 Using the data of Table 16–6, find the mean output torque and flywheel inertia

required for a three-cylinder in-line engine corresponding to a nominal speed of 2400 rev/min. Use Cs = 0.30.

16–26 When a motor armature inertia, a pinion inertia, and a motor torque reside on a motor

shaft, and a gear inertia, a load inertia, and a load torque exist on a second shaft, it is useful to reflect all the torques and inertias to one shaft, say, the armature shaft. We need some rules to make such reflection easy. Consider the pinion and gear as disks of pitch radius. ∙ A torque on a second shaft is reflected to the motor shaft as the load torque divided by the negative of the stepdown ratio. ∙ An inertia on a second shaft is reflected to the motor shaft as its inertia divided by the stepdown ratio squared. ∙ The inertia of a disk gear on a second shaft in mesh with a disk pinion on the motor shaft is reflected to the pinion shaft as the pinion inertia multiplied by the stepdown ratio squared. (a) Verify the three rules. (b) Using the rules, reduce the two-shaft system in the figure to a motor-shaft shish kebab equivalent. Correctly done, the dynamic response of the shish kebab and the real system are identical. (c) For a stepdown ratio of n = 10 compare the shish kebab inertias.

Clutches, Brakes, Couplings, and Flywheels     879 Load torque reflection Load inertia reflection n Gear inertia reflection

IG

IP Problem 16–26

IP

Dimensions in millimeters. IM

IL

T(ω1)

IM T(ω1)

T(ω2 ) 1

Shish kebab equivalent

2 (a)

(b)

16–27 Apply the rules of Problem 16–26 to the three-shaft system shown in the figure to create a motor shaft shish kebab. (a) Show that the equivalent inertia Ie is given by Ie = IM + IP + n2IP +

IP

+

2

n

m2IP 2

n

+

IL m2 n2

(b) If the overall gear reduction R is a constant nm, show that the equivalent inertia becomes Ie = IM + IP + n2IP +

IP 2

n

+

R 2IP n4

+

IL R2

(c) If the problem is to minimize the gear-train inertia, find the ratios n and m for the values of IP = 1, IM = 10, IL = 100, and R = 10. n IP

IG1 m

IM

IP

IG2

Problem 16–27 TM

IL

R = nm

16–28 For the conditions of Problem 16–27, make a plot of the equivalent inertia Ie as ordinate and the stepdown ratio n as abscissa in the range 1 ≤ n ≤ 10. How does the minimum inertia compare to the single-step inertia?

16–29 A punch-press geared 10:1 is to make six punches per minute under circumstances

where the torque on the crankshaft is 1300 lbf · ft for 12 s. The motor's nameplate reads 3 bhp at 1125 rev/min for continuous duty. Design a satisfactory flywheel for use on

880      Mechanical Engineering Design

the motor shaft to the extent of specifying material and rim inside and outside diameters as well as its width. As you prepare your specifications, note ωmax, ωmin, the coefficient of speed fluctuation Cs, energy transfer, and peak power that the flywheel transmits to the punch-press. Note power and shock conditions imposed on the gear train because the flywheel is on the motor shaft.

16–30 The punch-press of Problem 16–29 needs a flywheel for service on the crankshaft of

the punch-press. Design a satisfactory flywheel to the extent of specifying material, rim inside and outside diameters, and width. Note ωmax, ωmin, Cs, energy transfer, and peak power the flywheel transmits to the punch. What is the peak power seen in the gear train? What power and shock conditions must the gear train transmit?

16–31 Compare the designs resulting from the tasks assigned in Problems 16–29 and 16–30. What have you learned? What recommendations do you have?

17

Flexible Mechanical Elements

©Stason4ik/Shutterstock

Chapter Outline 17–1

Belts  882

17–5

Roller Chain   909

17–2

Flat- and Round-Belt Drives   885

17–6

Wire Rope   917

17–3

V Belts   900

17–7

Flexible Shafts   926

17–4

Timing Belts   908 881

882      Mechanical Engineering Design

Belts, ropes, chains, and other similar elastic or flexible machine elements are used in conveying systems and in the transmission of power over comparatively long distances. It often happens that these elements can be used as a replacement for gears, shafts, bearings, and other relatively rigid power-transmission devices. In many cases their use simplifies the design of a machine and substantially reduces the cost. In addition, since these elements are elastic and usually quite long, they play an important part in absorbing shock loads and in damping out and isolating the effects of vibration. This is an important advantage as far as machine life is concerned. Most flexible elements do not have an infinite life. When they are used, it is important to establish an inspection schedule to guard against wear, aging, and loss of elasticity. The elements should be replaced at the first sign of deterioration.

17–1  Belts The four principal types of belts are shown, with some of their characteristics, in Table 17–1. Crowned pulleys are used for flat belts, and grooved pulleys, or sheaves, for round and V belts. Timing belts require toothed wheels, or sprockets. In all cases, the pulley axes must be separated by a certain minimum distance, depending upon the belt type and size, to operate properly. Other characteristics of belts are: ∙ They may be used for long center distances. ∙ Except for timing belts, there is some slip and creep, and so the angular-velocity ratio between the driving and driven shafts is neither constant nor exactly equal to the ratio of the pulley diameters. ∙ In some cases an idler or tension pulley can be used to avoid adjustments in center distance that are ordinarily necessitated by age or the installation of new belts. Figure 17–1 illustrates the geometry of open and crossed flat-belt drives. For a flat belt with this drive the belt tension is such that the sag or droop is visible in Figure 17–2a, when the belt is running. Although the top is preferred for the loose side of the belt, for other belt types either the top or the bottom may be used, because their installed tension is usually greater. Table 17–1  Characteristics of Some Common Belt Types (Figures are cross sections except for the timing belt, which is a side view). Belt Type

Figure

Joint

Flat Yes

Size Range

Center Distance

0.03 to 0.20 in t={ 0.75 to 5 mm

No upper limit

t

Round Yes d = 18 to 34 in

No upper limit

V None 0.31 to 0.91 in b={ b 8 to 19 mm

Limited

Timing None p = 2 mm and up

Limited

d

p

Flexible Mechanical Elements     883 sin–1

sin–1

D–d 2C

ϕd

– 2 4C – (D

1 2

Figure 17–1

D–d 2C

Flat-belt geometry. (a) Open belt. (b) Crossed belt.

2

d)

ϕD

d

D

D–d 2C –1 D – d ϕD = π + 2 sin 2C ϕd = π – 2 sin–1

L=

C

4C 2 – (D – d )2 +

1 2

(DϕD + dϕd)

1 2

(D + d)ϕ

(a) sin–1 sin–1

D+d 2C

D+d 2C

d ϕ

ϕ D 1 2

4C 2 – (D + d)2

ϕ = π + 2 sin–1 L=

C

D+d 2C

4C 2 – (D + d)2 +

(b)

Figure 17–2

Driver

(a)

(b)

(c)

Nonreversing and reversing belt drives. (a) Nonreversing open belt. (b) Reversing crossed belt. Crossed belts must be separated to prevent rubbing if highfriction materials are used. (c) Reversing open-belt drive.

884      Mechanical Engineering Design Midpoint

Figure 17–3 Quarter-twist belt drive; an idler guide pulley must be used if motion is to be in both directions.

Two types of reversing drives are shown in Figure 17–2. Notice that both sides of the belt contact the driving and driven pulleys in Figures 17–2b and 17–2c, and so these drives cannot be used with V belts or timing belts. Figure 17–3 shows a flat-belt drive with out-of-plane pulleys. The shafts need not be at right angles as in this case. Note the top view of the drive in Figure 17–3. The pulleys must be positioned so that the belt leaves each pulley in the midplane of the other pulley face. Other arrangements may require guide pulleys to achieve this condition. Another advantage of flat belts is shown in Figure 17–4, where clutching action is obtained by shifting the belt from a loose to a tight or driven pulley. Figure 17–5 shows two variable-speed drives. The drive in Figure 17–5a is commonly used only for flat belts. The drive of Figure 17–5b can also be used for V belts and round belts by using grooved sheaves. Flat belts are made of urethane and also of rubber-impregnated fabric reinforced with steel wire or nylon cords to take the tension load. One or both surfaces may have a friction surface coating. Flat belts are quiet, they are efficient at high speeds, and they can transmit large amounts of power over long center distances. Usually, flat belting is purchased by the roll and cut and the ends are joined by using special kits furnished by the manufacturer. Two or more flat belts running side by side, instead of a single wide belt, are often used to form a conveying system. A V belt is made of fabric and cord, usually cotton, rayon, or nylon, and impregnated with rubber. In contrast with flat belts, V belts are used with similar sheaves and at shorter center distances. V belts are slightly less efficient than flat belts, but a number of them can be used on a single sheave, thus making a multiple drive. V belts are made only in certain lengths and have no joints. Timing belts are made of rubberized fabric and steel wire and have teeth that fit into grooves cut on the periphery of the sprockets. The timing belt does not stretch or slip and consequently transmits power at a constant angular-velocity ratio. The fact that the belt is

Loose pulley

Driven Fork Shift fork

(a)

Driver

Figure 17–4 This drive eliminates the need for a clutch. Flat belt can be shifted left or right by use of a fork.

(b)

Figure 17–5 Variable-speed belt drives.

Flexible Mechanical Elements     885

toothed provides several advantages over ordinary belting. One of these is that no initial tension is necessary, so that fixed-center drives may be used. Another is the elimination of the restriction on speeds; the teeth make it possible to run at nearly any speed, slow or fast. Disadvantages are the first cost of the belt, the necessity of grooving the sprockets, and the attendant dynamic fluctuations caused at the belt-tooth meshing frequency.

17–2  Flat- and Round-Belt Drives Modern flat-belt drives consist of a strong elastic core surrounded by an elastomer. These drives have distinct advantages over gear drives or V-belt drives. A flat-belt drive has an efficiency of about 98 percent, which is about the same as for a gear drive. On the other hand, the efficiency of a V-belt drive ranges from about 70 to 96 percent.1 Flat-belt drives produce very little noise and absorb more torsional vibration from the system than either V-belt or gear drives. When an open-belt drive (Figure 17–1a) is used, the wrap angles are found to be

ϕd = π − 2 sin−1

−1

where D d C ϕ

D−d 2C D−d 2C

ϕD = π + 2 sin = = = =

(17–1)

diameter of large pulley diameter of small pulley center distance wrap angle

The length of the belt is found by summing the two arc lengths with twice the distance between the beginning and end of contact. The result is

L = [4C 2 − (D − d ) 2]1∕2 +

1 (DϕD + dϕd ) 2

(17–2)

A similar set of equations can be derived for the crossed belt of Figure 17–2b. For this belt, the angle of wrap is the same for both pulleys and is

ϕ = π + 2 sin−1

D+d 2C

(17–3)

The belt length for crossed belts is found to be

L = [4C 2 − (D + d) 2]1∕2 +

1 (D + d )ϕ 2

(17–4)

Firbank2 explains flat-belt-drive theory in the following way. A change in belt tension due to friction forces between the belt and pulley will cause the belt to elongate or contract and move relative to the surface of the pulley. This motion is caused by elastic creep and is associated with sliding friction as opposed to static friction. The action at the driving pulley, through that portion of the angle of wrap that is 1

A. W. Wallin, "Efficiency of Synchronous Belts and V-Belts," Proc. Nat. Conf. Power Transmission, vol. 5, Illinois Institute of Technology, Chicago, Nov. 7–9, 1978, pp. 265–271. 2 T. C. Firbank, Mechanics of the Flat Belt Drive, ASME paper no. 72-PTG-21.

886      Mechanical Engineering Design

r

F + dF dS dN

f dN

dθ θ

F

Figure 17–6 Free body of an infinitesimal element of a flat belt in contact with a pulley.

actually transmitting power, is such that the belt moves more slowly than the surface speed of the pulley because of the elastic creep. The angle of wrap is made up of the effective arc, through which power is transmitted, and the idle arc. For the driving pulley the belt first contacts the pulley with a tight-side tension F1 and a velocity V1, which is the same as the surface velocity of the pulley. The belt then passes through the idle arc with no change in F1 or V1. Then creep or sliding contact begins, and the belt tension changes in accordance with the friction forces. At the end of the effective arc the belt leaves the pulley with a loose-side tension F2 and a reduced speed V2. Firbank has used this theory to express the mechanics of flat-belt drives in mathematical form and has verified the results by experiment. His observations include the finding that substantially more power is transmitted by static friction than sliding friction. He also found that the coefficient of friction for a belt having a nylon core and leather surface was typically 0.7, but that it could be raised to 0.9 by employing special surface finishes. Our model will assume that the friction force on the belt is proportional to the normal pressure along the arc of contact. We seek first a relationship between the tight side tension and slack side tension, similar to that of band brakes but incorporating the consequences of movement, that is, centrifugal tension in the belt. In Figure 17–6 we see a free body of a small segment of the belt. The differential force dS is due to centrifugal force, dN is the normal force between the belt and pulley, and f dN is the shearing traction due to friction at the point of slip. The belt width is b and the thickness is t. The belt mass per unit length is m. The centrifugal force dS can be expressed as

dS = (mr dθ)rω2 = mr 2ω2 dθ = mV 2 dθ = Fc dθ

(a)

where V is the belt speed. Summing forces radially, with sin(dθ∕2) ≈ dθ∕2, gives

∑ Fr = −(F + dF)

dθ dθ −F + dN + dS = 0 2 2

Ignoring the higher-order term, we have

dN = F dθ − dS

(b)

Summing forces tangentially gives

∑ Ft = −f dN − F + (F + dF) = 0 from which, incorporating Equations (a) and (b), we obtain dF = f dN = f F dθ − f dS = f F dθ − f mr 2ω2 dθ or

dF − f F = −f mr 2ω2 dθ

(c)

The solution to this nonhomogeneous first-order linear differential equation is

F = A exp( f θ) + mr 2ω2

(d)

where A is an arbitrary constant. As shown in Figure 17–7, θ starts at the loose side, and the boundary condition that F at θ = 0 equals F2 gives A = F2 − mr2ω2. The solution is

F = (F2 − mr 2ω2 ) exp( f θ) + mr 2ω2

(17–5)

At the end of the angle of wrap ϕ, the tight side,

F∣ θ=ϕ = F1 = (F2 − mr 2ω2 ) exp( f ϕ) + mr 2ω2

(17–6)

Flexible Mechanical Elements     887

Now we can write F1 − mr 2ω2

2

2

F2 − mr ω

=

F1 − Fc = exp( f ϕ) F2 − Fc

(17–7)

where, from Equation (a), Fc = mr2ω2. Equation (17–7) is referred to as the belting equation and can be written as

F1 − F2 = (F1 − Fc )

exp ( f ϕ) − 1 exp( f ϕ)

(17–8)

Now Fc is found as follows: with n being the rotational speed, in rev/min, of the pulley of diameter d, in inches, the belt speed is V = π dn∕12

ft/min

The weight w of a foot of belt is given in terms of the weight density γ in lbf/in3 as w = 12γbt lbf/ft where b and t are in inches. Fc is written as

Fc =

w V 2 w V 2 = g ( 60 ) 32.17 ( 60 )

(e)

Figure 17–7 shows a free body diagram of a pulley and part of the belt. The tight side tension F1 and the loose side tension F2 have the following additive components:

F1 = Fi + Fc + ΔF∕2 = Fi + Fc + T∕d

(f )

F2 = Fi + Fc − ΔF∕2 = Fi + Fc − T∕d

(g)

where Fi Fc ΔF∕2 d

= = = =

initial tension hoop tension due to centrifugal force tension due to the transmitted torque T diameter of the pulley

The difference between F1 and F2 is related to the pulley torque. Subtracting Equation (g) from Equation ( f ) gives

F1 − F2 =

2T d

(h)

Adding Equations ( f ) and (g) gives F1 + F2 = 2Fi + 2Fc F2 = Fi + Fc – ΔF/2 = Fi + Fc – T/d θ

ϕ

d

T

F1 = Fi + Fc + ΔF/2 = Fi + Fc + T/d

Figure 17–7 Forces and torques on a pulley.

888      Mechanical Engineering Design

from which

Fi =

F1 + F2 − Fc 2

(i)

Dividing Equation (i) by Equation (h), manipulating, and using Equation (17–7) gives

(F1 + F2 )∕2 − Fc F1 + F2 − 2Fc (F1 − Fc ) + (F2 − Fc ) Fi = = = T∕d (F1 − F2 )∕2 F1 − F2 (F1 − Fc ) − (F2 − Fc )

=

(F1 − Fc )∕(F2 − Fc ) + 1 exp( f ϕ) + 1 = (F1 − Fc )∕(F2 − Fc ) − 1 exp( f ϕ) − 1

from which

Fi =

T exp( f ϕ) + 1 d exp( f ϕ) − 1

(17–9)

Equation (17–9) give us a fundamental insight into flat belting. If Fi equals zero, then T equals zero: no initial tension, no torque transmitted. The torque is in proportion to the initial tension. This means that if there is to be a satisfactory flat-belt drive, the initial tension must be (1) provided, (2) sustained, (3) in the proper amount, and (4) maintained by routine inspection. From Equation ( f ), incorporating Equation (17–9) gives F1 = Fi + Fc + = Fc +

exp( f ϕ) − 1 T = Fc + Fi + Fi d exp( f ϕ) + 1

Fi [exp( f ϕ) + 1] + Fi [exp( f ϕ) − 1] exp( f ϕ) + 1 F1 = Fc + Fi

2 exp( f ϕ) exp( f ϕ) + 1

(17–10)

From Equation (g), incorporating Equation (17–9) gives F2 = Fi + Fc − = Fc +

exp( f ϕ) − 1 T = Fc + Fi − Fi d exp( f ϕ) + 1

Fi [exp( f ϕ) + 1] − Fi [exp( f ϕ) − 1] exp( f ϕ) + 1 F2 = Fc + Fi

2 exp( f ϕ) + 1

(17–11)

Equation (17–7) is the belting equation, but Equations (17–9), (17–10), and (17–11) reveal how belting works. We plot Equations (17–10) and (17–11) as shown in Figure 17–8 against Fi as abscissa. The initial tension needs to be sufficient so that the difference between the F1 and F2 curve is 2T∕d. With no torque transmitted, the least possible belt tension is F1 = F2 = Fc. The transmitted horsepower is given by

H=

(F1 − F2 )V 33 000

( j)

where the forces are in lbf and V is in ft/min. Manufacturers provide specifications for their belts that include allowable tension Fa (or stress σall), the tension being expressed in units of force per unit width. Belt life is usually several years. The severity of flexing at the pulley and its effect on life is reflected in a pulley correction factor Cp. Speed in excess of

Flexible Mechanical Elements     889

(F1)a

Belt tension F1 or F2

F1 1.0

11 64

2F exp( fϕ) F1 = Fc + i exp( fϕ) + 1

F2 = Fc +

T d

Velocity factor C v

2

2Fi exp( fϕ) + 1

18 64

in

in and 20 in 64

0.9

13 64

0.8

25 64

F2 Fc Fi

(Fi )a

0.7

0

1

2

3

4

in

5

Belt velocity 10 –3V, ft /min

Initial tension Fi

Figure 17–8

Figure 17–9

Plot of initial tension Fi against belt tension F1 or F2, showing the intercept Fc, the equations of the curves, and where 2T∕d is to be found.

Velocity correction factor Cv for leather belts for various thicknesses. (Data source: Machinery's Handbook, 20th ed., Industrial Press, New York, 1976, p. 1047.)

600 ft/min and its effect on life is reflected in a velocity correction factor Cv. For polyamide and urethane belts use Cv = 1. For leather belts see Figure 17–9. A service factor Ks is used for excursions of load from nominal, applied to the nominal power as Hd = HnomKsnd, where nd is the design factor for exigencies. These effects are incorporated as follows: where (F1)a b Fa Cp Cv

(F1 ) a = bFaCpCv = = = = =

(17–12)

allowable largest tension, lbf belt width, in manufacturer's allowed tension, lbf/in pulley correction factor (Table 17–4) velocity correction factor

The steps in analyzing a flat-belt drive can include (see Example 17–1)   1 Find exp( f ϕ) from belt-drive geometry and friction from Table 17–2   2 From belt geometry, material (Table 17–2), and speed, find Fc from Eq. (e)   3 Using Eq. (3–42), with H = Hd = HnomKsnd, find the necessary torque from T = 63 025HnomKsnd∕n   4 From torque T find the necessary (F1)a − F2 = 2T∕d   5 From Tables 17–2 and 17–4, and Equation (17–12) determine (F1)a   6 Find F2 from (F1)a − [(F1)a − F2]   7 From Equation (i) find the necessary initial tension Fi   8 Check the friction development, f′ < f. Use Equation (17–7) solved for f ′: f′=

1 (F1 ) a − Fc ln ϕ F2 − Fc

 9 Find the factor of safety from nfs = Ha∕(HnomKs)

in

6

890      Mechanical Engineering Design

It is unfortunate that many of the available data on belting are from sources in which they are presented in a very simplistic manner. These sources use a variety of charts, nomographs, and tables to enable someone who knows nothing about belting to apply them. Little, if any, computation is needed for such a person to obtain valid results. Since a basic understanding of the process, in many cases, is lacking, there is no way this person can vary the steps in the process to obtain a better design. Incorporating the available belt-drive data into a form that provides a good understanding of belt mechanics involves certain adjustments in the data. Because of this, the results from the analysis presented here will not correspond exactly with those of the sources from which they were obtained. A moderate variety of belt materials, with some of their properties, are listed in Table 17–2. These are sufficient for solving a large variety of design and analysis problems. The design equation to be used is Equation ( j). Table 17–2  Properties of Some Flat- and Round-Belt Materials. (Diameter = d, thickness = t, width = w) Minimum Allowable Tension Pulley per Unit Width Size, Diameter, at 600 ft/min, Material Specification in in lbf/in Leather

1 ply

30

0.035–0.045 0.4

t=

312

33

0.035–0.045 0.4

1 t = 18 64 4 2

2 ply

13 64

41

0.035–0.045 0.4

t=

20 a 64  6

50

0.035–0.045 0.4

a t = 23 64  9

60

0.035–0.045 0.4

Polyamideb F–0c

t = 0.03    0.60

10

0.035

0.5

t = 0.05   1.0

35

0.035

0.5

F–1c c

F–2

t = 0.07   2.4

60

0.051

0.5

A–2c

t = 0.11   2.4

60

0.037

0.8

c

A–3

t = 0.13   4.3

100

0.042

0.8

A–4c

t = 0.20   9.5

175

0.039

0.8

t = 0.25

275

0.039

0.8

c

A–5 d

Urethane

w = 0.50 in

t = 0.062

w = 0.75 in

t = 0.078

w = 1.25 in

t = 0.090

13.5

e

See 5.2 Table 9.8e 17–3 18.9e

Round d=

1 4 See

d=

3 8

d=

1 2

d = 34

e

8.3 Table 18.6e 17–3 33.0e

Add 2 in to pulley size for belts 8 in wide or more. Source: Habasit Engineering Manual, Habasit Belting, Inc., Chamblee (Atlanta), Ga. c Friction cover of acrylonitrile-butadiene rubber on both sides. d Source: Eagle Belting Co., Des Plaines, Ill. e At 6% elongation; 12% is maximum allowable value. b

Coefficient of Friction

3

a

t = 11 64

Specific Weight, lbf/in3

74.3e

0.038–0.045 0.7 0.038–0.045 0.7 0.038–0.045 0.7 0.038–0.045 0.7 0.038–0.045 0.7 0.038–0.045 0.7 0.038–0.045 0.7

Flexible Mechanical Elements     891

Table 17–3  Minimum Pulley Sizes for Flat and Round Urethane Belts (Listed are the pulley diameters in inches).

Ratio of Pulley Speed to Belt Length, rev/(ft · min)

Belt Belt Style

Size, in

Up to 250

250 to 499

500 to 1000

Flat

0.50 × 0.062

0.38

0.44

0.50

0.75 × 0.078

0.50

0.63

0.75

1.25 × 0.090

0.50

0.63

0.75

1 4 3 8 1 2 3 4

Round

1.50 1.75 2.00 2.25 2.62 3.00 3.00 3.50 4.00 5.00 6.00 7.00

Source: Eagle Belting Co., Des Plaines, Ill.

The values given in Table 17–2 for the allowable belt tension are based on a belt speed of 600 ft/min. For higher speeds, use Figure 17–9 to obtain Cv values for leather belts. For polyamide and urethane belts, use Cv = 1.0. The service factors Ks for V-belt drives, given in Table 17–15 in Section 17–3, are also recommended here for flat- and round-belt drives. Minimum pulley sizes for the various belts are listed in Tables 17–2 and 17–3. The pulley correction factor accounts for the amount of bending or flexing of the belt and how this affects the life of the belt. For this reason it is dependent on the size and material of the belt used. See Table 17–4. Use Cp = 1.0 for urethane belts. Flat-belt pulleys should be crowned to keep belts from running off the pulleys. If only one pulley is crowned, it should be the larger one. Both pulleys must be crowned whenever the pulley axes are not in a horizontal position. Use Table 17–5 for the crown height.

Table 17–4  Pulley Correction Factor CP for Flat Belts*

Small-Pulley Diameter, in Material

1.6 to 4

4.5 to 8

9 to 12.5

14, 16

18 to 31.5

Over 31.5

Leather

0.5 0.6 0.7 0.8 0.9

1.0

Polyamide, F–0

0.95

1.0

1.0

1.0

1.0

1.0

F–1 0.70

0.92

0.95 1.0

1.0

1.0

F–2 0.73

0.86

0.96 1.0

1.0

1.0

A–2 0.73 0.86

0.96 1.0

1.0

1.0

A–3 — 0.70 0.87 0.94 0.96

1.0

A–4 —

0.71 0.80 0.85

0.92

A–5 —

— 0.72 0.77

0.91

*Average values of CP for the given ranges were approximated from curves in the Habasit Engineering Manual, Habasit Belting, Inc., Chamblee (Atlanta), Ga.

892      Mechanical Engineering Design

Table 17–5  Crown Height and ISO Pulley Diameters for Flat Belts* ISO Pulley Diameter, in

Crown Height, in

ISO Pulley Diameter, in

Crown Height, in w ≤ 10 in

w > 10 in

1.6, 2, 2.5

0.012

12.5, 14

0.03

0.03

2.8, 3.15

0.012

12.5, 14

0.04

0.04

3.55, 4, 4.5

0.012

22.4, 25, 28

0.05

0.05

5, 5.6

0.016

31.5, 35.5

0.05

0.06

6.3, 7.1

0.020

40

0.05

0.06

8, 9

0.024

45, 50, 56

0.06

0.08

10, 11.2

0.030

63, 71, 80

0.07

0.10

*Crown should be rounded, not angled; maximum roughness is Ra  =  AA 63 μin.

EXAMPLE 17–1 A polyamide A-3 flat belt 6 in wide is used to transmit 15 hp under light shock conditions where Ks = 1.25, and a factor of safety equal to or greater than 1.1 is appropriate. The pulley rotational axes are parallel and in the horizontal plane. The shafts are 8 ft apart. The 6-in driving pulley rotates at 1750 rev/min in such a way that the loose side is on top. The driven pulley is 18 in in diameter. See Figure 17–10. The factor of safety is for unquantifiable exigencies (see the steps outlined earlier). (a) Estimate the centrifugal tension Fc and the torque T. (b) Estimate the allowable F1, F2, Fi and allowable power Ha. (c) Estimate the factor of safety. Is it satisfactory? Solution

ϕd = π − 2 sin−1 [

(a) Eq. (17–1):

γ = 0.042 lb/in3

Table 17–2: Step 1:

18 − 6 = 3.0165 rad 2(8)12 ] f = 0.8

Fa = 100 lbf/in

exp( f ϕ) = exp[0.8(3.0165)] = 11.17

V = π(6)1750∕12 = 2749 ft/min

w = 12γbt = 12(0.042)6(0.130) = 0.393 lbf/ft

Answer  Step 2:

Fc =

w V 2 0.393 2749 2 = = 25.6 lbf g ( 60 ) 32.17 ( 60 )

T=

63 025Hnom Ks nd 63 025(15)1.25(1.1) = n 1750

Step 3:

= 742.8 lbf · in

Answer Figure 17–10 The flat-belt drive of Ex. 17–1. (Drawing is not to scale)

1750 rpm

Belt 6 in × 0.130 in 18 in

6 in

96 in

15 hp γ = 0.042

lbf in3

Flexible Mechanical Elements     893

(b) From step 4, the necessary (F1)a − F2 to transmit the torque T, from Eq. (h), is Step 4:

(F1 ) a − F2 =

2T 2(742.8) = = 247.6 lbf d 6

For polyamide belts Cv = 1, and from Table 17–4 Cp = 0.70. From Eq. (17–12) the allowable largest belt tension (F1)a is Answer  Step 5:

(F1 ) a = bFaCpCv = 6(100)0.70(1) = 420 lbf

then Answer  Step 6:

F2 = (F1 ) a − [(F1 ) a − F2] = 420 − 247.6 = 172.4 lbf

and from Eq. (i) Answer  Step 7:

Fi =

(F1 ) a + F2 420 + 172.4 − Fc = − 25.6 = 270.6 lbf 2 2

Answer The combination (F1)a, F2, and Fi will transmit the design power of Ha = HnomKsnd = 15(1.25)(1.1) = 20.6 hp and protect the belt. From step 8, we check the friction development by solving Eq. (17–7) for f ′: Step 8:

f′ =

1 (F1 ) a − Fc 1 420 − 25.6 ln = ln = 0.328 ϕ F2 − Fc 3.0165 172.4 − 25.6

As determined earlier, f = 0.8. Since f ′ < f, there is no danger of slipping. (c) From step 9, Answer  Step 9:

nfs =

Ha 20.6 = = 1.1 Hnom Ks 15(1.25)

(as expected)

Answer The belt is satisfactory and the maximum allowable belt tension exists. If the initial tension is maintained, the capacity is the design power of 20.6 hp.

Initial tension is the key to the functioning of the flat belt as intended. There are ways of controlling initial tension. One way is to place the motor and drive pulley on a pivoted mounting plate so that the weight of the motor, pulley, and mounting plate and a share of the belt weight induces the correct initial tension and maintains it. A second way is the use of a spring-loaded idler pulley, adjusted to the same task. Both of these methods accommodate to temporary or permanent belt stretch. See Figure 17–11. Because flat belts were used for long center-to-center distances, the weight of the belt itself can provide the initial tension. The static belt deflects to an approximate catenary curve, and the dip from a straight belt can be measured against a stretched music wire. This provides a way of measuring and adjusting the dip. From catenary theory the dip is related to the initial tension by

dip =

12(C∕12) 2 w C 2w = 8Fi 96Fi

(17–13)

894      Mechanical Engineering Design

Figure 17–11

W

Belt-tensioning schemes. (a) Weighted idler pulley. (b) Pivoted motor mount. (c) Catenary-induced tension.

(a) Slack side

F2

Tight side

F1 W

(b) C dip

Fi

Fi

(c)

where dip C w Fi

= = = =

dip, in center-to-center distance, in weight per foot of the belt, lbf/ft initial tension, lbf

In Example 17–1 the dip corresponding to a 270.6-lbf initial tension is dip = ∙ ∙ ∙ ∙ ∙ ∙ ∙

(962 )0.393 = 0.14 in 96(270.6)

A decision set for a flat belt can be Function: power, speed, durability, reduction, service factor, center distance Design factor: nd Initial tension maintenance Belt material Drive geometry, d, D Belt thickness: t Belt width: b

Depending on the problem, some or all of the last four could be design variables. Belt cross-sectional area is really the design decision, but available belt thicknesses and widths are discrete choices. Available dimensions are found in s­ uppliers' catalogs.

Flexible Mechanical Elements     895

EXAMPLE 17–2 Design a flat-belt drive to connect horizontal shafts on 16-ft centers. The velocity ratio is to be 2.25:1. The angular speed of the small driving pulley is 860 rev/min, and the nominal power transmission is to be 60 hp under very light shock. Solution ∙ Function: Hnom = 60 hp, 860 rev/min, 2.25:1 ratio, Ks = 1.15, C = 16 ft ∙ Design factor: nd = 1.05 ∙ Initial tension maintenance: catenary ∙ Belt material: polyamide ∙ Drive geometry, d, D ∙ Belt thickness: t ∙ Belt width: b The last four could be design variables. Let's make a few more a priori decisions. Decision

d = 16 in, D = 2.25d = 2.25(16) = 36 in.

Decision Use polyamide A-3 belt; therefore t = 0.13 in and Cv = 1. Now there is one design decision remaining to be made, the belt width b. Table 17–2:   γ = 0.042 lbf/in3  f = 0.8   Fa = 100 lbf/in at 600 rev/min Table 17–4: Eq. (17–12):

Cp = 0.94 (F1)a = b(100)0.94(1) = 94.0b lbf

(1)

Hd = Hnom Ks n d = 60(1.15)1.05 = 72.5 hp

T=

63 025Hd 63 025(72.5) = = 5310 lbf · in n 860

Estimate exp( f ϕ) for full friction development: ϕd = π − 2 sin−1

Eq. (17–1):

36 − 16 = 3.037 rad 2(16)12

exp ( f ϕ) = exp[0.80(3.037)] = 11.35

Estimate centrifugal tension Fc in terms of belt width b:

w = 12γbt = 12(0.042)b(0.13) = 0.0655b lbf/ft

V = πdn∕12 = π(16) 860∕12 = 3602 ft/min

Eq. (e):

Fc =

w V 2 0.0655b 3602 2 = = 7.34b lbf g ( 60 ) 32.17 ( 60 )

(2)

For design conditions, that is, at Hd power level, using Eq. (h) gives

(F1 ) a − F2 = 2T∕d = 2(5310)∕16 = 664 lbf

(3)

F2 = (F1 ) a − [(F1 ) a − F2] = 94.0b − 664 lbf

(4)

896      Mechanical Engineering Design

Using Eq. (i) gives

Fi =

(F1 ) a + F2 94.0b + 94.0b − 664 − Fc = − 7.34b = 86.7b − 332 lbf 2 2

(5)

Place friction development at its highest level, using Eq. (17–7):

f ϕ = ln

(F1 ) a − Fc 94.0b − 7.34b 86.7b = ln = ln F2 − Fc 94.0b − 664 − 7.34b 86.7b − 664

Solving the preceding equation for belt width b at which friction is fully developed gives

664 exp( f ϕ) 664 11.38 = = 8.40 in 86.7 exp( f ϕ) − 1 86.7 11.38 − 1

b=

A belt width greater than 8.40 in will develop friction less than f = 0.80. The manufacturer's data indicate that the next available larger width is 10 in. Decision Use 10-in-wide belt. It follows that for a 10-in-wide belt Eq. (2):

Fc = 7.34(10) = 73.4 lbf

Eq. (1):

(F1 ) a = 94(10) = 940 lbf

Eq. (4):

F2 = 94(10) − 664 = 276 lbf

Eq. (5):

Fi = 86.7(10) − 332 = 535 lbf

The transmitted power, from Eq. (3), is

Ht =

[(F1 ) a − F2]V 664(3602) = = 72.5 hp 33 000 33 000

and the level of friction development f′, from Eq. (17–7) is

f′ =

1 (F1 ) a − Fc 1 940 − 73.4 ln = ln = 0.479 ϕ F2 − Fc 3.037 276 − 73.4

which is less than f = 0.8, and thus is satisfactory. Had a 9-in belt width been available, the analysis would show (F1)a = 846 lbf, F2 = 182 lbf, Fi = 448 lbf, and f ′ = 0.63. With a figure of merit available reflecting cost, thicker belts (A-4 or A-5) could be examined to ascertain which of the satisfactory alternatives is best. From Eq. (17–13) the catenary dip is

dip =

C 2w [16(12)]2 0.0655(10) = = 0.470 in 96Fi 96(535)

Figure 17–12 illustrates the variation of flexible flat-belt tensions at some cardinal points during a belt pass. Flat Metal Belts Thin flat metal belts with their attendant strength and geometric stability could not be fabricated until laser welding and thin rolling technology made possible belts as

Flexible Mechanical Elements     897

F1 T

B

+

A

+ C

F2

Figure 17–12

F

Flat-belt tensions.

+ D

E

(a)

ΔF/2

ΔF/2

F1 Fi

A

F2

Fc

Fc B

C

D

E

F

A

(b)

thin as 0.002 in and as narrow as 0.026 in. The introduction of perforations allows no-slip applications. Thin metal belts exhibit ∙ ∙ ∙ ∙ ∙

High strength-to-weight ratio Dimensional stability Accurate timing Usefulness to temperatures up to 700°F Good electrical and thermal conduction properties

In addition, stainless steel alloys offer "inert," nonabsorbent belts suitable to hostile (corrosive) environments, and can be made sterile for food and pharmaceutical applications. Thin metal belts can be classified as friction drives, timing or positioning drives, or tape drives. Among friction drives are plain, metal-coated, and perforated belts. Crowned pulleys are used to compensate for tracking errors. Figure 17–13 shows a thin flat metal belt with the tight tension F1 and the slack side tension F2 revealed. The relationship between F1 and F2 and the driving torque T is the same as in Equation (h). Equations (17–9), (17–10), and (17–11) also apply, with the hoop tension due to centrifugal force typically neglected for the very thin Figure 17–13 Metal-belt tensions and torques.

F1 D2

F2 D1

TL

D1

TM TM (a)

(b)

898      Mechanical Engineering Design

metal belts. The largest allowable tension, as in Equation (17–12), is posed in terms of stress in metal belts. A bending stress σb is created by making the belt conform to the pulley of radius, D∕2. Bending to a radius of curvature, ρ, is given by Equation (4–8), 1∕ρ = M∕EI. Substituting ρ = D∕2 and M∕I = σb∕ymax with ymax = t∕2 gives

σb = Et∕D

(k)

This equation applies to a narrow beam that is classified as a plane stress condition (see Section 3–5). However, a wide belt of narrow thickness is classified as a plane strain condition (see Section 3–7). It can be shown that converting a plane stress equation to a plane strain equation, E should be replaced by E∕(1−v2).3 Thus, Equation (k) becomes σb =

where

E t ν D

= = = =

Et E = 2 (1 − ν )D (1 − ν2 )(D∕t)

(17–14)

Young's modulus belt thickness Poisson's ratio pulley diameter

The tensile stresses (σ)1 and (σ)2 imposed by the belt tensions F1 and F2 are (σ) 1 = F1∕(bt)

and

(σ) 2 = F2∕(bt)

The largest tensile stress is (σb)1 + F1∕(bt) and the smallest is (σb)2 + F2∕(bt). During a belt pass both levels of stress appear. Although the belts are of simple geometry, the method of Marin is not used because the condition of the butt weldment (to form the loop) is not accurately known, and the testing of weld coupons is difficult. The belts are run to failure on two equalsized pulleys. Information concerning fatigue life, as shown in Table 17–6, is obtainable. Tables 17–7 and 17–8 give additional information. Table 17–6  Belt Life for Stainless Steel Friction Drives* D Belt Passes t

Table 17–7  Minimum Pulley Diameter* Belt Thickness, in

Minimum Pulley Diameter, in

0.002 1.2 0.003 1.8

6

0.005 3.0

625 ≥10

400 0.500 · 106

0.008 5.0

333 0.165 · 10

6

0.010 6.0

200 0.085 · 106

0.015 10.0

*Data courtesy of Belt Technologies, Agawam, Mass.

0.020 12.5 0.040 25.0 *Data courtesy of Belt Technologies, Agawam, Mass.

3

Richard G. Budynas, Advanced Strength and Applied Stress Analysis, 2nd ed., McGraw-Hill, New York, 1999, Table 4.1.1, p. 224.

Flexible Mechanical Elements     899

Table 17–8  Typical Material Properties, Metal Belts* Alloy

Yield Young's Strength, Modulus, Poisson's kpsi Mpsi Ratio

301 or 302 stainless steel

175

28

0.285

BeCu

170 17 0.220

1075 or 1095 carbon steel

230

30

0.287

Titanium 150 15 — Inconel

160 30 0.284

*Data courtesy of Belt Technologies, Agawam, Mass.

Table 17–6 shows metal belt life expectancies for a stainless steel belt. From Equation (17–14) with E = 28 Mpsi and ν = 0.29, the bending stresses corresponding to the four entries of the table are 48 914, 76 428, 91 805, and 152 855 psi. Using a natural log transformation on stress and passes shows that the regression line (r 2 = 0.9303) is

σ = 15 161 723 Np−0.412 = 15.16(106 )Np−0.412

(17–15)

where Np is the number of belt passes. The selection of a metal flat belt can consist of the following steps:   1 Find exp( f ϕ) from geometry and friction   2 Find endurance strength Sf = 15.16(106 )Np−0.412

301, 302 stainless

Sf = Sy∕3

others

  3 Allowable tension Et (F1 ) a = [ Sf − tb = ab (1 − ν2 ) D ]   4 ΔF = 2T∕D   5 F2 = (F1 ) a − ΔF = ab − ΔF (F1 ) a + F2 ab + ab − ΔF ΔF   6 Fi = = = ab − 2 2 2 exp( f ϕ) ΔF   7 bmin = a exp( f ϕ) − 1   8 Choose b > bmin, (F1)a = ab, F2 = ab − ΔF, Fi = ab − ΔF∕2, T = ΔFD∕2   9 Check frictional development f′: f′ =

1 (F1 ) a ln ϕ F2

f′ < f

900      Mechanical Engineering Design

EXAMPLE 17–3 A friction-drive stainless steel metal belt runs over two 4-in metal pulleys ( f = 0.35). The belt thickness is to be 0.003 in. For a life exceeding 106 belt passes with smooth torque (Ks = 1), (a) select the belt if the torque is to be 30 lbf · in, and (b) find the initial tension Fi. Solution (a) From step 1, ϕd = π, therefore exp(0.35π) = 3.00. From step 2,

(Sf ) 106 = 15.16(106 ) (106 ) −0.412 = 51 100 psi

From steps 3, 4, 5, and 6,

28(106 )0.003 (F1 ) a = [ 51 100 − 0.003b = 84.7b (1 − 0.2852 ) 4 ]

ΔF = 2T∕D = 2(30)∕4 = 15 lbf

F2 = (F1 ) a − ΔF = 84.7b − 15

Fi =

(F1 ) a + F2 15 = 84.7b − 2 2

(1)

(2)

(3)

From step 7,

ΔF exp( f ϕ) 15 3.00 = = 0.266 in a exp( f ϕ) − 1 84.7 3.00 − 1

bmin =

Decision From step 8, select an available 0.75-in-wide belt 0.003 in thick. Eq. (1):

(F1 ) a = 84.7(0.75) = 63.5 lbf

Eq. (2):

F2 = 84.7(0.75) − 15 = 48.5 lbf

Eq. (3):

Fi = 63.5 − 15∕2 = 56.0 lbf

From step 9,

f′ =

1 (F1 ) a 1 63.5 ln = ln = 0.0858 π 48.5 ϕ F2

Note f ′ < f, that is, 0.0858 < 0.35.

17–3  V Belts The cross-sectional dimensions of V belts have been standardized by manufacturers, with each section designated by a letter of the alphabet for sizes in inch dimensions. Metric sizes are designated in numbers. Though these have not been included here, the procedure for analyzing and designing them is the same as presented here. Dimensions, minimum sheave diameters, and the horsepower range for each of the lettered sections are listed in Table 17–9. To specify a V belt, give the belt-section letter, followed by the inside circumference in inches (standard circumferences are listed in Table 17–10). For example, B75 is a B-section belt having an inside circumference of 75 in.

Flexible Mechanical Elements     901

Table 17–9  Standard V-Belt Sections Belt Width a, Thickness b, Section in in

a

Minimum Sheave Diameter, in

hp Range, One or More Belts

A

1 2

11 1 32 3.0 4 –10

B

21 32

7 16 5.4

C

7 8

17 32 9.0 15–100

D

114

3 4 13.0

E

112

1

21.6

1–25

50–250

100 and up

Table 17–10  Inside Circumferences of Standard V Belts Section

Circumference, in

A 26, 31, 33, 35, 38, 42, 46, 48, 51, 53, 55, 57, 60, 62, 64, 66, 68, 71, 75, 78, 80, 85, 90, 96, 105, 112, 120, 128 B 35, 38, 42, 46, 48, 51, 53, 55, 57, 60, 62, 64, 65, 66, 68, 71, 75, 78, 79, 81, 83, 85, 90, 93, 97, 100, 103, 105, 112, 120, 128, 131, 136, 144, 158, 173, 180, 195, 210, 240, 270, 300 C 51, 60, 68, 75, 81, 85, 90, 96, 105, 112, 120, 128, 136, 144, 158, 162, 173, 180, 195, 210, 240, 270, 300, 330, 360, 390, 420 D 120, 128, 144, 158, 162, 173, 180, 195, 210, 240, 270, 300, 330, 360, 390, 420, 480, 540, 600, 660 E

180, 195, 210, 240, 270, 300, 330, 360, 390, 420, 480, 540, 600, 660

Table 17–11  Length Conversion Dimensions (Add the listed quantity to the inside circumference to obtain the pitch length in inches). Belt section

A B C D E

Quantity to be added 1.3 1.8 2.9 3.3 4.5

Calculations involving the belt length are usually based on the pitch length. For any given belt section, the pitch length is obtained by adding a quantity to the inside circumference (Tables 17–10 and 17–11). For example, a B75 belt has a pitch length of 76.8 in. Similarly, calculations of the velocity ratios are made using the pitch diameters of the sheaves, and for this reason the stated diameters are usually understood to be the pitch diameters even though they are not always so specified. The groove angle of a sheave, αs, is made somewhat smaller than the belt-section angle, αb. This causes the belt to wedge itself into the groove, thus increasing friction. The exact value of this angle depends on the belt section, the sheave diameter, and the angle of wrap. If it is made too much smaller than the belt, the force required to pull the belt out of the groove as the belt leaves the pulley will be excessive. Optimum values are given in the commercial literature. The minimum sheave diameters have been listed in Table 17–9. For best results, a V belt should be run quite fast: 4000 ft/min is a good speed. Trouble

b αb = 40°

902      Mechanical Engineering Design

may be encountered if the belt runs much faster than 5000 ft/min or much slower than 1000 ft/min. The pitch length Lp and the center-to-center distance C are

Lp = 2C + π(D + d)∕2 + (D − d) 2∕(4C)

2 π π   C = 0.25{[ L p − (D + d) ] + √ [ Lp − (D + d) ] − 2(D − d) 2} 2 2

(17–16a) (17–16b)

where D = pitch diameter of the large sheave and d = pitch diameter of the small sheave. In the case of flat belts, there is virtually no limit to the center-to-center distance. Long center-to-center distances are not recommended for V belts because the excessive vibration of the slack side will shorten the belt life materially. In general, the center-to-center distance should be no greater than three times the sum of the sheave diameters and no less than the diameter of the larger sheave. Link-type V belts have less vibration, because of better balance, and hence may be used with longer centerto-center distances. The basis for power ratings of V belts depends somewhat on the manufacturer; it is not often mentioned quantitatively in vendors' literature but is available from vendors. The basis may be a number of hours, 24 000 h, for example, or a life of 108 or 109 belt passes. Since the number of belts must be an integer, an undersized belt set that is augmented by one belt can be substantially oversized. Table 17–12 gives power ratings of standard V belts. The rating, whether in terms of hours or belt passes, is for a belt running on equal-diameter sheaves (180° of wrap), of moderate length, and transmitting a steady load. Deviations from these laboratory test conditions are acknowledged by multiplicative adjustments. If the tabulated power of a belt for a C-section belt is 9.46 hp for a 12-in-diameter sheave at a peripheral speed of 3000 ft/min (Table 17–12), then, when the belt is used under other conditions, the tabulated value Htab is adjusted as follows:

Ha = K1 K2 Htab

(17–17)

where Ha = allowable power, per belt K1 = angle-of-wrap (ϕ) correction factor, Table 17–13 K2 = belt length correction factor, Table 17–14 The allowable power can be near to Htab, depending upon circumstances. In a V belt the effective coefficient of friction f ′ is f∕sin(αs∕2), which amounts to an augmentation by a factor of about 3 due to the grooves. The effective coefficient of friction f ′ is sometimes tabulated against sheave groove angles of αs = 30°, 34°, and 38°. The corresponding tabulated values are f ′ = 0.50, 0.45, and 0.40, respectively, revealing a belt material-on-metal coefficient of friction of f = 0.13 for each case. The Gates Rubber Company declares its effective coefficient of friction to be 0.5123 for grooves. Thus,

F1 − Fc = exp(0.5123ϕ) F2 − Fc

(17–18)

The design power is given by Hd = Hnom K s nd

(17–19)

Flexible Mechanical Elements     903

Table 17–12  Horsepower Ratings of Standard V Belts Belt Section

Sheave Pitch Diameter, in

Belt Speed, ft/min 1000

2000

3000

4000

5000

A

2.6 0.47 0.62 0.53 0.15 3.0 0.66 1.01 1.12 0.93 0.38 3.4 0.81 1.31 1.57 1.53 1.12 3.8 0.93 1.55 1.92 2.00 1.71 4.2 1.03 1.74 2.20 2.38 2.19 4.6 1.11 1.89 2.44 2.69 2.58 5.0 and up 1.17 2.03 2.64 2.96 2.89

B

4.2 1.07 1.58 1.68 1.26 0.22 4.6 1.27 1.99 2.29 2.08 1.24 5.0 1.44 2.33 2.80 2.76 2.10 5.4 1.59 2.62 3.24 3.34 2.82 5.8 1.72 2.87 3.61 3.85 3.45 6.2 1.82 3.09 3.94 4.28 4.00 6.6 1.92 3.29 4.23 4.67 4.48 7.0 and up 2.01 3.46 4.49 5.01 4.90

C

6.0 1.84 2.66 2.72 1.87 7.0 2.48 3.94 4.64 4.44 3.12 8.0 2.96 4.90 6.09 6.36 5.52 9.0 3.34 5.65 7.21 7.86 7.39 10.0 3.64 6.25 8.11 9.06 8.89 11.0 3.88 6.74 8.84 10.0 10.1 12.0 and up 4.09 7.15 9.46 10.9 11.1

D

10.0 11.0 12.0 13.0 14.0 15.0 16.0 17.0 and up

E

16.0 8.68 14.0 17.5 18.1 15.3 18.0 9.92 16.7 21.2 23.0 21.5 20.0 10.9 18.7 24.2 26.9 26.4 22.0 11.7 20.3 26.6 30.2 30.5 24.0 12.4 21.6 28.6 32.9 33.8 26.0 13.0 22.8 30.3 35.1 36.7 28.0 and up 13.4 23.7 31.8 37.1 39.1

4.14 6.13 6.55 5.09 1.35 5.00 7.83 9.11 8.50 5.62 5.71 9.26 11.2 11.4 9.18 6.31 10.5 13.0 13.8 12.2 6.82 11.5 14.6 15.8 14.8 7.27 12.4 15.9 17.6 17.0 7.66 13.2 17.1 19.2 19.0 8.01 13.9 18.1 20.6 20.7

where Hnom is the nominal power, Ks is the service factor given in Table 17–15, and nd is the design factor. The number of belts, Nb, is usually the next higher integer to Hd∕Ha. That is,

Nb ≥

Hd Ha

Designers work on a per-belt basis.

Nb = 1, 2, 3, . . .

(17–20)

Table 17–13  Angle of Wrap Correction Factor K1 for VV and V-Flat Drives† K1

D−d C

ϕ, deg

VV*

V-Flat

0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90

180 174.3 166.5 162.7 156.9 151.0 145.1 139.0 132.8 126.5

1.00 0.75 0.99 0.76 0.97 0.78 0.96 0.79 0.94 0.80 0.93 0.81 0.91 0.83 0.89 0.84 0.87 0.85 0.85 0.85

1.00 1.10 1.20 1.30 1.40 1.50

120.0 113.3 106.3 98.9 91.1 82.8

0.82 0.82 0.80 0.80 0.77 0.77 0.73 0.73 0.70 0.70 0.65 0.65

*A curve fit for the VV column in terms of ϕ is K1 = 0.143 543 + 0.007 468 ϕ − 0.000 015 052 ϕ2 in the range 90° ≤ ϕ ≤ 180°. † VV drives are when the driver and driven pulleys are both V pulleys. A V-flat drive uses a V pulley, typically the smaller on the driver shaft, and a flat pulley on the driven shaft.

Table 17–14  Belt-Length Correction Factor K*2 Length Factor

A Belts

Nominal Belt Length, in B Belts

C Belts

D Belts

E Belts

0.85 Up to 35 Up to 46 Up to 75 Up to 128 0.90 38–46   48–60   81–96 144–162 Up to 195 0.95 48–55   62–75 105–120 173–210 210–240 1.00 60–75  78–97 128–158 240 270–300 1.05 78–90 105–120 162–195 270–330 330–390 1.10 96–112 128–144 210–240 360–420 420–480 1.15 120 and up 158–180 270–300 480 540–600 1.20 195 and up 330 and up 540 and up 660 *Multiply the rated horsepower per belt by this factor to obtain the corrected horsepower.

Table 17–15  Suggested Service Factors KS for V-Belt Drives Driven Machinery

904

Uniform Light shock Medium shock Heavy shock

Source of Power Normal Torque Characteristic

High or Nonuniform Torque

1.0 to 1.2 1.1 to 1.3 1.2 to 1.4 1.3 to 1.5

1.1 to 1.3 1.2 to 1.4 1.4 to 1.6 1.5 to 1.8

Flexible Mechanical Elements     905

F1

B

+

T

F

A

+ C

F2

+ D

E

(a)

(Fb )1

ΔF/2

T1 F1

T2 ΔF/2

Fi Fc A

Table 17–16  Some V-Belt Parameters*

(Fb ) 2

F2

Fc B

C

D

E

F

A

(b)

Figure 17–14

Belt Section

Kb Kc

A

220 0.561

B

576 0.965

C

1600 1.716

D

5680 3.498

E

10850 5.041

3V

230 0.425

5V

1098 1.217

8V

4830 3.288

*Data courtesy of Gates Rubber Co., Denver, Colo.

V-belt tensions.

The flat-belt tensions shown in Figure 17–12 ignored the tension induced by bending the belt about the pulleys. This is more pronounced with V belts, as shown in Figure 17–14. The centrifugal tension Fc is given by

V 2 Fc = Kc ( 1000 )

(17–21)

where Kc is from Table 17–16. The power that is transmitted per belt is based on ΔF = F1 − F2, where

ΔF =

63 025Hd∕Nb n(d∕2)

(17–22)

then from Equation (17–8) the larger tension F1 is given by

F1 = Fc +

ΔF exp( f′ϕ) exp( f′ϕ) − 1

(17–23)

From the definition of ΔF, the smaller tension F2 is

F2 = F1 − ΔF

(17–24)

From Equation ( j) in Section 17–2

Fi =

F1 + F2 − Fc 2

(17–25)

Ha Nb Hnom Ks

(17–26)

The factor of safety is

nfs =

Durability (life) correlations are complicated by the fact that the bending induces flexural stresses in the belt; the corresponding belt tension that induces the same

906      Mechanical Engineering Design

maximum tensile stress is (Fb)1 at the driving sheave and (Fb)2 at the driven pulley. These equivalent tensions are added to F1 as T1 = F1 + (Fb ) 1 = F1 +

Kb d

T2 = F1 + (Fb ) 2 = F1 +

Kb D

where Kb is given in Table 17–16. The equation for the tension versus pass trade-off used by the Gates Rubber Company is of the form T bNP = K b where NP is the number of passes and b is approximately 11. See Table 17–17. The Miner rule is used to sum damage incurred by the two tension peaks: 1 K −b K −b =( ) + ( ) NP T1 T2 or

K −b K −b −1 NP = [( ) + ( ) ] T1 T2

(17–27)

The lifetime t in hours is given by

t=

NP L p 720V

(17–28)

The constants K and b have their ranges of validity. If NP > 109, report that NP = 109 and t > NPLp∕(720V ) without placing confidence in numerical values beyond the validity interval. See the statement about NP and t near the conclusion of Example 17–4. The analysis of a V-belt drive can consist of the following steps:  1 Find  2 Find  3 Find  4 Find

V, Lp, C, ϕ, and exp(0.5123ϕ) Hd, Ha, and Nb from Hd∕Ha and round up Fc, ΔF, F1, F2, and Fi, and nfs belt life in number of passes, or hours, if possible

Table 17–17  Durability Parameters for Some V-Belt Sections Belt Section

108 to 109 Force Peaks

109 to 1010 Minimum Force Peaks Sheave

K b K b

Diameter, in

A

674 11.089

3.0

B

1193 10.926

5.0

C

2038 11.173

8.5

D

4208 11.105

13.0

E

6061 11.100

21.6

3V

728 12.464 1062 10.153

2.65

5V

1654 12.593 2394 10.283

7.1

8V

3638 12.629 5253 10.319

12.5

Source: M. E. Spotts, Design of Machine Elements, 6th ed. Prentice Hall, Englewood Cliffs, N.J., 1985.

Flexible Mechanical Elements     907

EXAMPLE 17–4 A 10-hp split-phase motor running at 1750 rev/min is used to drive a rotary pump, which operates 24 hours per day. An engineer has specified a 7.4-in small sheave, an 11-in large sheave, and three B112 belts. The service factor of 1.2 was augmented by 0.1 because of the continuous-duty requirement. Analyze the drive and estimate the belt life in passes and hours. Solution For step 1, the peripheral speed V of the belt is

V = π dn∕12 = π(7.4)1750∕12 = 3390 ft/min

Table 17–11:  L p = L + Lc = 112 + 1.8 = 113.8 in π Eq. (17–16b): C = 0.25{[ 113.8 − (11 + 7.4) ] 2 Eq. (17–1):

+ √ [ 113.8 −

2 π (11 + 7.4) ] − 2(11 − 7.4) 2} 2

= 42.4 in ϕd = π − 2 sin−1 (11 − 7.4)∕[2(42.4)] = 3.057 rad

exp[0.5123(3.057)] = 4.788

For step 2, interpolating in Table 17–12 for V = 3390 ft/min gives Htab = 4.693 hp. The wrap angle in degrees is 3.057(180)∕π = 175°. From Table 17–13, K1 = 0.99. From Table 17–14, K2 = 1.05. Thus, from Eq. (17–17), Ha = K1K 2 Htab = 0.99(1.05)4.693 = 4.878 hp Eq. (17–19):

Hd = Hnom K s nd = 10(1.2 + 0.1) (1) = 13 hp

Eq. (17–20):

Nb ≥ Hd ∕Ha = 13∕4.878 = 2.67 → 3

For step 3, from Table 17–16, Kc = 0.965. Thus, from Eq. (17–21), Eq. (17–22):

Fc = 0.965(3390∕1000) 2 = 11.1 lbf ΔF =

63 025(13)∕3 = 42.2 lbf 1750(7.4∕2) 42.2(4.788) = 64.4 lbf 4.788 − 1

Eq. (17–23):

F1 = 11.1 +

Eq. (17–24):

F2 = F1 − ΔF = 64.4 − 42.2 = 22.2 lbf

Eq. (17–25):

Fi =

64.4 + 22.2 − 11.1 = 32.2 lbf 2

Eq. (17–26):

nfs =

Ha Nb 4.878(3) = = 1.13 Hnom Ks 10(1.3)

908      Mechanical Engineering Design

Life: For step 4, from Table 17–16, Kb = 576.

(Fb ) 1 =

Kb 576 = = 77.8 lbf d 7.4

(Fb ) 2 =

576 = 52.4 lbf 11

T1 = F1 + (Fb ) 1 = 64.4 + 77.8 = 142.2 lbf

T2 = F1 + (Fb ) 2 = 64.4 + 52.4 = 116.8 lbf

From Table 17–17, K = 1193 and b = 10.926. Eq. (17–27):

NP = [(

1193 −10.926 1193 −10.926 −1 9 + ( 116.8 ) ] = 11(10 ) passes 142.2 )

Answer Since NP is out of the validity range of Eq. (17–27), life is reported as greater than 109 passes. Then Answer  Eq. (17–28):

t >

109 (113.8) = 46 600 h 720(3390)

17–4  Timing Belts A timing belt is made of a rubberized fabric coated with a nylon fabric, and has steel wire within to take the tension load. It has teeth that fit into grooves cut on the periphery of the pulleys (Figure 17–15). A timing belt does not stretch appreciably or slip and consequently transmits power at a constant angular-velocity ratio. No initial tension is needed. Such belts can operate over a very wide range of speeds, have efficiencies in the range of 97 to 99 percent, require no lubrication, and are quieter than chain drives. There is no chordal-speed variation, as in chain drives (see Section 17–5), and so they are an attractive solution for precision-drive requirements. The steel wire, the tension member of a timing belt, is located at the belt pitch line (Figure 17–15). Thus the pitch length is the same regardless of the thickness of the backing. Figure 17–15

Belt pitch

Timing-belt drive showing portions of the pulley and belt. Note that the pitch diameter of the pulley is greater than the diametral distance across the top lands of the teeth.

Belt pitch line

Pitch circle of pulley

Root diameter Outside diameter

Flexible Mechanical Elements     909

Table 17–18  Standard Pitches of Timing Belts Service

Designation

Pitch p, in

XL

1 5

Light

L

3 8

Heavy

H

1 2

XH

7 8

XXH

114

Extra light

Extra heavy Double extra heavy

The five standard inch-series pitches available are listed in Table 17–18 with their letter designations. Standard pitch lengths are available in sizes from 6 to 180 in. Pulleys come in sizes from 0.60 in pitch diameter up to 35.8 in and with groove numbers from 10 to 120. The design and selection process for timing belts is so similar to that for V belts that the process will not be presented here. As in the case of other belt drives, the manufacturers will provide an ample supply of information and details on sizes and strengths.

17–5  Roller Chain Basic features of chain drives include a constant ratio, since no slippage or creep is involved; long life; and the ability to drive a number of shafts from a single source of power. Roller chains have been standardized as to sizes by the ANSI. Figure 17–16 shows the nomenclature. The pitch is the linear distance between the centers of the rollers. The width is the space between the inner link plates. These chains are manufactured in single, double, triple, and quadruple strands. The dimensions of standard sizes are listed in Table 17–19. Figure 17–17 shows a sprocket driving a chain and rotating in a counterclockwise direction. Denoting the chain pitch by p, the pitch angle by γ, and the pitch diameter of the sprocket by D, from the trigonometry of the figure we see

sin

p∕2 γ = 2 D∕2

or

D=

p sin(γ∕2)

(a)

Figure 17–16

Roller diameter

Portion of a double-strand roller chain. Strand spacing Width

Pitch p

910      Mechanical Engineering Design

Table 17–19  Dimensions of American Standard Roller Chains—Single Strand ANSI Chain Number

Pitch, in (mm)

Width, in (mm)

Minimum Tensile Strength, lbf (N)

25

0.250

0.125

   780

0.09

0.130

0.252

(6.35)

(3.18)

  (3 470)

(1.31)

(3.30)

(6.40)

0.375

0.188

  1 760

0.21

0.200

0.399

(9.52)

(4.76)

  (7 830)

(3.06)

(5.08)

(10.13)

0.500

0.25

  1 500

0.25

0.306

(12.70)

(6.35)

  (6 670)

(3.65)

(7.77)

0.500

0.312

  3 130

0.42

0.312

0.566

(12.70)

(7.94)

  (13 920)

(6.13)

(7.92)

(14.38)

35 41 40 50 60

0.625

0.375

  4 880

(15.88)

(9.52)

  (21 700)

0.750

0.500

  7 030

(19.05) 80 100 120 140 160 180 200 240

(12.7)

1.000

0.625

(25.40)

(15.88)

1.250

0.750

(31.75)

(19.05)

1.500

1.000

(38.10)

(25.40)

1.750

1.000

(44.45)

(25.40)

2.000

1.250

(50.80)

(31.75)

2.250

1.406

(57.15)

(35.71)

2.500

1.500

(63.50)

(38.10)

3.00 (76.70)

  (31 300)

Average Weight, lbf/ft (N/m)

0.400

0.713

(10.16)

(18.11)

1.00

0.469

0.897

(14.6)

(11.91)

(22.78)

1.71

0.625

1.153

(25.0)

(15.87)

(29.29)

2.58

0.750

1.409

(37.7)

(19.05)

(35.76)

3.87

0.875

1.789

(56.5)

(22.22)

(45.44)

4.95

1.000

1.924

(72.2)

(25.40)

(48.87)

6.61

1.125

2.305

(96.5)

(28.57)

(58.55)

9.06

1.406

2.592

(132.2)

(35.71)

(65.84)

  28 000 (124 500)   38 000 (169 000)   50 000 (222 000)   63 000 (280 000)   78 000

1.875 (47.63)

0.69

  19 500   (86 700)

MultipleStrand Spacing, in (mm)

(10.1)

  12 500   (55 600)

Roller Diameter, in (mm)

10.96

1.562

2.817

(347 000)

(159.9)

(39.67)

(71.55)

112 000

16.4

1.875

3.458

(47.62)

(87.83)

(498 000)

(239)

Source: Compiled from ANSI B29.1-1975.

Since γ = 360°∕N, where N is the number of sprocket teeth, Equation (a) can be written

D=

p sin(180°∕N)

(17–29)

The angle γ∕2, through which the link swings as it enters contact, is called the angle of articulation. It can be seen that the magnitude of this angle is a function of the number of teeth. Rotation of the link through this angle causes impact between

Flexible Mechanical Elements     911 p e

Figure 17–17

A B γ/2 γ

Variable D

the rollers and the sprocket teeth and also wear in the chain joint. Since the life of a properly selected drive is a function of the wear and the surface fatigue strength of the rollers, it is important to reduce the angle of articulation as much as possible. The number of sprocket teeth also affects the velocity ratio during the rotation through the pitch angle γ. At the position shown in Figure 17–17, the chain AB is tangent to the pitch circle of the sprocket. However, when the sprocket has turned an angle of γ∕2, the chain line AB moves closer to the center of rotation of the sprocket. This means that the chain line AB is moving up and down, and that the lever arm varies with rotation through the pitch angle, all resulting in an uneven chain exit velocity. You can think of the sprocket as a polygon in which the exit velocity of the chain depends upon whether the exit is from a corner, or from a flat of the polygon. Of course, the same effect occurs when the chain first enters into engagement with the sprocket. The chain velocity V is defined as the number of feet coming off the sprocket per unit time. Thus the chain velocity in feet per minute is

V=

Npn 12

(17–30)

where N = number of sprocket teeth p = chain pitch, in n = sprocket speed, rev/min The maximum exit velocity of the chain is

vmax =

π np π Dn = 12 12 sin(γ∕2)

(b)

where Equation (a) has been substituted for the pitch diameter D. The minimum exit velocity occurs at a diameter d, smaller than D. Using the geometry of Figure 17–17, we find γ d = D cos 2

(c)

Thus the minimum exit velocity is

vmin =

π dn π np cos(γ∕2) = 12 12 sin(γ∕2)

(d)

Engagement of a chain and sprocket.

912      Mechanical Engineering Design

Chordal speed variation, %

Figure 17–18 20

10

0

0

10

20

30

40

Number of teeth, N

Now substituting γ∕2 = 180°∕N and employing Equations (17–30), (b), and (d), we find the speed variation to be

ΔV vmax − vmin π 1 1 = = [ − V V N sin(180°∕N) tan(180°∕N) ]

(17–31)

This is called the chordal speed variation and is plotted in Figure 17–18. When chain drives are used to synchronize precision components or processes, due consideration must be given to these variations. For example, if a chain drive synchronized the cutting of photographic film with the forward drive of the film, the lengths of the cut sheets of film might vary too much because of this chordal speed variation. Such variations can also cause vibrations within the system. Although a large number of teeth is considered desirable for the driving sprocket, in the usual case it is advantageous to obtain as small a sprocket as possible, and this requires one with a small number of teeth. For smooth operation at moderate and high speeds it is considered good practice to use a driving sprocket with at least 17 teeth; 19 or 21 will, of course, give a better life expectancy with less chain noise. Where space limitations are severe or for very slow speeds, smaller tooth numbers may be used by sacrificing the life expectancy of the chain. Driven sprockets are not made in standard sizes over 120 teeth, because the pitch elongation will eventually cause the chain to "ride" high long before the chain is worn out. The most successful drives have velocity ratios up to 6:1, but higher ratios may be used at the sacrifice of chain life. Roller chains seldom fail because they lack tensile strength; they more often fail because they have been subjected to a great many hours of service. Actual failure may be due either to wear of the rollers on the pins or to fatigue of the surfaces of the rollers. Roller-chain manufacturers have compiled tables that give the horsepower capacity corresponding to a life expectancy of 15 kh for various sprocket speeds. These capacities are tabulated in Table 17–20 for 17-tooth sprockets. Table 17–21 displays available tooth counts on sprockets of one supplier. Table 17–22 lists the tooth correction factors for other than 17 teeth. Table 17–23 shows the multiple-strand factors K2. The capacities of chains are based on the following: ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙

15 000 h at full load Single strand ANSI proportions Service factor of unity 100 pitches in length Recommended lubrication Elongation maximum of 3 percent Horizontal shafts Two 17-tooth sprockets

Flexible Mechanical Elements     913

The fatigue strength of link plates governs capacity at lower speeds. The American Chain Association (ACA) publication Chains for Power Transmission and Materials Handling (1982) gives, for single-strand chain, the nominal power H1, link-plate limited, as (3−0.07p) H1 = 0.004N 11.08 n0.9 1 p

hp

(17–32)

and the nominal power H2, roller-limited, as

H2 =

0.8 1000Kr N 1.5 1 p

n1.5 1

hp

(17–33)

where N1 = number of teeth in the smaller sprocket n1 = sprocket speed, rev/min p = pitch of the chain, in Kr = 29 for chain numbers 25, 35; 3.4 for chain 41; and 17 for chains 40–240

Table 17–20  Rated Horsepower Capacity of Single-Strand Single-Pitch Roller Chain for a 17-Tooth Sprocket Sprocket Speed, rev/min

ANSI Chain Number 25

35

40

41

50

60

 50

0.05

0.16

0.37

0.20

 0.72

 1.24

100

0.09

0.29

0.69

0.38

 1.34

 2.31

150

0.13*

0.41*

0.99*

0.55*

 1.92*

 3.32

200

0.16*

0.54*

1.29

0.71

 2.50

 4.30

300

0.23

0.78

1.85

1.02

 3.61

 6.20

400

0.30*

1.01*

2.40

1.32

 4.67

 8.03

500

0.37

1.24

2.93

1.61

 5.71

 9.81

600

0.44*

1.46*

3.45*

1.90*

 6.72*

11.6

700

0.50

1.68

3.97

2.18

 7.73

13.3

800

0.56*

1.89*

4.48*

2.46*

 8.71*

15.0

900

0.62

2.10

4.98

2.74

 9.69

16.7

1000

0.68*

2.31*

5.48

3.01

10.7

18.3

1200

0.81

2.73

6.45

3.29

12.6

21.6

1400

0.93*

3.13*

7.41

2.61

14.4

18.1

1600

1.05*

3.53*

8.36

2.14

12.8

14.8

1800

1.16

3.93

8.96

1.79

10.7

12.4

2000

1.27*

4.32*

7.72*

1.52*

 9.23*

10.6

2500

1.56

5.28

5.51*

1.10*

 6.58*

 7.57

3000

1.84

5.64

4.17

0.83

 4.98

 5.76

Type A

Type B

Type C

*Estimated from ANSI tables by linear interpolation. Note: Type A—manual or drip lubrication; type B—bath or disk lubrication; type C—oil-stream lubrication. Source: Compiled from ANSI B29.1-1975 information only section, and from B29.9-1958.

(Continued)

Table 17–20  Rated Horsepower Capacity of Single-Strand Single-Pitch Roller Chain for a 17-Tooth Sprocket (Continued) Sprocket Speed, rev/min

80

120

140

160

180

200

240

 2.88

 5.52

 9.33

 14.4

 20.9

 28.9

 38.4

 61.8

 5.38

10.3

17.4

 26.9

 39.1

 54.0

 71.6

115

150

 7.75

14.8

25.1

 38.8

 56.3

 77.7

103

166

200

10.0

19.2

32.5

 50.3

 72.9

101

134

215

300

14.5

27.7

46.8

 72.4

105

145

193

310

400

18.7

35.9

60.6

 93.8

136

188

249

359

22.9

43.9

74.1

115

166

204

222

 0

27.0

51.7

87.3

127

141

155

169

700

31.0

59.4

89.0

101

112

123

  0

800

35.0

63.0

72.8

 82.4

 91.7

101

900

39.9

52.8

61.0

   69.1

 76.8

 84.4

1000

37.7

45.0

52.1

 59.0

 65.6

 72.1

1200

28.7

34.3

39.6

 44.9

 49.9

  0

1400

22.7

27.2

31.5

 35.6

  0

1600

18.6

22.3

25.8

 0

1800

15.6

18.7

21.6

2000

13.3

15.9

 0

2500

 9.56

 0.40

3000

 7.25

 0

500 600

Type A

100

100

Type B

 50

ANSI Chain Number

Type C

Type C′

Note: Type A—manual or drip lubrication; type B—bath or disk lubrication; type C—oil-stream lubrication; type C′—type C, but this is a galling region; submit design to manufacturer for evaluation.

Table 17–21  Single-Strand Sprocket Tooth Counts Available from One Supplier* No.

Available Sprocket Tooth Counts

 25  35  41  40  50  60  80 100 120 140 160 180 200 240

8-30, 32, 34, 35, 36, 40, 42, 45, 48, 54, 60, 64, 65, 70, 72, 76, 80, 84, 90, 95, 96, 102, 112, 120 4-45, 48, 52, 54, 60, 64, 65, 68, 70, 72, 76, 80, 84, 90, 95, 96, 102, 112, 120 6-60, 64, 65, 68, 70, 72, 76, 80, 84, 90, 95, 96, 102, 112, 120 8-60, 64, 65, 68, 70, 72, 76, 80, 84, 90, 95, 96, 102, 112, 120 8-60, 64, 65, 68, 70, 72, 76, 80, 84, 90, 95, 96, 102, 112, 120 8-60, 62, 63, 64, 65, 66, 67, 68, 70, 72, 76, 80, 84, 90, 95, 96, 102, 112, 120 8-60, 64, 65, 68, 70, 72, 76, 78, 80, 84, 90, 95, 96, 102, 112, 120 8-60, 64, 65, 67, 68, 70, 72, 74, 76, 80, 84, 90, 95, 96, 102, 112, 120 9-45, 46, 48, 50, 52, 54, 55, 57, 60, 64, 65, 67, 68, 70, 72, 76, 80, 84, 90, 96, 102, 112, 120 9-28, 30, 31, 32, 33, 34, 35, 36, 37, 39, 40, 42, 43, 45, 48, 54, 60, 64, 65, 68, 70, 72, 76, 80, 84, 96 8-30, 32–36, 38, 40, 45, 46, 50, 52, 53, 54, 56, 57, 60, 62, 63, 64, 65, 66, 68, 70, 72, 73, 80, 84, 96 13-25, 28, 35, 39, 40, 45, 54, 60 9-30, 32, 33, 35, 36, 39, 40, 42, 44, 45, 48, 50, 51, 54, 56, 58, 59, 60, 63, 64, 65, 68, 70, 72 9-30, 32, 35, 36, 40, 44, 45, 48, 52, 54, 60

*Morse Chain Company, Ithaca, NY, Type B hub sprockets. 914

Flexible Mechanical Elements     915

Table 17–22  Tooth Correction Factors, K1 Number of Teeth on Driving Sprocket

K1 Pre-extreme Horsepower

K1 Post-extreme Horsepower

11

0.62

0.52

12

0.69

0.59

13

0.75

0.67

14

0.81

0.75

15

0.87

0.83

16

0.94

0.91

17

1.00

1.00

18

1.06

1.09

19

1.13

1.18

20

1.19

Number of Strands

K2

1

1.0

2

1.7

3

2.5

4

3.3

5

3.9

6

4.6

8

6.0

1.28 1.08

(N1/17)1.5

(N1/17)

N

Table 17–23  Multiple-Strand Factors, K2

The constant 0.004 in Equation (17–32) becomes 0.0022 for no. 41 lightweight chain. The nominal horsepower in Table 17–20 is Hnom = min(H1, H2). For example, for N1 = 17, n1 = 1000 rev/min, no. 40 chain with p = 0.5 in, from Equation (17–32),

H1 = 0.004(17) 1.081000 0.9 0.5[3−0.07(0.5)] = 5.48 hp

From Equation (17–33),

H2 =

1000(17)171.5 (0.50.8 ) 10001.5

= 21.64 hp

The tabulated value in Table 17–20 is Htab = min(5.48, 21.64) = 5.48 hp. It is preferable to have an odd number of teeth on the driving sprocket (17, 19, . . .) and an even number of pitches in the chain to avoid a special link. The approximate length of the chain L in pitches is

L 2C N1 + N2 (N2 − N1 ) 2 ≈ + + p p 2 4π2C∕p

(17–34)

The center-to-center distance C is given by

C=

p N2 − N1 2 2 −A + A − 8 √ ( 2π ) ] 4[

(17–35)

N1 + N2 L − p 2

(17–36)

where

A=

The allowable power Ha is given by

Ha = K1 K2 Htab

where K1 = correction factor for tooth number other than 17 (Table 17–22) K2 = strand correction (Table 17–23)

(17–37)

916      Mechanical Engineering Design

The horsepower that must be transmitted Hd is given by Hd = Hnom Ks nd

(17–38)

where Ks is a service factor to account for nonuniform loads, and nd is a design factor. Equation (17–32) is the basis of the pre-extreme power entries (vertical entries) of Table 17–20, and the chain power is limited by link-plate fatigue. Equation (17–33) is the basis for the post-extreme power entries of these tables, and the chain power performance is limited by impact fatigue. The entries are for chains of 100 pitch length and 17-tooth sprocket. For a deviation from this H2 = 1000 [ K r (

N1 1.5 0.8 L p 0.4 15 000 0.4 p ( n1 ) 100 ) ( h ) ]

(17–39)

where Lp is the chain length in pitches and h is the chain life in hours. Viewed from a deviation viewpoint, Equation (17–39) can be written as a trade-off equation in the following form: H 2.5 2 h

N 3.75 1 Lp

= constant

(17–40)

If tooth-correction factor K1 is used, then omit the term N13.75. In Equation (17–40) one would expect the h∕Lp term because doubling the hours can require doubling the chain length, other conditions constant, for the same number of cycles. Our experience with contact stresses leads us to expect a load (tension) life relation of the form F aL = constant. In the more complex circumstance of rollerbushing impact, the Diamond Chain Company has identified a = 2.5. The maximum speed (rev/min) for a chain drive is limited by galling between the pin and the bushing. Tests suggest

1∕(1.59 log p+1.873) 82.5 n1 ≤ 1000 [ 7.95 p (1.0278) N1 (1.323) F∕1000 ]

rev/min

where F is the chain tension in lbf.

EXAMPLE 17–5 Select drive components for a 2:1 reduction, 90-hp input at 300 rev/min, moderate shock, an abnormally long 18-hour day, poor lubrication, cold temperatures, dirty surroundings, short drive C∕p = 25. Solution Function: Hnom = 90 hp, n1 = 300 rev/min, C∕p = 25 Design factor: Choose nd = 1.5 Service factor: Choose Ks = 1.3 for moderate shock Sprocket teeth: N1 = 17 teeth, N2 = 34 teeth, K1 = 1, K2 = 1, 1.7, 2.5, 3.3 Chain number of strands: From Equations (17–37) and (17–38), with Ha = Hd,

Htab =

nd Ks Hnom 1.5(1.3)90 176 = = K1K2 (1)K2 K2

Flexible Mechanical Elements     917

Form a table: Number of Strands 1 2 3 4

176∕K2 (Table 17–23) 176∕1 176∕1.7 176∕2.5 176∕3.3

= = = =

176 104 70.4 53.3

Chain Number (Table 17–20)

Lubrication Type

200 160 140 140

C′ C B B

Decision 3 strands of number 140 chain (from Table 17–20, Htab is 72.4 hp). Number of pitches in the chain:

L 2C N1 + N2 (N2 − N1 ) 2 = + + p p 2 4π2C∕p = 2(25) +

17 + 34 (34 − 17) 2 + = 75.79 pitches 2 4π2 (25)

Decision Use 76 pitches. Then L∕p = 76. Identify the center-to-center distance: From Equations (17–35) and (17–36),

A=

N1 + N2 L 17 + 34 − = − 76 = −50.5 p 2 2

C=

p N2 − N1 2 2 −A + A − 8 √ ( 2π ) ] 4[

=

p 34 − 17 2 2 50.5 + 50.5 − 8 √ ( 2π ) ] = 25.104p 4[

For a 140 chain, p = 1.75 in. Thus, C = 25.104p = 25.104(1.75) = 43.93 in Lubrication: Type B Comment: This is operating on the pre-extreme portion of the power, so durability estimates other than 15 000 h are not available. Given the poor operating conditions, life will be much shorter. Lubrication of roller chains is essential in order to obtain a long and trouble-free life. Either a drip feed or a shallow bath in the lubricant is satisfactory. A medium or light mineral oil, without additives, should be used. Except for unusual conditions, heavy oils and greases are not recommended, because they are too viscous to enter the small clearances in the chain parts.

17–6  Wire Rope Wire rope is made with two types of winding, as shown in Figure 17–19. The regular lay, which is the accepted standard, has the wire twisted in one direction to form the strands, and the strands twisted in the opposite direction to form the rope. In the completed rope the visible wires are approximately parallel to the axis of the rope. Regular-lay ropes do not kink or untwist and are easy to handle.

918      Mechanical Engineering Design

Figure 17–19 Types of wire rope; both lays are available in either right or left hand.

(a) Regular lay

(c) Section of 6 × 7 rope (b) Lang lay

Lang-lay ropes have the wires in the strand and the strands in the rope twisted in the same direction, and hence the outer wires run diagonally across the axis of the rope. Lang-lay ropes are more resistant to abrasive wear and failure due to fatigue than are regular-lay ropes, but they are more likely to kink and untwist. Standard ropes are made with a hemp core, which supports and lubricates the strands. When the rope is subjected to heat, either a steel center or a wire-strand center must be used. Wire rope is designated as, for example, a 118 -in 6 × 7 haulage rope. The first figure is the diameter of the rope (Figure 17–19c). The second and third figures are the number of strands and the number of wires in each strand, respectively. Table 17–24 lists some of the various ropes that are available, together with their Table 17–24  Wire-Rope Data

Rope

Weight per Foot, lbf 2

Minimum Sheave Diameter, in

Standard Sizes d, in

Size of Outer Wires

Modulus of Elasticity,* Mpsi

Strength,† kpsi

Monitor steel Plow steel Mild plow steel

d∕9 d∕9 d∕9

14 14 14

100  88  76

d∕13–d∕16 d∕13–d∕16 d∕13–d∕16

12 12 12

106  93  80

Material

6 × 7 haulage

1.50d

42d

1 1 4 –12

6 × 19 standard hoisting

1.60d2

26d–34d

1 3 4 –24

Monitor steel Plow steel Mild plow steel

6 × 37 special flexible

1.55d2

18d

1 1 4 –32

Monitor steel Plow steel

d∕22 d∕22

11 11

100  88

8 × 19 extra flexible

1.45d2

21d–26d

1 1 4 –12

Monitor steel Plow steel

d∕15–d∕19 d∕15–d∕19

10 10

 92  80

7 × 7 aircraft

1.70d2

1 3 16 – 8

Corrosion-resistant steel Carbon steel

124

124

7 × 9 aircraft

1.75d2

1 3 8 –18

Corrosion-resistant steel Carbon steel

135

143

19-wire aircraft

2.15d2

1 5 32 –16

Corrosion-resistant steel Carbon steel

165

165

*The modulus of elasticity is only approximate; it is affected by the loads on the rope and, in general, increases with the life of the rope. † The strength is based on the nominal area of the rope. The figures given are only approximate and are based on 1-in rope sizes and 14 -in aircraft-cable sizes. Source: Compiled from American Steel and Wire Company Handbook.

Flexible Mechanical Elements     919

characteristics and properties. The area of the metal in standard hoisting and haulage rope is about Am = 0.38d2. When a wire rope passes around a sheave, there is a certain amount of readjustment of the elements. Each of the wires and strands must slide on several others, and presumably some individual bending takes place. It is probable that in this complex action there exists some stress concentration. The stress in one of the wires of a rope passing around a sheave may be calculated as follows. From solid mechanics, we have

M=

EI ρ

and

M=

σI c

(a)

where the quantities have their usual meaning. Eliminating M and solving for the stress gives

σ=

Ec ρ

(b)

For the radius of curvature ρ, we can substitute the sheave radius D∕2. Also, c = dw∕2, where dw is the wire diameter. These substitutions give

σ = Er

dw D

(c)

where Er is the modulus of elasticity of the rope, not the wire. To understand this equation, observe that the individual wire makes a corkscrew figure in space and if you pull on it to determine E it will stretch or give more than its native E would suggest. Therefore E is still the modulus of elasticity of the wire, but in its peculiar configuration as part of the rope, its modulus is smaller. For this reason we say that Er in Equation (c) is the modulus of elasticity of the rope, not the wire, recognizing that one can quibble over the name used. Equation (c) gives the tensile stress σ in the outer wires. The sheave diameter is represented by D. This equation reveals the importance of using a largediameter sheave. The suggested minimum sheave diameters in Table 17–24 are based on a D∕dw ratio of 400. If possible, the sheaves should be designed for a larger ratio. For elevators and mine hoists, D∕dw is usually taken from 800 to 1000. If the ratio is less than 200, heavy loads will often cause a permanent set in the rope. A wire rope tension giving the same tensile stress as the sheave bending is called the equivalent bending load Fb, given by

Fb = σAm =

Er d w A m D

(17–41)

A wire rope may fail because the static load exceeds the ultimate strength of the rope. Failure of this nature is generally not the fault of the designer, but rather that of the operator in permitting the rope to be subjected to loads for which it was not designed. The first consideration in selecting a wire rope is to determine the static load. This load is composed of the following items: ∙ ∙ ∙ ∙

The known or dead weight Additional loads caused by sudden stops or starts Shock loads Sheave-bearing friction

920      Mechanical Engineering Design

When these loads are summed, the total can be compared with the ultimate strength of the rope to find a factor of safety. However, the ultimate strength used in this determination must be reduced by the strength loss that occurs when the rope passes over a curved surface such as a stationary sheave or a pin; see Figure 17–20. For an average operation, use a factor of safety of 5. Factors of safety up to 8 or 9 are used if there is danger to human life and for very critical situations. Table 17–25 lists minimum factors of safety for a variety of design situations. Here, the factor of safety is defined as

n=

Fu Ft

where Fu is the ultimate wire load and Ft is the largest working tension. Once you have made a tentative selection of a rope based upon static strength, the next consideration is to ensure that the wear life of the rope and the sheave or Figure 17–20

40 Percent strength loss

Percent strength loss due to different D∕d ratios; derived from standard test data for 6 × 19 and 6 × 17 class ropes. (Materials provided by the Wire Rope Technical Board (WRTB), Wire Rope Users Manual Third Edition, Second printing. Reprinted by permission.)

50

30 20 10 0

0

10

20

30

40

D /d ratio

Table 17–25  Minimum Factors of Safety for Wire Rope* Track cables

3.2

Guys 3.5 Mine shafts, ft:    Up to 500   1000–2000   2000–3000   Over 3000

8.0 7.0 6.0 5.0

Hoisting 5.0 Haulage 6.0 Cranes and derricks

6.0

Electric hoists

7.0

Hand elevators

5.0

Private elevators

7.5

Hand dumbwaiter

4.5

Grain elevators

7.5

Passenger elevators, ft/min: 50 300 800 1200 1500

7.60 9.20 11.25 11.80 11.90

Freight elevators, ft/min: 50 300 800 1200 1500

6.65 8.20 10.00 10.50 10.55

Powered dumbwaiters, ft/min: 50 300 500

*Use of these factors does not preclude a fatigue failure. Source: Compiled from a variety of sources, including ANSI A17.1-1978.

4.8 6.6 8.0

Flexible Mechanical Elements     921

sheaves meets certain requirements. When a loaded rope is bent over a sheave, the rope stretches like a spring, rubs against the sheave, and causes wear of both the rope and the sheave. The amount of wear that occurs depends upon the pressure of the rope in the sheave groove. This pressure is called the bearing pressure; a good estimate of its magnitude is given by

P=

2F dD

(17–42)

where F = tensile force on rope d = rope diameter D = sheave diameter The allowable pressures given in Table 17–26 are to be used only as a rough guide; they may not prevent a fatigue failure or severe wear. They are presented here because they represent past practice and furnish a starting point in design. A fatigue diagram not unlike an S-N diagram can be obtained for wire rope. Such a diagram is shown in Figure 17–21. Here the ordinate is the pressure-strength ratio p∕Su, and Su is the ultimate tensile strength of the wire. The abscissa is the number of bends that occur in the total life of the rope. The curve implies that a wire rope has a fatigue limit; but this is not true at all. A wire rope that is used over sheaves will eventually fail in fatigue or in wear. However, the graph does show that the rope will have a long life if the ratio p∕Su is less than 0.001. Substitution of this ratio in Equation (17–42) gives

Su =

2000F dD

(17–43)

where Su is the ultimate strength of the wire, not the rope, and the units of Su are related to the units of F. This interesting equation contains the wire strength, the load, Table 17–26  Maximum Allowable Bearing Pressures of Ropes on Sheaves (in psi)

Sheave Material

Wooda

Cast Ironb

Cast Steelc

Chilled Cast Ironsd

Manganese Steele

   6 × 7

150

300

 550

 650

1470

   6 × 19

250

480

 900

1100

2400

   6 × 37

300

585

1075

1325

3000

   8 × 19

350

680

1260

1550

3500

   6 × 7

165

350

 600

 715

1650

   6 × 19

275

550

1000

1210

2750

   6 × 37

330

660

1180

1450

3300

Rope Regular lay:

Lang lay:

a

On end grain of beech, hickory, or gum. For HB (min.) = 125. c 30–40 carbon; HB (min.) = 160. d Use only with uniform surface hardness. e For high speeds with balanced sheaves having ground surfaces. Source: Wire Rope Users Manual, AISI, 1979. b

922      Mechanical Engineering Design

Figure 17–21

6 Pressure-strength ratio, 1000 p /Su

Experimentally determined relation between the fatigue life of wire rope and the sheave pressure.

7

5 4 3

2

24

6 × 37

6 × 19

1 0

6 × 12 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Number of bends to failure, in millions

the rope diameter, and the sheave diameter—all four variables in a single equation! Dividing both sides of Equation (17–42) by the ultimate strength of the wires Su and solving for F gives

Ff =

(p∕Su )Su dD 2

(17–44)

where Ff is interpreted as the allowable fatigue tension as the wire is flexed a number of times corresponding to p∕Su selected from Figure 17–21 for a particular rope and life expectancy. The factor of safety can be defined in fatigue as

nf =

Ff − Fb Ft

(17–45)

where Ff is the rope tension strength under flexing and Ft is the tension at the place where the rope is flexing. Unfortunately, the designer often has vendor information that tabulates ultimate rope tension and gives no ultimate-strength Su information concerning the wires from which the rope is made. Some guidance in strength of individual wires is Improved plow steel (monitor) Plow steel Mild plow steel

240 < Su < 280 kpsi 210 < Su < 240 kpsi 180 < Su < 210 kpsi

In wire-rope usage, the factor of safety is defined for static loading as ns =

Fu − Fb Ft

(17–46)

where Fb is the rope tension that would induce the same outer-wire stress as that given by Equation (c). Be careful when comparing recommended static factors of safety to Equation (17–46), as ns is sometimes defined as Fu∕Ft. The factor of safety in fatigue loading can be defined as in Equation (17–45), or by using a static analysis and compensating with a large factor of safety applicable to static loading, as in Table 17–25. When using factors of safety expressed in codes, standards, corporate

Flexible Mechanical Elements     923

design manuals, or wire-rope manufacturers' recommendations or from the literature, be sure to ascertain upon which basis the factor of safety is to be evaluated, and proceed accordingly. If the rope is made of plow steel, the wires are probably hard-drawn AISI 1070 or 1080 carbon steel. Referring to Table 10–3, we see that this lies somewhere between hard-drawn spring wire and music wire. But the constants m and A needed to solve Equation (10–14), for Su are lacking. Practicing engineers who desire to solve Equation (17–43) should determine the wire strength Su for the rope under consideration by unraveling enough wire to test for the Brinell hardness. Then Su can be found using Equation (2–21). Fatigue failure in wire rope is not sudden, as in solid bodies, but progressive, and shows as the breaking of an outside wire. This means that the beginning of fatigue can be detected by periodic routine inspection. Figure 17–22 is another graph showing the gain in life to be obtained by using large D∕d ratios. In view of the fact that the life of wire rope used over sheaves is only finite, it is extremely important that the designer specify and insist that periodic inspection, lubrication, and maintenance procedures be carried out during the life of the rope. Table 17–27 gives useful properties of some wire ropes. For a mine-hoist problem we present here the working equations, some repeated from the preceding presentation. The wire rope tension Ft due to load and acceleration/ deceleration is W a Ft = ( + wl) (1 + ) g m

(17–47)

Figure 17–22

100

Service-life curve based on bending and tensile stresses only. This curve shows that the life corresponding to D∕d = 48 is twice that of D∕d = 33. (Materials provided by the Wire Rope Technical Board (WRTB), Wire Rope Users Manual Third Edition, Second printing. Reprinted by permission.)

Relative service life, %

80

60

40

20

0

0

10

20

30

40

50

60

D /d ratio

Table 17–27  Some Useful Properties of 6 × 7, 6 × 19, and 6 × 37 Wire Ropes Weight per Foot Including Core w, lbf/ft

Wire Rope

Weight per Foot w, lbf/ft

6×7

1.50d2

6 × 19

1.60d2

1.76d2

6 × 37

2

2

1.55d

1.71d

Minimum Sheave Diameter D, in

Better Sheave Diameter D, in

Diameter of Wires dw, in

Area of Metal Am, in2

Rope Young's Modulus Er, psi

42d

72d

0.111d

0.38d2

13 × 106

30d

45d

0.067d

0.40d2

12 × 106

0.048d

2

12 × 106

18d

27d

0.40d

924      Mechanical Engineering Design

where

W m w l a g

= = = = = =

weight at the end of the rope (cage and load), lbf number of wire ropes supporting the load weight/foot of the wire rope, lbf/ft maximum suspended length of rope, ft maximum acceleration/deceleration experienced, ft/s2 acceleration of gravity, ft/s2

The fatigue tensile strength in pounds for a specified life is

( p∕Su ) Su Dd 2 specified life, from Figure 17–21 ultimate tensile strength of the wires, psi sheave or winch drum diameter, in nominal wire rope size, in Ff =

where (p∕Su) Su D d

= = = =

(17–44)

The equivalent bending load Fb is where Er dw Am D

Er d w Am (17–41) D Young's modulus for the wire rope, Table 17–24 or 17–27, psi diameter of the wires, in metal cross-sectional area, Table 17–27, in2 sheave or winch drum diameter, in Fb =

= = = =

The static factor of safety ns is Fu − Fb Ft sometimes defined as Fu∕Ft. The fatigue factor of safety nf is Ff − Fb nf = Ft

ns =

(17–46)

(17–45)

EXAMPLE 17–6 Given a 6 × 19 monitor steel (Su = 240 kpsi) wire rope. (a) Develop the expressions for rope tension Ft, fatigue tension Ff, equivalent bending tensions Fb, and fatigue factor of safety nf for a 531.5-ft, 1-ton cage-and-load mine hoist with a starting acceleration of 2 ft/s2 as depicted in Fig. 17–23. The sheave diameter is 72 in. (b) Using the expressions developed in part (a), examine the variation in factor of safety nf for various wire rope diameters d and number of supporting ropes m. Solution (a) Rope tension Ft from Eq. (17–47), using Table 17–24 for w, is given by Answer

W a 2000 2 + 1.60d 2 (531.5) 1 + Ft = ( + wl) (1 + ) = [ ]( g m m 32.2 ) =

2124 + 903d 2 m

Flexible Mechanical Elements     925

Figure 17–23 Geometry of the mine hoist of Example 17–6.

72 in

Ft

531.5 ft

wl = 531.5(1.6)d 2 lbf

a = 2 ft/s2

W = 2000 lbf

From Fig. 17–21, use p∕Su = 0.0014. Fatigue tension Ff from Eq. (17–44) is given by Ff =

Answer

( p∕Su )Su Dd 0.0014(240 000)72d = = 12 096d 2 2

Equivalent bending tension Fb from Eq. (17–41) and Table 17–27 is given by Fb =

Answer

Er d w Am 12(106 ) 0.067d(0.40d 2 ) = = 4467d 3 D 72

Factor of safety nf in fatigue from Eq. (17–45) is given by nf =

Answer

Ff − Fb Ft

=

12 096d − 4467d 3 2124∕m + 903d 2

(b) Using a spreadsheet program, form a table as follows: nf d

m=1

m=2

m=3

m=4

0.25

1.355

2.641

3.865

5.029

0.375

1.910

3.617

5.150

6.536

0.500

2.336

4.263

5.879

7.254

0.625

2.612

4.573

6.099

7.331

0.750

2.731

4.578

5.911

6.918

0.875

2.696

4.330

5.425

6.210

1.000

2.520

3.882

4.736

5.320

Wire rope sizes are discrete, as is the number of supporting ropes. Note that for each m the factor of safety exhibits a maximum. Predictably the largest factor of safety increases with m. If the required factor of safety were to be 6, only three or four ropes could meet the requirement. The sizes are different: 58 -in ropes with three ropes or 38 -in ropes with four ropes. The costs include not only the wires, but the grooved winch drums.

926      Mechanical Engineering Design

17–7  Flexible Shafts One of the greatest limitations of the solid shaft is that it cannot transmit motion or power around corners. It is therefore necessary to resort to belts, chains, or gears, together with bearings and the supporting framework associated with them. The flexible shaft may often be an economical solution to the problem of transmitting motion around corners. In addition to the elimination of costly parts, its use may reduce noise considerably. There are two main types of flexible shafts: the power-drive shaft for the transmission of power in a single direction, and the remote-control or manual-control shaft for the transmission of motion in either direction. The construction of a flexible shaft is shown in Figure 17–24. The cable is made by winding several layers of wire around a central core. For the power-drive shaft, rotation should be in a direction such that the outer layer is wound up. Remotecontrol cables have a different lay of the wires forming the cable, with more wires in each layer, so that the torsional deflection is approximately the same for either direction of rotation. Flexible shafts are rated by specifying the torque corresponding to various radii of curvature of the casing. A 15-in radius of curvature, for example, will give from 2 to 5 times more torque capacity than a 7-in radius. When flexible shafts are used in a drive in which gears are also used, the gears should be placed so that the flexible shaft runs at as high a speed as possible. This permits the transmission of the maximum amount of horsepower. Figure 17–24 Flexible shaft: (a) construction details; (b) a variety of configurations. (Courtesy of S. S. White Technologies, Inc.)

Mandrel

First Layer (4 Wires) (a)

(b)

Last Layer (7 Wires)

Flexible Mechanical Elements     927

PROBLEMS 17–1

Example 17–2 resulted in selection of a 10-in-wide A-3 polyamide flat belt. Show that the value of F1 restoring f to 0.80 is

F1 =

(ΔF + Fc ) exp f ϕ − Fc exp f ϕ − 1

and compare the initial tensions.

17–2 A 6-in-wide polyamide F-1 flat belt is used to connect a 2-in-diameter pulley to drive

a larger pulley with an angular velocity ratio of 0.5. The center-to-center distance is 9 ft. The angular speed of the small pulley is 1750 rev/min as it delivers 2 hp. The service is such that a service factor Ks of 1.25 is appropriate. (a) Find Fc, Fi, (F1)a, and F2, assuming operation at the maximum tension limit. (b) Find Ha, nfs, and belt length. (c) Find the dip.

17–3 Perspective and insight can be gained by doubling all geometric dimensions and

observing the effect on problem parameters. Take the drive of Problem 17–2, double the dimensions, and compare.

17–4 A flat-belt drive is to consist of two 4-ft-diameter cast-iron pulleys spaced 16 ft apart. Select a belt type to transmit 60 hp at a pulley speed of 380 rev/min. Use a service factor of 1.1 and a design factor of 1.0.

17–5 In solving problems and examining examples, you probably have noticed some recurring forms:

w = 12γbt = (12γt)b = a1b

(F1 ) a = Fa bCpCv = (FaCpCv ) b = a0 b

Fc =

a1b wV 2 V 2 = a 2b = ( g 32.174 60 )

(F1 ) a − F2 = 2T∕d = 33 000Hd∕V = 33 000Hnom Ks n d∕V

F2 = (F1 ) a − [(F1 ) a − F2] = a 0 b − 2T∕d

f ϕ = ln

(a 0 − a2 )b (F1 ) a − Fc = ln F2 − Fc (a0 − a2 ) b − 2T∕d

Show that

b=

33 000Hd exp ( f ϕ) 1 a0 − a2 V exp ( f ϕ) − 1

17–6 Return to Example 17–1 and complete the following.

(a) What is the minimum initial tension, Fi, that would put the drive as built at the point of slip? (b) With the tension from part a, find the belt width b that exhibits nfs = nd = 1.1. (c) For the belt width from part b find the corresponding (F1)a, Fc, Fi, F2, power, and nfs, assuming operation at the tension limit. (d) What have you learned?

928      Mechanical Engineering Design

17–7 Take the drive of Example 17–1 and double the belt width. Compare Fc, Fi, (F1)a, F2, Ha, nfs, and dip, assuming operation at the tension limit.

17–8 Belted pulleys place loads on shafts, inducing bending and loads on bearings. Examine

Figure 17–7 and develop an expression for the load the belt places on the pulley, and then apply it to Example 17–2.

17–9 The line shaft illustrated in the figure is used to transmit power from an electric motor

by means of flat-belt drives to various machines. Pulley A is driven by a vertical belt from the motor pulley. A belt from pulley B drives a machine tool at an angle of 70° from the vertical and at a center-to-center distance of 9 ft. Another belt from pulley C drives a grinder at a center-to-center distance of 11 ft. Pulley C has a double width to permit belt shifting as shown in Figure 17–4. The belt from pulley D drives a dustextractor fan whose axis is located horizontally 8 ft from the axis of the lineshaft. Additional data are

Speed, rev/min

Power, hp

Lineshaft Pulley

Diameter, in

Machine tool

400

12.5

B

16

Grinder

300

 4.5

C

14

Dust extractor

500

 8.0

D

18

Machine

A

D

From D

A

8 ft m

From

B

C

(Courtesy of Dr. Ahmed F. Abdel Azim, Zagazig University, Cairo.)

C

Fro

Problem 17–9

B

70°

60°

10 ft

Motor pulley: Dia. = 12 in Speed = 900 rev/min

The power requirements, listed above, account for the overall efficiencies of the equipment. The two line-shaft bearings are mounted on hangers suspended from two overhead wide-flange beams. Select the belt types and sizes for each of the four drives. Make provision for replacing belts from time to time because of wear or permanent stretch.

Flexible Mechanical Elements     929

17–10 Two shafts 20 ft apart, with axes in the same horizontal plane, are to be connected

with a flat belt in which the driving pulley, powered by a six-pole squirrel-cage induction motor with a 100 brake hp rating at 1140 rev/min, drives the second shaft at half its angular speed. The driven shaft drives light-shock machinery loads. Select a flat belt.

17–11 The mechanical efficiency of a flat-belt drive is approximately 98 percent. Because

of its high value, the efficiency is often neglected. If a designer should choose to include it, where would he or she insert it in the flat-belt protocol?

17–12 In metal belts, the centrifugal tension Fc is ignored as negligible. Convince yourself that this is a reasonable problem simplification.

17–13 A designer has to select a metal-belt drive to transmit a power of Hnom under circumstances where a service factor of Ks and a design factor of nd are appropriate. The design goal becomes Hd = HnomKsnd. Use Equation (17–8) with negligible centrifugal force to show that the minimum belt width is given by

bmin =

exp f ϕ 1 33 000Hd ) exp f ϕ − 1 a( V

where a is the constant from (F1)a = ab.

17–14 Design a friction metal flat-belt drive to connect a 1-hp, four-pole squirrel-cage motor turning at 1750 rev/min to a shaft 15 in away, running at half speed. The circumstances are such that a service factor of 1.2 and a design factor of 1.05 are appropriate. The life goal is 106 belt passes, f = 0.35, and the environmental considerations require a stainless steel belt.

17–15 A beryllium-copper metal flat belt with Sf = 56.67 kpsi is to transmit 5 hp at 1125 rev/min with a life goal of 106 belt passes between two shafts 20 in apart whose centerlines are in a horizontal plane. The coefficient of friction between belt and pulley is 0.32. The conditions are such that a service factor of 1.25 and a design factor of 1.1 are appropriate. The driven shaft rotates at one-third the motor-pulley speed. Specify your belt, pulley sizes, and initial tension at installation.

17–16 For the conditions of Problem 17–15 use a 1095 plain carbon-steel heat-treated belt.

Conditions at the driving pulley hub require a pulley outside diameter of 3 in or more. Specify your belt, pulley sizes, and initial tension at installation.

17–17 A single Gates Rubber V belt is to be selected to deliver engine power to the wheel-

drive transmission of a riding tractor. A 5-hp single-cylinder engine is used. At most, 60 percent of this power is transmitted to the belt. The driving sheave has a diameter of 6.2 in, the driven, 12.0 in. The belt selected should be as close to a 92-in pitch length as possible. The engine speed is governor-controlled to a maximum of 3100 rev/min. Select a satisfactory belt and assess the factor of safety and the belt life in passes.

17–18 Two B85 V belts are used in a drive composed of a 5.4-in driving sheave, rotating at

1200 rev/min, and a 16-in driven sheave. Find the power capacity of the drive based on a service factor of 1.25, and find the center-to-center distance.

930      Mechanical Engineering Design

17–19 A 60-hp four-cylinder internal combustion engine is used to drive a medium-shock

brick-making machine under a schedule of two shifts per day. The drive consists of two 26-in sheaves spaced about 12 ft apart, with a sheave speed of 400 rev/min. Select a Gates Rubber V-belt arrangement. Find the factor of safety, and estimate the life in passes and hours.

17–20 A reciprocating air compressor has a 5-ft-diameter flywheel 14 in wide, and it oper-

ates at 170 rev/min. An eight-pole squirrel-cage induction motor has nameplate data 50 bhp at 875 rev/min. (a) Design a Gates Rubber V-belt drive to transmit power from the motor to the compressor flywheel. (b) Can cutting the V-belt grooves in the flywheel be avoided by using a V-flat  drive?

17–21 The geometric implications of a V-flat drive are interesting.

(a) If the earth's equator was an inextensible string, snug to the spherical earth, and you spliced 6 ft of string into the equatorial cord and arranged it to be concentric to the equator, how far off the ground is the string? (b) Using the solution to part a, formulate the modifications to the expressions for mG, ϕd and ϕD, Lp, and C for a V-flat drive. (c) As a result of this exercise, how would you revise your solution to part b of Problem 17–20?

17–22 A 2-hp electric motor running at 1720 rev/min is to drive a blower at a speed of

240 rev/min. Select a V-belt drive for this application and specify standard V belts, sheave sizes, and the resulting center-to-center distance. The motor size limits the center distance to at least 22 in.

17–23 The standard roller-chain number indicates the chain pitch in inches, construction proportions, series, and number of strands as follows: 10 0 H-2

two strands

heavy series

standard proportions

pitch is 10∕8 in

This convention makes the pitch directly readable from the chain number. In Example 17–5 ascertain the pitch from the selected chain number and confirm from Table 17–19.

17–24 Equate Equations (17–32) and (17–33) to find the rotating speed n1 at which the power equates and marks the division between the premaximum and the postmaximum power domains. (a) Show that

n1 = [

0.25(106 ) Kr N 10.42 p

(2.2−0.07p)

]

1∕2.4

Flexible Mechanical Elements     931

(b) Find the speed n1 for a no. 60 chain, p = 0.75 in, N1 = 17, Kr = 17, and confirm from Table 17–20. (c) At which speeds is Equation (17–40) applicable?

17–25 A double-strand no. 60 roller chain is used to transmit power between a 13-tooth driv-

ing sprocket rotating at 300 rev/min and a 52-tooth driven sprocket. (a) What is the allowable horsepower of this drive? (b) Estimate the center-to-center distance if the chain length is 82 pitches. (c) Estimate the torque and bending force on the driving shaft by the chain if the actual horsepower transmitted is 30 percent less than the corrected (allowable) power.

17–26 A four-strand no. 40 roller chain transmits power from a 21-tooth driving sprocket to

an 84-tooth driven sprocket. The angular speed of the driving sprocket is 2000 rev/min. A life of 20 000 hours is desired. (a) Estimate the chain length if the center-to-center distance is to be about 20 in. (b) Estimate the allowable horsepower from Table 17–20, adjusting with K1 and K2, and neglecting the deviations from the table's life and length assumptions. (c) Estimate the allowable horsepower from Equations (17–32) and (17–33) (which are the source of Table 17–20). Compare answers with part (b). (d) Estimate the allowable horsepower H2 from Equation (17–39), which is a broader version of Equation (17–33) that allows for adjustment for life and length. Compare the answer with part (c). (e) Estimate the average chain velocity in feet/minute. ( f ) Estimate the torque that is transmitted from the driving sprocket when operating at the maximum allowable power from part (d). (g) Estimate the tension in the chain when operating at the maximum allowable power from part (d).

17–27 A 700 rev/min 25-hp squirrel-cage induction motor is to drive a two-cylinder recipro-

cating pump, out-of-doors under a shed. A service factor Ks of 1.5 and a design factor of 1.1 are appropriate. The pump speed is 140 rev/min. Select a suitable chain and sprocket sizes.

17–28 A centrifugal pump is driven by a 50-hp synchronous motor at a speed of 1800 rev/min.

The pump is to operate at 900 rev/min. Despite the speed, the load is smooth (Ks = 1.2). For a design factor of 1.1 specify a chain and sprockets that will realize a 50 000-h life goal. Let the sprockets be 19T and 38T.

17–29 A mine hoist uses a 2-in 6 × 19 monitor-steel wire rope. The rope is used to haul

loads of 4 tons from the shaft 480 ft deep. The drum has a diameter of 6 ft, the sheaves are of good-quality cast steel, and the smallest is 3 ft in diameter. (a) Using a maximum hoisting speed of 1200 ft/min and a maximum acceleration of 2 ft/s2, estimate the stresses in the rope. (b) Estimate the various factors of safety.

17–30 A temporary construction elevator is to be designed to carry workers and materials

to a height of 90 ft. The maximum estimated load to be hoisted is 5000 lbf at a velocity not to exceed 2 ft/s. For minimum sheave diameters and acceleration of 4  ft/s2, specify the number of ropes required if the 12 -in plow-steel 6 × 19 hoisting strand is used.

932      Mechanical Engineering Design

17–31 A 2000-ft mine hoist operates with a 72-in drum using 6 × 19 monitor-steel wire

rope. The cage and load weigh 8000 lbf, and the cage is subjected to an acceleration of 2  ft/s2 when starting. (a) For a single-strand hoist how does the factor of safety nf = Ff ∕Ft, neglecting bending, vary with the choice of rope diameter? (b) For four supporting strands of wire rope attached to the cage, how does the factor of safety vary with the choice of rope diameter?

17–32 Generalize the results of Problem 17–31 by representing the factor of safety n as ad (b∕m) + cd 2

nf =

where m is the number of ropes supporting the cage, and a, b, and c are constants. Show that the optimal diameter is d* = [b∕(mc)]1∕2 and the corresponding maximum attainable factor of safety is n*f = a[m∕(bc)]1∕2∕2.

17–33 From your results in Problem 17–32, show that to meet a fatigue factor of safety n1 the optimal solution is

m=

4bcn1 a2

ropes

having a diameter of d=

a 2cn1

Solve Problem 17–31 if a factor of safety of 2 is required. Show what to do in order to accommodate to the necessary discreteness in the rope diameter d and the number of ropes m.

17–34 For Problem 17–29 estimate the elongation of the rope if a 7000-lbf loaded mine cart is placed on the cage which weighs 1000 lbf. The results of Problem 4–7 may be useful.

Computer Programs In approaching the ensuing computer problems, the following suggestions may be helpful: ∙ Decide whether an analysis program or a design program would be more useful. In problems as simple as these, you will find the programs similar. For maximum instructional benefit, try the design problem. ∙ Creating a design program without a figure of merit precludes ranking alternative designs but does not hinder the attainment of satisfactory designs. Your instructor can provide the class design library with commercial catalogs, which not only have price information but define available sizes. ∙ Quantitative understanding and logic of interrelations are required for programming. Difficulty in programming is a signal to you and your instructor to increase your understanding. The following programs can be accomplished in 100 to 500 lines of code.

Flexible Mechanical Elements     933

∙ Make programs interactive and user-friendly. ∙ Let the computer do what it can do best; the user should do what a human can do  best. ∙ Assume the user has a copy of the text and can respond to prompts for information. ∙ If interpolating in a table is in order, solicit table entries in the neighborhood, and let the computer crunch the numbers. ∙ In decision steps, allow the user to make the necessary decision, even if it is undesirable. This allows learning of consequences and the use of the program for analysis. ∙ Display a lot of information in the summary. Show the decision set used up-front for user perspective. ∙ When a summary is complete, adequacy assessment can be accomplished with ease, so consider adding this feature.

17–35 Your experience with Problems 17–1 through 17–11 has placed you in a position to

write an interactive computer program to design/select flat-belt drive components. A possible decision set is A Priori Decisions ∙ Function: Hnom, rev/min, velocity ratio, approximate C ∙ Design factor: nd ∙ Initial tension maintenance: catenary ∙ Belt material: t, dmin, allowable tension, density, f ∙ Drive geometry: d, D ∙ Belt thickness: t (in material decision) Design Decisions ∙ Belt width: b

17–36 Problems 17–12 through 17–16 have given you some experience with flat metal

friction belts, indicating that a computer program could be helpful in the design/ selection process. A possible decision set is A Priori Decisions ∙ Function: Hnom, rev/min, velocity ratio approximate C ∙ Design factor: nd ∙ Belt material: Sy, E, ν, dmin ∙ Drive geometry: d, D ∙ Belt thickness: t Design Decisions ∙ Belt width: b ∙ Length of belt (often standard loop periphery)

934      Mechanical Engineering Design

17–37 Problems 17–17 through 17–22 have given you enough experience with V belts to convince you that a computer program would be helpful in the design/selection of V-belt drive components. Write such a program.

17–38 Experience with Problems 17–23 through 17–28 can suggest an interactive computer program to help in the design/selection process of roller-chain elements. A possible decision set is A Priori Decisions ∙ Function: power, speed, space, Ks, life goal ∙ Design factor: nd ∙ Sprocket tooth counts: N1, N2, K1, K2 Design Decisions ∙ Chain number ∙ Strand count ∙ Lubrication system ∙ Chain length in pitches (center-to-center distance for reference)

18

Power Transmission Case Study

©hanoiphotography/123RF

Chapter Outline 18–1  Design Sequence for Power Transmission  937 18–2

Power and Torque Requirements   938

18–3

Gear Specification   938

18–4

Shaft Layout   945

18–5

Force Analysis   947

18–6

Shaft Material Selection   947

18–7

Shaft Design for Stress   948

18–8

Shaft Design for Deflection   948

18–9

Bearing Selection   949

18–10

Key and Retaining Ring Selection   950

18–11

Final Analysis   953

935

936      Mechanical Engineering Design

Transmission of power from a source, such as an engine or motor, through a machine to an output actuation is one of the most common machine tasks. An efficient means of transmitting power is through rotary motion of a shaft that is supported by bearings. Gears, belt pulleys, or chain sprockets may be incorporated to provide for torque and speed changes between shafts. Most shafts are cylindrical (solid or hollow), and include stepped diameters with shoulders to accommodate the positioning and support of bearings, gears, etc. The design of a system to transmit power requires attention to the design and selection of individual components (gears, bearings, shafts, etc.). However, as is often the case in design, these components are not independent. For example, in order to design the shafts for stress and deflection, it is necessary to know the applied forces. If the forces are transmitted through gears, it is necessary to know the gear specifications in order to determine the forces that will be transmitted to the shafts. But stock gears come with certain bore sizes, requiring knowledge of the necessary shaft diameters. It is no surprise that the design process is interdependent and iterative, but where should a designer start? The nature of machine design textbooks is to focus on each component separately. This chapter will focus on an overview of a power transmission system design, demonstrating how to incorporate the details of each component into an overall design process. A typical two-stage gear reduction such as shown in Figure 18–1 will be assumed for this discussion. The design sequence is similar for variations of this particular transmission system. The following outline will help clarify a logical design sequence. Discussion of how each part of the outline affects the overall design process will be given in sequence in this chapter. Details on the specifics for designing and selecting major components are covered in separate chapters, particularly Chapter 7 on shaft design, Chapter 11 on bearing selection, and Chapters 13 and 14 on gear specification. A complete case study is presented as a specific vehicle to demonstrate the process. Figure 18–1

5

A compound reverted gear train.

2

2

5

Y

4

4 3

3

CASE STUDY EXAMPLE 18–1, PART 1 PROBLEM SPECIFICATION Section 1–18 presents the background for this case study involving a speed reducer. A two-stage, compound reverted gear train such as shown in Figure 18–1 will be designed. In this chapter, the design of the intermediate shaft and its components is presented, taking into account the other shafts as

Power Transmission Case Study     937

necessary. A subset of the pertinent design specifications that will be needed for this part of the design are given here. Power to be delivered: 20 hp Input speed: 1750 rpm Output speed: 82–88 rev/min Usually low shock levels, occasional moderate shock Input and output shafts extend 4 in outside gearbox Maximum gearbox size: 14-in × 14-in base, 22-in height Output shaft and input shaft in-line Gear and bearing life > 12 000 hours; infinite shaft life

18–1  Design Sequence for Power Transmission There is not a precise sequence of steps for any design process. By nature, design is an iterative process in which it is necessary to make some tentative choices, and to build a skeleton of a design, and to determine which parts of the design are critical. However, much time can be saved by understanding the dependencies between the parts of the problem, allowing the designer to know what parts will be affected by any given change. In this section, only an outline is presented, with a short explanation of each step. Further details will be discussed in the following sections. ∙ Power and torque requirements. Power considerations should be addressed first, as this will determine the overall sizing needs for the entire system. Any necessary speed or torque ratio from input to output must be determined before addressing gear/pulley sizing. ∙ Gear specification. Necessary gear ratios and torque transmission issues can now be addressed with selection of appropriate gears. Note that a full force analysis of the shafts is not yet needed, as only the transmitted loads are required to specify the gears. ∙ Shaft layout. The general layout of the shafts, including axial location of gears and bearings, must now be specified. Decisions on how to transmit the torque from the gears to each shaft need to be made (keys, splines, etc.), as well as how to hold gears and bearings in place (retaining rings, press fits, nuts, etc.). However, it is not necessary at this point to size these elements, since their standard sizes allow estimation of stress-concentration factors. ∙ Force analysis. Once the gear/pulley diameters are known, and the axial locations of the gears and bearings are known, the free-body, shear force, and bending moment diagrams for the shafts can be produced. Forces at the bearings can be determined. ∙ Shaft material selection. Since fatigue design depends so heavily on the material choice, it is usually easier to make a reasonable material selection first, then check for satisfactory results. ∙ Shaft design for stress ( fatigue and static). At this point, a stress design of each shaft should look very similar to a typical design problem from the shaft chapter (Chapter 7). Shear force and bending moment diagrams are known, critical locations can be predicted, approximate stress concentrations can be used, and estimates for shaft diameters can be determined.

938      Mechanical Engineering Design

∙ Shaft design for deflection. Since deflection analysis is dependent on the entire shaft geometry, it is saved until this point. With all shaft geometry now estimated, the critical deflections at the bearing and gear locations can be checked by analysis. ∙ Bearing selection. Specific bearings from a catalog may now be chosen to match the estimated shaft diameters. The diameters can be adjusted slightly as necessary to match the catalog specifications. ∙ Key and retaining ring selection. With shaft diameters settling in to stable values, appropriate keys and retaining rings can be specified in standard sizes. This should make little change in the overall design if reasonable stress-concentration factors were assumed in previous steps. ∙ Final analysis. Once everything has been specified, iterated, and adjusted as necessary for any specific part of the task, a complete analysis from start to finish will provide a final check and specific safety factors for the actual system.

18–2  Power and Torque Requirements Power transmission systems will typically be specified by a power capacity, for example, a 40-horsepower gearbox. This rating specifies the combination of torque and speed that the unit can endure. Remember that, in the ideal case, power in equals power out, so that we can refer to the power being the same throughout the system. In reality, there are small losses due to factors like friction in the bearings and gears. In many transmission systems, the losses in the rolling bearings will be negligible. Gears have a reasonably high efficiency, with about 1 to 2 percent power loss in a pair of meshed gears. Thus, in the double-reduction gearbox in Figure 18–1, with two pairs of meshed gears the output power is likely to be about 2 to 4 percent less than the input power. Since this is a small loss, it is common to speak of simply the power of the system, rather than input power and output power. Flat belts and timing belts have efficiencies typically in the mid to upper 90 percent range. V belts and worm gears have efficiencies that may dip much lower, requiring a distinction between the necessary input power to obtain a desired output power. Torque, on the other hand, is typically not constant throughout a transmission system. Remember that power equals the product of torque and speed. Since power in = power out, we know that for a gear train

H = Tiωi = Toωo

(18–1)

With a constant power, a gear ratio to decrease the angular velocity will simultaneously increase torque. The gear ratio, or train value, for the gear train is

e = ωo∕ωi = Ti∕To

(18–2)

A typical power transmission design problem will specify the desired power capacity, along with either the input and output angular velocities, or the input and output torques. There will usually be a tolerance specified for the output values. After the specific gears are specified, the actual output values can be determined.

18–3  Gear Specification With the gear train value known, the next step is to determine appropriate gears. As a rough guideline, a train value of up to 10 to 1 can be obtained with one pair of gears. Greater ratios can be obtained by compounding additional pairs of gears (see Section 13–13). The compound reverted gear train in Figure 18–1 can obtain a train value of up to 100 to 1.

Power Transmission Case Study     939

Since numbers of teeth on gears must be integers, it is best to design with teeth numbers rather than diameters. See Examples 13–3, 13–4, and 13–5 for details on designing appropriate numbers of teeth to satisfy the gear train value and any necessary geometry condition, such as in-line condition of input and output shaft. Care should be taken at this point to find the best combination of teeth numbers to minimize the overall package size. If the train value only needs to be approximate, use this flexibility to try different options of tooth numbers to minimize the package size. A difference of one tooth on the smallest gear can result in a significant increase in size of the overall package. If designing for large production quantities, gears can be purchased in large enough quantities that it is not necessary to worry about preferred sizes. For small lot production, consideration should be given to the trade-offs between smaller gearbox size and extra cost for odd gear sizes that are difficult to purchase off the shelf. If stock gears are to be used, their availability in prescribed numbers of teeth with anticipated diametral pitch should be checked at this time. If necessary, iterate the design for numbers of teeth that are available.

CASE STUDY EXAMPLE 18–1, PART 2 SPEED, TORQUE, AND GEAR RATIOS Continue the case study by determining appropriate tooth counts to reduce the input speed of ωi = 1750 rev/min to an output speed within the range

82 rev/min < ωo < 88 rev/min

Once final tooth counts are specified, determine values of (a) Speeds for the intermediate and output shafts (b) Torques for the input, intermediate and output shafts, to transmit 20 hp.

Solution Use the notation for gear numbers from Figure 18–1. Choose mean value for initial design, ω5 = 85 rev/min. ω5 85 1 e= = =  ω2 1750 20.59

Equation (18–2)

For a compound reverted gear train, N2 N4 1 e= =  20.59 N3 N5

Equation (13–30)

For smallest package size, let both stages be the same reduction. Also, by making the two stages identical, the in-line condition on the input and output shaft will automatically be satisfied. N 2 N4 1 1 = =√ = N3 N5 20.59 4.54

940      Mechanical Engineering Design

For this ratio, assuming a pressure angle of ϕ = 20°, the minimum number of teeth from Equation (13–11) is N=

2(1)

[1 + 2(4.54) ]sin2 20o

[ 4.54 +

Thus,

√4.542 + [1 + 2(4.54) ]sin2 20o ] = 15.6 = 16 teeth

N2 = N4 = 16 teeth N3 = 4.54(N2 ) = 72.64

Try rounding down and check if ω5 is within limits. ω5 = (

16 16 (1750) = 86.42 rev/min 72 )( 72 )

Acceptable

Proceed with N2 = N4 = 16 teeth N3 = N5 = 72 teeth e=(

16 16 1 = 72 )( 72 ) 20.25

ω5 = 86.42 rev/min ω3 = ω4 = (

16 (1750) = 388.9 rev/min 72 )

To determine the torques, return to the power relationship, H = T2 ω2 = T5 ω5 

T2 = H∕ω2 = (

Equation (18–1)

20 hp ft-lbf/s 1 rev s 550 60 1750 rev/min )( hp )( 2π rad )( min )

T2 = 60.0 lbf · ft T3 = T2

ω2 1750 = 60.0 = 270 lbf · ft ω3 388.9

T5 = T2

ω2 1750 = 60.0 = 1215 lbf · ft ω5 86.42

If a maximum size for the gearbox has been specified in the problem specification, a minimum diametral pitch (maximum tooth size) can be estimated at this point by writing an expression for the gearbox size in terms of gear diameters, and converting to numbers of teeth through the diametral pitch. For example, from Figure 18–1, the overall height of the gearbox is Y = d3 + d2∕2 + d5∕2 + 2∕P + clearances + wall thicknesses where the 2∕P term accounts for the addendum height of the teeth on gears 3 and 5 that extend beyond the pitch diameters. Substituting di = Ni∕P gives Y = N3∕P + N2∕(2P) + N5∕(2P) + 2∕P + clearances + wall thicknesses

Power Transmission Case Study     941

Solving this for P, we find

P = (N3 + N2∕2 + N5∕2 + 2)∕(Y − clearances − wall thicknesses)

(18–3)

This is the minimum value that can be used for diametral pitch, and therefore the maximum tooth size, to stay within the overall gearbox constraint. It should be rounded up to the next standard diametral pitch, which reduces the maximum tooth size. The AGMA approach, as described in Chapter 14, for both bending and contact stress should be applied next to determine suitable gear parameters. The primary design parameters to be specified by the designer include material, diametral pitch, and face width. A recommended procedure is to start with an estimated diametral pitch. This allows determination of gear diameters (d = N∕P), pitch-line velocities [Equation (13–34)], and transmitted loads [Equation (13–35) or (13–36)]. Typical spur gears are available with face widths from 3 to 5 times the circular pitch p. Using an average of 4, a first estimate can be made for face width F = 4p = 4π∕P. Alternatively, the designer can simply perform a quick search of online gear catalogs to find available face widths for the diametral pitch and number of teeth. Next, the AGMA equations in Chapter 14 can be used to determine appropriate material choices to provide desired safety factors. It is generally most efficient to attempt to analyze the most critical gear first, as it will determine the limiting values of diametral pitch and material strength. Usually, the critical gear will be the smaller gear, on the high-torque (low-speed) end of the gearbox. If the required material strengths are too high, such that they are either too expensive or not available, iteration with a smaller diametral pitch (larger tooth) will help. Of course, this will increase the overall gearbox size. Often the excessive stress will be in one of the small gears. Rather than increase the tooth size for all gears, it is sometimes better to reconsider the design of tooth counts, shifting more of the gear ratio to the pair of gears with less stress, and less ratio to the pair of gears with the excessive stress. This will allow the offending gear to have more teeth and therefore larger diameter, decreasing its stress. If contact stress turns out to be more limiting than bending stress, consider gear materials that have been heat treated or case hardened to increase the surface strength. Adjustments can be made to the diametral pitch if necessary to achieve a good balance of size, material, and cost. If the stresses are all much lower than the material strengths, a larger diametral pitch is in order, which will reduce the size of the gears and the gearbox. Everything up to this point should be iterated until acceptable results are obtained, as this portion of the design process can usually be accomplished independently from the next stages of the process. The designer should be satisfied with the gear selection before proceeding to the shaft. Selection of specific gears from catalogs at this point will be helpful in later stages, particularly in knowing overall width, bore size, recommended shoulder support, and maximum fillet radius.

CASE STUDY EXAMPLE 18–1, PART 3 GEAR SPECIFICATION Continue the case study by specifying appropriate gears, including pitch diameter, diametral pitch, face width, and material. Achieve safety factors of at least 1.2 for wear and bending.

942      Mechanical Engineering Design

Solution

Estimate the minimum diametral pitch for overall gearbox height = 22 in. From Equation (18–3) and Figure 18–1, N2 N5 (N3 + 2 + 2 + 2) Pmin = (Y − clearances − wall thickness)

Allow 1.5 in for clearances and wall thickness: 16 72 (72 + 2 + 2 + 2) Pmin = = 5.76 teeth/in (22 − 1.5)

Start with P = 6 teeth/in d2 = d4 = N2∕P = 16∕6 = 2.67 in d3 = d5 = 72∕6 = 12.0 in

Shaft speeds were previously determined to be

ω2 = 1750 rev/min

ω3 = ω4 = 388.9 rev/min

ω5 = 86.4 rev/min

Get pitch-line velocities and transmitted loads for later use.

V23 =

πd2ω2 π(2.67) (1750) = = 1223 ft /min 12 12

V45 =

πd5ω5 = 271.5 ft /min 12



Equation (13–34)

t W23 = 33 000

H 20 = 33 000( = 540.0 lbf  V23 1223 )

Equation (13–35)

t W45 = 33 000

H 20 = 33 000( = 2431 lbf V45 271.5 )

Start with gear 4, since it is the smallest gear, transmitting the largest load. It will likely be critical. Start with wear by contact stress, since it is often the limiting factor.

Gear 4 Wear

I=

cos 20° sin 20° 4.5 = 0.1315 ( 4.5 + 1 ) 2(1)

For Kv, assume Qv = 7. B = 0.731, A = 65.1

Kv = (

65.1 + √271.5 ) 65.1

0.731

= 1.18

Equation (14–23) Equation (14–29) Equation (14–27)

Face width F is typically from 3 to 5 times circular pitch. Try

π π F = 4( ) = 4( ) = 2.09 in. P 6

Since gear specifications are readily available on the Internet, we might as well check for commonly available face widths. On www.globalspec.com, entering P = 6 teeth/in and d = 2.67 in, stock spur gears from several sources have face widths of 1.5 in or 2.0 in. These are also available for the meshing gear 5 with d = 12 in. Choose F = 2.0 in.

Power Transmission Case Study     943

For Km,

Cpf = 0.0624



Equation (14–32)

Cmc = 1 uncrowned teeth



Equation (14–31)

Cpm = 1 straddle-mounted



Equation (14–33)

Cma = 0.15 commercial enclosed unit

Equation (14–34)

Ce = 1



Equation (14–35)

Km = 1.21



Equation (14–30)

Cp = 2300



Table 14–8

K o = Ks = Cf = 1

σc = 2300 √

2431(1.18) (1.21) = 161 700 psi 2.67(2) (0.1315)

Equation (14–16)

Get factors for σc,all. For life factor ZN, get number of cycles for specified life of 12 000 h. L 4 = (12 000 h)(60

Z N = 0.9

K R = KT = CH = 1

min rev 389 = 2.8 × 108 rev h )( min )



Figure 14–15

For a design factor of 1.2, σc,all = Sc ZN∕SH = σc Sc =



Equation (14–18)

SH σc 1.2(161 700) = = 215 600 psi ZN 0.9

From Table 14–6 this strength is achievable with Grade 2 carburized and hardened with Sc = 225 000 psi. To find the achieved factor of safety, nc = σc,all∕σc with SH = 1. The factor of safety for wear of gear 4 is nc =

σc,all σc

=

Sc Z N 225 000(0.9) = = 1.25 σc 161 700

Gear 4 Bending J = 0.27

Figure 14–6

KB = 1

Everything else is the same as before.

σ = Wt Kv

Pd K m 6 1.21 = (2431) (1.18)( )(  F J 2 0.27 )

Equation (14–15)

σ = 38 570 psi YN = 0.9

Figure 14–14

944      Mechanical Engineering Design

Using Grade 2 carburized and hardened, same as chosen for wear, find St = 65 000 psi (Table 14–3). σall = StYN = 58 500 psi

The factor of safety for bending of gear 4 is n=

σall 58 500 = = 1.52 σ 38 570

Gear 5 Bending and Wear

Everything is the same as for gear 4, except J, YN, and ZN. J = 0.41Figure 14–6 L5 = (12 000 h) (60 min/h) (86.4 rev/min) = 6.2 × 107 rev YN = 0.97

Figure 14–14

ZN = 1.0

Figure 14–15

σc = 2300 √

2431(1.18) (1.21) = 161 700 psi 2.67(2) (0.1315)

6 1.21 σ = (2431) (1.18)( )( = 25 400 psi 2 0.41 )

Choose Grade 2 carburized and hardened, the same as gear 4 σc,all

nc =

σc

=

225 000 = 1.39 161 700

σall 65 000(0.97) = = 2.48 σ 25 400

n=

Gear 2 Wear Gears 2 and 3 are evaluated similarly. Only selected results are shown. Kv = 1.37

Try F = 1.5 in, since the loading is less on gears 2 and 3. Km = 1.19

All other factors are the same as those for gear 4. σc = 2300 √

(539.7) (1.37) (1.19) = 94 000 psi 2.67(1.5) (0.1315)

L 2 = (12 000 h) (60 min/h) (1750 rev/min) = 1.26 × 109 rev

Try grade 1 flame-hardened, Sc = 170 000 psi nc =

σc,all σc

=

170 000(0.8) = 1.40 94 000

Gear 2 Bending J = 0.27

YN = 0.88

σ = 539.7(1.37)

(6) (1.19) = 13 040 psi (1.5) (0.27)

ZN = 0.8

Power Transmission Case Study     945

n=

σall 45 000(0.88) = = 3.04 σ 13 040

Gear 3 Wear and Bending

J = 0.41

σc = 2300√

YN = 0.9

ZN = 0.9

(539.7) (1.37) (1.19) = 94 000 psi 2.67(1.5) (0.1315)

σ = 539.7(1.37)

(6) (1.19) = 8584 psi 1.5(0.41)

Try Grade 1 steel, through-hardened to 300 HB. From Figure 14–2, St = 36 000 psi and from Figure 14–5, Sc = 126 000 psi. 126 000(0.9) = 1.21 94 000 σall 36 000(0.9) n= = = 3.77 σ 8584

nc =

In summary, the resulting gear specifications are: All gears, P = 6 teeth/in Gear 2, Grade 1 flame-hardened, Sc = 170 000 psi and St = 45 000 psi d2 = 2.67 in, face width = 1.5 in Gear 3, Grade 1 through-hardened to 300 HB, Sc = 126 000 psi and St = 36 000 psi d3 = 12.0 in, face width = 1.5 in Gear 4, Grade 2 carburized and hardened, Sc = 225 000 psi and St = 65 000 psi d4 = 2.67 in, face width = 2.0 in Gear 5, Grade 2 carburized and hardened, Sc = 225 000 psi and St = 65 000 psi d5 = 12.0 in, face width = 2.0 in

18–4  Shaft Layout The general layout of the shafts, including axial location of gears and bearings, must now be specified in order to perform a free-body force analysis and to obtain shear force and bending moment diagrams. If there is no existing design to use as a starter, then the determination of the shaft layout may have many solutions. Section 7–3 discusses the issues involved in shaft layout. In this section the focus will be on how the decisions relate to the overall process. A free-body force analysis can be performed without knowing shaft diameters, but can not be performed without knowing axial distances between gears and bearings. It is extremely important to keep axial distances small. Even small forces can create large bending moments if the moment arms are large. Also, recall that beam deflection equations typically include length terms raised to the third power. It is worth examining the entirety of the gearbox at this time, to determine what factors drive the length of the shaft and the placement of the components. A rough sketch, such as shown in Figure 18–2, is sufficient for this purpose.

946      Mechanical Engineering Design

CASE STUDY EXAMPLE 18–1, PART 4 SHAFT LAYOUT Continue the case study by preparing a sketch of the gearbox sufficient to determine the axial dimensions. In particular, estimate the overall length, and the distance between the gears of the intermediate shaft, in order to fit with the mounting requirements of the other shafts.

Solution Figure 18–2 shows the rough sketch. It includes all three shafts, with consideration of how the bearings are to mount in the case. The gear widths are known at this point. Bearing widths are guessed, allowing a little more space for larger bearings on the intermediate shaft where bending moments will be greater. Small changes in bearing widths will have minimal effect on the force analysis, since the location of the ground reaction force will change very little. The 4-in distance between the two gears on the countershaft is dictated by the requirements of the input and output shafts, including the space for the case to mount the bearings. Small allotments are given for the retaining rings, and for space behind the bearings. Adding it all up gives the intermediate shaft length as 11.5 in.

1 2

3 4

1

12

3 4

1 2

Gear 2 Gear 3

Gear 5

Gear 4

1 4

1

3 4

1

12

4

Figure 18–2 Sketch for shaft layout. Dimensions are in inches.

2

3 4

1

1 4

Power Transmission Case Study     947

Wider face widths on gears require more shaft length. Originally, gears with hubs were considered for this design to allow the use of set screws instead of high-stressconcentration retaining rings. However, the extra hub lengths added several inches to the shaft lengths and the gearbox housing. Several points are worth noting in the layout in Figure 18–2. The gears and bearings are positioned against shoulders, with retaining rings to hold them in position. While it is desirable to place gears near the bearings, a little extra space is provided between them to accommodate any housing that extends behind the bearing, and to allow for a bearing puller to have space to access the back of the bearing. The extra change in diameter between the bearings and the gears allows the shoulder height for the bearing and the bore size for the gear to be different. This diameter can have loose tolerances and a large fillet radius. Each bearing is restrained axially on its shaft, but only one bearing on each shaft is axially fixed in the housing, allowing for slight axial thermal expansion of the shafts.

18–5  Force Analysis Once the gear diameters are known, and the axial locations of the components are set, the free-body diagrams and shear force and bending moment diagrams for the shafts can be produced. With the known transmitted loads, determine the radial and axial loads transmitted through the gears (see Sections 13–14 through 13–17). From summation of forces and moments on each shaft, ground reaction forces at the bearings can be determined. For shafts with gears and pulleys, the forces and moments will usually have components in two planes along the shaft. For rotating shafts, usually only the resultant magnitude is needed, so force components at bearings are summed as vectors. Shear force and bending moment diagrams are usually obtained in two planes, then summed as vectors at any point of interest. A torque diagram should also be generated to clearly visualize the transfer of torque from an input component, through the shaft, and to an output component. See the beginning of Example 7–2 for the force analysis portion of the case study for the intermediate shaft. The bending moment is largest at gear 4. This is predictable, since gear 4 is smaller, and must transmit the same torque that entered the shaft through the much larger gear 3. While the force analysis is not difficult to perform manually, if beam software is to be used for the deflection analysis, it will necessarily calculate reaction forces, along with shear force and bending moment diagrams in the process of calculating deflections. The designer can enter guessed values for diameters into the software at this point, just to get the force information, and later enter actual diameters to the same model to determine deflections.

18–6  Shaft Material Selection A trial material for the shaft can be selected at any point before the stress design of the shaft, and can be modified as necessary during the stress design process. Section 7–2 provides details for decisions regarding material selection. For the case study, Example 7–2 provides the stress analysis. Initially, an inexpensive steel, 1020 CD, is selected. After the stress analysis, a slightly higher strength 1050 CD is chosen to avoid increasing the shaft diameters.

948      Mechanical Engineering Design

18–7  Shaft Design for Stress The critical shaft diameters are to be determined by stress analysis at critical locations. Section 7–4 provides a detailed examination of the issues involved in shaft design for stress.

CASE STUDY EXAMPLE 18–1, PART 5 DESIGN FOR STRESS Proceed with the next phase of the case study design, in which appropriate diameters for each section of the shaft are estimated, based on providing sufficient fatigue and static stress capacity for infinite life of the shaft, with minimum safety factor of 1.5.

Solution The solution to this phase of the design is presented in Example 7–2.

Since the bending moment is highest at gear 4, potentially critical stress points are at its shoulder, keyway, and retaining ring groove. It turns out that the keyway is the critical location. It seems that shoulders often get the most attention. This example demonstrates the danger of neglecting other stress concentration sources, such as keyways. The material choice was changed in the course of this phase, choosing to pay for a higher strength to limit the shaft diameter to 2 in. If the shaft were to get much bigger, the small gear would not be able to provide an adequate bore size. If it becomes necessary to increase the shaft diameter any more, the gearing specification will need to be redesigned.

18–8  Shaft Design for Deflection Section 7–5 provides a detailed discussion of deflection considerations for shafts. Typically, a deflection problem in a shaft will not cause catastrophic failure of the shaft, but will lead to excess noise and vibration, and premature failure of the gears or bearings.

CASE STUDY EXAMPLE 18–1, PART 6 DEFLECTION CHECK Proceed with the next phase of the case study by checking that deflections and slopes at the gears and bearings on the intermediate shaft are within acceptable ranges.

Solution The solution to this phase of the design is presented in Example 7–3.

Power Transmission Case Study     949

It turns out that in this problem all the deflections are within recommended limits for bearings and gears. This is not always the case, and it would be a poor choice to neglect the deflection analysis. In a first iteration of this case study, with longer shafts due to using gears with hubs, the deflections were more critical than the stresses.

18–9  Bearing Selection Bearing selection is straightforward now that the bearing reaction forces and the approximate bore diameters are known. See Chapter 11 for general details on bearing selection. Rolling-contact bearings are available with a wide range of load capacities and dimensions, so it is usually not a problem to find a suitable bearing that is close to the estimated bore diameter and width.

CASE STUDY EXAMPLE 18–1, PART 7 BEARING SELECTION Continue the case study by selecting appropriate bearings for the intermediate shaft, with a reliability of 99 percent. The problem specifies a design life of 12 000 h. The intermediate shaft speed is 389 rev/min. The estimated bore size is 1 in, and the estimated bearing width is 1 in.

Solution From the free-body diagram (see Example 7–2),

RAz = 115.0 lbf

RAy = 356.7 lbf

RA = 375 lbf

RBz = 1776.0 lbf

RBy = 725.3 lbf

RB = 1918 lbf

At the shaft speed of 389 rev/min, the design life of 12 000 h correlates to a bearing life of LD = (12 000 h)(60 min/h)(389 rev/min) = 2.8 × 108 rev. Start with bearing B since it has the higher loads and will likely raise any lurking ­problems. From Equation (11–10), assuming a ball bearing with a = 3 and L = 2.8 × 108 rev, 1∕3 2.8 × 108∕106 FRB = (1)1918 [ = 20 820 lbf 1∕1.483 ] 0.02 + (4.459 − 0.02) (1 − 0.99)

A check on the Internet for available bearings (www.globalspec.com is one good starting place) shows that this load is relatively high for a ball bearing with bore size in the neighborhood of 1 in. Try a cylindrical roller bearing. Recalculating FRB with the exponent a = 10∕3 for roller bearings, we obtain FRB = 16 400 lbf

Cylindrical roller bearings are available from several sources in this range. A specific one is chosen from SKF, a common supplier of bearings, with the following specifications: Cylindrical roller bearing at right end of shaft C = 18 658 lbf, ID = 1.181 1 in, OD = 2.834 6 in, W = 1.063 in

Shoulder diameter range = 1.45 in to 1.53 in, and maximum fillet radius = 0.043 in

950      Mechanical Engineering Design

For bearing A, again assuming a ball bearing, FRA = 375 [

0.02 + 4.439(1 − 0.99) 1∕1.483 ] 2.8 × 108∕106

1∕3

= 4070 lbf

A specific ball bearing is chosen from the SKF Internet catalog. Deep-groove ball bearing at left end of shaft C = 5058 lbf, ID = 1.000 in, OD = 2.500 in, W = 0.75 in

Shoulder diameter range = 1.3 in to 1.4 in, and maximum fillet radius = 0.08 in At this point, the actual bearing dimensions can be checked against the initial assumptions. For bearing B the bore diameter of 1.1811 in is slightly larger than the original 1.0 in. There is no reason for this to be a problem as long as there is room for the shoulder diameter. The original estimate for shoulder support diameters was 1.4 in. As long as this diameter is less than 1.625 in, the next step of the shaft, there should not be any problem. In the case study, the recommended shoulder support diameters are within the acceptable range. The original estimates for stress concentration at the bearing shoulder assumed a fillet radius such that r∕d = 0.02. The actual bearings selected have ratios of 0.036 and 0.080. This allows the fillet radii to be increased from the original design, decreasing the stress-concentration factors. The bearing widths are close to the original estimates. Slight adjustments should be made to the shaft dimensions to match the bearings. No redesign should be necessary.

18–10  Key and Retaining Ring Selection The sizing and selection of keys is discussed in Section 7–7, with an example in Example 7–6. The cross-sectional size of the key will be dictated to correlate with the shaft size (see Tables 7–6 and 7–8), and must certainly match an integral keyway in the gear bore. The design decision includes the length of the key, and if necessary an upgrade in material choice. The key could fail by shearing across the key, or by crushing due to bearing stress. For a square key, it turns out that checking only the crushing failure is adequate, since the shearing failure will be less critical according to the distortion energy failure theory, and equal according to the maximum shear stress failure theory. Check Example 7–6 to investigate why.

CASE STUDY EXAMPLE 18–1, PART 8 KEY DESIGN Continue the case study by specifying appropriate keys for the two gears on the intermediate shaft to provide a factor of safety of 2. The gears are to be custom bored and keyed to the required specifications. Previously obtained information includes the following: Transmitted torque: T = 3240 lbf-in Bore diameters: d3 = d4 = 1.625 in Gear hub lengths: l3 = 1.5 in, l4 = 2.0 in

Power Transmission Case Study     951

Solution

From Table 7–6, for a shaft diameter of 1.625 in, choose a square key with side dimension t = 38 in. Choose 1020 CD material, with Sy = 57 kpsi. The force on the key at the surface of the shaft is

F=

T 3240 = 3988 lbf = r 1.625∕2

Checking for failure by crushing, we find the area of one-half the face of the key is used. n=

Sy σ

=

Sy F∕(tl∕2)

Solving for l gives l=

2(3988) (2) 2Fn = = 0.75 in tSy (0.375) (57000)

Since both gears have the same bore diameter and transmit the same torque, the same key specification can be used for both.

Retaining ring selection is simply a matter of checking catalog specifications. The retaining rings are listed for nominal shaft diameter, and are available with different axial load capacities. Once selected, the designer should make note of the depth of the groove, the width of the groove, and the fillet radius in the bottom of the groove. The catalog specification for the retaining ring also includes an edge margin, which is the  minimum distance to the next smaller diameter change. This is to ensure support for the axial load carried by the ring. It is important to check stress-concentration factors with actual dimensions, as these factors can be rather large. In the case study, a specific retaining ring was already chosen during the stress analysis (see Example 7–2) at the potentially critical location of gear 4. The other locations for retaining rings were not at points of high stress, so it is not necessary to worry about the stress concentration due to the retaining rings in these locations. Specific retaining rings should be selected at this time to complete the dimensional specifications of the shaft. For the case study, retaining rings specifications are entered into globalspec, and specific rings are selected from Truarc Co., with the following specifications: Both Gears

Left Bearing

Right Bearing

Nominal shaft diameter

1.625 in

1.000 in

1.181 in

Groove diameter

1.529 ± 0.005 in

0.940 ± 0.004 in

1.118 ± 0.004 in

Groove width

0.068

Nominal groove depth

0.048 in

0.030 in

0.035 in

Max groove fillet radius

0.010 in

0.010 in

0.010 in

Minimum edge margin

0.144 in

0.105 in

0.105 in

Allowable axial thrust

11 850 lbf

6000 lbf

7000 lbf

+0.004 in − 0.000

0.046

+ 0.004 in −0.000

These are within the estimates used for the initial shaft layout, and should not require any redesign. The final shaft should be updated with these dimensions.

0.046

+ 0.004 in −0.000

952

Figure 18–3

Power Transmission Case Study     953

18–11  Final Analysis At this point in the design, everything seems to check out. Final details include determining dimensions and tolerances for appropriate fits with the gears and bearings. See Section 7–8 for details on obtaining specific fits. Any small changes from the nominal diameters already specified will have negligible effect on the stress and deflection analysis. However, for manufacturing and assembly purposes, the designer should not overlook the tolerance specification. Improper fits can lead to failure of the design. Lack of attention to tolerance specification can make the part nonfunctional or overly expensive to manufacture. Further insight on tolerance specification is given in Section 1–14. The final drawing for the intermediate shaft is shown in Figure 18–3. This drawing shows the important dimensions and dimensional tolerances in a form that is generally considered satisfactory for small production quantities where direct attention is given to manufacturing methods. A more robust method of part specification that also addresses the allowed variations from perfect form (e.g., straightness or concentricity) is known as Geometric Dimensioning and Tolerancing and is introduced in Chapter 20. For documentation purposes, and for a check on the design work, the design process should conclude with a complete analysis of the final design. Remember that analysis is much more straightforward than design, so the investment of time for the final analysis will be relatively small.

PROBLEMS 18–1

For the case study problem, design the input shaft, including complete specification of the gear, bearings, key, retaining rings, and shaft.

18–2 For the case study problem, design the output shaft, including complete specification of the gear, bearings, key, retaining rings, and shaft.

18–3 For the case study problem, use helical gears and design the intermediate shaft. Compare your results with the spur gear design presented in this chapter.

18–4 Perform a final analysis for the resulting design of the intermediate shaft of the case study problem presented in this chapter. Produce a final drawing with dimensions and tolerances for the shaft. Does the final design satisfy all the requirements? Identify the critical aspects of the design with the lowest factor of safety.

18–5 For the case study problem, change the power requirement to 40 horsepower. Design the intermediate shaft, including complete specification of the gears, bearings, keys, retaining rings, and shaft.

E O

D C

2

2

A xy

x G

part

Courtesy of Dee Dehokenanan

Special Topics Chapter 19 Finite-Element Analysis   955 Chapter 20 Geometric Dimensioning and Tolerancing   977

4

19

Finite-Element Analysis

©Mathew Alexander/Shutterstock

Chapter Outline 19–1

The Finite-Element Method   957

19–7

Modeling Techniques   967

19–2

Element Geometries   959

19–8

Thermal Stresses   970

19–3

The Finite-Element Solution Process   961

19–9

Critical Buckling Load   972

19–4

Mesh Generation   964

19–10

Vibration Analysis   973

19–5

Load Application   966

19–11

Summary  974

19–6

Boundary Conditions   967

955

956      Mechanical Engineering Design

Mechanical components in the form of simple bars, beams, etc., can be analyzed quite easily by basic methods of mechanics that provide closed-form solutions. Actual components, however, are rarely so simple, and the designer is forced to less effective approximations of closed-form solutions, experimentation, or numerical methods. There are a great many numerical techniques used in engineering applications for which the digital computer is very useful. In mechanical design, where computer-aided design (CAD) software is heavily employed, the analysis method that integrates well with CAD is finite-element analysis (FEA). The mathematical theory and applications of the method are vast. There are also a number of commercial FEA software packages that are available, such as ANSYS, MSC/NASTRAN, ALGOR, etc. The purpose of this chapter is only to expose the reader to some of the fundamental aspects of FEA, and therefore the coverage is extremely introductory in nature. For further detail, the reader is urged to consult the many references cited at the end of this chapter. Figure 19–1 shows a finite-element model of a connecting rod that was developed to study the effects of dynamic elastohydrodynamic lubrication on bearing and structural performance.1 There are a multitude of FEA applications such as static and dynamic, linear and nonlinear, stress and deflection analysis; free and forced vibrations; heat transfer (which can be combined with stress and deflection analysis to provide thermally

Figure 19–1 Model of a connecting rod using ANSYS finite-element software. (a) Meshed model; (b) stress contours. Courtesy of Dr. Stephen Boedo (see footnote 1).

Z X

Y

(a)

Z X

Y

(b)

1

S. Boedo, "Elastohydrodynamic Lubrication of Conformal Bearing Systems," Proceedings of 2002 ANSYS Users Conference, Pittsburgh, PA, April 22–24, 2002.

Finite-Element Analysis     957

induced stresses and deflections); elastic instability (buckling); acoustics; electrostatics and magnetics (which can be combined with heat transfer); fluid dynamics; piping analysis; and multiphysics. For the purposes of this chapter, we will limit ourselves to basic mechanics analyses. An actual mechanical component is a continuous elastic structure (continuum). FEA divides (discretizes) the structure into small but finite, well-defined, elastic substructures (elements). By using polynomial functions, together with matrix operations, the continuous elastic behavior of each element is developed in terms of the element's material and geometric properties. Loads can be applied within the element (gravity, dynamic, thermal, etc.), on the surface of the element, or at the nodes of the element. The element's nodes are the fundamental governing entities of the element, as it is the node where the element connects to other elements, where elastic properties of the element are eventually established, where boundary conditions are assigned, and where forces (contact or body) are ultimately applied. A node possesses degrees of freedom (dof's). Degrees of freedom are the independent translational and rotational motions that can exist at a node. At most, a node can possess three translational and three rotational degrees of freedom. Once each element within a structure is defined locally in matrix form, the elements are then globally assembled (attached) through their common nodes (dof's) into an overall system matrix. Applied loads and boundary conditions are then specified and through matrix operations the values of all unknown displacement degrees of freedom are determined. Once this is done, it is a simple matter to use these displacements to determine strains and stresses through the constitutive equations of elasticity.

19–1  The Finite-Element Method The modern development of the finite-element method began in the 1940s in the field of structural mechanics with the work of Hrennikoff,2 McHenry,3 and Newmark,4 who used a lattice of line elements (rods and beams) for the solution of stresses in continuous solids. In 1943, from a 1941 lecture, Courant5 suggested piecewise polynomial interpolation over triangular subregions as a method to model torsional problems. With the advent of digital computers in the 1950s it became practical for engineers to write and solve the stiffness equations in matrix form.6,7,8 A classic paper by Turner, Clough, Martin, and Topp published in 1956 presented the matrix stiffness equations 2

A. Hrennikoff, "Solution of Problems in Elasticity by the Frame Work Method," Journal of Applied Mechanics, Vol. 8, No. 4, pp. 169–175, December 1941. 3 D. McHenry, "A Lattice Analogy for the Solution of Plane Stress Problems," Journal of Institution of Civil Engineers, Vol. 21, pp. 59–82, December 1943. 4 N. M. Newmark, "Numerical Methods of Analysis in Bars, Plates, and Elastic Bodies," Numerical Methods in Analysis in Engineering (ed. L. E. Grinter), Macmillan, 1949. 5 R. Courant, "Variational Methods for the Solution of Problems of Equilibrium and Vibrations," Bulletin of the American Mathematical Society, Vol. 49, pp. 1–23, 1943. 6 S. Levy, "Structural Analysis and Influence Coefficients for Delta Wings," Journal of Aeronautical Sciences, Vol. 20, No. 7, pp. 449–454, July 1953. 7 J. H. Argyris, "Energy Theorems and Structural Analysis," Aircraft Engineering, October, November, December 1954 and February, March, April, May 1955. 8 J. H. Argyris and S. Kelsey, Energy Theorems and Structural Analysis, Butterworths, London, 1960 (reprinted from Aircraft Engineering, 1954–55).

958      Mechanical Engineering Design

for the truss, beam, and other elements.9 The expression finite element is first attributed to Clough.10 Since these early beginnings, a great deal of effort has been expended in the development of the finite element method in the areas of element formulations and computer implementation of the entire solution process. The major advances in computer technology include the rapidly expanding computer hardware capabilities, efficient and accurate matrix solver routines, and computer graphics for ease in the visual preprocessing stages of model building, including automatic adaptive mesh generation, and in the postprocessing stages of reviewing the solution results. A great abundance of literature has been presented on the subject, including many textbooks. A partial list of some textbooks, introductory and more comprehensive, is given at the end of this chapter. Since the finite-element method is a numerical technique that discretizes the domain of a continuous structure, errors are inevitable. These errors are:  1 Computational errors. These are due to round-off errors from the computer floating-point calculations and the formulations of the numerical integration schemes that are employed. Most commercial finite-element codes concentrate on reducing these errors, and consequently the analyst generally is concerned with discretization factors.  2 Discretization errors. The geometry and the displacement distribution of a true structure continuously vary. Using a finite number of elements to model the structure introduces errors in matching geometry and the displacement distribution due to the inherent mathematical limitations of the elements. For an example of discretization errors, consider the constant thickness, thin plate structure shown in Figure 19–2a. Figure 19–2b shows a finite-element model

(a)

(b)

Figure 19–2 Structural problem. (a) Idealized model; (b) finite-element model. 9

M. J. Turner, R. W. Clough, H. C. Martin, and L. J. Topp, "Stiffness and Deflection Analysis of Complex Structures," Journal of Aeronautical Sciences, vol. 23, no. 9, pp. 805–824, September 1956. 10 R. W. Clough, "The Finite Element Method in Plane Stress Analysis," Proceedings of the Second Conference on Electronic Computation, American Society of Civil Engineers, Pittsburgh, PA, pp. 345–378, September 1960.

Finite-Element Analysis     959

of the structure where three-node, plane stress, simplex triangular elements are employed. This element type has a flaw that creates two basic problems. The element has straight sides that remain straight after deformation. The strains throughout the plane stress triangular element are constant. The first problem, a geometric one, is the modeling of curved edges. Note that the surface of the model with a large curvature appears poorly modeled, whereas the surface of the hole seems to be reasonably modeled. The second problem, which is much more severe, is that the strains in various regions of the actual structure are changing rapidly, and the constant strain element will provide only an approximation of the average strain at the center of the element. So, in a nutshell, the results predicted by this model will be extremely poor. The results can be improved by significantly increasing the number of elements (increased mesh density). Alternatively, using a better element, such as an eight-node quadrilateral, which is more suited to the application, will provide the improved results. Because of higher-order interpolation functions, the eight-node quadrilateral element can model curved edges and provides a higher-order function for the strain distribution. In Figure 19–2b, the triangular elements are shaded and the nodes of the elements are represented by the black dots. Forces and constraints can be placed only at the nodes. The nodes of the simplex triangular plane stress elements have only two degrees of freedom, corresponding to translation in the plane. Thus, the solid black, simple support triangles on the left edge represent the fixed support of the model. Also, the distributed load can be applied only to three nodes as shown. The modeled load has to be statically consistent with the actual load.

19–2  Element Geometries Many geometric shapes of elements are used in finite-element analysis for specific applications. The various elements used in a general-purpose commercial FEA software code constitute what is referred to as the element library of the code. Elements can be placed in the following categories: line elements, surface elements, solid elements, and special-purpose elements. Table 19–1 provides some, but not all, of the types of elements available for finite-element analysis for structural problems. Not all elements support all degrees of freedom. For example, the 3-D truss element supports Table 19–1  Sample Finite-Element Library Element Type None Shape

Line

Number of Nodes

Applications

Truss

2

Pin-ended bar in tension or compression

Beam

2

Bending

Frame

2

Axial, torsional, and bending. With or without load stiffening. (Continued)

Table 19–1  Sample Finite-Element Library  (Continued) Element Type None Shape

Number of Nodes

Applications

4-node quadri­ lateral

4

Plane stress or strain, axisymmetry, shear panel, thin flat plate in bending

8-node quadri­ lateral

8

Plane stress or strain, thin plate or shell in bending

3-node triangular

3

Plane stress or strain, axisymmetry, shear panel, thin flat plate in bending. Prefer quad where possible. Used for transitions of quads.

6-node triangular

6

Plane stress or strain, axisymmetry, thin plate or shell in bending. Prefer quad where possible. Used for transitions of quads.

8-node hexagonal (brick)

8

Solid, thick plate

6-node pentagonal (wedge)

6

Solid, thick plate. Used for transitions.

4-node tetrahedron (tet)

4

Solid, thick plate. Used for transitions.

Gap

2

Free displacement for prescribed compressive gap

Hook

2

Free displacement for prescribed extension gap

Rigid

Vari­abl­e

Surface

Solid†

Special purpose

These elements are also available with midside nodes. 960

Rigid constraints between nodes

Finite-Element Analysis     961

only three translational degrees of freedom at each node. Connecting elements with differing dof's generally requires some manual modification. For example, consider connecting a truss element to a frame element. The frame element supports all six dof's at each node. A truss member, when connected to it, can rotate freely at the connection.

19–3  The Finite-Element Solution Process We will describe the finite-element solution process on a very simple one-dimensional problem, using the linear truss element. A truss element is a bar loaded in tension or compression and is of constant cross-sectional area A, length l, and elastic modulus E. The basic truss element has two nodes, and for a one-dimensional problem, each node will have only one degree of freedom. A truss element can be modeled as a simple linear spring with a spring rate, given by Equation (4–4), as

k=

AE l

(19–1)

Consider a spring element (e) of spring rate ke, with nodes i and j, as shown in Figure 19–3. Nodes and elements will be numbered. So, to avoid confusion as to what a number corresponds to, elements will be numbered within parentheses. Assuming all forces f and displacements u directed toward the right as positive, the forces at each node can be written as

fi, e = ke (ui − u j ) = ke ui − ke u j

fj,e = ke (uj − ui ) = −ke u i + ke uj

(19–2)

The two equations can be written in matrix form as

fi, e ke −ke ui {fi, e} = [ −ke ke ]{uj}

(19–3)

Next, consider a two-spring system as shown in Figure 19–4a. Here we have numbered the nodes and elements. We have also labeled the forces at each node. However, these forces are the total external forces at each node, F1, F2, and F3. If we draw separate free-body diagrams we will expose the internal forces as shown in Figure 19–4b. Using Equation (19–3) for each spring gives Element 1

f1,1 k1 {f2,1} = [ −k1

−k1 u1 k1 ]{u2}

(19–4a)

Element 2

f2,2 k2 {f3,2} = [ −k2

−k2 u2 k2 ]{u3}

(19–4b)

Figure 19–3

ke fi,e

j

i

ui

(e)

fj,e

uj

A simple spring element.

962      Mechanical Engineering Design F2

k1 F1

1

k2

2

(1)

F3

3 u2

(2)

u3

(a)

f1,1

k2

k1

1

2

f2,1

2

f2,2

3

(1)

f3,2

(2)

u1

u2

u3

u2 (b)

Figure 19–4 A two-element spring system. (a) System model; (b) separate free-body diagrams.

The total force at each node is the external force, F1 = f1,1, F2 = f2,1 + f2,2, and F3 = f3,2. Combining the two matrices in terms of the external forces gives  f1,1   f2,1   f3

 

+ f2,2  =  

 F1     F2  =    F3 

 k1   −k1   0

−k1 (k1 + k2 ) −k2

u1  0     −k2   u2    k2   u3 

(19–5)

If we know the displacement of a node, then the force at the node will be unknown. For example, in Figure 19–4a, the displacement of node 1 at the wall is zero, so F1 is the unknown reaction force (note, up to this point, we have not applied a static solution of the system). If we do not know the displacement of a node, then we know the force. For example, in Figure 19–4a, the displacements at nodes 2 and 3 are unknown, and the forces F2 and F3 are to be specified. To see how the remainder of the solution process can be implemented, let us consider the following example.

EXAMPLE 19–1 Consider the aluminum step-shaft shown in Figure 19–5a. The areas of sections AB and BC are 0.100 in2 and 0.150 in2, respectively. The lengths of sections AB and BC are 10 in and 12 in, respectively. A force F = 1000 lbf is applied to B. Initially, a gap of ε = 0.002 in exists between end C and the right rigid wall. Determine the wall reactions, the internal forces in the members, and the deflection of point B. Let E = 10 Mpsi and assume that end C hits the wall. Check the validity of the assumption. Figure 19–5 (a) Step shaft; (b) spring model.

(a)

A

F

B

C

ε F (b)

2

1 k1

3

u2

k2

u3

Finite-Element Analysis     963

Solution The step-shaft is modeled by the two-spring system of Figure 19–5b where 0.1(10) 106 AE k1 = ( ) = = 1 (105 ) lbf/in l AB 10

0.15(10) 106 AE k2 = ( ) = = 1.25 (105 ) lbf/in l BC 12

With u1 = 0, F2 = 1000 lbf and the assumption that u3 = ε = 0.002 in, Equation (19–5) becomes 0  0   1 −1  F1         1000  = 105  −1 2.25 −1.25   u2         0 −1.25  F3  1.25   0.002 

(1)

For large problems, there is a systematic method of solving equations like Equation (1), called partitioning or the elimination approach.11 However, for this simple problem, the solution is quite simple. From the second equation of the matrix equation 1000 = 105[−1(0) + 2.25 u2 − 1.25(0.002)]

or,

uB = u2 =

Answer

1000∕105 + 1.25(0.002) = 5.556 (10−3 ) in 2.25

Since uB > ε, it is verified that point C hits the wall. The reactions at the walls are F1 and F3. From the first and third equations of matrix Equation (1), F1 = 105[−1(u2 )] = 105[−1(5.556)10−3] = −555.6 lbf

Answer and Answer

F3 = 10 5[−1.25u 2 + 1.25(0.002)] = 10 5[−1.25(5.556)10−3 + 1.25(0.002)] = −444.4 lbf

Since F3 is negative, this also verifies that C hits the wall. Note that F1 + F3 = −555.6 − 444.4 = −1000 lbf, balancing the applied force (with no statics equations necessary). For internal forces, it is necessary to return to the individual (local) equations. From Equation (19–4a), f1,1 k1 −k1 u1 1 −1 0 −555.6 = 105 [ = lbf { f2,1} = [ −k1 k1 ]{u2} −1 1 ]{5.556(10−3 )} { 555.6}

Answer Since f1,1 is directed to the left and f2,1 is directed to the right, the element is in tension, with a force of 555.6 lbf. If the stress is desired, it is simply σAB = f2,1∕AAB = 555.6∕0.1 = 5556 psi. For element BC, from Equation (19.4b), f2,2 k2 {f3,2} = [ −k2

−k2 u2 1.25 = 105 [ k2 ]{u3} −1.25

−1.25 5.556(10−3 ) 444.5 = lbf 1.25 ]{ 0.002 } {−444.5}

Answer Since f2,2 is directed to the right and f3,2 is directed to the left, the element is in compression, with a force of 444.5 lbf. If the stress is desired, it is simply σBC = −f2,2∕ABC = −444.5∕0.15 = −2963 psi. 11

See T. R. Chandrupatla and A. D. Belegundu, Introduction to Finite Elements in Engineering, 4th ed., Prentice Hall, Upper Saddle River, NJ, 2012, pp. 71–75.

964      Mechanical Engineering Design

19–4  Mesh Generation The network of elements and nodes that discretize a region is referred to as a mesh. The mesh density increases as more elements are placed within a given region. Mesh refinement is when the mesh is modified from one analysis of a model to the next analysis to yield improved results. Results generally improve when the mesh density is increased in areas of high stress gradients and/or when geometric transition zones are meshed smoothly. Generally, but not always, the FEA results converge toward the exact results as the mesh is continuously refined. To assess improvement, in regions where high stress gradients appear, the structure can be remeshed with a higher mesh density at this location. If there is a minimal change in the maximum stress value, it is reasonable to presume that the solution has converged. There are three basic ways to generate an element mesh—manually, semiautomatically, or fully automatically.  1 Manual mesh generation. This is how the element mesh was created in the early days of the finite-element method. This is a very labor intensive method of creating the mesh, and except for some quick modifications of a model is it rarely done. Note: Care must be exercised in editing an input text file. With some FEA software, other files such as the preprocessor binary graphics file may not change. Consequently, the files may no longer be compatible with each other.  2 Semiautomatic mesh generation. Over the years, computer algorithms have been developed that enable the modeler to automatically mesh regions of the structure that he or she has divided up, using well-defined boundaries. Since the modeler has to define these regions, the technique is deemed semiautomatic. The development of the many computer algorithms for mesh generation emanates from the field of computer graphics. If the reader desires more information on this subject, a review of the literature available from this field is recommended.  3 Fully automatic mesh generation. Many software vendors have concentrated their efforts on developing fully automatic mesh generation, and in some instances, with automatic self-adaptive mesh refinement. The obvious goal is to significantly reduce the modeler's preprocessing time and effort to arrive at a final well-constructed FEA mesh. Once the complete boundary of the structure is defined, without subdivisions as in semiautomatic mesh generation and with a minimum of user intervention, various schemes are available to discretize the region with one element type. For plane elastic problems the boundary is defined by a series of internal and external geometric lines and the element type to be automeshed would be the plane elastic element. For thinwalled structures, the geometry would be defined by three-dimensional surface representations and the automeshed element type would be the three-dimensional plate element. For solid structures, the boundary could be constructed by using constructive solid geometry (CSG) or boundary representation (B-rep) techniques. The finite-element types for automeshing would be the brick and/or tetrahedron element(s). Automatic self-adaptive mesh refinement programs estimate the error of the FEA solution. On the basis of the error, the mesh is automatically revised and reanalyzed. The process is repeated until some convergence or termination criterion is satisfied. Returning to the thin-plate model of Figure 19–2, the boundaries of the structure are constructed as shown in Figure 19–6a. The boundaries were then automeshed as

Finite-Element Analysis     965

Figure 19–6 Automatic meshing the thin-plate model of Figure 19–2. (a) Model boundaries; (b) automesh with 294 elements and 344 nodes; (c) deflected (exaggerated scale) with stress contours; (d) automesh with 1008 elements and 1096 nodes, (e) deflected (exaggerated scale) with stress contours.

(a)

Von Mises 4110.4 3524.8 2939.2 2353.6 1768.1 1182.5 596.91 11.341

4110.4 psi

(b)

(c) Von Mises 4184.9 3588.2 2991.6 2394.9 1798.2 1201.6 604.91 8.2392

4184.9 psi

(d)

(e)

shown in Figure 19–6b, where 294 elements and 344 nodes were generated. Note the uniformity of the element generation at the boundaries. The finite-element solver then generated the deflections and von Mises stresses shown in Figure 19–6c. The maximum von Mises stress at the location shown is 4110.4 psi. The model was then automeshed with an increased mesh density as shown in Figure 19–6d, where the model has 1008 elements and 1096 nodes. The results are shown in Figure 19–6e where the maximum von Mises stress is found to be 4184.9 psi, which is only 1.8 percent higher. In all likelihood, the solution has nearly converged. Note: The stress contours of Figures 19–6c and e are better visualized in color. When stress concentrations are present, it is necessary to have a very fine mesh at the stress-concentration region in order to get realistic results. What is important is that the mesh density needs to be increased only in the region around the stress

966      Mechanical Engineering Design

concentration and that the transition mesh from the rest of the structure to the stressconcentration region be gradual. An abrupt mesh transition, in itself, will have the same effect as a stress concentration. Stress concentration will be discussed further in Section 19–7, Modeling Techniques.

19–5  Load Application There are two basic forms of specifying loads on a structure—nodal and element loading. However, element loads are eventually applied to the nodes by using equivalent nodal loads. One aspect of load application is related to Saint-Venant's principle. If one is not concerned about the stresses near points of load application, it is not necessary to attempt to distribute the loading very precisely. The net force and/or moment can be applied to a single node, provided the element supports the dof associated with the force and/or moment at the node. However, the analyst should not be surprised, or concerned, when reviewing the results and the stresses in the vicinity of the load application point are found to be very large. Concentrated moments can be applied to the nodes of beam and most plate elements. However, concentrated moments cannot be applied to truss, two-dimensional plane elastic, axisymmetric, or brick elements. They do not support rotational degrees of freedom. A pure moment can be applied to these elements only by using forces in the form of a couple. From the mechanics of statics, a couple can be generated by using two or more forces acting in a plane where the net force from the forces is zero. The net moment from the forces is a vector perpendicular to the plane of the forces and is the summation of the moments from the forces taken about any common point. Element loads include ∙ ∙ ∙ ∙

Static loads due to gravity (weight), Thermal effects, Surface loads such as uniform and hydrostatic pressure, and Dynamic loads due to constant acceleration and steady-state rotation (centrifugal acceleration).

As stated earlier, element loads are converted by the software to equivalent nodal loads and in the end are treated as concentrated loads applied to nodes. For gravity loading, the gravity constant in appropriate units and the direction of gravity must be supplied by the modeler. If the model length and force units are inches and lbf, g = 386.1 ips2. If the model length and force units are meters and Newtons, g = 9.81 m/s2. The gravity direction is normally toward the center of the earth. For thermal loading, the thermal expansion coefficient α must be given for each material, as well as the initial temperature of the structure, and the final nodal temperatures. Most software packages have the capability of first performing a finiteelement heat transfer analysis on the structure to determine the final nodal temperatures. The temperature results are written to a file, which can be transferred to the static stress analysis. Here the heat transfer model should have the same nodes and element type the static stress analysis model has. Surface loading can generally be applied to most elements. For example, uniform or linear transverse line loads (force/length) can be specified on beams. Uniform and linear pressure can normally be applied on the edges of two-dimensional plane and

Finite-Element Analysis     967

axisymmetric elements. Lateral pressure can be applied on plate elements, and pressure can be applied on the surface of solid brick elements. Each software package has its unique manner in which to specify these surface loads, usually in a combination of text and graphic modes.

19–6  Boundary Conditions The simulation of boundary conditions and other forms of constraint is probably the single most difficult part of the accurate modeling of a structure for a finiteelement analysis. In specifying constraints, it is relatively easy to make mistakes of omission or misrepresentation. It may be necessary for the analyst to test different approaches to model esoteric constraints such as bolted joints, welds, etc., which are not as simple as the idealized pinned or fixed joints. Testing should be confined to simple problems and not to a large, complex structure. Sometimes, when the exact nature of a boundary condition is uncertain, only limits of behavior may be possible. For example, we have modeled shafts with bearings as being simply supported. It is more likely that the support is something between simply supported and fixed, and we could analyze both constraints to establish the limits. However, by assuming simply supported, the results of the solution are conservative for stress and deflections. That is, the solution would predict stresses and deflections larger than the actual. For another example, consider beam 16 in Table A–9. The horizontal beam is uniformly loaded and is fixed at both ends. Although not explicitly stated, tables such as these assume that the beams are not restrained in the horizontal direction. That is, it is assumed that the beam can slide horizontally in the supports. If the ends were completely or partially restrained, a beam-column solution would be necessary.12 With a finite-element analysis, a special element, a beam with stiffening, could be used. Multipoint constraint equations are quite often used to model boundary conditions or rigid connections between elastic members. When used in the latter form, the equations are acting as elements and are thus referred to as rigid elements. Rigid elements can rotate or translate only rigidly. Boundary elements are used to force specific nonzero displacements on a structure. Boundary elements can also be useful in modeling boundary conditions that are askew from the global coordinate system.

19–7  Modeling Techniques With today's CAD packages and automatic mesh generators, it is an easy task to create a solid model and mesh the volume with finite elements. With today's computing speeds and with gobs of computer memory, it is very easy to create a model with extremely large numbers of elements and nodes. The finite-element modeling techniques of the past now seem passé and unnecessary. However, much unnecessary time can be spent on a very complex model when a much simpler model will do. The complex model may not even provide an accurate solution, whereas a simpler one 12

See R. G. Budynas, Advanced Strength and Applied Stress Analysis, 2nd ed., McGraw-Hill, New York, 1999, pp. 471–482.

968      Mechanical Engineering Design 600 lbf y 1.750 dia

1.500 dia

1.000 dia A

C

B

1.000 dia

E

D

F x

0.5 R1

8 8.5

R2

19.5 20 (a) Dimensions in inches

1

2 (1)

4

3 (2)

5

(3)

(4)

6 (5)

(b)

Displacements/rotations (degrees) of nodes Node No.

x Translation

y Translation

z Translation

1

0.0000 E + 00

0.0000 E + 00

0.0000 E + 00

0.0000 E + 00

0.0000 E + 00

−9.7930 E − 02

2

0.0000 E + 00

−8.4951 E − 04

0.0000 E + 00

0.0000 E + 00

0.0000 E + 00

−9.6179 E − 02

3

0.0000 E + 00

−9.3649 E − 03

0.0000 E + 00

0.0000 E + 00

0.0000 E + 00

−7.9874 E − 03

4

0.0000 E + 00

−9.3870 E − 03

0.0000 E + 00

0.0000 E + 00

0.0000 E + 00

2.8492 E − 03

5

0.0000 E + 00

−6.0507 E − 04

0.0000 E + 00

0.0000 E + 00

0.0000 E + 00

6.8558 E − 02

6

0.0000 E + 00

0.0000 E + 00

0.0000 E + 00

0.0000 E + 00

0.0000 E + 00

6.9725 E − 02

Figure 19–7

θx θy θz Rotation (deg) Rotation (deg) Rotation (deg)

(c)

(a) Steel step shaft of Example 4–7; (b) finite-element model using five beam elements; (c) displacement results for FEA model.

will. What is important is what solution the analyst is looking for: deflections, stresses, or both? For example, consider the steel step shaft of Example 4–7, repeated here as Figure 19–7a. Let the fillets at the steps have a radius of 0.02 in. If only deflections and slopes were sought at the steps, a highly meshed solid model would not yield much more than the simple five-element beam model, shown in Figure 19–7b, would. The fillets at the steps, which could not be modeled easily with beam elements, would not contribute much to a difference in results between the two models. Nodes are necessary wherever boundary conditions, applied forces, and changes in cross section and/or material occur. The displacement results for the FEA model are shown in Figure 19–7c.

Finite-Element Analysis     969 x

Figure 19–8 (a) Solid model of the step-shaft of Example 4–7 using 56384 brick and tetrahedron elements; (b) view of stress contours at step, rotated 180° about x axis, showing maximum tension.

y

(a)

z

σmax= 23.9 kpsi

(b)

The FE model of Figure 19–7b is not capable of providing the stress at the fillet of the step at D. Here, a full-blown solid model would have to be developed and meshed, using solid elements with a high mesh density at the fillet as shown in Figure 19–8a. Here, the steps at the bearing supports are not modeled, as we are concerned only with the stress concentration at x = 8.5 in. The brick and tetrahedron elements do not support rotational degrees of freedom. To model the simply supported boundary condition at the left end, nodes along the z axis were constrained from translating in the x and y directions. Nodes along the y axis were constrained from translating in the z direction. Nodes on the right end on an axis parallel with the z axis through the center of the shaft were constrained from translating in the y  direction, and nodes on an axis parallel with the y axis through the center of the shaft were constrained from translating in the z direction. This ensures no rigid-body translation or rotation and no overconstraint at the ends. The maximum tensile stress at the fillet at the beam bottom is found to be σmax = 23.9 kpsi. Performing an analytical check at the step yields D∕d = 1.75∕1.5 = 1.167, and r∕d = 0.02∕1.5 = 0.0133.

970      Mechanical Engineering Design

Figure A–15–9 is not very accurate for these values. Resorting to another source,13 the stress-concentration factor is found to be Kt = 3.00. The reaction at the right support is RF = (8∕20)600 = 240 lbf. The bending moment at the start of the fillet is M = 240(11.52) = 2765 lbf · in = 2.765 kip · in. The analytical prediction of the maximum stress is thus σmax = Kt(

32(2.765) 32M = 3.00[ = 25.03 kpsi 3 ) πd π(1.53 ) ]

The finite-element model is 4.5 percent lower. If more elements were used in the fillet region, the results would undoubtedly be closer. However, the results are within engineering acceptability. If we want to check deflections, we should compare the results with the threeelement beam model, not the five-element model. This is because we did not model the bearing steps in the solid model. The vertical deflection, at x = 8.5 in, for the solid model was found to be −0.00981 in. This is 4.6 percent higher in magnitude than the −0.00938 in deflection for the three-element beam model. For slopes, the brick element does not support rotational degrees of freedom, so the rotation at the ends has to be computed from the displacements of adjacent nodes at the ends. This results in the slopes at the ends of θA = −0.103° and θF = 0.0732°; these are 6.7 and 6.6 percent higher in magnitude than the three-element beam model, respectively. However, the point of this exercise is, if deflections were the only result desired, which model would you use? There are countless modeling situations which could be examined. The reader is urged to read the literature, and peruse the tutorials available from the software vendors.14

19–8  Thermal Stresses A heat transfer analysis can be performed on a structural component including the effects of heat conduction, convection, and/or radiation. After the heat transfer analysis is completed, the same model can be used to determine the resulting thermal stresses. For a simple illustration, we will model a 10 in × 4 in, 0.25-in-thick steel plate with a centered 1.0-in-diameter hole. The plate is supported as shown in Figure 19–9a, and the temperatures of the ends are maintained at temperatures of 100°F and 0°F. Other than at the walls, all surfaces are thermally insulated. Before placing the plate between the walls, the initial temperature of the plate was 0°F. The thermal coefficient of expansion for steel is αs = 6.5 × 10−6 °F−1. The plate was meshed with 1312 two-dimensional elements, with the mesh refined along the border of the hole. Figure 19–9b shows the temperature contours of the steady-state temperature distribution obtained by the FEA. Using the same elements for a linear stress

13

See W. D. Pilkey and D. F. Pilkey, Peterson's Stress-Concentration Factors, 3rd ed. John Wiley & Sons, New York, 2008, Chart 3.11. 14 See, for example, R. D. Cook, Finite Element Modeling for Stress Analysis, Wiley & Sons, New York, 1995; and R. G. Budynas, Advanced Strength and Applied Stress Analysis, 2nd ed., McGraw-Hill, New York, 1999, Chapter 10.

Finite-Element Analysis     971

0°F

100°F

(a)

Temperature 99.711 85.508 71.305 57.102 42.898 28.695 14.492 0.28899

(b)

Von Mises 31888 27569 23249 18930 14611 10292 5972.2 1652.9

(c)

Figure 19–9 (a) Plate supported at ends and maintained at the temperatures shown; (b) steady-state temperature contours; (c) thermal von Mises stress contours where the initial temperature of the plate was 0°F.

972      Mechanical Engineering Design

analysis, where the temperatures were transferred from the heat transfer analysis, Figure 19–9c shows the resulting stress contours. As expected, the maximum compressive stresses occurred at the top and bottom of the hole, with a magnitude of 31.9 kpsi.

19–9  Critical Buckling Load Finite elements can be used to predict the critical buckling load for a thin-walled structure. An example was shown in Figure 4–25. Another example can be seen in Figure 19–10a, which is a thin-walled aluminum beverage can. A specific pressure was applied to the top surface. The bottom of the can was constrained in translation vertically, the center node of the bottom of the can was constrained in translation in all three directions, and one outer node on the can bottom was constrained in translation tangentially. This prevents rigid-body motion, and provides vertical support for the bottom of the can with unconstrained motion of the bottom of the can horizontally. The finite element software returns a value of the load multiplier, which, when multiplied with the total applied force, indicates the critical buckling load. Buckling analysis is an eigenvalue problem, and a reader who reviews a basic mechanics of materials textbook would find there is a deflection mode shape associated with the critical load. The buckling mode shape for the buckled beverage can is shown in Figure 19–10b.

(b)

(a)

Figure 19–10 (a) Thin-walled aluminum beverage container loaded vertically downward on the top surface; (b) isometric view of the buckled can (deflections greatly exaggerated).

Finite-Element Analysis     973

19–10  Vibration Analysis The design engineer may be concerned as to how a component behaves relative to dynamic input, which results in vibration. For vibration, most finite element packages start with a modal analysis of the component. This provides the natural frequencies and mode shapes that the component naturally vibrates at. These are called the eigenvalues and eigenvectors of the component. Next, this solution can be transferred (much the same as for thermal stresses) to solvers for forced vibration analyses, such as frequency response, transient impact, or random vibration, to see how the component's modes behave to dynamic input. The mode shape analysis is primarily based on ­stiffness and the resulting deflections. Thus, similar to static stress analysis, simpler models will suffice. However, if, when solving forced response problems, stresses are desired, a more detailed model is necessary (similar to the shaft illustration given in Section 19–7). A modal analysis of the beam model without the bearing steps was performed for a 20-element beam model,15 and the 56 384-element brick and tetrahedron model. Needless to say, the beam model took less than 9 seconds to solve, whereas the solid model took considerably longer. The first (fundamental) vibration mode was bending and is shown in Figure 19–11 for both models, together with the respective frequencies. The difference between the frequencies is about 1.9 percent. Further note that the mode shape is just that, a shape. The actual magnitudes of the deflections are unknown, only their relative values are known. Thus, any scale factor can be used to exaggerate the view of the deflection shape. The convergence of the 20-element beam model was checked by doubling the number of elements. This resulted in no change.

y

x (a)

y

x

(b)

Figure 19–11 First free vibration mode of step beam. (a) Twenty-element beam model, f1 = 322 Hz; (b) 56 384-element brick and tetrahedron model, f1 = 316 Hz.

15

For static deflection analysis, only three beam elements were necessary. However, because of mass distribution for the dynamics problem, more beam elements are necessary.

974      Mechanical Engineering Design y

x

(a) y

x

(b)

Figure 19–12

Second free-vibration mode of step beam. (a) Twenty-element beam model, f2 = 1296 Hz; (b) 56 384-element brick and tetrahedron model, f2 = 1249 Hz.

Figure 19–12 provides the frequencies and shapes for the second mode.16 Here, the difference between the models is 3.6 percent. As stated earlier, once the mode shapes are obtained, the response of the structure to various dynamic loadings, such as harmonic, transient, or random input, can be obtained. This is accomplished by using the mode shapes together with modal superposition. The method is called modal analysis.17

19–11  Summary As stated in Section 1–4, the mechanical design engineer has many powerful computational tools available today. Finite-element analysis is one of the most important and is easily integrated into the computer-aided engineering environment. Solid-modeling CAD software provides an excellent platform for the easy creation of FEA models. Several types of analysis have been described in this chapter, using some fairly simple illustrative problems. The purpose of this chapter, however, was to discuss some basic considerations of FEA element configurations, parameters, modeling considerations, and solvers, and not to necessarily describe complex geometric situations. Finiteelement theory and applications is a vast subject, and will take years of experience before one becomes knowledgeable and skilled with the technique. There are many sources of information on the topic in various textbooks; FEA software suppliers (such as ANSYS, MSC/NASTRAN, and ALGOR) provide case studies, user's guides, user's group newsletters, tutorials, etc.; and the Internet provides many sources. Footnotes 11, 12, and 14 referenced some textbooks on FEA. Additional references are cited below. 16

Note: Both models exhibited repeated frequencies and mode shapes for each bending mode. Since the beam and the bearing supports (boundary conditions) are axisymmetric, the bending modes are the same in all transverse planes. So, the second mode shown in Figure 19–12 is the next unrepeated mode. 17 See S. S. Rao, Mechanical Vibrations, 5th ed., Pearson Prentice Hall, Upper Saddle River, NJ, 2010, Section 6.14.

Finite-Element Analysis     975

Additional FEA References K. J. Bathe, Finite Element Procedures, Prentice Hall, Englewood Cliffs,   NJ, 1996. R. D. Cook, D. S. Malkus, M. E. Plesha, and R. J. Witt, Concepts and   Applications of Finite Element Analysis, 4th ed., Wiley, New York, 2001. D. L. Logan, A First Course in the Finite Element Method, 4th ed., Nelson, a   division of Thomson Canada Limited, Toronto, 2007. J. N. Reddy, An Introduction to the Finite Element Method, 3rd ed.,   McGraw-Hill, New York, 2002. O. C. Zienkiewicz and R. L. Taylor, The Finite Element Method, 4th ed.,   Vols. 1 and 2, McGraw-Hill, New York, 1989 and 1991.

PROBLEMS The following problems are to be solved by FEA. It is recommended that you also solve the problems analytically, compare the two results, and explain any differences.

19–1

Solve Example 3–6.

19–2 For Example 3–10, apply a torque of 23 730 lbf · in, and determine the maximum shear stress and angle of twist. Use 18 -in-thick plate elements.

19–3 The steel tube with the cross section shown is transmitting a torsional moment of

100 N · m. The tube wall thickness is 2.5 mm, all radii are r = 6.25 mm, and the tube is 500 mm long. For steel, let E = 207 GPa and ν = 0.29. Determine the average shear stress in the wall and the angle of twist over the given length. Use 2.5-mm-thick plate elements.

y

r r Problem 19–3 z

19–4 For Figure A–15–1, let w = 2 in, d = 0.3 in, and estimate Kt. Use 1∕4 symmetry and 1∕8-in-thick 2-D elements.

19–5 For Figure A–15–3, let w = 1.5 in, d = 1.0 in, r = 0.10 in, and estimate Kt. Use 1∕4 symmetry and 1∕8-in-thick 2-D elements.

976      Mechanical Engineering Design

19–6 For Figure A–15–5, let D = 3 in, d = 2 in, r = 0.25 in, and estimate Kt. Use 1∕2 symmetry and 1∕8-in-thick 2-D elements.

19–7 Solve Problem 3–136, using solid elements. Note: You may omit the top part of the eyebolt above the applied force.

19–8 Solve Problem 3–146, using solid elements. Note: Since there is a plane of symmetry,

a one-half model can be constructed. However, be very careful to constrain the plane of symmetry properly to assure symmetry without overconstraint.

19–9 Solve Example 4–11, with F = 10 lbf, d = 1∕8 in, a = 0.5 in, b = 1 in, c = 2 in, E = 30 Mpsi, and ν = 0.29, using beam elements.

19–10 Solve Example 4–13, modeling Figure 4–14b with 2-D elements of 2-in thickness.

Since this example uses symmetry, be careful to constrain the boundary conditions of the bottom horizontal surface appropriately.

19–11 Solve Problem 4–12, using beam elements. 19–12 Solve Problem 4–47, using beam elements. Pick a diameter, and solve for the slopes. Then, use Equation 7–18 to readjust the diameter. Use the new diameter to verify.

19–13 Solve Problem 4–69, using beam elements. 19–14 Solve Problem 4–94, using solid elements. Use a one-half model with symmetry. Be

very careful to constrain the plane of symmetry properly to assure symmetry without overconstraint.

19–15 Solve Problem 4–95, using beam elements. Use a one-half model with symmetry. At the plane of symmetry, constrain translation and rotation.

19–16 Solve Problem 4–96, using beam elements. Model the problem two ways: (a) Model

the entire wire form, using 200 elements. (b) Model half the entire wire form, using 100 elements and symmetry. That is, model the form from point A to point C. Apply half the force at the top, and constrain the top horizontally and in rotation in the plane.

19–17 Solve Problem 4–108, using beam elements. 19–18 Solve Problem 10–47, using beam elements. 19–19 An aluminum cylinder (Ea = 70 MPa, νa = 0.33) with an outer diameter of 150 mm

and inner diameter of 100 mm is to be press-fitted over a stainless-steel cylinder (Es = 190 MPa, νs = 0.30) with an outer diameter of 100.20 mm and inner diameter of 50 mm. Determine (a) the interface pressure p and (b) the maximum tangential stresses in the cylinders. Note: Solve the press-fit problem, using the following procedure. Using the planestress two-dimensional element, utilizing symmetry, create a quarter model meshing elements in the radial and tangential directions. The elements for each cylinder should be assigned their unique material properties. The interface between the two cylinders should have common nodes. To simulate the press fit, the inner cylinder will be forced to expand thermally. Assign a coefficient of expansion and temperature increase, α and ΔT, respectively, for the inner cylinder. Do this according to the relation δ = αΔTb, where δ and b are the radial interference and the outer radius of the inner member, respectively. Nodes along the straight edges of the quarter model should be fixed in the tangential directions, and free to deflect in the radial direction.

20

Geometric Dimensioning and Tolerancing

20 ± 0.1 4 × ∅ 10 +0.1 –0.0

∅ 0 M A BM

∅ 20 ± 0.05 ∅ 0.3 A B

0.1 A

0.05 0.00 ∅ 40 +–0.04

A

∅ 0.05 A B

Chapter Outline 20–1  Dimensioning and Tolerancing Systems  978 20–2  Definition of Geometric Dimensioning and Tolerancing   979 20–3

Datums  983

20–4

Controlling Geometric Tolerances   989

20–5

Geometric Characteristic Definitions   992

20–6

Material Condition Modifiers   1002

20–7

Practical Implementation   1004

20–8

GD&T in CAD Models   1009

20–9

Glossary of GD&T Terms   1010

977

978      Mechanical Engineering Design

Early in the design process the designer works largely from a macro point of view, in which components are nominally sized to satisfy the design requirements, typically including control of stresses and deflections. But when it comes to issues of manufacturability, fits between components, and assembly of components, the designer must take a much closer look at the precise specification of the parts. Section 1–14 addressed some of the basic issues of dimensioning and tolerancing that a designer must consider. In this chapter, the focus is on a standardized method of defining part geometry that takes into consideration that no part is perfectly formed—nor should it need to be. The method known as Geometric Dimensioning and Tolerancing allows clarity for functionality, flexibility for manufacturing, and level of precision for inspection.

20–1  Dimensioning and Tolerancing Systems The traditional method of dimensioning and tolerancing is referred to as the coordinate dimensioning system. In this system, every dimension is associated with a plus/minus tolerance, either directly specified immediately adjacent to the dimension, or implicitly specified with a general tolerance notation. This method of tolerancing has been used for many generations. It works reasonably well for parts that are not mass produced and that do not need to assemble with other parts. In general, this method is acceptable when a high level of precision is not needed. However, this system is lacking in many respects, particularly in its inability to address geometric issues of form and orientation. As an example, consider the part shown in Figure 20–1. This simple part is fully dimensioned and toleranced according to the traditional coordinate dimensioning system. In general, the designer's intent is clear, and most machine shops could manufacture such a part. But suppose during inspection of an actual part, it is found that the bar stock is not perfectly flat, the corners are not perfectly square, the hole is not perfectly perpendicular to the face of the bar stock, and the hole is not perfectly round. In fact, this will always be the case since manufacturing can never achieve perfect form. Figure 20–2 shows an exaggerated view of the imperfections of the manufactured part. The problem is that every dimension may be within its tolerance, but the part may be 4 2

2 1 DIA.

8 6

All dimensions +/− 0.01

Figure 20–1

Figure 20–2

Part dimensioned with traditional coordinate dimensioning system.

Exaggerated view of imperfections of a manufactured part.

Geometric Dimensioning and Tolerancing     979

unusable for its application owing to one of the geometric imperfections. A more challenging problem is that it is not even clear how to measure some of the dimensions. For example, the center of the hole is to be 2 inches from the edge of the part. If the hole is not perfectly round, how is its center defined? If the corner of the part is not square, from where should the 2 inches be measured? Bottom corner? Top corner? Closest edge? There is no defined correct answer with this dimensioning system. For many applications these issues can be overlooked because typical manufacturing methods are deemed to be good enough. However, mass production calls for the most efficient and inexpensive operation allowable. The manufacturer legitimately needs to be able to cut corners as much as possible. This requires the part specification to define precisely how good is good enough. In fact, the designer should always view the task of geometry specification as simultaneously restricting and freeing the manufacturer—restricting within the necessary limits for functional requirements, and freeing from unnecessary levels of perfectness. Proper balance of these two aspects provides for cost-effective and functional parts. Clearly, to address issues of functionality, manufacturability, interchangeability, and quality control requires parts that can be uniquely defined and consistently measured. This requires a dimensioning and tolerancing method that takes into account not only size, but geometric location, orientation, and form as well. This system is known as Geometric Dimensioning and Tolerancing (GD&T).

20–2  Definition of Geometric Dimensioning and Tolerancing Geometric Dimensioning and Tolerancing (GD&T) is a comprehensive system of symbols, rules, and definitions for defining the nominal (theoretically perfect) geometry of parts and assemblies, along with the allowable variation in size, location, orientation, and form of the features of a part. It serves as a means of accurately representing the part for the purposes of design, manufacture, and quality control. GD&T is not new. It has been developing as a standard within industry since the 1940s. Today, most of the major manufacturing companies utilize GD&T. The part previously considered in Figure 20–1 is shown again in Figure 20–3 using GD&T 0.005 A B

4 ± 0.1

Figure 20–3

∅ 1.002 1.000 ∅ 0.003 M A B C

2

Part dimensioned and toleranced with GD&T terminology.

2 ± 0.1

8 ± 0.1

A 6 0.003

C

0.005 A B

ion

ation

980      Mechanical Engineering Design

terminology. Unfortunately, many mechanical engineers are not able to interpret the drawing. During the time when the standard was becoming most prevalent in manufacturing, most engineering schools were phasing out comprehensive drafting courses in favor of computerized CAD instruction. Consequently, GD&T is often missing from the engineering curriculum. A full understanding of GD&T is usually obtained through an intensive course or training program, readily available to practicing engineers. Some mechanical engineers will benefit from such a rigorous training. All mechanical engineers, however, should at least be familiar with the basic concepts and notation. The purpose of the coverage of GD&T in this chapter is to provide this foundational exposure that is essential for all machine designers. The coverage is not comprehensive. The focus is on the foundational concepts and the most commonly used notation. A first exposure to GD&T can feel overwhelming, as there are numerous concepts and terminology that all depend upon each other. This chapter is organized to help a beginner gradually build the most important concepts first, adding detail as needed, and culminating with a section on practical applications. Section 20–9 includes a glossary of some of the most important vocabulary terms used in GD&T, and should be used as a reference for clarification of terms while reading the rest of the chapter. (a) size

(b) location

(c) orientation

GD&T Standards (b) location (c) orientation (d) form GD&T is defined and controlled by standards to provide uniformity and clarity on a global scale. One widely utilized standard is published by the American Society of Mechanical Engineers as ASME Y14.5–2009 Dimensioning and Tolerancing. It is part of the broader set of ASME Y14 standards that cover all aspects of engineering drawings and terminology. The International Organization of Standards (ISO) also publishes a series of standards that have been more commonly used in European countries. The two standards have developed in parallel and are mostly similar in concept and (c) orientation form terminology. The ASME (d) standard tends to put more emphasis on design intent while the ISO standards have a greater emphasis on metrology, or the measurement of the resulting part. According to the ASME approach, the parts are defined primarily in a way to ensure that they will perform the desired function, without specifying what equipment or processes should be used to manufacture or inspect the parts. The ASME Y14.5–2009 standard is utilized in this textbook. Four Geometric Attributes of Features A feature of a part is a general term referring to a clearly identifiable physical portion of a part. Examples include a hole, pin, slot, surface, or cylinder. There are four (d) form attributes of every feature that must be considered to define the geometry geometric of the feature:

(d) form

Figure 20–4 The four geometric attributes of a feature. (a) Size; (b) location; (c) orientation; (d) form.

1 2 3 4

Size Location Orientation Form

These attributes are illustrated in Figure 20–4. It is important in understanding GD&T to distinguish these four geometric attributes. They are each briefly described here, followed by further elaboration in the following sections. The term feature of size refers to a feature that has a size that can be measured across two opposing points, such as a hole, cylinder, or slot. A helpful rule of thumb is that a feature of size can usually be measured with the head of a caliper tool, such

Geometric Dimensioning and Tolerancing     981

Figure 20–5

Internal feature of size containing opposed points 40

Measuring opposing points of features of size with a caliper head.

±0

.01

∅ 40 ± 0.1

Not a feature of size since there are no opposed points

40

±0

.1

External feature of size containing opposed points

as illustrated in Figure 20–5. A dimension that is measured with the probe end of the caliper does not have opposed points, and is therefore not a feature of size. Such a dimension would be a locating dimension rather than a size dimension. Location refers to the location of a feature with respect to some origin of measurement. Orientation refers to the angle of a feature or centerline of a feature with respect to some origin of measurement. It includes parallelism, perpendicularity, and angularity. Form refers to imperfections in the shape of a feature, and includes straightness, flatness, circularity, and cylindricity. According to the GD&T standard, plus/minus tolerancing should only be directly applied to the size dimension of a feature of size. The other attributes (location, orientation, and form) are controlled through geometric tolerancing, described in Section 20–4. Symbolic Language The ASME Y14.5–2009 standard utilizes an international language of symbols that minimizes the need for, and potential confusion from, written notes on machine drawings. The symbols are classified into two categories—geometric characteristic and modifying. For reference, a brief introductory summary of the symbols will be given here. The usage of the symbols will be clarified in later sections. Table 20–1 shows the 14 geometric characteristics and their symbols. These geometric characteristics are refinements of the geometric attributes (size, location, orientation, and form), and are each used to control a geometric tolerance of a feature. Each geometric characteristic has a symbol that is used on the drawing to specify a tolerance zone associated with a geometric characteristic. The geometric characteristic symbols are directly associated with the geometry of a feature, not with a size dimension. This is why they are referred to as geometric tolerancing. As shown in the table, the geometric characteristics are further subdivided into the types of tolerance defined by GD&T (form, profile, orientation, location, and runout), as well as the broader geometric attribute being controlled (size, location, orientation, and form). The table also indicates whether the symbol can be associated with a datum reference and any material condition modifiers, which will be clarified in later sections.

982      Mechanical Engineering Design

Table 20–1  Geometric Characteristic Controls and Symbols Type of Tolerance

Geometric Characteristic

Geometric Attribute Controlled

Symbol

Datum Referencing?

Straightness Form

⏥ ○

Flatness Circularity

Form

No

M

L

or RFS

M

L

or RFS

RFS

Cylindricity Profile

Material Condition Modifier Allowed

RFS

Profile of a line Profile of a surface

Location, orientation, size, and form

Optional

M

L

or RFS

Orientation

Required

M

L

or RFS

M

L

or RFS

Angularity Orientation

Perpendicularity

Parallelism

⫽ Location and orientation of feature of size

Position Location

Concentricity

Location of derived median points or planes

Symmetry Runout

Circular runout

Location of cylinder

Total runout

Required

RFS RFS

Required

RFS RFS

Tables 20–2 and 20–3 show the modifying symbols. Table 20–2 includes the dimensioning modifiers and their symbols. These are used to modify or clarify the meaning of a dimension on the drawing. Table 20–3 includes the tolerance modifiers and their symbols. These are used in a feature control frame (to be defined later) to modify or clarify the tolerance specification.

Table 20–2  Dimensioning Modifiers and Symbols Description

Symbol

Basic Dimension

98

Description

Symbol

Dimension Origin

Diameter

All Around

Spherical Diameter

S∅

All Over

Radius

R

Independency

I

Spherical Radius

SR

Continuous Feature

CF

Controlled Radius

CR

Counterbore

Square

Countersink

Reference

()

Spotface

Arc Length

Depth

SF

Geometric Dimensioning and Tolerancing     983

Table 20–3  Tolerance Modifiers and Symbols Description Maximum Material Condition (applied to tolerance) Maximum Material Boundary (applied to datum) Least Material Condition (applied to tolerance) Least Material Boundary (applied to datum) Translation

Symbol M

L

Projected Tolerance Zone

P

Free State

F

Tangent Plane

T

Statistical Tolerance

ST

Between Unequally Disposed Profile

U

20–3  Datums Several concepts are foundational to the implementation of GD&T. In this section, the concept of datums will be explained, followed by specific methods of implementation in GD&T. The geometric characteristics of features are defined and measured by relating them to clearly defined datums. A datum is an origin from which location or orientation of part features is established. Note that size dimensions and form control do not require an origin for measurement, and therefore do not need to be referenced to a datum. For a feature of a part to be manufactured or inspected, the entire part is located with respect to a datum reference frame. A datum reference frame is a set of up to three mutually perpendicular planes that are defined as the origin of measurement for locating the features of a part. The datum reference frame is idealized and geometrically perfect. It is necessary to consider its relationship to the nonideal physical part and the processing equipment. To do so, it is necessary to distinguish between several related terms, namely datum, datum reference frame, datum feature, and datum feature simulator. A datum feature is a nonideal physical surface of the part that is specified in order to establish a theoretically exact datum. A datum feature is always a surface of the part that can be physically touched, not a centerline or other theoretical entity. Since the datum feature is not perfect, it is not directly used for measurements. Suppose a flat surface of a part is selected as a datum feature. The surface is an imperfect plane with localized hills and valleys, and is not perfectly flat. If the part is placed on a polished granite surface plate, a minimum of three high points on the surface of the datum feature will contact the nearly perfect plane of the polished surface plate. The surface plate serves as a datum feature simulator of the actual datum feature. A datum feature simulator is a precision embodiment, such as a surface plate, gauge pin, or machine tool bed, of the datum described by an imperfect datum feature. The datum feature simulator is often a physical gauging surface, but may also be simulated by "soft gauging" optical or probing methods. The datum itself is a theoretically exact point, axis, or plane derived from the datum feature simulator. A recap of the relationship between the various datum terms may be helpful, using the example shown in Figure 20–6. An imperfect actual surface of a part, such

984      Mechanical Engineering Design

Figure 20–6 Example demonstrating datum terminology.

Part

Datum

Datum feature simulator

Datum feature

Granite surface plate

as the bottom surface, is designated as a datum feature. The datum feature (bottom surface) is placed in contact with a nearly perfect datum feature simulator (granite surface plate). A theoretical datum (true plane) is defined in association with the datum feature simulator. The process is repeated as necessary to define enough datums to obtain a three-plane datum reference frame. For example, if the back surface and one side surface are also selected as datum features, then the three-plane datum reference frame in Figure 20–7 may be obtained. How, then, are locations and orientations of features handled through the process of design, manufacture, and inspection? The designer specifies datum features that are best suited for the functionality, manufacture, and inspection of the part. Locations and orientations are defined by the designer on the drawing with respect to the datum reference frame. They are actually manufactured with respect to a datum feature simulator inherent in the manufacturing equipment, such as the table surface of a milling machine. They are measured for quality control with respect to a datum feature simulator, such as a granite surface plate. Note that measurements of location and orientation are not made with respect to the actual surface of the datum feature, but from the datum feature simulator. Figure 20–7

Z

Example demonstrating a threeplane datum reference frame.

w

v Y

X

u

Geometric Dimensioning and Tolerancing     985 Second datum plane constrains 2 DOF 1. Translation in X 2. Rotation in w

First datum plane constrains 3 DOF 1. Translation in Z 2. Rotation in u 3. Rotation in v

Third datum plane constrains 1 DOF 1. Translation in Y

w

Figure 20–8 Immobilization of a part by sequential application of datum planes.

Z v u

Y X (a)

(b)

(c)

Immobilization of the Part The selection of datum features can be thought of as selecting which surfaces of the part will be put into contact with datum feature simulators in order to immobilize the part for manufacture and inspection. The part floating in space has six degrees of freedom (three translations and three rotations). Each datum constrains some of the degrees of freedom in order to immobilize the part in a precise, repeatable location. Consider the process of immobilizing a part with three datum planes, demonstrated in Figure 20–8. First, let the bottom surface of the part be selected as a datum feature to be constrained by the first datum plane, as shown in Figure 20–8a. Remember that the datum feature is imperfect, so it may touch the datum plane in only a few places. Specifically, a minimum of three points of contact are required to prevent the part from rocking on the datum plane. This contact with the datum plane will constrain three degrees of freedom of motion of the part: translation in Z, rotation in u, and rotation in v. Next, let the back surface of the part be designated as the second datum feature, from which the second datum plane is derived, as shown in Figure 20–8b. Imagine holding the part in contact with the first datum plane and sliding it into contact with the second datum plane. It must touch in a minimum of two points to stabilize it with respect to the second datum plane. This will constrain an additional two degrees of freedom of motion: translation in X and rotation in w. Finally, let a side surface be designated as the third datum feature to define the third datum plane, as shown in Figure 20–8c. Maintaining contact of the part with the first two datum planes, and sliding it into contact with the third datum plane will result in a minimum of one point of contact with the third datum plane. This constrains the final degree of freedom: translation in Y. The part is now fully constrained in a precise, repeatable location. Order of Datums Notice that the order of application of the datum planes is important. Suppose the part from Figure 20–8 is constrained by the first datum plane as before, but then the order of application of the second and third datum planes is reversed. Figure 20–9a shows a top view of the part that has been constrained by the YZ plane first, then by the XZ plane. Figure 20–9b shows the same part with the order of application of the two datum planes reversed. The final position of the two parts is not the same. Because measurements are made from the datum planes, not from the edges of the part itself, the measured locations of features on the part are clearly dependent on the choice of datum features and the order of application of the resulting datum planes. It is necessary for the part drawings to specify clearly, for each feature to be located on the part,

986      Mechanical Engineering Design Datum planes of datum reference frame

2 point contact Z

1 point contact Z

Y

1 point contact

2 point contact

X

Datum feature

Y

Datum axis

X (a)

(b)

Figure 20–9

Figure 20–10

Comparison of order of application of datum planes. (a) YZ plane constrained first; (b) XZ plane constrained first.

An example of a hole as a datum feature.

the datum features as well as the order of application of the resulting datum planes. The features do not all have to use the same datums and order of application. Nonplanar Datum Features So far, only planar datum features have been presented since they are the easiest to visualize the progression from the datum feature to the datum reference frame. Several other datum features are also provided for in the Y14.5 standard. In particular, cylindrical features such as shafts, bosses, and holes are often useful as datum features. Suppose in the part shown in Figure 20–10, the hole is selected as a datum feature. The actual surface of the hole is the datum feature; the center axis of the hole is the datum. The center axis defines the intersection of two perpendicular datum planes. In conjunction with another datum feature, say the back surface, the part is constrained and a datum reference frame is defined. Actual Mating Envelopes In the previous paragraph, it was stated that the center axis of the hole is the datum. This is a simplified statement of a more detailed concept that warrants a better explanation. Since the hole feature is imperfect in form (that is, it does not have a perfectly circular cross section, or a perfectly straight centerline, or a perfectly smooth surface), how is its theoretically perfect datum axis determined? To answer this question, a few GD&T terms will be introduced. An actual mating envelope is a perfectly shaped counterpart of an imperfect feature of size, which can be contracted about an external feature, or expanded within an internal feature, so that it contacts the high points of the feature's surface. For example, Figure 20–11a shows an imperfect dowel pin (the feature of size) that is circumscribed by the smallest possible perfect cylinder (the actual mating envelope). The imperfect pin does not technically have a center axis. Instead, it has a collection of derived median points that represent the centroids of each cross section. When referring to the center axis of an imperfect feature such as the pin, what is actually meant is the theoretically perfect center axis of the theoretically perfect actual mating envelope of the pin. The same concept can be applied to a feature of size with an internal surface, such as the hole feature shown in Figure 20–11b. Actual mating envelopes are categorized as related or unrelated to a datum. An unrelated actual mating envelope is sized to fit the feature without any constraint to any datum. In other words, it is free to float to find the best fit. A related actual

Geometric Dimensioning and Tolerancing     987 Actual mating envelope (smallest circumscribed cylinder, shown in section)

Imperfect hole (feature of size) Actual mating envelope (largest gauge pin)

Imperfect pin (feature of size)

Figure 20–11 Definition of terms for an actual mating envelope. (a) External feature; (b) internal feature.

Derived median points of pin Derived median points of hole "Center axis" of hole, defined by the theoretically perfect axis of the actual mating envelope

"Center axis" of pin, defined by the theoretically perfect axis of the actual mating envelope (a)

(b)

mating envelope is sized to fit the feature while maintaining some constraint in orientation or location with respect to a datum. For example, for the hole feature in Figure 20–10, the related actual mating envelope with respect to the back plane datum surface is the largest pin that can fit in the hole while being held perpendicular to the back datum plane. Now, returning to the datum example in Figure 20–10, the datum axis corresponding to the datum feature (the hole) is defined by the actual mating envelope of the hole, that is, the largest cylinder that can fit within the hole. In practical implementation, this largest cylinder can be determined by physically inserting very precisely manufactured gauge pins of increasing size until the largest one is found. Alternatively, an expanding mandrel can be used. The largest gauge pin serves as the datum feature simulator (previously defined). In the case of an external datum feature, such as the surface of a shaft, the datum feature simulator is typically the jaws of a chuck or collet that is closed onto the surface. The center axis of the chuck is then the datum axis. Datum Feature Symbol On a drawing, a datum feature is defined symbolically by a capital letter enclosed in a square frame, attached to a leader line that terminates at the datum feature with a triangle. The triangle can be filled or unfilled. Any letter can be used, except I, O, and Q, which may be confused with numbers. The letters need not be assigned alphabetically as the precedence of the datums will be specified later as needed for each feature to be controlled. The triangle may attach directly to the datum feature surface outline, point to it with a leader line, or attach to an extension line to the surface. All three of these methods are illustrated in Figure 20–12, where the datum features correlate to those previously demonstrated for the part in Figure 20–8. Figure 20–12

50 ± 0.1 C

B

A

Three methods of designating a datum feature.

Datum axis defined by B Datum feature B

988      Mechanical Engineering Design

(a) ∅ 15 ∅ 0.1 M A B

Datum feature A A

D

E

Datum axis defined by A

∅ 15

A

B ∅ 30

40

10

∅ 10

B Datum axis defined by B

∅ 15

Datum feature B

C (b)

(a)

Figure 20–13

∅ 15

0.1 axis M Aor Bcenter plane of a feature of size as a datum. (a) Datums A and B are defined as the center axes of two Methods of designating∅ an different cylindrical surface features. (b) Datum symbols A and B are A attached to width dimensions, defining the center planes as the datums. D E ∅ 15 Datum symbols C, D, and E are attached to features with axes, defining the center axes as the datums.

10 B

(b)

If the datum is to be an axis or center plane of a feature of size, then the datum 40 triangle is placed in-line with the dimension line of the feature of size. In the case of a cylinder it may be attached directly to the surface of the cylinder. The triangle may optionally replace one of the dimension arrowheads if both will not fit. Several exam15 ples∅ are shown in Figure 20–13. A datum triangle always indicates the datum feature C (a physical surface) from which the datum (a theoretical axis or center plane) is derived. Accordingly, the triangle is never placed directly on an axis, centerline, or center plane. The datum triangle may also be attached to a feature control frame (to be defined in a later section) that controls the geometric tolerance of the datum feature. An example is included in Figure 20–13b. Note that in the case of a feature of size, the subtle difference in placement of the triangle in Figures 20–14a and 20–14b has a significant difference in meaning. When the triangle is placed in-line with the dimension line, as in Figure 20–14a, the datum is the center plane of the feature of size. When the triangle is placed away from the dimension line, as in Figure 20–14b, the datum is the plane defined by the edge of the part. The center plane datum would likely be used if it was important that the hole be centered in the part, regardless of fluctuations in the overall width of the part. The edge datum has the advantage of simpler setup for manufacturing and inspecting, but because the hole location is controlled from one edge, the part may not be symmetric.

Figure 20–14 Different datum plane designation due to placement of datum symbol. (a) Symbol in-line with dimension; (b) symbol out-of-line with dimension.

Datum plane A A

Datum plane A

A

60 ± 0.1

60 ± 0.1

(a)

(b)

Geometric Dimensioning and Tolerancing     989

20–4  Controlling Geometric Tolerances Tolerance Zones GD&T generally only uses direct application of plus/minus tolerances when dimensioning a feature of size. A fundamental concept of GD&T is that in addition to any dimensional tolerance on a feature of size, the geometric shape and location of the surfaces must be controlled to stay within tolerance zones. Tolerance zones are defined in a variety of ways, such as by two parallel planes or concentric cylinders, to define limiting boundaries for the physical surfaces of the parts. The tolerance zones are defined relative to a theoretically exact location or shape. The actual shape and location of the part surfaces may vary from the theoretically exact location and shape, so long as they stay within the limiting boundaries of the tolerance zones. There are special terms defined to represent the maximum and minimum boundaries of a feature of size. The maximum material condition (MMC) is the condition in which a feature of size contains the maximum amount of material within the stated limits of size. For an external feature, such as the outer surface of a shaft, the MMC is when the shaft diameter is at its maximum value allowed by the tolerance. For an internal feature, such as a hole, the MMC is when the hole is at its smallest diameter allowed by the tolerance. Similarly, the least material condition (LMC) is the condition in which a feature of size contains the least amount of material within the stated limits of size. This would correlate with the smallest allowed shaft diameter, or the largest allowed hole diameter. These terms are used extensively in GD&T. The Y14.5 standard specifies a default tolerance zone for features of size through what is referred to as Rule #1, also known as the envelope principle. This rule states that when only a tolerance of size (i.e., a plus/minus tolerance) is specified for a feature of size, the limits of size prescribe the extent of which variation in its geometric form, as well as size, are allowed. Specifically, the envelope principle states that the surface of a feature of size may not extend beyond an envelope of perfect form at MMC. Consider a simple example of a dowel pin in Figure 20–15a. The limiting envelope is a perfectly formed pin (e.g., perfectly straight, perfectly circular cross section, etc.) that is at its largest allowed size, as shown in Figure 20–15b. An imperfect dowel pin would still meet the specifications so long as its diameter at any Figure 20–15

∅ 20 ± 0.5

(a) Pin with imperfect form, always within the MMC envelope ∅ 20.5

∅ 20.5

Envelope of perfect form at MMC (b)

Envelope of perfect form at MMC Pin diameter is between 19.5 and 20.5 at every cross section (c)

The envelope principle (Rule #1). (a) Size and tolerance specification on the drawing; (b) the envelope of perfect form at MMC; (c) an acceptable imperfect pin within the envelope.

990      Mechanical Engineering Design

location is within the allowed tolerance, and its surface does not exceed the envelope. An implication of this is that if the pin is manufactured at its MMC, then it must have perfect form. As the diameter is reduced from the MMC, the pin may deviate from perfect form, as shown in Figure 20–15c. Other geometric controls (described in Section 20–5) may be specified when the default tolerance zone provided by Rule #1 is not sufficient to meet the requirements of the application. Basic Dimensions A theoretically exact location is specified with a basic dimension. A basic dimension is a theoretically exact dimension which does not have a tolerance directly associated with it, but is instead associated with a geometric control of a tolerance zone. When a basic dimension is used to locate a part feature, the feature itself must include a geometric control that defines a tolerance zone specifying the permissible variation from perfect form and location. Basic dimensions are indicated on a drawing by enclosing the dimension within a rectangular box, or by a general note indicating that all untoleranced dimensions are basic. Feature Control Frames A geometric control is specified on a drawing in a feature control frame. A feature control frame is a rectangular box attached to a feature on a drawing, containing the necessary information to define the tolerance zone of the specified feature. The frame is subdivided into compartments in a specific order, as shown in the example in Figure 20–16. The first compartment always contains one of the geometric control symbols from Table 20–1 to indicate what aspect of the feature is being controlled by the tolerancing information to follow. The second compartment always contains a numerical value designating the total allowed tolerance. The tolerance value specified is always a total tolerance, that is, the entire range of the tolerance, not a plus/minus tolerance value from a midpoint. If the tolerance is circular or cylindrical, the diameter symbol will precede the specified tolerance. The third and following compartments are used as needed to define the datum(s) necessary to immobilize the part. The order of the datum letters from left to right defines the precedence of the application of the datums. The number of datum letters may vary from zero to three, depending on the datum reference frame required for the particular tolerance being controlled. Tolerances of form affect only the designated feature, independent of any other feature or datum (as indicated in Table 20–1), and therefore never include datum reference letters in the feature control frame. Tolerances of location, orientation, and runout always relate the Figure 20–16

Geometric control symbol (from Table 20 –1) Tolerance

Example of a feature control frame. ∅ 0.25

M

A

Datum references

B

C M Datum modifier

Tolerance modifier Total tolerance Diameter symbol

Geometric Dimensioning and Tolerancing     991 This feature must be positioned within a cylindrical

Figure 20–17 tolerance zone of 0.25 total

∅ 0.25

M

Reading a feature control frame. when the feature is produced at its maximum material condition

A

B

C M

with respect to the datum reference frame established by datum features A (primary), B (secondary), and C (tertiary), where datum feature C is at maximum material condition.

designated feature to some other feature or datum, and therefore always require the specification of datum reference letters. Modifiers from Table 20–3 can be included in a compartment immediately following a tolerance value, or after a datum specification. The effect of these modifiers is discussed in Section 20–6. The feature control frame is read from left to right, as illustrated in Figure 20–17. If more than one geometric control is to be applied to the same feature, the feature control frames can be stacked, and are applied from top to bottom. A feature control frame controls the feature to which it is attached. The four attachment methods are as follows: 1 A leader from the feature control frame points directly to the feature. See Figure 20–18a. 2 The feature control frame is attached to an extension line from a planar feature. See Figure 20–18b. 3 The feature control frame is placed below a leader-directed dimension or note pertaining to the feature. See Figure 20–18c. 4 The feature control frame is attached to an extension of the dimension line pertaining to a feature of size. See Figure 20–18d. .003

Figure 20–18

A

Four methods of attaching a feature control frame to a feature.

∅ 10 ± .005 ∅ 10 ± .005 .002 A (a)

(b)

∅ 10 ± .005 ∅ 10 ± .005 ∅ .003 (c)

∅ .003 (d)

992      Mechanical Engineering Design

A feature control frame associated with a feature of size may control either the actual surface of the feature, or the axis or centerline of the feature. To control the axis or centerline of a feature of size, the feature control frame is associated with the dimension of the feature of size using either attachment method 3 or 4. The dimension and feature control frame should be shown in a drawing view in which the axis or centerline appears as a line. Further, if the feature dimension is for a diameter, the diameter symbol precedes the tolerance value in the feature control frame. As an example, in Figure 20–18a, the straightness control applies to the surface of the small cylinder, whereas in Figures 20–18c and 20–18d, the straightness control applies to the axis of the small cylinder. The significance of the difference between controlling the surface or the axis is described later.

20–5  Geometric Characteristic Definitions Each of the geometric characteristic symbols in Table 20–1 is used to define a tolerance zone particular to a certain geometric characteristic. The geometric controls are categorized as controls of form, orientation, profile, location, and runout. Some of the controls are quite general and encompass most of the common needs, while some are very specific to a particular geometric need. For reference, a basic description will be given for each geometric control, followed by a broader discussion on practical implementation. The reader may find it helpful to only skim this section the first time through, then refer back to it as a reference when the details are needed for practical applications. Form Controls The four geometric characteristics that provide form control are

1 Straightness 2 Flatness 3 Circularity 4 Cylindricity.

These control the form of an individual feature, independent of that feature's location or relationship to any other feature. Consequently, form controls never include reference to datums. Note that the form controls are a further refinement of any size tolerance, which must also be satisfied. Straightness — The straightness control specifies a tolerance zone within which line elements of a surface or an axis must lie. When applied to the surface of a feature with either a leader line or an extension line to the surface, the straightness applies to all lines within the surface that appear as straight lines in the drawing view. See Figure 20–19. When the feature control frame is applied beneath the size dimension of a feature of size, the straightness control is on the axis or derived median line of the feature. The diameter symbol is included with the straightness tolerance to apply a cylindrical tolerance zone to the axis of a cylinder. See Figure 20–20. Flatness ⏥ The flatness control specifies a tolerance zone of a specified distance between two parallel planes within which all points of a surface (or derived median plane) must

Geometric Dimensioning and Tolerancing     993

Translation: Line elements in this surface, that are seen as lines in this view, must be straight within 0.05.

0.05

0.05 tolerance zone for each line in the surface.

or

Figure 20–19 Straightness control applied to a surface. (a) Drawing specification; (b) Interpretation.

0.05

20 ± 0.1 (b)

(a)

∅ 20 ± 0.1 ∅ 0.05

Translation: The derived median line of the feature must be straight within a cylindrical tolerance zone with a diameter of 0.05.

∅ 0.05

Straightness control applied to an axis of a feature. (a) Drawing specification; (b) Interpretation.

(b)

(a)

Translation: The entire surface seen as a line in this view, must be flat within 0.05, i.e. all points must lie between two parallel planes separated by 0.05. 0.05

or 0.05

20 ± 0.1 (a)

Figure 20–20

0.05

(b)

lie. The feature control frame is applied with a leader line or an extension line to the surface in a drawing view in which the surface appears as a line. See Figure 20–21. If the feature control frame is applied beneath the size dimension of a feature of size, the flatness control is on the derived median plane rather than on the surface. The flatness control is often used to provide additional control of the primary datum feature to improve reproducibility of measurement. Circularity ◯ The circularity control is used to control the periphery of the circular cross sections of a cylinder, cone, or sphere. The tolerance zone is the annulus between two concentric circles within which each circular element of the surface must lie. See

Figure 20–21 Flatness control. (a) Drawing specification; (b) Interpretation.

994      Mechanical Engineering Design

Figure 20–22

Translation: Every circular element (slice) in this surface must lie between two concentric circles with a radial difference of 0.01.

Circularity control. (a) Drawing specification; (b) Interpretation. 0.01

0.01

∅ 30 ± 0.1 (b)

(a)

Figure 20–22. The circularity control should be used sparingly as it is difficult to inspect, since every circular cross section of the surface must be evaluated independently from one another and independently from any datum. The runout or profile controls provide alternate methods that are usually sufficient to ensure circularity with easier inspection methods. Cylindricity The cylindricity control is used to control a combination of circularity and straightness of a cylinder. The tolerance zone is the space bounded by two concentric cylinders with a difference of radius equal to the tolerance. See Figure 20–23. Orientation Controls ⟂⫽ The three geometric characteristics that provide orientation control are 1 Angularity 2 Parallelism 3 Perpendicularity. These control the orientation of a feature with respect to one or more datums, therefore mandating the inclusion of at least one datum reference in the feature control frame. The parallelism and perpendicularity controls are essentially convenient subsets of the angularity control in which the desired angle is 0° or 90°, respectively. The Figure 20–23

Translation: The entire surface of the cylinder must lie between two concentric cylinders with a radial difference of 0.01.

Cylindricity control. (a) Drawing specification; (b) Interpretation. 0.01

0.01

∅ 30 ± 0.1 (a)

(b)

0.01

Geometric Dimensioning and Tolerancing     995

orientation controls may be applied to surfaces, axes, or center planes. In the case of surfaces and planes, the tolerance zone is defined by two parallel planes oriented at the specified basic angle from the datum reference. See Figures 20–24 and 20–25. When controlling axes or center planes, the feature control frame is placed beneath the size dimension. When controlling axes, the diameter symbol is included before the tolerance, and the tolerance zone is cylindrical, as demonstrated in Figure 20–26. Translation: The entire surface must lie between two parallel planes 0.2 apart which are inclined at a basic angle of 30° to datum plane A.

Figure 20–24 Angularity control. (a) Drawing specification; (b) Interpretation.

0.2 A 30°

30°

0.2 A

Datum plane A (b)

(a)

Translation: The entire surface must lie between two parallel planes 0.2 apart which are perpendicular to datum plane A. Note that this orients datum plane B with respect to datum A.

Figure 20–25 Perpendicularity control. (a) Drawing specification; (b) Interpretation.

0.2 A 0.2

B

90°

A

∅ 30 ± 0.1 ∅ 0.05 A

Datum plane A

Figure 20–26 Cylindrical tolerance zone

An orientation control applied to the axis of a feature. (a) Drawing specification; (b) Interpretation.

A

∅ 0.05

Axis of hole Translation: The axis of the hole feature must lie within a cylindrical tolerance zone with 0.05 diameter and perpendicular to datum plane A.

Datum plane A

996      Mechanical Engineering Design

Orientation tolerances are constrained only in rotational degrees of freedom with respect to the referenced datum. Since the translational degrees of freedom are not constrained by orientation tolerances, an orientation tolerance zone cannot be used to locate a feature. It should only be used as a refinement of a tolerance that is doing the locating, such as position or profile of a surface. The most common use of the orientation controls is to orient a secondary or tertiary datum with respect to the primary datum plane. Profile Controls Profile controls are used to define a tolerance zone around a desired true profile that is defined with basic dimensions. The two geometric characteristics that provide profile control are 1 Profile of a line 2 Profile of a surface. Profile of a line is a two-dimensional tolerance zone that controls each line within the feature's surface, similar to straightness or circularity controls. Profile of a surface applies a three-dimensional control, similar to flatness or cylindricity. Profile controls are often used for irregularly shaped features and for castings, forgings, or stampings where it is desired to provide a tolerance zone for the overall surface of the part. A profile tolerance is implied to be an overall tolerance that is centered on the true profile. A nonsymmetrical tolerance zone can be specified by following the overall tolerance value with the unequally disposed profile modifier (an uppercase U in a circle), followed by the amount of tolerance that is in the direction that would allow additional material to be added to the true profile. The feature control frame is attached with a leader line in a drawing view where the true profile is shown. The profile tolerance applies only to the individual feature surface, unless modified with the "between," "all around," or "all over" symbols, as shown in Figure 20–27. The profile controls are the only geometric characteristics that have the option of including or not including a datum reference. If the profile tolerance does not Figure 20–27 Application of the profile control. (a) Applied to a single surface; (b) applied between two designated points; (c) applied all around; (d) applied all over.

0.2

0.2

G

G

H (b)

(a)

0.2

0.2

(c)

H

(d)

Geometric Dimensioning and Tolerancing     997

reference a datum, then the tolerance zone "floats" around the true profile, providing a form control of the surface, but not a location control. This option should be used sparingly, as it usually makes the inspection of the part more difficult. If a datum is referenced, then the profile tolerance can simultaneously control size, form, orientation, and location of a feature. This general capability makes this control extremely useful as an overall default tolerance control. When the profile tolerance feature control frame is placed in a general note on the drawing, the tolerance applies to all features in the drawing unless otherwise specified. With this general note, other controls (e.g., flatness, perpendicularity, etc.) are needed only if a tighter control is desired than provided by the general profile tolerance. Consider the example in Figure 20–28. The bottom surface is the datum feature that defines datum A. The flatness control establishes that when this surface is in contact with datum plane A, all points on the surface must be within the 0.05 tolerance zone. Datum plane B is defined to be exactly perpendicular to datum plane A, but the datum feature B (the actual surface), may vary in form and orientation so long as it stays within the 0.2 tolerance zone. The profile control on the curved surface indicates that the part is first placed into contact with datum plane A, then into contact with datum plane B. Then the ideal curved surface is defined with the basic dimensions. Then a tolerance zone is defined as the space between two curved surfaces centered around the true surface with a space of 0.1 between them. In addition, Translation: This surface must lie within a tolerance zone defined as the space between two surfaces separated by 0.1 and centered on the ideal surface. The ideal surface is defined with the basic dimensions, while the part is in contact with datum A and datum B. 0.2 tolerance zone 0.1 tolerance zone

0.1 A B R4

Datum B

B These surfaces are controlled by the default surface profile of 0.3 (tolerance zones not shown).

2

0.05 A

90°

0.05 tolerance zone 3

2 0.2 A

UNLESS OTHERWISE SPECIFIED UNTOLERANCED DIMENSIONS ARE BASIC Default surface profile specification.

Datum A

0.3 A B (a)

(b)

Figure 20–28 An example using a profile control note as an overall default tolerance control, with tighter controls added as needed. (a) Drawing specification; (b) Interpretation.

998      Mechanical Engineering Design

owing to the profile tolerance of 0.3 in the general note at the bottom of the drawing, all other surfaces that do not have a tighter tolerance specified will be within a profile tolerance zone of 0.3 centered on the basic (ideal) shape. Any deviation of size, form, orientation, and location is allowable so long as the surfaces remain within these tolerance zones. Location Controls There are three location controls: 1 Position 2 Concentricity 3 Symmetry. The location controls are specified to control the location of a feature of size, such as a hole, slot, boss, or tab, with respect to a datum or another feature. Position The position control is one of the most effective and oft-used controls, as it incorporates most of the advantages of GD&T. The position control defines the allowed location (and orientation) of the axis, centerline, or center plane of a feature of size. It does not control the size or form of the feature. The application of the position control is interpreted in Figure 20–29. The true position of a feature of size is first located with respect to datums by specifying basic dimensions to the axis, centerline, or center plane of the feature of size. Then the size of the feature of size is directly dimensioned, along with a plus/minus tolerance. Finally, the position control is applied with a feature control frame placed beneath the feature's dimension. The position control specifies a tolerance zone centered around the theoretically exact location of the feature's axis, centerline, or center plane. The tolerance Figure 20–29

Translation: The axis of the hole may be located and oriented anywhere within a 0.15-diameter cylindrical tolerance zone which is ideally located by the basic dimensions, with the part located by datums A, B, and C.

Position control. (a) Drawing specification; (b) Interpretation.

B

First datum plane A

4 ± 0.1 2

8 ± 0.1

Second datum plane B

∅ 1 ± 0.1 ∅ 0.15 A B C

Tolerance zone cylinder, with 0.15 diameter centered on true axis

2 ± 0.1

Axis of tolerance zone, exactly located with respect to datums

A

6

Axis of hole, located anywhere within the tolerance zone

2

6

C

Third datum plane C (a)

(b)

Geometric Dimensioning and Tolerancing     999

Figure 20–30

60 ± 0.2 10

20

C

B

20

Drawing specification for position control applied to a group of features. The interpretation is shown in Figure 20–31.

10 ± 0.2

10

30 ± 0.2 A

Position control applied to all 3 holes.

3 × ∅10 ± 0.1 First row is PLTZF, controlling the overall location of the pattern. ∅5 A B C ∅2 A Second row is FRTZF, controlling the feature-to-feature relationship within the pattern.

zone is cylindrical if the diameter symbol precedes the tolerance; otherwise, it is the space between two parallel planes. The specification for the part shown in the machine drawing in Figure 20–29a leads to the interpretation shown in Figure 20–29b. Note that the hole's diameter may vary within its specified dimensional tolerance, while the hole's axis may be at any position and orientation within the cylindrical tolerance zone. The position control also provides an excellent means of controlling the location of a group of features of size. A group of features is indicated to share the same dimension, dimensional tolerance, and position control by including the number of intended features preceding the specification of the size dimension, such as shown by the 3 × in Figure 20–30. Figure 20–30 also demonstrates the use of a composite control frame where there are two rows associated with the position control specification. This allows different tolerance specifications for the overall location of the pattern and the interrelation of features within the pattern. This is commonly needed when the feature pattern must mate with similarly spaced features on another part. The first row in the composite control frame is referred to as a Pattern-Locating Tolerance Zone Framework (PLTZF), pronounced "plahtz." The PLTZF applies to the overall location of the pattern as a group with respect to the datums. The PLTZF defines tolerance zones for the centerline of each feature just like previously described for the position control. The second row of the composite control frame adds additional constraint to the feature-to-feature relationship within the pattern, and is referred to as the FeatureRelating Tolerance Zone Framework (FRTZF), pronounced "fritz." The FRTZF is applicable to the basic dimensions between the features, but not to the basic dimensions locating the features with respect to the datums. The FRTZF defines another smaller tolerance zone for each feature, centered around the exact locations as defined by the basic dimensions between the features. The FRTZF tolerance zones may float anywhere within the PLTZF tolerance zones, as long as they maintain their relative positions with respect to each other. The actual feature centerlines must lie within the FRTZF tolerance zones, whose centerlines must in turn lie within the PLTZF tolerance zones. Consider the example specified in Figure 20–30, where the tolerance specifications are larger than typical in order to allow the tolerance zones to be more easily visualized. Figure 20–31 shows the tolerance zones, without showing the actual holes, to minimize the clutter. The first row of the composite control frame specifies a

1000      Mechanical Engineering Design

Figure 20–31 Interpretation of the PLTZF and FRTZF for a pattern of features located with a composite control frame, as specified in Figure 20–30.

10 10

10

20

20

10

20

Feature relating tolerance zones, with axes constrained to a rigid pattern with separation of 20. The rigid pattern of the axes is free to rotate and translate within the PLTZF tolerance zones. Pattern locating tolerance zones, with axes located by basic dimensions, with the part located by datums A, B, and C.

20

For clarity, the actual holes are not shown. The axes of the holes may be anywhere within the FRTZF (smaller) cylindrical tolerance zones.

PLTZF consisting of cylindrical tolerance zones each having a diameter of 5, centered around axes located by the basic dimensions from the datums. These are shown in Figure 20–31 as the larger cylinders. The second row specifies a FRTZF consisting of cylindrical tolerance zones each having a diameter of 2. The FRTZF tolerance zones form a rigid pattern that must maintain the relative distances of 20 with respect to each other, but which is free to rotate and translate within the cylindrical zones defined by the PLTZF. Think of the FRTZF tolerance zones as being rigidly connected to each other by a frame that maintains the basic dimensions between them. The entire frame may be translated and rotated to any position that keeps all of the FRTZF centerlines within the PLTZF tolerance zones. When datum references are included in the FRTZF specification, they govern only the rotation of the FRTZF relative to the specified datums. In the example in Figure 20–30, since the second row references datum A, the FRTZF must be aligned with datum plane A, that is, the tolerance zone axes will be perpendicular to datum plane A. Likewise, if datum B had been specified, the tolerance zone axes would be required to lie in a plane that was parallel to datum plane B. Concentricity Nominally, concentricity is the condition of the center axis of a surface of revolution, such as a cylinder, being congruent with a datum axis. The Y14.5 standard defines it more precisely as the condition where the median points of all diametrically opposed elements of a surface of revolution are within a cylindrical tolerance zone centered around a datum axis. This means the feature's center is not determined as a single straight line, but rather as a collection of all points obtained by finding the medians of all diameter measurements across the surface. This is extremely difficult and expensive to measure, so it is recommended that the concentricity control be rarely used. The preferred options for controlling concentric features are position, profile, and runout. The concentricity control might be warranted in cases where it is critical to control the axis rather than the surface of a feature, such as in dynamic balancing of a high-speed rotating part. Symmetry Symmetry is the condition where a feature has the same profile on either side of the center plane of a datum feature. In the Y14.5 standard, it is defined similarly to

Geometric Dimensioning and Tolerancing     1001

concentricity, except it applies to the center plane of a feature of size instead of the center axis of a surface of revolution. Because it is based on controlling a collection of all median points measured across the feature, rather than a single center plane of the feature, symmetry suffers from the same difficulties of measurement as concentricity. Consequently, the symmetry control should rarely be used. In most cases, symmetrical features are best controlled with position or profile controls. Runout Controls Runout controls the surface variation on a feature as it rotates around a datum axis. A simple example is the surface of a rotating shaft. There are two runout controls: 1 Circular runout 2 Total runout. Circular runout measures the radial variation of each circular section of a feature independently from each other. Total runout measures the runout of the entire surface of a cylindrical feature simultaneously. Both types of runout are demonstrated in Figure 20–32. The cylinder on the left is defined as datum feature A. Chuck jaws clamped onto the surface of this datum feature serve as the datum feature simulator to define the centerline as the actual datum. Thus, when the chuck rotates, the part necessarily rotates around the datum centerline. The circular runout specification on the tapered feature requires that an indicator at any location along the feature must not move more than 0.01 units during a complete rotation of the part. Each point where the indicator is located should independently satisfy the runout tolerance control. The total runout specification on the right cylindrical feature requires that an indicator must not move more than 0.02 for all locations along the cylinder, measured in one setup. Another way of stating this is that the entire surface of the controlled feature must lie within the zone between two concentric cylinders that are separated radially by the stated tolerance of 0.02. Circular runout can be applied to any surface of revolution since the measurements are made independently at each cross section. It is inherently controlling both concentricity and circularity. Total runout applies only to cylindrical features, since the diameter at each cross section must fit within the same tolerance zone. It is inherently controlling cylindricity, circularity, straightness, and surface profile. Total runout is a Translation: An indicator must not move more than 0.01 units when placed at any location along the tapered feature, when the part is rotated around datum axis A. 0.01 A

Datum axis defined by chuck jaws

0.01

0.01 at each circular element

A

Rotate ∅ 20 ± 0.1

∅ 15 ± 0.2 0.02 A

Translation: An indicator must not move more than 0.02 units for all locations along the cylinder, measured in one setup, when the part is rotated around datum axis A. (a)

Chuck jaws 0.02 all elements together (b)

Figure 20–32 Circular runout and total runout. (a) Drawing specification; (b) Interpretation.

1002      Mechanical Engineering Design

particularly useful control for rotating shafts that carry bearings or gears that are sensitive to misalignment. Total runout can also be used effectively to control the coaxiality of multiple cylindrical surfaces by relating each cylinder to the same datum centerline. The runout controls can also be applied to features constructed at right angles to a datum axis, such as the end-face of a cylinder. In this case, runout controls variations of perpendicularity (such as wobble) and flatness, measured while rotating the feature about the datum axis.

20–6  Material Condition Modifiers Maximum material condition (MMC) and least material condition (LMC) can be applied as modifiers to most of the geometric controls that deal with a feature of size. The symbols for the modifiers, M and L , can be included in a feature control frame directly following the geometric tolerance value, and/or immediately following a datum reference. When a geometric tolerance is modified with the maximum material condition modifier, it indicates that the stated tolerance value applies when the feature is produced at its MMC. If the feature is produced at a size with less material than its MMC, the deviation from the MMC is added to the allowed geometric tolerance value. The implication is that the value of the geometric tolerance is not constant, as it depends on the size of the actual produced part. As an example, consider the part in Figure 20–33a, which contains both an external feature of size (the cylindrical boss) and an internal feature of size (the hole), each of which is located with a position control that is toleranced at MMC. The position Figure 20–33

C

Application of MMC to a position tolerance. (a) Drawing specification; (b) summary of bonus tolerance for external feature; (c) summary of bonus tolerance for internal feature.

70 ± 0.1

150 ± 0.8

50 ± 0.1

50

200 ± 0.8

170

∅ 30.3 30.0

100

A

∅ 0.2 M A B C

∅ 50.3 50.0 ∅ 0.2 M A B C

B

(a)

External Feature ϕ

Additional Tolerance (Bonus)

Tolerance Zone Cylinder ϕ

Internal Feature ϕ

Additional Tolerance (Bonus)

Tolerance Zone Cylinder ϕ

50.3 (MMC)

0.0

0.2

30.3 (LMC)

0.3

0.5

50.2

0.1

0.3

30.2

0.2

0.4

50.1

0.2

0.4

30.1

0.1

0.3

50.0 (LMC)

0.3

0.5

30.0 (MMC)

0.0

0.2

(b)

(c)

Geometric Dimensioning and Tolerancing     1003

control on the external feature is interpreted as follows: "The center axis of this external cylindrical feature must be in position within a cylindrical tolerance zone of 0.2 diameter if the feature is produced at its maximum material condition of 50.3, where the position is specified by the basic dimensions with respect to the datum reference frame established by datum features A, B, and C." If the cylinder is produced with a diameter less than its MMC of 50.3, say at 50.2, then the amount of the deviation from the MMC, that is 0.1, is added to the specified geometric tolerance, providing a realized tolerance of 0.3. This increase in tolerance is traditionally referred to as a "bonus" tolerance. The table in Figure 20–33b shows how this bonus tolerance accumulates as the produced diameter of the cylinder is reduced from its MMC of 50.3 to its LMC of 50.0. Note that the bonus tolerance is applied to the tolerance associated with the geometric control (position, in this example), and not to the direct tolerance on the size of the feature. This bonus tolerance is one of the significant advantages afforded by GD&T over the fixed tolerances provided by the traditional coordinate dimensioning. It is particularly useful for applications where mating parts need to fit with each other with a clearance for assembly. A minimum tolerance is guaranteed, but a bonus tolerance is made available to the manufacturer to reduce costs. This modifier would not be appropriate for an application where the fit between the mating parts is important, such as a press fit between a bearing and a shaft. The same bonus tolerance concept is applicable for the hole in Figure 20–33a, except that the MMC condition for the internal feature is the smallest hole size. The table in Figure 20–33c shows how the bonus tolerance adds to the stated MMC tolerance as the hole size gets larger, moving from the MMC to the LMC. The least material condition modifier works similarly to the MMC modifier, but in the opposite direction. When a geometric tolerance is modified with the LMC modifier, it indicates that the stated tolerance value applies when the feature is produced at its LMC. If the feature is produced at a size with more material than its LMC, the deviation from the LMC is added to the allowed geometric tolerance value. The LMC modifier is typically used for applications in which it is critical to maintain a minimum amount of material. Examples include the material between a hole and the edge of a part, or a wall thickness. LMC is also useful for specifying features on a casting that will later be machined, to ensure that enough material is left in the casting for the finish machining operation. The MMC and LMC material condition modifiers can be similarly applied to most of the geometric tolerances, in particular to those that control features of size with a central axis or plane. If no material condition modifier is specified, the default material condition is known as regardless of feature size (RFS). RFS means that the stated tolerance is applicable regardless of the feature size. In other words, no matter what size the feature is produced (within its tolerance), the geometric tolerance is fixed at the stated value. This is much more restrictive for manufacturing. It is warranted in applications where variable play between mating parts is not desirable, such as the press fit between a shaft and components like bearings and gears. A fine detail to be noted is that in all the preceding discussion, the produced diameter of the feature is determined by the unrelated actual mating envelope, as defined in Section 20–3, as it is about the only practical way to determine a single value for the diameter of an imperfect feature. The M and L symbols can also be applied immediately following a datum reference in the feature control frame, in particular when the datum is based on a feature of size. When applied to a datum, the symbols are referred to as maximum material

1004      Mechanical Engineering Design

boundary (MMB) and least material boundary (LMB). A full explanation of the effect of applying MMB or LMB to a datum is beyond the scope of this chapter. In essence, it allows the part to float or shift relative to the datum reference frame as the produced datum feature of size deviates from its maximum or minimum material condition. Consequently, it does not change the tolerance on the considered feature, but just the relative position of the feature with respect to the datum reference frame.

20–7  Practical Implementation The basic concept of GD&T is to define the ideal part, then specify the amount of variation that is acceptable. The allowable variation includes all four of the geometric attributes: size, location, orientation, and form. The size of a feature of size is directly dimensioned and toleranced with a plus/minus tolerance. The other three geometric attributes, broken down into more specific geometric characteristics, are controlled with geometric controls (Table 20–1). Some of the geometric controls are broad and inherently provide control of multiple characteristics. Anytime a boundary envelope is defined by any geometric control, it constrains all geometric variations from the ideal size and shape to fit within the envelope. For example, by applying profile of a surface to a feature, the bounding envelope that defines the allowable variation from the ideal profile also automatically controls orientation characteristics (e.g., parallelism and perpendicularity) and form characteristics (e.g., flatness and cylindricity). Consequently, most of the geometric characteristics can be controlled by a few controls, and refinements are only added as necessary. A suggested general framework for implementation of GD&T comprises the following five steps:

1 2 3 4 5

Select the datum features. Control the datum features. Locate the features. Size and locate the features of size. Refine the orientation and the form of features, if needed.

Each step is elaborated in the following sections. 1 Select the datum features. The datum features should be selected based on the functional use of the part first, rather than on the anticipated manufacturing method. The primary datum is usually the most critical to the part function, and of sufficient surface size to assure a stable setup for establishing the remaining datums. For mating parts, the corresponding interfacing features are usually selected as datum features. Remember that datum features identified with the datum symbol are only utilized when they are called out in the feature control frame of a feature being controlled. Though it is not unusual for every feature of a part to reference the same set of datums, it is not a requirement that they do so. 2 Control the datum features. Though a datum is considered theoretically perfect, the physical datum feature is not. Consequently, the imperfect datum features need to be geometrically controlled just like any other feature. Sometimes the default controls are sufficient. However, since the datum features are used to stabilize the part for manufacturing and inspection,

Geometric Dimensioning and Tolerancing     1005

they may warrant additional consideration for that purpose, beyond what is needed just for the functionality of the part. If the primary datum feature is a flat surface, it may be helpful to consider a flatness or surface profile control on it. The secondary and tertiary datum features establish datum planes perpendicular to the primary datum plane. Consequently, it might be useful to use an orientation control (such as perpendicularity) on the secondary and tertiary datum features with respect to the primary datum. A default surface profile applied by a general note to the entire part might be sufficient for this. When a feature of size is used as a datum feature, the size tolerance automatically provides form control through Rule #1 (see Section 20–4). Also, when the position control is used to locate a datum feature of size, it will automatically provide orientation control of the datum. 3 Locate the features. All features have surfaces that need to be located with respect to appropriate datums. The best strategy for most situations is to use basic dimensions to locate the true position of each feature, accompanied by one or more appropriate geometric controls. A default surface profile control can be established with a general note. 4 Size and locate the features of size. Features of size need to be sized as well as located. The following three steps are typical: (1) Locate the true position of the center axis or center plane of each feature of size with basic dimensions. (2) Directly dimension the feature's size, including a plus/minus tolerance. (3) Attach a position control to the feature's size callout to establish limits for location and orientation. For cylindrical features of size that are coaxial to a datum axis, the runout or surface profile controls may be used instead of the position control. The size tolerance on the feature of size automatically provides form control through Rule #1. 5 Refine the orientation and the form of features, if needed. If any feature needs a tighter control in orientation or form than is provided by the previous steps, additional controls can be added.

The following example demonstrates this process.

EXAMPLE 20–1 Interpret and explain the GD&T notation for the part shown in Figure 20–3. Solution Since this part has already been drawn, the five steps will be used to organize the explanation rather than to make the decisions. 1 Select the datum features. The datum features that have been identified with the datum symbols are the back face and the left and bottom edges. The functional needs of the part are not specified, but this choice of datums is pretty common for this type of simple rectangular plate. Note that the callout for datum B is not placed in-line with the dimension, thus indicating the datum is at the edge of the part rather than the center plane of the part. Since the features

1006      Mechanical Engineering Design

are located with respect to the datum, the choice of datum indicates that it is more important to maintain the distance of the hole from the edge rather than to ensure that it is centered. The back face makes a good primary datum (as called out in the position control of the hole), as it is of sufficient size to stabilize the part with respect to three degrees of freedom while establishing the other datums. It is also likely that the back surface will be in contact with a mating part. 2 Control the datum features. A flatness specification ensures that the surface of datum feature A does not vary more than 0.003. This is a common control for a primary datum that is a planar surface, especially if it must fit with a mating surface. Datum feature B, the left surface of the part, is required to be perpendicular to the back face, within a tolerance zone between two parallel planes separated by a distance of 0.005. Note that the envelope generated by this orientation control also constrains the form of this surface (e.g., flatness and straightness). Datum feature C must be perpendicular to datum features A and B, within a tolerance zone of 0.005. 3 Locate the features. All features of this part are features of size, which are handled in the next step. 4 Size and locate the features of size. There are four features of size: the hole, and the height, width, and thickness of the plate. The plate features do not need to be located, since each dimension starts at one of the datum features. Consequently, the three plate dimensions simply need a directly toleranced dimension. The hole feature is located by basic dimensions from datums B and C. The hole diameter is specified to be within the range from 1.000 to 1.002. The position control on the hole stipulates that the center axis of the hole must be within a cylindrical tolerance zone that has a diameter of 0.003 if the hole is produced at its MMC of 1.000. The tolerance zone may increase to as much as 0.005 as the produced hole diameter increases from 1.000 to 1.002. The tolerance zone is determined with the part immobilized with respect to the three datums, applied in order of the back surface datum A, then the side edge datum B, then the lower edge datum C. Changing this order would change the location and orientation of the hole's tolerance zone. Note that the hole's tolerance zone, in addition to providing a tolerance on the location of the axis, also limits the orientation of the hole's axis. A separate orientation control could have been stipulated if the same tolerance zone was not suitable for both position and orientation. 5 Refine the orientation and the form of features, if needed. No additional controls are specified, so apparently no further refinement is necessary. It is always a good idea to consider how far from the true geometry the part could be produced and still be within specifications. For example, consider how out-of-flat the front face (opposite of datum A) could be. Since the thickness of the plate is a feature of size, the front face is controlled by Rule #1, that is, the size tolerance also establishes an envelope for form control. The size tolerance allows the thickness to vary between 1.9 and 2.1. Since the back datum face may vary only by 0.003, the majority of the size tolerance can be realized by a front face that is curved, bent, or wavy. The plate could be 1.9 thick on one edge and 2.1 thick on the other. If this is not acceptable, the Rule #1 control should be overruled by another more specific refinement.

There are many concepts of GD&T that have not been addressed in this brief introductory chapter. Significant training is necessary to gain competence in defining appropriate geometric controls to achieve the desired function. However, a fundamental ability to "read" a drawing is not out of reach. The next example will provide an opportunity to practice.

Geometric Dimensioning and Tolerancing     1007

EXAMPLE 20–2 Interpret and explain the GD&T notation for the part shown in Figure 20–34. Solution Various aspects of the drawing are circled and annotated with note numbers that correlate with the explanations given here. Note 1.  The front face is defined as a datum feature and is used as the primary datum for several of the geometric controls on the drawing. Since the front face, rather than the back face, is selected as a datum, it is likely that this surface will fit against a mating surface. Functionally, this fit is apparently more important than the back face. The flatness control is common for a surface that is a primary datum. The entire surface must fit within a tolerance zone defined by two parallel planes separated by a distance of 0.05. Note 2.  Consider the outer surface of the protruding cylinder, which is identified as a datum feature. Since it is a feature of size, the datum is the center axis. More precisely, the datum is the theoretically perfect center axis of the imperfect outer surface, as determined by the unrelated actual mating envelope. Note that if the bore through the center of the part had been chosen as the datum feature, the datum would also have been a center axis, but not precisely the same one. The fact that the outer surface is selected indicates it may have greater functional precedent over the bore. Perhaps the outer cylindrical surface will fit within the bore of a mating part. Even though the function of the part is not specified, the geometric requirements of the functional needs are clear, and the part should be manufactured and inspected accordingly. Every feature of size must be controlled in all four geometric attributes: size, location, orientation, and form. Consider each in turn for the outer surface of the protruding cylinder. The cylinder's size is directly specified. The diameter at any cross section must be within the stated range of 39.06 to 40.00. The center axis of a feature of size must be located, usually by basic dimensions. In this case, since this is a datum feature, its own center axis becomes part of the definition of the origin of the datum reference frame. The feature's location is at the origin, so no further basic dimensions are needed to locate it. Figure 20–34

4. 1.

+0.1 4 × ∅ 10 –0.0

0.05

∅0 M A BM

5.

∅ 90

A

R3

∅ 60

A

R1

6.

2.

3.

∅ 20 ± 0.05 ∅ 0.3 A B

A

∅40+0.00 –0.04 ∅ 0.05 A

0.1 A 20 20 ± 0.1

B

SECTION A–A UNLESS OTHERWISE SPECIFIED: 0.2 A B

1008      Mechanical Engineering Design

The orientation of this cylinder feature is controlled by the perpendicularity control. This control requires the center axis of the feature to lie within a cylindrical tolerance zone that has a diameter of 0.05. The tolerance zone is exactly perpendicular to datum plane A; the center axis of the feature is allowed to tip within the tolerance zone. No specific form controls are specified for this feature. The form of its surface is controlled by default through Rule #1 for features of size. Rule #1 requires that the surface of the cylinder must be within the cylindrical envelope of perfect form at the MMC diameter of 40.00. This feature of size is controlled in all aspects: size, location, orientation, and form. Note 3.  Now consider the center bore. This is a feature of size, so its size is directly specified and toleranced. Its form is controlled by Rule #1. It is located by implication (since no basic dimensions are given) at the origin of the datum reference frame. Its orientation and location is controlled by the position control. The position control defines a cylindrical tolerance zone that is first held perpendicular to datum plane A, then centered at the datum axis defined from datum feature B. Since there is no material modifier specified on the tolerance, the default condition of regardless of feature size (RFS) applies. Therefore, the tolerance zone diameter is a constant 0.3, regardless of what size the feature is actually produced. Note 4.  The four holes are defined collectively as a pattern. They are features of size, so their diameters are directly dimensioned and toleranced, and their surfaces must not exceed the envelope of perfect form at the MMC diameter of 10 (Rule #1). Their location is specified by the basic dimension of 60 for the bolt circle, as well as by the implication that they are spaced at 90 degrees around the bolt circle. Control of orientation and location is provided by the position control. Specifically, the position control requires the center axis of each hole to be within a cylindrical tolerance zone. The datum ­callouts require this tolerance zone to be first held perpendicular to datum A, then centered at the true location as measured with respect to datum axis B. The M modifier that accompanies datum B allows a little more leeway on the overall shift of the hole pattern if datum feature B is produced at a ­diameter less than its MMC. The zero tolerance value in the position control does not mean that no tolerance is allowed, as it is accompanied by the M modifier. This is known as zero tolerancing at MMC. The meaning is that the tolerance zone diameter is zero if the hole is produced at its maximum material diameter of 10, but grows to a diameter of 0.1 as the hole's diameter increases to its least material diameter of 10.1. Consequently, the axis of the hole would need to be perfectly located and oriented if the hole is ­produced at its MMC of 10, but can deviate from perfection if the hole is produced at a larger size. Note 5.  This cylindrical feature is a feature of size, and could have been directly dimensioned and ­toleranced. Instead, it is specified with a basic dimension of 90 with no tolerance or geometric control in sight. This does not mean it has to be perfect. It means that it is controlled by the default profile tolerance that is stated at the bottom of the drawing. The ideal cylindrical surface is first sized at the basic diameter of 90, and located with respect to datums A and B. Then a tolerance zone is centered around this ideal surface with a total tolerance of 0.2. In this case, the tolerance zone is the space between two concentric cylinders with diameters of 89.8 and 90.2 (a radial difference of 0.2). This ­tolerance zone controls all four geometric attributes of size, location, orientation, and form. Note 6.  The basic dimension of 20 locates the ideal surface of the front face of the protruding ­cylinder. The surface profile control requires that the actual front face lie within a tolerance zone consisting of the space of 0.1 between two planes centered around the ideal surface, where each plane is parallel to datum A. This effectively establishes the location of the face (a distance of 19.95 to 20.05 from datum A), as well as the flatness of the face, and the parallelism of the face with respect to datum A.

Geometric Dimensioning and Tolerancing     1009

20–8  GD&T in CAD Models Many industries are utilizing 3D computer-aided design (CAD) data for some or all of the engineering, manufacturing, and inspection phases of the product life cycle. The prominent standard regulating this is ASME Y14.41-2003, Digital Production Definition Data Practices. The standard addresses many aspects of the practices, requirements, and interpretation of CAD data. The standard defines the use of GD&T in a digital environment where specifications are embedded directly into the data set—not just visually, but functionally as well. Most of the concepts of GD&T apply directly to digital models. The significant difference in CAD models versus 2D drawings is that the CAD model represents the ideal geometry of the part. Any dimension can be queried from the model and the exact (ideal) dimension can be obtained. In fact, the ideal data can be directly utilized in computer-aided manufacturing operations. This leads to a misconception that a part manufactured directly from the model on a computer numerical controlled (CNC) machine will be perfect. In fact, CNC manufacturing has the same requirements as manual manufacturing methods with regard to the need to specify and inspect geometric tolerances. The Y14.41-2003 standard allows for all geometry data to be considered basic, unless superseded by a toleranced dimension or defined as a reference dimension. Geometric controls are applied to the 3D data models to control the features just the same as on the 2D drawings. Direct tolerancing is recommended only for features of size. Essentially, GD&T works the same on 3D models as in 2D drawings, except that in 3D models the basic dimensions do not have to be shown, since the default is that all dimensions are basic. Figure 20–35 shows the 3D solid model of the part used in Example 20–2, with appropriate GD&T elements embedded into the CAD data. The ongoing transition to 3D digital representation brings with it the possibilities of tighter integration of the various processes of design, analysis, and manufacturing. For example, the embedded geometric dimensioning and tolerancing information can be directly accessed for tolerance analysis and process planning. Figure 20–35

20 ± 0.1 4 × ∅ 10 +0.1 –0.0

∅ 0 M A BM

∅ 20 ± 0.05 ∅ 0.3 A B

0.1 A

0.05 ∅ 40 +0.00 –0.04 ∅ 0.05 A B

A

An example of GD&T applied to a 3D CAD model.

1010      Mechanical Engineering Design

20–9  Glossary of GD&T Terms Most of the concepts in GD&T are simple, but the vocabulary to describe them can seem overwhelming at first. That is because the vocabulary needs to be precise enough to be consistent and unambiguous. For convenient reference, some of the most commonly used terms in GD&T are summarized in this section. Actual Mating Envelope—a perfectly shaped counterpart of an imperfect feature of size, which can be contracted about an external feature, or expanded within an internal feature, so that it contacts the high points of the feature's surface. Actual Mating Envelope, Related—an actual mating envelope that is sized to fit the feature while maintaining some constraint in orientation or location with respect to a datum. Actual Mating Envelope, Unrelated—an actual mating envelope that is sized to fit the feature without any constraint to any datum. Axis—a line defining the center of a cylindrical feature, established from the theoretical axis of the unrelated actual mating envelope of the cylindrical feature's extremities. Basic Dimension—a theoretically exact dimension that ideally locates and/or orients the tolerance zone of a feature. It does not have a tolerance directly associated with it, but is instead associated with a geometric control of a tolerance zone. Basic dimensions are denoted by a box around the dimension, or by a general note. Bonus Tolerance—additional tolerance that applies to a feature as its size shifts from a stated material condition of MMC or LMC. Center Plane—the theoretical plane located at the center of a noncylindrical feature of size, established from the center plane of the unrelated actual mating envelope of the feature's extremities. Combined Feature Control Frame—a feature control frame made of two or more feature control frames, each with a geometric characteristic symbol. The geometric controls are applied to the feature, in order from top to bottom. Composite Feature Control Frame—a feature control frame made of two or more feature control frames sharing a common geometric characteristic symbol. Datum—a theoretically exact point, axis, line, or plane derived from a datum feature simulator, used as an origin for repeatable measurements. Datum Axis—the theoretical axis of a cylindrical datum feature, established from the axis of the unrelated actual mating envelope of the cylindrical feature's extremities. Datum Feature—an actual physical surface of the part that is specified in order to establish a theoretically exact datum. Datum Feature Simulator—a precision embodiment, such as a surface plate, gauge pin, or machine tool bed, of the datum described by an imperfect datum feature. Datum of Size—a datum feature that is a feature of size, and therefore subject to size variation based on plus/minus tolerances.

Geometric Dimensioning and Tolerancing     1011

Datum Reference Frame—a set of up to three mutually perpendicular planes that are defined as the origin of measurement for locating the features of a part. Derived Median Line—an imperfect "line" formed by the center points of all cross sections of a feature, where the cross sections are normal to the axis of the Unrelated Actual Mating Envelope of the feature. Derived Median Plane—an imperfect "plane" formed by the center points of all line segments bounded by a feature, where the line segments are normal to the center plane of the unrelated actual mating envelope of the feature. Envelope Principle—See definition for Rule #1, given later in this section. Feature—a general term referring to a clearly identifiable physical portion of a part, such as a hole, pin, slot, surface, or cylinder. Feature Control Frame—a rectangular box attached to a feature on a drawing, containing the necessary information to define the tolerance zone of the specified feature. Feature of Size, Irregular—a directly toleranced feature or collection of features that may contain or be contained by an actual mating envelope. Feature of Size, Regular—a cylindrical surface, a spherical surface, a circular element, or a set of two opposed parallel elements or surfaces that are associated with a directly toleranced dimension. A regular feature of size has a size that can be measured across two opposing points, and has a reproducible center point, axis, or center plane. Feature-Relating Tolerance Zone Framework (FRTZF)—a tolerance zone framework that governs the positional relationship from feature to feature within a pattern of features. It is specified in the bottom row of a composite feature control frame. Geometric Attributes—the four broad attributes (size, location, orientation, and form) that must be considered to geometrically define a feature. This term is not strictly defined by GD&T. Geometric Characteristics—the 14 characteristics defined in Table 20–1 that are available to control some aspect of a geometric tolerance of a feature. A geometric characteristic symbol is the first item in any feature control frame. Least Material Boundary (LMB)—the limit defined by a tolerance or combination of tolerances that exists on or inside the material of a feature(s). When applied as a modifier to a datum reference in a feature control frame with the L symbol, it establishes the datum feature simulator at a boundary determined by the combined effects of size (least material), and all applicable geometric tolerances. Least Material Condition (LMC)—the condition in which a feature of size contains the least amount of material within the stated limits of size (e.g., maximum hole diameter or minimum shaft diameter). This condition may be specified as a tolerance modifier in a feature control frame with the L symbol. Material Condition Modifier—a modifier symbol, M or L , applied to a geometric tolerance to indicate that the tolerance applies at maximum material condition or least material condition, respectively. The absence of a material condition modifier indicates that the tolerance applies at all material conditions, that is, regardless of feature size (RFS).

1012      Mechanical Engineering Design

Maximum Material Boundary (MMB)—the limit defined by a tolerance or combination of tolerances that exists on or outside the material of a feature(s). When applied as a modifier to a datum reference in a feature control frame with the M symbol, it establishes the datum feature simulator at a boundary determined by the combined effects of size (maximum material), and all applicable geometric tolerances. Maximum Material Condition (MMC)—the condition in which a feature of size contains the maximum amount of material within the stated limits of size (e.g., minimum hole diameter or maximum shaft diameter). This condition may be specified as a tolerance modifier in a feature control frame with the M symbol. Pattern-Locating Tolerance Zone Framework (PLTZF)—a tolerance zone framework that governs the positional relationship between a pattern of features and the datum features. It is specified in the top row of a composite feature control frame. Regardless of Feature Size (RFS)—indicates that the stated tolerance applies, regardless of the actual size to which the feature is produced. This  is the default condition for a tolerance with no modifier symbol (i.e., M or L ). Regardless of Material Boundary (RMB)—indicates that a datum feature simulator progresses from MMB toward LMB until it makes maximum contact with the extremities of a feature(s). This is the default condition for a datum reference with no modifier symbol (i.e., M or L ). Rule #1—when only a tolerance of size (i.e., a plus/minus tolerance) is specified for a feature of size, the limits of size prescribe the extent of which variation in its geometric form, as well as size, are allowed. The surface of a feature of size may not extend beyond an envelope of perfect form at MMC. This rule is also referred to as the envelope principle. Tolerance—the total amount a specific dimension is permitted to vary, between the maximum and minimum limits. Tolerance Zone—a limiting boundary in which the actual feature must be contained. Virtual Condition—a constant "worst case" boundary defined by the collective effects of a feature's size, geometric tolerance, and material condition.

PROBLEMS 20–1 In the traditional coordinate dimensioning system, which of the following is true? (Select one.)   i. Only "features of size" need to include a tolerance.   ii. Only dimensions that are important need to include a tolerance. iii. Only dimensions that need to be tightly controlled need to include a tolerance.   iv. All dimensions should include a tolerance.

20–2 In GD&T, which type of dimension should generally be directly toleranced with a plus/minus tolerance?

Geometric Dimensioning and Tolerancing     1013

20–3 What underlying purpose is emphasized by the ASME Y14.5–2009 standard in the dimensioning and tolerancing of a part? (Select one.)   i. The method of manufacturing.   ii. The design intent. iii. The inspection process.   iv. Equal attention to all of the above.

20–4 What is the term that refers to a feature which has a size that can be measured across two opposing points?

20–5 What are the four geometric attributes that must be considered to define the geometry of a feature of a part?

20–6 What are the four geometric characteristics that provide form control? 20–7 What are the three geometric characteristics that provide orientation control? 20–8 What are the three geometric characteristics that provide location control? Which of the three is most prominently used?

20–9 How is a basic dimension toleranced? (Select one.)

  i. Basic dimensions receive the default tolerance specified in the title block.   ii. Basic dimensions are not toleranced. iii. Basic dimensions receive the tolerance from an associated feature control frame.

20–10 For the part shown, identify all of the features of size. C

70 ± 0.1

150 ± 0.8

50 ± 0.1

50

Problems 20–10 to 20–14

200 ± 0.8

170 100

∅ 30 ± 0.1 ∅ 0.3 M A B C B

∅ 50 ± 0.3 ∅ 0.2 M A B C

20–11 For the part shown, clearly identify each of the following, with labels and sketches on the drawing. (a) Datum features A, B, and C. (b) Datums A, B, and C. (c) Datum reference frame based on datum features A, B, and C.

20–12 For the part shown, the ideal position of the cylindrical boss is located with the basic dimensions of 100 and 50. These basic dimensions are measured from which of the following? (Select one.)   i. The physical edges of the part.   ii. The high points on the physical edges of the part. iii. The low points on the physical edges of the part.   iv. The theoretical datum planes B and C.

A

1014      Mechanical Engineering Design

The basic dimensions are locating which of the following? (Select one.)   i. The physical location of the center axis of the boss.   ii. The center axis of the boss as determined by the actual mating envelope of the  boss. iii. The ideal location of the center axis of the tolerance zone specified by the position control. If the part is produced and is being inspected, the location of the boss will be measured from which of the following? (Select one.)   i. The physical edges of the part.   ii. The high points on the physical edges of the part. iii. The low points on the physical edges of the part.   iv. The theoretical datum planes B and C. v. The datum feature simulators for datum features B and C.

20–13 For the part shown, answer the following questions with regard to the cylindrical boss.

(a) What are the maximum and minimum diameters allowed for the boss? (b) What is the effect of the position tolerance of 0.2 on the diameters specified in part (a)? (c) The position control defines a tolerance zone. Specifically what must stay within that tolerance zone? (d) What is the diameter of the tolerance zone if the boss is produced with a diameter of 50.3? (e) What is the diameter of the tolerance zone if the boss is produced with a diameter of 49.7? ( f ) Describe the significance of the datum references to the determination of the position tolerance zone. (g) What is the perpendicularity tolerance with respect to datum A? (Select one.)  i. Not defined.  ii. Controlled by the position tolerance; 0.2 at MMC to 0.8 at LMC. iii. Controlled by the size tolerance; 0.3.  iv. Must be perfectly perpendicular; 0. (h) What controls the cylindricity? (Select one.)  i. There is no control on the cylindricity.  ii. From Rule #1, the envelope of a perfect cylinder with diameter of 50. iii. From Rule #1, the envelope of a perfect cylinder with diameter of 50.3.  iv. From the position control, the center axis of each cross section must be within the 0.2 cylindrical tolerance zone.

20–14 For the part shown, answer the following questions with regard to the hole.

(a) What are the maximum and minimum diameters allowed for the hole? (b) What is the effect of the position tolerance of 0.3 on the diameters specified in part (a)? (c) The position control defines a tolerance zone. Specifically what must stay within that tolerance zone? (d) What is the diameter of the tolerance zone if the hole is produced with a diameter of 30.1? (e) What is the diameter of the tolerance zone if the hole is produced with a diameter of 29.9? ( f ) Describe the significance of the datum references to the determination of the position tolerance zone.

Geometric Dimensioning and Tolerancing     1015

(g) What is the perpendicularity tolerance with respect to datum A? (Select one.)  i. Not defined.  ii. Controlled by the position tolerance; 0.3 at MMC to 0.5 at LMC. iii. Controlled by the size tolerance; 0.1.  iv. Must be perfectly perpendicular; 0. (h) What controls the cylindricity? (Select one.)  i. There is no control on the cylindricity.  ii. From Rule #1, the envelope of a perfect cylinder with diameter of 30. iii. From Rule #1, the envelope of a perfect cylinder with diameter of 29.9.  iv. From the position control, the center axis of each cross section must be within the 0.3 cylindrical tolerance zone.

20–15 Describe how a center axis is determined for a physical hole that is not perfectly formed.

20–16 According to the envelope principle (Rule #1), the size tolerance applied to a feature of size controls the size and   i. location   ii. orientation iii. form   iv. runout v. All of the above.

of the feature. (Select one.)

20–17 If a shaft diameter is dimensioned 20 ± 0.2, determine the diameter of the shaft at MMC and at LMC.

20–18 If a hole diameter is dimensioned 20 ± 0.2, determine the diameter of the hole at MMC and at LMC.

20–19 A hole diameter is dimensioned 20 ± 0.2. According to the form control provided by

the envelope principle (Rule #1), is the limiting envelope a perfect cylinder with diameter of 19.8, 20.2, or both? Explain your answer.

20–20 A shaft diameter is dimensioned 20 ± 0.2. According to the form control provided by

the envelope principle (Rule #1), is the limiting envelope a perfect cylinder with diameter of 19.8, 20.2, or both? Explain your answer.

20–21 The diameter of a cylindrical boss is dimensioned 25 ± 0.2. A position control is used

to control the basic location of the boss. Specify the diameters allowed for the position tolerance zone if the boss is produced with diameters of 24.8, 25.0, and 25.2, for each of the following position tolerance specifications: (a)  ∅0.1 (b)  ∅0.1 M  (c)  ∅0.1 L

20–22 A hole diameter is dimensioned 32 +0.4 −0.0 . A position control is used to control the basic location of the hole. Specify the diameters allowed for the position tolerance zone if the hole is produced with diameters of 32.0, 32.2, and 32.4, for each of the following position tolerance specifications: (a)  ∅0.3 (b)  ∅0.3 M  (c)  ∅0.3 L

20–23 What is the name of the geometric characteristic that effectively controls a combination of circularity and straightness of a cylinder?

20–24 What is the name of the geometric characteristic that can be specified in a note to

provide a default tolerance zone to control size, form, orientation, and location of all features that are not otherwise controlled?

1016      Mechanical Engineering Design

20–25 Which geometric characteristics never reference datums? Why? 20–26 Answer the following questions regarding material condition modifiers.

(a) What are the three material condition modifiers? (b) Which one is the default if nothing is specified? (c) Which one(s) can provide "bonus" tolerance? (d) Which of the following is increased by a bonus tolerance? (Select one.)  i. A size dimension.  ii. A ± tolerance of a size dimension. iii. A basic dimension locating a feature.  iv. A size of a tolerance zone controlling a feature. (e) To which of the following can a material condition modifier symbol be applied? (Select one.)  i. A size dimension.  ii. A ± tolerance of a size dimension. iii. A tolerance of a geometric characteristic controlling a feature of size.  iv. A tolerance of a geometric characteristic controlling any feature. ( f ) Which material condition modifier should be considered if the goal is to ensure a minimum clearance fit for a bolt in a hole, but to give greater manufacturing flexibility if the hole is produced with a greater clearance? (g) Which material condition modifier should be considered if the goal is to provide a consistent press fit between interchangeable parts? (h) Which material condition modifier should be considered if the goal is to ensure a minimum wall thickness for a casting, but to give greater manufacturing flexibility if the wall is produced with a greater thickness?

20–27 The drawing shown is of a mounting fixture to locate and orient a rod (not shown)

through the large bore. The fixture will be bolted to a frame through the four bolt holes that are countersunk to recess the bolt heads. The bolt holes have too much clearance to properly align the rod, so the fixture will be aligned with two locating pins in the frame that will fit in the ∅6 hole and slot. (a) Determine the minimum diameter allowed for the countersink. (b) Determine the maximum depth allowed for the countersink. (c) Determine the diameter of the bolt holes at MMC. (d) Identify every feature that qualifies as a feature of size. (e) The width of the base is specified with a basic dimension of 60, with no tolerance. (Note that as a feature of size, it could have had a tolerance directly specified.) What are the minimum and maximum allowed dimensions for the base width? Explain how they are determined. ( f ) Describe the datum features A, B, and C. Describe their corresponding datums. Describe the datum reference frame that is defined by applying A, B, and C in that order. Describe how the part is stabilized by these datums. Explain why this is more appropriate for this application than using the edges of the base for datums B and C. (Notice that the basic dimensions are either measured from, or implied to be centered on, the datums of the datum reference frame.) (g) If datum feature B is produced with a diameter of ∅6.00, what is the diameter of the tolerance zone in which its axis must lie? What if it is produced at ∅6.05? (h) If the bolt holes are produced at ∅6.0, what is the diameter of the tolerance zones locating the bolt hole pattern with respect to the true position specified by the basic dimensions? What if the bolt holes are produced at ∅6.1?

Geometric Dimensioning and Tolerancing     1017 12

+ 0.1 4 × ∅ 6 –0.0 THRU

68 56

∅ 12 ± 0.2 3 ± 0.1 ∅ 0.3 M A B M C M ∅ 0.1 M A

28

+ 0.05 6 –0.00 0.05 A B M

18 36

R3

+ 0.05 ∅6 – 0.00 THRU ∅ 0.05 A

C

Problem 20–27 B

R15

∅15 ± 0.1 ∅ 0.05 A B M C M 4 × R4

30 0.1

A

8 ± 0.1

UNLESS OTHERWISE SPECIFIED: 0.5 A B C

(i) If the bolt holes are produced at ∅6.0, what is the diameter of the tolerance zones locating the position of the bolt holes with respect to one another? What if the bolt holes are produced at ∅6.1? ( j) Explain why the M modifier is appropriate for the bolt hole position tolerance. (k) For the large bore, explain what provides control of each of the following: orientation, straightness of its center axis, and cylindricity of its surface. (l) Assume the part is cast, and the casting operation can provide a surface profile tolerance of less than 0.5. Which surfaces can likely be left in the as-cast condition without compromising any of the requirements of the drawing? How would this change if the drawing were modified to use the edges of the base as datum features B and C, while still maintaining the functional goals for the alignment of the rod?

20–28 Answer the following questions for the drawing shown in Figure 20–34.

(a) Which of the following is datum feature B? (Select one.) i. The surface of the bore hole. ii. The center axis of the bore hole. iii. The center axis of the largest gauge pin that will fit into the bore hole. iv. The outer surface of the protruding cylinder. v. The center axis of the outer surface of the protruding cylinder. (b) What is the maximum diameter allowed for a bolt hole? (c) What is the diameter of a bolt hole at its maximum material condition (MMC)?

60

1018      Mechanical Engineering Design

(d) If a bolt hole is produced at ∅10.1, what is the diameter of the tolerance zone in which its axis must lie? (e) If the center bore is produced at ∅20.05, what is the diameter of the tolerance zone in which its axis must lie? ( f ) The outer diameter of the flange is specified with a basic dimension of 90, with no tolerance. What are the minimum and maximum allowed dimensions for the diameter? Explain how they are determined. (g) The position control associated with the large bore hole defines a tolerance zone. Specifically what must stay within that tolerance zone? (Select one.)  i. The center axis of the hole, as determined by the largest gauge pin that fits into the hole. ii. The center axis of the hole, as determined by the center of each cross section of the hole. iii. The ideal location of the center axis of the tolerance zone, as specified by the basic dimensions. iv. The surface of the hole, when manufactured at its maximum material condition. (h) What controls the cylindricity of the large bore hole? (Select one.)  i. There is no control on the cylindricity. ii. From Rule #1, the envelope of a perfect cylinder with diameter of 20.05. iii. From Rule #1, the envelope of a perfect cylinder with diameter of 19.95. iv. From the position control, the center axis of each cross section must be within the 0.3 cylindrical tolerance zone. (i) The position control on the bolt holes has a tolerance of 0. Does this mean that the bolt holes must be perfectly located? Explain.

Useful TablesAppendix

Appendix Outline

A–1

Standard SI Prefixes

A–2

Conversion Factors

A–3

Optional SI Units for Bending, Torsion, Axial, and Direct Shear Stresses

A–4

Optional SI Units for Bending and Torsional Deflections

A–5

Physical Constants of Materials

A–6

Properties of Structural-Steel Angles

A–7

Properties of Structural-Steel Channels

A–8

Properties of Round Tubing

A–9

Shear, Moment, and Deflection of Beams

A–10

Cumulative Distribution Function of Normal (Gaussian) Distribution

A–11

A Selection of International Tolerance Grades—Metric Series

A–12

Fundamental Deviations for Shafts—Metric Series

A–13

A Selection of International Tolerance Grades—Inch Series

A–14

Fundamental Deviations for Shafts—Inch Series

A–15

Charts of Theoretical Stress-Concentration Factors Kt

A

1021 1022 1023

1023

1023 1024 1026

1028 1029 1037

1038

1039 1040

1041 1042

A–16 Approximate Stress-Concentration Factors Kt or Kts of a Round Bar or Tube with a Transverse Round Hole and Loaded in Bending or Torsion  A–17

Preferred Sizes and Renard (R-Series) Numbers

A–18

Geometric Properties

A–19

American Standard Pipe

1049

1051

1052 1055

A–20 Deterministic ASTM Minimum Tensile and Yield Strengths for HR and CD Steels  1056 A–21

Mean Mechanical Properties of Some Heat-Treated Steels

A–22

Results of Tensile Tests of Some Metals

1057

1059

A–23 Mean Monotonic and Cyclic Stress-Strain Properties of Selected Steels  1060 A–24

Mechanical Properties of Three Non-Steel Metals

A–25

Stochastic Yield and Ultimate Strengths for Selected Materials

1062 1064 1019

1020      Mechanical Engineering Design

A–26 Stochastic Parameters for Finite Life Fatigue Tests in Selected Metals  1065 A–27

Finite Life Fatigue Strengths of Selected Plain Carbon Steels

A–28

Decimal Equivalents of Wire and Sheet-Metal Gauges

A–29

Dimensions of Square and Hexagonal Bolts

A–30

Dimensions of Hexagonal Cap Screws and Heavy Hexagonal Screws

A–31

Dimensions of Hexagonal Nuts

A–32

Basic Dimensions of American Standard Plain Washers

A–33

Dimensions of Metric Plain Washers

A–34

Gamma Function

1074

1067

1069

1071

1073

1066

1072

1070

Useful Tables     1021

Table A–1  Standard SI Prefixes*† Name Symbol

Factor

exa

E

1 000 000 000 000 000 000 = 1018

peta

P

1 000 000 000 000 000 = 1015

tera

T

giga

G

1 000 000 000 = 109

mega

M

1 000 000 = 106

kilo

k

1 000 = 103

hecto

h

100 = 102

deka‡

da

10 = 101

deci

d

0.1 = 10−1

centi‡

c

0.01 = 10−2

milli

m

0.001 = 10−3

micro

μ

0.000 001 = 10−6

nano

n

0.000 000 001 = 10−9

pico

p

0.000 000 000 001 = 10−12

femto

f

0.000 000 000 000 001 = 10−15

atto

a

0.000 000 000 000 000 001 = 10−18

1 000 000 000 000 = 1012

*If possible use multiple and submultiple prefixes in steps of 1000. †

Spaces are used in SI instead of commas to group numbers to avoid confusion with the practice in some European countries of  using commas for decimal points. ‡

Not recommended but sometimes encountered.

1022      Mechanical Engineering Design

Table A–2  Conversion Factors A to Convert Input X to Output Y Using the Formula Y = AX* Multiply Input X

By Factor To Get Output A Y

Multiply Input X

British thermal unit, Btu

1055

joule, J

mass, lbf · s2/in

By Factor To Get Output A Y 175

kilogram, kg

mile, mi

1.610

kilometer, km

Btu/second, Btu/s

1.05

kilowatt, kW

mile/hour, mi/h

1.61

kilometer/hour, km/h

calorie

4.19

joule, J

mile/hour, mi/h

0.447

meter/second, m/s

centimeter of mercury (0°C)

1.333

kilopascal, kPa

moment of inertia, 0.0421 kilogram-meter2, 2 lbm · ft kg · m2

centipoise, cP 0.001 pascal-second, Pa · s

moment of inertia, 293 kilogram-millimeter2, lbm · in2 kg · mm2

degree (angle)

0.0174

radian, rad

foot, ft

0.305

meter, m

foot , ft

0.0929

meter2, m2

moment of section (second moment of area), in4

foot/minute, ft/min

0.0051

meter/second, m/s

foot-pound, ft · lbf

1.35

joule, J

foot-pound/ second, ft · lbf/s

1.35

foot/second, ft/s

2

2

41.6

centimeter4, cm4

ounce-force, oz

0.278

newton, N

ounce-mass

0.0311

kilogram, kg

pound, lbf

4.45

newton, N

watt, W

pound-foot, lbf · ft

1.36

newton-meter, N · m

0.305

meter/second, m/s

pound-inch, lbf · in

gallon (U.S.), gal

3.785

liter, L

pound/inch, lbf/in

horsepower, hp

0.746

kilowatt, kW

6.89

kilopascal, kPa

0.0254

meter, m

pound/inch2, psi (lbf/in2) pound-mass, lbm

0.454

kilogram, kg

inch, in inch, in 2

2

inch , in

25.4 645

pound/foot2, lbf/ft2

millimeter, mm 2

millimeter , mm

2

0.113 175

pascal, Pa newton-meter, N · m newton/meter, N/m

pound-mass/ 0.454 kilogram/second, second, lbm/s kg/s

inch of mercury (32°F)

3.386

inch-pound, in · lbf

0.113

joule, J

kilopound, kip

4.45

kilonewton, kN

slug

megapascal, MPa (N/mm2)

yard, yd

kilopound/inch2, 6.89 kpsi (ksi)

kilopascal, kPa

47.9

quart (U.S. liquid), qt

946

milliliter, mL

3

section modulus, in 16.4 centimeter3, cm3 ton (short 2000 lbm)

14.6 907 0.914

kilogram, kg kilogram, kg meter, m

*Approximate. †

The U.S. Customary system unit of the pound-force is often abbreviated as lbf to distinguish it from the pound-mass, which is abbreviated as lbm.

Useful Tables     1023

Bending and Torsion

M, T

I, J

c, r

Axial and Direct Shear

σ, τ

F

N · m* m4 m Pa N · m cm

4

N* m2

mm GPa

N · mm mm

m2

kN

MPa (N/mm2) kPa

kN mm2

kN · m cm cm GPa 4

Pa

MPa (N/mm ) N† mm2

4

σ, τ

2

cm

N · m† mm4

A

GPa

Table A–3  Optional SI Units for Bending Stress σ = Mc∕l, Torsion Stress τ = Tr∕J, Axial Stress σ = F∕A, and Direct Shear Stress τ = F∕A

2

mm

MPa (N/mm )

*Basic relation. †

Often preferred.

Bending Deflection

F, w l

l

I

E

Torsional Deflection y

T

N* m m4 Pa m N · m* †

kN

4

J

G

m

m4 Pa

mm mm GPa mm N · m mm mm4

kN m m4 N

l

mm mm

GPa 4

μm N · mm

kPa m

N · m

mm cm

mm4 cm

4

θ rad

GPa rad MPa (N/mm2) rad MPa (N/mm2) rad

Table A–4  Optional SI Units for Bending Deflection y = f (Fl 3∕El) or y = f (wl 4∕El) and Torsional Deflection θ = Tl∕GJ

*Basic relation. †

Often preferred.

Table A–5  Physical Constants of Materials Modulus of Modulus of Elasticity E Rigidity G GPa

Material

Mpsi

Aluminum (all alloys)

10.4

71.7

3.9

26.9

0.333

0.098

169

26.6

Beryllium copper

18.0

124.0

7.0

48.3

0.285

0.297

513

80.6

Brass

15.4 106.0

0.324

0.309

534 83.8

Carbon steel

30.0

207.0

0.292

0.282

487

76.5

Cast iron (gray)

14.5

100.0

70.6

Copper

17.2 119.0

Douglas fir

1.6

Glass

6.7 46.2

Inconel Lead Magnesium

11.0

31.0 214.0 5.3 36.5 6.5 44.8

Mpsi

GPa

Unit Weight w 3 Poisson's Ratio ν lbf/in lbf/ft 3 kN/m3

5.82 40.1 11.5

79.3

6.0

41.4

6.49 44.7 0.6

4.1

2.7 18.6 11.0 75.8

0.211

0.260

450

0.326

0.322

556 87.3

0.33

0.016

0.245

0.094 162 25.4

28

4.3

0.290

0.307

1.9 13.1

0.425

0.411 710 111.5

2.4 16.5

0.350

0.065 112 17.6

0.307

0.368

636 100.0

17.0 117.0

530 83.3

Molybdenum

48.0 331.0

Monel metal

26.0

179.0

9.5

65.5

0.320

0.319

551

86.6

Nickel silver

18.5

127.0

7.0

48.3

0.322

0.316

546

85.8

Nickel steel

30.0

207.0

11.5

79.3

0.291

0.280

484

76.0

Phosphor bronze

16.1

111.0

6.0

41.4

0.349

0.295

510

80.1

Stainless steel (18-8)

27.6

190.0

10.6

73.1

0.305

0.280

484

76.0

Titanium alloys

16.5

114.0

6.2

42.4

0.340

0.160

276

43.4

1024      Mechanical Engineering Design

Table A–6  Properties of Structural-Steel Equal Legs Angles*† w m A I k y Z

= = = = = = =

weight per foot, lbf/ft mass per meter, kg/m area, in2 (cm2) second moment of area, in4 (cm4) radius of gyration, in (cm) centroidal distance, in (cm) section modulus, in3, (cm3)

Size, in

1×1×

×

112 × 112 ×

×

2×2×

×

× 212

×

212

×

×

3×3×

×

×

312 × 312 ×

×

×

4×4×

×

×

×

6×6×

×

×

×

w

A

3

2

1

1 y 2

I1−1

k1−1

3

Z1−1

y

k3−3

1 0.80 0.234 0.021 0.298 0.029 0.290 0.191 8 1 1.49 0.437 0.036 0.287 0.054 0.336 0.193 4 1 1.23 0.36 0.074 0.45 0.068 0.41 0.29 8 1 0.135 0.44 0.130 0.46 0.29 4 2.34 0.69 1 1.65 0.484 0.190 0.626 0.131 0.546 0.398 8 1 3.19 0.938 0.348 0.609 0.247 0.592 0.391 4 3 4.7 1.36 0.479 0.594 0.351 0.636 0.389 8 1 4.1 1.19 0.703 0.769 0.394 0.717 0.491 4 3 5.9 1.73 0.984 0.753 0.566 0.762 0.487 8 1 4.9 1.44 1.24 0.930 0.577 0.842 0.592 4 3 7.2 2.11 1.76 0.913 0.833 0.888 0.587 8 1 2.22 0.898 1.07 0.932 0.584 2 9.4 2.75 1 5.8 1.69 2.01 1.09 0.794 0.968 0.694 4 3 8.5 2.48 2.87 1.07 1.15 1.01 0.687 8 1 11.1 3.25 3.64 1.06 1.49 1.06 0.683 2 1 6.6 1.94 3.04 1.25 1.05 1.09 0.795 4 3 9.8 2.86 4.36 1.23 1.52 1.14 0.788 8 1 12.8 3.75 5.56 1.22 1.97 1.18 0.782 2 5 15.7 4.61 6.66 1.20 2.40 1.23 0.779 8 3 14.9 4.36 15.4 1.88 3.53 1.64 1.19 8 1 19.6 5.75 19.9 1.86 4.61 1.68 1.18 2 5 24.2 7.11 24.2 1.84 5.66 1.73 1.18 8 3 28.7 8.44 28.2 1.83 6.66 1.78 1.17 4

Useful Tables     1025

Table A–6  Properties of Structural-Steel Equal Legs Angles*† (Continued) Size, mm 25 × 25 × 3

m

A

1.11

1.42

I1−1 0.80

k1−1

Z1−1

y

k3−3

0.75

0.45

0.72

0.48

× 4  1.45 1.85 1.01 0.74 0.58 0.76 0.48

× 5  1.77 2.26 1.20 0.73 0.71 0.80 0.48

40 × 40 × 4

2.42

3.08

4.47

1.21

1.12

0.78

× 5  2.97 3.79 5.43 1.20 1.91 1.16 0.77

× 6  3.52 4.48 6.31 1.19 2.26 1.20 0.77

50 × 50 × 5

3.77

4.80

11.0

1.51

3.05

1.40

0.97

× 6  4.47 5.59 12.8 1.50 3.61 1.45 0.97

× 8  5.82 7.41 16.3 1.48 4.68 1.52 0.96

60 × 60 × 5

4.57

5.82

19.4

1.82

4.45

1.64

1.17

× 6  5.42 6.91 22.8 1.82 5.29 1.69 1.17

× 8  7.09 9.03 29.2 1.80 6.89 1.77 1.16

× 10 8.69 11.1 34.9 1.78 8.41 1.85 1.16

80 × 80 × 6

7.34

9.35

55.8

2.44

9.57

2.17

1.57

× 8  9.63 12.3 72.2 2.43 12.6 2.26 1.56

× 10 11.9 15.1 87.5 2.41 15.4 2.34 1.55

100 × 100 × 8

12.2

15.5

145

3.06

19.9

2.74

1.96

× 12

17.8

22.7

207

3.02

29.1

2.90

1.94

× 15

21.9

27.9

249

2.98

35.6

3.02

1.93

150 × 150 × 10

23.0

29.3

624

4.62

56.9

4.03

2.97

× 12

27.3

34.8

737

4.60

67.7

4.12

2.95

× 15

33.8

43.0

898

4.57

83.5

4.25

2.93

× 18

40.1

51.0

1050

4.54

98.7

4.37

2.92

*Metric sizes also available in sizes of 45, 70, 90, 120, and 200 mm. †

1.55

These sizes are also available in aluminum alloy.

1026      Mechanical Engineering Design

Table A–7  Properties of Structural-Steel Channels* a, b w m t A I k x Z

= = = = = = = = =

a, in

size, in (mm) weight per foot, lbf/ft mass per meter, kg/m web thickness, in (mm) area, in2 (cm2) second moment of area, in4 (cm4) radius of gyration, in (cm) centroidal distance, in (cm) section modulus, in3 (cm3) b, in

t

A

2 t 1

1 a

x 2

w

I1−1

k1−1

b

Z1−1

I2−2

k2−2

Z2−2

x

3 1.410 0.170 1.21 4.1

1.66 1.17 1.10 0.197 0.404 0.202 0.436

3 1.498 0.258 1.47 5.0

1.85 1.12 1.24 0.247 0.410 0.233 0.438

3 1.596 0.356 1.76 6.0

2.07 1.08 1.38 0.305 0.416 0.268 0.455

4 1.580 0.180 1.57 5.4

3.85 1.56 1.93 0.319 0.449 0.283 0.457

4 1.720 0.321 2.13 7.25

4.59 1.47 2.29 0.433 0.450 0.343 0.459

5 1.750 0.190 1.97 6.7

7.49 1.95 3.00 0.479 0.493 0.378 0.484

5 1.885 0.325 2.64 9.0

8.90 1.83 3.56 0.632 0.489 0.450 0.478

6 1.920 0.200 2.40 8.2

13.1 2.34 4.38 0.693 0.537 0.492 0.511

6 2.034 0.314 3.09 10.5

15.2 2.22 5.06 0.866 0.529 0.564 0.499

6 2.157 0.437 3.83 13.0

17.4 2.13 5.80 1.05 0.525 0.642 0.514

7 2.090 0.210 2.87 9.8

21.3 2.72 6.08 0.968 0.581 0.625 0.540

7 2.194 0.314 3.60 12.25 24.2 2.60 6.93 1.17 0.571 0.703 0.525 7 2.299 0.419 4.33 14.75 27.2 2.51 7.78 1.38 0.564 0.779 0.532 8 2.260 0.220 3.36 11.5

32.3 3.10 8.10 1.30 0.625 0.781 0.571

8 2.343 0.303 4.04 13.75 36.2 2.99 9.03 1.53 0.615 0.854 0.553 8 2.527 0.487 5.51 18.75 44.0 2.82 11.0 1.98 0.599 1.01 0.565 9 2.430 0.230 3.91 13.4

47.7 3.49 10.6 1.75 0.669 0.962 0.601

9 2.485 0.285 4.41 15.0

51.0 3.40 11.3 1.93 0.661 1.01 0.586

9 2.648 0.448 5.88 20.0

60.9 3.22 13.5 2.42 0.647 1.17 0.583

10 2.600 0.240 4.49 15.3

67.4 3.87 13.5 2.28 0.713 1.16 0.634

10 2.739 0.379 5.88 20.0

78.9 3.66 15.8 2.81 0.693 1.32 0.606

10 2.886 0.526 7.35 25.0

91.2 3.52 18.2 3.36 0.676 1.48 0.617

10 3.033 0.673 8.82 30.0 103

3.43 20.7 3.95 0.669 1.66 0.649

12 3.047 0.387 7.35 25.0 144

4.43 24.1 4.47 0.780 1.89 0.674

12 3.170 0.510 8.82 30.0 162

4.29 27.0 5.14 0.763 2.06 0.674

Useful Tables     1027

Table A–7  Properties of Structural-Steel Channels*  (Continued) a × b, mm

m

t

  76 × 38

6.70

5.1

8.53

102 × 51

10.42

6.1

13.28

127 × 64

14.90

6.4

18.98

152 × 76

17.88

6.4

22.77

152 × 89

23.84

7.1

30.36

1166

178 × 76

20.84

6.6

26.54

178 × 89

26.81

7.6

34.15

203 × 76

23.82

7.1

203 × 89

29.78

8.1

229 × 76

26.06

229 × 89

32.76

254 × 76 254 × 89

I1−1

k1−1

Z1−1

I2−2

k2−2

Z2−2

x

74.14

2.95

19.46

10.66

1.12

4.07

1.19

207.7

3.95

40.89

29.10

1.48

8.16

1.51

482.5

5.04

75.99

67.23

1.88

15.25

1.94

851.5

6.12

111.8

113.8

2.24

21.05

2.21

6.20

153.0

215.1

2.66

35.70

2.86

1337

7.10

150.4

134.0

2.25

24.72

2.20

1753

7.16

197.2

241.0

2.66

39.29

2.76

30.34

1950

8.02

192.0

151.3

2.23

27.59

2.13

37.94

2491

8.10

245.2

264.4

2.64

42.34

2.65

7.6

33.20

2610

8.87

228.3

158.7

2.19

28.22

2.00

8.6

41.73

3387

9.01

296.4

285.0

2.61

44.82

2.53

28.29

8.1

36.03

3367

9.67

265.1

162.6

2.12

28.21

1.86

35.74

9.1

45.42

4448

9.88

350.2

302.4

2.58

46.70

2.42

305 × 89

41.69

10.2

53.11

7061

11.5

463.3

325.4

2.48

48.49

2.18

305 × 102

46.18

10.2

58.83

8214

11.8

539.0

499.5

2.91

66.59

2.66

A

*These sizes are also available in aluminum alloy.

1028      Mechanical Engineering Design

Table A–8  Properties of Round Tubing wa ws m A I J k Z d × t

= = = = = = = = =

unit weight of aluminum tubing, lbf/ft unit weight of steel tubing, lbf/ft unit mass, kg/m area, in2 (cm2) second moment of area, in4 (cm4) second polar moment of area, in4 (cm4) radius of gyration, in (cm) section modulus, in3 (cm3) size (OD) and thickness, in (mm)

Size, in

wa ws A I k Z J

  1 ×

1 8

0.416 1.128 0.344 0.034 0.313 0.067 0.067

  1 ×

1 4

0.713 2.003 0.589 0.046 0.280 0.092 0.092

1 8

0.653 1.769 0.540 0.129 0.488 0.172 0.257

112 × 14

1.188 3.338 0.982 0.199 0.451 0.266 0.399

 2 ×

1 8

0.891 2.670 0.736 0.325 0.664 0.325 0.650

  2×

1 4

1.663 4.673 1.374 0.537 0.625 0.537 1.074

112

212

×

1 8

1.129 3.050 0.933 0.660 0.841 0.528 1.319

212 × 14

2.138 6.008 1.767 1.132 0.800 0.906 2.276

  3 × 14

2.614 7.343 2.160 2.059 0.976 1.373 4.117

  3 ×

3 8

3.742 10.51 3.093 2.718 0.938 1.812 5.436

  4×

3 16

×

2.717 7.654 2.246 4.090 1.350 2.045 8.180

  4 × 38

5.167 14.52 4.271 7.090 1.289 3.544 14.180

Size, mm

m A I k Z J

12 × 2 0.490 0.628 0.082 0.361 0.136 0.163 16 × 2 0.687 0.879 0.220 0.500 0.275 0.440 16 × 3 0.956 1.225 0.273 0.472 0.341 0.545 20 × 4 1.569 2.010 0.684 0.583 0.684 1.367 25 × 4 2.060 2.638 1.508 0.756 1.206 3.015 25 × 5 2.452 3.140 1.669 0.729 1.336 3.338 30 × 4 2.550 3.266 2.827 0.930 1.885 5.652 30 × 5 3.065 3.925 3.192 0.901 2.128 6.381 42 × 4 3.727 4.773 8.717 1.351 4.151 17.430 42 × 5 4.536 5.809 10.130 1.320 4.825 20.255 50 × 4 4.512 5.778 15.409 1.632 6.164 30.810 50 × 5 5.517 7.065 18.118 1.601 7.247 36.226

Useful Tables     1029

Table A–9  Shear, Moment, and Deflection of Beams (Note: Force and moment reactions are positive in the directions shown; equations for shear force V and bending moment M follow the sign conventions given in Sec. 3–2.)  1 Cantilever—end load y

R1 = V = F l

M = F(x − l ) F

x M1

M1 = Fl

R1

y=

Fx 2 (x − 3l ) 6EI

ymax = −

Fl 3 3EI

V

+ x M x –

 2 Cantilever—intermediate load y

R1 = V = F l

a

MAB = F(x − a)

b F

A

C x

M1

B

M1 = Fa

R1 V

MBC = 0

2

yAB =

Fx (x − 3a) 6EI

yBC =

Fa2 (a − 3x) 6EI

ymax =

Fa 2 (a − 3l ) 6EI

+ x M –

x

(Continued)

1030      Mechanical Engineering Design

Table A–9  Shear, Moment, and Deflection of Beams  (Continued) (Note: Force and moment reactions are positive in the directions shown; equations for shear force V and bending moment M follow the sign conventions given in Sec. 3–2.)  3 Cantilever—uniform load y

R1 = wl

l

M1 =

w

x

V = w(l − x)

M1 R1

y=

w M = − (l − x) 2 2

wx 2 (4lx − x2 − 6l 2 ) 24EI

V

wl 2 2

ymax = −

wl 4 8EI

+ x M x

 4 Cantilever—moment load

R1 = V = 0

y l

M1

y=

MB A x

B R1 V x M

x

MB x 2 2EI

M1 = M = MB ymax =

MB l 2 2EI

Useful Tables     1031

Table A–9  Shear, Moment, and Deflection of Beams  (Continued) (Note: Force and moment reactions are positive in the directions shown; equations for shear force V and bending moment M follow the sign conventions given in Sec. 3–2.)   5  Simple supports—center load y

l/2

F

A

B

R1

MAB =

R2

V

yAB = +

Fx 2

VBC = −R2 MBC =

F (l − x) 2

Fx (4x 2 − 3l 2 ) 48EI

ymax = −

x

F 2

VAB = R1

C x

R 1 = R2 =

l

Fl 3 48EI

M

+ x

  6  Simple supports—intermediate load

y a

A

b

F B

R1

MAB =

R2

V +

Fb l

VAB = R1 C x

R1 =

l

x

Fbx l

R2 =

Fa l

VBC = −R2 MBC =

Fa (l − x) l

yAB =

Fbx 2 (x + b2 − l 2 ) 6EI l

yBC =

Fa (l − x) 2 (x + a2 − 2lx) 6EI l

M

+ x

(Continued)

1032      Mechanical Engineering Design

Table A–9  Shear, Moment, and Deflection of Beams  (Continued) (Note: Force and moment reactions are positive in the directions shown; equations for shear force V and bending moment M follow the sign conventions given in Sec. 3–2.)   7  Simple supports—uniform load

y

R 1 = R2 =

l w

M=

x R1

R2

y=

V

V=

wl − wx 2

wx (l − x) 2 wx (2lx 2 − x 3 − l 3 ) 24EI

ymax = − +

wl 2

5wl 4 384EI

x

M + x

  8  Simple supports—moment load

y

R 1 = R2 =

l b

a

A

C

x

B

MAB =

R2

MB

R1 V

+ x M

+ –

x

MB x l

MB l

V= MBC =

MB l

MB (x − l) l

yAB =

MB x 2 (x + 3a2 − 6al + 2l 2 ) 6EI l

yBC =

MB 3 [x − 3lx 2 + x (2l 2 + 3a2 ) − 3a2l] 6EI l

Useful Tables     1033

Table A–9  Shear, Moment, and Deflection of Beams  (Continued) (Note: Force and moment reactions are positive in the directions shown; equations for shear force V and bending moment M follow the sign conventions given in Sec. 3–2.)   9  Simple supports—twin loads

y

R 1 = R2 = F

l

F

a

A

F C

MAB = Fx

D x

R1

R2

V

+

VBC = 0

VCD = −F

a

B

VAB = F

x

MBC = Fa

yAB =

Fx 2 (x + 3a2 − 3la) 6EI

yBC =

Fa (3x 2 + a2 − 3lx) 6EI

ymax =

Fa (4a2 − 3l 2 ) 24EI

MCD = F(l − x)

M + x

10  Simple supports—overhanging load

y

R1 =

a

l

Fa l

F

R1 B

A

C x

R2

VAB = −

Fa l

MAB = −

Fax l

V

+

x

– M

F (l + a) l

VBC = F MBC = F(x − l − a)

yAB =

Fax 2 (l − x2 ) 6EI l

yBC =

F(x − l) [(x − l) 2 − a(3x − l )] 6EI

yC = −

R2 =

Fa2 (l + a) 3EI

x

(Continued)

1034      Mechanical Engineering Design

Table A–9  Shear, Moment, and Deflection of Beams  (Continued) (Note: Force and moment reactions are positive in the directions shown; equations for shear force V and bending moment M follow the sign conventions given in Sec. 3–2.) 11  One fixed and one simple support—center load y

F

l/2

A

R1 =

l

VAB = R1

C

B

x

MAB =

R2

M1

R1

V

11F 16

+

R2 =

5F 16

M1 =

3Fl 16

VBC = −R2

F (11x − 3l ) 16

MBC =

5F (l − x) 16

yAB =

Fx 2 (11x − 9l) 96EI

yBC =

F(l − x) (5x 2 + 2l 2 − 10lx) 96EI

x

M

+ x

12  One fixed and one simple support—intermediate load y

l

F

a

A

R1

V

+ –

x R2

M1

Fb (3l 2 − b2 ) 2l 3

M1 =

Fb 2 (l − b2 ) 2l 2

b C

B

R1 =

x

M

VAB = R1

x

Fa2 (3l − a) 2l 3

VBC = −R2

MAB =

Fb 2 [b l − l 3 + x (3l 2 − b2 )] 2l 3

MBC =

Fa2 2 (3l − 3lx − al + ax) 2l 3

yAB =

Fbx 2 [3l(b2 − l 2 ) + x(3l 2 − b2 )] 12EI l 3

yBC = yAB − +

R2 =

F(x − a) 3 6EI

Useful Tables     1035

Table A–9  Shear, Moment, and Deflection of Beams  (Continued) (Note: Force and moment reactions are positive in the directions shown; equations for shear force V and bending moment M follow the sign conventions given in Sec. 3–2.) 13  One fixed and one simple support—uniform load y

R1 =

l w

R2

M1

V=

x

5wl 8

R2 =

3wl 8

M1 =

wl2 8

5wl − wx 8

w M = − (4x 2 − 5lx + l 2 ) 8

R1 V

y=

wx 2 (l − x) (2x − 3l) 48EI

+ x

M + x

14  Fixed supports—center load y

l/2

F

A

B

C x

M1

R 1 = R2 =

l

R1

R2

V

M2

VAB = −VBC = MAB = yAB =

+

x

F 2

M1 = M2 =

Fl 8

F 2

F (4x − l ) 8

MBC =

F (3l − 4x) 8

Fx 2 (4x − 3l) 48EI

ymax = −

Fl 3 192EI

M + –

x

(Continued)

1036      Mechanical Engineering Design

Table A–9  Shear, Moment, and Deflection of Beams  (Continued) (Note: Force and moment reactions are positive in the directions shown; equations for shear force V and bending moment M follow the sign conventions given in Sec. 3–2.) 15  Fixed supports—intermediate load y

l a

Fb2 (3a + b) l3

M1 =

Fab2 l2

b

F

A

R1 =

B

C x

M1

R1

R2

MAB =

V

M2 =

Fa2 (3b + a) l3

Fa2b l2

VBC = −R2 2

Fb [x(3a + b) − al] l3

MBC = MAB − F(x − a)

+

VAB = R1

M2

R2 =

yAB =

x

yBC =

M

Fb2x 2 [x(3a + b) − 3al] 6EI l 3 Fa2 (l − x) 2 6EI l 3

[(l − x) (3b + a) − 3bl]

+ –

x

16  Fixed supports—uniform load y

R 1 = R2 =

l w

x M1

R1

R2

w (l − 2x) 2

M=

w (6lx − 6x2 − l 2 ) 12

M2

y=− + –

x

M

+ –

x

M1 = M2 =

V=

V

wl 2

ymax = −

wx 2 (l − x) 2 24EI wl 4 384EI

wl2 12

Useful Tables     1037

Table A–10  Cumulative Distribution Function of Normal (Gaussian) Distribution Φ(z α ) =

1 u2 exp(− )du √2π 2 −∞

α ={ 1−α Z α 0.00

zα ≤ 0 zα > 0 0.01

f(z) Φ(zα) α 0 zα

0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

0.0 0.5000 0.4960 0.4920 0.4880 0.4840 0.4801 0.4761 0.4721 0.4681 0.4641 0.1 0.4602 0.4562 0.4522 0.4483 0.4443 0.4404 0.4364 0.4325 0.4286 0.4247 0.2 0.4207 0.4168 0.4129 0.4090 0.4052 0.4013 0.3974 0.3936 0.3897 0.3859 0.3 0.3821 0.3783 0.3745 0.3707 0.3669 0.3632 0.3594 0.3557 0.3520 0.3483 0.4 0.3446 0.3409 0.3372 0.3336 0.3300 0.3264 0.3238 0.3192 0.3156 0.3121 0.5 0.3085 0.3050 0.3015 0.2981 0.2946 0.2912 0.2877 0.2843 0.2810 0.2776 0.6 0.2743 0.2709 0.2676 0.2643 0.2611 0.2578 0.2546 0.2514 0.2483 0.2451 0.7 0.2420 0.2389 0.2358 0.2327 0.2296 0.2266 0.2236 0.2206 0.2177 0.2148 0.8 0.2119 0.2090 0.2061 0.2033 0.2005 0.1977 0.1949 0.1922 0.1894 0.1867 0.9 0.1841 0.1814 0.1788 0.1762 0.1736 0.1711 0.1685 0.1660 0.1635 0.1611 1.0 0.1587 0.1562 0.1539 0.1515 0.1492 0.1469 0.1446 0.1423 0.1401 0.1379 1.1 0.1357 0.1335 0.1314 0.1292 0.1271 0.1251 0.1230 0.1210 0.1190 0.1170 1.2 0.1151 0.1131 0.1112 0.1093 0.1075 0.1056 0.1038 0.1020 0.1003 0.0985 1.3 0.0968 0.0951 0.0934 0.0918 0.0901 0.0885 0.0869 0.0853 0.0838 0.0823 1.4 0.0808 0.0793 0.0778 0.0764 0.0749 0.0735 0.0721 0.0708 0.0694 0.0681 1.5 0.0668 0.0655 0.0643 0.0630 0.0618 0.0606 0.0594 0.0582 0.0571 0.0559 1.6 0.0548 0.0537 0.0526 0.0516 0.0505 0.0495 0.0485 0.0475 0.0465 0.0455 1.7 0.0446 0.0436 0.0427 0.0418 0.0409 0.0401 0.0392 0.0384 0.0375 0.0367 1.8 0.0359 0.0351 0.0344 0.0336 0.0329 0.0322 0.0314 0.0307 0.0301 0.0294 1.9 0.0287 0.0281 0.0274 0.0268 0.0262 0.0256 0.0250 0.0244 0.0239 0.0233 2.0 0.0228 0.0222 0.0217 0.0212 0.0207 0.0202 0.0197 0.0192 0.0188 0.0183 2.1 0.0179 0.0174 0.0170 0.0166 0.0162 0.0158 0.0154 0.0150 0.0146 0.0143 2.2 0.0139 0.0136 0.0132 0.0129 0.0125 0.0122 0.0119 0.0116 0.0113 0.0110 2.3 0.0107 0.0104 0.0102 0.00990 0.00964 0.00939 0.00914 0.00889 0.00866 0.00842 2.4 0.00820 0.00798 0.00776 0.00755 0.00734 0.00714 0.00695 0.00676 0.00657 0.00639 2.5 0.00621 0.00604 0.00587 0.00570 0.00554 0.00539 0.00523 0.00508 0.00494 0.00480 2.6 0.00466 0.00453 0.00440 0.00427 0.00415 0.00402 0.00391 0.00379 0.00368 0.00357 2.7 0.00347 0.00336 0.00326 0.00317 0.00307 0.00298 0.00289 0.00280 0.00272 0.00264 2.8 0.00256 0.00248 0.00240 0.00233 0.00226 0.00219 0.00212 0.00205 0.00199 0.00193 2.9 0.00187 0.00181 0.00175 0.00169 0.00164 0.00159 0.00154 0.00149 0.00144 0.00139 (Continued)

1038      Mechanical Engineering Design

Table A–10  Cumulative Distribution Function of Normal (Gaussian) Distribution*  (Continued) Zα 0.0

0.1

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

3

3 0.00135 0.0 968 0.03687 0.03483 0.03337 0.03233 0.03159 0.03108 0.04723 0.04481 4 0.04317 0.04207 0.04133 0.05854 0.05541 0.05340 0.05211 0.05130 0.06793 0.06479 5 0.06287 0.06170 0.07996 0.07579 0.07333 0.07190 0.07107 0.08599 0.08332 0.08182 6 0.09987 0.09530 0.09282 0.09149 0.010777 0.010402 0.010206 0.010104 0.011523 0.011260 z α −1.282 −1.643 −1.960 −2.326 −2.576 −3.090 −3.291 −3.891

−4.417

F(z α) 0.10 0.05 0.025 0.010 0.005 0.001 0.0005 0.0001 0.000005 R(z α) 0.90 0.95 0.975 0.990 0.995 0.999 0.9995 0.9999 0.999995 *The superscript on a zero after the decimal point indicates how many zeros there are after the decimal point. For example, 0.04481 = 0.000 048 1.

Table A–11  A Selection of International Tolerance Grades—Metric Series (Size Ranges Are for Over the Lower Limit and Including the Upper Limit. All Values Are in Millimeters) Basic Sizes

Tolerance Grades IT6

IT7

IT8

IT9

IT10

IT11

0–3

0.006 0.010 0.014 0.025 0.040 0.060

3–6

0.008 0.012 0.018 0.030 0.048 0.075

6–10 0.009 0.015 0.022 0.036 0.058 0.090 10–18 0.011 0.018 0.027 0.043 0.070 0.110 18–30 0.013 0.021 0.033 0.052 0.084 0.130 30–50 0.016 0.025 0.039 0.062 0.100 0.160 50–80 0.019 0.030 0.046 0.074 0.120 0.190 80–120 0.022 0.035 0.054 0.087 0.140 0.220 120–180 0.025 0.040 0.063 0.100 0.160 0.250 180–250 0.029 0.046 0.072 0.115 0.185 0.290 250–315 0.032 0.052 0.081 0.130 0.210 0.320 315–400 0.036 0.057 0.089 0.140 0.230 0.360 Source: Preferred Metric Limits and Fits, ANSI B4.2-1978. See also BSI 4500.

Useful Tables     1039

Table A–12  Fundamental Deviations for Shafts—Metric Series (Size Ranges Are for Over the Lower Limit and Including the Upper Limit. All Values Are in Millimeters) Basic Sizes

Upper-Deviation Letter c

d

f

g

h

Lower-Deviation Letter k

0

n

p

s

u

0–3

−0.060 −0.020 −0.006 −0.002 0

+0.004 +0.006 +0.014 +0.018

3–6

−0.070 −0.030 −0.010 −0.004 0 +0.001 +0.008 +0.012 +0.019 +0.023

6–10 −0.080 −0.040 −0.013 −0.005 0 +0.001 +0.010 +0.015 +0.023 +0.028 10–14 −0.095 −0.050 −0.016 −0.006 0 +0.001 +0.012 +0.018 +0.028 +0.033 14–18 −0.095 −0.050 −0.016 −0.006 0 +0.001 +0.012 +0.018 +0.028 +0.033 18–24 −0.110 −0.065 −0.020 −0.007 0 +0.002 +0.015 +0.022 +0.035 +0.041 24–30 −0.110 −0.065 −0.020 −0.007 0 +0.002 +0.015 +0.022 +0.035 +0.048 30–40 −0.120 −0.080 −0.025 −0.009 0 +0.002 +0.017 +0.026 +0.043 +0.060 40–50 −0.130 −0.080 −0.025 −0.009 0 +0.002 +0.017 +0.026 +0.043 +0.070 50–65 −0.140 −0.100 −0.030 −0.010 0 +0.002 +0.020 +0.032 +0.053 +0.087 65–80 −0.150 −0.100 −0.030 −0.010 0 +0.002 +0.020 +0.032 +0.059 +0.102 80–100 −0.170 −0.120 −0.036 −0.012 0 +0.003 +0.023 +0.037 +0.071 +0.124 100–120 −0.180 −0.120 −0.036 −0.012 0 +0.003 +0.023 +0.037 +0.079 +0.144 120–140 −0.200 −0.145 −0.043 −0.014 0 +0.003 +0.027 +0.043 +0.092 +0.170 140–160 −0.210 −0.145 −0.043 −0.014 0 +0.003 +0.027 +0.043 +0.100 +0.190 160–180 −0.230 −0.145 −0.043 −0.014 0 +0.003 +0.027 +0.043 +0.108 +0.210 180–200 −0.240 −0.170 −0.050 −0.015 0 +0.004 +0.031 +0.050 +0.122 +0.236 200–225 −0.260 −0.170 −0.050 −0.015 0 +0.004 +0.031 +0.050 +0.130 +0.258 225–250 −0.280 −0.170 −0.050 −0.015 0 +0.004 +0.031 +0.050 +0.140 +0.284 250–280 −0.300 −0.190 −0.056 −0.017 0 +0.004 +0.034 +0.056 +0.158 +0.315 280–315 −0.330 −0.190 −0.056 −0.017 0 +0.004 +0.034 +0.056 +0.170 +0.350 315–355 −0.360 −0.210 −0.062 −0.018 0 +0.004 +0.037 +0.062 +0.190 +0.390 355–400 −0.400 −0.210 −0.062 −0.018 0 +0.004 +0.037 +0.062 +0.208 +0.435 Source: Preferred Metric Limits and Fits, ANSI B4.2-1978. See also BSI 4500.

1040      Mechanical Engineering Design

Table A–13  A Selection of International Tolerance Grades—Inch Series (Size Ranges Are for Over the Lower Limit and Including the Upper Limit. All Values Are in Inches, Converted from Table A–11) Basic Sizes

IT6

IT7

Tolerance Grades IT8 IT9

IT10

IT11

  0–0.12

0.0002 0.0004 0.0006 0.0010 0.0016 0.0024

0.12–0.24

0.0003 0.0005 0.0007 0.0012 0.0019 0.0030

0.24–0.40

0.0004 0.0006 0.0009 0.0014 0.0023 0.0035

0.40–0.72

0.0004 0.0007 0.0011 0.0017 0.0028 0.0043

0.72–1.20

0.0005 0.0008 0.0013 0.0020 0.0033 0.0051

1.20–2.00

0.0006 0.0010 0.0015 0.0024 0.0039 0.0063

2.00–3.20

0.0007 0.0012 0.0018 0.0029 0.0047 0.0075

3.20–4.80

0.0009 0.0014 0.0021 0.0034 0.0055 0.0087

4.80–7.20

0.0010 0.0016 0.0025 0.0039 0.0063 0.0098

7.20–10.00

0.0011 0.0018 0.0028 0.0045 0.0073 0.0114

10.00–12.60

0.0013 0.0020 0.0032 0.0051 0.0083 0.0126

12.60–16.00

0.0014 0.0022 0.0035 0.0055 0.0091 0.0142

1041

0

0.40–0.72 −0.0037 −0.0020 −0.0006 −0.0002 0

u

+0.0005 +0.0007 +0.0011 +0.0013

+0.0004 +0.0006 +0.0009 +0.0011

+0.0003 +0.0005 +0.0007 +0.0009

+0.0002 +0.0002 +0.0006 +0.0007

Lower-Deviation Letter n p s

14.20–16.00 −0.0157 −0.0083 −0.0024 −0.0007 0 +0.0002 +0.0015 +0.0024 +0.0082 +0.0171

12.60–14.20 −0.0142 −0.0083 −0.0024 −0.0007 0 +0.0002 +0.0015 +0.0024 +0.0075 +0.0154

11.20–12.60 −0.0130 −0.0075 −0.0022 −0.0007 0 +0.0002 +0.0013 +0.0022 +0.0067 +0.0130

10.00–11.20 −0.0118 −0.0075 −0.0022 −0.0007 0 +0.0002 +0.0013 +0.0022 +0.0062 +0.0124

9.00–10.00 −0.0110 −0.0067 −0.0020 −0.0006 0 +0.0002 +0.0012 +0.0020 +0.0055 +0.0112

8.00–9.00 −0.0102 −0.0067 −0.0020 −0.0006 0 +0.0002 +0.0012 +0.0020 +0.0051 +0.0102

7.20–8.00 −0.0094 −0.0067 −0.0020 −0.0006 0 +0.0002 +0.0012 +0.0020 +0.0048 +0.0093

6.40–7.20 −0.0091 −0.0057 −0.0017 −0.0006 0 +0.0001 +0.0011 +0.0017 +0.0043 +0.0083

5.60–6.40 −0.0083 −0.0057 −0.0017 −0.0006 0 +0.0001 +0.0011 +0.0017 +0.0039 +0.0075

4.80–5.60 −0.0079 −0.0057 −0.0017 −0.0006 0 +0.0001 +0.0011 +0.0017 +0.0036 +0.0067

4.00–4.80 −0.0071 −0.0047 −0.0014 −0.0005 0 +0.0001 +0.0009 +0.0015 +0.0031 +0.0057

3.20–4.00 −0.0067 −0.0047 −0.0014 −0.0005 0 +0.0001 +0.0009 +0.0015 +0.0028 +0.0049

2.60–3.20 −0.0059 −0.0039 −0.0012 −0.0004 0 +0.0001 +0.0008 +0.0013 +0.0023 +0.0040

2.00–2.60 −0.0055 −0.0039 −0.0012 −0.0004 0 +0.0001 +0.0008 +0.0013 +0.0021 +0.0034

1.60–2.00 −0.0051 −0.0031 −0.0010 −0.0004 0 +0.0001 +0.0007 +0.0010 +0.0017 +0.0028

1.20–1.60 −0.0047 −0.0031 −0.0010 −0.0004 0 +0.0001 +0.0007 +0.0010 +0.0017 +0.0024

0.96–1.20 −0.0043 −0.0026 −0.0008 −0.0003 0 +0.0001 +0.0006 +0.0009 +0.0014 +0.0019

0.72–0.96 −0.0043 −0.0026 −0.0008 −0.0003 0 +0.0001 +0.0006 +0.0009 +0.0014 +0.0016

0

0.24–0.40 −0.0031 −0.0016 −0.0005 −0.0002 0

k 0

g h 0

Upper-Deviation Letter d f

0–0.12 −0.0024 −0.0008 −0.0002 −0.0001 0

c

0.12–0.24 −0.0028 −0.0012 −0.0004 −0.0002 0

Basic Sizes

(Size Ranges Are for Over the Lower Limit and Including the Upper Limit. All Values Are in Inches, Converted from Table A–12)

Table A–14  Fundamental Deviations for Shafts—Inch Series

1042      Mechanical Engineering Design

Table A–15  Charts of Theoretical Stress-Concentration Factors K*t 3.0

Figure A–15–1 Bar in tension or simple compression with a transverse hole. σ0 = F∕A, where A = (w − d)t and t is the thickness.

d 2.8

w

F

F

2.6 Kt 2.4

2.2

2.0

Figure A–15–2

0

0.1

0.2

0.3

0.4 d/w

0.5

0.6

3.0

Rectangular bar with a transverse hole in bending. σ0 = Mc∕I, where I = (w − d)h3∕12.

0.7

0.8

d d /h = 0

w

2.6 0.25

M

M

0.5

2.2

h

1.0

Kt

2.0

1.8

1.4

1.0

0

0.1

0.2

0.3

0.4 d/w

0.5

3.0

Figure A–15–3 Notched rectangular bar in tension or simple compression. σ0 = F∕A, where A = dt and t is the thickness.

0.6

0.7

0.8

r w/d = 3 F

2.6

w

F

d

1.5

2.2

1.2

Kt

1.1

1.8

1.05

1.4

1.0

0

0.05

0.10

0.15 r /d

0.20

0.25

0.30

Useful Tables     1043

Table A–15  Charts of Theoretical Stress-Concentration Factors K*t  (Continued) 3.0

Figure A–15–4

1.10

Notched rectangular bar in bending. σ0 = Mc∕I, where c = d∕2, I = td3∕12, and t is the thickness.

w/d = ∞

2.6

r M

1.5

1.05

w

M

d

2.2 1.02

Kt 1.8

1.4

1.0

Figure A–15–5

3.0

Rectangular filleted bar in tension or simple compression. σ0 = F∕A, where A = dt and t is the thickness.

2.6

0

0.05

0.15 r /d

0.20

0.25

0.30

r

D/d = 1.50 F

F

d

D

1.10

2.2 Kt

1.05

1.8

1.02

1.4

1.0

0.10

0

0.05

0.10

0.15 r/d

0.20

0.25

0.30

3.0

Figure A–15–6 Rectangular filleted bar in bending. σ0 = Mc∕I, where c = d∕2, I = td3∕12, t is the thickness.

r 2.6

M

1.05

d

D

M

3

2.2

1.1 1.3

Kt 1.8

D/d = 1.02

1.4

1.0

0

0.05

0.10

0.15 r/d

0.20

0.25

0.30

(Continued)

*Factors from R. E. Peterson, "Design Factors for Stress Concentration," Machine Design, vol. 23, no. 2, February 1951, p. 169; no. 3, March 1951, p. 161, no. 5, May 1951, p. 159; no. 6, June 1951, p. 173; no. 7, July 1951, p. 155. Reprinted with permission from Machine Design, a Penton Media Inc. publication.

1044      Mechanical Engineering Design

Table A–15  Charts of Theoretical Stress-Concentration Factors K*t  (Continued) 2.6

Figure A–15–7 Round shaft with shoulder fillet in tension. σ0 = F∕A, where A = πd 2∕4.

r 2.2

F

Kt 1.8

D/d = 1.0

d

D

F

1.50

1.10

5

1.4

1.02 1.0

Figure A–15–8

0

0.05

0.10

0.15 r/d

0.20

0.25

0.30

3.0

Round shaft with shoulder fillet in torsion. τ0 = Tc∕J, where c = d∕2 and J = πd 4∕32.

r 2.6

d

D

T

T

2.2 Kts 1.8

1.4

1.0

Figure A–15–9

3.0

Round shaft with shoulder fillet in bending. σ0 = Mc∕I, where c = d∕2 and I = πd 4∕64.

2.6

1.09

0

0.05

1.20 1.33

D/d = 2

0.10

0.15 r/d

0.20

0.25

0.30

r M

d

D

M

2.2 Kt 1.8

D/d =

3

1.5 1.4

1.10

1.02

1.05 1.0

0

0.05

0.10

0.15 r/d

0.20

0.25

0.30

Useful Tables     1045

Table A–15  Charts of Theoretical Stress-Concentration Factors K*t  (Continued) 4.0

Figure A–15–10 Round shaft in torsion with transverse hole.

T

3.6 Kts, A

Figure A–15–11

3.0

Round shaft in bending with a transverse hole. σ0 = M∕[(πD3∕32) − (dD2∕6)], approximately.

2.6

B A 3

J πD dD2 c = 16 – 6 (approx)

Kts, B

2.8

2.4

D

T

Kts 3.2

d

0

0.05

0.10

0.15 d/D

0.20

0.25

0.30

d D M

M

2.2 Kt 1.8

1.4

1.0

Figure A–15–12 Plate loaded in tension by a pin through a hole. σ0 = F∕A, where A = (w − d)t. When clearance exists, increase Kt 35 to 50 percent. (M. M. Frocht and H. N. Hill, "Stress-Concentration Factors around a Central Circular Hole in a Plate Loaded through a Pin in Hole," J. Appl. Mechanics, vol. 7, no. 1, March 1940, p. A-5.)

0

0.05

0.10

0.15 d/D

0.20

0.25

0.30

11 t h

9

d

h/w = 0.35

F/2

7

w

F/2

Kt 5

F h/w = 0.50

3

1

h/w ≥1.0

0

0.1

0.2

0.3

0.4 d/w

0.5

0.6

0.7

0.8

(Continued)

*Factors from R. E. Peterson, "Design Factors for Stress Concentration," Machine Design, vol. 23, no. 2, February 1951, p. 169; no. 3, March 1951, p. 161, no. 5, May 1951, p. 159; no. 6, June 1951, p. 173; no. 7, July 1951, p. 155. Reprinted with permission from Machine Design, a Penton Media Inc. publication.

1046      Mechanical Engineering Design

Table A–15  Charts of Theoretical Stress-Concentration Factors K*t  (Continued) 3.0

Figure A–15–13

r

1.15

Grooved round bar in tension. σ0 = F∕A, where A = πd 2∕4.

2.6

F

1.05

D

d

F

2.2 Kt

1.02

1.8

D/d = 1.50

1.4

1.0

Figure A–15–14

3.0

Grooved round bar in bending. σ0 = Mc∕I, where c = d∕2 and I = πd 4∕64.

2.6

0

0.05

0.10

0.15 r/d

0.20

0.25

0.30

r

M

D

M

d

1.05

2.2 Kt

D/d = 1.50

1.02

1.8

1.4

1.0

0

Figure A–15–15

2.6

Grooved round bar in torsion. τ0 = Tc∕J, where c = d∕2 and J = πd 4∕32.

2.2

0.05

0.10

0.15 r/d

0.20

0.25

r T

Kts

T D

1.8

d

1.05 D/d = 1.30

1.4

1.0

0.30

1.02 0

0.05

0.10

0.15 r/d

0.20

0.25

0.30

*Factors from R. E. Peterson, "Design Factors for Stress Concentration," Machine Design, vol. 23, no. 2, February 1951, p. 169; no. 3, March 1951, p. 161, no. 5, May 1951, p. 159; no. 6, June 1951, p. 173; no. 7, July 1951, p. 155. Reprinted with permission from Machine Design, a Penton Media Inc. publication.

Useful Tables     1047

Table A–15  Charts of Theoretical Stress-Concentration Factors K*t  (Continued) Figure A–15–16 Round shaft with flat-bottom groove in bending and/or tension. σ0 =

4F πd 2

+

32M

r r 9.0

t

F M

d

M

a 8.0

πd 3

Source: Adapted from W. D. Pilkey and D. F. Pilkey, Peterson's Stress-Concentration Factors, 3rd ed. John Wiley & Sons, Hoboken, NJ, 2008, p. 115.

F

D

0.03 7.0

0.04 0.05

Kt 6.0 0.07 0.10

5.0

0.15 0.20

4.0

0.40 0.60

3.0

2.0 0.5 0.6 0.7 0.8 0.91.0

r t 2.0

a/t

3.0

1.00 4.0

5.0 6.0

(Continued)

1048      Mechanical Engineering Design

Table A–15  Charts of Theoretical Stress-Concentration Factors K*t  (Continued) Figure A–15–17

r

T

t

Round shaft with flat-bottom groove in torsion. τ0 =

d

16T πd

r D T

a

3

Source: Adapted from W. D. Pilkey and D. F. Pilkey, Peterson's Stress-Concentration Factors, 3rd ed. John Wiley & Sons, Hoboken, NJ, 2008, p. 133.

5.0 0.03 0.04

4.0

0.06 Kts 3.0

0.10 0.20

2.0

r t

1.0 0.5 0.6 0.7 0.8 0.91.0

2.0 a/t

3.0

4.0

5.0

6.0

Useful Tables     1049

Table A–16  Approximate Stress-Concentration Factor Kt of a Round Bar or Tube with a Transverse Round Hole and Loaded in Bending a

D M

d M

The nominal bending stress is σ0 = M∕Znet where Znet is a reduced value of the section modulus and is defined by Z net =

πA (D 4 − d 4 ) 32D

Values of A are listed in the table. Use d = 0 for a solid bar.

d∕D

0.9 0.6 0 a∕D A Kt A Kt A Kt 0.050

0.92 2.63 0.91 2.55 0.88 2.42

0.075

0.89 2.55 0.88 2.43 0.86 2.35

0.10

0.86 2.49 0.85 2.36 0.83 2.27

0.125

0.82 2.41 0.82 2.32 0.80 2.20

0.15

0.79 2.39 0.79 2.29 0.76 2.15

0.175

0.76 2.38 0.75 2.26 0.72 2.10

0.20

0.73 2.39 0.72 2.23 0.68 2.07

0.225

0.69 2.40 0.68 2.21 0.65 2.04

0.25

0.67 2.42 0.64 2.18 0.61 2.00

0.275

0.66 2.48 0.61 2.16 0.58 1.97

0.30

0.64 2.52 0.58 2.14 0.54 1.94

Source: Data from R. E. Peterson, Stress-Concentration Factors, Wiley, New  York, 1974, pp. 146, 235.

(Continued)

1050      Mechanical Engineering Design

Table A–16  Approximate Stress-Concentration Factors Kts for a Round Bar or Tube Having a Transverse Round Hole and Loaded in Torsion  (Continued)

D T

a

T

d

The maximum stress occurs on the inside of the hole, slightly below the shaft surface. The nominal shear stress is τ0 = TD∕2Jnet, where Jnet is a reduced value of the second polar moment of area and is defined by Jnet =

πA(D 4 − d 4 ) 32

Values of A are listed in the table. Use d = 0 for a solid bar.

d∕D

0.9 0.8 0.6 0.4 0 a∕D A Kts A Kts A Kts A Kts A Kts 0.05 0.96 1.78 0.95 1.77 0.075 0.95 1.82 0.93 1.71 0.10 0.94 1.76 0.93 1.74 0.92 1.72 0.92 1.70 0.92 1.68 0.125 0.91 1.76 0.91 1.74 0.90 1.70 0.90 1.67 0.89 1.64 0.15 0.90 1.77 0.89 1.75 0.87 1.69 0.87 1.65 0.87 1.62 0.175 0.89 1.81 0.88 1.76 0.87 1.69 0.86 1.64 0.85 1.60 0.20 0.88 1.96 0.86 1.79 0.85 1.70 0.84 1.63 0.83 1.58 0.25 0.87 2.00 0.82 1.86 0.81 1.72 0.80 1.63 0.79 1.54 0.30 0.80 2.18 0.78 1.97 0.77 1.76 0.75 1.63 0.74 1.51 0.35 0.77 2.41 0.75 2.09 0.72 1.81 0.69 1.63 0.68 1.47 0.40 0.72 2.67 0.71 2.25 0.68 1.89 0.64 1.63 0.63 1.44 Source: Data from R. E. Peterson, Stress-Concentration Factors, Wiley, New York, 1974, pp. 148, 244.

Useful Tables     1051

Table A–17  Preferred Sizes and Renard (R-Series) Numbers (When a choice can be made, use one of these sizes; however, not all parts or items are available in all the sizes shown in the table.) Fraction of Inches 1 1 1 3 1 5 3 1 5 3 7 1 9 5 11 3 7 64 , 32 , 16 , 32 , 8 , 32 , 16 , 4 , 16 , 8 , 16 , 2 , 16 , 8 , 16 , 4 , 8 ,

1, 114 , 112 , 134 , 2, 214 , 212 , 234 , 3,

314 , 312 , 334 , 4, 414 , 412 , 434 , 5, 514 , 512 , 534 , 6, 612 , 7, 712 , 8, 812 , 9, 912 , 10, 1012 , 11, 1112 , 12, 1212 , 13, 1312 , 14, 1412 , 15, 1512 , 16, 1612 , 17, 1712 , 18, 1812 , 19, 1912 , 20 Decimal Inches 0.010, 0.012, 0.016, 0.020, 0.025, 0.032, 0.040, 0.05, 0.06, 0.08, 0.10, 0.12, 0.16, 0.20, 0.24, 0.30, 0.40, 0.50, 0.60, 0.80, 1.00, 1.20, 1.40, 1.60, 1.80, 2.0, 2.4, 2.6, 2.8, 3.0, 3.2, 3.4, 3.6, 3.8, 4.0, 4.2, 4.4, 4.6, 4.8, 5.0, 5.2, 5.4, 5.6, 5.8, 6.0, 7.0, 7.5, 8.5, 9.0, 9.5, 10.0, 10.5, 11.0, 11.5, 12.0, 12.5, 13.0, 13.5, 14.0, 14.5, 15.0, 15.5, 16.0, 16.5, 17.0, 17.5, 18.0, 18.5, 19.0, 19.5, 20 Millimeters 0.05, 0.06, 0.08, 0.10, 0.12, 0.16, 0.20, 0.25, 0.30, 0.40, 0.50, 0.60, 0.70, 0.80, 0.90, 1.0, 1.1, 1.2, 1.4, 1.5, 1.6, 1.8, 2.0, 2.2, 2.5, 2.8, 3.0, 3.5, 4.0, 4.5, 5.0, 5.5, 6.0, 6.5, 7.0, 8.0, 9.0, 10, 11, 12, 14, 16, 18, 20, 22, 25, 28, 30, 32, 35, 40, 45, 50, 60, 80, 100, 120, 140, 160, 180, 200, 250, 300 Renard Numbers* 1st choice, R5: 1, 1.6, 2.5, 4, 6.3, 10 2d choice, R10: 1.25, 2, 3.15, 5, 8 3d choice, R20: 1.12, 1.4, 1.8, 2.24, 2.8, 3.55, 4.5, 5.6, 7.1, 9 4th choice, R40: 1.06, 1.18, 1.32, 1.5, 1.7, 1.9, 2.12, 2.36, 2.65, 3, 3.35, 3.75, 4.25, 4.75, 5.3, 6, 6.7, 7.5, 8.5, 9.5 *May be multiplied or divided by powers of 10.

1052      Mechanical Engineering Design

Table A–18  Geometric Properties Part 1 Properties of Sections A = area G = location of centroid

∫ I = ∫ x dA = second moment of area about y axis I = ∫ xy dA = mixed moment of area about x and y axes J = ∫ r dA = ∫ (x + y ) dA = I + I Ix = y 2 dA = second moment of area about x axis 2

y

xy

2

G

2

2

x

y

= second polar moment of area about axis through G k 2x = Ix∕A = squared radius of gyration about x axis Rectangle

y b 2 h

h 2

G

x

b

A = bh

Ix =

bh3 12

Iy =

b3h 12

Circle

Ixy = 0

y

D x

G

A=

πD 2 4

I x = Iy =

πD 4 64

Ixy = 0

Hollow circle

JG =

y

d

D

G

A=

π 2 (D − d 2 ) 4

πD 4 32

I x = Iy =

π (D 4 − d 4 ) 64

x

Ixy = 0

JG =

π (D4 − d 4 ) 32

Useful Tables     1053

Table A–18  Geometric Properties  (Continued) y

Right triangles

y

b 3

b h 3

G

h h 3

G

x b 3

b

A=

bh 2

Ix =

bh3 36

Iy =

b3h 36

Ixy =

Right triangles

−b2h2 72 y

y b 3

b h 3 h

h 3

h

bh 2

Ix =

bh3 36

Iy =

G

G

b 3

b3h 36

Ixy =

b2h2 72 y

y

Quarter-circles

x

x

b

A=

x

h

4r 3π

r 4r 3π G

4r 3π

r

A=

πr 2 4

Ix = Iy = r4(

π 4 − 16 9π )

Quarter-circles

x

G

x

4r 3π

1 4 Ixy = r4( − ) 8 9π y

y

4r 3π

r 4r 3π G

x 4r 3π

4r 3π

A=

πr 2 4

Ix = Iy = r4(

π 4 − 16 9π )

G

x

r

Ixy = r4(

4 1 − 9π 8 ) (Continued)

1054      Mechanical Engineering Design

Table A–18  Geometric Properties  (Continued) Part 2 Properties of Solids ( ρ = Mass Density, Mass per Unit Volume) Rods

y

d z

l x

m=

πd 2lρ 4

I y = Iz =

ml 2 12

Round disks

y t d

x

z

m=

πd 2tρ 4

Ix =

md 2 8

I y = Iz =

md 2 16

Rectangular prisms

y

b

c

z

m = abcρ

Ix =

m 2 (a + b2 ) 12

Iy =

Cylinders

a

x

m 2 (a + c2 ) 12

Iz =

m 2 (b + c2 ) 12

y

d z

m=

πd 2lρ 4

Ix =

md 2 8

l

I y = Iz =

x

m (3d 2 + 4l 2 ) 48

Hollow cylinders

y di z

m=

π(d o2 − d 2i ) lρ 4

Ix =

m 2 (d o + d 2i ) 8

l

do x

I y = Iz =

m (3do2 + 3d i2 + 4l 2 ) 48

Useful Tables     1055

Table A–19  American Standard Pipe Nominal Outside Size, Diameter, Threads in in per inch

Wall Thickness, in Extra Double Standard Strong Extra No. 40 No. 80 Strong

1 8

0.405

27

0.070 0.098

1 4

0.540

18

0.090 0.122

3 8

0.675

18

0.093 0.129

1 2

0.840

14

0.111 0.151 0.307

3 4

1.050

14

0.115 0.157 0.318

1 1.315 1112

0.136 0.183 0.369

1 14 1.660

1112

0.143 0.195 0.393

1 12 1.900

1112

0.148 0.204 0.411

2 2.375 1112

0.158 0.223 0.447

2 12

2.875

8

0.208 0.282 0.565

3

3.500

8

0.221 0.306 0.615

3 12

4.000

8

0.231 0.325

4

4.500

8

0.242 0.344 0.690

5

5.563

8

0.263 0.383 0.768

6

6.625

8

0.286 0.441 0.884

8

8.625

8

0.329 0.510 0.895

1056      Mechanical Engineering Design

Table A–20  Deterministic ASTM Minimum Tensile and Yield Strengths for Some Hot-Rolled (HR) and Cold-Drawn (CD) Steels [The strengths listed are estimated ASTM minimum values in the size range 18 to 32 mm ( 34 to 114 in). These strengths are suitable for use with the design factor defined in Sec. 1–10, provided the materials conform to ASTM A6 or A568 requirements or are required in the purchase specifications. Remember that a numbering system is not a specification.] 1 2 3 4 5 6 7 8 Tensile Yield SAE and/or Process- Strength, Strength, Elongation in Reduction in Brinell UNS No. AISI No. ing MPa  (kpsi) MPa  (kpsi) 2 in, % Area, % Hardness G10060

1006

HR

300 (43)

170 (24)

30

55

86

CD

330 (48)

280 (41)

20

45

95

G10100

1010

HR

320 (47)

180 (26)

28

50

95

CD

370 (53)

300 (44)

20

40

105

G10150 1015

HR

340 (50)

190 (27.5)

28

50

101

CD

390 (56)

320 (47)

18

40

111

G10180 1018

HR

400 (58)

220 (32)

25

50

116

CD

440 (64)

370 (54)

15

40

126

G10200 1020

HR

380 (55)

210 (30)

25

50

111

CD

470 (68)

390 (57)

15

40

131

G10300 1030

HR

470 (68)

260 (37.5)

20

42

137

CD

520 (76)

440 (64)

12

35

149

G10350 1035

HR

500 (72)

270 (39.5)

18

40

143

CD

550 (80)

460 (67)

12

35

163

G10400 1040

HR

520 (76)

290 (42)

18

40

149

CD

590 (85)

490 (71)

12

35

170

G10450 1045

HR

570 (82)

310 (45)

16

40

163

CD

630 (91)

530 (77)

12

35

179

G10500 1050

HR

620 (90)

340 (49.5)

15

35

179

CD

690 (100)

580 (84)

10

30

197

G10600

1060

HR

680 (98)

370 (54)

12

30

201

G10800

1080

HR

770 (112)

420 (61.5)

10

25

229

G10950

1095

HR

830 (120)

460 (66)

10

25

248

Source: Data from 1986 SAE Handbook, p. 2.15.

Useful Tables     1057

Table A–21  Mean Mechanical Properties of Some Heat-Treated Steels [These are typical properties for materials normalized and annealed. The properties for quenched and tempered (Q&T) steels are from a single heat. Because of the many variables, the properties listed are global averages. In all cases, data were obtained from specimens of diameter 0.505 in, machined from 1-in rounds, and of gauge length 2 in. Unless noted, all specimens were oil-quenched.] 1 2 3 4 5 6 7 8 Tensile Yield Temperature Strength Strength, Elongation, Reduction Brinell AISI No. Treatment °C  (°F) MPa  (kpsi) MPa  (kpsi) % in Area, % Hardness 1030

Q&T*

205 (400)

848 (123)

648 (94)

17

47

495

Q&T*

315 (600)

800 (116)

621 (90)

19

53

401

Q&T*

425 (800)

731 (106)

579 (84)

23

60

302

Q&T*

540 (1000)

669 (97)

517 (75)

28

65

255

Q&T*

650 (1200)

586 (85)

441 (64)

32

70

207

Normalized

925 (1700)

521 (75)

345 (50)

32

61

149

Annealed

870 (1600)

430 (62)

317 (46)

35

64

137

1040

Q&T

205 (400)

779 (113)

593 (86)

19

48

262

Q&T

425 (800)

758 (110)

552 (80)

21

54

241

Q&T

650 (1200)

634 (92)

434 (63)

29

65

192

Normalized

900 (1650)

590 (86)

374 (54)

28

55

170

Annealed

790 (1450)

519 (75)

353 (51)

30

57

149

Q&T*

205 (400)

1120 (163)

807 (117)

9

27

514

Q&T*

425 (800)

1090 (158)

793 (115)

13

36

444

Q&T*

650 (1200)

717 (104)

538 (78)

28

65

235

Normalized

900 (1650)

748 (108)

427 (62)

20

39

217

Annealed

790 (1450)

636 (92)

365 (53)

24

40

187

1050

1060

Q&T

425 (800)

1080 (156)

765 (111)

14

41

311

Q&T

540 (1000)

965 (140)

669 (97)

17

45

277

Q&T

650 (1200)

800 (116)

524 (76)

23

54

229

Normalized

900 (1650)

776 (112)

421 (61)

18

37

229

Annealed

790 (1450)

626 (91)

372 (54)

22

38

179

Q&T

315 (600)

1260 (183)

813 (118)

10

30

375

Q&T

425 (800)

1210 (176)

772 (112)

12

32

363

Q&T

540 (1000)

1090 (158)

676 (98)

15

37

321

Q&T

650 (1200)

896 (130)

552 (80)

21

47

269

Normalized

900 (1650)

1010 (147)

500 (72)

9

13

293

1095

1141

Annealed

790 (1450)

658 (95)

380 (55)

13

21

192

Q&T

315 (600)

1460 (212)

1280 (186)

9

32

415

Q&T

540 (1000)

896 (130)

765 (111)

18

57

262 (Continued)

1058      Mechanical Engineering Design

Table A–21  Mean Mechanical Properties of Some Heat-Treated Steels  (Continued) [These are typical properties for materials normalized and annealed. The properties for quenched and tempered (Q&T) steels are from a single heat. Because of the many variables, the properties listed are global averages. In all cases, data were obtained from specimens of diameter 0.505 in, machined from 1-in rounds, and of gauge length 2 in. Unless noted, all specimens were oil-quenched.] 1 2 3 4 5 6 7 8 Tensile Yield Temperature Strength Strength, Elongation, Reduction Brinell AISI No. Treatment °C  (°F) MPa  (kpsi) MPa  (kpsi) % in Area, % Hardness 4130

Q&T*

205 (400)

1630 (236)

1460 (212)

10

41

467

Q&T*

315 (600)

1500 (217)

1380 (200)

11

43

435

Q&T*

425 (800)

1280 (186)

1190 (173)

13

49

380

Q&T*

540 (1000)

1030 (150)

910 (132)

17

57

315

Q&T*

650 (1200)

814 (118)

703 (102)

22

64

245

Normalized

870 (1600)

670 (97)

436 (63)

25

59

197

Annealed

865 (1585)

560 (81)

361 (52)

28

56

156

4140

Q&T

205 (400)

1770 (257)

1640 (238)

8

38

510

Q&T

315 (600)

1550 (225)

1430 (208)

9

43

445

Q&T

425 (800)

1250 (181)

1140 (165)

13

49

370

Q&T

540 (1000)

951 (138)

834 (121)

18

58

285

Q&T

650 (1200)

758 (110)

655 (95)

22

63

230

Normalized

870 (1600)

1020 (148)

655 (95)

18

47

302

Annealed

815 (1500)

655 (95)

417 (61)

26

57

197

4340

Q&T

315 (600)

1720 (250)

1590 (230)

10

40

486

Q&T

425 (800)

1470 (213)

1360 (198)

10

44

430

Q&T

540 (1000)

1170 (170)

1080 (156)

13

51

360

Q&T

650 (1200)

965 (140)

855 (124)

19

60

280

*Water-quenched Source: Data from ASM Metals Reference Book, 2d ed., American Society for Metals, Metals Park, Ohio, 1983.

1059

Steel

Steel

Steel

Steel

1045

1045

4142

4340

HR

Q&T 600°F

Q&T 600°F

HR

HR

T4 T6

2024 Aluminum alloy

7075 Aluminum alloy

542 (78.6)

296 (43.0)

169 (24.5)

276 (40.0)

241 (35.0)

910 (132)

1720 (250)

1520 (220)

414 (60.0)

193 (28.0)

358 (52.0)

290 (42.0)

220 (32.0)

593 (86.0)

446 (64.8)

324 (47.0)

568 (82.4)

601 (87.3)

1041 (151)

1930 (280)

1580 (230)

638 (92.5)

424 (61.5)

646 (93.7)

456 (66.2)

341 (49.5)

992 (144)

1448 (210) 1410 (205) 1270 (185) 620 (90) 689 (100) 882 (128)

1600 (233)† 325 (47.2)† 533 (77.3)† 706 (102)†

0.13

0.15

0.28

0.45

0.51

0.09

Derived value.

Source: Data from J. Datsko, "Solid Materials," chap. 32 in Joseph E. Shigley, Charles R. Mischke, and Thomas H. Brown, Jr. (eds.-in-chief), Standard Handbook of Machine Design, 3rd ed., McGraw-Hill, New York, 2004, pp. 32.49–32.52.

0.18

0.18

0.10

1.67

1.16

0.45

0.43

1344 (195)

0.81

0.58

0.85

0.49

0.90

1.05

1760 (255)† 0.048

0.14

0.24

0.14

0.22

0.25

1880 (273) 0.041

1520 (221)†

2340 (340)

2380 (345)

965 (140)

896 (130)†

758 (110)

729 (106)

793 (115)

898 (130)†

620 (90.0)

772 (112)

628 (91.1)†

*Values from one or two heats and believed to be attainable using proper purchase specifications. The fracture strain may vary as much as 100 percent.

T6

2011 Aluminum alloy

Annealed

Steel

1212

Annealed

HR

304 Stainless steel

Steel

1144

Annealed

Steel

1020

Annealed

303 Stainless steel

Steel

1018

True Strain Yield Ultimate Fracture Strengthening Strength Strength Strength Coefficient Strain- True Sy Su σ˜f σ0 Strengthening Fracture Number Material Condition MPa (kpsi) MPa (kpsi) MPa (kpsi) MPa (kpsi) Exponent m Strain ε˜f

Table A–22  Results of Tensile Tests of Some Metals*

1060

L

L

L

L

L

L

L

L

L

L

L

L

L

1005-1009

1015

1020

1040

1045

1045

1045

1045

1045

1045

1144

1144

DAT

CDSR

Q&T

Q&T

Q&T

Q&T

Q&T

Q&T

As forged

HR plate

Normalized

HR sheet

CD sheet

LT CD sheet

1005-1009

Q&T

HR plate

HR plate

1005-1009

L

RQC-100 (c)

L

LT

RQC-100 (c)

Ausformed

510

74

77

930 135

940 136

90

64

725 105

620

440

415 60

930 135

305 1035 150

265

595 2240 325

500 1825 265

450 1585 230

390 1345 195

410 1450 210

225

225

108

80

90 345 50

125 415 60

125 470 68

90 360 52

430 1640 238

290

290

660 2585 375

HR sheet

LT HR sheet

L

H-11

1005-1009

L

Gainex (c)

530

496 1905 276

HR sheet

CD

1315 191

480 2000 290

460 1860 270

405 1515 220

HR, A

STA

STA

STA

10B62

L

L

AM-350 (c)

LT

L

A538C (b)

Gainex (c)

L

A538B (b)

AM-350 (c)

L

A538A (b)

25

33

41

51

55

59

51

65

60

62

68

80

64

66

73

38

67

43

33

64

58

20

52

55

56

67

0.29

0.51

0.52

0.71

0.81

0.89

0.72

1.04

0.93

0.96

1.14

1.6

1.02

1.09

1.3

0.89

1.02

0.56

0.40

1.02

0.86

0.23

0.74

0.81

0.82

1.10

28

29.2

29.2

30

30

29

29.5

200 28.8

195 28.5

205 30

205 30

205 30

205 30

200 29

200 29

200

205

205 30

200 29

200 29

205 30

205 30

195 28

205

205

205 30

200

200

180 26

195

180 26

185 27

185 27

−0.14

−0.071

−0.07

−0.07

−0.07

−0.14

−0.12

1585 230 −0.09

1000 145 −0.08

2725 395 −0.081

2275 330 −0.08

1795 260 −0.07

1585 230 −0.074

1860 270 −0.073

1225 178 −0.095

1540 223

895 130

825 120 −0.11

640 93 −0.109

540 78 −0.073

515 75 −0.059

580 84 −0.09

1780 258 −0.067

1240 180

1240 180

3170 460 −0.077

805 117

805 117

2690 390 −0.102

2800 406

2240 325 −0.07

2135 310 −0.071

1655 240 −0.065

0.27

0.32

0.07

0.25

0.35

0.45

0.60

1.00

0.61

0.41

0.95

0.10

0.11

0.30

0.15

0.32

0.66

0.66

0.08

0.86

0.86

0.10

0.33

0.60

0.80

0.30

−0.53

−0.58

−0.60

−0.68

−0.69

−0.68

−0.70

−0.66

−0.57

−0.51

−0.64

−0.39

−0.41

−0.51

−0.43

−0.56

−0.69

−0.69

−0.74

−0.68

−0.65

−0.42

−0.84

−0.75

−0.71

−0.62

Fatigue Tensile True Strength Fatigue Fatigue Fatigue Strength Hard- Reduction Fracture Modulus of Coefficient Strength Ductility Ductility Sut Elasticity E σ′f Orienta- Description ness in Area Strain Exponent Coefficient Exponent Grade (a) tion (e) (f) HB MPa ksi % ε˜f GPa 106 psi MPa ksi b ε′f c

Table A–23  Mean Monotonic and Cyclic Stress-Strain Properties of Selected Steels

1061

L

L

L

L

4140

4142

4142

4142

Q&T

DAT

DAT

Q&T, DAT

Q&T

895 130

890 129

475 1930 280

L

L

L

L

L

L

L

L

L

LT

L

L

L

L

4142

4340

4340

4340

5160

52100

9262

9262

9262

950C (d)

950C (d)

950X (d)

950X (d)

950X (d)

825 120

925 134

225

156

150

150

159

410

77

64

82

82

695 101

530

440

565

565

565 227

280 1000 145

260

518 2015 292

430 1670 242

350 1240 180

409 1470 213

243

560 2240 325

68

72

65

69

64

32

33

14

11

42

57

38

43

27

35

37

20

42

47

48

28

29

60

55

67

60

49

1.15

1.24

1.06

1.19

1.03

0.38

0.41

0.16

0.12

0.87

0.84

0.48

0.57

0.31

0.43

0.46

0.22

0.54

0.63

0.66

0.34

0.35

0.69

0.79

1.12

0.93

0.68

29

29

29

195

205

205

205

205

28.2

29.5

30

30

29.6

200 29

195 28

205 30

205 30

195 28

195 28

200 29

195 28

205 30

205 30

200

200

205 30

200

205 30

200 28.9

200 29

200 29.2

200 29

220 32

205 29.9

205 29.9

−0.09

−0.09

−0.082

91 1055 153

1005 146

625

970 141

1170 170

−0.08

−0.10

−0.075

−0.11

−0.12

1855 269 −0.057

1220 177 −0.073

1040 151 −0.071

2585 375 −0.09

1930 280 −0.071

1655 240 −0.076

2000 290 −0.091

1200 174 −0.095

2655 385 −0.089

2170 315 −0.081

2105 305

2070 300

2000 290 −0.08

1895 275

1825 265 −0.08

1250 181 −0.08

1450 210 −0.10

1825 265 −0.08

1695 246 −0.081

1275 185 −0.083

1275 185 −0.071

1275 185 −0.076

0.21

0.85

0.35

0.85

0.95

0.38

0.41

0.16

0.18

0.40

0.73

0.48

0.45

0.07

0.09

0.60

0.20

0.40

0.50

0.45

0.06

0.22

1.2

0.89

0.92

0.93

0.68

−0.53

−0.61

−0.54

−0.59

−0.61

−0.65

−0.60

−0.47

−0.56

−0.57

−0.62

−0.60

−0.54

−0.76

−0.61

−0.76

−0.77

−0.73

−0.75

−0.75

−0.62

−0.51

−0.59

−0.69

−0.63

−0.65

−0.65

Source: Data from ASM Metals Reference Book, 2nd ed., American Society for Metals, Metals Park, Ohio, 1983, p. 217.

Notes: (a) AISI/SAE grade, unless otherwise indicated. (b) ASTM designation. (c) Proprietary designation. (d) SAE HSLA grade. (e) Orientation of axis of specimen, relative to rolling direction; L is longitudinal (parallel to rolling direction); LT is long transverse (perpendicular to rolling direction). (f) STA, solution treated and aged; HR, hot rolled; CD, cold drawn; Q&T, quenched and tempered; CDSR, cold drawn strain relieved; DAT, drawn at temperature; A, annealed.

Plate channel

HR plate

Plate channel

HR bar

HR plate

Q&T

Q&T

A

SH, Q&T

Q&T

Q&T

Q&T

HR, A

Q&T

Q&T

L

4142

1930 280

450

2035 295

4142 L Q&T and deformed

Q&T 475

L

4142 L Q&T and deformed

1550 225

380 1415 205

335 1250 181

310 1060 154

310 1075 156

365 1425 207

258

260

450 1760 255

L

4130

Q&T

Q&T forging

950 138

4142

L

4130

290

400

L

1541F

Q&T forging

4142 L Q&T and deformed

L

1541F

1062

26

31

36.5

42.5

52.5

62.5

25

30

35

40

50

60

88.5

73

57

48.5

40

32

26

20.4–23.5 7.8–8.5

18.8–22.8 7.2–8.0

16–20 6.4–7.8

14.5–17.2 5.8–6.9

13–16.4 5.2–6.6

11.5–14.8 4.6–6.0

9.6–14 3.9–5.6

24.5

21.5

18.5

16

14

11.5

10

302

262

235

212

201

174

156

1.50

1.35

1.25

1.15

1.10

1.05

1.00

The modulus of elasticity of cast iron in compression corresponds closely to the upper value in the range given for tension and is a more constant value than that for tension.

187.5

164

140

124

109

97

83

Note: The American Society for Testing and Materials (ASTM) numbering system for gray cast iron is such that the numbers correspond to the minimum tensile strength in kpsi. Thus an ASTM No. 20 cast iron has a minimum tensile strength of 20 kpsi. Note particularly that the tabulations are typical of several heats.

*Polished or machined specimens.

22

20

Fatigue Shear StressModulus of Tensile Compressive Modulus Endurance Brinell Concentration Elasticity, Mpsi ASTM Strength Strength of Rupture Limit* Hardness Factor Number Sut, kpsi Suc, kpsi Ssu, kpsi Tension† Torsion Se, kpsi HB Kf

(a) Typical Properties of Gray Cast Iron

Table A–24  Mechanical Properties of Three Non-Steel Metals

Useful Tables     1063

Table A–24  Mechanical Properties of Three Non-Steel Metals  (Continued) (b) Mechanical Properties of Some Aluminum Alloys [These are typical properties for sizes of about 12 in; similar properties can be obtained by using proper purchase specifications. The values given for fatigue strength correspond to 50(107) cycles of completely reversed stress. Alluminum alloys do not have an endurance limit. Yield strengths were obtained by the 0.2 percent offset method.] Yield Tensile Fatigue Aluminum Strength Strength Strength Elongation Brinell Association Sy Su Sf in 2 in Hardness Number Temper MPa (kpsi) MPa (kpsi) MPa (kpsi) % HB Wrought: 2017

O

70 (10)

179 (26)

90 (13)

22

45

2024

O

76 (11)

186 (27)

90 (13)

22

47

3003 3004 5052

T3

345 (50)

482 (70)

138 (20)

16

120

H12

117 (17)

131 (19)

55 (8)

20

35

H16

165 (24)

179 (26)

65 (9.5)

14

47

H34

186 (27)

234 (34)

103 (15)

12

63

H38

234 (34)

276 (40)

110 (16)  6

77

H32

186 (27)

234 (34)

117 (17)

18

62

H36

234 (34)

269 (39)

124 (18)

10

74

T6

165 (24)

248 (36)

69 (10)

2.0

80

Cast: 319.0* †

333.0

T5

172 (25)

234 (34)

83 (12)

1.0

100

T6

207 (30)

289 (42)

103 (15)

1.5

105

335.0*

T6

172 (25)

241 (35)

62 (9)

3.0

80

T7

248 (36)

262 (38)

62 (9)

0.5

85

*Sand casting. †

Permanent-mold casting.

(c) Mechanical Properties of Some Titanium Alloys Yield Tensile Strength Strength Elongation Hardness Sy Sut in 2 in (Brinell or Titanium Alloy Condition MPa (kpsi) MPa (kpsi) % Rockwell)

Ti-35A†

Annealed

210 (30)

275 (40)

30

135 HB

Ti-50A†

Annealed

310 (45)

380 (55)

25

215 HB

Ti-0.2 Pd

Annealed

280  (40)

340  (50)

28

200 HB

Ti-5 Al-2.5 Sn

Annealed

760  (110)

790  (115)

16

36 HRC

Ti-8 Al-1 Mo-1 V

Annealed

900  (130)

965  (140)

15

39 HRC

Ti-6 Al-6 V-2 Sn

Annealed

970  (140)

1030  (150)

14

38 HRC

Ti-6Al-4V

Annealed

830 (120)

900 (130)

14

36 HRC

Ti-13 V-11 Cr-3 Al

Sol. + aging

1207  (175)

1276  (185)

8

40 HRC

Commercially pure alpha titanium.

1064

HR

CD

CD

CD

CD

HT bolts

1035

1045

1117

1137

12L14

1038

78.4 5.90 56

80.6 4.29 0.0655 0.0753

CSy

3.38

122.3

134.6

3.64

Malleable 53.4 2.68 44.7 54.3 3.61 34.9 1.47 30.1 35.5 3.67 0.0502 0.0421

175.4 7.91 141.8 178.5 4.85 163.7 9.03 101.5 167.4 8.18 0.0451 0.0552 28.1 1.73 24.2 28.7 2.43

0

T4

T6

T6 .025″ 75.5 2.10 68.8 76.2 3.53 63.7 1.98 58.9 64.3 2.63 0.0278 0.0311

2024

2024

7075

Source: Data compiled from "Some Property Data and Corresponding Weibull Parameters for Stochastic Mechanical Design," Trans. ASME Journal of Mechanical Design, vol. 114 (March 1992), pp. 29–34.

67.5 1.50 55.9 68.1 9.26 53.4 1.17 51.2 53.6 1.91 0.0222 0.0219

64.9 1.64 60.2 65.5 3.16 40.8 1.83 38.4 41.0 1.32 0.0253 0.0449

0.0616

149.1 8.29 101.8 152.4 6.68 63.0 5.05 38.0 65.0 5.73 0.0556 0.0802

0.0499

Ti-6AL-4V

84.8 4.23 71.6 86.3 3.45

85.0 4.14 66.6 86.6 5.11 37.9 3.76 30.2 38.9 2.17 0.0487 0.0992

AM350SS A

A

310SS

198.8 9.51 163.3 202.3 4.21 189.4 11.49 144.0 193.8 4.48 0.0478 0.0607

A

304SS

105.0 5.68 92.3 106.6 2.38 46.8 4.70 26.3 48.7 4.99 0.0541 0.1004

191.2 5.82 151.9 193.6 8.00 166.8 9.37 139.7 170.0 3.17 0.0304 0.0562

105.3 3.09 95.7 106.4 3.44 78.5 3.91 64.8 79.9 3.93 0.0293 0.0498

A

0.0396

17-7PSS

CD

301SS

195.9 7.76 180.7 197.9 2.06

403SS

CD

201SS

100-70-04 Nodular 122.2 7.65 47.6 125.6 11.84 79.3 4.51 64.1 81.0 3.77 0.0626 0.0569

604515 Nodular 64.8 3.77 53.7 66.1 3.23 49.0 4.20 33.8 50.5 4.06 0.0582 0.0857

Pearlitic Malleable 93.9 3.83 80.1 95.3 4.04 60.2 2.78 50.2 61.2 4.02 0.0408 0.0462

Malleable 53.3 1.59 48.7 53.8 3.18 38.5 1.42 34.7 39.0 2.93 0.0298 0.0369

0.0975

0.0253

32510

44.5 4.34 27.7 46.2 4.38

133.4

79.6 6.92 70.3 80.4 1.36 78.1 8.27 64.3 78.8 1.72 0.0869 0.1059

106.5 6.15 96.2 107.7 1.72 98.1 4.24 92.2 98.7 1.41 0.0577 0.0432

83.1 5.25 73.0 84.4 2.01 81.4 4.71 72.4 82.6 2.00 0.0632 0.0579

117.7 7.13 90.2 120.5 4.38 95.5 6.59 82.1 97.2 2.14 0.0606 0.0690

86.2 3.92 72.6 87.5 3.86 49.6 3.81 39.5 50.8 2.88 0.0455 0.0768

87.6 5.74 30.8 90.1 12

σˆSut x0 θ b µSy σˆSy x0 θ b CSut

35018

ASTM40

CD

1018

Material µSut

Table A–25  Stochastic Yield and Ultimate Strengths for Selected Materials

1065

799 (116)

661 (96)

W  x0

599 (87)

W  x0

HT-46

1040 (151)

992 (144)

N   σ

712 (108)

39.6 (5.75)

143 (20.7)

26.3 (3.82)

684 (99.3)

38.1 (5.53)

116 (16.9)

657 (95.4)

36.6 (5.31)

95 (13.8)

17.4 (2.53)

5.5

463 (67.2)

393 (57)

4.1

496 (72.0)

420 (61)

2.85

425 (61.7)

391 (56.7)

493 (71.6)

35.1 (5.10)

77 (11.2)

14.0 (2.03)

Source: Data from E. B. Haugen, Probabilistic Mechanical Design, Wiley, New York, 1980, Appendix 10–B.

Note: Statistical parameters from a large number of fatigue tests are listed. Weibull distribution is denoted W and the parameters are x0, "guaranteed" fatigue strength; θ, characteristic fatigue strength; and b, shape factor. Normal distribution is denoted N and the ­parameters are µ, mean fatigue strength; and σ, standard deviation of the fatigue strength. The life is in stress-cycles-to-failure. TS = tensile strength, YS = yield strength. All testing by rotating-beam specimen.

μ

Ti-6A1-4V

μ

N   σ

Aluminum

365 (53)

21.4 (3.11)

489 (71)

5.0

T-4

2024

528 (76.7)

b 5.2

604 (87.7)

510 (74)

455 (66)

744 (108)

θ

OQ&T, 1300°F

3140

588 (85.4) 3.4

699 (101.5)

579 (84)

b 4.3

θ

OQ&T 1200°F

2.75

462 (67)

510 (74)

544 (79)

2340

W  x0

b 2.60

565 (82)

503 (73.0)

723 (105) 594 (86.2)

WQ&T, 1210°F

θ

1046

3

4 5 6 7 8 9 Stress Cycles to Failure Tensile Strength Yield Strength Distri- MPa (kpsi) MPa (kpsi) bution 104 105 106 107

2

Number Condition

1

Table A–26  Stochastic Parameters for Finite Life Fatigue Tests in Selected Metals

1066

180

117

115

84

134

111

98

0.20

65

60

55

50

48

48

48

130 0.15 102 95 91 91 91 91

59 0.12 60 56 51 50 50 50

65 0.40 77 68 64 57 56 56 56 56

33 0.37 50 43 40 34 31 30 30 30

65

84 0.57 94 81 73 62 57 55 55 55

47 0.42 61 55 51 47 43 41 41 41

70 0.58 60 57 52 50 50 50 50

47 0.40 50 48 46 40 38 34 34

63 0.49 80 70 56 47 47 47 47

53 0.23 40 47 33 33

87 0.65 80 72 65 60 57 57 57

35 0.54 44 40 37 34 33 33

45 0.62 51 47 42 38 38 38

30 0.63 37 34 30 28 25

Source: Compiled from Table 4 in H. J. Grover, S. A. Gordon, and L. R. Jackson, Fatigue of Metals and Structures, Bureau of Naval Weapons Document NAVWEPS 00-25-534, 1960.

*BHN = Brinell hardness number; RA = fractional reduction in area.

860

369

OQT

224

10120

1200

227

OQT

67 Rb 162

As Rec.

1095

1060

277

WQT

1200

193

.56 MN N

196

1200

97

92

1050 N, AC 164

WQT

1045 HR, N

92 107

1040 Forged 195

72 103

209

WQT

80

1030 Air-cooled 135

1035 Normal 132

58

1020 Furnace cooled

Tensile Yield Stress Cycles to Failure Strength Strength 4 4 5 Material Condition BHN* kpsi kpsi RA* 10 4(10 ) 10 4(105) 106 4(106) 107 108

Table A–27  Finite Life Fatigue Strengths of Selected Plain Carbon Steels

1067

0.090 0.080 0.071 0.064 0.057

0.050 82 0.045 26

11 12 13 14 15

16 17

74 81 96 08 07

0 3 5 4 9

0.162 0.144 0.128 0.114 0.101

6 7 8 9 10

9 3 6 4 3 9

0.324 0.289 0.257 0.229 0.204 0.181

0 1 2 3 4 5

0.065 0.058

0.120 0.109 0.095 0.083 0.072

0.203 0.180 0.165 0.148 0.134

0.340 0.300 0.284 0.259 0.238 0.220

7∕0 6∕0 0.580 0 5∕0 0.516 5 4∕0 0.460 0 0.454 3∕0 0.409 6 0.425 2∕0 0.364 8 0.380

125 5 875 25 625

0.062 5 0.056 25

6 6 7 7 3

3 3 4 5 5

0.059 8 0.053 8

0.119 0.104 0.089 0.074 0.067

0.194 0.179 0.164 0.149 0.134

5 25 625 0.239 1 375 0.224 2 75 0.209 2

0.125 0.109 357 0.093 75 0.078 125 0.070 312 5

0.203 0.187 0.171 0.156 0.140

0.312 0.281 0.265 0.25 0.234 0.218

0.500 0.468 75 0.437 5 0.406 25 0.375 0.343 75

5 5 5 0 0

0 0 0 3 0

5 0 5 7 3 0

5 5 8 5 0

0.062 5 0.054 0

0.120 0.105 0.091 0.080 0.072

0.192 0.177 0.162 0.148 0.135

0.306 0.283 0.262 0.243 0.225 0.207

0.490 0.461 0.430 0.393 0.362 0.331

0.037 0.039

0.026 0.029 0.031 0.033 0.035

0.016 0.018 0.020 0.022 0.024

0.009 0.010 0.011 0.012 0.013 0.014

0.004 0.005 0.006 0.007 0.008

Tubing, Ferrous Ferrous Ferrous Nonferrous Strip, Flat Sheet and Wire Principal Sheet, Wire, Wire, and Plate, Ferrous Except Music Use: and Rod Spring Steel 480 lbf/ft3 Sheet Music Wire Wire

0.175 0.172

0.188 0.185 0.182 0.180 0.178

0.201 0.199 0.197 0.194 0.191

0.227 0.219 0.212 0.207 0.204

Steel Drill Rod

Steel Wire Name American Birmingham United Manu- or Stubs of or Brown or Stubs States facturers Washburn Music Steel Gauge: & Sharpe Iron Wire Standard† Standard & Moen Wire Wire

Table A–28  Decimal Equivalents of Wire and Sheet-Metal Gauges* (All Sizes Are Given in Inches)

0 0 0 0 0

0 0 0 0 5

0 0 0 0 5

(Continued)

0.177 0 0.173 0

0.191 0.189 0.185 0.182 0.180

0.204 0.201 0.199 0.196 0.193

0.228 0.221 0.213 0.209 0.205

Twist Drills and Drill Steel

Twist Drill

1068

0.028 0.025 0.022 0.020 0.017

0.015 0.014 0.012 0.011 0.010

0.008 0.007 0.007 0.006 0.005

0.005 0.004 0.003 0.003 0.003

21 22 23 24 25

26 27 28 29 30

31 32 33 34 35

36 37 38 39 40

0.010 0.009 0.008 0.007 0.005

0.018 0.016 0.014 0.013 0.012

0.032 0.028 0.025 0.022 0.020

0.049 0.042 0.035

0.010 0.010 0.009 0.008 0.007

0.018 0.017 0.015 0.014 0.012 937 5 156 25 375 593 75 812 5

75 187 5 625 062 5 5

0.034 375 0.031 25 0.028 125 0.025 0.021 875

0.05 0.043 75 0.037 5

0.010 0.009 0.009 0.008 0.007

0.017 0.016 0.014 0.013 0.012

0.032 0.029 0.026 0.023 0.020

5 7 0 2 5

9 4 9 5 0

9 9 9 9 9

0.047 8 0.041 8 0.035 9

0.009 0.008 0.008 0.007 0.007

0.013 0.012 0.011 0.010 0.009

0.018 0.017 0.016 0.015 0.014

0.031 0.028 0.025 0.023 0.020

Reflects present average and weights of sheet steel.

0.063 0.067 0.071 0.075 0.080

0.047 0.049 0.051 0.055 0.059

0.041 0.043 0.045

0 5 0 5 0

2 0.085 8 0.090 8 0.095 4 5

1 3 2 0 0

7 6 8 0 4

0.047 5 0.041 0 0.034 8

*Specify sheet, wire, and plate by stating the gauge number, the gauge name, and the decimal equivalent in parentheses.

000 0.004 0.007 031 25 0.006 7 453 0.006 640 625 0.006 4 965 0.006 25 0.006 0 531 145

928 950 080 305 615

94 20 64 26 03

46 35 57 10 90

0.040 30 0.035 89 0.031 96

18 19 20

Tubing, Ferrous Ferrous Ferrous Nonferrous Strip, Flat Sheet and Wire Principal Sheet, Wire, Wire, and Plate, Ferrous Except Music Use: and Rod Spring Steel 480 lbf/ft3 Sheet Music Wire Wire

0.106 0.103 0.101 0.099 0.097

0.120 0.115 0.112 0.110 0.108

0.146 0.143 0.139 0.134 0.127

0.157 0.155 0.153 0.151 0.148

0.168 0.164 0.161

Steel Drill Rod

Steel Wire Name American Birmingham United Manu- or Stubs of or Brown or Stubs States facturers Washburn Music Steel Gauge: & Sharpe Iron Wire Standard† Standard & Moen Wire Wire

Table A–28  Decimal Equivalents of Wire and Sheet-Metal Gauges* (All Sizes Are Given in Inches)  (Continued)

0.106 0.104 0.101 0.099 0.098

0.120 0.116 0.113 0.111 0.110

0.147 0.144 0.140 0.136 0.128

0.159 0.157 0.154 0.152 0.149

5 0 5 5 0

0 0 0 0 0

0 0 5 0 5

0 0 0 0 5

0.169 5 0.166 0 0.161 0

Twist Drills and Drill Steel

Twist Drill

Useful Tables     1069

Table A–29  Dimensions of Square and Hexagonal Bolts H

W R

Head Type

Nominal Size, in

Square

Regular Hexagonal

W H W H Rmin

Heavy Hexagonal W H

Structural Hexagonal

Rmin

W

H

Rmin

14

3 8

11 64

7 16

11 64 0.01

165

1 2

13 64

1 2

7 32 0.01

38

9 16

1 4

9 16

1 4 0.01

167

5 8

19 64

5 8

19 64 0.01

12

3 4

21 64

3 4

11 7 32 0.01 8

11 32 0.01

7 8

5 16 0.009

58

15 16

27 64

15 16

27 1 64 0.02 116

27 64 0.02

1161

25 64 0.021

34

118

1 2

118

1 1 2 0.02 14

1 1 2 0.02 14

15 32 0.021

1

112

21 32

112

43 5 64 0.03 18

43 5 64 0.03 18

39 64 0.062

118

111 16

3 4

111 16

3 13 4 0.03 116

3 13 4 0.03 116

11 16 0.062

114

178

27 32

178

27 27 32 0.03 2 32 0.03

2 25 32 0.062

138

2161

29 32

2161

29 3 32 0.03 216

3163

112

214 1 214 1 0.03 238 1 0.03 238

29 32 0.03

27 32 0.062 15 16 0.062

Nominal Size, mm M5

8 3.58 8 3.58 0.2

M6

10 4.38 0.3

M8

13 5.68 0.4

M10

16 6.85 0.4

M12

18 7.95 0.6 21 7.95 0.6

M14

21 9.25 0.6 24 9.25 0.6

M16

24 10.75 0.6

27 10.75 0.6

27 10.75 0.6

M20

30 13.40 0.8

34 13.40 0.8

34 13.40 0.8

M24

36 15.90 0.8

41 15.90 0.8

41 15.90 1.0

M30

46 19.75 1.0

50 19.75 1.0

50 19.75 1.2

M36

55 23.55 1.0

60 23.55 1.0

60 23.55 1.5

1070      Mechanical Engineering Design

Table A–30  Dimensions of Hexagonal Cap Screws and Heavy Hexagonal Screws (W = Width across Flats; H = Height of Head; See Figure in Table A–29) Type of Screw Minimum Nominal Fillet Cap Heavy Height Size, in Radius W W H 7 14 0.015 16

5 32

5 1 16 0.015 2

13 64

9 3 8 0.015 16

15 64

5 7 16 0.015 8

9 32

1 3 7 2 0.015 4   8

5 16

5 15 8 0.020 16

1161

25 64

3 118 4 0.020

114

15 32

7 1165 8 0.040

1167

35 64

118

39 64

1

0.060

112

114 0.060 178

2

25 32

138 0.060 2161

2163

27 32

112 0.060 214

238

15 16

Nominal Size, mm M5

0.2  8

3.65

M6

0.3 10

4.15

M8

0.4 13

5.50

M10 0.4 16

6.63

M12 0.6 18 21 7.76 M14 0.6 21 24 9.09 M16 0.6 24 27 10.32 M20 0.8 30 34 12.88 M24 0.8 36 41 15.44 M30 1.0 46 50 19.48 M36 1.0 55 60 23.38

Useful Tables     1071

Table A–31  Dimensions of Hexagonal Nuts

Height H

Nominal Size, in

Width W

Regular Thick or Hexagonal Slotted JAM

1 4

7 16

7 32

9 32

5 32

5 16

1 2

17 64

21 64

3 16

3 8

9 16

21 64

13 32

7 32

7 16

11 16

3 8

29 64

1 4

1 2

3 4

7 16

9 16

5 16

9 16

7 8

31 64

39 64

5 16

5 8

15 16

35 64

23 32

3 8

3 4

118

41 64

13 16

27 64

7 8

1165

3 4

29 32

31 64

112

55 35 64 1 64 31 32

1325

39 64

1

118

111 16

114

178

1161

114

23 32

138

2161

111 64

138

25 32

112

214

1329

112

27 32

Nominal Size, mm M5

8

4.7

5.1 2.7

M6

10

5.2

5.7 3.2

M8

13

6.8

7.5 4.0

M10 16

8.4

9.3 5.0

M12 18

10.8

12.0 6.0

M14 21

12.8

14.1 7.0

M16 24

14.8

16.4 8.0

M20

30

18.0

20.3 10.0

M24

36

21.5

23.9 12.0

M30

46

25.6

28.6 15.0

M36

55

31.0

34.7 18.0

1072      Mechanical Engineering Design

Table A–32  Basic Dimensions of American Standard Plain Washers (All Dimensions in Inches)

Fastener Washer Diameter Size Size ID OD Thickness # 6

0.138

0.156

0.375

0.049

# 8

0.164

0.188

0.438

0.049

#10

0.190

0.219

0.500

0.049

#12

0.216

0.250

0.562

0.065

1 4 N 1 4 W 5 16 N 5 16 W 3 8 N 3 8 W 7 16 N 7 16 W 1 2 N 1 2 W 9 16 N 9 16 W 5 8 N 5 8 W 3 4 N 3 4 W 7 8 N 7 8 W

0.250 0.281 0.625

0.065

0.250 0.312 0.734

0.065

0.312 0.344 0.688

0.065

0.312 0.375 0.875

0.083

0.375 0.406 0.812

0.065

0.375 0.438 1.000

0.083

0.438 0.469 0.922

0.065

0.438 0.500 1.250

0.083

0.500 0.531 1.062

0.095

0.500 0.562 1.375

0.109

0.562 0.594 1.156

0.095

0.562 0.625 1.469

0.109

0.625 0.656 1.312

0.095

0.625 0.688 1.750

0.134

0.750 0.812 1.469

0.134

0.750 0.812 2.000

0.148

0.875 0.938 1.750

0.134

0.875 0.938 2.250

0.165

1 N

1.000

1.062

2.000

0.134

1 W

1.000

1.062

2.500

0.165

118 N 118 W 114 N 114 W 138 N 138 W 112 N 112 W 158 134 178

1.125 1.250 2.250

0.134

1.125 1.250 2.750

0.165

1.250 1.375 2.500

0.165

1.250 1.375 3.000

0.165

1.375 1.500 2.750

0.165

1.375 1.500 3.250

0.180

1.500 1.625 3.000

0.165

1.500 1.625 3.500

0.180

1.625 1.750 3.750

0.180

1.750 1.875 4.000

0.180

1.875 2.000 4.250

0.180

2

2.000

0.180

214

2.250 2.375 4.750

0.220

212 234

2.500 2.625 5.000

0.238

2.750 2.875 5.250

0.259

3.000

0.284

3

2.125

3.125

N = narrow; W = wide; use W when not specified.

4.500

5.500

Useful Tables     1073

Table A–33  Dimensions of Metric Plain Washers (All Dimensions in Millimeters) Washer Minimum Maximum Maximum Washer Minimum Maximum Maximum Size* ID OD Thickness Size* ID OD Thickness 1.6 N

1.95

4.00

0.70

10 N

10.85

20.00

2.30

1.6 R

1.95

5.00

0.70

10 R

10.85

28.00

2.80

1.6 W

1.95

6.00

0.90

10 W

10.85

39.00

3.50

2 N

2.50

5.00

0.90

12 N

13.30

25.40

2.80

2 R

2.50

6.00

0.90

12 R

13.30

34.00

3.50

2 W

2.50

8.00

0.90

12 W

13.30

44.00

3.50

2.5 N

3.00

6.00

0.90

14 N

15.25

28.00

2.80

2.5 R

3.00

8.00

0.90

14 R

15.25

39.00

3.50

2.5 W

3.00

10.00

1.20

14 W

15.25

50.00

4.00

3 N

3.50

7.00

0.90

16 N

17.25

32.00

3.50

3 R

3.50

10.00

1.20

16 R

17.25

44.00

4.00

3 W

3.50

12.00

1.40

16 W

17.25

56.00

4.60

3.5 N

4.00

9.00

1.20

20 N

21.80

39.00

4.00

3.5 R

4.00

10.00

1.40

20 R

21.80

50.00

4.60

3.5 W

4.00

15.00

1.75

20 W

21.80

66.00

5.10

4 N

4.70

10.00

1.20

24 N

25.60

44.00

4.60

4 R

4.70

12.00

1.40

24 R

25.60

56.00

5.10

4 W

4.70

16.00

2.30

24 W

25.60

72.00

5.60

5 N

5.50

11.00

1.40

30 N

32.40

56.00

5.10

5 R

5.50

15.00

1.75

30 R

32.40

72.00

5.60

5 W

5.50

20.00

2.30

30 W

32.40

90.00

6.40

6 N

6.65

13.00

1.75

36 N

38.30

66.00

5.60

6 R

6.65

18.80

1.75

36 R

38.30

90.00

6.40

6 W

6.65

25.40

2.30

36 W

38.30

110.00

8.50

8 N

8.90

18.80

2.30

8 R

8.90

25.40

2.30

8 W

8.90

32.00

2.80

N = narrow; R = regular; W = wide. *Same as screw or bolt size.

1074      Mechanical Engineering Design

Table A–34  Gamma Function* Values of Γ(n) =

e−xx n−1 dx

0

n

Γ(n)

n

Γ(n)

n

Γ(n)

n

Γ(n)

1.00

1.000 00

1.25

.906 40

1.50

.886 23

1.75

.919 06

1.01

.994 33

1.26

.904 40

1.51

.886 59

1.76

.921 37

1.02

.988 84

1.27

.902 50

1.52

.887 04

1.77

.923 76

1.03

.983 55

1.28

.900 72

1.53

.887 57

1.78

.926 23

1.04

.978 44

1.29

.899 04

1.54

.888 18

1.79

.928 77

1.05

.973 50

1.30

.897 47

1.55

.888 87

1.80

.931 38

1.06

.968 74

1.31

.896 00

1.56

.889 64

1.81

.934 08

1.07

.964 15

1.32

.894 64

1.57

.890 49

1.82

.936 85

1.08

.959 73

1.33

.893 38

1.58

.891 42

1.83

.939 69

1.09

.955 46

1.34

.892 22

1.59

.892 43

1.84

.942 61

1.10

.951 35

1.35

.891 15

1.60

.893 52

1.85

.945 61

1.11

.947 39

1.36

.890 18

1.61

.894 68

1.86

.948 69

1.12

.943 59

1.37

.889 31

1.62

.895 92

1.87

.951 84

1.13

.939 93

1.38

.888 54

1.63

.897 24

1.88

.955 07

1.14

.936 42

1.39

.887 85

1.64

.898 64

1.89

.958 38

1.15

.933 04

1.40

.887 26

1.65

.900 12

1.90

.961 77

1.16

.929 80

1.41

.886 76

1.66

.901 67

1.91

.965 23

1.17

.936 70

1.42

.886 36

1.67

.903 30

1.92

.968 78

1.18

.923 73

1.43

.886 04

1.68

.905 00

1.93

.972 40

1.19

.920 88

1.44

.885 80

1.69

.906 78

1.94

.976 10

1.20

.918 17

1.45

.885 65

1.70

.908 64

1.95

.979 88

1.21

.915 58

1.46

.885 60

1.71

.910 57

1.96

.983 74

1.22

.913 11

1.47

.885 63

1.72

.912 58

1.97

.987 68

1.23

.910 75

1.48

.885 75

1.73

.914 66

1.98

.991 71

1.24

.908 52

1.49

.885 95

1.74

.916 83

1.99

.995 81

2.00 1.000 00

*For n > 2, use the recursive formula Γ(n) = (n − 1) Γ(n − 1) For example, Γ(5.42) = 4.42(3.42) (2.42) Γ(1.42) = 4.42(3.42) (2.42) (0.886 36) = 32.4245 †

For large positive values of x, Γ(x) can be expressed by the asymptotic series based on Stirling's approximation 2π 1 1 139 571 Γ(x) ≈ xxe−x√ (1 + + − − x 12 x 288 x 2 51 840 x 3 2 488 320 x 4 ) Source: Data from William H. Beyer (ed.), Handbook of Tables for Probability and Statistics, 2nd ed., 1966. Copyright CRC Press, Boca Raton, Florida.

Answers to Selected Problems B–1 Chapter 1 1–8  P = 100 units 1–11  (a) e1 = 0.005 751 311 1, e2 = 0.008 427 124 7, e = 0.014 178 435 8, (b) e1 = −0.004 248 688 9, e2 = −0.001 572 875 3, e = −0.005 821 564 2 1–13  (a) x = 122.9 kcycles, sˆx = 30.3 kcycles, (b) 27 1–14  x = 198.61 kpsi, sˆx = 9.68 kpsi 1–15  L10 = 84.1 kcycles 1–16  x0.01 = 88.3 kpsi 1–18  n = 1.32, d = 31.9 mm 1–20  n = 1.17, R = 94.9 percent 1–21  (a) w = 0.020 ± 0.018 in, (b) d = 6.528 in 1–23  a = 1.569 ± 0.016 in 1–24  Do = 4.012 ± 0.036 in 1–31  (a) σ = 1.90 kpsi, (b) σ = 397 psi, (c) y = 0.609 in, (d) θ = 4.95°

B–2 Chapter 2 2–6  (b) E = 30.5 Mpsi, Sy = 45.6 kpsi, Sut = 85.6 kpsi, area reduction = 45.8 percent 2–9  (a) Before: Sy = 32 kpsi, Su = 49.5 kpsi, After: S′y = 61.8 kpsi, 93% increase, S′u = 61.9 kpsi, 25% increase, (b) Before: Su∕Sy = 1.55, After: S′u∕S′y = 1.002 2–15  Su = 177 kpsi, sSu = 1.27 kpsi 2–17  (a) uR ≈ 34.7 in • lbf/in3, (b) uT ≈ 66.7 (103) in • lbf/in3 2–25  Aluminum alloys have greatest potential followed closely by high carbon heat-treated steel. Warrants further discussion. 2–33  Steel, titanium, aluminum alloys, and composites

B–3 Chapter 3 3–1  RB = 33.3 lbf, RO = 66.7 lbf, RC = 33.3 lbf 3–6  RO = 740 lbf, MO = 8080 lbf • in

Appendix

B

3–14  (a) Mmax = 253 lbf • in, (b) amin = 2.07 in, Mmin = 214 lbf • in 3–15  (a) σ1 = 22 kpsi, σ2 = −12 kpsi, σ3 = 0 kpsi, ϕp = 14.0° cw, τ1 = 17 kpsi, σave = 5 kpsi, ϕs = 31.0° ccw, (b) σ1 = 18.6 kpsi, σ2 = 6.4 kpsi, σ3 = 0 kpsi, ϕp = 27.5° ccw, τ1 = 6.10 kpsi, σave = 12.5 kpsi, ϕs = 17.5° cw, (c) σ1 = 26.2 kpsi, σ2 = 7.78 kpsi, σ3 = 0 kpsi, ϕp = 69.7° ccw, τ1 = 9.22 kpsi, σave = 17 kpsi, ϕs = 24.7° ccw, (d) σ1 = 25.8 kpsi, σ2 = −15.8 kpsi, σ3 = 0 kpsi, ϕp = 72.4° cw, τ1 = 20.8 kpsi, σave = 5 kpsi, ϕs = 27.4° ccw 3–20  σ1 = 24.0 kpsi, σ2 = 0.819 kpsi, σ3 = −24.8 kpsi, τmax = 24.4 kpsi 3–23  σ = 34.0 kpsi, δ = 0.0679 in, ε1 = 1.13(10−3), ε2 = −3.30(10−4), Δd = −2.48(10−4) in 3–27  δ = 5.86 mm 3–29  σx = 382 MPa, σy = −37.4 MPa 3–36  σmax = 84.3 MPa, τmax = 5.63 MPa 3–41  model c: σ = 17.8 kpsi, τ = 3.4 kpsi, model d: σ = 25.5 kpsi, τ = 3.4 kpsi, model e: σ = 17.8 kpsi, τ = 3.4 kpsi 3–47  (b) (τmax)A = 75.1 MPa, (τmax)B = 53.1 MPa, (τmax)C = 116.8 MPa 3–53  (a) r = 6 mm, 126.25° CCW from the y axis, (b) σx = 112.3 MPa, τxz = 35.4 MPa, (d) σ1 = 122.6 MPa, σ2 = 0 MPa, σ3 = −10.2 MPa, τmax = 66.4 MPa 3–56  (a) τmax = 5.85 kpsi, (b) θ = 0.562°, (c) ny = 3.59 3–60  (a) T = 204 N • m, (b) T = 52.5 N • m 3–62  (a) T = 1318 lbf • in, θ = 4.59°, (b) T = 1287 lbf • in, θ = 4.37° 3–64  (a) T1 = 1.47 N • m, T2 = 7.45 N • m, T3 = 0 N • m, T = 8.92 N • m, (b) θ1 = 0.348 rad/m 3–70  H = 55.5 kW

1075

1076      Mechanical Engineering Design

3–77  dc = 1.4 in 3–80  (a) T1 = 2880 N, T2 = 432 N, (b) RC = 1794 N, RO = 3036 N, (d) σ = 263 MPa, τ = 57.7 MPa, (e) σ1 = 275 MPa, σ2 = −12.1 MPa, τmax = 144 MPa 3–83  (a) FB = 750 lbf, (b) RCy = 183.1 lbf, RCz = 861.5 lbf, ROy = 208.5 lbf, ROz = 259.3 lbf, (d) σ = 35.2 kpsi, τ = 7.35 kpsi, (e) σ1 = 36.7 kpsi, σ2 = −1.47 kpsi, τmax = 19.1 kpsi 3–91  (a) Critical at the wall at top or bottom of rod. (b) σx = 16.3 kpsi, τxz = 5.09 kpsi, (c) σ1 = 17.8 kpsi, σ2 = −1.46 kpsi, τmax = 9.61 kpsi 3–95  (a) Critical at the top or bottom. (b) σx = 28.0 kpsi, τxz = 15.3 kpsi, (c) σ1 = 34.7 kpsi, σ2 = −6.7 kpsi, τmax = 20.7 kpsi 3–108  xmin = 8.3 mm 3–110  xmax = 1.9 kpsi 3–113  po = 82.8 MPa 3–117  σl = −254 psi, σt = 5710 psi, σr = −23.8 psi, τ1/3 = 2980 psi, τ1∕2 = 2870 psi, τ2∕3 = 115 psi 3–120  τmax = 630 psi, (σr)max = 490 psi 3–124  δmax = 0.021 mm, δmin = 0.0005 mm, pmax = 65.2 MPa, pmin = 1.55 MPa 3–130  δ = 0.001 in, p = 8.33 kpsi, (σt)i = −8.33 kpsi, (σt)o = 21.7 kpsi 3–134  σi = 300 MPa, σo = −195 MPa 3–140  (a) σ = ±8.02 kpsi, (b) σi = −10.1 kpsi, σo = 6.62 kpsi, (c) Ki = 1.26, Ko = 0.825 3–143  σi = 64.6 MPa, σo = −21.7 MPa 3–148  σmax = 352F1∕3 MPa, τmax = 106F1∕3 MPa 3–153  F = 117.4 lbf 3–156  σx = −35.0 MPa, σy = −22.9 MPa, σz = −96.9 MPa, τmax = 37.0 MPa

B–4 Chapter 4 πd 4G 1 1 + , ( x l − x) 32 l−x x   T1 = 1500 , T2 = 1500 , l l 3   (b) k = 28.2 (10 ) lbf • in/rad, T1 = T2 = 750 lbf • in, τmax = 30.6 kpsi 4–7  δ = 5.262 in, % elongation due to weight = 3.21% 4–10  ymax = −25.4 mm, σmax = −163 MPa 4–3  (a) k =

4–13  yO = yC = −3.72 mm, y∣x=550mm = 1.11 mm 4–16  dmin = 32.3 mm 4–24  yA = −7.99 mm, θA = −0.0304 rad 4–27  yA = 0.0805 in, zA = −0.1169 in, (θA)y = 0.00115 rad, (θA)z = 8.06(10−5) rad 4–30  (θO)z = 0.0131 rad, (θC)z = −0.0191 rad 4–33  (θO)y = 0.0104 rad, (θO)z = 0.00751 rad, (θC)y = −0.0193 rad, (θC)z = −0.0109 rad 4–36  d = 62.0 mm 4–39  d = 2.68 in 4–41  y = 0.1041 in 4–43  Stepped bar: θ = 0.026 rad, simplified bar: θ = 0.0345 rad, 1.33 times greater, 0.847 in 4–46  d = 38.1 mm, ymax = −0.0678 mm 4–51  yB = −0.0155 in 4–54  yA = −5.56 (103) in 4–58  k = 8.10 N/mm 4–76  δ = 0.0102 in 4–80  Stepped bar: δ = 0.706 in, uniform bar: δ = 0.848 in, 1.20 times greater 4–88  (a) (yD)AB = 0.0121 in, (b) (yD)BC = 0 in, (c) (yD)BD = 3.494(10−3) in, (d) yD = 0.0156 in 4–92  δ = 0.0338 mm 4–94  δ = 0.0226 in 4–97  δ = 0.689 in 4–103  δ = 0.1217 in 4–105  δ = 6.067 mm 4–110  (a) σb = 48.8 kpsi, σc = −13.9 kpsi, (b) σb = 50.5 kpsi, σc = −12.0 kpsi 4–112  RB = 1.6 kN, RO = 2.4 kN, δA = 0.0223 mm 4–117  RC = 1.33 kips, RO = 4.67 kips, δA = 0.00622 in, σAB = −14.7 kpsi 4–121  σBE = 20.2 kpsi, σDF = 10.3 kpsi, yB = −0.0255 in, yC = −0.0865 in, yD = −0.0131 in 4–127  (a) t = 11 mm, (b) No 4–135  Fmax = 143.6 lbf, δmax = 1.436 in

B–5 Chapter 5 5–1  (a) MSS: n = 3.5, DE: n = 3.5, (b) MSS: n = 3.5, DE: n = 4.04, (c) MSS: n = 1.94, DE: n = 2.13, (d) MSS: n = 3.07, DE: n = 3.21, (e) MSS: n = 3.34, DE: n = 3.57

Answers to Selected Problems     1077

5–3  (a) MSS: n = 1.5, DE: n = 1.72, (b) MSS: n = 1.25, DE: n = 1.44, (c) MSS: n = 1.33, DE: n = 1.42, (d) MSS: n = 1.16, DE: n = 1.33, (e) MSS: n = 0.96, DE: n = 1.06 5–7  (a) n = 3.03 5–12  (a) n = 2.40, (b) n = 2.22, (c) n = 2.19, (d) n = 2.04, (e) n = 1.92 5–17  (a) n = 1.81 5–19  (a) BCM: n = 1.2, MM: n = 1.2, (b) BCM: n = 1.5, MM: n = 2.0, (c) BCM: n = 1.18, MM: n = 1.24, (d) BCM: n = 1.23, MM: n = 1.60, (e) BCM: n = 2.57, MM: n = 2.57 5–24  (a) BCM: n = 3.63, MM: n = 3.63 5–29  (a) n = 1.54 5–34  (a) n = 1.54 5–39  (a) n = 1.22, (b) n = 1.41 5–51  MSS: n = 1.29, DE: n = 1.32 5–59  MSS: n = 13.9, DE: n = 14.3 5–64  MSS: n = 1.30, DE: n = 1.40 5–69  For yielding: p = 934 psi, For rupture: p = 1.11 kpsi 5–76  d = 0.892 in 5–78  Model c: n = 1.80, Model d: n = 1.25, Model e: n = 1.80 5–80  Fx = 2π f T∕(0.2d) 5–81  (a) Fi = 16.7 kN, (b) pi = 111.3 MPa, (c) σt = 185.5 MPa, σr = −111.3 MPa (d) τmax = 148.4 MPa, σ′ = 259.7 MPa, (e) MSS: n = 1.52, DE: n = 1.73 5–87  no = 1.84, ni = 1.80 5–89  n = 1.91 5–97  (a) F = 958 kN, (b) F = 329.4 kN

B–6 Chapter 6 6–1  Se = 429 MPa 6–3  N = 117 000 cycles 6–5  Sf = 117.0 kpsi 6–15  nf = 0.69, ny = 1.51 6–17  nf = 0.45, N = 4100 cycles 6–20  ny = 1.66, (a) nf = 1.05, (b) nf = 1.31, (c) nf = 1.21

6–24  ny = 2.0, (a) nf = 1.19, (b) nf = 1.43, (c) nf = 1.31 6–25  ny = 3.32, using Goodman: nf = 0.59, N = 25 000 cycles 6–28  (a) nf = 0.99, N = 929 000 cycles, (b) nf = 1.23 for infinite life 6–30  The design is controlled by fatigue at the hole, nf = 1.33 6–33  (a) T = 21.8 lbf • in, (b) T = 23.8 lbf • in, (c) ny = 2.54 6–35  nf = 0.81, N = 716 000 cycles, yielding is not predicted 6–38  nf = 0.51 6–46  nf = 5.5 6–47  nf = 1.3 6–51  nf = 0.77, N = 61 000 cycles 6–57  P = 4.1 kips, yielding is not predicted 6–59  (a) n2 = 7 000 cycles, (b) n2 = 10 000 cycles

B–7 Chapter 7 7–1  (a) DE-Goodman: d = 27.27 mm, (b) DE-Morrow: d = 26.68 mm, (c) DE-Gerber: d = 25.85 mm, (d) DE-SWT: d = 27.99 mm 7–3  d = 0.983 in, D = 1.31 in, r = 0.0655 in 7–11  These answers are a partial assessment of potential failure. Deflections: θO = 5.47(10)−4 rad, θA = 7.09(10)−4 rad, θB = 1.10(10)−3 rad. Compared to Table 7–2 recommendations, θB is high for an uncrowned gear. Strength: Using DE-Goodman at the shoulder at A, nf = 3.25 7–23  (a) Fatigue strength using DE-Goodman: Left keyway nf = 2.9, right bearing shoulder nf = 4.4. Yielding: Left keyway ny = 4.3, right keyway ny = 2.7, (b) Deflection factors compared to minimum recommended in Table 7–2: Left bearing n = 3.5, right bearing n = 1.8, gear slope n = 1.6 7–33  (a) ω = 883 rad/s, (b) d = 50 mm, (c) ω = 1766 rad/s (doubles) 7–35  (b) ω = 466 rad/s = 4450 rev/min 7–39  14 -in square key, 78 -in long, AISI 1020 CD 7–41  dmin = 14.989 mm, dmax = 15.000 mm, Dmin = 15.000 mm, Dmax = 15.018 mm

1078      Mechanical Engineering Design

7–47  (a) dmin = 35.043 mm, dmax = 35.059 mm, Dmin = 35.000 mm, Dmax = 35.025 mm, (b) pmin = 35.1 MPa, pmax = 115 MPa, (c) Shaft: ny = 3.4, hub: ny = 1.9, (d) Assuming f = 0.8, T = 2700 N • m

B–8 Chapter 8 8–1  (a) Thread depth 2.5 mm, thread width 2.5 mm, dm = 22.5 mm, dr = 20 mm, l = p = 5 mm 8–4  TR = 15.85 N • m, TL = 7.83 N • m, e = 0.251 8–8  F = 182 lbf 8–11  (a) L = 45 mm, (b) kb = 874.6 MN/m, (c) km = 3116.5 MN/m 8–14  (a) L = 3.5 in, (b) kb = 1.79 Mlbf/in, (c) km = 7.67 Mlbf/in 8–19  (a) L = 60 mm, (b) kb = 292.1 MN/m, (c) km = 692.5 MN/m 8–25  From Eqs. (8–20) and (8–22), km = 2762 MN/m. From Eq. (8–23), km = 2843 MN/m 8–29  (a) np = 1.10, (b) nL = 1.60, (c) n0 = 1.20 8–33  L = 55 mm, np = 1.29, nL = 11.1, n0 = 11.8 8–37  np = 1.29, nL = 10.7, n0 = 12.0 8–41  Bolt sizes of diameters 8, 10, 12, and 14 mm were evaluated and all were found acceptable. For d = 8 mm, km = 854 MN/m, L = 50 mm, kb = 233.9 MN/m, C = 0.215, N = 20 bolts, Fi = 6.18 kN, P = 2.71 kN/bolt, np = 1.22, nL = 3.53, n0 = 2.90 8–46  (a) T = 823 N • m, (b) np = 1.10, nL = 17.7, n0 = 57.7 8–52  (a) Goodman: nf = 7.55, (b) Gerber: nf = 11.4, (c) ASME-elliptic: nf = 9.73 8–56  Goodman: nf = 11.9 8–61  (a) np = 1.16, (b) nL = 2.96, (c) n0 = 6.70, (d) nf = 4.56 8–64  np = 1.24, nL = 4.62, n0 = 5.39, nf = 4.75 8–68  Bolt shear, n = 2.30; bolt bearing, n = 4.06; member bearing, n = 1.31; member tension, n = 3.68 8–71  Bolt shear, n = 1.70; bolt bearing, n = 4.69; member bearing, n = 2.68; member tension, n = 6.68 8–76  F = 2.32 kN based on channel bearing 8–78  Bolt shear, n = 4.78; bolt bearing, n = 10.55; member bearing, n = 5.70; member bending, n = 4.13

B–9 Chapter 9 9–1  F = 49.5 kN 9–5  F = 51.2 kN 9–9  F = 31.1 kN 9–14  τ = 22.6 kpsi 9–20  (a) F = 2.71 kips, (b) F = 1.19 kips 9–26  F = 5.41 kips 9–34  F = 4.67 kips 9–37  F = 12.5 kips 9–40  F = 5.04 kN 9–43  All-around square, four beads each h = 6 mm, 75 mm long, Electrode E6010 9–54  τmax = 25.6 kpsi 9–56  τmax = 45.3 MPa 9–57  n = 3.48 9–60  F = 61.2 kN

B–10 Chapter 10 10–3  (a) L0 = 162.8 mm, (b) Fs = 167.9 N, (c) k = 1.314 N/mm, (d) (L0)cr = 149.9 mm, spring needs to be supported 10–5  (a) Ls = 2.6 in, (b) Fs = 67.2 lbf, (c) ns = 2.04 10–7  (a) L0 = 1.78 in, (b) p = 0.223 in, (c) Fs = 18.78 lbf, (d) k = 16.43 lbf/in, (e) (L0)cr = 4.21 in 10–11  Spring is solid safe, ns = 1.28 10–17  Spring is not solid safe, (ns < 1.2), L0 = 68.2 mm 10–20  (a) Na = 12 turns, Ls = 1.755 in, p = 0.396 in, (b) k = 6.08 lbf/in, (c) Fs = 18.2 lbf, (d) τs = 38.5 kpsi 10–28  With d = 2 mm, L0 = 48 mm, k = 4.286 N/mm, D = 13.25 mm, Na = 15.9 coils, ns = 2.63 > 1.2, ok. No other d works. 10–33  (a) d = 0.2375 in, (b) D = 1.663 in, (c) k = 150 lbf/in, (d) Nt = 8.69 turns, (e) L0 = 3.70 in 10–35  Use A313 stainless wire, d = 0.0915 in, OD = 0.971 in, Nt = 15.59 turns, L0 = 3.606 in 10–41  (a) L0 = 16.12 in, (b) τi = 14.95 kpsi, (c) k = 4.855 lbf/in, (d) F = 85.8 lbf, (e) y = 14.4 in 10–44  Σ = 31.3° (see Fig. 10–9), Fmax = 87.3 N 10–47  (a) k = 12 EI{4l3 + 3R[2πl2 + 4(π − 2) lR + (3π − 8) R2]}−1, (b) k = 36.3 lbf/in, (c) F = 3.25 lbf

Answers to Selected Problems     1079

B–11 Chapter 11 11–2  xD = 525, FD = 3.0 kN, C10 = 24.3 kN, 02–35 mm deep-groove ball bearing, R = 0.920 11–8  xD = 456, C10 = 145 kN 11–10  C10 = 20 kN 11–17  C10 = 26.1 kN 11–23  (a) Fe = 5.34 kN, (b) ℒD = 444 h 11–26  60 mm deep-groove 11–29  (a) C10 = 12.8 kips 11–35  C10 = 5.7 kN, 02–12 mm deep-groove ball bearing 11–36  RO = 112 lbf, RC = 298 lbf, deep-groove 02–17 mm at O, deep-groove 02–35 mm at C 11–44  l2 = 0.267(106) rev 11–49  FRA = 35.4 kN, FRB = 17.0 kN

B–12 Chapter 12 12–1  cmin = 0.015 mm, r = 12.5 mm, r∕c = 833, Nj = 18.3 rev/s, S = 0.182, h0∕c = 0.3, rf∕c = 5.4, Q∕(rcNl) = 5.1, Qs∕Q = 0.81, h0 = 0.0045 mm, Hloss = 11.2 W, Q = 219 mm3/s, Qs = 177 mm3/s 12–3  SAE 10: h0 = 0.000 275 in, pmax = 847 psi, cmin = 0.0025 in 12–7  h0 = 0.00069 in, f = 0.007 87, Q = 0.0833 in3/s 12–9  h0 = 0.011 mm, H = 48.1 W, Q = 1426 mm3/s, Qs = 1012 mm3/s 12–10  Tav = 154°F, h0 = 0.00113 in, Hloss = 0.0750 Btu /s, Qs = 0.0802 in3/s 12–19  Approx: μ = 45.7 mPa • s, Fig. 12–3: μ = 39 mPa • s

B–13 Chapter 13 13–1  35 teeth, 3.25 in 13–2  400 rev/min, p = 3π mm, C = 112.5 mm 13–4  a = 0.3333 in, b = 0.4167 in, c = 0.0834 in, p = 1.047 in, t = 0.523 in, d1 = 7 in, d1b = 6.578 in, d2 = 9.333 in, d2b = 8.77 in, pb = 0.984 in, mc = 1.55 13–5  dP = 2.333 in, dG = 5.333 in, γ = 23.63°, Γ = 66.37°, A0 = 2.910 in, F = 0.873 in

13–10  (a) 13, (b) 15, 45, (c) 18 13–12  10:20 and higher 13–15  (a) pn = 3π mm, pt = 10.40 mm, px = 22.30 mm, (b) mt = 3.310 mm, ϕt = 21.88°, (c) dP = 59.58 mm, dG = 105.92 mm 13–17  e = 4∕51, nd = 47.06 rev/min cw 13–24  N2 = N4 = 15 teeth, N3 = N5 = 44 teeth 13–30  nA = 68.57 rev/min cw 13–42  (a) d2 = d4 = 2.5 in, d3 = d5 = 7.33 in, (b) Vi = 1636 ft/min, Vo = 558 ft/min, (c) Wti = 504 lbf, Wri = 184 lbf, Wi = 537 lbf, Wto = 1478 lbf, Wro = 538 lbf, Wo = 1573 lbf, (d) Ti = 630 lbf • in, To = 5420 lbf • in 13–44  (a) NPmin = 15 teeth, (b) P = 1.875 teeth/in, (c) FA = 311 lbf, FB = 777.6 lbf 13–47  (a) NF = 30 teeth, NC = 15 teeth, (b) P = 3 teeth/in, (c) T = 900 lbf • in, (d) Wr = 65.5 lbf, Wt = 180 lbf, W = 192 lbf 13–49  FA = 94.4 i + 64.2 j + 421.3 k lbf, FB = −222.8 i − 815.6 k lbf 13–56  FC = 1565 i + 672 j lbf, FD = 1610 i − 425 j + 154 k lbf

B–14 Chapter 14 14–1  σ = 7.63 kpsi 14–4  σ = 32.6 MPa 14–7  F = 2.5 in 14–10  m = 2 mm, F = 25 mm 14–15  σc = −617 MPa 14–18  W t = 16 390 N, H = 94.3 kW (pinion bending); W t = 3469 N, H = 20.0 kW (pinion and gear wear) 14–19  W t = 1283 lbf, H = 32.3 hp (pinion bending); W t = 1633 lbf, H = 41.1 hp (gear bending); W t = 265 lbf, H = 6.67 hp (pinion and gear wear) 14–23  W t = 775 lbf, H = 19.5 hp (pinion bending); W t = 300 lbf, H = 7.55 hp (pinion wear), AGMA method accounts for more conditions 14–25  Rating power = min(157.5, 192.9, 53.0, 59.0) = 53 hp 14–29  Rating power = min(270, 335, 240, 267) = 240 hp 14–35  H = 69.7 hp

1080      Mechanical Engineering Design

B–15 Chapter 15

B–17 Chapter 17

15–1  W Pt = 690 lbf, H1 = 16.4 hp, W Gt = 620 lbf, H2 = 14.8 hp 15–2  W Pt = 464 lbf, H3 = 11.0 hp, W Gt = 531 lbf, H4 = 12.6 hp 15–8  Pinion core 300 Bhn, case, 373 Bhn; gear core 339 Bhn, case, 345 Bhn 15–9  All four W t= 690 lbf 15–11  Pinion core 180 Bhn, case, 266 Bhn; gear core, 180 Bhn, case, 266 Bhn

17–2  (a) Fc = 0.913 lbf, Fi = 101.1 lbf, (F1)a = 147 lbf, F2 = 57 lbf, (b) Ha = 2.5 hp, nfs = 1.0, (c) 0.151 in 17–4  A-3 polyamide belt, b = 6 in, Fc = 77.4 lbf, T = 10 946 lbf • in, F1 = 573.7 lbf, F2 = 117.6 lbf, Fi = 268.3 lbf, dip = 0.562 in 17–6  (a) T = 742.8 lbf • in, Fi = 148.1 lbf, (b) b = 4.13 in, (c) (F1)a = 289.1 lbf, Fc = 17.7 lbf, Fi = 147.6 lbf, F2 = 41.5 lbf, H = 20.6 hp, nfs = 1.1 17–8  Rx = (F1 + F2){1 − 0.5[(D − d)∕(2C)]2}, Ry = (F1 − F2)(D − d)∕(2C). From Ex. 17–2, Rx = 1214.4 lbf, Ry = 34.6 lbf

B–16 Chapter 16 16–1  (a) Right shoe: pa = 734.5 kPa cw rotation, (b) Right shoe: T = 277.6 N • m; left shoe: 144.4 N • m; total T = 422 N • m, (c) RH shoe: Rx = −1.007 kN, Ry = 4.13 kN, R = 4.25 kN, LH shoe: Rx = 570 N, Ry = 751 N, R = 959 N 16–3  LH shoe: T = 2.265 kip • in, pa = 133.1 psi, RH shoe: T = 0.816 kip • in, pa = 47.93 psi, Ttotal = 3.09 kip • in 16–5  pa = 27.4 psi, T = 348.7 lbf • in 16–8  a′ = 1.209r, a = 1.170r 16–10  P = 1.25 kips, T = 25.52 kip • in 16–14  (a) T = 8200 lbf • in, P = 504 lbf, H = 26 hp, (b) R = 901 lbf, (c) p∣θ=0 = 70 psi, p∣θ=270° = 27.3 psi 16–17  (a) F = 1885 lbf, T = 7125 lbf • in, (c) torque capacity exhibits a stationary point maximum 16–18  (a) d* = D∕ √3, (b) d* = 3.75 in, T* = 7173 lbf • in, (c) (d∕D)* = 1∕ √3 = 0.577 16–19  (a) Uniform wear: pa = 14.04 psi, F = 243 lbf, (b) Uniform pressure: pa = 13.42 psi, F = 242 lbf 16–23  Cs = 0.08, t = 143 mm 16–26  (b) Ie = IM + IP + n2IP + IL∕n2, (c) Ie = 10 + 1 + 102(1) + 100∕102 = 112 16–27  (c) n* = 2.430, m* = 4.115, which are independent of IL

17–14  With d = 2 in, D = 4 in, life of 106 passes, b = 4.5 in, nfs = 1.05 17–17  Select one B90 belt 17–20  Select nine C270 belts, life > 109 passes, life > 150 000 h 17–24  (b) n1 = 1227 rev/min. Table 17–20 confirms this point occurs in the range 1200 ± 200 rev/min, (c) Eq. (17–40) applicable at speeds exceeding 1227 rev/min for No. 60 chain 17–25  (a) Ha = 7.91 hp; (b) C = 18 in, (c) T = 1164 lbf • in, F = 744 lbf 17–27  Four-strand No. 60 chain, N1 = 17 teeth, N2 = 84 teeth, rounded L = 100 in, C = 30.0 in nfs = 1.17, life 15 000 h (pre-extreme)

B–20 Chapter 20 20–13  Partial answers: (a) 50.3, 49.7, (b) No effect, (c) the center axis of the boss, as determined by the related actual mating envelope, (d) 0.2, (e) 0.8 20–15  Hint: Read about actual mating envelopes. 20–17  20.2, 19.8 20–21  (a) 0.1, 0.1, 0.1, (b) 0.5, 0.3, 0.1, (c) 0.1, 0.3, 0.5

Index A

Abrasion, 749 Absolute stability, 530 Absolute tolerance system, 31 Absolute viscosity, 626 Acme threads, 424, 426 Actual mating envelope, 986–987, 1010 Actual strain, 44 Actual stress, 44 Addendum a, 684 Adhesive bonds application of, 508–509 joint design and, 513–515 references for, 516 stress distributions and, 510–513 types of adhesives and, 509–510 Admiralty metal, 77–78 AGMA equation factors. See also American Gear Manufacturers Association (AGMA) allowable bending stress numbers, 753, 805–806 allowable contact stress, 754, 804–806 bending-strength geometry factor, 758, 799, 800 bevel and worm gears, 797–808 contact stress and, 941 crowning factor for pitting, 799 dynamic factor, 744–749, 763, 764, 798 elastic coefficient for pitting resistance, 750, 761–762 804 geometry factors, 757–761 hardness-ratio factor, 767–768, 801–803 lengthwise curvature factor for bending strength, 799 load-distribution factor, 765–767, 799 overload factor, 764, 797 pitting resistance geometry factor, 752, 799, 800 reliability factors, 769–770, 803–804 reversed loading, 806 rim-thickness factor, 770–771

safety factor, 771, 797 size factor, 765, 799 stress, 751–757 stress-cycle factor, 768–769, 801 surface condition factor, 764 surface-strength geometry factor, 758–761 temperature factor, 758, 770, 803 worm gearing, 814–818 AGMA quality numbers, 763 ALGOR, 956 Alignment (bearings), 576, 578, 612 Allowable bending stress numbers, 753, 805–806 Allowable contact stress, 754, 804–806 Allowable stress numbers, 752 Allowance, 28 Alloy cast irons, 75 Alloy steels, 72–73, 375, 531 Alternating stress, 302, 326 Aluminum characteristics of, 75–76 numbering system for, 65 Aluminum alloys, 76 Aluminum brass, 78 Aluminum bronze, 78 American Bearing Manufacturers Association (ABMA), 579, 586, 587 American Chain Association (ACA), 913 American Gear Manufacturers Association (AGMA), 359. See also AGMA equation factors approach of, 740, 745 nomenclature, 740–742 standards, 744–745 strength equations, 752–757 stress equations, 751–752 American Institute of Steel Construction (AISC) building construction code, 500 American Iron and Steel Institute (AISI), 64, 65 American National (Unified) thread standard, 423, 425

American Society for Testing and Materials (ASTM), 65–66, 500 bolt strength standards, 443, 445 lap joint stress, 510, 513 viscosity standards, 627 American Society of Mechanical Engineers (ASME), 11, 626, 980 American Welding Society (AWS), 486, 500 Anaerobic adhesives, 510 Angle of action, 687 Angle of approach, 687 Angle of articulation, 910 Angle of recess, 687 Angle of wrap, 902, 904 Angularity control, 994–995 Angular-velocity ratio, 685, 688, 882, 884, 906 Annealing, 69 ANSI/AGMA standards, 771, 818 ANSYS, 956 Antifriction bearings, 576 Arc of action, 689, 690 Arc of approach, 689 Arc of recess, 689 Arrow side of joint, 487 Ashby, M. F., 81 ASME. See American Society of Mechanical Engineers (ASME) ASME-elliptic line, 335, 336, 459 ASME Y14.41-2003, 1009 ASME Y14.5-2009 Dimensioning and Tolerancing (American Society of Mechanical Engineers), 980 ASME Y14.5-2009 standard, 980, 981 Associated Spring, 562 ASTM. See American Society for Testing and Materials (ASTM) Austenitic chromium-nickel steels, 73 Automatic self-adaptive mesh refinement programs, 964–965 Automeshing, 964 Average life, 580 AWS. See American Welding Society (AWS)

1081

1082      Mechanical Engineering Design

Axial clutch, frictional contact, 849–852 Axial layout, for shaft components, 376–377 Axial loads, 377, 380–381 Axial pitch, 697, 700 Axial stresses, on shafts, 380 Axle, 374

B

Backlash, 684 Back to back mounting (DB), 611 Bainite, 70 Ball bearings. See also Bearings selection of, 593–596 thrust and, 576, 577, 585–590 types of, 577 Ball bushings, 579 Band-type clutches and brakes, 847–848 Barth, Carl G., 745 Barth equation, 745 Base circle, 685, 686 Base pitch, 687–688 Basic dimension, 990, 1010 Basic Dynamic Load Rating, 580 Basic static load rating, 586 Basquin's equation, 307, 308 Bauschinger effect, 57–58 Beach marks, 288, 289 Beam deflections due to bending, 176–178 methods for, 179 by singularity functions, 182–188 by superposition, 180–182 Beams with asymmetrical sections, 115–116 bending in curved, 141–145 bending moment and curvature of, 176–178 normal stresses and beams in bending, 111–116 shear force and bending moments in, 97–98 shear stresses for beams in bending, 116–122 two-plane bending and, 114–115 Bearing alloys, 661–662 Bearing characteristic number, 632 Bearing life, 576, 579–585, 662 Bearing pressure, 468, 921 Bearing reliability, 582–583 Bearings. See also Journal bearings; Rolling-contact bearings; specific types of bearings alignment of, 576, 578, 612 alloys for, 661–662

antifriction, 576 ball and cylindrical roller, 577, 585–590, 593–596 big-end connecting rod, 667–670 boundary-lubricated, 670–677 collar, 429 combined radial and thrust loading in, 585–590 double-row, 578 life of, 576, 579–585, 662 load life at rated reliability and, 580–581 lubrication of, 608–609 material choice for, 661–662 mounting and enclosure of, 609–613 needle, 578 overview of, 576 related load, life, and reliability and, 583–585 reliability vs. life of, 582–583 rolling-contact, 576, 604–608 screw, 426, 430, 433 selection of, 949–950 self-aligning, 586 self-contained, 649–653 sleeve, 624, 661 tapered roller, 579, 596–604 thrust and, 576, 577 types of, 576–579 variable loading in, 590–593 Bearing stress, 464 Belleville springs, 526, 564–565 Belting equation, 887 Belts flat metal, 896–899 timing, 882, 884–885, 908–909 types of, 882–885 V, 882, 884, 900–908 Bending moments in beams, 97–98, 114–115, 176–178, 195–196 (See also Beams) fundamental equations for, 751 on shafts, 380, 381 in springs, 566 Bending-strength geometry factor, 758, 799, 800 Bending stress, 111–122, 141–145, 497–499, 559–560, 740, 753, 797. See also Lewis bending equation Bergsträsser factor, 527, 545 Beryllium bronze, 78 Beryllium copper, 531 Bevel and worm gears. See also Gears; Worm gears AGMA equation factors for, 797–808, 814–817

AGMA symbols for bevel gear rating equations, 795–796 Buckingham wear load and, 825–826 classification of, 792–794 straight-bevel gear analysis and, 808–811 straight-bevel gear mesh design and, 811–814 stresses and strengths of, 794–797 worm-gear mesh design and, 822–826 Bevel gears. See also Bevel and worm gears analysis of straight, 808–811 description of, 682, 683 force analysis and, 713–716 hypoid, 793 spiral, 792, 794 straight, 695–696, 792 stresses and strengths of, 794–797 zerol, 792–793 Big-end connecting rod bearings, 667–670 Bilateral tolerance, 27 Blake, J. C., 450 Blanking, 69 Bolted joints loaded in shear, 463–467 tension-loaded, 436 Bolts. See also Joints function of, 434 preload and, 436, 452 relating bolt torque to bolt tension, 448–451 stiffness and, 437, 448 strength of, 443–446 thread length of, 434 torquing of, 450 Bonds, adhesive. See Adhesive bonds Bonus tolerance, 1003, 1010 Booker, J. F., 627 Bottom land, 684 Boundary conditions, 967 Boundary elements, 967 Boundary-lubricated bearings bushing wear and, 673–677 description of, 670–671 linear sliding wear and, 672 temperature rise and, 675 Boundary lubrication, 625, 670–677 Boundary representation (B-rep), 964 Bowman Distribution, 453 Boyd, John, 640, 641 Brakes band-type, 847–848 cone, 856–858

Index     1083

disk, 852–856 energy considerations for, 858–860 external contracting rim, 844–847 friction materials for, 863–866 internal expanding rim, 836–844 linings for, 864 static analysis of, 831–835 temperature rise for, 860–863 Brake shoe, 831 Brass, 77–78 Breakeven points, 14, 15 Brinell hardness, 61 Brittle-Coulomb-Mohr (BCM) theory, 263 Brittle-ductile transition temperature, 62 Brittle materials Brittle-Coulomb-Mohr (BCM) theory, 263 classification of, 247 failure of, 265–266 fatigue failure criteria for, 345–347 modifications of Mohr theory for, 263–265 relatively, 267 Smith-Dolan fracture criteria for, 345–346 stress-concentration factor and, 133 Broghamer, E. I., 748 Bronze, 77, 78 Bubble chart, 83 Buckingham, E., 357, 825 Buckingham load-stress factor, 357, 358 Burnishing, gear, 695 Bushings, 662–663 Bushing wear, 673–675 Button pad caliper brake, 855–856 Butt welds, 487–491

C

CAD software applications for, 8–9 GD&T and, 980, 1009 Caliper brakes, 852–853 Cap screws, 434–436 Carbon steels, 65, 375, 531 Cartesian stress components, 101–102 Cartridge brass, 77 Case hardening, 70–71 Case study (power transmission). See Power transmission (case study) Castigliano's theorem, 179, 190–195, 202–204, 528, 560, 566 Casting centrifugal, 67, 693 die, 67

of gear teeth, 693 investment, 67, 693 materials for, 73–75 permanent-mold, 693 sand, 66, 693 Casting alloys, 76 Cast irons hardness, 61 numbering system for, 65–66 shafts and, 375 stress concentration and, 133 types of, 73–75 Cast steels, 75 Catalog load rating, 580 Centrifugal casting, 67, 693 Centrifugal clutch, 836 Cermet pads, 864 CES Edupack software, 81, 87 Chain dimensioning, 29–30 Chains, 882 Chains for Power Transmissions and Materials Handling (American Chain Association), 913 Charpy notched-bar test, 62, 63 Chevron lines, 289 Chordal speed variation, 912 Choudury, M., 440 Chrome-silicon wire, 531 Chrome-vanadium wire, 531 Chromium, 72, 75 Circularity control, 993–994 Circular pad caliper brake, 855–856 Circular pitch p, 684 Circular runout, 1001–1002 Classical method of design, 18 Clearance, 27 journal bearings and, 653–655 Clearance c, 684 Clearance circle, 684 Close-wound extension springs, 551 Clough, R. W., 957, 958 Clutches band-type, 847–848 cone, 856–858 energy considerations for, 858–860 external contracting rim, 844–847 friction, 830 frictional-contact axial, 849–852 friction materials for, 863–866 internal expanding rim, 836–844 miscellaneous, 866–867 static analysis of, 831–835 temperature rise for, 860–863 uniform pressure in, 858 uniform wear in, 857–858

Codes, 12–13 Coefficient of friction journal bearings and, 645 in screw threads, 433 torque and, 449, 450 Coefficients of variance, 26 Coining, 69 Cold-drawn steel, 375 Cold rolling, 68 Cold-work factor, 56 Cold working, 44, 51–53, 55–57 Cold-working processes, 68–69 Collins, J. A., 356 Columns classification of, 207 with eccentric loading, 212–215 intermediate-length with central loading, 210–211 long with central loading, 207–210 Commercial bronze, 77 Commercial seal, 612 Complete journal bearing, 633, 634 Completely reversed stress, 302 Completely reversing simple loading, 359–361 Composite materials, 80 Compound gear train, 705 Compound reverted gear train, 706 Compression, 175 Compression members, analysis and design of, 207 Compression springs description of, 528–529 extension springs vs., 550, 557 helical, 535–541 Compression tests, stress-strain relationships from, 47 Compressive stress, 101, 217 Computational errors, 958 Computational fluid dynamics (CFD) programs, 9 Computation frame, 664 Computer-aided design (CAD) software, 8–9, 956 Computer-aided engineering (CAE), 9 Concentrated force function, 98 Concentrated moment function, 98 Concentricity control, 1000 Concept design, 6–7 Cone angle, 856–857 Cone clutch description of, 856–857 uniform pressure in, 858 uniform wear in, 857–858 Conical springs, 566

1084      Mechanical Engineering Design

Conjugate action, 684–685 Constant amplitude loading, 302 Constant-force springs, 565 Constant-life approach, 544 Constant-life curves, 328, 342–345 Constructive solid geometry (CSG), 964 Contact adhesives, 509 Contact ratio, 689–690 Contact strength (contact fatigue strength), 358 Contact stresses cylindrical, 147–149 description of, 145–146, 941 spherical, 146–147 Continuing education, 11 Coordinate dimensioning system, 978. See also Geometric Dimensioning and Tolerancing (GD&T) Copper-base alloys, 77 Corrosion (endurance limit), 318 Corrosion-resistant steels, 73, 531 Cost considerations. See Economics Cost estimates, 15 Coulomb-Mohr theory, 255–259, 265, 275 Couplings, 866–867 Courant, R., 957 Crack growth, 268 Crack modes, stress intensity factor and, 268–272 Crack nucleation, fatigue failure from, 289–293 Crack propagation, fatigue failure from, 293 Cracks, 266. See also Fracture mechanics Creep, 64, 882, 885 Creep test, 64 Creep-time curve, 64 Critical buckling load, 972 Critical deflection, of springs, 529 Critical speeds, for shafts, 395–400 Critical stress intensity factor, 270 Critical unit load, 208 Crossed belts, 882, 883 Crowned pulleys, 882 Crowning factor for pitting, 799 Cumulative fatigue damage, 351–356 Curvature effect, 527 Curve-beam theory, 559 Curved beams, bending in, 141–145 Curved members, deflection in, 195–201 Curve-fit equations, 322 Curve-fit polynomials, 315–316

Cyclic frequency, 319 Cyclic hardening, 58 Cyclic-minimum film thickness, 668 Cyclic Ramberg-Osgood, 60 Cyclic softening, 58 Cyclic strain strengthening exponent, 60 Cyclic strength coefficient, 60 Cyclic stress-strain curve, 59–60 Cyclic stress-strain properties, 57–60 Cyclic yield strength, 60 Cylindrical contact stresses, 147–149 Cylindrical roller bearings, 577, 585–590, 593–596 Cylindricity control, 994

D

Damage-tolerant design, 295 Datum features, 983–985, 987–988, 1004–1006, 1010 Datum feature simulator, 983, 984, 1010 Datum reference frame, 983, 984 Datums. See also Geometric Dimensioning and Tolerancing (GD&T) actual mating envelopes and, 986–987 description of, 983–984 feature symbol for, 987–988 immobilization of part and, 985 nonplanar features of, 986 order of, 985–986 Dedendum b, 684 Deflection analysis, 174 shafts and, 391–395, 948–949 in springs, 528, 529, 566–567 Deflection and stiffness, 174, 384. See also Stiffness beam deflection methods and, 179 beam deflections by singularity functions and, 182–188 beam deflections by superposition and, 180–182 bending and, 176–178 Castigliano's theorem and, 179, 190–195 columns with eccentric loading and, 212–215 compression members and, 207 deflection of curved members and, 195–201 elastic stability and, 217–218 helical springs and, 528 intermediate-length columns with central loading and, 210–211

long columns with central loading and, 207–210 shock and impact and, 218–220 spring rates and, 174–175 statically indeterminate problems, 201–207 strain energy and, 188–190 struts or short compression members and, 215–217 tension, compression, and torsion and, 175 Deflection equations, 201 Deformation equation, 202 DE-Gerber criteria, 381, 382 Degrees of freedom (dof's), 957 DE-Morrow criteria, 283 Derived unit, 31 Design basics calculations and significant figures, 32–33 case study specifications, 34–36 considerations, 8 design factor/factor of safety, 18–20 dimensions and tolerances, 27–31 economics, 13–15 in general, 4–5 information sources, 9–10 phases and interactions of, 5–8 relating design factor to reliability, 24–27 reliability and probability of failure, 20–24 safety/product liability, 15 standards and codes, 12–13 stress and strength, 16 tools and resources, 8–10 topic interdependencies, 33 uncertainty in, 16–17 units, 31–32 Design engineer communication and, 5, 10–11 professional responsibilities of, 10–12 Design factor, 17, 18 Design Manual for Cylindrical Wormgearing, 814 DE-SWT, 382 Deterministic design factor method, 17 Deviation, 406 Diametral interference, between shaft and hub, 410 Diametrical pitch P, 684 Die castings, 67, 693

Index     1085

Dimensioning. See Geometric Dimensioning and Tolerancing (GD&T) Dimensions choice of, 28–29 terminology of, 27–28 Dimension-series code, 587–588 Direct load, 468 Direct mounting, 597, 608 Discrete distributions, 23 Discretization errors, 958–959 Disk brakes description of, 852–853 uniform pressure in, 854 uniform wear in, 854 Disk clutches, 852. See also Disk brakes Dislocations, 292, 304 Distortion-energy (DE) theory, 249– 255, 259, 275, 314, 337, 348, 362 yield strength in shear and, 533–534 Dolan, T. J., 748 Doorstops, 831, 832 Double-row bearings, 578 Dowling, N., 301, 336 Drawing (tempering), 70 Drum brake, 836 Ductile cast iron, 74 Ductile materials classification of, 247 Coulomb-Mohr theory for, 255–258 distortion-energy theory for, 249–255 failure of, 258–262 maximum-shear-stress theory for, 247–249 selection of failure criteria for, 266 static loading and, 246–262 stress-strain diagram of, 51–52 Ductility, 56 Dunkerley's equation, 398 Duplexing, 611 Duplex mounting, 611 Dynamically loaded journal bearings, 663–670 Dynamic factor, spur and helical gears and, 744–749, 763, 764 Dynamic loads as element loads, 966 journal bearings and, 663–670 stress concentration effect and, 134 Dynamic viscosity, 626, 627 Dyne, 626

E

Eccentricity, centrifugal force deflection in shafts and, 395 Eccentricity rate, 633, 664–665

Eccentricity ratio, 213, 641 Eccentric loading columns with, 212–215 shear joints with, 467–469 Economics breakeven points, 14, 15 cost estimates, 15 large tolerances, 13–14 standard sizes, 13 Edge shearing, 464–465 Effective arc, 886 Effective slenderness ratio, 530 Effective stress, 250 Effective viscosity, 649 Eigenvalues, 973 Eigenvectors, 973 Elastic coefficient, 750, 761–762, 804 Elastic creep, 885 Elastic deformation, 43 Elastic instability, 217 Elasticity, 110, 174 Elastic limit, 43 Elastic stability, 217–218 Elastic strain, 109–110 Elastic-strain Basquin equation, 301 Elastohydrodynamic lubrication (EHL), 625 Electrolytic plating, 318 Element geometries, 958–961 Element library, 959 Element loads, 966 Elimination approach, 963 Enclosures (bearings), 612–613 End-condition constant, 208, 209, 530 Endurance limit modifying factors application of, 319–320 corrosion, 318 cyclic frequency, 319 electrolytic plating, 318 frettage corrosion, 318 loading factor, 314 metal spraying, 319 miscellaneous-effects factor, 317–318 reliability factor, 316–317 size factor, 312–314 surface factor, 309–311 temperature factor, 314–316 Endurance limits estimation of, 305–307 flexural, 357 Energy absorption, properties of, 47–48 Energy considerations, for brakes and clutches, 858–860 Engineering stress-strain diagrams, 43, 45

Engineers' Creed (NSPE), 12 Engraver's brass, 77 Envelope principle, 989, 1011 Epicyclic gear trains, 708 Equilibrium, 94 unstable, 208 Equivalent bending load, 919, 924 Equivalent radial load, 585 Euler column formula, 207, 209 Euler equation, 210, 871 Euler's method, 209–210, 213–215, 665 Evaluation, 7 Expanding-ring clutch, 836 Extension springs close-wound, 552 compression springs vs., 550, 557 correction stresses for, 553–555 description of, 550–552 ends for, 550 fatigue and, 555–556 function of, 557 initial tension in, 552 Extreme-pressure (EP) lubricants, 670–671 Extrusion, 68

F

Face-contact ratio, 700 Face load distribution factor, 765–766 Face-to-face mounting (DF), 611 Factor of safety, 18–19, 329, 330, 362, 452, 458, 771, 797, 920, 922. See also Safety Fail-safe design, 295 Failure. See also Fatigue failure; Fatigue failure from variable loading; Static loading, failures resulting from examples of, 242–244 meaning of, 242 Mohr theory of, 255–256, 265, 266 probability of, 20–23 selection of criteria for, 266 theories of, 247, 259, 265, 266 Failure prevention, knowledge of, 242 Failure theory flowchart, 276 Fasteners. See also specific types of fasteners eccentric loading of, 468 overview of, 422 stiffness of, 436–443 threaded, 434–436 Fatigue damage, cumulative, 351–356 Fatigue ductility coefficient, 301 Fatigue ductility exponent, 301

1086      Mechanical Engineering Design

Fatigue failure background on, 286 examples of, 289–292 low-cycle, 303–304 mechanisms of, 287–288 shaft materials and, 375 stages of, 288–289 Fatigue failure criteria applications of, 337–342 ASME-Elliptic, 335 for brittle materials, 345–347 Gerber, 334, 381, 544 Goodman, 333–334, 342–343 Morrow, 334, 343 in pure shear case, 337 recommendations, 336, 343–344 Smith-Watson-Topper, 335, 343 Soderberg, 334–335 Walker, 336, 343 Fatigue failure from variable loading characterizing fluctuating stresses and, 325–327 combinations of loading modes and, 347–350 constant-life curves and, 342–345 crack formation and propagation and, 288–293 cumulative fatigue damage and, 351–356 endurance limit modifying factors and, 309–320 fatigue failure criteria and, 333–342 fatigue failure criterion for brittle materials and, 345–347 fatigue-life methods and, 294–295 fluctuating-stress diagram and, 327–333 idealized S-N diagram for steels and, 304–308 linear-elastic fracture mechanics method and, 294–299 road maps and important design equations for the stress-life method and, 359–362 strain-life method and, 294, 299–301 stress concentration and notch sensitivity and, 320–325 stress-life method and, 294, 302–304 surface fatigue strength and, 356–359 Fatigue-life methods, 294–295 Fatigue loading of helical compression springs, 543–547 of tension joints, 456–463 of welded joints, 505–507

Fatigue strength data for, 296–297 estimated at 103 cycles, 306 nature of, 303 of springs, 543, 562–564 Fatigue strength coefficient, 301, 308 Fatigue strength exponent, 308 Fatigue stress-concentration factor, 320, 323–327, 456, 500, 748, 758 Fazekas, G. A., 856 FEA. See Finite-element analysis (FEA) Feature, 980, 1011 Feature control frame, 990–992, 1011 Feature of size, 980, 1011 Felt seals, 612 Ferritic chromium steels, 73 Field, J., 71 Filler, 80 Fillet welds, 487–491 bending properties of, 497–499 steady loads and size of, 500, 501 torsional properties of, 494 Filling notch, 577 Film pressure, 646–647 Finite element, 958 Finite-element analysis (FEA) applications for, 149, 201, 956–957 beam deflections and, 179 boundary conditions and, 967 critical buckling load and, 972 element geometries and, 956–961 load application and, 966–967 mesh generation and, 964–966 method of, 957–959 modeling techniques and, 967–970 software programs for, 9, 956, 964, 967, 974 solution process and, 961–963 stress-concentration factors and, 246 stress interactions and, 410 summary of, 974 thermal stresses and, 970–972 vibration analysis and, 973–974 Firbank, T. C., 885, 886 Fits description of shaft, 406–411 interference, 409–411 using basic hole system, 407, 408 Fitted bearing, 634 Flat-belt and round-belt drives analysis of, 889–892 description of, 882, 884, 885 flat metal belts and, 896–900 function of, 886–889 materials for, 890 pulley size and, 891–893

tension and, 893–896 theory of, 885–886 Flat belts, 882, 883, 893 Flat metal belts, 896–899 Flatness control, 992–993 Flat springs, 526. See also Springs Flexible mechanical elements belts as, 882–885 flat-and round-belt drives as, 885–900 flexible shafts as, 926 roller chains as, 909–917 timing belts as, 882, 884–885, 908–909 V belts as, 882, 884, 900–908 wire rope as, 917–925 Flexible shafts, 926 Flexural endurance limit, 357 Floating caliper brake, 852–853 Fluctuating simple loading, 361–362 Fluctuating-stress diagram, 327–333 Fluctuating stresses fatigue cracks and, 296 fatigue failure criterion and, 342 issues characterizing, 325–327 Fluctuating-stress values, 544 Fluid lubrication, 624 Flywheels, 830, 868–873 Foot-pound-second system (fps), 31 Force analysis of bevel gearing, 713–716 free-body diagrams for, 95–96 guidelines for, 947 of helical gearing, 716–718 of spur gearing, 710–713 of worm gearing, 719–724 Forging, 68 Forming, 69 Fracture mechanics background on, 266–267 crack modes and stress intensity factor and, 268–272 equation for, 276 fatigue failure and, 294–299 fracture toughness and, 270–274 quasi-static fracture and, 267–268 Fracture toughness, 270, 271, 273 Free-body diagrams, for force analysis, 95–96 Free-body force analysis, 945 Free-cutting brass, 77 Frettage corrosion, 319 Frictional-contact axial clutches description of, 849 uniform pressure in, 850–852 uniform wear in, 849–850 Friction clutch, 830

Index     1087

Friction materials, 863–866 Friction variable, 645 Full-film lubrication, 624 Functions, singularity, 94, 98–101 Fundamental contact stress equation, 794–797 Fundamental deviation, for shafts, 408

G

Gamma function, 582 Gasketed joints, 456 Gaskets, soft, 439 Gates Rubber Company, 902 Gaussian (normal) distribution, 21 GD&T (Geometric Dimensioning and Tolerancing). See Geometric Dimensioning and Tolerancing (GD&T) Gear bending strength, 752–755 Gear mesh analysis of, 771–781 design of, 781–786, 822–826 Gears. See also Bevel and worm gears; Spur and helical gears; specific types of gears AGMA approach and, 740–742 conjugate action in, 684–685 contact ratio and, 689–690 fundamentals of, 686–689 interference in, 690–693 involute properties of, 685–686 nomenclature for, 683–684 parallel helical, 696–700 selecting appropriate, 938–945 straight bevel, 695–696, 702 tooth systems for, 701–703 types of, 682, 683 worm, 682, 683, 700–701, 703 Gear strength, 752–757 Gear teeth bending stress in, 740, 742–749 conjugate action and, 684–685 contact ratio and, 689–690 formation of, 693–695 helical, 696–700 interference and, 690–693 terminology of, 683–684 Gear tooth bending, 776, 778, 785 Gear tooth wear, 776, 779, 785 Gear train force analysis bevel gearing, 713–716 helical gearing, 716–718 notation for, 710 spur gearing, 710–713 worm gearing, 719–724

Gear trains axes rotation and, 708 description of, 703–710 train value and, 704–705 two-stage compound, 706 Gear wheel, 815 Generating line, 687 Geometric characteristic controls and symbols, 982 Geometric controls form, 992–994 location, 998–1001 orientation, 994–996 profile, 995–998 runout, 1001–1002 Geometric Dimensioning and Tolerancing (GD&T), 28, 953 CAD models and, 1009 control of, 989–992 datums and, 983–988 defined, 978–979 geometric attributes of features for, 980–981 geometric characteristic definitions and, 992–1002 glossary of terms for, 1010–1012 material condition modifiers and, 1002–1004 overview of, 28, 953, 978–979 practical implementation of, 1004–1008 standards for, 980 symbolic language for, 981–983 Geometric stress-concentration factor, 133 Geometry factor, 757–761 Gerber fatigue-failure criterion, 334, 381, 544, 562 Gerber line, 334, 459 Gib-head key, 402, 403 Gilding brass, 77 Glasses, 82 Global instabilities, 217 Goodman, John, 329 Goodman diagram, 328 Goodman failure criterion, 544 Goodman failure line, 329, 354 Goodman-Haigh diagram, 328 Goodman line, 328–330, 333, 334, 343 Government information sources, 10 Gravitational mass of units, 31 Gravity loading, 966 Gray cast iron, 73–74, 133 Green, I., 440 Griffith, A. A., 268 Grip l, 437

Grooved pulleys, 882 Grossman, M. A., 71 Guest theory. See Maximumshear-stress theory (MSS)

H

Hagen-Poiseuille law, 627 Haigh diagram, 328 Ham, C. W., 433 Hard-drawn steel spring wire, 531 Hardness, 61 Hardness-ratio factor, 767–768, 801–803 Hardness testing, 61 Haringx, J. A., 530 Haugen, E. B., 317 Heading, 69 Heat transfer analysis, 970–972 Heat transfer rate, 861 Heat treatment, of steel, 69–71 Helical coil torsion springs bending stress in, 559–560 deflection and spring rate in, 560–561 description of, 557–558 end location of, 558–559 fatigue strength in, 562–564 static strength in, 561–562 Helical compression spring design, for static service, 535–541 Helical compression springs design for fatigue loading, 547–550 fatigue loading of, 543–547 Helical gears. See also Gears; Spur and helical gears crossed, 814 description of, 682 force analysis and, 716–718 parallel, 696–700 standard tooth proportions for, 702 Helical rollers, 578 Helical springs critical frequency of, 542–543 deflection of, 528 fatigue loading of, 543–550 maximum allowable torsional stresses for, 534 for static service, 535–541 stresses in, 526–527 Hertz, H., 146, 148 Hertzian contact pressure, 356, 357 Hertzian endurance strength, 358 Hertzian field, 148–149 Hertzian stress, 146, 357, 749 Hertz theory, 749 Hexagonal-head bolts, 434, 435

1088      Mechanical Engineering Design

Hexagon-head cap screws, 434, 435 Hexagon nuts, 434, 436 Hobbing, 694 Holding power, 400 Hole basis, 407 Hooke's law, 43, 110, 215 Hoop stress, 136 Hot melts, 509–510 Hot rolling, 67–68, 310 Hot-working processes, 67–68 Hrennikoff, A., 957 Hydraulic clutches, 836 Hydrodynamic lubrication, 624, 634–638 Hydrostatic lubrication, 625 Hypoid gears, 793

I

Idle arc, 886 Impact, shock and, 218–220 Impact load, 62 Impact properties, 62–63 Impact value, 62 Inch-pound-second system (ips), 31 Indirect mounting, 597 Infinite-life design, 295 Influence coefficients, 396 Information sources, 9–10 Injection molding, 693 Interference diametral, 410 gear teeth and, 690–693 nature of, 27, 690 Interference fits, 409–411 Interference of strength, 25 Interference of stress, 25 Internal-friction theory, 256 Internal gear, 688 Internal-shoe brake, 836–837 International Committee of Weights and Measures (CIPM), 626 International System of Units (SI), 32 International tolerance grade numbers (IT), 407 Internet information sources, 10 Interpolation equation, 649 Investment casting, 67, 693 Involute helicoid, 696, 697 Involute profile, 684 Involute properties, of gears, 685–686 Irons alloy cast, 75 ductile and nodular cast, 74 gray cast, 73–75

malleable cast, 74–75 white cast, 74 Ito, Y., 439 Izod notched-bar test, 62, 63

J

J. B. Johnson formula, 210–211, 429 Joerres, R. E., 533, 534 Joints. See also specific types of joints arrow side of, 487 bolted, 447, 452 bolted and riveted, 463–467 eccentric loading in shear, 467–469 fastener stiffness and, 436–437 fatigue loading of, 505–507 gasketed, 456 lap, 510–514 member stiffness and, 437–443 statically loaded, 452–456, 502–505 stiffness constant of, 448 strength of welded, 499–501 tension, 446–448 tension-loaded bolted, 436 welded, 492–499, 505–507 Jominy test, 71 Journal bearings. See also Bearings; Lubrication and journal bearings alloys for, 661–662 big-end connecting rod, 667–670 boundary-lubricated, 670–677 design variables for, 639–640 dynamically loaded, 663–670 fitted, 634 material choice for, 661–662 nomenclature of complete, 633, 634 partial, 634 pressure-fed, 655–660 self-contained, 649–653 Trumpler's design criteria for, 639–640 types of, 662–663 Journal orbit, 663–664 Journal translational velocity, 664–665

K

Karelitz, G. B., 650 Keys (shafts), 401–405, 950–952 Kinematic viscosity, 627 Kuhn, P., 322 Kurtz, H. J., 450

L

Labyrinth seal, 612 Lamont, J. L., 71 Langer lines, 330 Lang-lay rope, 918

Lap joints, 510–514 Law of action and reaction (Newton), 95 Least material condition (LMC), 1002–1004, 1011 Leibensperger, R. L., 609 Lengthwise curvature factor for bending strength, 799 Lewis, Wilfred, 740 Lewis bending equation, 740, 742–749, 757 Lewis form factor, 743 Limits, 27 Limits (shaft), 406–407 Linear-elastic fracture mechanics (LEFM) method, 266, 294–299 Linear sliding wear, 672 Linear springs, 174 Linear stress analysis, 970, 972 Linear transverse line loads, 966 Lined bushing, 662 Line elements, 959 Line of action, 685, 687 Line of contact, 148 Lipson, C., 310, 311 Little, R. E., 439 Load and stress analysis Cartesian stress components and, 101–102 contact stresses and, 145–149 curved beams in bending, 141–145 elastic strain and, 109–110 general three-dimensional stress and, 108–109 Mohr's circle for plane stress and, 94, 102–109 normal stresses for beams in bending and, 111–116 press and shrink fits and, 139–140 shear stresses for beams in bending and, 116–122 singularity functions and, 94, 99–101 stress and, 16, 101, 148–149 stress concentration and, 132–135 stresses in pressurized cylinders and, 135–137 stresses in rotating rings and, 137–139 temperature effects and, 140–141 torsion and, 123–132 uniformly distributed stresses and, 110–111 Load-application factor, 588, 589 Load-distribution factor, 765–767, 799 Load factor, 452

Index     1089

Loading factor (endurance limit), 314, 348 Loading modes, 347–350, 362 Load intensity, 97, 179 Load-life-reliability relationship, 576 Load-sharing ratio, 758 Loads/loading critical unit, 208 dynamic, 966 element, 966 fatigue, 505–507 impact, 62 nodal, 966 reversed, 806 static, 133, 242, 286, 502–505, 966 (See also Static loading, failures resulting from; Static loads/ loading) surface, 966–967 variable, 590–593 (See also Fatigue failure from variable loading) Load-stress factor, 357, 358 Load zone, 600 Location controls concentricity, 1000 position, 998–1000 symmetry, 1000–1001 Loose-side tension, 886 Loss-of-function parameter, 18 Low brass, 77 Low-contact-ratio (LCR) helical gears, 758 Low-cycle fatigue, 303–304 Low-leaded brass, 77 Lubricant flow, 646 Lubricant temperature rise, 647–649, 655, 675 Lubrication boundary, 625, 670–677 elastohydrodynamic, 625 function of, 624 hydrodynamic, 624, 634–638 hydrostatic, 625 mathematical theory of, 635 mixed-film, 671 roller chain, 917 rolling bearing, 478, 576, 579, 608–609 solid-film, 625 splash, 608 stable, 632–633 temperature rise and, 647–649, 655, 675 thick-film, 633–634 unstable, 633

Lubrication and journal bearings bearing types and, 663 boundary-lubricated bearings and, 670–677 clearance and, 653–655 design variables for, 639–640 dynamically loaded journal bearings and, 663–670 hydrodynamic theory and, 624, 634–638 loads and materials and, 661–662 lubrication types and, 624–625 Petroff's equation and, 627–632 pressure-fed bearings and, 655–660 relationship between variables and, 640–649 stable lubrication and, 632–633 steady-state conditions in selfcontained bearings and, 649–653 thick-film lubrication and, 633–634 viscosity and, 625–627 Lüder lines, 247–248

M

Mabie, H. H., 748 Macaulay functions. See Singularity functions Magnesium, 76 Magnetic clutches, 836 Major diameter, of screw threads, 422 Malleable cast iron, 74–75 Manganese, 72 Manson, S. S., 355 Manson-Coffin equation, 301 Map frame, 665 Margin of safety, 25 Marin, Joseph, 258, 309 Marin's equation, 309, 317 Martensite, 70, 306 Martensitic stainless steel, 73 Martin, H. C., 957 Master fatigue diagram, 328 Material condition modifiers (MMC), 1002–1004, 1011 Material efficiency coefficient, 85 Material index, 85–86 Materials. See also specific materials alloy steel and, 72–73 casting, 73–75 cold-working processes and, 68–69 composite, 80 corrosion-resistant steel and, 73 cyclic stress-strain properties and, 57–60 energy absorption properties of, 47–48 hardness and, 61, 62

heat treatment of steel and, 69–71 hot-working processes and, 67–68 impact properties and, 62–63 investment casting and, 67 nonferrous metals and, 75–78 numbering systems and, 64–66 plastic deformation and cold work and, 50–57 plastics and, 78, 79 powder-metallurgy process and, 67 sand casting and, 66 selection of, 42, 81–87 for shafts, 374–375 shell molding and, 66, 693 statistical significance of properties of, 48–50 strength and stiffness, 42–48 temperature effects and, 63–64 Materials selection charts, 81–82 Matrix, 80 Maximum load, 247–249, 262–263, 265 Maximum material condition (MMC), 989, 990, 1012 Maximum-normal-stress theory for brittle materials, 262–263, 265 Maximum shear stress, 47, 247, 248, 526 Maximum-shear-stress theory (MSS), ductile materials and, 247–249, 252, 257 Maximum shear theory, 275 Maxwell's reciprocity theorem, 396 McHenry, D., 957 McKee, S. A., 632 McKee, T. R., 632 McKee abscissa, 632 McKelvey, S. A., 311 Mean coil diameter, 526 Mean stress, 302, 307 fatigue failure and, 333–336, 351 fluctuating stresses and, 325–330 loading mode and, 347, 348 nonzero, 301 Mechanical springs. See Springs Mesh, 964 Mesh density, 964 Mesh design for straight-bevel gears, 811–814 for worm gears, 822–826 Mesh generation fully automatic, 964 manual, 964 semiautomatic, 964 Mesh refinement, 964 Metal-mold castings, 67 Metals, nonferrous, 75–78

1090      Mechanical Engineering Design

Metals Handbook (ASM), 289 Metal spraying, 319 Metric threads, 423–424 Milling, gear teeth, 693 Miner's rule, 352, 353, 355 Minimum film thickness, 641, 642 Minimum life, 580 Minor diameter, of screw threads, 422 Mischke, Charles R., 312, 316 Mitchiner, R. G., 748 Mixed-film lubrication, 671 MMC. See Material condition modifiers (MMC) Mobility map, 665 Mobility method, 664–665 Mobility vector, 665, 666 Modal analysis, 973, 974 Mode I, plane strain fracture toughness, 270 Modeling techniques, 967–970 Moderate applications, 746 Modern Steels and Their Properties Handbook (Bethlehem Steet), 71 Modified-Goodman diagram, 328 Modified-Goodman line, 328–329 Modified Mohr (MM) theory, 263–265 Modified Mohr (plane stress), 275 Module m, 684 Modulus of elasticity, 43, 109 Modulus of elasticity of rope, 919 Modulus of resilience, 47–48 Modulus of rigidity, 47, 110 Modulus of rupture, 47 Modulus of toughness, 48 Mohr's circle diagram, 104, 256, 259 Mohr's circle for plane stress, 94, 102–109, 147 Mohr's circle shear convention, 104–106 Mohr theory for brittle materials, 263–265 Mohr theory of failure, 255–256, 266. See also Coulomb-Mohr theory Molded-asbestos linings, 864 Molded-asbestos pads, 864 Molding, 66, 693 Molybdenum, 72, 75 Moment connection, 492 Moment load, 468 Monte Carlo computer simulations, 31 Morrow fatigue-failure criterion, 334, 336, 382 Morrow line, 334, 353 MSC/NASTRAN, 956 Multiple of rating life, 581

Multiple-threaded product, 423 Multipoint constraint equations, 967 Muntz metal, 78 Music wire, 531, 532

N

Naval brass, 78 Necking, 45, 46 Needle bearings, 578 Neuber, H., 322 Neuber constant, 322, 323 Neutral plane, 112 Newmark, N. M., 957 Newtonian fluids, 626 Newton (N), 32 Newton's cooling model, 860–861 Newton's third law, 95 Nickel, 72, 75 Nodal loads, 966 Nodes, 957, 961, 962 Nodular cast iron, 74 Noll, C. J., 310, 311 Nominal size, 27 Nominal stress, 133 Nonferrous metals, 75–78 Nonlinear softening spring, 175 Nonlinear stiffening spring, 174 Nonplanar datum features, 986 Nonprecision bearings, 579 Normal circular pitch, 697 Normal diametrical pitch, 697 Normalizing process, 69 Normal strain, 43 Normal stress, 101 Norris, C. H., 490 Notched-bar tests, 62, 63 Notch sensitivity, stress concentration and, 320–325 Numbering systems, 64–66 Nuts grade of, 444 hexagon, 434, 436

O

Octahedral shear stress, 251–252 Octahedral-shear-stress theory, 251, 252 Offset method, 44 Oil outlet temperature, 649 Oil-tempered wire, 531 Opening crack propagation mode, 268, 269 Orientation control, 994–996 Osgood, C. C., 439 Overconstrained systems, 201

Overload factor, 764, 797 Overrunning clutch or coupling, 867

P

Palmgren linear damage rule, 356 Palmgren-Miner cycle-ratio summation rule (Miner's rule), 352, 353, 355 Parabolic formula, 210–211 Parallel-axis theorem, 113 Parallel helical gears, 696–700 Parallelism control, 994–995 Paris equation, 297 Partial journal bearing, 634 Partitioning approach, 963 Pedestal bearings, 649 Performance factors, 639 Permanent joint design adhesive bonding and, 508–516 butt and fillet welds and, 487–494 resistance welding and, 507–508 welded joints and, 492–499, 505–506 welding symbols and, 486–488 Permanent-mold casting, 693 Permissible contact stress number (strength) equation, 797 Peterson, R. E., 246 Petroff's equation, 631–632 Phosphor bronze, 77, 78, 531 Pilkey, W. D., 404 Pillow-block bearings, 649 Pinion, 683, 691, 693 Pinion bending, 782 Pinion cutter, 693, 694 Pinion tooth bending, 775, 778, 784, 785 Pinion tooth wear, 776, 779 Pins (shafts), 401–405 Pitch of bevel gears, 695 of screw threads, 422, 423, 426 timing belt, 908–909 Pitch circle, 683, 685, 686, 688, 693, 798 Pitch diameter of screw threads, 422 of spur-gear teeth, 683 Pitch length, 902 Pitch-line velocity, 686, 711, 712, 745 Pitch point, 685 Pitch radius, 685 Pitting, 749 Pitting resistance, 750, 752, 761–762, 804 Pitting-resistance geometry factor, 752, 799, 800 Plane slider bearing, 635

Index     1091

Plane strain, 297 Plane stress, 275 Mohr's circle for, 94, 102–109, 256–257 Plane-stress transformation equations, 102 Planetary gear trains, 708 Planet carrier (arm), 708 Planet gears, 708 Plastic deformation, 43, 44, 51–53 Plastics, 78, 79 Plastic strain, 289, 292 Plastic-strain Manson-Coffin equation, 301 Pneumatic clutches, 836 Poisson's ratio, 80, 109, 749 Position control, 998–1000 Positive-contact clutch, 866, 867 Potential energy. See Strain energy Pound-force (lbf), 31 Powder-metallurgy process, 67 Power curve equation, 311 Power screws, 426–433 Power transmission (case study) background of, 936 bearing selection, 949–950 design sequence for, 937–938 final analysis, 953 force analysis, 947 gear specification, 938–945 key and retaining ring selection, 950–953 power and torque requirements, 938 shaft design for deflection, 948–949 shaft design for stress, 948 shaft layout, 945–947 shaft material selection, 947 Precipitation-hardenable stainless steels, 73 Preload bolt, 436, 452 considerations for, 460 statically loaded tension joint with, 452–456 Presetting, 529 Press fits, 139–140, 378–379 Pressure angle, 687, 698, 703 Pressure-fed bearings, 655–660 Pressure line, 687 Pressure-sensitive adhesives, 509 Pressure-strength ratio, 921 Pressurized cylinders, stresses in, 135–137 Pretension, 436, 452–453

Primary shear, 468 Principal stresses, 103 Probability density function (PDF), 20 Probability of failure, 20–24 Product liability, 15 Professional societies, 10, 11 Profile control, 996–998 Profile of a line, 996 Profile of a surface, 996 Proof load, bolt, 443 Proof strength, bolt, 443 Propagation of dispersion, 24 Propagation of error, 24 Propagation of uncertainty, 24 Proportional limit, 43 Puck pad caliper brake, 855–856 Pulleys flat-belt and round-belt, 891–892 forces and torque on, 887 Pure shear, 337

Q

Quality numbers (AGMA), 763 Quasi-static fracture, 267–268 Quenching, 69–70

R

R. R. Moore high-speed rotating-beam machine, 302–303 Radial clearance ratio, 632 Radial loading/thrust loading combined, 585–590 Radial stress, 135–137, 410 Radius of gyration, 208 Raimondi, Albert A., 640, 641 Raimondi-Boyd analysis, 640–647 Rain-flow counting technique, 352 Ramberg-Osgood relationship, 53 Ramp function, 98, 99 Rating life, 579–580 Rayleigh's equation, 396 Redundant supports, 201 Reemsnyder, Harold S., 297 Regular-lay rope, 917 Related actual mating envelope, 986–987 Relative velocity, 859 Reliability factor, 316–317, 360, 769–770, 803–804 Reliability method of design, 20, 24–26 Repeated stress, 302 Residual stress, 317, 487 Resilience, 47–48

Resistance welding, 507–508 Retaining rings, 405–406, 950–952 Reynolds, Osborne, 626, 635, 636, 638 Reynolds equation, 636, 638, 665, 666 Right-hand rule, 423, 703, 722 Rim clutches external contracting, 844–847 internal expanding, 836–844 Rim-thickness factor, 770–771 Ring gear, 688 Riveted joints, loaded in shear, 463–467 Roark's formulas, 179 Roark's Formulas for Stress and Strain (Young, Budynas & Sadegh), 179, 180 Rockwell hardness, 61, 806 Roller chains, 909–917 Rolling bearings description of, 576, 578 lubrication and, 478, 576, 579, 608–609 types of, 578–579 Rolling-contact bearings ball and cylindrical roller bearing selection and, 577, 585–590, 593–596 bearing load life at rated reliability and, 580–581 combined radial and thrust loading of, 585–590 description of, 576 design assessment for, 604–608 fit and, 608 life of, 579–580 lubrication of, 608–609 mounting and enclosure of, 609–613 relating load, life and reliability and, 583–585 reliability of, 605–608 reliability vs. life of, 582–583 tapered roller bearings selection and, 596–604 types of, 576–579 variable loading and, 590–593 Roll threading, 69 Ropes, 882 Rotary fatigue, 356 Rotating-beam specimen, 305, 309 Rotating-beam test, 303 Rotating rings, stresses in, 137–139 Rotscher's pressure-cone method, 439 Round-belt drives. See Flat-belt and round-belt drives Round belts, 882

1092      Mechanical Engineering Design

Runge-Kutta method, 665 Runout controls, 1001–1002 Russell, Burdsall & Ward Inc. (RB&W), 453–454 Ryan, D. G., 433

S

Safe-life design, 295 Safety, 15, 18–20, 771, 797. See also Factor of safety Salakian, A. G., 490 Samónov, C., 530, 534 Sand casting, 66, 693 Saybolt Universal viscosity (SUV), 627 Screws cap, 434–436 elongation of, 449 power, 426–433 self-locking, 428 stiffness and, 437 Screw threads efficiency of, 428 square, 430 terminology of and standards for, 422–426 Sealants. See Adhesive bonds Sealings, 612–613 Seam welding, 507, 508 Secant column formula, 213, 216 Secondary shear, 468, 492 Self-acting (self-locking) phenomenon, 833 Self-aligning bearings, 586 Self-contained bearings, 649–653 Self-deenergizing brake shoe, 831 Self-energizing brake shoe, 831 Self-locking screw, 428 Series system, 24 Set removal, 529 Setscrews, 400–401 Shaft basis, 407 Shaft components keys and pins as, 401–405 retaining rings as, 405–406 setscrews as, 400–401 Shaft coupling, 866–867 Shaft design, for deflection, 948–949 Shaft design for stress critical locations and, 380 estimating stress concentrations and, 384–390 first iteration estimates for, 386 shaft stresses and, 380–384, 948 Shaft layout about, 375–376, 945–947

assembly and disassembly and, 379–380 axial, 376 supporting axial loads, 377 torque transmission provisions and, 377–379 Shaft material fatigue failure and, 375 selection of, 947 Shafts, 286 bearings in, 585 bending moments on, 380 couplings, 866–867 critical speeds for, 395–400 defined, 374 deflection analysis and, 391–395 flexible, 926 fundamental deviation for, 408 layout of, 375–380 limits and fits for, 406–411 materials for, 374–375 misalignment in, 611 Shallow drawing, 69 Shear-energy theory, 251 Shear force bending moments in beams and, 97–98 welded joints in torsion and, 492, 493 Shearing edge, 464–465 Shear-lag model, 511 Shear loading, bolted and riveted joints and, 463–467 Shear modulus, 47, 110 Shear stress for beams in bending, 116–122 maximum, 47, 124 tangential, 101, 135, 410, 430 torsion in shaft and, 410–411 transverse, 117–122 Shear stress-correction factor, 527 Shear yield strength, 275 Sheaves, 882, 901–902, 919–921 Shell molding, 66, 693 Shock, 218 Short compression members, struts as, 215–217 Shrink fits, 139–140, 378–379 Significant figures, 32–33 Silicon, 72 Silicon bronze, 78 Simple loading completely reversing, 359–361 fluctuating, 361–362 Sines failure criterion, 544 Single-row bearings, 577, 578

Singularity functions application of, 99–101, 185 beam deflections by, 182–188 description of, 94, 98–99 Sintered-metal pads, 864 Size factor, 312–314, 765, 799 Sleeve bearings, 624, 661 Slenderness ratio, 208 Sliding fit, 407 Sliding mode, 268, 269 Slip, 44, 882 Slip lines, 247–248 Slip planes, 44 Slug, 31 Smith, G. M., 543 Smith-Dolan locus, 345–346 Smith-Watson-(SWT) criterion, 335, 336 Smith-Watson-Topper (SWT) fatiguefailure criterion, 335, 336, 343, 344 S-N diagram. See Stress-life (S-N) diagram Snug-tight condition, 449 torque and, 449 Society of Automotive Engineers (SAE), 11, 64 bolt strength standards, 443, 444 Society of Manufacturing Engineers (SME), 11 Socket setscrews, 400–401 Soderberg line, 334–336 Software CAD, 8–9, 980, 1009 engineering-based, 9 engineering-specific, 9 Solid elements, 959, 960 Solid-film lubricant, 625 Sommerfeld, A., 638 Sommerfeld number, 632, 641, 645, 659 Special-purpose elements, 959, 960 Specific modulus (specific stiffness), 83 Speed ratio, 759 Spherical contact stresses, 146–147 Spherical-roller thrust bearing, 578 Spinning, 69 Spiral bevel gears, 792, 794 Spiroid gearing, 794 Splash lubrication, 608 Splines, 378 Spot welding, 507, 508 Spring brass, 531 Spring constant, 175

Index     1093

Spring index, 526, 535, 552, 559 Spring rate, 174, 175, 437, 440, 528, 560–561 Springs. See also specific types of springs Belleville, 526, 564–565 classification of, 526 compression, 528–529 conical, 566 constant-force, 565 critical frequency of helical, 542–543 curvature effect in, 527 deflection in helical, 528 extension, 550–557 fatigue loading of helical compression, 543–547 function of, 174 helical coil torsion, 557–564 helical compression spring design for fatigue loading, 547–550 helical compression spring design for static service, 535–541 linear, 174 materials used for, 531–535 nonlinear softening in, 175 nonlinear stiffening in, 174 scale of, 528 stability of, 529–530 stresses in helical, 526–527 volute, 565 Spring surge, 542 Sprockets, 882 Spur and helical gears. See also Gears; Helical gears AGMA nomenclature, 741–742 AGMA strength equations and, 752–757 AGMA stress equations and, 751–752 analysis of, 771–781 description of, 682 dynamic factor and, 744–749, 763, 764 elastic coefficient and, 750, 761–762 force analysis and, 710–713 gear mesh analysis and, 771–781 gear mesh design and, 781–786 geometry factors and, 757–761 hardness-ratio factor and, 767–768 Lewis bending equation and, 740, 742–749 load-distribution factor and, 765–767

overload factor and, 764 reliability factor and, 769–770 rim-thickness factor and, 770–771 safety factors and, 771 size factor and, 765 spur gears analysis and, 774–776 stress-cycle factors and, 768–769 surface condition factor and, 764 surface durability of, 749–751 temperature factor and, 770 Square-jaw clutch, 866, 867 Square threads, 424 Squeeze-film action, 663 Stable cyclic hysteresis loop, 59, 300 Stable lubrication, 632–633 Stainless steels, 73 Stamping, 69 Standard Handbook of Machine Design (Shigley et al.), 71 Standards and codes, 12–13 defined, 12 organizations with specific, 12–13 Standard sizes, 13 Statically indeterminate problems, 201–207 Statically indeterminate systems, 201 Static equilibrium, 94, 201 Static loading, failures resulting from Brittle-Coulomb-Mohr (BCM) and modified Mohr (MM) theories, 263–265 brittle materials failure and, 265–266 Coulomb-Mohr theory for ductile materials and, 255–258 design equations and, 275–276 distortion-energy theory for ductile materials, 249–255 ductile materials and, 258–262 failure theories and, 247 fracture mechanics and, 266–274 maximum-normal-stress theory for brittle materials and, 262–263, 265 maximum-shear-stress theory for ductile materials and, 247–249 selection of failure criteria and, 266 static strength and, 244–245 stress concentration and, 245–246 Static loads/loading background of, 286 basic rating, 586 defined, 242, 966 helical coil compression spring design flowchart for, 537–538 stress-concentration factors and, 133 welding and, 502–505

Static strength in helical coil torsion springs, 561–562 static loading and, 244–245 Statistical tolerance system, 31 Steels alloy, 72–73, 375 cast, 75 cold-drawn, 375 corrosion-resistant, 73 hardness, 61 heat treatment of, 69–71 modulus of elasticity, 43 numbering system for, 64–65 quantitative estimation of properties of heat-treated, 71 for springs, 531, 533, 534 stress-life (S-N) diagram for, 304–308 temperature and, 315–316 Stereolithography (STL), 9 Stiffness. See also Deflection and stiffness bolt, 437, 448 fastener, 436–443 joint, 437–443 procedure to find fastener, 437, 438 Stiffness constant of the joint, 448 Stochastic methods, 17 Stock sizes, 13 Straight-bevel gears. See also Bevel and worm gears; Bevel gears analysis of, 808–811 description of, 695–696, 792 mesh design for, 811–814 standard tooth proportions for, 702 Straightness control, 992 Straight roller bearings, 578 Straight-tooth bevel gears, 702 Strain elastic, 109–110 shear, 110 true, 44–47 Strain energy, 188–190 Strain hardening, 44 Strain-life method, 294, 299–301 Strain-life relation, 301 Strain-strengthening equation, 53 Strain-strengthening exponent, 53 Strength nature of, 48–50 static loading and, 242 (See also Static loading, failures resulting from) Strength equations (AGMA), 752–757

1094      Mechanical Engineering Design

Stress-concentration factor (SCF) application of elastic, 267 estimation of, 386–387 fatigue, 320, 323–327, 456, 500, 748, 758 gear teeth and, 748 geometric, 133 static loading and, 133–135, 245, 246 theoretical, 133 Stress concentrations estimation of, 384–390 load and stress analysis and, 132–135 mesh and, 965–966 notch sensitivity and, 320–325 shafts and, 381 static loading and, 245–246 Stress-correction factor, 758 Stress-cycle factor, 768–769, 801 Stress distributions, adhesives and, 510–513 Stress equations (AGMA), 751–752 Stresses/stress analysis bearing, 464 bending, 111–122, 141–145, 497– 499, 559–560, 740, 797 Cartesian stress components and, 101–102 compressive, 101 contact, 145–149 elastic strain and, 109–110 fluctuating, 296, 325–333, 342 in helical springs, 526–527 hoop, 136 in interference fits, 409–411 Mohr's circle for plane, 94, 102–109 nature of, 16, 101, 148–149 nominal, 133 normal, 101, 111–116 press and shrink fits and, 139–140 in pressurized cylinders, 135–137 principal, 103 radial, 135–137 residual, 317, 487 in rotating rings, 137–139 shaft design for, 380–390, 948 shear, 47, 101, 116–122, 124 temperature effects on, 140–141 tensile, 101 three-dimensional, 108–109 torsion and, 123–132, 492–496 true, 44–47 uniformly distributed, 110–111

Stress intensity factor, 268–270 Stress intensity modification factor, 270 Stress-life method, 294, 302–304, 359–362 Stress-life (S-N) diagram description of, 303–304 high-cycle, 307–308 mean stress and, 327 for steels, 304–308 Stress numbers, 751 Stress raisers, 132 Stress ratio, 326 Stress relieving, 70 Stress-strain diagram, 43–45, 50 of ductile materials, 51–52 for hot-rolled and cold-drawn steel, 68 stable cycle, 58–59 strain rate in, 62–63 Stress yield envelope, 248 Strict liability concept, 15 Structural adhesives, 509 Structural instabilities (buckling), 217–218 Struts, 215–217 Studs, 436–437 Sun gear, 708 Superposition, beam deflections by, 180–182, 202, 203 Surface compression stress, 749 Surface condition factor, 764 Surface elements, 959, 960 Surface endurance shear, 357 Surface endurance strength, 358 Surface factor (endurance limit), 309–3311 Surface fatigue strength, 356–359 Surface loads, 966–967 Surface-strength geometry factor, 758–761 Symmetry control, 1000–1001

T

Tandem mounting (DT), 611 Tangential stress, 101, 135–137, 410, 430 Tapered fits, 379 Tapered roller bearings advantages of, 579 components of, 596–597 nomenclature for, 596–597 radial load on, 600 selection of, 596–604 single-row straight bore, 598–599

Taper pins, 401, 402 Tearing mode, 268, 269 Temperature effects on impact values, 62 load and stress analysis and, 140–141 on materials, 63–64 Temperature factor, 314–316, 758, 770, 803 Temperature rise boundary-lubricated bearings and, 675 brakes and, 860–863 clutches and, 830, 860–863 lubricant, 647–649, 655, 675 Tempered martensite, 70 Tempering, 70 Tensile strength, 44, 45, 305 Tensile stress, 101, 248, 317 Tensile tests application of, 42, 54–55 stress-strain relationships from, 42–47 of threaded rods, 423 Tension, 175 loose-side, 886 tight-side, 886 Tension joints external load and, 446–448 fatigue loading of, 456–463 statically loaded, 452–456 Tension-loaded bolted joint, 436 Tension-test specimen, 42–43 Theoretical stress-concentration factor, 133 Thermal loading, 966 Thermal stresses, 970–972 Thermoplastics, 78, 79 Thermoset, 78, 79 Thin-film lubrication, 670–671 Threaded fasteners, 434–436 Threads, screw, 422–426. See also Screws Three-dimensional printing, 996 Three-dimensional stress, 108–109 3-D truss element, 959, 961 Three-parameter Weibull distribution, 582 Thrust load, 577, 697 Thrust loading/radial loading combined, 585–590 Tight-side tension, 886 Timing belts, 882, 884–885, 908–909 Timken Company, 579, 580, 597–601 Tin bronze, 77

Index     1095

Titanium, 76–77 Tolerances/tolerancing. See also Geometric Dimensioning and Tolerancing (GD&T) bilateral, 27 choice of, 28 cost considerations, 13–14 defined, 27 modifiers and symbols, 982, 983 position letters, 407 shaft, 407 tolerance stack-up, 29–31 unilateral, 27 Tolerance zones, 989–990, 1012 Toothed wheels, 882 Tooth systems, 701–703 Tooth thickness, 684 Top land, 684 Topp, L. J., 957 Torque, 448–451, 938 Torque capacity, 409–411 Torque transmission, 377–379 Torque-twist diagram, 47 Torque vector, 123–125 Torque wrench, 449 Torsion closed thin-walled tubes in, 129–131 function of, 123–132 open thin-walled sections in, 131–132 stresses in welded joints in, 492–496 Torsional loading, 175 Torsional strengths, 47 Torsional yield strength, 47 Torsion springs, helical coil, 557–564 Torsion tests, stress-strain relationships from, 47 Total runout, 1001–1002 Toughness, 48 Tower, Beauchamp, 634–635 Train value, 704–705 Transmission accuracy umber, 798 Transmission error, 763 Transmitted load, 711 Transverse circular pitch, 697, 700 Transverse shear stress, 117–122 Tredgold's approximation, 696 Tresca theory. See Maximum-shearstress theory (MSS) True fracture strain, 47 True fracture strength, 47 True strain, 44–47 True stress, 44–47 True stress-strain diagram, 45 Trumpler, P. R., 639–640 Truss element, 961

Tungsten, 73 Turner, M. J., 957 Turn-of-the-nut method, 449 Two-bearing mountings, 610, 611 Two-plane bending, 114–115

Vanadium, 73 Variable loading, 590–593. See also Fatigue failure from variable loading V belts, 882, 884, 900–908 Velocity journal translational, 664–665 pitch-line, 686, 711, 712 relative, 859 Velocity factor, 744–745, 750 Vibration analysis, 973–974 Virtual number of teeth, 696, 698 Viscosity, 625–627, 640, 649 Viscosity-temperature trends, 627, 628 Volkersen, O., 511, 513 Volute springs, 565 von Mises, R., 250 von Mises stresses, 250–251, 275, 314, 348, 362, 380, 382, 431, 490, 965

Welded joints fatigue loading of, 505–507 strength of, 499–501 stresses in bending, 497–499 stresses in torsion, 492–496 Welds/welding butt and fillet, 487–494 resistance, 507–508 symbols for, 486–488 Wheel, 815 White cast iron, 74 Whole depth h, 684 Width of space, 684 Wileman, J., 440 Wire diameter, 526 Wire rope selection of, 919–921 types of, 917–919 Wire springs, 526. See also Springs Wirsching, P. H., 317 Wöhler, Albert, 286, 293, 295 Wöhler curve, 303 Wolford, J. C., 543 Woodruff key, 403, 404 Work hardening, 44. See also Cold working Worm gears. See also Bevel and worm gears; Gears AGMA equation factors for, 814–817 analysis of, 818–822 description of, 682, 683 force analysis and, 719–724 mesh design for, 822–826 nomenclature of, 700 pitch diameter of, 700–701 pressure angle and tooth depth for, 703 Worms, 682 Worm wheel, 815 Woven-asbestos lining, 864 Woven-cotton lining, 864 Wrought alloys, 76

W

Y

U

Ultimate strength, 44 Ultimate tensile strength, 45 Uncertainty, 16–17 Undercutting, 691 Unified Numbering System for Metals and Alloys (UNS), 64–65, 499 Unified thread series, 423, 425 Uniform transverse line loads, 966 Unilateral tolerance, 27 Unit step function, 98, 99 UNS. See Unified Numbering System for Metals and Alloys (UNS) Unstable equilibrium, 208 Unstable lubrication, 633

V

Wahl, A. M., 530, 559 Wahl factor, 545 Walker fatigue-failure criterion, 335, 336, 343, 344 Washers, 434 Wave equation, 542 Wear, 749 Wear factor, 357 Weibull distribution, 21, 576, 582, 584 Weibull probability density function, 582 Weld bonding, 515

Yellow brass, 77 Yield design equation, 275 Yield point, 43, 53 Yield strength, 44, 56 Young's modulus, 43, 80, 81, 83, 84, 86, 109, 440

Z

Zerol bevel gear, 792 Zimmerli, F. P., 543, 544, 547, 557, 562 Zinc, brass with, 77–78

Source: https://dokumen.pub/shigleys-mechanical-engineering-design-11nbsped-9780073398211-0073398217-9781260407648-1260407640-2018023098.html

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