For courses in Machine Design. An integrated, casebased approach to machine design Machine Design: An Integrated Appr
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Table of contents :
Frontmatter
Preface
Video Contents
Part I. Fundamentals
Chapter 1. Introduction to Design
1.1 Design
Machine Design
1.2 A Design Process
1.3 Problem Formulation and Calculation
Definition Stage
Preliminary Design Stage
Detailed Design Stage
Documentation Stage
1.4 The Engineering Model
Estimation and FirstOrder Analysis
The Engineering Sketch
1.5 ComputerAided Design and Engineering
ComputerAided Design (CAD)
ComputerAided Engineering (CAE)
Computational Accuracy
1.6 The Engineering Report
1.7 Factors of Safety and Design Codes
Factor of Safety
Choosing a Safety Factor
Design and Safety Codes
1.8 Statistical Considerations
1.9 Units
1.10 Summary
1.11 References
1.12 Web References
1.13 Bibliography
1.14 Problems
Chapter 2. Materials and Processes
2.0 Introduction
2.1 MaterialProperty Definitions
The Tensile Test
Ductility and Brittleness
The Compression Test
The Bending Test
The Torsion Test
Fatigue Strength and Endurance Limit
Impact Resistance
Fracture Toughness
Creep and Temperature Effects
2.2 The Statistical Nature of Material Properties
2.3 Homogeneity and Isotropy
2.4 Hardness
Heat Treatment
Surface (Case) Hardening
Heat Treating Nonferrous Materials
Mechanical Forming and Hardening
2.5 Coatings and Surface Treatments
Galvanic Action
Electroplating
Electroless Plating
Anodizing
PlasmaSprayed Coatings
Chemical Coatings
2.6 General Properties of Metals
Cast Iron
Cast Steels
Wrought Steels
Steel Numbering Systems
Aluminum
Titanium
Magnesium
Copper Alloys
2.7 General Properties of Nonmetals
Polymers
Ceramics
Composites
2.8 Selecting Materials
2.9 Summary
2.10 References
2.11 Web References
2.12 Bibliography
2.13 Problems
Chapter 3. Kinematics and Load Determination
3.0 Introduction
3.1 Degree of Freedom
3.2 Mechanisms
3.3 Calculating Degree of Freedom (Mobility)
3.4 Common 1DOF Mechanisms
Fourbar Linkage and the Grashof Condition
Sixbar Linkage
Cam and Follower
3.5 Analyzing Linkage Motion
Types of Motion
Complex Numbers as Vectors
The Vector Loop Equation
3.6 Analyzing the Fourbar Linkage
Solving for Position in the Fourbar Linkage
Solving for Velocity in the Fourbar Linkage
Angular Velocity Ratio and Mechanical Advantage
Solving for Acceleration in the Fourbar Linkage
3.7 Analyzing the Fourbar CrankSlider
Solving for Position in the Fourbar CrankSlider
Solving for Velocity in the Fourbar CrankSlider
Solving for Acceleration in the Fourbar CrankSlider
Other Linkages
3.8 Cam Design and Analysis
The Timing Diagram
The svaj Diagram
Polynomials for the DoubleDwell Case
Polynomials for the SingleDwell Case
Pressure Angle
Radius of Curvature
3.9 Loading Classes For Force Analysis
3.10 Freebody Diagrams
3.11 Load Analysis
ThreeDimensional Analysis
TwoDimensional Analysis
Static Load Analysis
3.12 TwoDimensional, Static Loading Case Studies
3.13 ThreeDimensional, Static Loading Case Study
3.14 Dynamic Loading Case Study
3.15 Vibration Loading
Natural Frequency
Dynamic Forces
3.16 Impact Loading
Energy Method
3.17 Beam Loading
Shear and Moment
Singularity Functions
Superposition
3.18 Summary
3.19 References
3.20 Web References
3.21 Bibliography
3.22 Problems
Chapter 4. Stress, Strain, and Deflection
4.0 Introduction
4.1 Stress
4.2 Strain
4.3 Principal Stresses
4.4 Plane Stress and Plane Strain
Plane Stress
Plane Strain
4.5 Mohr’s Circles
4.6 Applied Versus Principal Stresses
4.7 Axial Tension
4.8 Direct Shear Stress, Bearing Stress, and Tearout
Direct Shear
Direct Bearing
Tearout Failure
4.9 Beams and Bending Stresses
Beams in Pure Bending
Shear Due to Transverse Loading
4.10 Deflection in Beams
Deflection by Singularity Functions
Statically Indeterminate Beams
4.11 Castigliano’s Method
Deflection by Castigliano’s Method
Finding Redundant Reactions with Castigliano’s Method
4.12 Torsion
4.13 Combined Stresses
4.14 Spring Rates
4.15 Stress Concentration
Stress Concentration Under Static Loading
Stress Concentration Under Dynamic Loading
Determining Geometric StressConcentration Factors
Designing to Avoid Stress Concentrations
4.16 Axial Compression  Columns
Slenderness Ratio
Short Columns
Long Columns
End Conditions
Intermediate Columns
4.17 Stresses in Cylinders
ThickWalled Cylinders
ThinWalled Cylinders
4.18 Case Studies in Static Stress and Deflection Analysis
4.19 Summary
4.20 References
4.21 Bibliography
4.22 Problems
Chapter 5. Static Failure Theories
5.0 Introduction
5.1 Failure of Ductile Materials Under Static Loading
The von MisesHencky or DistortionEnergy Theory
The Maximum ShearStress Theory
The Maximum NormalStress Theory
Comparison of Experimental Data with Failure Theories
5.2 Failure of Brittle Materials Under Static Loading
Even and Uneven Materials
The CoulombMohr Theory
The ModifiedMohr Theory
5.3 Fracture Mechanics
FractureMechanics Theory
Fracture Toughness Kc
5.4 Using The Static Loading Failure Theories
5.5 Case Studies in Static Failure Analysis
5.6 Summary
5.7 References
5.8 Bibliography
5.9 Problems
Chapter 6. Fatigue Failure Theories
6.0 Introduction
History of Fatigue Failure
6.1 Mechanism of Fatigue Failure
Crack Initiation Stage
Crack Propagation Stage
Fracture
6.2 FatigueFailure Models
Fatigue Regimes
The StressLife Approach
The StrainLife Approach
The LEFM Approach
6.3 MachineDesign Considerations
6.4 Fatigue Loads
Rotating Machinery Loading
Service Equipment Loading
6.5 Measuring Fatigue Failure Criteria
Fully Reversed Stresses
Combined Mean and Alternating Stress
FractureMechanics Criteria
Testing Actual Assemblies
6.6 Estimating Fatigue Failure Criteria
Estimating the Theoretical Fatigue Strength S or Endurance Limit S
Correction Factors—Theoretical Fatigue Strength or Endurance Limit
Corrected Fatigue Strength S or Corrected Endurance Limit S
Creating Estimated SN Diagrams
6.7 Notches and Stress Concentrations
Notch Sensitivity
6.8 Residual Stresses
6.9 Designing for HighCycle Fatigue
6.10 Designing for Fully Reversed Uniaxial Stresses
Design Steps for Fully Reversed Stresses with Uniaxial Loading
6.11 Designing for Fluctuating Uniaxial Stresses
Creating the ModifiedGoodman Diagram
Applying StressConcentration Effects with Fluctuating Stresses
Determining the Safety Factor with Fluctuating Stresses
Design Steps for Fluctuating Stresses
6.12 Designing for Multiaxial Stresses in Fatigue
Frequency and Phase Relationships
Fully Reversed Simple Multiaxial Stresses
Fluctuating Simple Multiaxial Stresses
Complex Multiaxial Stresses
6.13 A General Approach to HighCycle Fatigue Design
6.14 A Case Study in Fatigue Design
6.15 Summary
6.16 References
6.17 Bibliography
6.18 Problems
7.0 Introduction
Chapter 7. Surface Failure
7.1 Surface Geometry
7.2 Mating Surfaces
7.3 Friction
Effect of Roughness on Friction
Effect of Velocity on Friction
Rolling Friction
Effect of Lubricant on Friction
7.4 Adhesive Wear
The AdhesiveWear Coefficient
7.5 Abrasive Wear
Abrasive Materials
AbrasionResistant Materials
7.6 Corrosion Wear
Corrosion Fatigue
Fretting Corrosion
7.7 Surface Fatigue
7.8 Spherical Contact
Contact Pressure and Contact Patch in Spherical Contact
Static Stress Distributions in Spherical Contact
7.9 Cylindrical Contact
Contact Pressure and Contact Patch in Parallel Cylindrical Contact
Static Stress Distributions in Parallel Cylindrical Contact
7.10 General Contact
Contact Pressure and Contact Patch in General Contact
Stress Distributions in General Contact
7.11 Dynamic Contact Stresses
Effect of a Sliding Component on Contact Stresses
7.12 Surface Fatigue Failure Models—Dynamic Contact
7.13 Surface Fatigue Strength
7.14 Summary
7.15 References
7.16 Problems
Chapter 8. Finite element Analysis
8.0 Introduction
Stress and Strain Computation
8.1 Finite Element Method
8.2 Element Types
Element Dimension and Degree of Freedom (DOF)
Element Order
HElements Versus PElements
Element Aspect Ratio
8.3 Meshing
Mesh Density
Mesh Refinement
Convergence
8.4 Boundary Conditions
8.5 Applying Loads
8.6 Testing the Model (Verification)
8.7 Modal Analysis
8.8 Case Studies
8.9 Summary
8.10 References
8.11 Bibliography
8.12 Web Resources
8.13 Problems
Part II. Machine Design
Chapter 9. Design Case Studies
9.0 Introduction
9.1 Case Study 8—A Portable Air Compressor
9.2 Case Study 9—A HayBale Lifter
9.3 Case Study 10—A CamTesting Machine
9.4 Summary
9.5 References
9.6 Design Projects
Chapter 10. Shafts, Keys, and Couplings
10.0 Introduction
10.1 Shaft Loads
10.2 Attachments and Stress Concentrations
10.3 Shaft Materials
10.4 Shaft Power
10.5 Shaft Loads
10.6 Shaft Stresses
10.7 Shaft Failure in Combined Loading
10.8 Shaft Design
General Considerations
Design for Fully Reversed Bending and Steady Torsion
Design for Fluctuating Bending and Fluctuating Torsion
10.9 Shaft Deflection
Shafts as Beams
Shafts as Torsion Bars
10.10 Keys and Keyways
Parallel Keys
Tapered Keys
Woodruff Keys
Stresses in Keys
Key Materials
Key Design
Stress Concentrations in Keyways
10.11 Splines
10.12 Interference Fits
Stresses in Interference Fits
Stress Concentration in Interference Fits
Fretting Corrosion
10.13 Flywheel Design
Energy Variation in a Rotating System
Determining the Flywheel Inertia
Stresses in Flywheels
Failure Criteria
10.14 Critical Speeds of Shafts
Lateral Vibration of Shafts and Beams—Rayleigh’s Method
Shaft Whirl
Torsional Vibration
Two Disks on a Common Shaft
Multiple Disks on a Common Shaft
Controlling Torsional Vibrations
10.15 Couplings
Rigid Couplings
Compliant Couplings
10.16 Case Study 8B
Designing Driveshafts for a Portable Air Compressor
10.17 Summary
10.18 References
10.19 Problems
Chapter 11. Bearings and Lubrication
11.0 Introduction
A Caveat
11.1 Lubricants
11.2 Viscosity
11.3 Types of Lubrication
FullFilm Lubrication
Boundary Lubrication
11.4 Material Combinations in Sliding Bearings
11.5 Hydrodynamic Lubrication Theory
Petroff’s Equation for NoLoad Torque
Reynolds’ Equation for Eccentric Journal Bearings
Torque and Power Losses in Journal Bearings
11.6 Design of Hydrodynamic Bearings
Design Load Factor—The Ocvirk Number
Design Procedures
11.7 Nonconforming Contacts
11.8 Rollingelement bearings
Comparison of Rolling and Sliding Bearings
Types of RollingElement Bearings
11.9 Failure of Rollingelement bearings
11.10 Selection of Rollingelement bearings
Basic Dynamic Load Rating C
Modified Bearing Life Rating
Basic Static Load Rating C0
Combined Radial and Thrust Loads
Calculation Procedures
11.11 Bearing Mounting Details
11.12 Special Bearings
11.13 Case Study 10B
11.14 Summary
Important Equations Used in This Chapter
11.15 References
11.16 Problems
Chapter 12. Spur Gears
12.0 Introduction
12.1 Gear Tooth Theory
The Fundamental Law of Gearing
The Involute Tooth Form
Pressure Angle
Gear Mesh Geometry
Rack and Pinion
Changing Center Distance
Backlash
Relative Tooth Motion
12.2 Gear Tooth Nomenclature
12.3 Interference and Undercutting
UnequalAddendum Tooth Forms
12.4 Contact Ratio
12.5 Gear Trains
Simple Gear Trains
Compound Gear Trains
Reverted Compound Trains
Epicyclic or Planetary Gear Trains
12.6 Gear Manufacturing
Forming Gear Teeth
Machining
Roughing Processes
Finishing Processes
Gear Quality
12.7 Loading on Spur Gears
12.8 Stresses in Spur Gears
Bending Stresses
Surface Stresses
12.9 Gear Materials
Material Strengths
BendingFatigue Strengths for Gear Materials
SurfaceFatigue Strengths for Gear Materials
12.10 Lubrication of Gearing
12.11 Design of Spur Gears
12.12 Case Study 8C
12.13 Summary
12.14 References
12.15 Problems
Chapter 13. Helical, Bevel, and Worm Gears
13.0 Introduction
13.1 Helical Gears
Helical Gear Geometry
HelicalGear Forces
Virtual Number of Teeth
Contact Ratios
Stresses in Helical Gears
13.2 Bevel Gears
BevelGear Geometry and Nomenclature
BevelGear Mounting
Forces on Bevel Gears
Stresses in Bevel Gears
13.3 Wormsets
Materials for Wormsets
Lubrication in Wormsets
Forces in Wormsets
Wormset Geometry
Rating Methods
A Design Procedure for Wormsets
13.4 Case Study 9B
13.5 Summary
13.6 References
13.7 Problems
Chapter 14. Spring Design
14.0 Introduction
14.1 Spring Rate
14.2 Spring Configurations
14.3 Spring Materials
Spring Wire
Flat Spring Stock
14.4 Helical Compression Springs
Spring Lengths
End Details
Active Coils
Spring Index
Spring Deflection
Spring Rate
Stresses in Helical Compression Spring Coils
Helical Coil Springs of Nonround Wire
Residual Stresses
Buckling of Compression Springs
CompressionSpring Surge
Allowable Strengths for Compression Springs
The TorsionalShear SN Diagram for Spring Wire
The ModifiedGoodman Diagram for Spring Wire
14.5 Designing Helical Compression Springs for Static Loading
14.6 Designing Helical Compression Springs for Fatigue Loading
14.7 Helical Extension Springs
Active Coils in Extension Springs
Spring Rate of Extension Springs
Spring Index of Extension Springs
Coil Preload in Extension Springs
Deflection of Extension Springs
Coil Stresses in Extension Springs
End Stresses in Extension Springs
Surging in Extension Springs
Material Strengths for Extension Springs
Design of Helical Extension Springs
14.8 Helical Torsion Springs
Terminology for Torsion Springs
Number of Coils in Torsion Springs
Deflection of Torsion Springs
Spring Rate of Torsion Springs
Coil Closure
Coil Stresses in Torsion Springs
Material Parameters for Torsion Springs
Safety Factors for Torsion Springs
Designing Helical Torsion Springs
14.9 Belleville Spring Washers
LoadDeflection Function for Belleville Washers
Stresses in Belleville Washers
Static Loading of Belleville Washers
Dynamic Loading
Stacking Springs
Designing Belleville Springs
14.10 Case Study 10C
14.11 Summary
14.12 References
14.13 Problems
Chapter 15. Screws and Fasteners
15.0 Introduction
15.1 Standard Thread Forms
Tensile Stress Area
Standard Thread Dimensions
15.2 Power Screws
Square, Acme, and Buttress Threads
Power Screw Application
Power Screw Force and Torque Analysis
Friction Coefficients
SelfLocking and BackDriving of Power Screws
Screw Efficiency
Ball Screws
15.3 Stresses in Threads
Axial Stress
Shear Stress
Torsional Stress
15.4 Types of Screw Fasteners
Classification by Intended Use
Classification by Thread Type
Classification by Head Style
Nuts and Washers
15.5 Manufacturing Fasteners
15.6 Strengths of Standard Bolts and Machine Screws
15.7 Preloaded Fasteners in Tension
Preloaded Bolts Under Static Loading
Preloaded Bolts Under Dynamic Loading
15.8 Determining the Joint Stiffness Factor
Joints With Two Plates of the Same Material
Joints With Two Plates of Different Materials
Gasketed Joints
15.9 Controlling Preload
The TurnoftheNut Method
TorqueLimited Fasteners
LoadIndicating Washers
Torsional Stress Due to Torquing of Bolts
15.10 Fasteners in Shear
Dowel Pins
Centroids of Fastener Groups
Determining Shear Loads on Fasteners
15.11 Case Study 8D
15.12 Summary
15.13 References
15.14 Bibliography
15.15 Problems
Chapter 16. Weldments
16.0 Introduction
16.1 Welding Processes
Types of Welding in Common Use
Why Should a Designer Be Concerned with the Welding Process?
16.2 Weld Joints and Weld Types
Joint Preparation
Weld Specification
16.3 Principles of Weldment Design
16.4 Static Loading of Welds
16.5 Static Strength of Welds
Residual Stresses in Welds
Direction of Loading
Allowable Shear Stress for Statically Loaded Fillet and PJP Welds
16.6 Dynamic Loading of Welds
Effect of Mean Stress on Weldment Fatigue Strength
Are Correction Factors Needed For Weldment Fatigue Strength?
Effect of Weldment Configuration on Fatigue Strength
Is There an Endurance Limit for Weldments?
Fatigue Failure in Compression Loading?
16.7 Treating a Weld as a Line
16.8 Eccentrically Loaded Weld Patterns
16.9 Design Considerations for Weldments in Machines
16.10 Summary
16.11 References
16.12 Problems
Chapter 17. Clutches and Brakes
17.0 Introduction
17.1 Types of Brakes and Clutches
17.2 Clutch/Brake Selection and Specification
17.3 Clutch and Brake Materials
17.4 Disk Clutches
Uniform Pressure
Uniform Wear
17.5 Disk Brakes
17.6 Drum Brakes
ShortShoe External Drum Brakes
LongShoe External Drum Brakes
LongShoe Internal Drum Brakes
17.7 Summary
17.8 References
17.9 Bibliography
17.10 Problems
Appendices
A. Material Properties
B. Beam Tables
C. StressConcentration Factors
D. Answers to Selected Problems
Index
Downloads Index
MACHINE DESIGN An Integrated Approach Sixth Edition
Robert L. Norton P.E. Milton P. Higgins II Distinguished Professor Emeritus Worcester Polytechnic Institute Worcester, Massachusetts
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Senior Vice President Courseware Portfolio Management: Engineering, Computer Science, Mathematics, Statistics, and Global Editions: Marcia J. Horton Director, Portfolio Management: Engineering, Computer Science, Mathematics, Statistics, and Global Editions: Julian Partridge Team Lead: Engineering and Computer Science Norrin Dias Portfolio Management Assistant: Emily Egan Managing Producer, Engineering, Computer Science, and Mathematics: Scott Disanno Senior Content Producer: Erin Ault Project Manager: Louise Capulli Manager, Rights and Permissions: Ben Ferrini Operations Specialist: Maura ZaldivarGarcia Inventory Manager: Bruce Boundy Product Marketing Manager: Yvonne Vannatta Field Marketing Manager: Demetrius Hall Marketing Assistant: Jon Bryant Cover Designer: Robert L. Norton Cover Printer: Phoenix Color/Hagerstown Printer/Binder: Lake Side Communications, Inc. (LSC) Copyright © 2020, 2014, 2011, 2006, 2000, 1998 Pearson Education, Inc., Hoboken, NJ 07030. All rights reserved. Manufactured in the United States of America. This publication is protected by copyright, and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise. For information regarding permissions, request forms and the appropriate contacts within the Pearson Education Global Rights & Permissions department, please visit www.pearsoned.com/permissions/. Many of the designations by manufacturers and seller to distinguish their products are claimed as trademarks. Where those designations appear in this book, and the publisher was aware of a trademark claim, the designations have been printed in initial caps or all caps. The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs. Mathcad is a registered trademark of PTC Inc. MATLAB™ is a registered trademark of The MathWorks, Inc. Microsoft is a registered trademark or trademark of Microsoft Corporation in the United States and/or other countries. TK Solver is a trademark of UTS Corporation in the U.S. and/or other countries. The cover photograph by the author is of Tom Norton, fivetime New England Trail Riders Association (NETRA) Hare Scrambles Champion (and the author’s son). The book, including all art and equations, was written and electronically typeset by the author using InDesign and Mathtype software. All nonphotographic figures were drawn in Freehand and Illustrator by the author. Library of Congress CataloginginPublication Data Norton, Robert L. Machine Design An Integrated Approach 6ed / Robert L. Norton p. cm. Includes bibliographical references and index. ISBN: 0135184231 (hard cover) 1. Machine design. I. Title TJ230.N64 2020 621.8’15dc22 9719522 CIP Data on file 1 19
ISBN10: 0135184231 ISBN13: 9780135184233
ABOUT THE AUTHOR Robert L. Norton earned undergraduate degrees in both mechanical engineering and industrial technology at Northeastern University, an MS in engineering design at Tufts University, and was awarded a Doctor of Engineering (h.c.) by Worcester Polytechnic Institute (WPI). He is a registered professional engineer in Massachusetts and Florida. He has over 50 years experience in engineering design and manufacturing and over 40 years experience teaching mechanical engineering, engineering design, computer science, and related subjects at Northeastern University, Tufts University, and WPI. Norton has been on the faculty of WPI since 1981, and is currently the Milton P. Higgins II Distinguished Professor Emeritus. He is also the founder and president of Norton Associates Engineering Consultants since 1970. At Polaroid Corporation for 10 years, he designed cameras, related mechanisms, and highspeed automated machinery. He spent three years at Jet Spray Cooler Inc., designing foodhandling machinery and products. For five years he helped develop artificialheart and noninvasive assistedcirculation devices at the Tufts New England Medical Center and Boston City Hospital. Since leaving industry to join academia in 1974, he has continued as an independent consultant on engineering projects ranging from disposable medical products to highspeed production machinery. He holds thirteen U.S. patents. He is the author of numerous technical papers and journal articles covering kinematics, dynamics of machinery, cam design and manufacturing, computers in education, engineering education, and of the texts Design of Machinery, Kinematics and Dynamics of Machinery, Machine Design: An Integrated Approach, and the Cam Design and Manufacturing Handbook. He is a Life Fellow of the American Society of Mechanical Engineers and a past member of the Society of Automotive Engineers and the American Society for Engineering Education. In 2007, he was selected as a U.S. Professor of the Year by the Council for the Advancement and Support of Education (CASE) and the Carnegie Foundation for the Advancement of Teaching, who jointly present the only national awards for highereducation teaching excellence given in the United States of America. WHAT USERS SAY ABOUT THE BOOK Your text is the best of all the texts I have used—the balance of fundamentals and practice is especially important, and you have achieved that with aplomb! —Professor John P. H. Steele, Colorado School of Mines This book is one of the bestwritten on any engineering subject that I have come across. We switched to the book because we felt that the students would get a lot out of reading the text. —Professor Ed Howard, Milwaukee School of Engineering (The) writing is clear and meaningful, (and the) text is much more accessible to students because of the practical nature of (its) case studies and problem sets. —Professor Douglas Walcerz, York College of Pennsylvania I’m doing my machine design homework now, and . . . your book is the single best text I have ever used in my educational career. It is straightforward, easy to understand and has good examples. —Josh, student, Colorado School of Mines
This book is dedicated to the memory of: Donald N. Zwiep 1924–2012 Provost, Department Head, and Professor Emeritus Worcester Polytechnic Institute A gentleman and a leader, without whose faith and foresight, this book would never have been written.
PREFACE Introduction This text is intended for the Design of Machine Elements courses typically given in the junior year of most mechanical engineering curricula. The usual prerequisites are a first course in Statics and Dynamics, and one in Strength of Materials. The purpose of this book is to present the subject matter in an uptodate manner with a strong design emphasis. The level is aimed at juniorsenior mechanical engineering students. A primary goal was to write a text that is very easy to read and that students will enjoy reading despite the inherent dryness of the subject matter. This textbook is designed to be an improvement over others currently available and to provide methods and techniques that take full advantage of computeraided analysis. It emphasizes design and synthesis as well as analysis. Example problems, case studies, and solution techniques are spelled out in detail and are selfcontained. All the illustrations are done in two colors. Short problems are provided in each chapter and, where appropriate, longer unstructured designproject assignments are given. The book is independent of any particular computer program. Computer files for the solution of all the examples and case studies written in several different languages (Mathcad, MATLAB, Excel, and TK Solver) are provided on the book’s website at http://www.pearsonhighered.com/norton. Several other programs written by the author are available from the author. These include a matrix solver (MATRIX.exe). An index of the website’s content is on the website. While this book attempts to be thorough and complete on the engineeringmechanics topics of failure theory and analysis, it also emphasizes the synthesis and design aspects of the subject to a greater degree than most other texts in print on this subject. It points out the commonality of the analytical approaches needed to design a wide variety of elements and emphasizes the use of computeraided engineering as an approach to the design and analysis of these classes of problems. The author’s approach to this course is based on over 50 years of practical experience in mechanical engineering design, both in industry and as a consultant to industry. He has taught mechanical engineering design at the university level for 40 of those years as well.
What’s in the Sixth Edition? • Twentyone Master Lecture videos on the topics of most chapters are provided on the website. These are taken from the author’s live lectures to classes at WPI. The student can watch these videos to review and enhance their understanding of the book’s topics. • Eight short videos are provided on the website in which the author demonstrates various principles of stress analysis and shows examples of common machine parts such as springs, gears, and bearings. • Six videos that show real machinery in operation are also provided on the website. • Over 80 problems are added with many being in SI units. • In addition to the printed version of the text, digital ebook versions are also available. These have hotlinks to all the videos and to the downloadable content provided. There
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MACHINE DESIGN

An Integrated Approach 
Sixth Edition
are 37 videos. All of these are marked in the print version as well, with their URLs provided, and they can be downloaded by printbook users. A Video Contents is provided, and all other downloadable items are listed in the Downloads Index. • The authorwritten programs that come with the book have been completely rewritten to improve their interface and usability, and they are now compatible with the latest operating systems and computers. Programs Linkages and Dynacam have been completely rewritten and are much improved. Program Matrix is updated. These computer programs undergo frequent revision to add features and enhancements. Once installed, the programs can be updated from the Start Menu. Users should occasionally check for updates. • All the downloadable files are accessible to digitalbook users through the publisher’s Mastering website via links in the digital book. Any instructor or student who uses the print book may register on my website, http://www.designofmachinery.com , either as a student or instructor, and I will send them a password to access a protected site where they can download the latest versions of my computer programs, Linkages, Dynacam, and Matrix, all videos, and all files listed in the Downloads Index. Note that I personally review each of these requests for access and approve only those that are filled out completely and correctly according to the provided instructions. I require complete information and only accept university email addresses.
Philosophy This is often the first course that mechanical engineering students see that presents them with design challenges rather than setpiece problems. Nevertheless, the type of design addressed in this course is that of detailed design, which is only one part of the entire designprocess spectrum. In detailed design, the general concept, application, and even general shape of the required device are typically known at the outset. We are not trying to invent a new device so much as define the shape, size, and material of a particular machine element such that it will not fail under the loading and environmental conditions expected in service. The traditional approach to the teaching of the Elements course has been to emphasize the design of individual machine parts, or elements, such as gears, springs, shafts, etc. One criticism that is sometimes directed at the Elements course (or textbook) is that it can easily become a “cookbook” collection of disparate topics that does not prepare the student to solve other types of problems not found in the recipes presented. There is a risk of this happening. It is relatively easy for the instructor (or author) to allow the course (or text) to degenerate into the mode “Well, it’s Tuesday, let’s design springs—on Friday, we’ll do gears.” If this happens, it may do the student a disservice because it doesn’t necessarily develop a fundamental understanding of the practical application of the underlying theories to design problems. However, many of the machine elements typically addressed in this course provide superb examples of the underlying theory. If viewed in that light, and if presented in a general context, they can be an excellent vehicle for the development of student understanding of complex and important engineering theories. For example, the topic of preloaded bolts is a perfect vehicle to introduce the concept of prestressing used as a foil against fatigue loading. The student may never be called upon in practice to design a preloaded bolt, but he or she may well utilize the understanding of prestressing gained from the experience. The design of helical gears to withstand timevarying loads provides an excellent vehicle to develop the student’s understanding of combined stresses, Hertzian stresses, and fatigue failure. Thus the elements approach is a valid and defensible one as long as the approach taken in the text is sufficiently global. That is, it should not be allowed to degenerate into a collection of apparently unrelated exercises, but rather provide an integrated approach.
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Another area in which the author has found existing texts (and Machine Elements courses) to be deficient is the lack of connection made between the dynamics of a system and the stress analysis of that system. Typically, these texts present their machine elements with (magically) predefined forces on them. The student is then shown how to determine the stresses and deflections caused by those forces. In real machine design, the forces are not always predefined and can, in large part, be due to the accelerations of the masses of the moving parts. However, the masses cannot be accurately determined until the geometry is defined and a stress analysis done to determine the strength of the assumed part. Thus, an impasse exists that is broken only by iteration, i.e., assume a part geometry and define its geometric and mass properties, calculate the dynamic loads due in part to the material and geometry of the part. Then calculate the stresses and deflections resulting from those forces, find out it fails, redesign, and repeat.
An Integrated Approach The text is divided into two parts. The first part presents the fundamentals of stress, strain, deflection, materials properties, failure theories, fatigue phenomena, fracture mechanics, FEA, etc. These theoretical aspects are presented in similar fashion to other texts. The second part presents treatments of specific, common design elements used as examples of applications of the theory but also attempts to avoid presenting a string of disparate topics in favor of an integrated approach that ties the various topics together via case studies. Most Elements texts contain many more topics and more content than can possibly be covered in a onesemester course. Before writing the first edition of this book, a questionnaire was sent to 200 U.S. university instructors of the Elements course to solicit their opinions on the relative importance and desirability of the typical set of topics in an Elements text. With each revision to second through fifth editions, users were again surveyed to determine what should be changed or added. The responses were analyzed and used to influence the structure and content of this book in all editions. One of the strongest desires originally expressed by the respondents was for case studies that present realistic design problems. We have attempted to accomplish this goal by structuring the text around a series of ten case studies. These case studies present different aspects of the same design problem in successive chapters, for example, defining the static or dynamic loads on the device in Chapter 3, calculating the stresses due to the static loads in Chapter 4, and applying the appropriate failure theory to determine its safety factor in Chapter 5. Later chapters present more complex case studies, with more design content. The case study in Chapter 6 on fatigue design is one such example a real problem taken from the author’s consulting practice. Chapter 8 presents FEA analyses of several of these case studies and compares those results to the classical solutions done in prior chapters. The case studies provide a series of machine design projects throughout the book that contain various combinations of the elements normally dealt with in this type of text. The assemblies contain some collection of elements such as links subjected to combined axial and bending loads, column members, shafts in combined bending and torsion, gearsets under alternating loads, return springs, fasteners under fatigue loading, rolling element bearings, etc. This integrated approach has several advantages. It presents the student with a generic design problem in context rather than as a set of disparate, unrelated entities. The student can then see the interrelationships and the rationales for the design decisions that affect the individual elements. These more comprehensive case studies are in Part II of the text. The case studies in Part I are more limited in scope and directed to the engineering mechanics topics of the chapter. In addition to the case studies, each chapter has a selection of workedout examples to reinforce particular topics.
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Chapter 9, Design Case Studies, is devoted to the setup of three design case studies that are used in the following chapters to reinforce the concepts behind the design and analysis of shafts, springs, gears, fasteners, etc. Not all aspects of these design case studies are addressed as workedout examples since another purpose is to provide material for studentproject assignments. The author has used these case study topics as multiweek or termlong project assignments for groups or individual students with good success. Assigning openended project assignments serves to reinforce the design and analysis aspects of the course much better than setpiece homework assignments.
Problem Sets Most of the 967 problem sets (767, or 79%) are independent within a chapter, responding to requests by users of the first edition to decouple them. The other 21% of the problem sets are built upon in succeeding chapters. These linked problems have the same dash number in each chapter and their problem number is boldface to indicate their commonality among chapters. For example, Problem 34 asks for a static force analysis of a trailer hitch; Problem 44 requests a stress analysis of the same hitch based on the forces calculated in Problem 34; Problem 54 asks for the static safety factor for the hitch using the stresses calculated in Problem 44; Problem 64 requests a fatiguefailure analysis of the same hitch, and Problem 74 requires a surface stress analysis. The same trailer hitch is used as an FEA case study in Chapter 8. Thus, the complexity of the underlying design problem is unfolded as new topics are introduced. An instructor who wishes to use this approach can assign problems with the same dash number in succeeding chapters. If one does not want to assign an earlier problem on which a later one is based, the solution manual data from the earlier problem can be provided to the students. Instructors who do not like interlinked problems can avoid them entirely and select from the 767 problems with nonbold problem numbers that are independent within their chapters.
Text Arrangement Chapter 1 provides an introduction to the design process, problem formulation, safety factors, and units. Material properties are reviewed in Chapter 2 since even the student who has had a first course in material science or metallurgy typically has but a superficial understanding of the wide spectrum of engineering material properties needed for machine design. Chapter 3 presents a discussion of the fundamentals of kinematic linkages and cams. It also provides a review of static and dynamic loading analysis, including beam, vibration, and impact loading, and sets up a series of case studies that are used in later chapters to illustrate the stress and deflection analysis topics with some continuity. The Design of Machine Elements course, at its core, is really an intermediatelevel, applied stressanalysis course. Accordingly, a review of the fundamentals of stress and deflection analysis is presented in Chapter 4. Static failure theories are presented in detail in Chapter 5 since the students have typically not yet fully digested these concepts from their first stressanalysis course. Fracturemechanics analysis for static loads is also introduced. The Elements course is typically the student’s first exposure to fatigue analysis since most introductory stressanalysis courses deal only with statically loaded problems. Accordingly, fatiguefailure theory is presented at length in Chapter 6 with the emphasis on stresslife approaches to highcycle fatigue design, which is commonly used in the design of rotating machinery. Fracturemechanics theory is further discussed with regard to crack propagation under cyclic loading. Strainbased methods for lowcycle fatigue analysis are not presented
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but their application and purpose are introduced to the reader and bibliographic references are provided for further study. Residual stresses are also addressed. Chapter 7 presents a thorough discussion of the phenomena of wear mechanisms, surface contact stresses, and surface fatigue. Chapter 8 provides an introduction to Finite Element Analysis (FEA). Many instructors are using the machine elements course to introduce students to FEA as well as to instruct them in the techniques of machine design. The material presented in Chapter 8 is not intended as a substitute for education in FEA theory. That material is available in many other textbooks devoted to that subject and the student is urged to become familiar with FEA theory through coursework or selfstudy. Instead, Chapter 8 presents proper techniques for the application of FEA to practical machine design problems. Issues of element selection, mesh refinement, and the definition of proper boundary conditions are developed in some detail. These issues are not usually addressed in books on FEA theory. Many engineers in training today will, in their professional practice, use CAD solid modeling software and commercial finite element analysis code. It is important that they have some knowledge of the limitations and proper application of those tools. This chapter can be taken up earlier in the course if desired, especially if the students are expected to use FEA to solve assigned tasks. It is relatively independent of the other chapters. Many of various chapters’ problem assignments have Solidworks models of their geometry provided on the website. These eight chapters comprise Part I of the text and lay the analytical foundation needed for design of machine elements. They are arranged to be taken up in the order presented and build upon each other with the exception of Chapter 8 on FEA. Part II of the text presents the design of machine elements in context as parts of a whole machine. The chapters in Part II are essentially independent of one another and can be taken (or skipped) in any order that the instructor desires (except that Chapter 12 on spur gears should be studied before Chapter 13 on helical, bevel, and worm gears). It is unlikely that all topics in the book can be covered in a oneterm or one semester course. Uncovered chapters will still serve as a reference for engineers in their professional practice. Chapter 9 presents a set of design case studies to be used as assignments and as example case studies in the following chapters and also provides a set of suggested design project assignments in addition to the detailed case studies as described above. Chapter 10 investigates shaft design using the fatigueanalysis techniques developed in Chapter 6. Chapter 11 discusses fluidfilm and rollingelement bearing theory and application using the theory developed in Chapter 7. Chapter 12 gives a thorough introduction to the kinematics, design and stress analysis of spur gears using the latest AGMA recommended procedures. Chapter 13 extends gear design to helical, bevel, and worm gearing. Chapter 14 covers spring design including helical compression, extension and torsion springs, as well as a thorough treatment of Belleville springs. Chapter 15 deals with screws and fasteners including power screws and preloaded fasteners. Chapter 16 presents an uptodate treatment of the design of weldments for both static and dynamic loading. Chapter 17 presents an introduction to the design and specification of disk and drum clutches and brakes. The appendices contain materialstrength data, beam tables, and stressconcentration factors, as well as answers to selected problems.
Supplements A Solutions Manual is available to instructors from the publisher and PowerPoint slides of all figures and tables in the text are available on the publisher’s website (password protected) at: http://www.pearsonhighered.com/
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To download these resources, choose the Instructor Support tab to register as an instructor and follow instructions on the site to obtain the resources provided. Mathcad files for all the problem solutions are available with the solutions manual. This computerized approach to problem solutions has significant advantages to the instructor who can easily change any assigned problem’s data and instantly solve it. Thus, an essentially infinite supply of problem sets is available, going far beyond those defined in the text. The instructor also can easily prepare and solve exam problems by changing data in the supplied files. As errata are discovered they will be posted on the author’s personal website at: http://www.designofmachinery.com/MD/errata.html
Professors who adopt the book may register at the author’s personal website to obtain additional information relevant to the subject (syllabi, master lectures, project assignments, etc.) and the text and to download web content and updated software (password protected). Go to: http://designofmachinery.com/books/machinedesign/professorsusingourbooksmd/ Anyone who purchases the book may register at the author’s personal website to request downloads and updated software for the current edition (password protected). Go to: http://designofmachinery.com/books/machinedesign/machinedesign6thedforstudentslogin/
Acknowledgments The author expresses his sincere appreciation to all those who reviewed the first edition of the text in various stages of development including Professors J. E. Beard, Michigan Tech; J. M. Henderson, U. California, Davis; L. R. Koval, U. Missouri, Rolla; S. N. Kramer, U. Toledo; L. D. Mitchell, Virginia Polytechnic; G. R. Pennock, Purdue; D. A. Wilson, Tennessee Tech; Mr. John Lothrop; and Professor J. AriGur, Western Michigan University, who also taught from a classtest version of the book. Robert Herrmann (WPIME ‘94) provided some problems and Charles Gillis (WPIME ‘96) solved most of the problem sets for the first edition. Professors John R. Steffen of Valparaiso University, R. Jay Conant of Montana State, Norman E. Dowling of Virginia Polytechnic, and Francis E. Kennedy of Dartmouth made many useful suggestions for improvement and caught many errors. Special thanks go to Professor Hartley T. Grandin of WPI, who provided much encouragement and many good suggestions and ideas throughout the book’s gestation, and also taught from various classtest versions. Two former and the current Prentice Hall editors deserve special mention for their efforts in developing this book: Doug Humphrey, who wouldn’t take no for an answer in persuading me to write it and Bill Stenquist, who usually said yes to my requests and expertly shepherded the book through to completion in its first edition. Norrin Dias’ support has helped marshall the sixth edition into print. Since the book’s first printing in 1995, several users have kindly pointed out errors and suggested improvements. My thanks go to Professors R. Boudreau of U. Moncton, Canada, V. Glozman of Cal Poly Pomona, John Steele of Colorado School of Mines, Burford J. Furman of San Jose State University, and Michael Ward of California State University, Chico.
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Several other faculty have been kind enough to point out errors and offer constructive criticisms and suggestions for improvement in the later editions. Notable among these are: Professors Cosme Furlong of Worcester Polytechnic Institute, Joseph Rencis of University of Arkansas, Annie Ross of Universite de Moncton, Andrew Ruina of Cornell University, Douglas Walcerz of York College, and Thomas Dresner of Mountain City, CA. Dr. Duane Miller of Lincoln Electric Company provided invaluable help with Chapter 16 on weldments and reviewed several drafts. Professor Stephen Covey of St. Cloud State University, and engineers Gregory Aviza and Charles Gillis of P&G Gillette also provided valuable feedback on the weldment chapter. Professor Robert Cornwell of Seattle University reviewed the discussion in Chapter 15 of his new method for the calculation of bolted joint stiffness and his method for computing stress concentration in rectangular wire springs discussed in Chapter 14. Professors Fabio Marcelo Peña Bustos of Universidad Autónoma de Manizales, Caldas, Colombia, and Juan L. BalsevichPrieto of Universidad Católica Nuestra Señora de la Asunción, Asunción, Paraguay, were kind enough to point out errata in the Spanish translation. Special thanks are due to William Jolley of The Gillette Company who created the FEA models for the examples and reviewed Chapter 8, and to Edwin Ryan, retired Vice President of Engineering at Gillette, who provided invaluable support. Donald A. Jacques of the UTC Fuel Cells division of the United Technologies Company also reviewed Chapter 8 on Finite Element Analysis and made many useful suggestions. Professor Eben C. Cobb of Worcester Polytechnic Institute and his student Thomas Watson created the Solidworks models of many problem assignments and case studies and solved the FEA for the case studies that are on the website. Thanks are due several people who responded to surveys for the fifth edition and made many good suggestions: Steven J. Covey of St. Cloud State University, Yesh P. Singh of University of Texas at San Antonio, César Augusto Álvarez Vargas of Universidad Autonoma de Manizales, Caldas, Colombia, Ardeshir Kianercy of University of Southern California, Kenneth W. Miller of St. Cloud State University, Yannis Korkolis of University of New Hampshire, J. Alex Thomasson of Texas A&M, Timothy Dewhurst of Cedarville University, Jon Svenninggaard of VIA University College, Horsens, Denmark, John D. Landes of University of Tennessee, Peter Schuster of California Polytechnic State University, Amanuel Haque of Penn State University, J. Brian Jordon of University of Alabama, Yabin Liao of Arizona State University, Raymond K. Yee of San Jose State University, JeanMichel Dhainaut of EmbryRiddle Aeronautical University, Pal Molian of Iowa State University, and John Strenkowski of North Carolina State University. The author is greatly indebted to Thomas A. Cook, Professor Emeritus, Mercer University, who did the Solutions Manual for this book, updated the Mathcad examples, and contributed most of the new problem sets for this edition. Thanks also to Dr. Adriana Hera of Worcester Polytechnic Institute who updated the MATLAB and Excel models of all the examples and case studies and thoroughly vetted their correctness. Finally, Nancy Norton, my infinitely patient wife for the past fiftynine years, deserves renewed kudos for her unfailing support and encouragement during many summers of “book widowhood.” I could not have done it without her. Every effort has been made to eliminate errors from this text. Any that remain are the author’s responsibility. He will greatly appreciate being informed of any errors that still remain so they can be corrected in future printings. An email to [email protected] will be sufficient.
Robert L. Norton Mattapoisett, MA. June 1, 2019
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Contents Preface
_______________________________________________________ vii
video contents __________________________________________________xxix
Part i
fundamentals
chaPter 1 introduction 1.1
to
1
design_________________________________ 3
Design ..........................................................................................................3 Machine Design
3
1.2
A Design Process ..........................................................................................5
1.3
Problem Formulation and Calculation ...........................................................7 Definition Stage Preliminary Design Stage Detailed Design Stage Documentation Stage
1.4
The Engineering Model .................................................................................9 Estimation and FirstOrder Analysis The Engineering Sketch
1.5
8 8 8 9 9 10
ComputerAided Design and Engineering ...................................................10 ComputerAided Design (CAD) ComputerAided Engineering (CAE) Computational Accuracy
11 12 15
1.6
The Engineering Report...............................................................................16
1.7
Factors of Safety and Design Codes ............................................................16 Factor of Safety Choosing a Safety Factor Design and Safety Codes
16
17 19
1.8
Statistical Considerations............................................................................20
1.9
Units ...........................................................................................................20
1.10
Summary .....................................................................................................25
1.11
References ..................................................................................................26
1.12
Web References ..........................................................................................26
1.13
Bibliography ...............................................................................................27
1.14
Problems .....................................................................................................27
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chaPter 2 materials
and
Sixth Edition
Processes ______________________________29
2.0
Introduction ................................................................................................29
2.1
MaterialProperty Definitions .....................................................................29 The Tensile Test Ductility and Brittleness The Compression Test The Bending Test The Torsion Test Fatigue Strength and Endurance Limit Impact Resistance Fracture Toughness Creep and Temperature Effects
31 33 35 35 35 37 38 40 40
2.2
The Statistical Nature of Material Properties...............................................41
2.3
Homogeneity and Isotropy..........................................................................42
2.4
Hardness .....................................................................................................42 Heat Treatment Surface (Case) Hardening Heat Treating Nonferrous Materials Mechanical Forming and Hardening
2.5
Coatings and Surface Treatments ................................................................48 Galvanic Action Electroplating Electroless Plating Anodizing PlasmaSprayed Coatings Chemical Coatings
2.6
49 50 50 51 51 51
General Properties of Metals ......................................................................52 Cast Iron Cast Steels Wrought Steels Steel Numbering Systems Aluminum Titanium Magnesium Copper Alloys
2.7
44 45 46 46
52 53 53 54 56 58 59 59
General Properties of Nonmetals ................................................................60 Polymers Ceramics Composites
60 62 62
2.8
Selecting Materials .....................................................................................63
2.9
Summary .....................................................................................................64
2.10
References ..................................................................................................68
2.11
Web References ..........................................................................................68
2.12
Bibliography ...............................................................................................68
2.13
Problems .....................................................................................................69
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chaPter 3 Kinematics
and
load determination ___________________73
3.0
Introduction ................................................................................................73
3.1
Degree of Freedom .....................................................................................73
3.2
Mechanisms................................................................................................74
3.3
Calculating Degree of Freedom (Mobility) ..................................................75
3.4
Common 1DOF Mechanisms ....................................................................76 Fourbar Linkage and the Grashof Condition Sixbar Linkage Cam and Follower
3.5
Analyzing Linkage Motion ..........................................................................80 Types of Motion Complex Numbers as Vectors The Vector Loop Equation
3.6
82 84 86 87
Analyzing the Fourbar CrankSlider ............................................................89 Solving for Position in the Fourbar CrankSlider Solving for Velocity in the Fourbar CrankSlider Solving for Acceleration in the Fourbar CrankSlider Other Linkages
3.8
80 81 82
Analyzing the Fourbar Linkage....................................................................82 Solving for Position in the Fourbar Linkage Solving for Velocity in the Fourbar Linkage Angular Velocity Ratio and Mechanical Advantage Solving for Acceleration in the Fourbar Linkage
3.7
76 78 79
89 90 91 92
Cam Design and Analysis ...........................................................................92 The Timing Diagram The svaj Diagram Polynomials for the DoubleDwell Case Polynomials for the SingleDwell Case Pressure Angle Radius of Curvature
93 94 94 97 99 100
3.9
Loading Classes For Force Analysis ..........................................................101
3.10
Freebody Diagrams..................................................................................103
3.11
Load Analysis............................................................................................103 ThreeDimensional Analysis TwoDimensional Analysis Static Load Analysis
104 105 106
3.12
TwoDimensional, Static Loading Case Studies.........................................106
3.13
ThreeDimensional, Static Loading Case Study .........................................121
3.14
Dynamic Loading Case Study....................................................................126
3.15
Vibration Loading .....................................................................................130 Natural Frequency Dynamic Forces
3.16
130 132
Impact Loading .........................................................................................134 Energy Method
135
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Beam Loading ...........................................................................................139 Shear and Moment Singularity Functions Superposition
139 141 151
3.18
Summary ...................................................................................................151
3.19
References ................................................................................................154
3.20
Web References ........................................................................................155
3.21
Bibliography .............................................................................................155
3.22
Problems ...................................................................................................156
chaPter 4 stress, strain,
and
deflection ______________________ 173
4.0
Introduction ..............................................................................................173
4.1
Stress ........................................................................................................173
4.2
Strain ........................................................................................................177
4.3
Principal Stresses ......................................................................................177
4.4
Plane Stress and Plane Strain ....................................................................180 Plane Stress Plane Strain
180 180
4.5
Mohr’s Circles ...........................................................................................180
4.6
Applied Versus Principal Stresses..............................................................185
4.7
Axial Tension ............................................................................................186
4.8
Direct Shear Stress, Bearing Stress, and Tearout........................................186 Direct Shear Direct Bearing Tearout Failure
4.9
Beams and Bending Stresses.....................................................................188 Beams in Pure Bending Shear Due to Transverse Loading
4.10
198 204
Castigliano’s Method ................................................................................207 Deflection by Castigliano’s Method Finding Redundant Reactions with Castigliano’s Method
4.12
188 192
Deflection in Beams ..................................................................................196 Deflection by Singularity Functions Statically Indeterminate Beams
4.11
187 187 188
209 209
Torsion ......................................................................................................210
4.13
Combined Stresses....................................................................................216
4.14
Spring Rates ..............................................................................................218
4.15
Stress Concentration .................................................................................219 Stress Concentration Under Static Loading Stress Concentration Under Dynamic Loading Determining Geometric StressConcentration Factors Designing to Avoid Stress Concentrations
4.16
220 222 222 224
Axial Compression  Columns ..................................................................226 Slenderness Ratio Short Columns
226 227
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Long Columns End Conditions Intermediate Columns
4.17
227 229 230
Stresses in Cylinders .................................................................................237 ThickWalled Cylinders ThinWalled Cylinders
237 238
4.18
Case Studies in Static Stress and Deflection Analysis ...............................239
4.19
Summary ...................................................................................................255
4.20
References ................................................................................................261
4.21
Bibliography .............................................................................................261
4.22
Problems ...................................................................................................262
chaPter 5 static failure theories ______________________________ 279 5.0
Introduction ..............................................................................................279
5.1
Failure of Ductile Materials Under Static Loading ....................................281 The von MisesHencky or DistortionEnergy Theory The Maximum ShearStress Theory The Maximum NormalStress Theory Comparison of Experimental Data with Failure Theories
5.2
Failure of Brittle Materials Under Static Loading ......................................294 Even and Uneven Materials The CoulombMohr Theory The ModifiedMohr Theory
5.3
282 288 290 290 294 295 296
Fracture Mechanics ...................................................................................301 FractureMechanics Theory Fracture Toughness Kc
302 305
5.4
Using The Static Loading Failure Theories ................................................309
5.5
Case Studies in Static Failure Analysis ......................................................310
5.6
Summary ...................................................................................................321
5.7
References ................................................................................................324
5.8
Bibliography .............................................................................................325
5.9
Problems ...................................................................................................326
chaPter 6 fatigue failure theories ____________________________ 339 6.0
Introduction ..............................................................................................339 History of Fatigue Failure
6.1
Mechanism of Fatigue Failure ..................................................................342 Crack Initiation Stage Crack Propagation Stage Fracture
6.2
339 343 343 344
FatigueFailure Models .............................................................................345 Fatigue Regimes The StressLife Approach The StrainLife Approach The LEFM Approach
345 347 347 347
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6.3
MachineDesign Considerations ...............................................................348
6.4
Fatigue Loads............................................................................................349 Rotating Machinery Loading Service Equipment Loading
6.5
Measuring Fatigue Failure Criteria ............................................................351 Fully Reversed Stresses Combined Mean and Alternating Stress FractureMechanics Criteria Testing Actual Assemblies
6.6
349 350 352 358 360 363
Estimating Fatigue Failure Criteria ............................................................364 Estimating the Theoretical Fatigue Strength Sf’ or Endurance Limit Se’ Correction Factors—Theoretical Fatigue Strength or Endurance Limit Corrected Fatigue Strength Sf or Corrected Endurance Limit Se Creating Estimated SN Diagrams
364 366 373 373
6.7
Notches and Stress Concentrations ...........................................................378
6.8
Residual Stresses ......................................................................................383
6.9
Designing for HighCycle Fatigue ............................................................388
6.10
Designing for Fully Reversed Uniaxial Stresses.........................................388
Notch Sensitivity
Design Steps for Fully Reversed Stresses with Uniaxial Loading
6.11
389
Designing for Fluctuating Uniaxial Stresses .............................................396 Creating the ModifiedGoodman Diagram Applying StressConcentration Effects with Fluctuating Stresses Determining the Safety Factor with Fluctuating Stresses Design Steps for Fluctuating Stresses
6.12
379
397 400 402 405
Designing for Multiaxial Stresses in Fatigue .............................................412 Frequency and Phase Relationships Fully Reversed Simple Multiaxial Stresses Fluctuating Simple Multiaxial Stresses Complex Multiaxial Stresses
413 413 414 415
6.13
A General Approach to HighCycle Fatigue Design ..................................417
6.14
A Case Study in Fatigue Design ................................................................422
6.15
Summary ...................................................................................................435
6.16
References ................................................................................................439
6.17
Bibliography .............................................................................................442
6.18
Problems ...................................................................................................443
7.0
Introduction ..............................................................................................457
chaPter 7 surface failure _____________________________________ 457 7.1
Surface Geometry .....................................................................................459
7.2
Mating Surfaces ........................................................................................461
7.3
Friction......................................................................................................462 Effect of Roughness on Friction Effect of Velocity on Friction Rolling Friction Effect of Lubricant on Friction
463 463 463 464
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7.4
Adhesive Wear ..........................................................................................464 The AdhesiveWear Coefficient
7.5
Abrasive Wear...........................................................................................468 Abrasive Materials AbrasionResistant Materials
7.6
467 471 471
Corrosion Wear .........................................................................................472 Corrosion Fatigue Fretting Corrosion
473 473
7.7
Surface Fatigue .........................................................................................474
7.8
Spherical Contact ......................................................................................476 Contact Pressure and Contact Patch in Spherical Contact Static Stress Distributions in Spherical Contact
7.9
Cylindrical Contact ...................................................................................482 Contact Pressure and Contact Patch in Parallel Cylindrical Contact Static Stress Distributions in Parallel Cylindrical Contact
7.10
482 483
General Contact ........................................................................................486 Contact Pressure and Contact Patch in General Contact Stress Distributions in General Contact
7.11
476 478
486 488
Dynamic Contact Stresses .........................................................................491 Effect of a Sliding Component on Contact Stresses
491
7.12
Surface Fatigue Failure Models—Dynamic Contact ..................................498
7.13
Surface Fatigue Strength ...........................................................................501
7.14
Summary ...................................................................................................508
7.15
References ................................................................................................512
7.16
Problems ...................................................................................................514
chaPter 8 finite 8.0
element
analysis ______________________________ 521
Introduction .............................................................................................521 Stress and Strain Computation
522
8.1
Finite Element Method..............................................................................523
8.2
Element Types ...........................................................................................525 Element Dimension and Degree of Freedom (DOF) Element Order HElements Versus PElements Element Aspect Ratio
8.3
525 526 527 527
Meshing....................................................................................................527 Mesh Density Mesh Refinement Convergence
528 528 528
8.4
Boundary Conditions ................................................................................532
8.5
Applying Loads .........................................................................................542
8.6
Testing the Model (Verification)................................................................543
8.7
Modal Analysis .........................................................................................546
8.8
Case Studies .............................................................................................548
8.9
Summary ...................................................................................................558
8.10
References ................................................................................................559
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8.11
Bibliography .............................................................................................559
8.12
Web Resources .........................................................................................559
8.13
Problems ...................................................................................................560
Part ii machine design ______________________________ 561 chaPter 9 design case studies _________________________________ 563 9.0
Introduction ..............................................................................................563
9.1
Case Study 8—A Portable Air Compressor................................................564
9.2
Case Study 9—A HayBale Lifter ..............................................................567
9.3
Case Study 10—A CamTesting Machine..................................................571
9.4
Summary ...................................................................................................577
9.5
References ................................................................................................577
9.6
Design Projects .........................................................................................578
chaPter 10 shafts, Keys,
and
couPlings _______________________ 589
10.0
Introduction ..............................................................................................589
10.1
Shaft Loads ...............................................................................................589
10.2
Attachments and Stress Concentrations ....................................................591
10.3
Shaft Materials..........................................................................................593
10.4
Shaft Power...............................................................................................593
10.5
Shaft Loads ...............................................................................................594
10.6
Shaft Stresses ............................................................................................594
10.7
Shaft Failure in Combined Loading ...........................................................595
10.8
Shaft Design .............................................................................................596 General Considerations Design for Fully Reversed Bending and Steady Torsion Design for Fluctuating Bending and Fluctuating Torsion
10.9
596 597 599
Shaft Deflection ........................................................................................606 Shafts as Beams Shafts as Torsion Bars
607 607
10.10 Keys and Keyways ....................................................................................610 Parallel Keys Tapered Keys Woodruff Keys Stresses in Keys Key Materials Key Design Stress Concentrations in Keyways
610 611 612 612 613 613 614
10.11 Splines ......................................................................................................618 10.12 Interference Fits ........................................................................................620 Stresses in Interference Fits Stress Concentration in Interference Fits Fretting Corrosion
620 621 622
xxiii
10.13 Flywheel Design .......................................................................................625 Energy Variation in a Rotating System Determining the Flywheel Inertia Stresses in Flywheels Failure Criteria
626 628 630 631
10.14 Critical Speeds of Shafts ...........................................................................633 Lateral Vibration of Shafts and Beams—Rayleigh’s Method Shaft Whirl Torsional Vibration Two Disks on a Common Shaft Multiple Disks on a Common Shaft Controlling Torsional Vibrations
635 637 639 640 641 642
10.15 Couplings .................................................................................................644 Rigid Couplings Compliant Couplings
645 646
10.16 Case Study 8B ...........................................................................................648 Designing Driveshafts for a Portable Air Compressor
648
10.17 Summary ...................................................................................................652 10.18 References ................................................................................................654 10.19 Problems ...................................................................................................655
chaPter 11 Bearings
and
luBrication ___________________________ 665
11.0
Introduction ..............................................................................................665
11.1
Lubricants ...............................................................................................667
11.2
Viscosity ...................................................................................................669
11.3
Types of Lubrication..................................................................................670
A Caveat
FullFilm Lubrication Boundary Lubrication
667
671 673
11.4
Material Combinations in Sliding Bearings ...............................................673
11.5
Hydrodynamic Lubrication Theory ............................................................674 Petroff’s Equation for NoLoad Torque Reynolds’ Equation for Eccentric Journal Bearings Torque and Power Losses in Journal Bearings
11.6
675 676 681
Design of Hydrodynamic Bearings ............................................................682 Design Load Factor—The Ocvirk Number Design Procedures
682 684
11.7
Nonconforming Contacts ..........................................................................688
11.8
Rollingelement bearings ........................................................................695 Comparison of Rolling and Sliding Bearings Types of RollingElement Bearings
11.9
696 696
Failure of Rollingelement bearings ..........................................................700
11.10 Selection of Rollingelement bearings ......................................................701 Basic Dynamic Load Rating C Modified Bearing Life Rating Basic Static Load Rating C0
701 702 703
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Combined Radial and Thrust Loads Calculation Procedures
704 705
11.11 Bearing Mounting Details .........................................................................707 11.12 Special Bearings ........................................................................................708 11.13 Case Study 10B ........................................................................................710 11.14 Summary ...................................................................................................712 Important Equations Used in This Chapter
713
11.15 References ................................................................................................715 11.16 Problems ...................................................................................................717
chaPter 12 sPur gears _________________________________________ 725 12.0
Introduction ..............................................................................................725
12.1
Gear Tooth Theory ...................................................................................727 The Fundamental Law of Gearing The Involute Tooth Form Pressure Angle Gear Mesh Geometry Rack and Pinion Changing Center Distance Backlash Relative Tooth Motion
727 728 729 730 731 731 733 733
12.2
Gear Tooth Nomenclature .........................................................................733
12.3
Interference and Undercutting ..................................................................736
12.4
Contact Ratio .............................................................................................738
12.5
Gear Trains ................................................................................................740
UnequalAddendum Tooth Forms
Simple Gear Trains Compound Gear Trains Reverted Compound Trains Epicyclic or Planetary Gear Trains
12.6
737
740 741 742 743
Gear Manufacturing ..................................................................................746 Forming Gear Teeth Machining Roughing Processes Finishing Processes Gear Quality
746 747 747 747 749
12.7
Loading on Spur Gears .............................................................................749
12.8
Stresses in Spur Gears ..............................................................................752 Bending Stresses Surface Stresses
12.9
752 761
Gear Materials ..........................................................................................765 Material Strengths BendingFatigue Strengths for Gear Materials SurfaceFatigue Strengths for Gear Materials
766 767 769
12.10 Lubrication of Gearing ..............................................................................775 12.11 Design of Spur Gears ................................................................................776
xxv
12.12 Case Study 8C ..........................................................................................777 12.13 Summary ...................................................................................................783 12.14 References ................................................................................................784 12.15 Problems ...................................................................................................785
chaPter 13 helical, Bevel,
and
Worm gears ___________________ 791
13.0
Introduction .............................................................................................791
13.1
Helical Gears ................................................................................................................................791 Helical Gear Geometry HelicalGear Forces Virtual Number of Teeth Contact Ratios Stresses in Helical Gears
13.2
Bevel Gears ...............................................................................................804 BevelGear Geometry and Nomenclature BevelGear Mounting Forces on Bevel Gears Stresses in Bevel Gears
13.3
793 794 795 796 796 804 806 806 806
Wormsets..................................................................................................810 Materials for Wormsets Lubrication in Wormsets Forces in Wormsets Wormset Geometry Rating Methods A Design Procedure for Wormsets
813 814 814 814 815 817
13.4
Case Study 9B ...........................................................................................817
13.5
Summary ...................................................................................................821
13.6
References ................................................................................................825
13.7
Problems ...................................................................................................825
chaPter 14 sPring design _____________________________________ 829 14.0
Introduction ..............................................................................................829
14.1
Spring Rate ...................................................................................................................................829
14.2
Spring Configurations .............................................................................................................832
14.3
Spring Materials........................................................................................834 Spring Wire Flat Spring Stock
14.4
834 837
Helical Compression Springs ....................................................................838 Spring Lengths End Details Active Coils Spring Index Spring Deflection Spring Rate Stresses in Helical Compression Spring Coils Helical Coil Springs of Nonround Wire Residual Stresses
838 839 839 840 840 840 840 842 843
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MACHINE DESIGN

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Sixth Edition
Buckling of Compression Springs CompressionSpring Surge Allowable Strengths for Compression Springs The TorsionalShear SN Diagram for Spring Wire The ModifiedGoodman Diagram for Spring Wire
844 844 846 847 849
14.5
Designing Helical Compression Springs for Static Loading .......................851
14.6
Designing Helical Compression Springs for Fatigue Loading ....................856
14.7
Helical Extension Springs ........................................................................863 Active Coils in Extension Springs Spring Rate of Extension Springs Spring Index of Extension Springs Coil Preload in Extension Springs Deflection of Extension Springs Coil Stresses in Extension Springs End Stresses in Extension Springs Surging in Extension Springs Material Strengths for Extension Springs Design of Helical Extension Springs
14.8
Helical Torsion Springs..............................................................................874 Terminology for Torsion Springs Number of Coils in Torsion Springs Deflection of Torsion Springs Spring Rate of Torsion Springs Coil Closure Coil Stresses in Torsion Springs Material Parameters for Torsion Springs Safety Factors for Torsion Springs Designing Helical Torsion Springs
14.9
863 864 864 864 864 865 865 866 866 866 875 875 875 876 876 876 877 878 878
Belleville Spring Washers .........................................................................881 LoadDeflection Function for Belleville Washers Stresses in Belleville Washers Static Loading of Belleville Washers Dynamic Loading Stacking Springs Designing Belleville Springs
882 884 885 885 885 886
14.10 Case Study 10C ........................................................................................888 14.11 Summary ...................................................................................................893 14.12 References ................................................................................................896 14.13 Problems ...................................................................................................897
chaPter 15 screWs
and
fasteners _____________________________ 903
15.0
Introduction ..............................................................................................903
15.1
Standard Thread Forms ...........................................................................906 Tensile Stress Area Standard Thread Dimensions
15.2
907 908
Power Screws ............................................................................................909 Square, Acme, and Buttress Threads Power Screw Application Power Screw Force and Torque Analysis
910 910 912
xxvii
Friction Coefficients SelfLocking and BackDriving of Power Screws Screw Efficiency Ball Screws
15.3
Stresses in Threads ...................................................................................918 Axial Stress Shear Stress Torsional Stress
15.4
913 914 915 915 919 919 920
Types of Screw Fasteners .........................................................................920 Classification by Intended Use Classification by Thread Type Classification by Head Style Nuts and Washers
921 921 921 923
15.5
Manufacturing Fasteners ..........................................................................923
15.6
Strengths of Standard Bolts and Machine Screws .....................................925
15.7
Preloaded Fasteners in Tension ................................................................925 Preloaded Bolts Under Static Loading Preloaded Bolts Under Dynamic Loading
15.8
Determining the Joint Stiffness Factor ......................................................939 Joints With Two Plates of the Same Material Joints With Two Plates of Different Materials Gasketed Joints
15.9
929 933 941 941 943
Controlling Preload ...................................................................................948 The TurnoftheNut Method TorqueLimited Fasteners LoadIndicating Washers Torsional Stress Due to Torquing of Bolts
949 949 949 950
15.10 Fasteners in Shear .....................................................................................950 Dowel Pins Centroids of Fastener Groups Determining Shear Loads on Fasteners
952 953 954
15.11 Case Study 8D ..........................................................................................956 15.12 Summary ...................................................................................................961 15.13 References ................................................................................................964 15.14 Bibliography .............................................................................................964 15.15 Problems ...................................................................................................965
chaPter 16 Weldments ________________________________________ 973 16.0
Introduction ..............................................................................................973
16.1
Welding Processes ....................................................................................975 Types of Welding in Common Use Why Should a Designer Be Concerned with the Welding Process?
16.2
Weld Joints and Weld Types .....................................................................977 Joint Preparation Weld Specification
16.3
976 977 979 979
Principles of Weldment Design .................................................................979
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An Integrated Approach
16.4
Static Loading of Welds ............................................................................981
16.5
Static Strength of Welds............................................................................982 Residual Stresses in Welds Direction of Loading Allowable Shear Stress for Statically Loaded Fillet and PJP Welds
16.6
983 983 983
Dynamic Loading of Welds .......................................................................986 Effect of Mean Stress on Weldment Fatigue Strength Are Correction Factors Needed For Weldment Fatigue Strength? Effect of Weldment Configuration on Fatigue Strength Is There an Endurance Limit for Weldments? Fatigue Failure in Compression Loading?
986 986 987 991 992
16.7
Treating a Weld as a Line ..........................................................................993
16.8
Eccentrically Loaded Weld Patterns...........................................................999
16.9
Design Considerations for Weldments in Machines ...............................1000
16.10 Summary .................................................................................................1001 16.11 References ..............................................................................................1002 16.12 Problems .................................................................................................1003
chaPter 17 clutches
and
BraKes ______________________________ 1007
17.0
Introduction ............................................................................................1007
17.1
Types of Brakes and Clutches .................................................................1009
17.2
Clutch/Brake Selection and Specification ................................................1014
17.3
Clutch and Brake Materials .....................................................................1016
17.4
Disk Clutches ..........................................................................................1016 Uniform Pressure Uniform Wear
1017 1017
17.5
Disk Brakes .............................................................................................1019
17.6
Drum Brakes ...........................................................................................1020 ShortShoe External Drum Brakes LongShoe External Drum Brakes LongShoe Internal Drum Brakes
1021 1023 1027
17.7
Summary .................................................................................................1027
17.8
References ..............................................................................................1030
17.9
Bibliography ...........................................................................................1030
17.10 Problems .................................................................................................1031
aPPendices ____________________________________________________ 1035 A Material Properties ............................................................................1035 B Beam Tables.......................................................................................1043 C StressConcentration Factors .............................................................1047 D Answers to Selected Problems...........................................................1055
index
____________________________________________________ 1065
doWnloads index _____________________________________________ 1078
VIDEO CONTENTS The Sixth Edition has a collection of Master Lecture Videos and Tutorials made by the author over a 31year period while teaching at Worcester Polytechnic Institute. The lectures were recorded in a classroom in front of students in 2011/2012. Tutorials were done in a recording studio and were intended as supplements to class lectures. There are 37 instructional videos in total. One is a short introduction to the master lecture series and 20 are “50minute” lectures. Eight are short tutorials and eight are demonstrations of machinery. The run times of all videos are noted in the tables. The sixth edition is available both as a print book and as digital media. The digital, ebook versions have active links that allow these videos to be run while reading the book. The print edition notes the names and URLs of all the videos in the text at their links. In addition to the lecture videos, all the digital content that was with the fifth and earlier editions is still available as downloads, including the authorwritten programs Linkages, Dynacam, and Matrix. An index of all the nonvideo downloadable files is in the Downloads Index. In the digital ebook versions, these are hotlinked to the text. The URL of each video is also provided for printbook readers to download them. Any instructor or student who uses the book may register on my website, http://www. designofmachinery.com , either as a student or instructor, and I will send them a password to access a protected site where they can download the latest versions of my computer programs, Linkages, Dynacam, and Matrix. They can also download the 33 videos and all the files listed in the Downloads Index. Note that I personally review each of these requests for access and will approve only those that are filled out completely and correctly according to the provided instructions. I require complete information and only accept university email addresses.
LECTURE VIDEOS Chapter Lecture 1 4 4 4 5 5 6 6 6 10 10 7 12 12 14 14 11 11 15 15 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
(Concatenate this URL with any filename below to run a video)
Topic Introduction Stress Review Stress Distribution Combined Stress, Stress Concentration, Columns Ductile Failure Theory Brittle Failure Theory Fatigue Failure Theory Fully Reversed Loads Fluctuating Loads Shaft Design I Shaft Design II Wear and Surface Fatigue Spur Gear Design I Spur Gear Design II Spring Design I Spring Design II Bearings and Lubrication Rolling Element Bearings Power Screws and Fasteners Preloaded Fasteners Finite Element Analysis
http://www.designofmachinery.com/MD/
Run Time
01_Introduction.mp4 02_Stress_Review.mp4 03_Stress_Distribution.mp4 04_Combined_stress_stress_concentration_columns.mp4 05_Ductile_Failure_Theory.mp4 06_Brittle_Failure_Theory.mp4 07_Fatigue_Failure_Theory.mp4 08_Fully_Reversed_Loads.mp4 09_Fluctuating_Loads.mp4 10_Shaft_Design_I.mp4 11_Shaft_Design_II.mp4 12_Wear_and_Surface_Fatigue.mp4 13_Spur_Gear_Design_I.mp4 14_Spur_Gear_Design_II.mp4 15_Spring_Design_I.mp4 16_Spring_Design_II.mp4 17_Bearings_and_Lubrication.mp4 18_Rolling_Element_Bearings.mp4 19_Power_Screws_and_Fasteners.mp4 20_Preloaded_Fasteners.mp4 21_Finite_Element_Analysis.mp4
03:30 53:40 50:52 54:11 46:02 51:02 55:49 52:59 54:14 44:44 47:20 52:25 51:31 50:37 52:20 47:46 50:07 46:54 44:42 48:22 52:28
xxx
MACHINE DESIGN

An Integrated Approach 
TUTORIAL VIDEOS Topic Bearings Bending Stress Columns Failure Modes Gears Springs Stress Cube Torsion
DEMONSTRATION VIDEOS Topic Boot Testing Machine Bottle Printing Machine Cam Machine Fourbar Machine Pick and Place Mechanism Spring Manufacturing Machinery Spring Surge and Spring Failure Vibration Testing
Sixth Edition
(Concatenate this URL with any filename below to run a video)
http://www.designofmachinery.com/MD/
Run Time 09:11 05:57 01:52 09:41 22:07 20:06 05:04 03:13
Bearings.mp4 Bending_Stress.mp4 Columns.mp4 Failure_Modes.mp4 Gears.mp4 Springs.mp4 Stress_Cube.mp4 Torsion.mp4
(Concatenate this URL with any filename below to run a video)
http://www.designofmachinery.com/MD/ Boot_Tester.mp4 Bottle_Printing_Machine.mp4 Cam_Machine.mp4 Fourbar_Machine.mp4 Pick_and_Place_Mechanism.mp4 Spring_Manufacturing.mp4 Fatigue_Failure.mp4 Vibration_Testing.mp4
Run Time 19:02 09:49 21:28 35:38 36:35 12:23 03:46 05:51
Note that you can download a PDF file containing hyperlinks to all the video content listed in the above tables. This allows printbook readers to easily access the videos without having to type in each URL as noted in the tables. Download the file: http://www.designofmachinery.com/MD/Video_Links_for_Machine_Design_6ed.pdf
I
Part
FUNDAMENTALS
INTRODUCTION TO DESIGN Learning without thought is labor lost; thought without learning is perilous. ConfuCius, 6th Century B.C.
1.1
DESIGN View the introductory video (03:30)†
What is design? Wallpaper is designed. You may be wearing “designer” clothes. Automobiles are “designed” in terms of their external appearance. The term design clearly encompasses a wide range of meaning. In the above examples, design refers primarily to the object’s aesthetic appearance. In the case of the automobile, all of its other aspects also involve design. Its mechanical internals (engine, brakes, suspension, etc.) must be designed, more likely by engineers than by artists, though even the engineer gets to exhibit some artistry when designing machinery. The word design is from the Latin word designare meaning to designate, or mark out. Design means many things. It can refer to the design of an artistic work or the appearance of a product. We are more concerned here with engineering design than with artistic design. Engineering design can be defined as The process of applying the various techniques and scientific principles for the purpose of defining a device, a process, or a system in sufficient detail to permit its realization. Machine Design This text is concerned with one aspect of engineering design—machine design. Machine design deals with the creation of machinery that works safely, reliably, and well. A machine can be defined as: A system of elements arranged to transmit motion and energy in a predetermined and controlled fashion, or even more simply as: A system to control force and motion. †
http://www.designofmachinery.com/MD/01_Introduction.mp4
3
1
4
1
Titlepage photograph courtesy of Boeing Commercial Airplane Co. Inc., Seattle, Wash.
MACHINE DESIGN

An Integrated Approach

Sixth Edition
The notion of useful work is basic to a machine’s function, as there is almost always some energy transfer involved. The mention of forces and motion is also critical to our concerns, as, in converting energy from one form to another, machines create motion and develop forces. It is the engineer’s task to define and calculate those motions, forces, and changes in energy in order to determine the sizes, shapes, and materials needed for each of the interrelated parts in the machine. This is the essence of machine design. While one must, of necessity, design a machine one part at a time, it is crucial to recognize that each part’s function and performance (and thus its design) are dependent on many other interrelated parts within the same machine. Thus, we are going to attempt to “design the whole machine” here, rather than simply designing individual elements in isolation from one another. To do this we must draw upon a common body of engineering knowledge encountered in previous courses, e.g., statics, dynamics, mechanics of materials (stress analysis), and material properties. Brief reviews and examples of these topics are included in the early chapters of this book. The ultimate goal in machine design is to size and shape the parts (machine elements) and choose appropriate materials and manufacturing processes so that the resulting machine can be expected to perform its intended function without failure. This requires that the engineer be able to calculate and predict the mode and conditions of failure for each element and then design it to prevent that failure. This in turn requires that a stress and deflection analysis be done for each part. Since stresses are a function of the applied and inertial loads, and of the part’s geometry, an analysis of the forces, moments, torques, and the dynamics of the system must be done before the stresses and deflections can be completely calculated. If the “machine” in question has no moving parts, then the design task becomes much simpler, because only a static force analysis is required. But if the machine has no moving parts, it is not much of a machine (and doesn’t meet the definition above); it is then a structure. Structures also need to be designed against failure, and, in fact, large external structures (bridges, buildings, etc.) are also subjected to dynamic loads from wind, earthquakes, traffic, etc., and thus must also be designed for these conditions. Structural dynamics is an interesting subject but one which we will not address in this text. We will concern ourselves with the problems associated with machines that move. If the machine’s motions are very slow and the accelerations negligible, then a static force analysis will suffice. But if the machine has significant accelerations within it, then a dynamic force analysis is needed and the accelerating parts become “victims of their own mass.” In a static structure, such as a building’s floor, designed to support a particular weight, the safety factor of the structure can be increased by adding appropriately distributed material to its structural parts. Though it will be heavier (more “dead” weight), if properly designed it may nevertheless carry more “live” weight (payload) than it did before, still without failure. In a dynamic machine, adding weight (mass) to moving parts may have the opposite effect, reducing the machine’s safety factor, its allowable speed, or its payload capacity. This is because some of the loading that creates stresses in the moving parts is due to the inertial forces predicted by Newton’s second law, F = ma. Since the accelerations of the moving parts in the machine are dictated by its kinematic design and by its running speed, adding mass to moving parts will increase the inertial loads on those same parts unless their kinematic accelerations are reduced by slowing its operation. Even though the added mass may increase the strength of the part, that benefit may be reduced or cancelled by the resultant increases in inertial forces.
Chapter 1
INTRODUCTION TO DESIGN
5
1
Iteration Thus, we face a dilemma at the initial stages of machine design. Generally, before reaching the stage of sizing the parts, the kinematic motions of the machine will have already been defined. External forces provided by the “outside world” on the machine are also often known. Note that in some cases, the external loads on the machine will be very difficult to predict—for example, the loads on a moving automobile. The designer cannot predict with accuracy what environmental loads the user will subject the machine to (potholes, hard cornering, etc.). In such cases, statistical analysis of empirical data gathered from actual testing can provide some information for design purposes. What remain to be defined are the inertial forces that will be generated by the known kinematic accelerations acting on the as yet undefined masses of the moving parts. The dilemma can be resolved only by iteration, which means to repeat, or to return to a previous state. We must assume some trial configuration for each part, use the mass properties (mass, CG location, and mass moment of inertia) of that trial configuration in a dynamic force analysis to determine the forces, moments, and torques acting on the part, and then use the crosssectional geometry of the trial design to calculate the resulting stresses. In general, accurately determining all the loads on a machine is the most difficult task in the design process. If the loads are known, the stresses can be calculated. Most likely, on the first trial, we will find that our design fails because the materials cannot stand the levels of stress presented. We must then redesign the parts (iterate) by changing shapes, sizes, materials, manufacturing processes, or other factors in order to reach an acceptable design. It is generally not possible to achieve a successful result without making several iterations through this design process. Note also that a change to the mass of one part will also affect the forces applied to parts connected to it and thus require their redesign also. It is truly the design of interrelated parts. 1.2
A DESIGN PROCESS*
The process of design is essentially an exercise in applied creativity. Various “design processes” have been defined to help organize the attack upon the “unstructured problem,” i.e., one for which the problem definition is vague and for which many possible solutions exist. Some of these design process definitions contain only a few steps and others a detailed list of 25 steps. One version of a design process is shown in Table 11, which lists ten steps.[2] The initial step, Identification of Need, usually consists of an illdefined and vague problem statement. The development of Background Research information (step 2) is necessary to fully define and understand the problem, after which it is possible to restate the Goal (step 3) in a more reasonable and realistic way than in the original problem statement. Step 4 calls for the creation of a detailed set of Task Specifications which bound the problem and limit its scope. The Synthesis step (5) is one in which as many alternative design approaches as possible are sought, usually without regard (at this stage) for their value or quality. This is also sometimes called the Ideation and Invention step, in which the largest possible number of creative solutions are generated. In step 6, the possible solutions from the previous step are Analyzed and either accepted, rejected, or modified. The most promising solution is Selected at step 7. Once an acceptable design is selected, the Detailed Design (step 8) can be done, in which all the loose ends are tied up, complete engineering drawings made, vendors identified,
*
Adapted from Norton, Design of Machinery, 5ed. McGrawHill, New York, 2012, with the publisher’s permission.
6
MACHINE DESIGN

An Integrated Approach

Sixth Edition
1 Table 11
A Design Process
1
Identification of need
2
Background research
3
Goal statement
4
Task specifications
5
Synthesis
6
Analysis
7
Selection
8
Detailed design
9
Prototyping and testing
10
Production
manufacturing specifications defined, etc. The actual construction of the working design is first done as a Prototype in step 9 and finally in quantity in Production at step 10. A more complete discussion of this design process can be found in reference 2, and a number of references on the topics of creativity and design are provided in the bibliography at the end of this chapter. The above description may give an erroneous impression that this process can be accomplished in a linear fashion as listed. On the contrary, iteration is required within the entire process, moving from any step back to any previous step, in all possible combinations, and doing this repeatedly. The best ideas generated at step 5 will invariably be discovered to be flawed when later analyzed. Thus, a return to at least the Ideation step will be necessary in order to generate more solutions. Perhaps a return to the Background Research phase may be necessary to gather more information. The Task Specifications may need to be revised if it turns out that they were unrealistic. In other words, anything is “fair game” in the design process, including a redefinition of the problem, if necessary. One cannot design in a linear fashion. It’s three steps forward and two (or more) back, until you finally emerge with a working solution. Theoretically, we could continue this iteration on a given design problem forever, continually creating small improvements. Inevitably, the incremental gains in function or reductions in cost will tend toward zero with time. At some point, we must declare the design “good enough” and ship it. Often someone else (most likely, the boss) will snatch it from our grasp and ship it over our protests that it isn’t yet “perfect.” Machines that have been around a long time and that have been improved by many designers reach a level of “perfection” that makes them difficult to improve upon. One example is the ordinary bicycle. Though inventors continue to attempt to improve this machine, the basic design has become fairly static after more than a century of development. In machine design, the early designprocess steps usually involve the Type Synthesis of suitable kinematic configurations, which can provide the necessary motions. Type synthesis involves the choice of the type of mechanism best suited to the problem. This is a difficult task for the student, as it requires some experience and knowledge of the various types of mechanisms that exist and that might be feasible from a performance and manufacturing standpoint. As an example, assume that the task is to design a device
Chapter 1
INTRODUCTION TO DESIGN
to track the constantspeed, straightline motion of a part on a conveyor belt and attach a second part to it as it passes by. This has to be done with good accuracy and repeatability and must be reliable and inexpensive. You might not be aware that this task could be accomplished by any of the following devices: • a straightline linkage • a cam and follower • an air cylinder • a hydraulic cylinder • a robot • a solenoid
Each of these solutions, while possible, may not be optimal or even practical. Each has good and bad points. The straightline linkage is large and may have undesirable accelerations; the cam and follower is expensive but is accurate and repeatable. The air cylinder is inexpensive but noisy and unreliable. The hydraulic cylinder and the robot are more expensive. The inexpensive solenoid has high impact loads and velocities. So, the choice of device type can have a big effect on design quality. A bad choice at the typesynthesis stage can create major problems later on. The design might have to be changed after completion at great expense. Design is essentially an exercise in tradeoffs. There is usually no clearcut solution to a real engineering design problem. Once the type of required mechanism is defined, its detailed kinematics must be synthesized and analyzed. The motions of all moving parts and their time derivatives through acceleration must be calculated in order to be able to determine the dynamic forces on the system. (See reference 2 for more information on this aspect of machine design.) In the context of machine design addressed in this text, we will not exercise the entire design process as described in Table 11. Rather, we will propose examples, problems, and case studies that already have had steps 1–4 defined. The type synthesis and kinematic analysis will already be done, or at least set up, and the problems will be structured to that degree. The tasks remaining will largely involve steps 5 through 8, with a concentration on synthesis (step 5) and analysis (step 6). Synthesis and analysis are the “two faces” of machine design, like two sides of the same coin. Synthesis means to put together and analysis means to decompose, to take apart, to resolve into its constituent parts. Thus, they are opposites, but they are symbiotic. We cannot take apart “nothing,” thus we must first synthesize something in order to analyze it. When we analyze it, we will probably find it lacking, requiring further synthesis, and then further analysis ad nauseam, finally iterating to a better solution. You will need to draw heavily upon your understanding of statics, dynamics, and mechanics of materials to accomplish this. 1.3
PROBLEM FORMULATION AND CALCULATION
It is extremely important for every engineer to develop good and careful computational habits. Solving complicated problems requires an organized approach. Design problems also require good recordkeeping and documentation habits in order to record the many assumptions and design decisions made along the way so that the designer’s thought process can be later reconstructed if redesign is necessary.
7
1
8
MACHINE DESIGN
*
A suggested procedure for the designer is shown in Table 12, which lists a set of subtasks appropriate to most machinedesign problems of this type. These steps should be documented for each problem in a neat fashion, preferably in a bound notebook in order to maintain their chronological order.*
1
If there is a possibility of a patentable invention resulting from the design, then the notebook should be permanently bound (not looseleaf), and its pages should be consecutively numbered, dated, and witnessed by someone who understands the technical content.

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Definition Stage In your design notebook, first Define the Problem clearly in a concise statement. The “givens” for the particular task should be clearly listed, followed by a record of the assumptions made by the designer about the problem. Assumptions expand upon the given (known) information to further constrain the problem. For example, one might assume the effects of friction to be negligible in a particular case, or assume that the weight of the part can be ignored because it will be small compared to the applied or dynamic loads expected. Preliminary Design Stage Once the general constraints are defined, some Preliminary Design Decisions must be made in order to proceed. The reasons and justifications for these decisions should be documented. For example, we might decide to try a solid, rectangular cross section for a connecting link and choose aluminum as a trial material. On the other hand, if we recognized from our understanding of the problem that this link would be subjected to significant accelerations of a timevarying nature that would repeat for millions of cycles, a better design decision might be to use a hollow or Ibeam section in order to reduce its mass and also to choose steel for its infinite fatigue life. Thus, these design decisions can have significant effect on the results and will often have to be changed or abandoned as we iterate through the design process. It has often been noted that 90% of a design’s characteristics may be determined in the first 10% of the total project time, during which these preliminary design decisions are made. If they are bad decisions, it may not be possible to save the bad design through later modifications without essentially starting over. The preliminary design concept should be documented at this stage with clearly drawn and labeled Design Sketches that will be understandable to another engineer or even to oneself after some time has passed. Detailed Design Stage With a tentative design direction established we can create one or more engineering (mathematical) models of the element or system in order to analyze it. These models will usually include a loading model consisting of freebody diagrams which show all forces, moments, and torques on the element or system and the appropriate equations for their calculation. Models of the stress and deflection states expected at locations of anticipated failure are then defined with appropriate stress and deflection equations. Analysis of the design is then done using these models and the safety or failure of the design determined. The results are evaluated in conjunction with the properties of the chosen engineering materials and a decision made whether to proceed with this design or iterate to a better solution by returning to an earlier step of the process.
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1 Table 12
Problem Formulation and Calculation
1
Define the problem
2
State the givens
3
Make appropriate assumptions
4
Preliminary design decisions
5
Design sketches
6
Mathematical models
7
Analysis of the design
8
Evaluation
9
Document results
Definition stage
Preliminary design stage
Detailed design stage Documentation stage
Documentation Stage Once sufficient iteration through this process provides satisfactory results, the documentation of the element’s or system’s design should be completed in the form of detailed engineering drawings, material and manufacturing specifications, etc. If properly approached, a great deal of the documentation task can be accomplished concurrent with the earlier stages simply by keeping accurate and neat records of all assumptions, computations, and design decisions made throughout the process. 1.4
THE ENGINEERING MODEL
The success of any design is highly dependent on the validity and appropriateness of the engineering models used to predict and analyze its behavior in advance of building any hardware. Creating a useful engineering model of a design is probably the most difficult and challenging part of the whole process. Its success depends a great deal on experience as well as skill. Most important is a thorough understanding of the first principles and fundamentals of engineering. The engineering model that we are describing here is an amorphous thing that may consist of some sketches of the geometric configuration and some equations that describe its behavior. It is a mathematical model that describes the physical behavior of the system. This engineering model invariably requires the use of computers to exercise it. Using computer tools for analyzing engineering models is discussed in the next section. A physical model or prototype usually comes later in the process and is needed to prove the validity of the engineering model through experiments. Estimation and FirstOrder Analysis The value of making even very simplistic engineering models of your preliminary designs cannot be overemphasized. Often, at the outset of a design, the problem is so loosely and poorly defined that it is difficult to develop a comprehensive and thorough model in the form of equations that fully describe the system. The engineering student is used to problems that are fully structured, of a form such as “Given A, B, and C, find D.” If one can identify the appropriate equations (model) to apply to such a problem, it is relatively easy to determine an answer (which might even match the one in the back of the book).
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*
Reallife engineering design problems are not of this type. They are very unstructured and must be structured by you before they can be solved. Also, there is no “back of the book” to refer to for the answer.* This situation makes most students and beginning engineers very nervous. They face the “blank paper syndrome,” not knowing where to begin. A useful strategy is to recognize that:
1 A student once commented that “Life is an oddnumbered problem.” This (slow) author had to ask for an explanation, which was: “The answer is not in the back of the book.”

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1. You must begin somewhere. 2. Wherever you begin, it will probably not be the “best” place to do so. 3. The magic of iteration will allow you to back up, improve your design, and eventually succeed.
With this strategy in mind, you can feel free to make some estimation of a design configuration at the outset, assume whatever limiting conditions you think appropriate, and do a “firstorder analysis,” one that will be only an estimate of the system’s behavior. These results will allow you to identify ways to improve the design. Remember that it is preferable to get a reasonably approximate but quick answer that tells you whether the design does or doesn’t work rather than to spend more time getting the same result to more decimal places. With each succeeding iteration, you will improve your understanding of the problem, the accuracy of your assumptions, the complexity of your model, and the quality of your design decisions. Eventually, you will be able to refine your model to include all relevant factors (or identify them as irrelevant) and obtain a higherorder, final analysis in which you have more confidence. The Engineering Sketch A sketch of the concept is often the starting point for a design. This may be a freehand sketch, but it should always be made reasonably to scale in order to show realistic geometric proportions. This sketch often serves the primary purpose of communicating the concept to other engineers and even to yourself. It is one thing to have a vague concept in mind and quite another to define it in a sketch. This sketch should, at a minimum, contain three or more orthographic views, aligned according to proper drafting convention, and may also include an isometric or trimetric view. Figure 11 shows a freehand sketch of a simple design for one subassembly of a trailer hitch for a tractor. While often incomplete in terms of detail needed for manufacture, the engineering sketch should contain enough information to allow the development of an engineering model for design and analysis. This may include critical, if approximate, dimensional information, some material assumptions, and any other data germane to its function that is needed for further analysis. The engineering sketch captures some of the givens and assumptions made, even implicitly, at the outset of the design process. 1.5
COMPUTERAIDED DESIGN AND ENGINEERING
The computer has created a true revolution in engineering design and analysis. Problems whose solution methods have literally been known for centuries but that only a generation ago were practically unsolvable due to their high computational demands can now be solved in minutes on inexpensive microcomputers. Tedious graphical solution methods were developed in the past to circumvent the lack of computational power available from slide rules. Some of these graphical solution methods still have value in that they can
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1
FIGURE 11 A Freehand Sketch of a Trailer Hitch Assembly for a Tractor
show the results in an understandable form. But one can no longer “do engineering” without using its latest and most powerful tool, the computer. ComputerAided Design (CAD) As the design progresses, the crude freehand sketches made at the earliest stages will be supplanted by formal drawings made either with conventional drafting equipment or, as is increasingly common, with computeraided design or computeraided drafting software. If the distinction between these two terms (both of which share the acronym CAD) was ever clear (a subject for debate that will be avoided here), then that distinction is fading as more sophisticated CAD software becomes available. The original CAD systems of a generation ago were essentially drafting tools that allowed the creation of computergenerated multiview drawings similar to those done for centuries before by hand on a drafting board. The data stored in these early CAD systems were strictly twodimensional representations of the orthographic projections of the part’s true 3D geometry. Only the edges of the part were defined in the database. This is called a wireframe model. Some 3D CAD packages use wireframe representation as well. Present versions of most CAD software packages allow (and sometimes require) that the geometry of the parts be encoded in a 3D data base as solid models. In a solid model the edges and the faces of the part are defined. From this 3D information, the conventional 2D orthographic views can be automatically generated if desired. The major advantage of creating a 3D solidmodel geometric data base for any design is that its massproperty information can be rapidly calculated. (This is not possible in a 2D or 3D wireframe model.) For example, in designing a machine part, we need to determine the location of its center of gravity (CG), its mass, its mass moment of inertia, and its crosssectional geometries at various locations. Determining this information from a 2D model must be done outside the CAD package. That is tedious to do and can only be approximate when the geometry is complex. But, if the part is designed in a solid modeling CAD system such as ProEngineer,[7] NX,[4] or one of many others, the mass properties can be calculated for the most complicated part geometries.
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Solid modeling systems usually provide an interface to one or more Finite Element Analysis (FEA) programs and allow direct transfer of the model’s geometry to the FEA package for stress, vibration, and heat transfer analysis. Some CAD systems include a meshgeneration feature that creates the FEA mesh automatically before sending the data to the FEA software. This combination of tools provides an extremely powerful means to obtain superior designs whose stresses are more accurately known than would be possible by conventional analysis techniques when the geometry is complex. While it is highly likely that the students reading this textbook will be using CAD tools including finite element or boundary element analysis (BEA) methods in their professional practice, it is still necessary that the fundamentals of applied stress analysis be thoroughly understood. That is the purpose of this text. FEA techniques will be discussed in Chapters 4 and 8 but will not be emphasized in this text. Rather, we will concentrate on the classical stressanalysis techniques in order to lay the foundation for a thorough understanding of the fundamentals and their application to machine design. FEA and BEA methods are rapidly becoming the methods of choice for the solution of complicated stressanalysis problems. However, there is danger in using those techniques without a solid understanding of the theory behind them. These methods will always give some results. Unfortunately, those results can be incorrect if the problem was not well formulated and well meshed with proper boundary conditions applied. Being able to recognize incorrect results from a computeraided solution is extremely important to the success of any design. Chapter 8 provides a brief introduction to FEA. The student should take courses in FEA and BEA to become familiar with these tools. Figure 12 shows a solid model of the ball bracket from Figure 11 that was created in a CAD software package. The shaded, isometric view in the upper right corner shows that the solid volume of the part is defined. The other three views show orthographic projections of the part. Figure 13 shows the massproperties data that are calculated by the software. Figure 14 shows a wireframe rendering of the same part generated from the solid geometry data base. A wireframe version is used principally to speed up the screendrawing time when working on the model. There is much less wireframe display information to calculate than for the solid rendering of Figure 12. Figure 15 shows a fully dimensioned, orthographic, multiview drawing of the ball bracket that was generated in the CAD software package. Another major advantage of creating a solid model of a part is that the dimensional and toolpath information needed for its manufacture can be generated in the CAD system and sent over a network to a computercontrolled machine on the manufacturing floor. This feature allows the production of parts without the need for paper drawings such as Figure 15. Figure 16 shows the same part after a finite element mesh was applied to it by the CAD software before sending it to the FEA software for stress analysis. ComputerAided Engineering (CAE) The techniques generally referred to above as CAD are a subset of the more general topic of computeraided engineering (CAE), which term implies that more than just the geometry of the parts is being dealt with. However, the distinctions between CAD and CAE continue to blur as more sophisticated software packages become available. In fact, the description of the use of a solid modeling CAD system and an FEA package together as
Chapter 1
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1
FIGURE 12 A CAD Solid Model of the Ball Bracket from the Trailer Hitch Assembly of Figure 11
described in the previous section is an example of CAE. When some analysis of forces, stresses, deflections, or other aspects of the physical behavior of the design is included, with or without the solid geometry aspects, the process is called CAE. Many commercial software packages do one or more aspects of CAE. The FEA and BEA software packages mentioned previously are in this category. See Chapter 8 for more information on FEA. Dynamic force simulations of mechanisms can be done with such packages as ADAMS[5] and Working Model.[6] Some software packages such as ProEngineer,[7] Solidworks,[12] NX,[4] and others combine aspects of a CAD system with general analysis capabilities.
FIGURE 13 Mass Properties of the Ball Bracket Calculated Within the CAD System from its Solid Model
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FIGURE 14 A Wireframe Representation of the Ball Bracket Generated from its Solid Model in a CAD System
These constraintbased programs allow constraints to be applied to the design which can control the part geometry as the design parameters are varied. Other classes of tools for CAE are equation solvers such as MATLAB[11], Mathcad,[9] TK Solver,[8] and spreadsheets such as Excel.[10] These are generalpurpose tools that will allow any combination of equations to be encoded in a convenient form and then will manipulate the equation set (i.e., the engineering model) for different trial data and conveniently display tabular and graphic output. Equation solvers are invaluable for the solution of force, stress, and deflection equations in machinedesign problems because they allow rapid “whatif” calculations to be done. The effects of dimensional or material
26.0 D 64 R
32 38.0
70 19 70
50
10.0 D TYP. 2 PL.
13 R 90° ± 2°
13
Widgets Perfected Inc. Title: BALL BRACKET
ASMILLED FINISH
Mat'l: AISI 1020 HotRolled Steel All dims in: mm By: RLN 4/12/04 xx. ± 1 xx.x ± 0.5 Dwg #: B–579328
FIGURE 15 A Dimensioned, 3View Orthographic Drawing Done in a 2D CAD Drawing Package
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1
FIGURE 16 An FEA Mesh Applied to the Solid Model of the Ball Bracket in the CAD System
changes on the stresses and deflections in the part can be seen instantly. In the absence of a true solid modeling system, an equation solver also can be used to approximate the part’s mass properties while iterating the geometry and material properties of trial part designs. Rapid iteration to an acceptable solution is thus enhanced. The website accompanying the text (see Preface for information on how to access this) contains a large number of models for various equation solvers that support the examples and case studies presented in the text. Introductions to the use of TK Solver and Mathcad along with examples of their use are provided as PDF files on the book’s website. In addition, some customwritten computer programs, Mohr, Contact, Linkages, Dynacam, and Matrix are provided on the book’s website to aid in the calculation of dynamic loads and stresses when solving the openended design problems assigned. However, one must be aware that these computer tools are just tools and are not a substitute for the human brain. Without a thorough understanding of the engineering fundamentals on the part of the user, the computer will not give good results. Garbage in, garbage out. Caveat Lector.* Computational Accuracy Computers and calculators make it very easy to obtain numerical answers having many significant figures. Before writing down all those digits, you are advised to recall the accuracy of your initial assumptions and given data. If, for example, your applied loads were known to only two significant figures, it is incorrect and misleading to express the calculated stresses to more significant figures than your input data possessed. However, it is valid and appropriate to make all intermediate calculations to the greatest accuracy available in your computational tools. This will minimize computational roundoff errors. But, when done, round off the results to a level consistent with your known or assumed data.
* Reader beware.
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*
1.6
Excerpted from Norton, Design of Machinery, 5ed McGrawHill, New York, 2012, with the publisher’s permission.

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THE ENGINEERING REPORT*
Communication of your ideas and results is a very important aspect of engineering. Many engineering students picture themselves in professional practice spending most of their time doing calculations of a nature similar to those they have done as students. Fortunately, this is seldom the case, as it would be very boring. Actually, engineers spend a large percentage of their time communicating with others, either orally or in writing. Engineers write proposals and technical reports, give presentations, and interact with support personnel. When your design is done, it is usually necessary to present the results to your client, peers, or employer. The usual form of presentation is a formal engineering report. In addition to a written description of the design, these reports will usually contain engineering drawings or sketches as described earlier, as well as tables and graphs of data calculated from the engineering model. It is very important for engineering students to develop their communication skills. You may be the cleverest person in the world, but no one will know that if you cannot communicate your ideas clearly and concisely. In fact, if you cannot explain what you have done, you probably don’t understand it yourself. The designproject assignments in Chapter 9 are intended to be written up in formal engineering reports to give you some experience in this important skill of technical communication. Information on writing engineering reports can be found in the suggested readings listed in the bibliography. 1.7
FACTORS OF SAFETY AND DESIGN CODES
The quality of a design can be measured by many criteria. It is always necessary to calculate one or more factors of safety to estimate the likelihood of failure. There may be legislated, or generally accepted, design codes that must be adhered to as well. †
Also called safety factor. We will use both terms interchangeably in this text.
Factor of Safety† A factor of safety or safety factor can be expressed in many ways. It is typically a ratio of two quantities that have the same units, such as strength/stress, critical load/applied load, load to fail part/expected service overload, maximum cycles/applied cycles, or maximum safe speed/operating speed. A safety factor is always unitless. The form of expression for a safety factor can usually be chosen based on the character of loading on the part. For example, consider the loading on the side wall of a cylindrical water tower that can never be “more than full” of a liquid of known density within a known temperature range. Since this loading is highly predictable over time, a comparison of the strength of the material to the stress in the wall of a full tank might be appropriate as a safety factor. Note in this example that the possibility of rust reducing the thickness of the wall over time must be considered. (See Section 4.17 for a discussion of stresses in cylinder walls and Section 7.6 for a discussion of corrosion.) If this cylindrical water tower is standing on legs loaded as columns, then a safety factor for the legs based on a ratio of the column’s critical buckling load over the applied load from a full water tower would be appropriate. (See Section 4.16 for a discussion of column buckling.)
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If a part is subjected to loading that varies cyclically with time, it may experience fatigue failure. The resistance of a material to some types of fatigue loading can be expressed as a maximum number of cycles of stress reversal at a given stress level. In such cases, it may be appropriate to express the safety factor as a ratio of the maximum number of cycles to expected material failure over the number of cycles applied to the part in service for its desired life. (See Chapter 6 for a discussion of fatiguefailure phenomena and several approaches to the calculation of safety factors in such situations.) The safety factor of a part such as a rotating sheave (pulley) or flywheel is often expressed as a ratio of its maximum safe speed over the highest expected speed in service. In general, if the stresses in the parts are a linear function of the applied service loads and those loads are predictable, then a safety factor expressed as strength/stress or failure load/applied load will give the same result. Not all situations fit these criteria. Some require a nonlinear ratio. A column is one example, because its stresses are a nonlinear function of the loading (see Section 4.16). Thus, a critical (failure) load for the particular column must be calculated for comparison to the applied load. Another complicating factor is introduced when the magnitudes of the expected applied loads are not accurately predictable. This can be true in virtually any application in which the use (and thus the loading) of the part or device is controlled by humans. For example, there is really no way to prevent someone from attempting to lift a 10ton truck with a jack designed to lift a 2ton automobile. When the jack fails, the manufacturer (and designer) may be blamed even though the failure was probably due more to the “nut behind the jack handle.” In situations where the user may subject the device to overloading conditions, an assumed overload may have to be used to calculate a safety factor based on a ratio of the load that causes failure over the assumed service overload. Labels warning against inappropriate use may be needed in these situations as well. Since there may be more than one potential mode of failure for any machine element, it can have more than one value of safety factor N. The smallest value of N for any part is of greatest concern, since it predicts the most likely mode of failure. When N becomes reduced to 1, the stress in the part is equal to the strength of the material (or the applied load is equal to the load that fails it, etc.) and failure occurs. Therefore, we desire N to be always greater than 1. Choosing a Safety Factor Choosing a safety factor is often a confusing proposition for the beginning designer. The safety factor can be thought of as a measure of the designer’s uncertainty in the analytical models, failure theories, and materialproperty data used, and should be chosen accordingly. How much greater than one N must be depends on many factors, including our level of confidence in the model on which the calculations are based, our knowledge of the range of possible inservice loading conditions, and our confidence in the available materialstrength information. If we have done extensive testing on physical prototypes of our design to prove the validity of our engineering model and of the design, and have generated test data on the particular material’s strengths, then we can afford to use a smaller safety factor. If our model is less well proven or the materialproperty information is less reliable, a larger N is in order. In the absence of any design codes that may specify N for particular cases, the choice of factor of safety involves engineering judgment. A reasonable approach is to determine the largest loads expected in service (including possible overloads) and the minimum expected material strengths and base
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the safety factors on these data. The safety factor then becomes a reasonable measure of uncertainty. If you fly, it may not give you great comfort to know that safety factors for commercial aircraft are in the range of 1.2 to 1.5. Military aircraft can have N < 1.1, but their crews wear parachutes. (Test pilots deserve their high salaries.) Missiles have N = 1 but have no crew and aren’t expected to return anyway. These small factors of safety in aircraft are necessary to keep weight low and are justified by sophisticated analytical modeling (usually involving FEA), testing of the actual materials used, extensive testing of prototype designs, and rigorous inservice inspections for incipient failures of the equipment. The opening photograph of this chapter shows an elaborate test rig used by the Boeing Aircraft Co. to mechanically test the airframe of fullscale prototype or production aircraft by applying dynamic forces and measuring their effects. It can be difficult to predict the kinds of loads that an assembly will experience in service, especially if those loads are under the control of the enduser, or Mother Nature. For example, what loads will the wheel and frame of a bicycle experience? It depends greatly on the age, weight, and recklessness of the rider, whether used on or offroad, etc. The same problem of load uncertainty exists with all transportation equipment, ships, aircraft, automobiles, etc. Manufacturers of these devices engage in extensive test programs to measure typical service loads. See Figures 341 and 67 for examples of such serviceload data. Some guidelines for the choice of a safety factor in machine design can be defined based on the quality and appropriateness of the materialproperty data available, the expected environmental conditions compared to those under which the material test data were obtained, and the accuracy of the loading and stressanalysis models developed for the analyses. Table 13 shows a set of factors for ductile materials which can be chosen in each of the three categories listed based on the designer’s knowledge or judgment of the quality of information used. The overall safety factor is then taken as the largest of the three factors chosen. Given the uncertainties involved, a safety factor typically should not be taken to more than 1 decimal place accuracy. N ductile ≅ MAX ( F1, F 2, F 3)
(1.1a)
The ductility or brittleness of the material is also a concern. Brittle materials are designed against the ultimate strength, so failure means fracture. Ductile materials under static loads are designed against the yield strength and are expected to give some visible warning of failure before fracture, unless cracks indicate the possibility of a fracturemechanics failure (see Sections 53 and 65). For these reasons, the safety factor for brittle materials is often made twice that which would be used for a ductile material in the same situation: N brittle ≅ 2 * MAX ( F1, F 2, F 3)
(1.1b)
This method of determining a safety factor is only a guideline to obtain a starting point and is obviously subject to the judgment of the designer in selecting factors in each category. The designer has the ultimate responsibility to ensure that the design is safe. A larger safety factor than any shown in Table 13 may be appropriate in some circumstances.
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1 Table 13
Factors Used to Determine a Safety Factor for Ductile Materials
Information
Quality of Information
Factor
The actual material used was tested
F1 1.3
Materialproperty data
Representative material test data are available
2
available from tests
Fairly representative material test data are available
3
Poorly representative material test data are available Are identical to material test conditions
5+ F2 1.3
Environmental conditions
Essentially roomambient environment
2
in which it will be used
Moderately challenging environment
3
Extremely challenging environment Models have been tested against experiments
5+ F3 1.3
Analytical models for
Models accurately represent system
2
loading and stress
Models approximately represent system
3
Models are crude approximations
5+
Design and Safety Codes Many engineering societies and government agencies have developed codes for specific areas of engineering design. Most are only recommendations, but some have the force of law. The ASME provides recommended guidelines for safety factors to be used in particular applications such as steam boilers and pressure vessels. Building codes are legislated in most U.S. states and cities and usually deal with publicly accessible structures or their components, such as elevators and escalators. Safety factors are sometimes specified in these codes and may be quite high. (The code for escalators in one state called for a factor of safety of 14.) Clearly, where human safety is involved, high values of N are justified. However, they come with a weight and cost penalty, as parts must often be made heavier to achieve large values of N. The design engineer must always be aware of these codes and standards and adhere to them where applicable. The following is a partial list of engineering societies and governmental, industrial, and international organizations that publish standards and codes of potential interest to the mechanical engineer. Addresses and data on their publications can be obtained in any technical library or from the Internet. American Gear Manufacturers Association (AGMA) http://www.agma.org/ American Institute of Steel Construction (AISC) http://www.aisc.org/ American Iron and Steel Institute (AISI) http://www.steel.org/ American National Standards Institute (ANSI) http://www.ansi.org/ American Society for Metals (ASM International) http://www.asmintl.org/ American Society of Mechanical Engineers (ASME) http://www.asme.org/
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American Society of Testing and Materials (ASTM) http://www.astm.org/ American Welding Society (AWS) http://www.aws.org/ American Bearing Manufacturers Association (ABMA) www.abmadc.org/ International Standards Organization (ISO) http://www.iso.ch/iso/en National Institute for Standards and Technology (NIST)* http://www.nist.gov/
*
Formerly the National Bureau of Standards (NBS).
Society of Automotive Engineers (SAE) http://www.sae.org/ Society of Plastics Engineers (SPE) http://www.4spe.org/ Underwriters Laboratories (UL) http://www.ul.com/
1.8
STATISTICAL CONSIDERATIONS
Nothing is absolute in engineering any more than in any other endeavor. The strengths of materials will vary from sample to sample. The actual size of different examples of the “same” part made in quantity will vary due to manufacturing tolerances. As a result, we should take the statistical distributions of these properties into account in our calculations. The published data on the strengths of materials may be stated either as minimum values or as average values of tests made on many specimens. If it is an average value, there is a 50% chance that a randomly chosen sample of that material will be weaker or stronger than the published average value. To guard against failure, we can reduce the materialstrength value that we will use in our calculations to a level that will include a larger percentage of the population. To do this requires some understanding of statistical phenomena and their calculation. All engineers should have this understanding and should include a statistics course in their curriculum. We briefly discuss some of the fundamental aspects of statistics in Chapter 2. †
Excerpted from Norton, Design of Machinery, 3ed, 2004, McGrawHill, New York, with the publisher’s permission.
1.9
UNITS†
Several different systems of units are used in engineering. The most common in the United States are the U.S. footpoundsecond system (fps), the U.S. inchpoundsecond system (ips), and the Systeme Internationale (SI). The metric centimeter, gram, second (cgs) system is being used more frequently in the United States, particularly in international companies, e.g., the automotive industry. All systems are created from the choice of three of the quantities in the general expression of Newton’s second law: F=
mL t2
(1.2a)
where F is force, m is mass, L is length, and t is time. The units for any three of these variables can be chosen and the other is then derived in terms of the chosen units. The three chosen units are called base units, and the remaining one is a derived unit. Most of the confusion that surrounds the conversion of computations between either one of the U.S. systems and the SI system is due to the fact that the SI system uses a different set of base units than the U.S. systems. Both U.S. systems choose force, length, and time as the base units. Mass is then a derived unit in the U.S. systems, which are referred to as gravitational systems because the value of mass is dependent on the local gravitational constant. The SI system chooses mass, length, and time as the base units,
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and force is the derived unit. SI is then referred to as an absolute system, since the mass is a base unit whose value is not dependent on local gravity. The U.S. footpoundsecond (fps) system requires that all lengths be measured in feet (ft), forces in pounds (lb), and time in seconds (sec). Mass is then derived from Newton’s law as m=
Ft 2 L
(1.2b)
and its units are pounds seconds squared per foot (lb sec2/ft) = slugs. The U.S. inchpoundsecond (ips) system requires that all lengths be measured in inches (in), forces in pounds (lb), and time in seconds (sec). Mass is still derived from Newton’s law, equation 1.2b, but the units are now pounds seconds squared per inch (lb sec2/in) = blobs†
†
This mass unit is not slugs! It is worth twelve slugs or one “blob”! Weight is defined as the force exerted on an object by gravity. Probably the most common units error that students make is to mix up these two unit systems (fps and ips) when converting weight units (which are pounds force) to mass units. Note that the gravitational acceleration constant (g or gc) on earth at sea level is approximately 32.17 feet per second squared, which is equivalent to 386 inches per second squared. The relationship between mass and weight is mass = weight / gravitational acceleration m=
W gc
(1.3)
It should be obvious that if you measure all your lengths in inches and then use g = gc = 32.17 feet/sec2 to compute mass, you will have an error of a factor of 12 in your results. This is a serious error, large enough to crash the airplane you designed. Even worse is the student who neglects to convert weight to mass at all. The results of this calculation will have an error of either 32 or 386, which is enough to sink the ship!* The value of mass is needed in Newton’s secondlaw equation to determine forces due to accelerations: F = ma
(1.4 a)
The units of mass in this equation are either g, kg, slugs, or blobs depending on the units system used. Thus, in either English system, the weight W (lbf) must be divided by the acceleration due to gravity gc as indicated in equation 1.3 to get the proper mass quantity for equation 1.4a. Adding further to the confusion is the common use of the unit of pounds mass (lbm). This unit, often used in fluid dynamics and thermodynamics, comes about through the use of a slightly different form of Newton’s equation: F=
ma gc
(1.4 b)
It is unfortunate that the mass unit in the ips system has never officially been given a name such as the term slug used for mass in the fps system. The author boldly suggests (with tongue only slightly in cheek) that this unit of mass in the ips system be called a blob (bl) to distinguish it more clearly from the slug (sl), and to help the student avoid some of the common errors listed below. Twelve slugs = one blob. Blob does not sound any sillier than slug, is easy to remember, implies mass, and has a convenient abbreviation (bl), which is an anagram for the abbreviation for pound (lb). Besides, if you have ever seen a garden slug, you know it looks just like a “little blob.”
22
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MACHINE DESIGN

An Integrated Approach

Sixth Edition
where m = mass in lbm, a = acceleration, and gc = the gravitational constant. On earth, the value of the mass of an object measured in pounds mass (lbm) is numerically equal to its weight in pounds force (lbf). However, the student must remember to divide the value of m in lbm by gc when using this form of Newton’s equation. Thus, the lbm will be divided either by 32.17 or by 386 when calculating the dynamic force. The result will be the same as when the mass is expressed in either slugs or blobs in the F = ma form of the equation. Remember that in round numbers at sea level on earth 1 lbm = 1 lbf
1 slug = 32.17 lbf
1 blob = 386 lbf
The SI system requires that lengths be measured in meters (m), mass in kilograms (kg), and time in seconds (sec). This is sometimes also referred to as the mks system. Force is derived from Newton’s law and the units are: kg m/sec2 = newtons * A 125milliondollar space probe was lost because NASA failed to convert data that had been supplied in ips units by its contractor, Lockheed Aerospace, into the metric units used in the NASA computer programs that controlled the spacecraft. It was supposed to orbit the planet Mars, but instead either burned up in the Martian atmosphere or crashed into the planet because of this units error. Source: The Boston Globe, October 1, 1999, p. 1. †
A valuable resource for information on the proper use of SI units can be found at the U.S. Government NIST site at https:// physics.nist.gov/cuu/Units Another excellent resource on the proper use of metric units in machine design can be found in a pamphlet “Metric Is Simple,” published and distributed by the fastener company Bossard International Inc., 235 Heritage Avenue, Portsmouth, NH 03801 http://www.bossard.com/
In the SI system there are distinct names for mass and force, which helps alleviate confusion.* When converting between SI and U.S. systems, be alert to the fact that mass converts from kilograms (kg) to either slugs (sl) or blobs (bl), and force converts from newtons (N) to pounds (lb). The gravitational constant (gc) in the SI system is approximately 9.81 m/sec2. The cgs system requires that lengths be measured in centimeters (cm), mass in grams (g), and time in seconds (sec). Force is measured in dynes. The SI system is generally preferred over the cgs system. The systems of units used in this textbook are the U.S. ips system and the SI system. Much of machine design in the United States is still done in the ips system, though the SI system is becoming more common.† Table 14 shows some of the variables used in this text and their units. Table 15 shows a number of conversion factors between commonly used units. The student is cautioned always to check the units in any equation written for a problem solution, whether in school or in professional practice. If properly written, an equation should cancel all units across the equal sign. If it does not, then you can be absolutely sure it is incorrect. Unfortunately, a unit balance in an equation does not guarantee that it is correct, as many other errors are possible. Always doublecheck your results. You might save a life.
EXAMPLE 11 Units Conversion Problem
The weight of an automobile is known in lbf. Convert it to mass units in the SI, cgs, fps, and ips systems. Also convert it to lbm.
Given
The weight = 4500 lbf.
Assumptions
The automobile is on earth at sea level.
Solution
1 Equation 1.4a is valid for the first four systems listed. For the fps system:
Chapter 1
INTRODUCTION TO DESIGN
23
1 Table 14
Variables and Units Base Units in Boldface—Abbreviations in ( )
Variable
Symbol
ips unit
Force
F
pounds (lb)
pounds (lb)
newtons (N)
Length
l
inches (in)
feet (ft)
meters (m)
Time
t
seconds (sec)
seconds (sec)
seconds (sec)
Mass
m
lbsec2/in (bl)
lbsec2/ft (sl)
kilograms (kg)
Weight
W
pounds (lb)
pounds (lb)
newtons (N)
Pressure
p
psi
psf
N/m2 = Pa
Velocity
v
in/sec
ft/sec
m/sec
a
in/sec2
ft/sec2
m/sec2
psi
psf
N/m2 = Pa
Acceleration
fps unit
SI unit
Stress
σ, τ
Angle
θ
degrees (deg)
degrees (deg)
degrees (deg)
Angular velocity
ω
radians/sec
radians/sec
radians/sec
Angular acceleration
α
radians/sec2
radians/sec2
radians/sec2
Torque
T
lbin
lbft
Nm
Mass moment of inertia
I
lbinsec2
lbftsec2
kgm2
Area moment of inertia
I
in4
ft4
m4
Energy
E
inlb
ftlb
joules = Nm
Power
P
inlb/sec
ftlb/sec
Nm/sec = watt
Volume
V
in3
ft3
m3
Specific weight
ν
lb/in3
lb/ft3
N/m3
ρ
bl/in3
sl/ft3
kg/m3
Mass density
m=
lb − sec 2 4500 lbf W = = 139.9 f = 139.9 slugs 2 ft g 32.17 ft sec
(a)
m=
lbf − sec 2 4500 lbf W 11.66 = = = 11.66 blobs in g 386 in sec 2
(b)
For the ips system:
For the SI system: W = 4500 lb m=
4.448 N = 20 016 N lb
20 016 N W N − sec 2 = = 2040 = 2040 kg 2 g 9.81 m sec m
(c)
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MACHINE DESIGN

An Integrated Approach

Sixth Edition
1 Table 15
Selected Units Conversion Factors Note That These Conversion Factors (and Others) are Built Into the TK Solver Files UNITMAST and STUDENT
Multiply this
by
this
to get
this
by
this
to get
this
mass moment of inertia
acceleration in/sec2
x
ft/sec2
x
0.0254 12
=
m/sec2
=
in/sec2
lbinsec2
x
57.2958
=
deg
area in2
x
645.16
=
mm2
ft2
x
144
=
in2
area moment of inertia
x x
0.1138 12
=
Nm
x
=
inlb
x
8.7873
=
inlb
Nm
x
0.7323
=
ftlb
power HP
x
550
=
ftlb/sec
HP
x
33 000
=
ftlb/min
6600
=
inlb/sec
=
watts
=
inlb/sec
x
=
mm4
HP
x
x
4.162E–07
=
m4
HP
x
m4
x
1.0E+12
=
mm4
Nm/sec
x
m4
x
1.0E+08
=
cm4
ft4
x
=
in4
density
Nmsec2
ftlb
in4
20 736
=
Nm
in4
416 231
0.1138
moments and energy inlb
angles radian
Multiply this
745.7 8.7873
pressure and stress psi
x
psi
x
6894.8 6.895E3
=
Pa
=
MPa
lb/in3
x
27.6805
=
g/cc
psi
x
144
=
psf
g/cc
x
0.001
=
g/mm3
kpsi
x
1000
=
psi
N/m2
x
1
=
Pa
N/mm2
x
1
=
MPa
175.126
lb/ft3
x
=
lb/in3
kg/m3
x
1.0E–06
=
g/mm3
lb
x
4.448
=
N
lb/in
x
=
N/m
N
x
1.0E+05
=
dyne
lb/ft
x
0.08333
=
lb/in
ton (short)
x
=
lb
x
0.909
=
ksiin0.5
0.0254
=
m/sec
=
in/sec
=
rpm
1728
spring rate
force
2000
length
MPam0.5
in
x
25.4
=
mm
ft
x
12
=
in
mass blob
x
=
lb
slug
x
32.17
=
lb
blob
x
12
=
slug
kg
x
2.205
=
lb
kg
x
9.8083
=
N
kg
stress intensity
x
386
1000
=
g
velocity in/sec
x
ft/sec
x
rad/sec
x
12 9.5493
volume in3
x
16 387.2
=
mm3
ft3
x
1728
=
in3
cm3
x
0.061 023 =
in3
m3
x
1.0E+9
mm3
=
Chapter 1
INTRODUCTION TO DESIGN
1
For the cgs system: W = 4500 lb m=
4.448 E 5 dynes = 2.002 E 9 dynes lb
W 2.002 E 9 dynes dynes − sec 2 = = 2.04 E 6 = 2.04 E 6 g 2 g cm 981 cm sec
(d )
2 For mass expressed in lbm, equation 1.4b must be used. m=W
386 in sec 2 gc = 4500 lbf = 4500 lbm g 386 in sec 2
(e)
Note that lbm is numerically equal to lbf and so must not be used as a mass unit unless you are using the form of Newton’s law expressed as equation 1.4b.
1.10
25
SUMMARY
Design can be fun and frustrating at the same time. Because design problems are very unstructured, a large part of the task is creating sufficient structure to make it solvable. This naturally leads to multiple solutions. To students used to seeking an answer that matches the one in the “back of the book” this exercise can be frustrating. There is no “one right answer” to a design problem, only answers that are arguably better or worse than others. The marketplace has many examples of this phenomenon. How many different makes and models of new automobiles are available? Don’t they all do more or less the same task? But you probably have your own opinion about which ones do the task better than others. Moreover, the task definition is not exactly the same for all examples. A fourwheeldrive automobile is designed for a slightly different problem definition than is a twoseat sports car (though some examples incorporate both those features). The message to the beginning designer then is to be openminded about the design problems posed. Don’t approach design problems with the attitude of trying to find “the right answer,” as there is none. Rather, be daring! Try something radical. Then test it with analysis. When you find it doesn’t work, don’t be disappointed; instead realize that you have learned something about the problem you didn’t know before. Negative results are still results! We learn from our mistakes and can then design a better solution the next time. This is why iteration is so crucial to successful design. The computer is a necessary tool to the solution of contemporary engineering problems. Problems can be solved more quickly and more accurately with proper use of computeraided engineering (CAE) software. However, the results are only as good as the quality of the engineering models and data used. The engineer should not rely on computergenerated solutions without also developing and applying a thorough understanding of the fundamentals on which the model and the CAE tools are based. Important Equations Used in This Chapter
See the referenced sections for information on the proper use of these equations.
26
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MACHINE DESIGN

An Integrated Approach

Sixth Edition
Mass (see Section 1.9):
m=
W gc
(1.3)
Dynamic Force—for use with standard mass units (kg, slugs, blobs) (see Section 1.9):
F = ma
(1.4 a)
Dynamic Force—for use with mass in lbm = lbf (see Section 1.9):
F=
1.11
ma gc
(1.4 b)
REFERENCES
1
Random House Dictionary of the English Language. 2e. unabridged, S.B. Flexner, ed., Random House: New York, 1987, p. 1151.
2
R. L. Norton, Design of Machinery: An Introduction to the Synthesis and Analysis of Mechanisms and Machines, 5e. McGrawHill: New York, 2012, pp. 7–14.
3
Autocad, Autodesk Inc., http://usa.autodesk.com
4
NX, Siemens Inc, http://www.plm.automation.siemens.com/en_us/products/nx/
5
ADAMS, MSC, http://www.mscsoftware.com/Products/CAETools/Adams.aspx
6
Working Model, DST, http://www.designsimulation.com/wm2d/index.php.
7
Pro/Engineer, PTC, http://www.ptc.com/community/landing/wf3.htm
8
TK Solver, Universal Technical Systems, http://www.uts.com
9
Mathcad, PTC, www.ptc.com/Mathcad
10
Excel, Microsoft Corp., http://office.microsoft.com/enus/excel/
11
MATLAB, Mathworks Inc, http://www.mathworks.com/products/matlab/
12
Solidworks, Dassault Systemes, http://www.solidworks.com
1.12
WEB REFERENCES
http://www.onlineconversion.com
Convert just about anything to anything else. Over 5000 units, and 50 000 conversions. http://www.katmarsoftware.com/uconeer.htm
Download a units converter program for engineers. http://global.ihs.com
Search a collection of technical standards with over 500 000 documents available for electronic download. http://www.thomasnet.com
An online resource for finding companies and products manufactured in North America.
Chapter 1
1.13
INTRODUCTION TO DESIGN
BIBLIOGRAPHY
For information on creativity and the design process, the following are recommended:
J. L. Adams, The Care and Feeding of Ideas. 3rd ed. Addison Wesley: Reading, Mass., 1986. J. L. Adams, Conceptual Blockbusting. 3rd ed. Addison Wesley: Reading, Mass., 1986. J. R. M. Alger and C. V. Hays, Creative Synthesis in Design. PrenticeHall: Englewood Cliffs, N.J., 1964. M. S. Allen, Morphological Creativity. PrenticeHall: Englewood Cliffs, N.J., 1962. H. R. Buhl, Creative Engineering Design. Iowa State University Press: Ames, Iowa, 1960. W. J. J. Gordon, Synectics. Harper and Row: New York, 1962. J. W. Haefele, Creativity and Innovation. Reinhold: New York, 1962. L. Harrisberger, Engineersmanship. 2nd ed. Brooks/Cole: Monterey, Calif., 1982. D. A. Norman, The Psychology of Everyday Things. Basic Books: New York, 1986. A. F. Osborne, Applied Imagination. Scribners: New York, 1963. C. W. Taylor, Widening Horizons in Creativity. John Wiley: New York, 1964. E. K. Von Fange, Professional Creativity. PrenticeHall: Englewood Cliffs, N.J., 1959. For information on writing engineering reports, the following are recommended:
R. Barrass, Scientists Must Write. Chapman and Hall: New York, 1978. W. G. Crouch and R. L. Zetler, A Guide to Technical Writing. 3rd ed. The Ronald Press Co.: New York, 1964. D. S. Davis, Elements of Engineering Reports. Chemical Publishing Co.: New York, 1963. D. E. Gray, So You Have to Write a Technical Report. Information Resources Press: Washington, D.C., 1970. H. B. Michaelson, How to Write and Publish Engineering Papers and Reports. ISI: Philadelphia, Pa., 1982. J. R. Nelson, Writing the Technical Report. 3rd ed. McGrawHill: New York, 1952.
1.14
PROBLEMS
11
It is often said, “Build a better mousetrap and the world will beat a path to your door.” Consider this problem and write a goal statement and a set of at least 12 task specifications that you would apply to its solution. Then suggest three possible concepts to achieve the goal. Make annotated, freehand sketches of the concepts.
12
A bowling machine is desired to allow quadriplegic youths, who can only move a joystick, to engage in the sport of bowling at a conventional bowling alley. Consider the factors involved, write a goal statement, and develop a set of at least 12 task specifications that constrain this problem. Then suggest three possible concepts to achieve the goal. Make annotated, freehand sketches of the concepts.
27
1
28
1 Table P10
MACHINE DESIGN
13
Topic/Problem Matrix

An Integrated Approach

Sixth Edition
A quadriplegic needs an automated pageturner to allow her to read books without assistance. Consider the factors involved, write a goal statement, and develop a set of at least 12 task specifications that constrain this problem. Then suggest three possible concepts to achieve the goal. Make annotated, freehand sketches of the concepts.
1.4 Engineering Model
*14
Convert a mass of 1000 lbm to (a) lbf, (b) slugs, (c) blobs, (d) kg.
11, 12, 13, 111, 113
*15
A 250lbm mass is accelerated at 40 in/sec2. Find the force in lb needed for this acceleration.
*16
Express a 100 kg mass in units of slugs, blobs, and lbm. How much does this mass weigh in lbf and in N?
1.9 Units 14, 15, 16, 17, 18, 112, 114, 115
17
Prepare an interactive computer program (using, for example, Excel, Mathcad, MATLAB, or TK Solver) from which the crosssectional properties for the shapes shown on the inside front cover can be calculated. Arrange the program to deal with both ips and SI units systems and convert the results between those systems.
18
Prepare an interactive computer program (using, for example, Excel, Mathcad, MATLAB, or TK Solver) from which the mass properties for the solids shown on the page opposite the inside front cover can be calculated. Arrange the program to deal with both ips and SI units systems and convert the results between those systems.
19
Convert the program written for Problem 17 to have and use a set of functions or subroutines that can be called from within any program in that language to solve for the crosssectional properties of the shapes shown on the inside front cover.
110
Convert the program written for Problem 18 to have and use a set of functions or subroutines that can be called from within any program in that language to solve for the mass properties for the solids shown on the page opposite the inside front cover.
111
A fledgling, nonprofit, recycling organization collects paper, cardboard, plastics, and metals from consumers in a rural county. After sorting plastics by grades 1 and 2, they are transported to a plastics recycler in a major city 150 miles away. In order to maximize the weight of the plastics that can be put into their trailer the organization desires to compact the plastic containers for transport. Consider this problem and write a goal statement and a set of at least 10 task specifications that you would apply to its solution. Then suggest three possible concepts to achieve the goal. Make annotated, freehand sketches of the concepts.
112
One square foot of ventilation is required for every 150 sq. ft. of floor space on a house with crawlspace under the floor. Prepare an interactive computer program (using, for example Excel, Mathcad, MATLAB, or TK Solver) from which the number of 40 cm by 20 cm vents can be determined that meet the ventilation requirement if only 75% of the nominal vent area is effective. Test the program using a house that is 13.5 m long by 8.25 m wide.
113
A startup company (husband and wife) manufactures over 50 different health and beauty aids that they sell at craft shows and also over the internet. They make their products in small batches and package them in small jars that they buy unlabeled in case lots. They design and print their own labels on peelable, adhesive sheets. It is very difficult to place the labels on to the containers so that they are in the proper position and properly aligned. Consider this problem and write a goal statement and at least 10 task specifications that you would apply to its solution.
114
A 360gram mass is accelerated at 250 cm/sec2. Find the force in N needed for this acceleration.
115
Express an 18blob mass in units of slugs, kilograms, and lbm. How much does this mass weigh?
*
Answers to these problems are provided in Appendix D.
MATERIALS AND PROCESSES There is no subject so old that something new cannot be said about it. Dostoevsky
2.0
INTRODUCTION
Whatever you design, you must make it out of some material and be able to manufacture it. A thorough understanding of material properties, treatments, and manufacturing processes is essential to good machine design. It is assumed that the reader has had a first course in material science. This chapter presents a brief review of some basic metallurgical concepts and a short summary of engineering material properties to serve as background for what follows. This is not intended as a substitute for a text on material science, and the reader is encouraged to review references such as those listed in the bibliography of this chapter for more detailed information. Later chapters of this text will explore some of the common materialfailure modes in more detail. Table 20 shows the variables used in this chapter and references the equations, figures, or sections in which they are used. At the end of the chapter, a summary section is provided which groups the significant equations from this chapter for easy reference and identifies the chapter section in which they are discussed. 2.1
MATERIALPROPERTY DEFINITIONS View a Video on Failure Modes (09:41)†
Mechanical properties of a material are generally determined through destructive testing of samples under controlled loading conditions. The test loadings do not accurately duplicate actual service loadings experienced by machine parts except in certain special cases. Also, there is no guarantee that the particular piece of material you purchase for your part will exhibit the same strength properties as the samples of similar materials tested previously. There will be some statistical variation in the strength of any particular † http://www.designofmachinery.com/MD/Failure_Modes.mp4
29
2
30
2
MACHINE DESIGN
Opening page photograph courtesy of the Aluminum Extruders Council.
Table 20

An Integrated Approach

Sixth Edition
Variables Used in This Chapter
Symbol
Variable
ips units
SI units
See
A A0
area
in2
m2
Sect. 2.1
original area, test specimen
in2
m2
Eq. 2.1a
E
Young's modulus
psi
Pa
Eq. 2.2
el
elastic limit
psi
Pa
Figure 22
f
fracture point
none
none
Figure 22
G
shear modulus, modulus of rigidity
psi
Pa
Eq. 2.4
HB
Brinell hardness
none
none
Eq. 2.10
HRB
Rockwell B hardness
none
none
Sect. 2.4
HRC
Rockwell C hardness
none
none
Sect. 2.4
HV
Vickers hardness
none
none
Sect. 2.4
polar second moment of area
in4
m4
Eq. 2.5
stress intensity
kpsiin0.5
MPam0.5
Sect. 2.1
Kc
fracture toughness
kpsiin0.5
MPam0.5
Sect. 2.1
l0
gage length, test specimen
in
m
Eq. 2.3
N
number of cycles
none
none
Figure 210
P
force or load
lb
N
Sect. 2.1
pl
proportional limit
psi
Pa
Figure 22
r Sd
radius
in
m
Eq. 2.5a
standard deviation
any
any
Eq. 2.9
Se
endurance limit
psi
Pa
Figure 210
Sel
strength at elastic limit
psi
Pa
Eq. 2.7
Sf
fatigue strength
psi
Pa
Figure 210
Sus
ultimate shear strength
psi
Pa
Eq. 2.5
Sut
ultimate tensile strength
psi
Pa
Figure 22
Sy
tensile yield strength
psi
Pa
Figure 22
Sys
shear yield strength
psi
Pa
Eq. 2.5c
T UR
torque
lbin
Nm
Sect. 2.1
modulus of resilience
psi
Pa
Eq. 2.7
UT
modulus of toughness
psi
Pa
Eq. 2.8
y
yield point
none
none
Figure 22 Eq. 2.1b
J K
ε
strain
none
none
σ
tensile stress
psi
Pa
Sect. 2.1
τ
shear stress
psi
Pa
Eq. 2.3
θ
angular deflection
rad
rad
Eq. 2.3
µ
arithmetic mean value
any
any
Eq. 2.9b
ν
Poisson's ratio
none
none
Eq. 2.4
sample compared to the average tested properties for that material. For this reason, many of the published strength data are given as minimum values. It is with these caveats that we must view all published materialproperty data, as it is the engineer’s responsibility to ensure the safety of his or her design.
Chapter 2
MATERIALS AND PROCESSES
31
The best materialproperty data will be obtained from destructive or nondestructive testing under actual service loadings of prototypes of your actual design, made from the actual materials by the actual manufacturing process. This is typically done only when the economic and safety risks are high. Manufacturers of aircraft, automobiles, motorcycles, snowmobiles, farm equipment, and other products regularly instrument and test finished assemblies under real or simulated service conditions. In the absence of such specific test data, the engineer must adapt and apply published materialproperty data from standard tests to the particular situation. The American Society for Testing and Materials (ASTM) defines standards for test specimens and test procedures for a variety of materialproperty measurements.* The most common material test used is the tensile test.
* ASTM, 1994 Annual Book of ASTM Standards, Vol. 03.01, Am. Soc. for Testing and Materials, Philadelphia, PA.
The Tensile Test A typical tensile test specimen is shown in Figure 21. This tensile bar is machined from the material to be tested in one of several standard diameters do and gage lengths lo. The gage length is an arbitrary length defined along the smalldiameter portion of the specimen by two indentations so that its increase can be measured during the test. The largerdiameter ends of the bar are threaded for insertion into a tensile test machine which is capable of applying either controlled loads or controlled deflections to the ends of the bar, and the gagelength portion is mirror polished to eliminate stress concentrations from surface defects. The bar is stretched slowly in tension until it breaks, while the load and the distance across the gage length (or alternatively the strain) are continuously monitored. The result is a stressstrain plot of the material’s behavior under load as shown in Figure 22a, which depicts a curve for a lowcarbon or “mild” steel. StreSS and Strain Note that the parameters measured are load and deflection, but those plotted are stress and strain. Stress (σ) is defined as load per unit area (or unit load) and for the tensile specimen is calculated from σ=
P Ao
(2.1a)
Strain is the change in length per unit length and is calculated from l − lo lo
(2.1b)
where lo is the original gage length and l is the gage length at any load P. The strain is unitless, being length divided by length. ModuluS of elaSticity This tensile stressstrain curve provides us with a number of useful material parameters. Point pl in Figure 22a is the proportional limit below which the stress is proportional to the strain, as expressed by the onedimensional form of Hooke’s law: E=
σ ε
do
FIGURE 21 A Tensile Test Specimen
where P is the applied load at any instant and Ao is the original crosssectional area of the specimen. The stress is assumed to be uniformly distributed across the cross section. The stress units are psi or Pa.
ε=
lo
(2.2)
32
MACHINE DESIGN

Stress σ
true
Sut
2
An Integrated Approach
y
Sy
u
f

Sixth Edition
Stress σ
true u
eng'g
y el pl
pl el (a)
(b)
f
eng'g
offset line E
E Elastic Range
Plastic Range
Strain ε
0.002
Strain ε
FIGURE 22 Engineering and True StressStrain Curves for Ductile Materials: (a) LowCarbon Steel (b) Annealed HighCarbon Steel
where E defines the slope of the stressstrain curve up to the proportional limit and is called Young’s modulus or the modulus of elasticity of the material. E is a measure of the stiffness of the material in its elastic range and has the units of stress. Most metals exhibit this linear stiffness behavior and also have elastic moduli that vary very little with heat treatment or with the addition of alloying elements. For example, the higheststrength steel has the same E as the loweststrength steel at about 30 Mpsi (207 GPa). For most ductile materials (defined below), the modulus of elasticity in compression is the same as in tension. This is not true for cast irons and other brittle materials (defined below) or for magnesium. elaStic liMit The point labeled el in Figure 22a is the elastic limit, or the point beyond which the material will take a permanent set, or plastic deformation. The elastic limit marks the boundary between the elasticbehavior and plasticbehavior regions of the material. Points el and pl are typically so close together that they are often considered to be the same. yield Strength At a point y slightly above the elastic limit, the material begins to yield more readily to the applied stress, and its rate of deformation increases (note the lower slope). This is called the yield point, and the value of stress at that point defines the yield strength Sy of the material. Materials that are very ductile, such as lowcarbon steels, will sometimes show an apparent drop in stress just beyond the yield point, as shown in Figure 22a. Many less ductile materials, such as aluminum and medium to highcarbon steels, will not exhibit this apparent drop in stress and will look more like Figure 22b. The yield strength of a material that does not exhibit a clear yield point has to be defined with an offset line, drawn parallel to the elastic curve and offset some small percentage along the strain axis. An offset of 0.2% strain is most often used. The yield strength is then taken at the intersection of the stressstrain curve and the offset line as shown in Figure 22b. ultiMate tenSile Strength The stress in the specimen continues to increase nonlinearly to a peak or ultimate tensile strength value Sut at point u. This is considered to be the largest tensile stress the material can sustain before breaking. However, for the ductile steel curve shown, the stress appears to fall off to a smaller value at the fracture
Chapter 2
FIGURE 23
MATERIALS AND PROCESSES
Copyright © 2018 Robert L. Norton: All Rights Reserved
A Tensile Test Specimen of Mild, Ductile Steel Before and After Fracture
point f. The drop in apparent stress before the fracture point (from u to f in Figure 22a) is an artifact caused by the “neckingdown” or reduction in area of the ductile specimen. The reduction of crosssectional area is nonuniform along the length of the specimen, as can be seen in Figure 23. Because the stress is calculated using the original area Ao in equation 2.1a, it understates the true value of stress after point u. It is difficult to accurately monitor the dynamic change in crosssectional area during the test, so these errors are accepted. The strengths of different materials can still be compared on this basis. When based on the uncorrected area Ao this is called the engineering stressstrain curve, as shown in Figure 22. The stress at fracture is actually larger than shown. Figure 22 also shows the true stressstrain curve that would result if the change in area were accounted for. The engineering stressstrain data from Figure 22 are typically used in practice. The most commonly used strength values for static loading are the yield strength Sy and the ultimate tensile strength Sut. The material stiffness is defined by Young’s modulus, E. In comparing the properties of different materials, it is quite useful to express those properties normalized to the material’s density. Since light weight is nearly always a goal in design, we seek the lightest material that has sufficient strength and stiffness to withstand the applied loads. The specific strength of a material is defined as the strength divided by the density. Unless otherwise specified, strength in this case is assumed to mean ultimate tensile strength, though any strength criterion can be so normalized. The strengthtoweight ratio (SWR) is another way to express the specific strength. Specific stiffness is the Young’s modulus divided by material density. Ductility and Brittleness The tendency for a material to deform significantly before fracturing is a measure of its ductility. The absence of significant deformation before fracture is called brittleness. ductility The stressstrain curve in Figure 22a is of a ductile material, mild steel. Take a common paper clip made of mildsteel wire. Straighten it out with your fingers. Bend it into some new shape. You are yielding this ductile steel wire but not
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Stress σ
2 Sut
f
Sy
y offset line E FIGURE 25 0.002
Strain ε
FIGURE 24 StressStrain Curve of a Brittle Material
Copyright © 2018 Robert L. Norton: All Rights Reserved
A Tensile Test Specimen of Brittle Cast Iron Before and After Fracture
fracturing it. You are operating between point y and point f on the stressstrain curve of Figure 22a. The presence of a significant plastic region on the stressstrain curve is evidence of ductility. Figure 23 shows a test specimen of ductile steel after fracture. The distortion called neckingdown can clearly be seen at the break. The fracture surface appears torn and is laced with hills and valleys, also indicating a ductile failure. The ductility of a material is measured by its percent elongation to fracture, or percent reduction in area at fracture. Materials with more than 5% elongation at fracture are considered ductile. BrittleneSS Figure 24 shows a stressstrain curve for a brittle material. Note the lack of a clearly defined yield point and the absence of any plastic range before fracture. Repeat your paperclip experiment, this time using a wooden toothpick or matchstick. Any attempt to bend it results in fracture. Wood is a brittle material. Brittle materials do not exhibit a clear yield point, so the yield strength has to be defined at the intersection of the stressstrain curve and an offset line, drawn parallel to the elastic curve and offset some small percentage such as 0.2% along the strain axis. Some brittle materials like cast iron do not have a linear elastic region, and the offset line is taken at the average slope of the region. Figure 25 shows a cast iron test specimen after fracture. The break shows no evidence of necking and has the finer surface contours typical of a brittle fracture. The same metals can be either ductile or brittle depending on the way they are manufactured, worked, and heat treated. Metals that are wrought (meaning drawn or pressed into shape in a solid form while either hot or cold) can be more ductile than metals that are cast by pouring molten metal into a mold or form. There are many exceptions to this broad statement, however. The cold working of metal (discussed below) tends to reduce its ductility and increase its brittleness. Heat treatment (discussed below) also has a marked effect on the ductility of steels. Thus, it is difficult to generalize about the relative ductility or brittleness of various materials. A careful look at all the mechanical properties of a given material will tell the story.
Chapter 2
(a)
FIGURE 26
MATERIALS AND PROCESSES
(b)
Copyright © 2018 Robert L. Norton: All Rights Reserved
Compression Test Specimens Before and After Failure (a) Ductile Steel (b) Brittle Cast Iron
The Compression Test The tensile test machine can be run in reverse to apply a compressive load to a specimen that is a constantdiameter cylinder as shown in Figure 26. It is difficult to obtain a useful stressstrain curve from this test because a ductile material will yield and increase its crosssectional area, as shown in Figure 26a, eventually stalling the test machine. The ductile sample will not fracture in compression. If enough force were available from the machine, it could be crushed into a pancake shape. Most ductile materials have compressive strengths similar to their tensile strengths, and the tensile stressstrain curve is used to represent their compressive behavior as well. A material that has essentially equal tensile and compressive strengths is called an even material. Brittle materials will fracture when compressed. A failed specimen of brittle cast iron is shown in Figure 26b. Note the rough, angled fracture surface. The reason for the failure on an angled plane is discussed in Chapter 4. Brittle materials generally have much greater strength in compression than in tension. Compressive stressstrain curves can be generated, since the material fractures rather than crushes and the crosssectional area doesn’t change appreciably. A material that has different tensile and compressive strengths is called an uneven material. The Bending Test A thin rod, as shown in Figure 27, is simply supported at each end as a beam and loaded transversely in the center of its length until it fails. If the material is ductile, failure is by yielding, as shown in Figure 27a. If the material is brittle, the beam fractures as shown in Figure 27b. Stressstrain curves are not generated from this test because the stress distribution across the cross section is not uniform. The tensile test’s σ–ε curve is used to predict failure in bending, since the bending stresses are tensile on the convex side and compressive on the concave side of the beam. The Torsion Test The shear properties of a material are more difficult to determine than its tensile properties. A specimen similar to the tensile test specimen is made with noncircular details on its ends so that it can be twisted axially to failure. Figure 28 shows two such samples, one of ductile steel and one of brittle cast iron. Note the painted lines along their lengths.
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(a )
(b)
FIGURE 27
Copyright © 2018 Robert L. Norton: All Rights Reserved
Bending Test Specimens Before and After Failure (a) Ductile Steel (b) Brittle Cast Iron
The lines were originally straight in both cases. The helical twist in the ductile specimen’s line after failure shows that it wound up for several revolutions before breaking. The brittle, torsiontest specimen’s line is still straight after failure as there was no significant plastic distortion before fracture. ModuluS of rigidity The stressstrain relation for pure torsion is defined by
Table 21
τ=
Poisson’s Ratio ν
Material
ν
Aluminum
0.34
Copper
0.35
Iron
0.28
Steel
0.28
Magnesium
0.33
Titanium
0.34
Gr θ
(2.3)
lo
where τ is the shear stress, r is the radius of the specimen, lo is the gage length, θ is the angular twist in radians, and G is the shear modulus or modulus of rigidity. G can be defined in terms of Young’s modulus E and Poisson’s ratio ν: G=
E 2 (1 + ν )
(2.4)
Poisson’s ratio (ν) is the ratio between lateral and longitudinal strain and for most metals is around 0.3 as shown in Table 21. ultiMate Shear Strength The breaking strength in torsion is called the ultimate shear strength or modulus of rupture Sus and is calculated from
(a)
FIGURE 28 Torsion Test Specimens Before and After Failure (a) Ductile Steel (b) Brittle Cast Iron
(b) Copyright © 2018 Robert L. Norton: All Rights Reserved
Chapter 2
Sus =
Tr J
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37
(2.5a)
where T is the applied torque necessary to break the specimen, r is the radius of the specimen, and J is the polar second moment of area of the cross section. The distribution of stress across the section loaded in torsion is not uniform. It is zero at the center and maximum at the outer radius. Thus, the outer portions have already plastically yielded while the inner portions are still below the yield point. This nonuniform stress distribution in the torsion test (unlike the uniform distribution in the tension test) is the reason for calling the measured value at failure of a solid bar in torsion a modulus of rupture. A thinwalled tube is a better torsiontest specimen than a solid bar for this reason and can give a better measure of the ultimate shear strength. In the absence of available data for the ultimate shear strength of a material, a reasonable approximation can be obtained from tension test data:* steels:
Sus ≅ 0.80 Sut
other ductile metals:
Sus ≅ 0.75Sut
(2.5b)
Note that the shear yield strength has a different relationship to the tensile yield strength: S ys ≅ 0.577 S y
(2.5c)
This relationship is derived in Chapter 5, where failure of materials under static loading is discussed in more detail. Fatigue Strength and Endurance Limit The tensile test and the torsion test both apply loads slowly and only once to the specimen. These are static tests and measure static strengths. While some machine parts may see only static loads in their lifetime, most will see loads and stresses that vary with time. Materials behave very differently in response to loads that come and go (called fatigue loads) than they do to loads that remain static. Most of machine design deals with the design of parts for timevarying loads, so we need to know the fatigue strength of materials under these loading conditions. One test for fatigue strength is the R. R. Moore rotatingbeam test in which a similar, but slightly smaller, test specimen than that shown in Figure 21 is loaded as a beam in bending while being rotated by a motor. Recall from your first course in strength of materials that a bending load causes tension on one side of a beam and compression on the other. (See Sections 49 and 410 for a review of beams in bending.) The rotation of the beam causes any one point on the surface to go from compression to tension to compression each cycle. This creates a loadtime curve as shown in Figure 29. The test is continued at a particular stress level until the part fractures, and the number of cycles N is then noted. Many samples of the same material are tested at various stress levels S until a curve similar to Figure 210 is generated. This is called a Wohler strengthlife diagram or an SN diagram. It depicts the breaking strength of a particular material at various numbers of repeated cycles of fully reversed stress.
* In Chapter 14 on helical spring design, an empirical relationship for the ultimate shear strength of small diameter steel wire, based on extensive testing of wire in torsion, is presented in equation 14.4 and is Sus = 0.67 Sut. This is obviously different than the general approximation for steel in equation 2.5b. The best data for material properties will always be obtained from tests of the same material, geometry, and loading as the part will be subjected to in service. In the absence of direct test data we must rely on approximations of the sort in equation 2.5b and apply suitable safety factors based on the uncertainty of these approximations.
+
stress time
– FIGURE 29 TimeVarying Loading
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log fatigue strength
Sf Sut
failure line An endurance limit Se exists for some ferrous metals and titanium alloys. Other materials show no endurance limit.
Se
Se
Sf 100 101
102
103
104
105
106
107
108
109
N
log number of cycles FIGURE 210 Wohler StrengthLife or SN Diagram Plots Fatigue Strength Against Number of Fully Reversed Stress Cycles
Note in Figure 210 that the fatigue strength Sf at one cycle is the same as the static strength Sut, and it decreases steadily with increasing numbers of cycles N (on a loglog plot) until reaching a plateau at about 106 cycles. This plateau in fatigue strength exists only for certain metals (notably steels and some titanium alloys) and is called the endurance limit Se. Fatigue strengths of other materials keep falling beyond that point. While there is considerable variation among materials, their raw (or uncorrected) fatigue strengths at about N = 106 cycles tend to be no more than about 40–50% of their static tensile strength Sut. This is a significant reduction and, as we will learn in Chapter 6, further reductions in the fatigue strengths of materials will be necessary due to other factors such as surface finish and type of loading. It is important at this stage to remember that the tensile stressstrain test does not tell the whole story and that a material’s static strength properties are seldom adequate by themselves to predict failure in a machinedesign application. This topic of fatigue strength and endurance limit is so important and fundamental to machine design that we devote Chapter 6 exclusively to a study of fatigue failure. The rotatingbeam test is now being supplanted by axialtension tests performed on modern test machines which can apply timevarying loads of any desired character to the axialtest specimen. This approach provides more testing flexibility and more accurate data because of the uniform stress distribution in the tensile specimen. The results are consistent with (but slightly lowervalued than) the historical rotatingbeam test data for the same materials. Impact Resistance The stressstrain test is done at very low, controlled strain rates, allowing the material to accommodate itself to the changing load. If the load is suddenly applied, the energy absorption capacity of the material becomes important. The energy in the differential element is its strain energy density (strain energy per unit volume U0), or the area under the stressstrain curve at any particular strain: U0 =
ε
∫0
σ dε
(2.6a)
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The strain energy U is equal to the strain energy density integrated over the volume v: U=
∫v
(2.6b)
U 0 dv
The resilience and toughness of the material are measures, respectively, of the strain energy present in the material at the elastic limit or at the fracture point. reSilience The ability of a material to absorb energy per unit volume without permanent deformation is called its resilience UR (also called modulus of resilience) and is equal to the area under the stressstrain curve up to the elastic limit, shown as the colorshaded area in Figure 22a. Resilience is defined as: UR = = UR ≅
εel
∫0
σdε =
1 Sel εel 2
S 1 1 Sel2 Sel el = E 2 2 E
(2.7)
2 1 Sy 2 E
where Sel and εel represent, respectively, the strength and strain at the elastic limit. Substitution of Hooke’s law from equation 2.2 expresses the relationship in terms of strength and Young’s modulus. Since the Sel value is seldom available, a reasonable approximation of resilience can be obtained by using the yield strength Sy instead. This relationship shows that a stiffer material of the same elastic strength is less resilient than a more compliant one. A rubber ball can absorb more energy without permanent deformation than one made of glass. toughneSS The ability of a material to absorb energy per unit volume without fracture is called its toughness UT (also called modulus of toughness) and is equal to the area under the stressstrain curve up to the fracture point, shown as the entire shaded area in Figure 22a. Toughness is defined as: UT =
εf
∫0
σdε
S y + Sut ≅ ε f 2
(2.8)
where Sut and εf represent, respectively, the ultimate tensile strength and the strain at fracture. Since an analytical expression for the stressstrain curve is seldom available for actual integration, an approximation of toughness can be obtained by using the average of the yield and ultimate strengths and the strain at fracture to calculate an area. The units of toughness and resilience are energy per unit volume (inlb/in3 or joules/m3). Note that these units are numerically equivalent to psi or Pa. A ductile material of similar ultimate strength to a brittle one will be much more tough. A sheetmetal automobile body will absorb more energy from a collision through plastic deformation than will a brittle, fiberglass body.*
* It is interesting to note that one of the toughest and strongest materials known is that of spider webs! These tiny arachnids spin a monofilament that has an ultimate tensile strength of 200 to 300 kpsi (1380 to 2070 MPa) and 35% elongation to fracture! It also can absorb more energy without rupture than any fiber known, absorbing 3 times as much energy as Kevlar, the manmade fiber used for bulletproof vests. According to the Boston Globe, January 18, 2002, researchers in Canada and the U.S. have synthesized a material with similar properties to spider silk in strands up to 10ft long with strengths of 1/4 to 1/3 that of natural silk fiber, “stronger than a steel wire of similar weight,” and one that has greater elasticity than organic silk fiber.
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iMpact teSting Various tests have been devised to measure the ability of materials to withstand impact loading. The Izod and the Charpy tests are two such procedures which involve striking a notched specimen with a pendulum and recording the kinetic energy needed to break the specimen at a particular temperature. While these data do not directly correlate with the area under the stressstrain curve, they nevertheless provide a means to compare the energy absorption capacity of various materials under controlled conditions. Materials handbooks, such as those listed in this chapter’s bibliography, give data on the impact resistance of various materials. Fracture Toughness Fracture toughness Kc (not to be confused with the modulus of toughness defined previously) is a material property that defines its ability to resist stress at the tip of a crack. The fracture toughness of a material is measured by subjecting a standardized, precracked test specimen to cyclical tensile loads until it breaks. Cracks create very high local stress concentrations which cause local yielding (see Section 4.15). The effect of the crack on the local stress is measured by a stress intensity factor K which is defined in Section 5.3. When the stress intensity K reaches the fracture toughness Kc, a sudden fracture occurs with no warning. The study of this failure phenomenon is called fracture mechanics and it is discussed in more detail in Chapters 5 and 6. Creep and Temperature Effects The tensile test, while slow, does not last long compared to the length of time an actual machine part may be subjected to constant loading. All materials will, under the right environmental conditions (particularly elevated temperatures), slowly creep (deform) under stress loadings well below the level (yield point) deemed safe in the tensile test. Ferrous metals tend to have negligible creep at room temperature or below. Their creep rates increase with increasing ambient temperature, usually becoming significant around 30–60% of the material’s absolute melting temperature. Lowmelttemperature metals such as lead, and many polymers, can exhibit significant creep at room temperature as well as increasing creep rates at higher temperatures. Creep data for engineering materials are quite sparse due to the expense and time required to develop the experimental data. The machine designer needs to be aware of the creep phenomenon and obtain the latest manufacturer’s data on the selected materials if high ambient temperatures are anticipated or if polymers are specified. The creep phenomenon is more complex than this simple description implies. See the bibliography to this chapter for more complete and detailed information on creep in materials. It is also important to understand that all material properties are a function of temperature, and published test data are usually generated at room temperature. Increased temperature usually reduces strength. Many materials that are ductile at room temperature can behave as brittle materials at low temperatures. Thus, if your application involves either elevated or low temperatures, you need to seek out relevant materialproperty data for your operating environment. Material manufacturers are the best source of uptodate information. Most manufacturers of polymers publish creep data for their materials at various temperatures.
Chapter 2
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MATERIALS AND PROCESSES
THE STATISTICAL NATURE OF MATERIAL PROPERTIES
f(x)
Some published data for material properties represent average values of many samples tested. (Other data are stated as minimum values.) The range of variation of the published test data is sometimes stated, sometimes not. Most material properties will vary about the average or mean value according to some statistical distribution such as the Gaussian or normal distribution shown in Figure 211. This curve is defined in terms of two parameters, the arithmetic mean µ and the standard deviation Sd. The equation of the Gaussian distribution curve is: f (x) =
( x − µ )2 1 , exp − 2 π Sd 2 Sd2
−∞ ≤ x ≤ ∞
41
(2.9a)
Sd Sd Sd Sd Sd Sd
x
µ FIGURE 211 The Gaussian (Normal) Distribution
where x represents some material parameter, f(x) is the frequency with which that value of x occurs in the population, and µ and Sd are defined as: µ=
Sd =
1 n
n
∑ xi
(2.9b)
i =1
1 n1
n
∑ ( xi − µ )
2
(2.9c)
i =1
The mean µ defines the most frequently occurring value of x at the peak of the curve, and the standard deviation Sd is a measure of the “spread” of the curve about the mean. A small value of Sd relative to µ means that the entire population is clustered closely about the mean. A large Sd indicates that the population is widely dispersed about the mean. We can expect to find 68% of the population within µ ± 1Sd, 95% within µ ± 2Sd, and 99% within µ ± 3Sd. There is considerable scatter in multiple tests of the same material under the same test conditions. Note that there is a 50% chance that the samples of any material that you buy will have a strength less than that material’s published mean value. Thus, you may not want to use the mean value alone as a predictor of the strength of a randomly chosen sample of that material. If the standard deviation of the test data is published along with the mean, we can “factor it down” to a lower value that is predictive of some larger percentage of the population based on the ratios listed previously. For example, if you want to have a 99% probability that all possible samples of material are stronger than your assumed material strength, you will subtract 3Sd from µ to get an allowable value for your design. This assumes that the material property’s distribution is Gaussian and not skewed toward one end or the other of the spectrum. If a minimum value of the material property is given (and used), then its statistical distribution is not of concern. Usually, no data are available on the standard deviation of the material samples tested. But you can still choose to reduce the published mean strength by a reliability factor based on an assumed Sd. One such approach assumes Sd to be some percentage of µ based on experience. Haugen and Wirsching[1] report that the standard deviations of strengths of steels seldom exceed 8% of their mean values. Table 22 shows reliability reduction factors based on an assumption of Sd = 0.08 µ for various reliabilities. Note that a 50% reliability has a factor of 1 and the factor reduces as you choose higher
Table 22 Reliability Factors for Sd = 0.08 µ Reliability %
Factor
50
1.000
90
0.897
95
0.868
99
0.814
99.9
0.753
99.99
0.702
99.999
0.659
99.9999
0.620
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reliability. The reduction factor is multiplied by the mean value of the relevant material property. For example, if you wish 99.99% of your samples to meet or exceed the assumed strength, multiply the mean strength value by 0.702. In summary, the safest approach is to develop your own materialproperty data for the particular materials and loading conditions relevant to your design. Since this approach is usually prohibitively expensive in both time and money, the engineer often must rely on published materialproperty data. Some published strength data are expressed as the minimum strength to be expected in a statistical sample, but other data may be given as the average value for the samples tested. In that case, some of the tested material samples failed at stresses lower than the average value, and your design strength may need to be reduced accordingly. 2.3
HOMOGENEITY AND ISOTROPY
All discussion of material properties so far has assumed that the material is homogeneous and isotropic. Homogeneous means that the material properties are uniform throughout its continuum, e.g., they are not a function of position. This ideal state is seldom attained in real materials, many of which are subject to the inclusion of discontinuities, precipitates, voids, or bits of foreign matter from their manufacturing process. However, most metals and some nonmetals can be considered, for engineering purposes, to be macroscopically homogeneous despite their microscopic deviations from this ideal. An isotropic material is one whose mechanical properties are independent of orientation or direction. That is, the strengths across the width and thickness are the same as along the length of the part, for example. Most metals and some nonmetals can be considered to be macroscopically isotropic. Other materials are anisotropic, meaning that there is no plane of materialproperty symmetry. Orthotropic materials have three mutually perpendicular planes of property symmetry and can have different material properties along each axis. Wood, plywood, fiberglass, and some coldrolled sheet metals are orthotropic. One large class of materials that is distinctly nonhomogeneous (i.e., heterogeneous) and nonisotropic is that of composites (also see below). Most composites are manmade, but some, such as wood, occur naturally. Wood is a composite of long fibers held together in a resinous matrix of lignin. You know from experience that it is easy to split wood along the grain (fiber) lines and nearly impossible to do so across the grain. Its strength is a function of both orientation and position. The matrix is weaker than the fibers, and it always splits between fibers. 2.4
HARDNESS
The hardness of a material can be an indicator of its resistance to wear (but is not a guarantee of wear resistance). The strengths of some materials such as steels are also closely correlated to their hardness. Various treatments are applied to steels and other metals to increase hardness and strength. These are discussed below. Hardness is most often measured on one of three scales: Brinell, Rockwell, or Vickers. These hardness tests all involve the forced impression of a small probe into the
Chapter 2
MATERIALS AND PROCESSES
surface of the material being tested. The Brinell test uses a 10mm hardened steel or tungstencarbide* ball impressed with either a 500 or 3000kg load depending on the range of hardness of the material. The diameter of the resulting indent is measured under a microscope and used to calculate the Brinell hardness number, which has the units of kgf/mm2. The Vickers test uses a diamondpyramid indenter and measures the width of the indent under the microscope. The Rockwell test uses a 1/16in ball or a 120° coneshaped diamond indenter and measures the depth of penetration. Hardness is indicated by a number followed by the letter H, followed by letter(s) to identify the method used, e.g., 375 HB or 396 HV. Several lettered scales (A, B, C, D, F, G, N, T) are used for materials in different Rockwell hardness ranges and it is necessary to specify both the letter and number of a Rockwell reading, such as 60 HRC. In the case of the Rockwell N scale, a narrowconeangle diamond indenter is used with loads of 15, 30, or 40 kg and the specification must include the load used as well as the letter specification, e.g., 84.6 HR15N. This Rockwell N scale is typically used to measure the “superficial” hardness of thin or casehardened materials. The smaller load and narrowangle Ntip give a shallow penetration that measures the hardness of the case without including effects of the soft core. All these tests are nondestructive in the sense that the sample remains intact. However, the indentation can present a problem if the surface finish is critical or if the section is thin, so they are actually considered destructive tests. The Vickers test has the advantage of having only one test setup for all materials. Both the Brinell and Rockwell tests require selection of the tip size or indentation load, or both, to match the material tested. The Rockwell test is favored for its lack of operator error, since no microscope reading is required and the indentation tends to be smaller, particularly if the Ntip is used. But, the Brinell hardness number provides a very convenient way to quickly estimate the ultimate tensile strength (Sut) of the material from the relationship Sut ≅ 500 HB ± 30 HB
psi
Sut ≅ 3.45 HB ± 0.2 HB
MPa
(2.10)
where HB is the Brinell hardness number. This gives a convenient way to obtain a rough experimental measure of the strength of any low or mediumstrength carbon or alloy steel sample, even one that has already been placed in service and cannot be truly destructively tested. Microhardness tests use a low force on a small diamond indenter and can provide a profile of microhardness as a function of depth across a sectioned sample. The hardness is computed on an absolute scale by dividing the applied force by the area of the indent. The units of absolute hardness are kgf/mm2. Brinell and Vickers hardness numbers also have these hardness units, though the values measured on the same sample can differ with each method. For example, a Brinell hardness of 500 HB is about the same as a Rockwell C hardness of 52 HRC and an absolute hardness of 600 kgf/mm2. Note that these scales are not linearly related, so conversion is difficult. Table 23 shows approximate conversions between the Brinell, Vickers, and Rockwell B and C hardness scales for steels and their approximate equivalent ultimate tensile strengths.
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* Tungsten carbide is one of the hardest substances known.
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Table 23
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Approximate Equivalent Hardness Numbers and Ultimate Tensile Strengths for Steels
Source: Table 510, p.185, in N. E. Dowling, Mechanical Behavior of Materials, Prentice Hall, Englewood Cliffs, N.J., 1993, with permission.
Heat Treatment The steel heattreatment process is quite complicated and is dealt with in detail in materials texts such as those listed in the bibliography at the end of this chapter. The reader is referred to such references for a more complete discussion. Only a brief review of some of the salient points is provided here. The hardness and other characteristics of many steels and some nonferrous metals can be changed by heat treatment. Steel is an alloy of iron and carbon. The weight percent of carbon present affects the alloy’s ability to be heattreated. A lowcarbon steel will have about 0.03 to 0.30% of carbon, a mediumcarbon steel about 0.35 to 0.55%, and a highcarbon steel about 0.60 to 1.50%. (Cast irons will have greater than 2% carbon.) Hardenability of steel increases with carbon content. Lowcarbon steel has too little carbon for effective throughhardening so other surfacehardening methods must be used (see below). Medium and highcarbon steels can be throughhardened by appropriate heat treatment. The depth of hardening will vary with alloy content. Quenching To harden a medium or highcarbon steel, the part is heated above a critical temperature (about 1400°F [760°C]), allowed to equilibrate for some time, and then suddenly cooled to room temperature by immersion in a water or oil bath. The rapid cooling creates a supersaturated solution of iron and carbon called martensite, which is
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45
extremely hard and much stronger than the original soft material. Unfortunately it is also very brittle. In effect, we have traded off the steel’s ductility for its increased strength. The rapid cooling also introduces strains to the part. The change in the shape of the stressstrain curve as a result of quenching a ductile, mediumcarbon steel is shown in Figure 212 (not to scale). While the increased strength is desirable, the severe brittleness of a fully quenched steel usually makes it unusable without tempering. teMpering Subsequent to quenching, the same part can be reheated to a lower temperature (400–1300°F [200–700°C]), heatsoaked, and then allowed to cool slowly. This will cause some of the martensite to convert to ferrite and cementite, which reduces the strength somewhat but restores some ductility. A great deal of flexibility is possible in terms of tailoring the resulting combination of properties by varying time and temperature during the tempering process. The knowledgeable materials engineer or metallurgist can achieve a wide variety of properties to suit any application. Figure 212 also shows a stressstrain curve for the same steel after tempering.
Stress σ
annealing The quenching and tempering process is reversible by annealing. The part is heated above the critical temperature (as for quenching) but now allowed to cool slowly to room temperature. This restores the solution conditions and mechanical properties of the unhardened alloy. Annealing is often used even if no hardening has been previously done in order to eliminate any residual stresses and strains introduced by the forces applied in forming the part. It effectively puts the part back into a “relaxed” and soft state, restoring its original stressstrain curve as shown in Figure 212. norMalizing Many tables of commercial steel data indicate that the steel has been normalized after rolling or forming into its stock shape. Normalizing is similar to annealing but involves a shorter soak time at elevated temperature and a more rapid cooling rate. The result is a somewhat stronger and harder steel than a fully annealed one but one that is closer to the annealed condition than to any tempered condition. Surface (Case) Hardening When a part is large or thick, it is difficult to obtain uniform hardness within its interior by through hardening. An alternative is to harden only the surface, leaving the core soft. This also avoids the distortion associated with quenching a large, throughheated part. If the steel has sufficient carbon content, its surface can be heated, quenched, and tempered as would be done for through hardening. For lowcarbon (mild) steels other techniques are needed to obtain a hardened condition. These involve heating the part in a special atmosphere rich in either carbon, nitrogen or both and then quenching it, a process called carburizing, nitriding, or cyaniding. In all situations, the result is a hard surface (i.e., case) on a soft core, referred to as being casehardened. Carburizing heats lowcarbon steel in a carbon monoxide gas atmosphere, causing the surface to take up carbon in solution. Nitriding heats lowcarbon steel in a nitrogengas atmosphere and forms hard iron nitrides in the surface layers. Cyaniding heats the part in a cyanide salt bath at about 1500°F (800°C), and the lowcarbon steel takes up both carbides and nitrides from the salt. For medium and highcarbon steels no artificial atmosphere is needed, as the steel has sufficient carbon for hardening. Two methods are in common use. Flame hardening
quenched tempered
annealed E 1 ε FIGURE 212 StressStrain Curves for Annealed, Quenched, and Tempered Steel
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passes an oxyacetylene flame over the surface to be hardened and follows it with a water jet for quenching. This results in a somewhat deeper hardened case than obtainable from the artificialatmosphere methods. Induction hardening uses electric coils to rapidly heat the part surface, which is then quenched before the core can get hot. Case hardening by any appropriate method is a very desirable hardening treatment for many applications. It is often advantageous to retain the full ductility (and thus the toughness) of the core material for better energy absorption capacity while also obtaining high hardness on the surface in order to reduce wear and increase surface strength. Large machine parts such as cams and gears are examples of elements that can benefit more from case hardening than from through hardening, as heat distortion is minimized and the tough, ductile core can better absorb impact energy. Heat Treating Nonferrous Materials Some nonferrous alloys are hardenable and others are not. Some of the aluminum alloys can be precipitation hardened, also called age hardening. An example is aluminum alloyed with up to about 4.5% copper. This material can be hotworked (rolled, forged, etc.) at a particular temperature and then heated and held at a higher temperature to force a random dispersion of the copper in the solid solution. It is then quenched to capture the supersaturated solution at normal temperature. The part is subsequently reheated to a temperature below the quenching temperature and held for an extended period of time while some of the supersaturated solution precipitates out and increases the material’s hardness. Other aluminum alloys, magnesium, titanium, and a few copper alloys are amenable to similar heat treatment. The strengths of the hardened aluminum alloys approach those of mediumcarbon steels. Since all aluminum is about 1/3 the density of steel, the stronger aluminum alloys can offer better strengthtoweight ratios than lowcarbon (mild) steels. Mechanical Forming and Hardening cold Working The mechanical working of metals at room temperature to change their shape or size will also workharden them and increase their strength at the expense of ductility. Cold working can result from the rolling process in which metal bars are progressively reduced in thickness by being squeezed between rollers, or from any operation that takes the ductile metal beyond the yield point and permanently deforms it. Figure 213 shows the process as it affects the material’s stressstrain curve. As the load is increased from the origin at O beyond the yield point y to point B, a permanent set OA is introduced. If the load is removed at that point, the stored elastic energy is recovered and the material returns to zero stress at point A along a new elastic line BA parallel to the original elastic slope E. If the load is now reapplied and brought to point C, again yielding the material, the new stressstrain curve is ABCf. Note that there is now a new yield point y’ which is at a higher stress than before. The material has strainhardened, increasing its yield strength and reducing its ductility. This process can be repeated until the material becomes brittle and fractures.
Chapter 2
Stress σ
MATERIALS AND PROCESSES
Stress σ
f B
Sy
47
C B
Sy
f
y'
y
(a)
E 1 O
A
1
O permanent set
Strain ε
elastic energy recovered
E
(b)
elastic energy recovered
A
D
permanent set
FIGURE 213 Strain Hardening a Ductile Material by Cold Working (a) First Working (b) Second Working
If significant plastic deformation is required for manufacture, such as in making deepdrawn metal pots or cylinders, it is necessary to cold form the material in stages and anneal the part between successive stages to avoid fracture. The annealing resets the material to more nearly the original ductile stressstrain curve, allowing further yielding without fracture. hot Working All metals have a recrystallization temperature below which the effects of mechanical working will be as described previously, e.g., cold worked. If the material is mechanically worked above its recrystallization temperature (hot working), it will tend to at least partially anneal itself as it cools. Thus, hot working reduces the strainhardening problem but introduces its own problems of rapid oxidation of the surface due to the high temperatures. Hotrolled metals tend to have higher ductility, lower strength and poorer surface finish than coldworked metals of the same alloy. Hot working does not increase the hardness of the material appreciably, though it can increase the strength by improving grain structure and aligning the “grain” of the metal with the final contours of the part. This is particularly true of forged parts. forging is an automation of the ancient art of blacksmithing. The blacksmith heats the part redhot in the forge, then beats it into shape with a hammer. When it cools too much for forming, it is reheated and the process is repeated. Forging uses a series of hammer dies shaped to gradually form the hot metal into the final shape. Each stage’s die shape represents an achievable change in shape from the original ingot form to the final desired part shape. The part is reheated between blows from the hammer dies which are mounted in a forging press. The large forces required to plastically deform the hot metal require massive presses for parts of medium to large size. Machining operations are required to remove the large “flash” belt at the die parting line and to machine holes, mounting surfaces, etc. The surface finish of a forging is as rough as any hotrolled part due to oxidation and decarburization of the heated metal. Virtually any wrought, ductile metal can be forged. Steel, aluminum, and titanium are commonly used. Forging has the advantage of creating stronger parts than casting or machining can. Casting alloys are inherently weaker in tension than wrought alloys. The hot forming of a wrought material into the final forged shape causes the material’s
Strain ε
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FIGURE 214
Copyright © 2018 Robert L. Norton: All Rights Reserved
Forged Steel Crankshaft for a DieselTruck Engine
internal “flow lines” or “grain” to approximate the contours of the part, which can result in greater strength than if a stock shape’s flow lines were severed by machining to the final contour. Forgings are used in highly stressed parts, such as aircraft wing and fuselage structures, engine crankshafts and connecting rods, and vehicle suspension links. Figure 214 shows a forged truck crankshaft. In the cross section, the grain lines can be seen to follow the crankshaft’s contours. The high cost of the multiple dies needed for forged shapes makes it an impractical choice unless production quantities are large enough to amortize the tooling cost. extruSion is used principally for nonferrous metals (especially aluminum) as it typically uses steel dies. The usual die is a thick, hardenedtoolsteel disk with a tapered “hole” or orifice ending in the crosssectional shape of the finished part. A billet of the extrudate is heated to a soft state and then rammed at fairly high speed through the die, which is clamped in the machine. The billet flows, or extrudes, into the die’s shape. The process is similar to the making of macaroni. A long strand of the material in the desired cross section is extruded from the billet. The extrusion then passes through a waterspray cooling station. Extrusion is an economical way to obtain custom shapes of constant cross section since the dies are not very expensive to make. Dimensional control and surface finish are good. Extrusion is used to make aluminum mill shapes such as angles, channels, Ibeams, and custom shapes for stormdoor and window frames, slidingdoor frames, etc. The extrusions are cut and machined as necessary to assemble them into the finished product. Some extruded shapes are shown in Figure 215. 2.5 FIGURE 215 Extrusions  Courtesy of The Aluminum Extruders Council
COATINGS AND SURFACE TREATMENTS
Many coatings and surface treatments are available for metals. Some have the prime purpose of inhibiting corrosion while others are intended to improve surface hardness and wear resistance. Coatings are also used to change dimensions (slightly) and to alter physical properties such as reflectance, color, and resistivity. For example, piston rings are chromeplated to improve wear resistance, fasteners are plated to reduce corrosion,
Chapter 2
MATERIALS AND PROCESSES
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(a) Nonmetallic Coatings
Polymer
Oxide
Chemical Conversion
Anodizing
GlassCeramic
Phosphate
Chromate
(b) Metallic Coatings
Diffusion
Mechanical
Plating
Electroplating
Hot Dip
Electroless
Immersion
Spray
Deposition
Chemical Vapor
Vacuum
FIGURE 216 Coating Methods Available for Metals
Table 24
and automobile trim is chromeplated for appearance and corrosion resistance. Figure 216 shows a chart of various types of coatings for machine applications. These divide into two major classes, metallic and nonmetallic, based on the type of coating, not substrate. Some of the classes divide into many subclasses. We will discuss only a few of these here. The reader is encouraged to seek more information from the references in the bibliography.
Galvanic Series of Metals in Seawater Least noble Magnesium Zinc Aluminum
Galvanic Action
Cadmium
When a coating of one metal is applied to another dissimilar metal, a galvanic cell may be created. All metals are electrolytically active to a greater or lesser degree and if sufficiently different in their electrolytic potential will create a battery in the presence of a conductive electrolyte such as seawater or even tap water. Table 24 lists some common metals ordered in terms of their galvanic action potential from the least noble (most electrolytically active) to the most noble (least active). Combinations of metals that are close to each other in the galvanic series, such as cast iron and steel, are relatively safe from galvanic corrosion. Combinations of metals far apart on this scale, such as aluminum and copper, will experience severe corrosion in an electrolyte or even in a moist environment.
Cast iron
In a conductive medium, the two metals become anode and cathode, with the lessnoble metal acting as the anode. The selfgenerated electrical current flow causes a loss of material from the anode and a deposition of material on the cathode. The lessnoble metal gradually disappears. This problem occurs whenever two metals sufficiently far apart in the galvanic series are present in an electrically conductive medium. Thus, not only coatings but fasteners and mating parts must be made of metal combinations that will not create this problem.
Steel Stainless steel Lead Tin Nickel Brass Copper Bronze Monel Silver Titanium Graphite Gold Platinum Most noble
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Electroplating Electroplating involves the deliberate creation of a galvanic cell in which the part to be plated is the cathode and the plating material is the anode. The two metals are placed in an electrolyte bath and a direct current applied from anode to cathode. Ions of the plating material are driven to the plating substrate through the electrolyte and cover the part with a thin coating of the plating material. Allowance must be made for the plating thickness, which is controllable. Plating thickness is uniform except at sharp corners or in holes and crevices. The plating builds up on the outside corners and will not go into holes or narrow crevices. Thus, grinding may be necessary after plating to restore dimensions. Worn parts (or mistakes) can sometimes be repaired by plating on a coating of suitable material, then regrinding to dimension.
2
* It is interesting to note that chromium in the pure form is softer than hardened steel but when electroplated onto steel, it becomes harder than the steel substrate. Nickel and iron also increase their hardness when electroplated on metal substrates. The mechanism is not well understood, but it is believed that internal microstrains are developed in the plating process that harden the coating. The hardness of the plating can be controlled by changes in process conditions.
Steels, nickel and copperbased alloys, as well as other metals are readily electroplatable. Two approaches are possible. If a more noble (less active) metal is plated onto the substrate, it can reduce the tendency to oxidize as long as the plating remains intact to protect the substrate from the environment. Tin, nickel, and chromium are often used to electroplate steel for corrosion resistance. Chrome plating also offers an increase in surface hardness to HRC 70, which is above that obtainable from many hardened alloy steels.* Unfortunately, any disruptions or pits in the plating can provide nodes for galvanic action if conductive media (such as rainwater) are present. Because the substrate is less noble than the plating, it becomes the sacrificial anode and rapidly corrodes. Electroplating with metals more noble than the substrate is seldom used for parts that will be immersed in water or other electrolytes. Alternatively, a lessnoble metal can be plated onto the substrate to serve as a sacrificial anode which will corrode instead of the substrate. The most common example of this is zinc coating of steel, also called galvanizing. (Cadmium can be used instead of zinc and will last longer in saltwater or saltair environments.) The zinc or cadmium coating will gradually corrode and protect the more noble steel substrate until the coating is used up, after which the steel will oxidize. Zinc coating can be applied by a process called “hot dipping” rather than by electroplating, which will result in a thicker and more protective coating recognizable by its “motherofpearl” appearance. Galvanizing is often applied by manufacturers to automobile body panels to inhibit corrosion. Sacrificial zinc anodes are also attached to aluminum outboard motors and aluminum boat hulls to shortcircuit corrosion of the aluminum in seawater. A caution about electroplated coatings is that hydrogen embrittlement of the substrate can occur, causing significant loss of strength. Electroplated finishes should not be used on parts that are fatigue loaded. Experience has shown that electroplating severely reduces the fatigue strength of metals and can cause early failure. Electroless Plating Electroless plating puts a coating of nickel on the substrate without any electric current needed. The substrate “cathode” in this case (there is no anode) acts as a catalyst to start a chemical reaction that causes nickel ions in the electrolyte solution to be reduced and deposited on the substrate. The nickel coating also acts as a catalyst and keeps the
Chapter 2
MATERIALS AND PROCESSES
reaction going until the part is removed from the bath. Thus, relatively thick coatings can be developed. Coatings are typically between 0.001 in and 0.002 in thick. Unlike electroplating, the electroless nickel plate is completely uniform and will enter holes and crevices. The plate is dense and fairly hard at around 43 HRC. Other metals can also be electroless plated but nickel is most commonly used. Anodizing While aluminum can be electroplated (with difficulty), it is more common to treat it by anodizing. This process creates a very thin layer of aluminum oxide on the surface. The aluminum oxide coating is selflimiting in that it prevents atmospheric oxygen from further attacking the aluminum substrate in service. The anodized oxide coating is naturally colorless but dyes can be added to color the surface and provide a pleasing appearance in a variety of hues. This is a relatively inexpensive surface treatment with good corrosion resistance and negligible distortion. Titanium, magnesium, and zinc can also be anodized. A variation on conventional anodizing of aluminum is socalled “hard anodizing.” Since aluminum oxide is a ceramic material, it is naturally very hard and abrasion resistant. Hard anodizing provides a thicker (but not actually harder) coating than conventional anodizing and is often used to protect the relatively soft aluminum parts from wear in abrasive contact situations. The hardness of this surface treatment exceeds that of the hardest steel, and hardanodized aluminum parts can be run against hardened steel though the somewhat abrasive aluminum oxide surface is not kind to the steel. PlasmaSprayed Coatings A variety of very hard ceramic coatings can be applied to steel and other metal parts by a plasmaspray technique. The application temperatures are high, which limits the choice of substrate. The coatings as sprayed have a rough “orangepeel” surface finish, which requires grinding or polishing to obtain a fine finish. The main advantage is a surface with extremely high hardness and chemical resistance. However, the ceramic coatings are brittle and subject to chipping under mechanical or thermal shock. Chemical Coatings The most common chemical treatments for metals range from a phosphoric acid wash on steel (or chromatic acid on aluminum) that provides limited and shortterm oxidation resistance, to paints of various types designed to give more lasting corrosion protection. Paints are available in a large variety of formulations for different environments and substrates. Onepart paints give somewhat less protection than twopart epoxy formulations, but all chemical coatings should be viewed as only temporary protection against corrosion, especially when used on corrosionprone materials such as steel. Baked enamel and porcelain finishes on steel have longer lives in terms of corrosion resistance, though they suffer from brittleness. New formulations of paints and protective coatings are continually being developed. The latest and best information will be obtained from vendors of these products.
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2.6 magnesium 6.5 (44.8)
2
aluminum 10.4 (71.8) gray cast iron 15 (104) brass, bronze 16 (110) titanium 16.5 (114) ductile cast iron

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GENERAL PROPERTIES OF METALS
The large variety of useful engineering materials can be confusing to the beginning engineer. There is not space enough in this book to deal with the topic of material selection in complete detail. Several references are provided in this chapter’s bibliography, which the reader is encouraged to use. Tables of mechanical property data are also provided for a limited set of materials in Appendix A of this book. Figure 217 shows the Young’s moduli for several engineering metals. The following sections attempt to provide some general information and guidelines for the engineer to help identify what types of materials might be suitable in a given design situation. It is expected that the practicing engineer will rely heavily on the expertise and help available from materials manufacturers in selecting the optimum material for each design. Many references are also published that list detailed property data for most engineering materials. Some of these references are listed in the bibliography to this chapter.
24 (166) stainless steel 27.5 (190) steel 30 (207) 0
10
20
0
70
140 210
30
Young's Modulus E Mpsi (GPa) FIGURE 217 Young's Moduli for Various Metals
Cast Iron Cast irons constitute a whole family of materials. Their main advantages are relatively low cost and ease of fabrication. Some are weak in tension compared to steels but, like most cast materials, have high compressive strengths. Their densities are slightly lower than steel at about 0.25 lb/in3 (6920 kg/m3). Most cast irons do not exhibit a linear stressstrain relationship below the elastic limit; they do not obey Hooke’s law. Their modulus of elasticity E is estimated by drawing a line from the origin through a point on the curve at 1/4 the ultimate tensile strength and is in the range of 14–25 Mpsi (97–172 MPa). Cast iron’s chemical composition differs from steel principally in its higher carbon content, being between 2 and 4.5%. The large amount of carbon, present in some cast irons as graphite, makes some of these alloys easy to pour as a casting liquid and also easy to machine as a solid. The most common means of fabrication is sand casting with subsequent machining operations. Cast irons are not easily welded, however. White caSt iron is a very hard and brittle material. It is difficult to machine and has limited uses, such as in linings for cement mixers where its hardness is needed. gray caSt iron is the most commonly used form of cast iron. Its graphite flakes give it its gray appearance and name. The ASTM grades gray cast iron into seven classes based on the minimum tensile strength in kpsi. Class 20 has a minimum tensile strength of 20 kpsi (138 MPa). The class numbers of 20, 25, 30, 35, 40, 50, and 60 then represent the tensile strength in kpsi. Cost increases with increasing tensile strength. This alloy is easy to pour, easy to machine, and offers good acoustical damping. This makes it the popular choice for machine frames, engine blocks, brake rotors and drums, etc. The graphite flakes also give it good lubricity and wear resistance. Its relatively low tensile strength recommends against its use in situations where large bending or fatigue loads are present, though it is sometimes used in lowcost engine crankshafts. It runs reasonably well against steel if lubricated. MalleaBle caSt iron has superior tensile strength to gray cast iron but does not wear as well. The tensile strength can range from 50 to 120 kpsi (345 to 827 MPa) depending on formulation. It is often used in parts where bending stresses are present.
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MATERIALS AND PROCESSES
nodular (ductile) caSt iron has the highest tensile strength of the cast irons, ranging from about 70 to 135 kpsi (480 to 930 MPa). The name nodular comes from the fact that its graphite particles are spheroidal in shape. Ductile cast iron has a higher modulus of elasticity (about 25 Mpsi {172 GPa}) than gray cast iron and exhibits a linear stressstrain curve. It is tougher, stronger, more ductile, and less porous than gray cast iron. It is the cast iron of choice for fatigueloaded parts such as crankshafts, pistons, and cams. Cast Steels Cast steel is similar to wrought steel in terms of its chemical content, i.e., it has much less carbon than cast iron. The mechanical properties of cast steel are superior to cast iron but inferior to wrought steel. Its principal advantage is ease of fabrication by sand or investment (lost wax) casting. Cast steel is classed according to its carbon content into low carbon ( 0.5%). Alloy cast steels are also made containing other elements for high strength and heat resistance. The tensile strengths of cast steel alloys range from about 65 to 200 kpsi (450 to 1380 MPa). Wrought Steels The term “wrought” refers to all processes that manipulate the shape of the material without melting it. Hot rolling and cold rolling are the two most common methods used though many variants exist, such as wire drawing, deep drawing, extrusion, and coldheading. The common denominator is a deliberate yielding of the material to change its shape either at room or at elevated temperatures. hotrolled Steel is produced by forcing hot billets of steel through sets of rollers or dies which progressively change their shape into Ibeams, channel sections, angle irons, flats, squares, rounds, tubes, sheets, plates, etc. The surface finish of hotrolled shapes is rough due to oxidation at the elevated temperatures. The mechanical properties are also relatively low because the material ends up in an annealed or normalized state unless deliberately heattreated later. This is the typical choice for lowcarbon structural steel members used for building and machineframe construction. Hotrolled material is also used for machine parts that will be subjected to extensive machining (gears, cams, etc.) where the initial finish of the stock is irrelevant and uniform, noncoldworked material properties are desired in advance of a planned heat treatment. A wide variety of alloys and carbon contents are available in hotrolled form. coldrolled Steel is produced from billets or hotrolled shapes. The shape is brought to final form and size by rolling between hardened steel rollers or drawing through dies at room temperature. The rolls or dies burnish the surface and cold work the material, increasing its strength and reducing its ductility as was described previously in the section on mechanical forming and hardening. The result is a material with good surface finish and accurate dimensions compared to hotrolled material. Its strength and hardness are increased at the expense of significant builtin strains, which can later be released during machining, welding, or heat treating, then causing distortion. Coldrolled shapes commonly available are sheets, strips, plates, round and rectangular bars, tubes, etc. Structural shapes, such as Ibeams, are typically available only as hotrolled.
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Steel Numbering Systems
2
* ASTM is the American Society for Testing and Materials, AISI is the American Iron and Steel Institute, and SAE is the Society of Automotive Engineers. AISI and SAE both use the same designations for steels.
Several steel numbering systems are in general use. The ASTM, AISI, and SAE* have devised codes to define the alloying elements and carbon content of steels. Table 25 lists some of the AISI/SAE designations for commonly used steel alloys. The first two digits indicate the principal alloying elements. The last two digits indicate the amount of carbon present, expressed in hundredths of a percent. ASTM and the SAE have developed a new Unified Numbering System for all metal alloys, which uses the prefix UNS followed by a letter and a 5digit number. The letter defines the alloy category, F for cast iron, G for carbon and lowalloy steels, K for specialpurpose steels, S for stainless steels, and T for tool steels. For the G series, the numbers are the same as the AISI/SAE designations in Table 25 with a trailing zero added. For example, SAE 4340 becomes UNS G43400. See reference 2 for more information on metal numbering systems. We will use the AISI/SAE designations for steels. plain carBon Steel is designated by a first digit of 1 and a second digit of 0, since no alloys other than carbon are present. The lowcarbon steels are those numbered AISI 1005 to 1030, mediumcarbon from 1035 to 1055, and highcarbon from 1060 to 1095. The AISI 11xx series adds sulphur, principally to improve machinability. These
Table 25
AISI/SAE Designations of Steel Alloys A partial list  other alloys are available  consult the manufacturers
Type
AISI/SAE Series
Principal Alloying Elements
Carbon Steels Plain
10xx
Carbon
Freecutting
11xx
Carbon plus Sulphur (resulphurized)
13xx
1.75% Manganese
15xx
1.00 to 1.65% Manganese
23xx
3.50% Nickel
25xx
5.00% Nickel
31xx
1.25% Nickel and 0.65 or 0.80% Chromium
33xx
3.50% Nickel and 1.55% Chromium
40xx
0.25% Molybdenum
Alloy Steels Manganese Nickel NickelChrome Molybdenum
44xx
0.40 or 0.52% Molybdenum
ChromeMoly
41xx
0.95% Chromium and 0.20% Molybdenum
NickelChromeMoly
43xx
1.82% Nickel, 0.50 or 0.80% Chromium, and 0.25% Molybdenum
47xx
1.45% Nickel, 0.45% Chromium, and 0.20 or 0.35% Molybdenum
46xx
0.82 or 1.82% Nickel and 0.25% Molybdenum
48xx
3.50% Nickel and 0.25% Molybdenum
50xx
0.27 to 0.65% Chromium
51xx
0.80 to 1.05% Chromium
52xx
1.45% Chromium
61xx
0.60 to 0.95% Chromium and 0.10 to 0.15% Vanadium minimum
NickelMoly Chrome
ChromeVanadium
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MATERIALS AND PROCESSES
are called freemachining steels and are not considered alloy steels as the sulphur does not improve the mechanical properties and also makes it brittle. The ultimate tensile strength of plain carbon steel can vary from about 60 to 150 kpsi (414 to 1034 MPa) depending on heat treatment. alloy SteelS have various elements added in small quantities to improve the material’s strength, hardenability, temperature resistance, corrosion resistance, and other properties. Any level of carbon can be combined with these alloying elements. Chromium is added to improve strength, ductility, toughness, wear resistance, and hardenability. Nickel is added to improve strength without loss of ductility, and it also enhances case hardenability. Molybdenum, used in combination with nickel and/or chromium, adds hardness, reduces brittleness, and increases toughness. Many other alloys in various combinations, as shown in Table 25, are used to achieve specific properties. Specialty steel manufacturers are the best source of information and assistance for the engineer trying to find the best material for any application. The ultimate tensile strength of alloy steels can vary from about 80 to 300 kpsi (550 to 2070 MPa), depending on its alloying elements and heat treatment. Appendix A contains tables of mechanical property data for a selection of carbon and alloy steels. Figure 218 shows approximate ultimate tensile strengths of some normalized carbon and alloy steels and Figure 219 shows engineering stressstrain curves from tensile tests of three steels. tool SteelS are medium to highcarbon alloy steels especially formulated to give very high hardness in combination with wear resistance and sufficient toughness to resist the shock loads experienced in service as cutting tools, dies and molds. There is a very large variety of tool steels available. Refer to the bibliography and to manufacturers’ literature for more information. StainleSS SteelS are alloy steels containing at least 10% chromium and offer much improved corrosion resistance over plain or alloy steels, though their name should not be taken too literally. Stainless steels will stain and corrode (slowly) in severe environments such as seawater. Some stainlesssteel alloys have improved resistance to high tempera
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AISI # 4340 4140 1095 6150 9255 3140 1050 1040 1030 111 1020 1015 Kpsi 0 MPa 0
100
200
700
1400
tensile strength FIGURE 218 Approximate Ultimate Tensile Strengths of Some Normalized Steels
FIGURE 219 Tensile Test StressStrain Curves of Three Steel Alloys (From Fig. 5.16, p. 160, in N. E. Dowling, Mechanical Behavior of Materials, PrenticeHall, Englewood Cliffs, N.J., 1993, with permission)
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ture. There are four types of stainless steel, called martensitic, ferritic, austenitic, and precipitation hardening.
2
Martensitic stainless steel contains 11.5 to 15% Cr and 0.15 to 1.2% C, is magnetic, can be hardened by heat treatment, and is commonly used for cutlery. Ferritic stainless steel has over 16% Cr and a low carbon content, is magnetic, soft, and ductile, but is not heat treatable though its strength can be increased modestly by cold working. It is used for deepdrawn parts, such as cookware, and has better corrosion resistance than the martensitic SS. The ferritic and martensitic stainless steels are both called 400 series stainless steel. Austenitic stainless steel is alloyed with 17 to 25% chromium and 10 to 20% nickel. It has better corrosion resistance due to the nickel, is nonmagnetic, and has excellent ductility and toughness. It cannot be hardened except by cold working. It is classed as 300 series stainless steel. Precipitationhardening stainless steels are designated by their alloy percentages followed by the letters PH, as in 174 PH, which contains 17% chromium and 4% nickel. These alloys offer high strength and high temperature and corrosion resistance. The 300 series stainless steels are very weldable but the 400 series are less so. All grades of stainless steel have poorer heat conductivity than regular steel and many of the stainless alloys are difficult to machine. All stainless steels are significantly more expensive than regular steel. See Appendix A for mechanical property data.
alloy 7075T6
Aluminum
2014T6 2024T4 6061T6 6063T6 1100H18 11000 0 kpsi
50
0 MPa 345
100 690
tensile strength FIGURE 220 Ultimate Tensile Strengths of Some Aluminum Alloys
Aluminum is the most widely used nonferrous metal, being second only to steel in world consumption. Aluminum is produced in both “pure” and alloyed forms. Aluminum is commercially available up to 99.8% pure. The most common alloying elements are copper, silicon, magnesium, manganese, and zinc, in varying amounts up to about 5%. The principal advantages of aluminum are its low density, good strengthtoweight ratio (SWR), ductility, excellent workability, castability, and weldability, corrosion resistance, high conductivity, and reasonable cost. Compared to steel it is 1/3 as dense (0.10 lb/ in3 versus 0.28 lb/in3), about 1/3 as stiff (E = 10.3 Mpsi [71 GPa] versus 30 Mpsi [207 GPa]), and generally less strong. If you compare the strengths of lowcarbon steel and pure aluminum, the steel is about three times as strong. Thus, the specific strength is approximately the same in that comparison. However, pure aluminum is seldom used in engineering applications. It is too soft and weak. Pure aluminum’s principal advantages are its bright finish and good corrosion resistance. It is used mainly in decorative applications. The aluminum alloys have significantly greater strengths than pure aluminum and are used extensively in engineering, with the aircraft and automotive industries among the largest users. The higherstrength aluminum alloys have tensile strengths in the 70 to 90 kpsi (480 to 620 MPa) range, and yield strengths about twice that of mild steel. They compare favorably to mediumcarbon steels in specific strength. Aluminum competes successfully with steel in some applications, though few materials can beat steel if very high strength is needed. See Figure 220 for tensile strengths of some aluminum alloys. Figure 221 shows tensiletest engineering stressstrain curves for three aluminum alloys. Aluminum’s strength is reduced at low temperatures as well as at elevated temperatures.
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MATERIALS AND PROCESSES
57
FIGURE 221 Tensile Test StressStrain Curves of Three Aluminum Alloys (From Fig. 5.17, p. 160, in N. E. Dowling, Mechanical Behavior of Materials, PrenticeHall, Englewood Cliffs, N.J., 1993, with permission)
Some aluminum alloys are hardenable by heat treatment and others by strain hardening or precipitation and aging. Highstrength aluminum alloys are about 1.5 times harder than soft steel, and surface treatments such as hard anodizing can bring the surface to a condition harder than the hardest steel. Aluminum is among the most easily worked of the engineering materials, though it tends to work harden. It casts, machines, welds,* and hot and cold forms† easily. It can also be extruded. Alloys are specially formulated for both sand and die casting as well as for wrought and extruded shapes and for forged parts. WroughtaluMinuM alloyS are available in a wide variety of stock shapes such as Ibeams, angles, channels, bars, strip, sheet, rounds, and tubes. Extrusion allows relatively inexpensive custom shapes as well. The Aluminum Association numbering system for alloys is shown in Table 26. The first digit indicates the principal alloying element and defines the series. Hardness is indicated by a suffix containing a letter and up to 3 numbers as defined in the table. The most commonly available and mostused aluminum alloys in machinedesign applications are the 2000 and 6000 series. The oldest aluminum alloy is 2024, which contains 4.5% copper, 1.5% magnesium, and 0.8% manganese. It is among the most machinable of the aluminum alloys and is heat treatable. In the higher tempers, such as T3 and T4, it has a tensile strength approaching 70 kpsi (483 MPa), which also makes it one of the strongest of the aluminum alloys. It also has high fatigue strength. However, it has poor weldability and formability compared to the other aluminum alloys. The 6061 alloy contains 0.6% silicon, 0.27% copper, 1.0% manganese, and 0.2% chromium. It is widely used in structural applications because of its excellent weldability. Its strength is about 40 to 45 kpsi (276 to 310 MPa) in the higher tempers. It has lower fatigue strength than 2024 aluminum. It is easily machined and is a popular alloy for extrusion, which is a hotforming process. The 7000 series is called aircraft aluminum and is used mostly in airframes. These are the strongest alloys of aluminum with tensile strengths up to 98 kpsi (676 MPa) and
*
The heat of welding causes localized annealing, which can remove the desirable strengthening effects of cold work or heat treatment in any metal. †
Some aluminum alloys will coldwork when formed to the degree that trying to bend them again (without first annealing) will cause fractures. Some bicycle racers prefer steel frames over aluminum despite their added weight because, once an aluminum frame is bent in a fall, it cannot be straightened without cracking. Damaged steel tube frames can be straightened and reused.
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Aluminum Association Designations of Aluminum Alloys A partial list  other alloys are available  consult the manufacturers
2 Series Alloys
Major Alloying Elements
Secondary
1xxx
Commercially pure (99%)
None
2xxx
Copper (Cu)
Mg, Mn, Si
3xxx
Manganese (Mn)
Mg, Cu
4xxx
Silicon (Si)
None
5xxx
Magnesium (Mg)
Mn, Cr
6xxx
Magnesium and Silicon
Cu, Mn
7xxx
Zinc (Zn)
Mg, Cu, Cr
Hardness Designations xxxxF
As fabricated
xxxxO
Annealed
xxxxHyyy
Work hardened
xxxxTyyy
Thermal/age hardened
the highest fatigue strength of about 22 kpsi (152 MPa) @ 108 cycles. Some alloys are also available in an alclad form that bonds a thin layer of pure aluminum to one or both sides to improve corrosion resistance. caStaluMinuM alloyS are differently formulated than the wrought alloys. Some of these are hardenable but their strength and ductility are less than those of the wrought alloys. Alloys are available for sand casting, die casting, or investment casting. See Appendix A for mechanical properties of wrought and castaluminum alloys. Titanium Though discovered as an element in 1791, commercially produced titanium has been available only since the 1940s, so it is among the newest of engineering metals. Titanium can be the answer to an engineer’s prayer in some cases. It has an upper servicetemperature limit of 1200 to 1400°F (650 to 750°C), weighs half as much as steel (0.16 lb/in3 [4429 kg/m3]), and is as strong as a mediumstrength steel (135 kpsi [930 MPa] typical). Its Young’s modulus is 16 to 18 Mpsi (110 to 124 GPa), or about 60% that of steel. Its specific strength approaches that of the strongest alloy steels and exceeds that of mediumstrength steels by a factor of 2. Its specific stiffness is greater than that of steel, making it as good or better in limiting deflections. It is also nonmagnetic. Titanium is very corrosion resistant and is nontoxic, allowing its use in contact with acidic or alkaline foodstuffs and chemicals, and in the human body as replacement heart valves and hip joints, for example. Unfortunately, it is expensive compared to aluminum and steel. It finds much use in the aerospace industry, especially in military aircraft structures and in jet engines, where strength, light weight, and high temperature and corrosion resistance are all required.
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MATERIALS AND PROCESSES
Titanium is available both pure and alloyed with combinations of aluminum, vanadium, silicon, iron, chromium, and manganese. Its alloys can be hardened and anodized. Limited stock shapes are available commercially. It can be forged and wrought, though it is quite difficult to cast, machine, and cold form. Like steel and unlike most other metals, some titanium alloys exhibit a true endurance limit, or leveling off of the fatigue strength, beyond about 106 cycles of repeated loading, as shown in Figure 210. See Appendix A for mechanical property data. Magnesium Magnesium is the lightest of commercial metals but is relatively weak. The tensile strengths of its alloys are between 10 and 50 kpsi (69 and 345 MPa). The most common alloying elements are aluminum, manganese, and zinc. Because of its low density (0.065 lb/in3 [1800 kg/m3]), its specific strength approaches that of aluminum. Its Young’s modulus is 6.5 Mpsi (45 GPa) and its specific stiffness exceeds those of aluminum and steel. It is very easy to cast and machine but is more brittle than aluminum and thus is difficult to cold form. It is nonmagnetic and has fair corrosion resistance, better than steel, but not as good as aluminum. Some magnesium alloys are hardenable, and all can be anodized. It is the most active metal on the galvanic scale and cannot be combined with most other metals in a wet environment. It is also extremely flammable, especially in powder or chip form, and its flame cannot be doused with water. Machining requires flooding with oil coolant to prevent fire. It is roughly twice as costly per pound as aluminum. Magnesium is used where light weight is of paramount importance such as in castings for chainsaw housings and other handheld items. See Appendix A for mechanical property data. Copper Alloys Pure copper is soft, weak, and malleable and is used primarily for piping, flashing, electrical conductors (wire) and motors. It cold works readily and can become brittle after forming, requiring annealing between successive draws. Many alloys are possible with copper. The most common are brasses and bronzes which themselves are families of alloys. Brasses, in general, are alloys of copper and zinc in varying proportions and are used in many applications, from artillery shells and bullet shells to lamps and jewelry. Bronzes were originally defined as alloys of copper and tin, but now also include alloys containing no tin, such as silicon bronze and aluminum bronze, so the terminology is somewhat confusing. Silicon bronze is used in marine applications such as ship propellers. Beryllium copper is neither brass nor bronze and is the strongest of the alloys, with strengths approaching those of alloy steels (200 kpsi [1380 MPa]). It is often used in springs that must be nonmagnetic, carry electricity, or exist in corrosive environments. Phosphor bronze is also used for springs but unlike beryllium copper, it cannot be bent along the grain or heat treated.
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Copper and its alloys have excellent corrosion resistance and are nonmagnetic. All copper alloys can be cast, hot or cold formed, and machined, but pure copper is difficult to machine. Some alloys are heat treatable and all will work harden. The Young’s modulus of most copper alloys is about 17 Mpsi (117 GPa) and their weight density is slightly higher than that of steel at 0.31 lb/in3 (8580 kg/m3). Copper alloys are expensive compared to other structural metals. See Appendix A for mechanical property data. 2.7
GENERAL PROPERTIES OF NONMETALS
The use of nonmetallic materials has increased greatly in the last 50 years. The usual advantages sought are light weight, corrosion resistance, temperature resistance, dielectric strength, and ease of manufacture. Cost can range from low to high compared to metals depending on the particular nonmetallic material. There are three general categories of nonmetals of general engineering interest: polymers (plastics), ceramics, and composites. Polymers have a wide variety of properties, principally low weight, relatively low strength and stiffness, good corrosion and electrical resistance, and relatively low cost per unit volume. Ceramics can have extremely high compressive (but not tensile) strengths, high stiffness, high temperature resistance, high dielectric strength (resistance to electrical current), high hardness, and relatively low cost per unit volume. Composites can have almost any combination of properties you want to build into them, including the highest specific strengths obtainable from any materials. Composites can be low or very high in cost. We will briefly discuss nonmetals and some of their applications. Space does not permit a complete treatment of these important classes of materials. The reader is directed to the bibliography for further information. Appendix A also provides some mechanical property data for polymers. Polymers The word polymers comes from poly = many and mers = molecules. Polymers are longchain molecules of organic materials or carbonbased compounds. (There is also a family of siliconbased polymeric compounds.) The source of most polymers is oil or coal, which contains the carbon or hydrocarbons necessary to create the polymers. While there are many natural polymer compounds (for example, wax, rubber, proteins), most polymers used in engineering applications are manmade. Their properties can be tailored over a wide range by copolymerization with other compounds or by alloying two or more polymers together. Mixtures of polymers and inorganic materials such as talc or glass fiber are also common. Because of their variety, it is difficult to generalize about the mechanical properties of polymers, but compared to metals they have low density, low strength, low stiffness, nonlinear elastic stressstrain curves as shown in Figure 222 (with a few exceptions), low hardness, excellent electrical and corrosion resistance, and ease of fabrication. Their apparent moduli of elasticity vary widely from about 10 kpsi (69 MPa) to about 400 kpsi (2.8 GPa), all much less stiff than any metals. Their ultimate tensile strengths range from about 4 kpsi (28 MPa) for the weakest unfilled polymer to about 22 kpsi (152 MPa) for the strongest glassfilled polymers. The specific gravities of most polymers range from about 0.95 to 1.8 compared to about 2 for magnesium, 3 for aluminum, 8 for steel, and
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MATERIALS AND PROCESSES
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FIGURE 222 Tensile Test StressStrain Curves of Three Thermoplastic Polymers (From Fig. 5.18, p. 161, in N. E. Dowling, Mechanical Behavior of Materials, PrenticeHall, Englewood Cliffs, N.J., 1993, with permission)
13 for lead. So, even though the absolute strengths of polymers are low, their specific strengths are respectable due to their low densities. Polymers are divided into two classes, thermoplastic and thermosets. Thermoplastic polymers can be repeatedly melted and solidified, though their properties can degrade due to the high melt temperatures. Thermoplastics are easy to mold and their rejects or leftovers can be reground and remolded. Thermosetting polymers become crosslinked when first heated and will burn, not melt, on reheating. Crosslinking creates connections (like the rungs of a ladder) between the longchain molecules that wind and twist through a polymer. These crossconnections add strength and stiffness. Another division among polymers can be made between filled and unfilled compounds. The fillers are usually inorganic materials, such as carbon black, graphite, talc, chopped glass fibers, and metal powders. Fillers are added to both thermoplastic and thermosetting resins, though they are more frequently used in the latter. These filled compounds have superior strength, stiffness, and temperature resistance over that of the raw polymers but are more difficult to mold and to fabricate. A confusing array of polymers is available commercially. The confusion is increased by a proliferation of brand names for similar compounds made by different manufacturers. The generic chemical names of polymers tend to be long, complex, and hard to remember. In some cases a particular polymer brand name has been so widely used that it has become generic. Nylon, plexiglass, and fiberglass are examples. Learning the generic chemical names and associated brand names of the main families of engineering polymers will eliminate some of the confusion. Table 27 shows a number of important polymer families. The mechanical properties of a few of these that have significant engineering applications are included in Appendix A.
Table 27 Families of Polymers Thermoplastics Cellulosics Ethylenics Polyamides Polyacetals Polycarbonates Polyphenyline oxides Polysulfones Thermosets Aminos Elastomers Epoxies Phenolics Polyesters Silicones Urethanes
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Ceramics Ceramic materials are finding increasing application in engineering, and a great deal of effort is being devoted to the development of new ceramic compounds. Ceramics are among the oldest known engineering materials; clay bricks are ceramic materials. Though still widely used in building, clay is not now considered an engineering ceramic. Engineering ceramics are typically compounds of metallic and nonmetallic elements. They may be single oxides of a metal, mixtures of metallic oxides, carbides, borides, nitrides, or other compounds such as Al2O3, MgO, SiC, and Si3N4, for example. The principal properties of ceramic materials are high hardness and brittleness, high temperature and chemical resistance, high compressive strength, high dielectric strength, and potentially low cost and weight. Ceramic materials are too hard to be machined by conventional techniques and are usually formed by compaction of powder, then fired or sintered to form bonds between particles and increase their strength. The powder compaction can be done in dies or by hydrostatic pressure. Sometimes, glass powder is mixed with the ceramic and the result is fired to melt the glass and fuse the two together. Attempts are being made to replace traditional metals with ceramics in such applications as cast engine blocks, pistons, and other engine parts. The low tensile strength, porosity, and low fracture toughness of most ceramics can be problems in these applications. Plasmasprayed ceramic compounds are often used as hard coatings on metal substrates to provide wear and corrosionresistant surfaces.
2
Composites Most composites are manmade, but some, such as wood, occur naturally. Wood is a composite of long cellulose fibers held together in a resinous matrix of lignin. Manmade composites are typically a combination of some strong, fibrous material such as glass, carbon, or boron fibers glued together in a matrix of resin such as epoxy or polyester. The fiberglass material used in boats and other vehicles is a common example of a glassfiber reinforced polyester (GFRP) composite. The directional material properties of a composite can be tailored to the application by arranging the fibers in different juxtapositions such as parallel, interwoven at random or particular angles, or wound around a mandrel. Custom composites are finding increased use in highly stressed applications such as airframes due to their superior strengthtoweight ratios compared to the common structural metals. Temperature and corrosion resistance can also be designed into some composite materials. These composites are typically neither homogeneous nor isotropic as was discussed in Section 2.3.
Table 28 Iron and Steel Strengths Form Theoretical
Sut kpsi (MPa) 2900 (20E3)
Whisker
1800 (12E3)
Fine wire
1400 (10E3)
Mild steel
60 (414)
Cast iron
40 (276)
It is interesting to note that if one calculates the theoretical strength of any “pure” elemental crystalline material based on the interatomic bonds of the element, the predicted strengths are orders of magnitude larger that those seen in any test of a “real” material, as seen in Table 28. The huge differences in actual versus theoretical strengths are attributed to disruptions of the atomic bonds due to crystal defects in the real material. That is, it is considered impossible to manufacture “pure anything” on any realistic superatomic scale. It is presumed that if we could make a “wire” of pure iron only one atom in diameter, it would exhibit its theoretical “super strength.” Crystal “whiskers” have been successfully made of some elemental materials and exhibit very high tensile strengths, which approach their theoretical values (Table 28).
Chapter 2
MATERIALS AND PROCESSES
Other empirical evidence for this theory comes from the fact that fibers of any material made in very small diameters exhibit much higher tensile strengths than would be expected from stressstrain tests of larger samples of the same material. Presumably, the very small cross sections are approaching a “purer” material state. For example, it is well known that glass has poor tensile strength. However, smalldiameter glass fibers show much larger tensile strength than sheet glass, making them a practical (and inexpensive) fiber for use in boat hulls, which are subjected to large tensile stresses in use. Smalldiameter fibers of carbon and boron exhibit even higher tensile strengths than glass fiber, which explains their use in composites for spacecraft and military aircraft applications, where their relatively high cost is not a barrier. 2.8
SELECTING MATERIALS
One of the most important design decisions is the proper choice of material. Materials limit design and new materials are still being invented that open new design possibilities. It would help if there were a systematic way to select a material for an application. M. F. Ashby has proposed such an approach that plots various material properties against one another to form “materials selection charts.”[3] Materials can be roughly divided into six classes: metals, ceramics, polymers (solid or foam), elastomers, glasses, and composites (which include wood). Members of these classes and subclasses tend to cluster together on a plot of this type. Figure 223 shows such a chart that plots Young’s modulus against density, which is called specific stiffness. By drawing lines of constant slope on such a chart, one can see which materials possess similar properties. A line of specific stiffness E / ρ = C has been drawn in color on Figure 223 and shows that some woods have equivalent specific stiffness to steel and some other metals. The line also passes through the lower range of the engineering composites’ “bubble” indicating that fiberglass (GFRP) has about the same specific stiffness as wood and steel, while the nonreinforced thermoplastics, such as nylon and polyester, have lower specific stiffness. So if you seek the stiffest/lightest material, you want to move up and to the left on the chart. Other lines are shown that have slopes equal to En / ρ = C where n is a fraction such as 1/2 or 1/3. These represent loading situations, such as beams in bending, for which the parameter of interest is a nonlinear function of specific stiffness. Since the chart is a loglog plot, exponential functions also plot as straight lines allowing simple comparisons to be made. Figure 224 shows a chart of strength versus density (called specific strength) for a number of materials. In this chart, the particular material strength used varies with the material depending on its character. For example, ductile metals and polymers show their yield strength, brittle ceramics their crushing compressive strength, and elastomers their tear strength. The vertical elongation of a material’s “bubble” indicates the range of strength values that can obtain due to thermal or work hardening, alloying elements, etc. The colored line drawn on the chart represents a particular value of specific strength or σ / ρ = C and shows that the strengthtoweight ratio of some woods are as good as highstrength steel and better than most other metals. It should be no surprise that wood is a popular material in building construction. Note also the high specific strength of engineering ceramics. Unfortunately, their tensile strengths are at best only about 10% of these compressive strengths, which is why you seldom see them used in structures where tensile stresses are commonly encountered.
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2
E/ρ=C
FIGURE 223
Copyright © 2018 Robert L. Norton: All Rights Reserved
Young's Modulus Plotted Against Density for Engineering Materials (Adapted from Fig. 43, p. 37 in M. F. Ashby, Materials Selection in Mechanical Design, 2ed, Elsevier, 2016)
Ashby’s book[3] is a very useful reference for the practicing engineer. It has dozens of charts of the type shown here that plot various properties against one another in a manner that enhances their comparison and develops good understanding. 2.9
SUMMARY
There are many different kinds of material strengths. It is important to understand which ones are important in particular loading situations. The most commonly measured and reported strengths are the ultimate tensile strength Sut and the tensile yield strength
Chapter 2
MATERIALS AND PROCESSES
65
σ/ρ=C
FIGURE 224
Copyright © 2018 Robert L. Norton: All Rights Reserved
Strength Plotted Against Density for Engineering Materials (Adapted from Fig. 44, p. 39 in M. F. Ashby, Materials Selection in Mechanical Design, 2ed, Elsevier, 2016)
Sy. The Sut indicates the largest stress that the material will accept before fracture, and Sy indicates the stress beyond which the material will take a permanent set. Many materials have compressive strengths about equal to their tensile strengths and are called even materials. Most wrought metals are in the even category. Some materials have significantly different compressive and tensile strengths and these are called uneven materials. Cast metals are usually in the uneven category, with compressive strengths much greater than their tensile strengths. The shear strengths of even materials tend to
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be about half their tensile strengths, while shear strengths of uneven materials tend to be between their tensile and compressive strengths.
2
One or more of these strengths may be of interest when the loading is static. If the material is ductile, then Sy is the usual criterion of failure, as a ductile material is capable of significant distortion before fracture. If the material is brittle, as are most cast materials, then the Sut is a more interesting parameter, because the material will fracture before any significant yielding distortion takes place. Yield strength values are nevertheless reported for brittle materials, but are usually calculated based on an arbitrary, small value of strain rather than on any measured yielding of the specimen. Chapter 5 deals with the mechanisms of material failure for both ductile and brittle materials in more detail than does this chapter. The tensile test is the most common measure of these static strength parameters. The stressstrain curve (σ–ε) generated in this test is shown in Figure 22. The socalled engineering σ–ε curve differs from the true σ–ε curve due to the reduction in area of a ductile test specimen during the failure process. Nevertheless, the engineering σ–ε curve is the standard used to compare materials, since the true σ–ε curve is more difficult to generate. The slope of the σ–ε curve in the elastic range, called Young’s modulus or the modulus of elasticity E, is a very important parameter as it defines the material’s stiffness or resistance to elastic deflection under load. If you are designing to control deflections as well as stresses, the value of E may be of more interest than the material’s strength. While various alloys of a given base material may vary markedly in terms of their strengths, they will have essentially the same E. If deflection is the prime concern, a lowstrength alloy is as good as a highstrength one of the same base material. When the loading on the part varies with time it is called dynamic or fatigue loading. Then the static strengths do not give a good indication of failure. Instead, the fatigue strength is of more interest. This strength parameter is measured by subjecting a specimen to dynamic loading until it fails. Both the magnitude of the stress and the number of cycles of stress at failure are reported as the strength criterion. The fatigue strength of a given material will always be lower than its static strength, and often is less than half its Sut. Chapter 6 deals with the phenomenon of fatigue failure of materials in more detail than does this chapter. Other material parameters of interest to the machine designer are resilience, which is the ability to absorb energy without permanent deformation, and toughness or the ability to absorb energy without fracturing (but with permanent deformation). Homogeneity is the uniformity of a material throughout its volume. Many engineering materials, especially metals, can be assumed to be macroscopically homogeneous even though at a microscopic level they are often heterogeneous. Isotropism means having properties that are the same regardless of direction within the material. Many engineering materials are reasonably isotropic in the macro and are assumed so for engineering purposes. However, other useful engineering materials, such as wood and composites, are neither homogeneous nor isotropic and their strengths must be measured separately in different directions. Hardness is important in wear resistance and is also related to strength. Heat treatment, both through and surface, as well as cold working, can increase the hardness and strength of some materials.
Chapter 2
MATERIALS AND PROCESSES
Important Equations Used in This Chapter See the referenced sections for information on the proper use of these equations. Axial tensile stress (Section 2.1):
P Ao
(2.1a)
l − lo lo
(2.1b)
σ= Axial tensile strain (Section 2.1):
ε=
Modulus of elasticity (Young’s modulus) (Section 2.1):
σ ε
(2.2)
E 2 (1 + ν )
(2.4)
E= Modulus of rigidity (Section 2.1):
G= Ultimate shear strength (Section 2.1):
steels:
Sus ≅ 0.80 Sut
other ductile metals:
Sus ≅ 0.75Sut
(2.5b)
Shear yield strength (Section 2.1):
U=
ε
∫0
σ dε
(2.6)
Modulus of resilience (Section 2.1):
UR ≅
2 1 Sy 2 E
(2.7)
Modulus of toughness (Section 2.1):
S y + Sut UT ≅ ε f 2
(2.8)
Arithmetic mean (Section 2.2):
µ=
1 n
n
∑ xi
(2.9b)
i =1
Standard deviation (Section 2.2):
Sd =
1 n1
n
∑ ( xi − µ )
2
(2.9c)
i =1
Ultimate tensile strength as a function of Brinell hardness (Section 2.4):
Sut ≅ 500 HB ± 30 HB
psi
Sut ≅ 3.45 HB ± 0.2 HB
MPa
(2.10)
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REFERENCES
1
E. B. Haugen and P. H. Wirsching, “Probabilistic Design.” Machine Design, v. 47, nos. 1014, Penton Publishing, Cleveland, Ohio, 1975.
2
H. E. Boyer and T. L. Gall, eds. Metals Handbook. Vol. 1. American Society for Metals: Metals Park, Ohio, 1985.
3
M. F. Ashby, Materials Selection in Mechanical Design, 2ed., Butterworth and Heinemann, 1999.
2.11
WEB REFERENCES The web is a useful resource for uptodate material property information at these and other sites that can be found with a search engine.
http://www.matweb.com
Properties data sheets for over 41,000 metals, plastics, ceramics, and composites. http://metals.about.com
Material properties and data.
2.12
BIBLIOGRAPHY
For general information on materials, consult the following:
Metals & Alloys in the Unified Numbering System. 6th ed. ASTM/SAE: Phila., Pa., 1994. H. E. Boyer, ed. Atlas of StressStrain Curves. Amer. Soc. for Metals: Metals Park, Ohio, 1987. Brady, ed. Materials Handbook. 13th ed. McGrawHill: New York, 1992. K. Budinski, Engineering Materials: Properties and Selection. 4th ed. RestonPrenticeHall: Reston, Va., 1992. M. M. Farag, Selection of Materials and Manufacturing Processes for Engineering Design. PrenticeHall International: Hertfordshire, U.K., 1989. I. Granet, Modern Materials Science. RestonPrenticeHall: Reston, Va., 1980. H. W. Pollack, Materials Science and Metallurgy. 2nd ed. RestonPrenticeHall: Reston, Va., 1977. M. M. Schwartz, ed. Handbook of Structural Ceramics. McGrawHill: New York, 1984. S. P. Timoshenko, History of Strength of Materials. McGrawHill: New York, 1983. L. H. V. Vlack, Elements of Material Science and Engineering. 6th ed. AddisonWesley: Reading, Mass., 1989. For specific information on material properties, consult the following:
H. E. Boyer and T. L. Gall, ed. Metals Handbook. Vol. 1. American Society for Metals: Metals Park, Ohio, 1985. R. Juran, ed. Modern Plastics Encyclopedia. McGrawHill: New York, 1988. J. D. Lubahn and R. P. Felgar, Plasticity and Creep of Metals. Wiley: New York, 1961. U. S. Department of Defense. Metallic Materials and Elements for Aerospace Vehicles and Structures MILHDBK5H, 1998.
Chapter 2
MATERIALS AND PROCESSES
69
For information on failure of materials, consult the following:
J. A. Collins, Failure of Materials in Mechanical Design. Wiley: New York, 1981. N. E. Dowling, Mechanical Behavior of Materials. PrenticeHall: Englewood Cliffs, N.J., 1992. R. C. Juvinall, Stress, Strain and Strength. McGrawHill: New York, 1967. For information on plastics and composites, consult the following:
ASM, Engineered Materials Handbook: Composites. Vol. 1. American Society for Metals: Metals Park, Ohio, 1987. ASM, Engineered Materials Handbook: Engineering Plastics. Vol. 2. American Society for Metals: Metals Park, Ohio, 1988. Harper, ed. Handbook of Plastics, Elastomers and Composites. 2nd ed. McGrawHill: New York, 1990. J. E. Hauck, “LongTerm Performance of Plastics.” Materials in Design Engineering, pp. 113128, November, 1965. M. M. Schwartz, Composite Materials Handbook. McGrawHill: New York, 1984. For information on manufacturing processes, see:
R. W. Bolz, Production Processes: The Productivity Handbook. Industrial Press: New York, 1974. J. A. Schey, Introduction to Manufacturing Processes. McGrawHill: New York, 1977.
2.13 21
PROBLEMS Figure P21 shows stressstrain curves for three failed tensiletest specimens. All are plotted on the same scale. (a) (b) (c) (d) (e)
Characterize each material as brittle or ductile. Which is the stiffest? Which has the highest ultimate strength? Which has the largest modulus of resilience? Which has the largest modulus of toughness?
22 Determine an approximate ratio between the yield strength and ultimate strength for each material shown in Figure P21. 23 Which of the steel alloys shown in Figure 219 would you choose to obtain (a) (b) (c) (d)
Maximum strength Maximum modulus of resilience Maximum modulus of toughness Maximum stiffness
24 Which of the aluminum alloys shown in Figure 221 would you choose to obtain (a) (b) (c) (d)
Maximum strength Maximum modulus of resilience Maximum modulus of toughness Maximum stiffness
25 Which of the thermoplastic polymers shown in Figure 222 would you choose in order to obtain (a) (b) (c) (d)
Maximum strength Maximum modulus of resilience Maximum modulus of toughness Maximum stiffness
Table P20 Topic/Problem Matrix 2.1 Material Properties 21, 22, 23, 24, 25, 26, 27, 28, 29, 210, 211, 212, 218, 219, 220, 221, 222, 223 2.4 Hardness 213, 214, 241, 242, 243, 244 2.6 General Properties 215, 216, 217, 224, 225, 226 2.8 Selecting Materials 237, 238, 239, 240
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*
Answers to these problems are provided in Appendix D.
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A metal has a strength of 414 MPa at its elastic limit and the strain at that point is 0.002. Assume the test specimen is 12.8mm dia and has a 50mm gage length. What is its modulus of elasticity? What is the strain energy at the elastic limit? Can you define the type of metal based on the given data?
27
A metal has a strength of 41.2 kpsi (284 MPa) at its elastic limit and the strain at that point is 0.004. Assume the test specimen is 0.505in dia and has a 2in gage length. What is the strain energy at the elastic limit? Can you define the type of metal based on the given data?
*28
A metal has a strength of 134 MPa at its elastic limit and the strain at that point is 0.003. What is its modulus of elasticity? Assume the test specimen is 12.8mm dia and has a 50mm gage length. What is its modulus of elasticity? What is the strain energy at the elastic limit? Can you define the type of metal based on the given data?
*29
A metal has a strength of 100 kpsi (689 MPa) at its elastic limit and the strain at that point is 0.006. What is its modulus of elasticity? What is the strain energy at the elastic limit? Assume the test specimen is 0.505in dia and has a 2in gage length. Can you define the type of metal based on the given data?
Strain ε (a )
210 A material has a yield strength of 689 MPa at an offset of 0.6% strain. What is its modulus of resilience?
Stress σ
211 A material has a yield strength of 60 kpsi (414 MPa) at an offset of 0.2% strain. What is its modulus of resilience? *212
A steel has a yield strength of 414 MPa, an ultimate tensile strength of 689 MPa, and an elongation at fracture of 15%. What is its approximate modulus of toughness? What is its approximate modulus of resilience?
213 The Brinell hardness of a steel specimen was measured to be 250 HB. What is the material’s approximate tensile strength? What is its hardness on the Vickers scale? The Rockwell scale? Strain ε
*214
(b )
Stress σ
The Brinell hardness of a steel specimen was measured to be 340 HB. What is the material’s approximate tensile strength? What is its hardness on the Vickers scale? The Rockwell scale?
215 What are the principal alloy elements of an AISI 4340 steel? How much carbon does it have? Is it hardenable? By what techniques? *216
What are the principal alloy elements of an AISI 1095 steel? How much carbon does it have? Is it hardenable? By what techniques?
217 What are the principal alloy elements of an AISI 6180 steel? How much carbon does it have? Is it hardenable? By what techniques? 218 Which of the steels in Problems 215, 216, and 217 is the stiffest? Strain ε
219 Calculate the specific strength and specific stiffness of the following materials and pick one for use in an aircraft wing spar. (a) (b) (c)
(c )
FIGURE P21 StressStrain Curves
Steel Aluminum Titanium
Sut = 80 kpsi (552 MPa) Sut = 60 kpsi (414 MPa) Sut = 90 kpsi (621 MPa)
220 If maximum impact resistance were desired in a part, which material properties would you look for? 221
Refer to the tables of material data in Appendix A and determine the strengthtoweight ratios of the following material alloys based on their tensile yield strengths: heattreated 2024 aluminum, SAE 1040 coldrolled steel, Ti75A titanium, type 302 coldrolled stainless steel.
Chapter 2
MATERIALS AND PROCESSES
222
Refer to the tables of material data in Appendix A and determine the strengthtoweight ratios of the following material alloys based on their ultimate tensile strengths: heattreated 2024 aluminum, SAE 1040 coldrolled steel, unfilled acetal plastic, Ti75A titanium, type 302 coldrolled stainless steel.
223
Refer to the tables of material data in Appendix A and calculate the specific stiffnesses of aluminum, titanium, gray cast iron, ductile iron, bronze, carbon steel, and stainless steel. Rank them in increasing order of this property and discuss the engineering significance of these data.
224
Call your local steel and aluminum distributors (consult the Yellow Pages or Internet) and obtain current costs per pound for round stock of consistent size in lowcarbon (SAE 1020) steel, SAE 4340 steel, 2024T4 aluminum, and 6061T6 aluminum. Calculate a strength/dollar ratio and a stiffness/dollar ratio for each alloy. Which would be your first choice on a costefficiency basis for an axialtensionloaded round rod (a) (b)
225
71
If maximum strength were needed? If maximum stiffness were needed?
Call your local plastic stockshapes distributors (consult the Yellow Pages or Internet) and obtain current costs per pound for round rod or tubing of consistent size in plexiglass, acetal, nylon 6/6, and PVC. Calculate a strength/dollar ratio and a stiffness/ dollar ratio for each alloy. Which would be your first choice on a costefficiency basis for an axialtensionloaded round rod or tube of particular diameters. (Note: material parameters can be found in Appendix A.) (a) (b)
If maximum strength were needed? If maximum stiffness were needed?
226 A part has been designed and its dimensions cannot be changed. To minimize its deflections under the same loading in all directions irrespective of stress levels, which of these materials would you choose: aluminum, titanium, steel, or stainless steel? Why? *227
Assuming that the mechanical properties data given in Appendix Table A9 for some carbon steels represents mean values, what is the value of the tensile yield strength for 1050 steel quenched and tempered at 400F if a reliability of 99.9% is required?
228 Assuming that the mechanical properties data given in Appendix Table A9 for some carbon steels represents mean values, what is the value of the ultimate tensile strength for 4340 steel quenched and tempered at 800F if a reliability of 99.99% is required? 229 Assuming that the mechanical properties data given in Appendix Table A9 for some carbon steels represents mean values, what is the value of the ultimate tensile strength for 4130 steel quenched and tempered at 400F if a reliability of 90% is required? 230 Assuming that the mechanical properties data given in Appendix Table A9 for some carbon steels represents mean values, what is the value of the tensile yield strength for 4140 steel quenched and tempered at 800F if a reliability of 99.999% is required? 231 A steel part is to be plated to give it better corrosion resistance. Two materials are being considered: cadmium and nickel. Considering only the problem of galvanic action, which would you choose? Why? 232 A steel part with many holes and sharp corners is to be plated with nickel. Two processes are being considered: electroplating and electroless plating. Which process would you choose? Why? 233
What is the common treatment used on aluminum to prevent oxidation? What other metals can also be treated with this method? What options are available with this method?
*234
Steel is often plated with a less noble metal that acts as a sacrificial anode that will corrode instead of the steel. What metal is commonly used for this purpose (when the
*
Answers to these problems are provided in Appendix D.
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finished product will not be exposed to saltwater), what is the coating process called, and what are the common processes used to obtain the finished product?
2
235
A lowcarbon steel part is to be heattreated to increase its strength. If an ultimate tensile strength of approximately 550 MPa is required, what mean Brinell hardness should the part have after treatment? What is the equivalent hardness on the Rockwell scale?
236
A lowcarbon steel part has been tested for hardness using the Brinell method and is found to have a hardness of 220 HB. What are the approximate lower and upper limits of the ultimate tensile strength of this part in MPa?
237
Figure 224 shows “guide lines” for minimum weight design when failure is the criterion. The guide line, or index, for minimizing the weight of a beam in bending is σf2/3 / ρ, where σf is the yield strength of a material and ρ is its mass density. For a given crosssection shape the weight of a beam with given loading will be minimized when this index is maximized. The following materials are being considered for a beam application: 5052 aluminum, cold rolled; CA170 beryllium copper, hard plus aged; and 4130 steel, [email protected] The use of which of these three materials will result in the lowestweight beam?
238
Figure 224 shows “guide lines” for minimum weight design when failure is the criterion. The guide line, or index, for minimizing the weight of a member in tension is σf / ρ, where σf is the yield strength of a material and ρ is its mass density. The weight of a member with given loading will be minimized when this index is maximized. For the three materials given in Problem 237, which will result in the lowest weight tension member?
239
Figure 223 shows “guide lines” for minimum weight design when stiffness is the criterion. The guide line, or index, for minimizing the weight of a beam in bending is E1/2 / ρ, where E is the modulus of elasticity of a material and ρ is its mass density. For a given crosssection shape the weight of a beam with given stiffness will be minimized when this index is maximized. The following materials are being considered for a beam application: 5052 aluminum, cold rolled; CA170 beryllium copper, hard plus aged; and 4130 steel, [email protected] The use of which of these three materials will result in the lowestweight beam?
240
Figure 224 shows “guide lines” for minimum weight design when stiffness is the criterion. The guide line, or index, for minimizing the weight of a member in tension is E / ρ, where E is the modulus of elasticity of a material and ρ is its mass density. The weight of a member with a given stiffness will be minimized when this index is maximized. For the three materials given in Problem 239, which will result in the lowestweight tension member?
241
Find a relationship between the Vickers and Brinell hardness scales that will give a Vickers number as a function of a Brinell number to an accuracy of at least 99%. Compare your computed Vickers number against that given in Table 23 for Brinell values of 159, 375, and 495 and calculate the percent variation for each.
242
Find a relationship between the Rockwell C and Brinell hardness scales over the range of 277 ≤ HB ≤ 578 that will give a Rockwell C number as a function of a Brinell number to an accuracy of at least 99%. Compare your computed Rockwell C number against that given in Table 23 for Brinell values of 311, 375, and 495 and calculate the percent variation for each.
243
After a part in service failed it was found to have a hardness of 34.2 HRC. What was its approximate ultimate tensile strength?
244
A part was tested and found to have a hardness of 437 HV. What was its approximate ultimate tensile strength?
KINEMATICS AND LOAD DETERMINATION
Look long on an engine. It is sweet to the eyes. Macknight black
3.0
INTRODUCTION
This chapter provides a brief introduction to the fundamentals of kinematics, which is defined as the study of motion without regard to forces. This approach, first suggested by Ampere, is intended to make it easier to understand motion analysis by temporarily assuming that all bodies are rigid and are not loaded by forces. This chapter also provides a review of the fundamentals of static and dynamic force analysis, impact forces, and beam loading. It also introduces singularity functions for beam calculations. The Newtonian solution method of force analysis is reviewed and a number of casestudy examples are presented to reinforce understanding of this subject. The case studies also set the stage for analysis of these same systems for stress, deflection, and failure modes in later chapters. At the end of the chapter, a summary section is provided that groups significant equations from this chapter for easy reference and identifies the chapter section in which their discussion can be found.
3.1
DEGREE OF FREEDOM
The degree of freedom (DOF) of a system is equal to the number of coordinates needed to define its position in space. In threespace, a single rigid body has six DOF, three translational (x, y, z) and three rotational (q, f, g). When multiple bodies are connected together with joints, then either more or fewer DOF may result. A system’s DOF also defines the number of inputs (motors or other actuators) needed to control its motion. If the DOF = 1, the system is a oneDOF mechanism, more than 1, a multiDOF mechanism, zero, a structure that cannot move, and less than 1, a preloaded structure. This chapter is Copyright © 2018 Robert L. Norton: All Rights Reserved
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Titlepage photograph is of a cutaway model of the valve train for a WWII radial aircraft engine taken by the author at the Bristol, England museum. Copyright 2018 Robert L. Norton: All Rights Reserved * A more complete presentation of the material in sections 3.1 to 3.8 inclusive can be found in reference 1.

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MECHANISMS*
3.2
Mechanisms come in many types but are all variants of a linkage, which is made of a collection of links and joints, one of which is grounded, and all are interconnected in a way to provide a controlled output in response to one or more inputs. A link, for the purpose of kinematic analysis, is a rigid body of any shape that has some number of attachment points called nodes (which are often holes) that allow multiple links to be connected by joints. Figure 31a shows some examples of links. Links with two attachment points or nodes are called binary, with three, ternary, with four quaternary. This naming and adding of nodes can continue indefinitely, but many mechanisms are made with combinations of the three links shown. Note that the links can be of any shape. The number of nodes alone determines the character of the link.
Nodes
Binary link
Quaternary link
Ternary link
(a) Some links—their names reflect the number of nodes
∆θ
∆θ ∆x
∆x
∆x ∆θ
Prismatic (P) joint—1 DOF (form closed)
Pin in slot—2 DOF (form closed)
Link against plane—2 DOF (force closed)
Revolute (R) joint—1 DOF (form closed)
∆φ
∆x
∆θ Cylindric (C) joint—2 DOF (form closed)
∆θ ∆ψ
∆y
∆φ ∆x
∆θ Helical (H) joint—1 DOF (form closed)
Spherical (S) joint—3 DOF (force closed)
Planar (F) joint—3 DOF (force closed)
(b) Some joint types—note their DOF and type of closure
FIGURE 31 Examples of Links and Joints
Copyright © 2018 Robert L. Norton: All Rights Reserved
Chapter 3
KINEMATICS AND LOAD DETERMINATION
Joints are characterized by their geometry, by the number of degrees of freedom they allow between the links they join, and by whether they are held together (closed) by a force or by their form (geometry). Figure 31b shows a number of joint types. Clockwise from the upper left of part b in the figure, two links joined with a pin have one DOF (Dq) and the pin joint is form closed because the pin is captured in the hole by its geometry. This is called a revolute joint. A block captive in a slot is a formclosed prismatic joint with one DOF (Dx). But a pin captive in a slot is a formclosed half joint with two DOF (Dx) and (Dq). Likewise, the link contacting a plane has two DOF but is force closed. A planar joint has three DOF and is force closed. The same is true for the spherical (joystick) joint. The helical joint (screw and nut) has one DOF. As you turn the screw it advances in the x direction. The cylindric joint has two DOF, Dx and Dq.
CALCULATING DEGREE OF FREEDOM (MOBILITY)
3.3
The degree of freedom of an assembly of links and joints can be calculated with a simple equation, though in some cases it may give a misleading result. We present the twodimensional version of this equation, which applies to any planar linkage, defined as one whose links move in parallel planes. The Kutzbach equation uses just the number of links and joints for the calculation: M = 3 ( L − 1) − 2 J1 − J2
(3.1)
where M = degree of freedom or mobility, L is the number of links, J1 is the number of full or oneDOF joints, and J2 is the number of half or twoDOF joints. This equation does not account for link lengths and so can predict an incorrect DOF when the link geometries have particular values. Figure 32 shows two linkages that each have 5 links, 6 full (J1) joints, and no half (J2) joints. Equation 3.1 says that they both have zero DOF, i.e., they are predicted to be structures, unable to move. This is true for the linkage of Figure 32a, but the linkage in Figure 32b can obviously move due to the fact that the three binary links are the same length and are also parallel due to the fact that the two ternary links (one of which—the ground link— is shown only as its three pin joints fixed to ground) have the same spacing between their nodes. This special geometry, not reflected in the Kutzbach equation, causes this linkage to have one DOF M = 3(5 – 1) – 2(6) = 0
M = 3(5 – 1) – 2(6) = 0 but is untrue 5
5 2 3
1
2
4
1
(a) A structure with DOF = 0
FIGURE 32 A Kutzbach Paradox
1 1
3
4
1
1
(b) A mechanism with DOF = 1 Copyright © 2018 Robert L. Norton: All Rights Reserved
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despite the prediction of the equation. This is called a paradox of the Kutzbach equation. Many other paradoxes to this equation exist, all due to unique geometries.
3
3.4
COMMON 1DOF MECHANISMS
Fourbar Linkage and the Grashof Condition The simplest 1DOF mechanism is the fourbar linkage: four binary links connected by four pin joints. This is an amazingly versatile mechanism and can solve many intricate motion control problems. It will behave differently depending on the relative lengths of its four links. Its behavior can be predicted by the Grashof criterion: S + L ≤ P+Q
(3.2)
where S is the length of the shortest link, L that of the longest link, and P, Q, the lengths of the other two. It matters not the order in which the links are assembled. If the Grashof inequality is false, then no link can make a full revolution and it is said to be a nonGrashof linkage. If true, then at least one link can revolve fully with respect to the others and it is called a Grashof linkage. But if the expression evaluates to equal, then it is a specialcase Grashof linkage that has “changepoint” positions at which the successive positions of the linkage are indeterminate. Such a linkage either has to be constrained to avoid these change points, or additional links added to carry it through the change points. Figure 33 shows the four inversions of a (nonspecialcase) Grashof linkage. An inversion results from the grounding of a different link in the chain, thus any linkage will have as many inversions as it has links. Note the link numbering. The ground link
ground 1
coupler 3
rocker 4
rocker 4 crank 2
crank 2 ground 1
coupler 3 (a) Crank rockers
rocker 2 crank 2
ground 1
coupler 3
coupler 3 ground 1
rocker 4 crank 4 (b) Double rocker
FIGURE 33 Inversions of the Grashof Fourbar Linkage
(c) Double crank
Copyright © 2018 Robert L. Norton: All Rights Reserved
Chapter 3
KINEMATICS AND LOAD DETERMINATION
is always 1, the input link is always 2, the coupler is 3, and the output link is 4. A crank can fully revolve, a rocker cannot. Each inversion has the potential to deliver different types of motion from the same set of links. Figure 33a shows the two crankrocker inversions of the Grashof fourbar in which the crank rotates fully, the rocker rocks or oscillates through an arc, and the link connecting crank and rocker, called the coupler, has complex motion. The crankrocker inversion results when either of the links adjacent to the shortest is grounded. This is a very useful mechanism used whenever one wishes to convert the continuous rotation of a motor to an oscillation. An example is a windshieldwiper mechanism on an automobile where the wiper blade is attached to the rocker. Figure 33b shows the doublerocker mechanism that results when the link opposite the shortest is grounded. Since it is always the shortest link of a Grashof mechanism that can fully revolve with respect to the others, this inversion is more difficult to motivate since a motor cannot be attached to either of the grounded rocker links, neither of which fully rotates. Figure 33c shows the doublecrank inversion in which the shortest link is grounded resulting in the other three revolving about it. This can be motor driven and if one of the cranks is driven at constant speed, the other will exhibit nonconstant velocity. This is also called a draglink mechanism and is sometimes used in automotive steering mechanisms. Figure 34a shows one inversion of the special case Grashof linkage: the parallelogram mechanism, which is quite useful as long as the links are not allowed to become colinear, which is a change point. How they will emerge from the colinear position is mathematically unpredictable (indeterminate). However, over its useful range of motion, the coupler remains parallel to the ground plane. Figure 34b shows a way that the parallelogram mechanism can be carried through the change points by adding a second stage that is out of phase with the first. Both stages have change points but each occurs at a different crank angle so the other stage carries it through. Note that this fivebar linkage is also a Kutzbach paradox like Figure 32b. All four inversions of a nonGrashof fourbar linkage are triple rockers and give similar motions. All three moving links just oscillate until they reach extreme positions called toggle positions. They are nevertheless useful as many applications do not require that any link make a full revolution. One example is the suspension linkage in an automobile that controls the up and down motion of the wheel assemblies over bumps coupler 3 crank 2 crank 2
coupler 3
crank 4 ground 1
(a) Parallelogram linkage
FIGURE 34 Specialcase Grashof linkages
crank 4 ground 1
crank 5
ground 1
(b) Able to make a full revolution Copyright © 2018 Robert L. Norton: All Rights Reserved
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3
Car frame
3
4
4 Springs
FIGURE 35
Copyright © 2018 Robert L. Norton: All Rights Reserved
NonGrashof Fourbar Linkages Used for Automotive Suspension
as shown in Figure 35. Here the wheel motion only requires the links to move through relatively small angles and never through a full revolution. So all forms of the fourbar linkage are very useful and are found in all sorts of machinery. It is the basic building block of all mechanisms. Sixbar Linkage The next simplest combination of links is the sixbar linkage, which comes in two forms, the Watt and the Stephenson. Figure 36 shows the two inversions of the Watt sixbar and the three inversions of the Stephenson sixbar. Note that these all contain four binary links to which a pair of ternary links has been added. These are all oneDOF linkages. 4
5
3
3
4
6 2
5
2
6
1
1
(a) Watt’s sixbar inversion I
(b) Watt’s sixbar inversion II
3
5
3
5 2
4
3
6
4 6
4
5
2
2
6
1 1 (c ) Stephenson’s sixbar inversion I
FIGURE 36 All Inversions of the Sixbar Linkage
1 (d) Stephenson’s sixbar inversion II
(e) Stephenson’s sixbar inversion III Copyright © 2018 Robert L. Norton: All Rights Reserved
Chapter 3
KINEMATICS AND LOAD DETERMINATION
Slider block
79
Connecting rod Pisto n
2
3
3
4
1
1
2
4
Crank 1
1 Cylinder
Effective link 4
Effective rocker pivot is at infinity
Gas pressure
∞ (a) Crankslider—crank drives slider
FIGURE 37
(b) Slidercrank—slider drives crank Copyright © 2018 Robert L. Norton: All Rights Reserved
CrankSlider and SliderCrank Mechanisms
The Watt II in Figure 36b is probably the most commonly seen and it can be thought of as two fourbar linkages in series. Fourbar CrankSlider and SliderCrank If the rocker of a fourbar crank rocker linkage is increased in length indefinitely, it effectively becomes infinite in length and the linkage is transformed into a fourbar crankslider as shown in Figure 37a. The rocker link becomes a slider block that oscillates in a straight line rather than along an arc. (A straight line can be thought of as an arc of infinite radius.) This is a very common linkage in machinery and is the heart of every internal combustion (IC) engine. There it is driven backward as a slidercrank with the piston (slider) driving the crank as shown in Figure 37b. In this mode, it has two change points at top and bottom dead center and its crank must be spun to start it moving continuously, which is why you must pull the starter rope on the lawn mower to start it and crank the car engine with a starter motor. The momentum of the rotating crank carries it through the change points once it is put in motion. The crankslider of Figure 37a is widely used to obtain straightline output motion from a continuously rotating crank. An example is a piston pump for pumping well water or compressing air. All inversions of the crankslider that can be physically assembled are Grashof linkages in which the crank can fully revolve. Cam and Follower A cam and follower is another variant of the fourbar linkage in which the crank takes on a contoured shape. It is then called a cam and drives a rocker or slider directly. There is no coupler link. Figures 38a and 38b show cam and follower arrangements that drive a translating slider and an oscillating rocker follower, respectively. Dotted lines show the effective links of an equivalent fourbar crankslider or fourbar crankrocker that would give the same motion as the cam for the instantaneous position shown. One major advantage of a camfollower mechanism over a pure linkage is the variety of follower motion functions possible. You can think of the camfollower mechanism as a fourbar linkage or crankslider in which the crank and coupler are able to change their lengths
3
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Instantaneous center of cam curvature Effective link 2
ω2
3
An Integrated Approach
Sixth Edition
Follower
Half joint
Half joint Vfollowe r 4
2
Effective link 2 Spring
Cam 2
Fo llower Spring parallel effect ive links [email protected] ∞

Effective link 3
Effective link 3
Cam Effect ive link 1

ω2
ω4
1
Effective link 4 Instantaneous center of cam curvature
(a) Cam with sliding follower—a variant of a crankslider
4
Effect ive link 4
(b) Cam with oscillating follower—a variant of a crankrocker
FIGURE 38
Copyright © 2018 Robert L. Norton: All Rights Reserved
Cams and Followers are Variants of Fourbar Linkages
as the cam rotates. The effective links are shown superposed on the mechanism in the figure. The effective coupler (link 3) runs from the instantaneous center of curvature of the cam contour (which changes for each cam position) to the center of curvature of the follower. The effective crank runs from the cam pivot to the instantaneous center of curvature of the cam contour. Thus, the lengths of the effective crank and effective coupler change with cam rotational position. The cam contour must be defined by appropriate mathematical functions, which will be addressed in a later section.
3.5
ANALYZING LINKAGE MOTION
Before we can determine the forces and torques in a linkage, which are needed for a stress analysis, we first need to know the accelerations to use in Newton’s second law, F = ma. In order to find the accelerations, we must first find the position and velocity of the links because acceleration is the derivative of velocity and velocity is the derivative of position. We will show one approach to this kinematic analysis problem and apply it to the fourbar linkage and crank slider. We will limit our discussion to two dimensional or planar mechanisms. Luckily these constitute the majority of practical mechanisms used in machinery. Types of Motion General motion in the plane is called complex motion. A link in the plane has 3DOF, call them x, y, and q. Complex motion of a link then involves a translation of a point on the link in x and y and a rotation q about a point in the link. So, complex motion then has two components, called translation and rotation. Each of these can also exist without the other, and then it becomes a special case. A crank spinning about a fixed pivot is in pure rotation and a slider block traveling along a slot has pure translation. These are the components of complex motion. In a fourbar crank slider, the crank is in pure rotation, the slider is in pure translation and the coupler is in complex motion, simultaneously rotating and translating.
Chapter 3
KINEMATICS AND LOAD DETERMINATION
81
Complex Numbers as Vectors Position, velocity, and acceleration are vector quantities and so we need a notation to represent a vector. There are several possibilities, but for twodimensional vectors, complex numbers provide some advantages. Figure 39a shows the complex plane with a vector drawn on it. The axes are labeled real and imaginary. Don’t be put off by the terminology; despite its name, the imaginary axis simply depicts the ydirected cartesian component of the vector with the xdirected component on the real axis. The imaginary name comes about because we use the notation j to represent the square root of minus one, which cannot be evaluated numerically. But think of j as an operator rather than as a value. When present on a term, it merely signifies that the term is ydirected.
3
A vector in this notation has a very compact and convenient form, rejq, where r is its magnitude, e is the Naperian base 2.718..., and q is the angle of the vector. This is the polar notation for the vector, i.e., magnitude and angle. We can easily convert from polar to cartesian coordinates with Euler’s identity: e ± jθ = cos θ ± j sin θ
(3.3)
The real advantage of this complex notation comes when we need to differentiate an expression, especially if we use the polar form. The differential of exdx is ex. When it has a constant multiplier in the exponent such as ejxdx, its differential is jejx. So the differential of our vector rejq dq is rjejq provided that r is a constant. Note in Figure 39b that each multiplication of the vector RA by the operator j results in a counterclockwise rotation of the vector through 90 degrees. The vector RB = jRA is directed along the positive imaginary or j axis. The vector RC = j2 RA is directed along the negative real axis because j2 = –1 and thus RC = –RA. In similar fashion, RD = j3 RA = –jRA and this component is directed along the negative j axis. j Polar form : R e θ
Imaginary
Cartesian form : R cos θ + j R sin θ R =
Imaginary j
j R sin θ
A
j
RA
B 2
RC = j R = – R
R B = jR +θ A
RA C
θ
O
Real RA
Real R cos θ (a) Complex number representation of a position vector
FIGURE 39 Complex Number Representation of Vectors in Two Dimensions
3
R D = j R = – jR D (b) Vector rotations in the complex plane Copyright © 2018 Robert L. Norton: All Rights Reserved
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Thus, the appearance of the j operator as a multiplier in the differential indicates that this vector has been rotated 90°. Two differentiations result in a multiplier j2 = –1, which rotates the second differential 180° from the original vector. By this means, the directions of the differentiated vectors are taken care of automatically. After the differentiation is completed, the expression can be converted to cartesian form with Euler’s equation 3.3.
3 The Vector Loop Equation Closed linkages such as those discussed here are conveniently represented with a vector loop in which each link becomes a vector. These vectors close around the loop making their vector sum zero. That addition provides an equation for the linkage that can be evaluated to find position, velocity and acceleration information for any position. The vectors can be arranged in any orientation, that is their heads and tails can be at either end of a link as convenient. Since the angle of a vector must be measured at its root, then the orientation should be chosen such that it defines the angle of most interest for each link. Accordingly the linkage in Figure 310 shows the vector R2 rooted at the fixed pivot of link 2, R3 rooted where link 2 joins link 3, and R4 arranged with its root at the fixed pivot of link 4. For convenience, the X axis is aligned with R1 and the origin is placed at O2, the driving link pivot. Any such arrangement can be used and the correct results obtained as long as the signs of the vectors are observed in the summation.
3.6
ANALYZING THE FOURBAR LINKAGE
Solving for Position in the Fourbar Linkage The first step is to sum the vectors around the loop, taking their orientations into account: R 2 + R 3 − R 4 − R1 = 0
(3.4 a)
Then substitute the polar complex number notation for each vector using the letters of Figure 310 to represent the link magnitudes:
Y
B
y A
R3 θ3
b x c
R2
R4 θ4
a θ2
O2 FIGURE 310 Vector Loop for a Fourbar Linkage
d R1
O4
X
Copyright © 2018 Robert L. Norton: All Rights Reserved
Chapter 3
KINEMATICS AND LOAD DETERMINATION
a e jθ2 + b e jθ3 − c e jθ4 − d e jθ1 = 0
83
(3.4 b)
Substitute equation 3.3 to convert the equation to cartesian form: a ( cos θ2 + j sin θ2 ) + b ( cos θ3 + j sin θ3 ) − c ( cos θ4 + j sin θ4 ) − d ( cos θ1 + j sin θ1 ) = 0
(3.4c)
Expand and separate the real and imaginary (x and y) terms into two equations. Note that terms without a j multiplier are x components and those with a j are y components: a cos θ2 + b cos θ3 − c cos θ4 − d cos θ1 = 0 but:
θ1 = 0, so: a cos θ2 + b cos θ3 − c cos θ4 − d = 0
(3.4 d )
ja sin θ2 + jb sin θ3 − jc sin θ4 − jd sin θ1 = 0 but:
θ1 = 0, and the j 's divide out, so: a sin θ2 + b sin θ3 − c sin θ4 = 0
(3.4e)
Equations 3.4d and 3.4e can be solved simultaneously to find angles q3 and q4. The link lengths a, b, c, d and the driver angle q2 for this position are known. The complete derivation of this solution is provided on the website as Fourbar Position Derivation.pdf. It contains the equations 3.4f through 3.4o that have been omitted here to save space. The final result is:
where:
− B ± B 2 − 4 AC θ41,2 = 2 arctan 2A A = cos θ2 − K1 − K 2 cos θ2 + K 3
(3.4 p)
B = −2 sin θ2
C = K1 − ( K 2 + 1) cos θ2 + K 3 and:
K1 =
d , a
K2 =
d , c
K3 =
a2 − b2 + c2 + d 2 2ac
Note that there are two solutions corresponding to the + / – conditions on the radical. These two solutions, as with any quadratic equation, may be of three types: real and equal, real and unequal, complex conjugate. If the discriminant under the radical is negative, then the solution is complex conjugate, which simply means that the link lengths chosen are not capable of connection for the chosen value of the input angle q2. This can occur either when the link lengths are completely incapable of connection in any position or, in a nonGrashof linkage, when the input angle is beyond a toggle limit position. Then there is no real solution for that value of input angle q2. Excepting this situation, the solution will usually be real and unequal, meaning there are two values of q4 corresponding to any one value of q2. These are referred to as the crossed and open configurations of the linkage and also as the two circuits of the linkage. In the fourbar linkage, the minus solution gives q4 for the open configuration and the positive solution gives q4 for the crossed configuration. Figure 311 shows both crossed and open solutions for a Grashof crankrocker linkage. The terms crossed and open are based on the assumption that the input link 2, for
3
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MACHINE DESIGN

An Integrated Approach

Sixth Edition
Y
Open B
y
3
θ3 '
3
b A
θ3
c
x
4
a 2 θ2 O2
θ4 '
1 3'
θ4
O4
d
X
4' Crossed B'
FIGURE 311
Copyright © 2018 Robert L. Norton: All Rights Reserved
The Two Solutions to θ4 from the Fourbar Position Equation
which q2 is defined, is placed in the first quadrant (i.e., 0 < q2 < π/2). A Grashof linkage is then defined as crossed if the two links adjacent to the shortest link cross one another, and as open if they do not cross one another in this position. Note that the configuration of the linkage, either crossed or open, is solely dependent upon the way that the links are assembled. You cannot predict, based on link lengths alone, which of the solutions will be the desired one. In other words, you can obtain either solution with the same linkage by simply taking apart the pin that connects links 3 and 4 in Figure 311, and moving those links to the only other positions at which the pin will again connect them with the same value of q2. In so doing, you will have switched from one position solution, or circuit, to the other. A similar derivation that eliminates q4 from the equations to solve for q3 gives the following solution:
where
− E ± E 2 − 4 DF θ 31,2 = 2 arctan 2D D = cos θ2 − K1 + K 4 cos θ2 + K 5 E = −2 sin θ2 F = K1 + ( K 4 − 1) cos θ2 + K 5
and
(3.5)
K1 =
d a
K4 =
d b
K5 =
c2 − d 2 − a2 − b2 2 ab
Solving for Velocity in the Fourbar Linkage We start with the vector loop equation 3.4a and its complex number form 3.4b: R 2 + R 3 − R 4 − R1 = 0
(3.4 a)
Chapter 3
KINEMATICS AND LOAD DETERMINATION
a e jθ2 + b e jθ3 − c e jθ4 − d e jθ1 = 0
85
(3.4 b)
Differentiate equation 3.4b with respect to time to get velocity: ja e jθ2
but, so:
dθ dθ2 dθ4 + jb e jθ3 3 − jc e jθ4 =0 dt dt dt dθ 3 = ω3; dt
dθ2 = ω2; dt
dθ4 = ω4 dt
ja ω 2 e jθ2 + jb ω 3e jθ3 − jc ω 4 e jθ4 = 0
(3.6a)
(3.6b) (3.6 c )
The angle q1 is constant so its derivative is zero and it has dropped out. Equation 3.6c is the velocity difference equation and can be written: VA + VBA − VB = 0 where:
VA = ja ω 2 e
(3.6d ) jθ2
VBA = jb ω 3e jθ3 VB = jc ω 4 e
(3.6e)
jθ4
VA is the linear velocity of point A in Figure 311, VB is the linear velocity of point B, and VBA is the velocity difference of point B versus point A. It is read, “the velocity of B with respect to A.” Equation 3.6c is solved for the angular velocities w3 and w4 with these expressions: ω3 =
aω 2 sin ( θ4 − θ2 ) b sin ( θ3 − θ4 )
(3.6l )
ω4 =
aω 2 sin ( θ2 − θ3 ) c sin ( θ4 − θ3 )
(3.6m)
A complete derivation of this solution is provided on the website as Fourbar Velocity Derivation.pdf. It contains equations 3.6f through 3.6k omitted here to save space. Note that the position analysis must be complete before the velocities can be found as they depend on those terms as well as the link lengths and w2, the known driving velocity. Once w3 and w4 are known, we can find the linear velocities from: VA = a ω 2 ( − sin θ2 + j cos θ2 )
VBA = b ω 3 ( − sin θ3 + j cos θ3 )
(3.6n)
VB = c ω 4 ( − sin θ4 + j cos θ4 )
where the real and imaginary terms are the x and y components of the vectors. There are two solutions for the velocities corresponding to the open and crossed position solutions. Find them using the values for q3 and q4 of each of those results. Figure 312 shows the velocity vectors on a fourbar linkage and a graphical solution to equation 3.6d. The angle m between links 3 and 4 is called the transmission angle.
3
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MACHINE DESIGN

An Integrated Approach
Y y
ω3
A ω2
VBA
b R3
δ3 θ3
VA θ4
c R2 θ2
O2
VB
x
a –
VB R4
µ
Sixth Edition
µ
B
P
RP p
3
+

VA
– d
R1
ω4
VBA
(b)
X
O4
(a)
FIGURE 312
Copyright © 2018 Robert L. Norton: All Rights Reserved
Vector Loop and Velocity Components in a Fourbar Linkage
Angular Velocity Ratio and Mechanical Advantage The angular velocity ratio mV is defined as the output angular velocity divided by the input angular velocity. For a fourbar mechanism this is expressed as: mV =
ω4 ω2
(3.6o)
For a rotating system, power P is the product of torque T and angular velocity w that, in two dimensions, have the same (z) direction: P = Tω
(3.6 p)
Linkage systems can be very efficient if they are well made with low friction bearings on all pivots. Losses are often less than 10%. For simplicity in the following analysis we will assume that the losses are zero (i.e., a conservative system). Then, letting Tin and win represent input torque and angular velocity, and Tout and wout represent output torque and angular velocity, Pin = Tin ω in Pout = Tout ω out Pout = Pin Tout ω out = Tin ω in Tout ω = in = mT Tin ω out
(3.6q)
(3.6r )
Note that the torque ratio (mT = Tout /Tin) is the inverse of the angular velocity ratio. Mechanical advantage (mA) can be defined as: mA =
Fout Fin
(3.6s)
Chapter 3
KINEMATICS AND LOAD DETERMINATION
t
ABA
α3 ω3 AB R3
t
AA R2
α2
n
b
θ3
AB
n
c
a
R4
d
ABA
ω4
–AA t
n
AA
ABA
O4 (b)
(a)
FIGURE 313
Copyright © 2018 Robert L. Norton: All Rights Reserved
Acceleration Vectors on the Linkage Vector Loop (a), and in an Acceleration Diagram (b)
Assuming that the input and output forces are applied at some radii rin and rout, perpendicular to their respective force vectors, Fout =
Tout rout
(3.6t )
T Fin = in rin
substituting equations 3.6t in 3.6s gives an expression in terms of torque: T r m A = out in Tin rout
(3.6u)
Substituting equation 3.6r in 3.6u gives ω r m A = in in ω out rout
(3.6 v )
These two ratios, angular velocity ratio and mechanical advantage, provide useful, dimensionless indices of merit by which we can judge the relative quality of various linkage designs that may be proposed as solutions. Solving for Acceleration in the Fourbar Linkage We start with the velocity vector loop equation 3.6c ja ω 2 e jθ2 + jb ω 3e jθ3 − jc ω 4 e jθ4 = 0
(3.6 c )
and differentiate it to get an expression for acceleration:
( j 2 aω 22 e jθ
2
n
AA
t
ABA X
O2
n
n
θ4
R1 AA
AB
AB
α4
ABA
x
A
θ2
ω2
t
AB
B
t
y
Y
87
) (
) (
)
+ jaα 2 e jθ2 + j 2 b ω 32 e jθ3 + jb α 3 e jθ3 − j 2 c ω 24 e jθ4 + jc α 4 e jθ4 = 0
(3.7a)
3
88
MACHINE DESIGN

An Integrated Approach

Sixth Edition
Simplifying and grouping terms:
(aα 2 je jθ 3
2
) (
) (
)
− aω 22 e jθ2 + b α 3 je jθ3 − b ω 32 e jθ3 − c α 4 je jθ4 − c ω 24 e jθ4 = 0
(3.7b)
Equation 3.7b is the acceleration difference equation and can be written as: A A + A BA − A B = 0
(3.7c)
( ) ( ) jθ t n 2 jθ A BA = ( A BA + A BA ) = ( b α 3 je − b ω 3 e ) A B = ( AtB + A nB ) = ( c α 4 je jθ − c ω 24 e jθ ) A A = AtA + A nA = aα 2 je jθ2 − aω 22 e jθ2
where:
3
4
3
(3.7d )
4
AA and AB are the linear accelerations of points A and B, respectively, and ABA is the acceleration difference between points B and A. Note that each of these vectors has two components denoted by the superscripts t and n, which refer to the tangential and normal directions, respectively. The normal (or centripetal) component is directed toward the center of rotation and the tangential component is directed as its name indicates. The acceleration difference component ABA has as its center of rotation its reference point A. Figure 313 shows these components on the linkage and in a vector diagram. Equations 3.7i are solved simultaneously to get: CD − AF AE − BD CE − BF α4 = AE − BD α3 =
where:
(3.7 k ) (3.7l )
A = c sin θ4 B = b sin θ3 C = aα 2 sin θ2 + aω 22 cos θ2 + bω 32 cos θ3 − cω 24 cos θ4 D = c cos θ4
(3.7m)
E = b cos θ3 F = aα 2 cos θ2 − aω 22 sin θ2 − bω 32 sin θ3 + cω 24 sin θ4
A complete derivation of these equations showing equations 3.7e through 3.7j is provided on the website as Fourbar Acceleration Derivation.pdf. Note that the angular accelerations a3 and a4 are functions of the link lengths, all the link angles, and all the link angular velocities, as well as the input angular acceleration a2. So, the position and velocity analyses must be done before the accelerations can be found. Once a3 and a4 are found, the linear accelerations can be found by substituting the Euler identity (equation 3.3) into equation 3.7d: A Ax
A A = aα 2 ( − sin θ2 + j cos θ2 ) − aω 22 ( cos θ2 + j sin θ2 ) = − aα 2 sin θ2 − aω 22 cos θ2 A Ay = aα 2 cos θ2 − aω 22 sin θ2
(3.7n)
A BAx
A BA = b α 3 ( − sin θ3 + j cos θ3 ) − b ω 32 ( cos θ3 + j sin θ3 ) = − b α 3 sin θ3 − b ω 32 cos θ3 A BAy = b α 3 cos θ3 − b ω 32 sin θ3
(3.7o)
Chapter 3
A Bx
KINEMATICS AND LOAD DETERMINATION
A B = c α 4 ( − sin θ4 + j cos θ4 ) − c ω 24 ( cos θ4 + j sin θ4 ) = − c α 4 sin θ4 − c ω 24 cos θ4 A By = c α 4 cos θ4 − c ω 24 sin θ4
89
(3.7 p)
where the real and imaginary terms are the x and y components, respectively. It is often the case that we need to know the linear acceleration of a point on a link such as its center of gravity (CG). Figure 312 shows the CG of link 3 at point P and defines its location with the position vector RP at angle d3. The acceleration of this point is found from the addition of two acceleration vectors, such as AA and APA. Vector AA is already defined from our analysis of the link accelerations. APA is the acceleration difference of point P with respect to point A. Point A is chosen as the reference point because angle q3 is defined at a local coordinate system whose origin is at A. Position vector RPA is defined using its magnitude p, the internal link offset angle d3 and the angle of link 3, q3. Differentiate this position vector twice to get acceleration: j θ +δ R PA = pe ( 3 3 ) j θ +δ VPA = jpe ( 3 3 )ω 3 j θ +δ j θ +δ A PA = pα 3 je ( 3 3 ) − pω 32 e ( 3 3 )
(3.7q)
= pα 3 − sin ( θ3 + δ 3 ) + j cos ( θ3 + δ 3 )
− pω 32 cos ( θ3 + δ 3 ) + j sin ( θ3 + δ 3 )
and use the acceleration difference equation to find AP: A P = A A + A PA
3.7
(3.7r )
ANALYZING THE FOURBAR CRANKSLIDER
Figure 314 shows an offset fourbar crankslider linkage. This is the general case in which the slider axis does not pass through the crank pivot. The vector loop uses four vectors, with R1 arranged parallel to the slider axis and through the crank pivot. Vector R4 is perpendicular to the slider axis and represents the offset, which has constant magnitude c. The variable magnitude d of R1 defines the position of the slider. R3 is rooted at the slider with constant magnitude b, and R2 is rooted at the crank pivot with constant magnitude a. Solving for Position in the Fourbar CrankSlider Write the vector loop equation from Figure 314: R 2 − R 3 − R 4 − R1 = 0
(3.8a)
Put it in complex polar form: a e jθ2 − b e jθ3 − c e jθ4 − d e jθ1 = 0
Substitute equation 3.3 to put it in complex cartesian form:
(3.8b)
3
90
MACHINE DESIGN

An Integrated Approach
Y

Sixth Edition
θ3
y
slider axis
3
b
offset
O2
R2
x
R3
A a
B
c
Rs
θ2
R4 θ4
d
X
R1 FIGURE 314
Copyright © 2018 Robert L. Norton: All Rights Reserved
Vector Loop for an Offset Fourbar CrankSlider Linkage
a ( cos θ2 + j sin θ2 ) − b ( cos θ3 + j sin θ3 )
− c ( cos θ4 + j sin θ4 ) − d ( cos θ1 + j sin θ1 ) = 0
(3.8c)
Separate the real and imaginary terms to get two scalar equations: a cos θ2 − b cos θ3 − c cos θ4 − d = 0
(3.8d )
a sin θ2 − b sin θ3 − c sin θ4 = 0
(3.8e)
Solve equations 3.8d and 3.8e simultaneously for slider position d and q3. Note that q4 is a known constant value: a sin θ2 − c θ31 = arcsin b d = a cos θ2 − b cos θ3
(3.8 f ) (3.8 g)
This solution is for only one circuit of the linkage. The second circuit has: a sin θ2 − c θ32 = arcsin − + π b
(3.8h)
and the value of d for the second circuit is found from equation 3.8g using this value of q3. Solving for Velocity in the Fourbar CrankSlider Start with the vector loop equation 3.8b for this linkage a e jθ2 − b e jθ3 − c e jθ4 − d e jθ1 = 0
(3.8b)
and differentiate it with respect to time to get velocity: ja ω 2 e jθ2 − jb ω 3e jθ3 − d = 0
(3.9a)
Substitute the Euler identity (equation 3.3) into equation 3.9a, separate into real and imaginary components and solve simultaneously to get:
Chapter 3
KINEMATICS AND LOAD DETERMINATION
θ3
Y
91
y VBA
ω3
+
R3 A ω2
R2
θ2
VBA θ4 (b)
d
O2
X
R1 to ∞
(a)
FIGURE 315
effective link 4 Copyright © 2018 Robert L. Norton: All Rights Reserved
Velocity Vectors and Vector Loop on a CrankSlider Linkage
ω3 =
a cos θ2 ω2 b cos θ3
(3.9 g)
d = − a ω 2 sin θ2 + b ω 3 sin θ3
(3.9h)
Note that a position analysis must be done before solving for velocity. The linear velocities are: VA = a ω 2 ( − sin θ2 + j cos θ2 )
(3.9i )
VAB = b ω 3 ( − sin θ3 + j cos θ3 )
(3.9 j )
VB = d
(3.9 k )
Figure 315 shows these vectors on the linkage. A complete derivation of these equations showing equations 3.9b through 3.9f can be found in the file Slider Velocity Derivation.pdf on the website. Solving for Acceleration in the Fourbar CrankSlider Start with the velocity equation 3.9a for this linkage ja ω 2 e jθ2 − jb ω 3e jθ3 − d = 0
(3.9a)
and differentiate it with respect to time to get acceleration:
( ja α 2e jθ
2
) (
)
+ j 2 a ω 22 e jθ2 − jb α 3e jθ3 + j 2 b ω 32 e jθ3 − d = 0
(3.10 a)
Substitute the Euler identity (equation 3.3) into equation 3.10a, separate into real and imaginary components and solve simultaneously to get: α3 =
3
VA c
VA
µ
R4
Rs
a
–
x
VB
µ
b
VB
4
B
a α 2 cos θ2 − a ω 22 sin θ2 + b ω 32 sin θ3 b cos θ3
d = − a α 2 sin θ2 − a ω 22 cos θ2 + b α 3 sin θ3 + b ω 32 cos θ3
(3.10i ) (3.10 j)
92
MACHINE DESIGN
θ3
α3 ω3
Y

t
AA
A
n
AA
a
b
R2
θ2
ω2
4
x AB
n
ABA c
AA
Sixth Edition
y B
R3

t
ABA
AB
3
An Integrated Approach
R4
AA
t
ABA
n
AA
θ4
α2 d
O2
t
n
AA
ABA X
R1
(b)
(a)
FIGURE 316
Copyright © 2018 Robert L. Norton: All Rights Reserved
Acceleration Vectors and Vector Loop on a CrankSlider Linkage
A position and velocity analysis must be done before solving for acceleration. The linear accelerations are:
( = (A
) ( + A ) = (b α
A A = AtA + A nA = aα 2 je A BA
t BA
n BA
3
j θ2
je
− aω 22 e
jθ3
j θ2
− b ω 32 e
)
jθ3
)
(3.10 k )
A B = AtB = d
Substitute the Euler identity (equation 3.3) into equation 3.10k to calculate the normal and tangential components of acceleration. Figure 316 shows the acceleration vectors on the linkage. A complete derivation of these equations showing equations 3.10b through 3.10h can be found in the file Slider Acceleration Derivation.pdf on the website. Other Linkages The vectorloop method can be used to analyze many other configurations of linkages. Space does not allow the presentation of derivations for other linkages such as the inverted crankslider, slidercrank, geared fivebar, or the sixbar linkage. These analyses can be found in reference 9. Also, the Student Edition of program Linkages, written by the author, is provided on the website and can be used to solve a number of linkage configurations including the fourbar linkage, fourbar slider, sixbar, and geared fivebar linkages.
† http://www.designofmachinery.com/MD/Cam_Machine.mp4
3.8
CAM DESIGN AND ANALYSIS View a Video of a Cam Machine
(21:28)†
Camfollower systems are a variant on the fourbar linkage as shown in Figure 38. Proper cam design requires that the higher derivatives of position be considered in defining the functions that control follower motion. The position function s defines the cam’s contour
Chapter 3
KINEMATICS AND LOAD DETERMINATION
93
but not the follower’s dynamic behavior. Velocity, acceleration, and its derivative, called jerk, need to be properly defined to control follower dynamics and avoid unwanted forces and vibrations in follower motion. Cams are particularly useful when a dwell in the follower motion is required. A dwell is defined as the cessation of follower motion while the input motion (cam rotation) continues for a portion of the cycle. This is very easy to obtain with a cam because a constant cam radius over some angle keeps the follower stationary. Any combination of motion and dwells is possible but two arrangements are the most common. A singledwell, or risefalldwell (RFD) cam is used to actuate the valves in an internal combustion engine (ICE). The risefall motion opens and closes the valve at the right point in the cycle to allow the airfuel mixture in or out of the cylinder, and the dwell holds the valve closed during the compression and power strokes of the cycle. A doubledwell, or risedwellfalldwell (RDFD) cam is common in automated assembly machines in which the follower picks up a part from a feeder during one dwell, moves the part into position during the rise, places it during the second dwell, and returns (falls) to pick up another part. Different mathematical functions are needed for singledwell and doubledwell motions but both must obey the Fundamental Law of Cam Design, which states that the motion function, including dwells, must be piecewise continuous over the entire cam through the second derivative of displacement (acceleration). In other words, there can be no discontinuities in the position, velocity, or acceleration functions over the full 360° of the cam. Over the last century, many functions have been proposed and used for double dwell cam motions. Some obey the fundamental law and some do not. These are dealt with in detail in references 9 and 10, but space does not permit their exposition here. Here we will show how to apply one type of function to cam motion, the polynomial, which is superior to the classical doubledwell functions in use and is also useful for the singledwell case. An advantage of polynomials for cam motion is their adaptability to different constraints. We can tailor a polynomial to any situation by properly applying a set of boundary conditions (BC) to the problem and then use those BC to calculate the necessary coefficients of a polynomial that will adhere to them. For situations in which the polynomial functions do not work well, BSpline functions will solve the problem in a superior way. These are addressed in detail in reference 10. The Timing Diagram A cam design problem typically starts with a timing diagram, which shows the timing of events such as rises, dwells, and falls over one cycle, defined as one full revolution of the cam. The circular cam is represented on a linear time (or angle) axis that corresponds to the circumference of a circle on the cam. The motions start at this prime circle and grow outward from it. Eventually, this linear axis will be “wrapped around” the circle to create the cam. This diagram does not indicate what functions will be used for the motions. That decision is left to the designer. Figure 317 shows an example of a timing diagram for a doubledwell cam. The durations, phasing, and maximum excursion (lift) of the various events (rise, fall, dwells) are defined but the rise and fall motions are not.
3
94
MACHINE DESIGN

An Integrated Approach

Sixth Edition
Motion mm or in High dwel l
1
3
Rise
Low dwel l
Fa ll
0 0
90
180
270
0
0.25
0.50
0.75
FIGURE 317
360
Cam angle θ deg
1.0
Time
t sec
Copyright © 2018 Robert L. Norton: All Rights Reserved
A Cam Timing Diagram
The svaj Diagram Cam functions are defined and displayed using svaj diagrams where s is displacement, v is velocity, a is acceleration, and j is jerk, each of the last three being the time derivative of the one preceding it. These are drawn one above the other to show their proper phase relationship and derivative character. The cam must be designed piece by piece. That is each motion (rise, fall, risefall, or dwell) must be handled as a separate function. They are then connected to create a piecewise continuous function over the entire interval that obeys the fundamental law of cam design. Figure 318 shows an example of an svaj diagram for a rise motion that uses a 345 polynomial displacement function.
s v
a
j
Polynomials for the DoubleDwell Case 0
cam angle θ
β
Copyright © 2018 Robert L. Norton: All Rights Reserved
F I G U R E 3  18 An svaj Diagram for a 345 Polynomial Displacement
The general form of a polynomial displacement function is: s = C0 + C1 x + C2 x 2 + C3 x 3 + C4 x 4 + C5 x 5 + C6 x 6 + + Cn x n
(3.11)
where s is follower displacement in length units, x is the independent variable, and the Cn are the coefficients to be determined based on our choice of boundary conditions. The order of a polynomial is equal to the number of terms used and its degree is the highest power present, which is always one less than the order because the first term has the variable to the zeroth power. The order needed for a particular design is equal to the number of boundary conditions (BC) chosen. The independent variable for cam work is usually set to x = q/b where q is the local cam angle for the interval and b is the duration angle of the motion interval. This makes x a normalized variable that runs from zero to one over the interval. To restore the proper units in the results, multiply velocity by w / b and acceleration by w2 / b2 with b in radians and w in rad/sec to get them in terms of time. Figure 319 shows the minimum number of BC needed for a doubledwell cam in order to satisfy the fundamental law. Because the dwells have zero velocity and acceleration, the rise and fall functions must have zero BC at the start (0°) and end (b1 or b2) to avoid any discontinuities through acceleration. The displacements must also match those of the dwells at each end of rise or fall. Thus, for a rise or fall between dwells, the minimum number of BC is six to obey the fundamental law. This means that the lowest order polynomial possible for this case is 6th order or 5th degree.
Chapter 3
s
0 v (b)
High dwell
Rise
h
(a)
KINEMATICS AND LOAD DETERMINATION
Low dwell
Fall
β1
0
β2
0
θ deg
0
(c)
β1
0
β2
0
0
θ deg
β1
j 0 (d)
3
θ deg
a
β2
0
0
θ deg
90
0
180
FIGURE 319
95
270
360
Copyright © 2018 Robert L. Norton: All Rights Reserved
The Minimum Set of Boundary Conditions for a Rise or Fall Between Dwells
EXAMPLE 31 The 345 Polynomial for a Double Dwell Cam Problem
Determine the coefficients of a 6th order, 5th degree polynomial to rise h mm over 90° between a low and high dwell.
Given
q is the local cam angle, b is the duration angle of the rise, and q/b is the normalized independent variable that runs from zero to one over the interval.
Assumptions
The fundamental law of cam design must be observed.
Solution
See Figures 318 and 319.
1 Figure 319 shows the desired boundary conditions for the rise to maintain continuity through accelerations. They are: when
θ = 0;
then
s = 0,
v = 0,
a=0
when
θ = β1;
then
s = h,
v = 0,
a=0
(a)
The corresponding set for the fall would be: when
θ = 0;
then
s = h,
v = 0,
a=0
when
θ = β2 ;
then
s = 0,
v = 0,
a=0
(b)
2 Write six terms of equation 3.11 in terms of the normalized variable, then differentiate it twice because our BC involve velocity and acceleration.
96
MACHINE DESIGN

An Integrated Approach
2

Sixth Edition
3
4
θ θ θ θ θ s = C0 + C1 + C2 + C3 + C4 + C5 β β β β β v=
θ θ θ θ 1 C1 + 2C2 + 3C3 + 4C4 + 5C5 β β β β β
a=
2 3 θ θ θ 1 2 6 12 20 + C + C C C + 2 3 4 5 β β β β 2
2
3
3
5
4
(c)
(d )
(e)
3 Substitute the boundary condition q = 0, s = 0 into equation (c): 0 = C0 + 0 + 0 + C0 = 0
(f )
4 Substitute q = 0, v = 0 into equation (d ): 1 C1 + 0 + 0 + β C1 = 0
(
0=
) (g)
5 Substitute q = 0, a = 0 into equation (e ): 0=
1
β2 C2 = 0
(C2 + 0 + 0 + ) (h)
6 Substitute q = b, s = h into equation (c): h = C3 + C4 + C5
(i )
7 Substitute q = b, v = 0 into equation (d ): 0=
1 (3C3 + 4C4 + 5C5 ) β
(j )
8 Substitute q = b, a = 0 into equation (e ): 0=
1 β2
(6C3 + 12C4 + 20C5 )
(k )
9 Three of the six coefficients are zero. Solve equations i, j, and k simultaneously to find the other three: C3 = 10 h;
C4 = −15h;
C5 = 6h
(l )
10 The equation for displacement is then: 4 5 θ3 θ θ s = h 10 − 15 + 6 β β β
(m)
Chapter 3
KINEMATICS AND LOAD DETERMINATION
11 The expressions for velocity and acceleration can be found by substituting the coefficients of equation f, g, h, and l into equations d and e. This function is called the 345 polynomial after its exponents. Figure 318 (repeated here) shows this function and its derivatives through jerk. The displacement is 5th degree, the velocity 4th degree, acceleration 3rd degree (a cubic), and jerk is 2nd degree (a parabola).
97
s v
3 We did not constrain the jerk in the previous example, but we could have. Polynomials offer much flexibility in design as we can change the functions simply by changing the boundary conditions. The advantage of making the jerk zero at the interfaces with the dwells is to reduce vibrations, as they are highly influenced by the jerk function. However, doing so will increase the peak acceleration and that may be undesirable. It is the designer’s choice and there are always tradeoffs. If we add two boundary conditions to the previous example to force j = 0 at q = 0 and q = b, we will then have eight BC and eight terms in the equation, which becomes 7th degree. Solving for the new coefficients with these constraints gives the following equation for displacement: 7
θ θ θ θ s = h 35 − 84 + 70 − 20 β β β β 4
5
6
a
j
0
cam angle θ
β
Copyright © 2018 Robert L. Norton: All Rights Reserved
F I G U R E 3  18 Repeated An svaj Diagram for a 345 Polynomial Displacement
(3.12) s
This is called the 4567 polynomial and is one of the lowest vibration double dwell functions available. See Figure 320. If the larger peak accelerations associated with this function do not cause the cam or follower to be too highly stressed, then this is an excellent choice, especially for high speed mechanisms. Otherwise, the 345 polynomial is a good allaround choice for doubledwell motions and has reasonably low vibration as compared to many other common doubledwell functions in use today.
v
a
j
Polynomials for the SingleDwell Case Cam functions that are good for double dwells are not a good choice for a single dwell case. If the specification calls for a symmetrical risefall motion, e.g., the same angle of rise as fall, then a sixth degree polynomial provides the best solution if jerk is unconstrained. Adding jerk constraints will raise it to eighth degree and is the best solution for that case as well. The flexibility of polynomials allows a risefall motion with a single expression. Consider the following set of boundary conditions for a singledwell cam: when q = 0,
s = 0,
when q = b/2,
s=h
when q = b,
s = 0,
v = 0,
a=0
v = 0,
a=0
The zero values at q = 0 and q = b are necessary to match the dwell at those points and the lift value at the midpoint (q = b / 2) creates the risefall. Because of symmetry, it is not necessary to specify the slope (velocity) to be zero at b / 2, as it will be so by default. The equation for this function is: θ θ θ θ s = h 64 − 192 + 192 − 64 β β β β 3
4
5
6
(3.13)
This function and its derivatives are shown in Figure 321 with its seven BC circled.
0
cam angle θ
β
Copyright © 2018 Robert L. Norton: All Rights Reserved
FIGURE 320 The 4567 Polynomial
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Note that the above solution did not create separate functions for rise and fall, but rather, used a single polynomial for both rise and fall. But, if the rise and fall are not the same duration, making the function asymmetrical, then you will have to use separate equations for rise and fall. Moreover, to make them work, you need to start with the motion that has the longer duration, whether rise or fall. Also, you need to leave the acceleration unspecified for the first segment calculated (the one of larger b) and determine its value from the function that results. That value is then used as a BC on the second segment to be calculated in order to match the accelerations at the joint between the segments and maintain mathematical continuity.
s
v
3 a
j 0
β/2
β
Copyright © 2018 Robert L. Norton: All Rights Reserved
FIGURE 321 3456 Polynomial for Symmetrical SingleDwell Cam
EXAMPLE 32 An Asymmetric RiseFall SingleDwell Cam Problem
Design a singledwell cam function to rise h units in 45°, fall h units in 135 and dwell for the remainder. Let h = 1 and cam w = 15 rad/sec.
Given
q is the local cam angle, b is the duration angle of the rise, and q/b is the normalized independent variable that runs from zero to one over an interval.
Assumptions
The fundamental law of cam design must be observed.
Solution
See Figure 322.
1 The rise and fall must be handled with separate polynomials. The longer duration motion (here the fall) must be done first. Set the boundary conditions for the fall to be: when
θ = 0;
then
s = h,
v=0
when
θ = β2 ;
then
s = 0,
v = 0,
a=0
(a)
Note that the acceleration is left unspecified at the start point. There are 5 BC for this segment, which will give a 4th degree polynomial. 2 Substitute the BC in the polynomial equation and its derivatives and solve them for the coefficients. The result is: θ θ θ s = h 1 − 6 + 8 − 3 β β β 2
3
4
(b)
3 Differentiate this equation twice to get the acceleration function and solve it at q = 0 for the given data. Multiply the acceleration equation by w2 / b2 with b in radians and w in rad/sec to get the result in terms of time. The acceleration at the start of the fall is –486.4 in/sec. This value must be used as a boundary condition on acceleration for the rise segment in order to maintain continuity. The BC for the rise then become: when
θ = 0;
then
s = h,
v = 0,
a=0
when
θ = β1;
then
s = 0,
v = 0,
a = −486.4
(c)
4 Substitute these BC in the polynomial equation and its derivatives and solve them for the coefficients for the rise. The result is:
Chapter 3
segment
1
KINEMATICS AND LOAD DETERMINATION
2
99
3
h = 1 at 45 ° s Segment 1 6 boundary conditions
Segment 2 5 boundary conditions
3 v
a
– 486. 4 in/sec 2
seg 1 only
– 2024 in/sec 2
j 0
45
180
FIGURE 322
360
Copyright © 2018 Robert L. Norton: All Rights Reserved
Asymmetric SingleDwell Polynomial Solution to Example 32
3 4 5 θ θ θ s = h 9.333 − 13.667 + 5.333 β β β
(d )
5 The result is shown in Figure 322.
Pressure Angle Figure 323 shows a cam and follower. The kinematic cam is defined by the pitch curve through the center of the roller. The cam surface is found as a normal offset from this curve by the roller radius. The prime circle radius defines the minimum cam radius and its circumference corresponds to the axis of the s diagram. The pressure angle is the angle between the velocity of the follower and the common normal at the contact point between cam and follower. The force between cam and follower lies along the common normal. So the pressure angle determines what percent of the force goes into motion and what percent is trying to jam the follower sideways in its guide. The rule of thumb is to keep the maximum pressure angle below about 30° for a translating follower. The eccentricity of the follower is defined as the offset of the follower axis from the cam centerline, shown as e in Figure 323. The pressure angle f varies around the cam and can be found from: φ = arctan
v−ε s + RP2 − ε 2
(3.14)
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Pressure angle
φ

Sixth Edition
Vfollower Follower
3
Common normal (axis of transmission)
Common tangent (axis of slip )
A
Pitch curve φ
Prime circle radius R p
ω cam
O2
Eccentricity FIGURE 323
ε
Fo llower axis of motion Copyright © 2018 Robert L. Norton: All Rights Reserved
Cam Pressure Angle
where v is the follower velocity in length/radian, e is the eccentricity, s is the displacement in length units, and Rp is the prime circle radius. Once the svaj functions are defined, the only things in this equation that the designer can change are Rp and e. Increasing Rp will reduce the pressure angle as it effectively increases the length of the s diagram. Eccentricity will shift the pressure angle function around zero but will have no effect on the peak to peak range. Radius of Curvature All mathematical functions possess a property called radius of curvature, denoted by r, and it can be positive (convex) or negative (concave). At any point on the function, in the limit, the function can be considered to be an infinitesimally short arc of particular radius. The relationship between this instantaneous radius of the cam contour and the radius of the roller follower is important to proper cam function. If the cam’s radius of curvature is smaller than that of the follower, then the follower will not be able to maintain proper contact with the cam. Two cases are possible. If the cam contour is concave (negative) and its r is smaller than the roller radius Rf, the condition shown in Figure 324 will obtain. If the cam
Chapter 3
KINEMATICS AND LOAD DETERMINATION
101
contour is convex (positive) and its r is equal to or smaller than the roller radius Rf, the conditions shown in Figure 325 will result. These conditions are called undercutting and result in a cusp on the cam surface that will cause the follower to jump off the cam. The radius of curvature of the cam pitch curve, which is the locus of the center of the roller as shown in Figure 323, can be calculated from this equation: ρ pitch =
( R + s )2 + v 2 P
3
32
( RP + s )2 + 2v 2 − a ( RP + s )
(3.15)
where s is displacement in length units, v is the follower velocity in length/radian, a is the follower acceleration in length/radian2, and Rp is the prime circle radius. Again, once svaj are defined, the designer can only control radius of curvature by changing Rp. The rule of thumb is to keep the absolute value of the minimum r of the pitch curve greater than the roller radius Rf by some comfort factor: ρmin >> R f
(3.16)
A factor of 2 is good. Less than 1.5 will be troublesome. See reference 10 for more information on cams and cam design. The Student Edition of program Dynacam, written by the author, is included on the book’s website. This program will calculate the polynomial coefficients for any set of boundary conditions and also provides a large number of additional cam functions. It calculates the pressure angle and radius of curvature and draws the cam profile.
3.9
LOADING CLASSES FOR FORCE ANALYSIS
The type of loading on a system can be divided into several classes based on the character of the applied loads and the presence or absence of system motion. Once the general configuration of a mechanical system is defined and its kinematic motions calculated, the next task is to determine the magnitudes and directions of all the forces and couples Rf >
ρmin
ρmin
Vfollower
ω cam
Cam FIGURE 324
Fo llower Copyright © 2018 Robert L. Norton: All Rights Reserved
Result of the Roller Radius Being Larger Than the Smallest Negative Radius of Curvature of the Cam
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Follower
3

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Cusp due to undercutting

Sixth Edition
Missing material and cusp due to undercutting Follower Pitch curve
Pitch curve
Cam surface (a) Radius of curvature of pitch curve equals the radius of the roller follower
FIGURE 325
(b) Radius of curvature of pitch curve is less than the radius of the roller follower Copyright © 2018 Robert L. Norton: All Rights Reserved
Undercutting Due to Cam Radius of Curvature Being Less Than or Equal to the Follower Radius
present on the various elements. These loads may be constant or may be varying over time. The elements in the system may be stationary or moving. The most general class is that of a moving system with timevarying loads. The other combinations are subsets of the general class. Table 31 shows the four possible classes. Class 1 is a stationary system with constant loads. One example of a Class 1 system is the base frame for an arbor press used in a machine shop. The base is required to support the dead weight of the arbor press, which is essentially constant over time, and the base frame does not move. The parts brought to the arbor press (to have something pressed into them) temporarily add their weight to the load on the base, but this is usually a small percentage of the dead weight. A static load analysis is all that is necessary for a Class 1 system. Class 2 describes a stationary system with timevarying loads. An example is a bridge which, though essentially stationary, is subjected to changing loads as vehicles drive over it and wind impinges on its structure. Class 3 defines a moving system with constant loads. Even though the applied external loads may be constant, any significant accelerations of the moving members can create timevarying reaction forces. An example might be a powered rotary lawn mower. Except for the case of mowing the occasional rock, the blades experience a nearly constant external load from mowing the grass. However, the accelerations of the spinning blades can create high loads at their fastenings. A dynamic load analysis is necessary for Classes 2 and 3.
Chapter 3
Table 31
KINEMATICS AND LOAD DETERMINATION
103
Load Classes Constant Loads
TimeVarying Loads
Stationary Elements
Class 1
Class 2
Moving Elements
Class 3
Class 4
3
Note however that, if the motions of a Class 3 system are so slow as to generate negligible accelerations on its members, it could qualify as a Class 1 system and then would be called quasistatic. An automobile scissors jack (see Figure 35) can be considered to be a Class 1 system since the external load (when used) is essentially constant, and the motions of the links are slow with negligible accelerations. The only complexity introduced by the motions of the elements in this example is that of determining in which position the internal loads on the jack’s elements will be maximal, since they vary as the jack is raised, despite the essentially constant external load. Class 4 describes the general case of a rapidly moving system subjected to timevarying loads. Note that even if the applied external loads are essentially constant in a given case, the dynamic loads developed on the elements from their accelerations will still vary with time. Most machinery, especially if powered by a motor or engine, will be in Class 4. An example of such a system is the engine in your car. The internal parts (crankshaft, connecting rods, pistons, etc.) are subjected to timevarying loads from the gasoline explosions, and also experience timevarying inertial loads from their own accelerations. A dynamic load analysis is necessary for Class 4.
3.10
FREEBODY DIAGRAMS
In order to correctly identify all potential forces and moments on a system, it is necessary to draw accurate freebody diagrams (FBDs) of each member of the system. These FBDs should show a general shape of the part and display all the forces and moments that are acting on it. There may be external forces and moments applied to the part from outside the system, and there will be interconnection forces and/or moments where each part joins or contacts adjacent parts in the assembly or system. In addition to the known and unknown forces and couples shown on the FBD, the dimensions and angles of the elements in the system are defined with respect to local coordinate systems located at the centers of gravity (CG) of each element.* For a dynamic load analysis, the kinematic accelerations, both angular and linear (at the CG), need to be known or calculated for each element prior to doing the load analysis.
3.11
LOAD ANALYSIS
This section presents a brief review of Newton’s laws and Euler’s equations as applied to dynamically loaded and statically loaded systems in both 3D and 2D. The method of
*
While it is not a requirement that the local coordinate system for each element be located at its CG, this approach provides consistency and simplifies the dynamic calculations. Further, most solid modeling CAD/CAE systems will automatically calculate the mass properties of parts with respect to their CGs. The approach taken here is to apply a consistent method that works for both static and dynamic problems and that is also amenable to computer solution.
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solution presented here may be somewhat different than that used in your previous statics and dynamics courses. The approach taken here in setting up the equations for force and moment analysis is designed to facilitate computer programming of the solution.
3
This approach assumes all unknown forces and moments on the system to be positive in sign, regardless of what one’s intuition or an inspection of the freebody diagram might indicate as to their probable directions. However, all known force components are given their proper signs to define their directions. The simultaneous solution of the set of equations that results will cause all the unknown components to have the proper signs when the solution is complete. This is ultimately a simpler approach than the one often taught in statics and dynamics courses, which requires that the student assume directions for all unknown forces and moments (a practice that does help the student develop some intuition, however). Even with that traditional approach, an incorrect assumption of direction results in a sign reversal on that component in the solution. Assuming all unknown forces and moments to be positive allows the resulting computer program to be simpler than would otherwise be the case. The simultaneous equation solution method used is extremely simple in concept, though it requires the aid of a computer to solve. Software is provided with the text to solve the simultaneous equations. See program Matrix on the website. Real dynamic systems are three dimensional and thus must be analyzed as such. However, many 3D systems can be analyzed by simpler 2D methods. Accordingly, we will investigate both approaches. ThreeDimensional Analysis Since three of the four cases potentially require dynamic load analysis, and because a static force analysis is really just a variation on the dynamic analysis, it makes sense to start with the dynamic case. Dynamic load analysis can be done by any of several methods, but the one that gives the most information about internal forces is the Newtonian approach based on Newton’s laws. NewtoN’s First Law A body at rest tends to remain at rest and a body in motion at constant velocity will tend to maintain that velocity unless acted upon by an external force. NewtoN’s secoNd Law The time rate of change of momentum of a body is equal to the magnitude of the applied force and acts in the direction of the force. Newton’s second law can be written for a rigid body in two forms, one for linear forces and one for moments or torques:
∑ F = ma
∑ MG = H G
(3.17a)
where F = force, m = mass, a = acceleration, MG = moment about the center of gravity and H G = the time rate of change of the moment of momentum, or the angular momentum about the CG. The left sides of these equations respectively sum all the forces and moments that act on the body, whether from known applied forces or from interconnections with adjacent bodies in the system.
Chapter 3
KINEMATICS AND LOAD DETERMINATION
105
For a threedimensional system of connected rigid bodies, this vector equation for the linear forces can be written as three scalar equations involving orthogonal components taken along a local x, y, z axis system with its origin at the CG of the body:
∑ Fx = max
∑ Fy = may
∑ Fz = maz
(3.17b)
If the x, y, z axes are chosen coincident with the principal axes of inertia of the body,* the angular momentum of the body is defined as HG = I x ω x ˆi + I y ω y ˆj + I z ω z kˆ
(3.17c)
where Ix, Iy, and Iz are the principal centroidal mass moments of inertia (second moments of mass) about the principal axes. This vector equation can be substituted into equation 3.17a to yield the three scalar equations known as Euler’s equations:
∑ M x = I x α x − ( I y − I z ) ω yω z ∑ M y = I yα y − ( I z − I x ) ω z ω x ∑ Mz = Izαz − (Ix − I y )ω xω y
(3.17d )
where Mx, My, Mz are moments about those axes and ax, ay, az are the angular accelerations about the axes. This assumes that the inertia terms remain constant with time, i.e., the mass distribution about the axes is constant. NewtoN’s third Law When two particles interact, a pair of equal and opposite reaction forces will exist at their contact point. This force pair will have the same magnitude and act along the same direction line, but have opposite sense. We will need to apply this relationship as well as applying the second law in order to solve for the forces on assemblies of elements that act upon one another. The six equations in equations 3.17b and 3.17d can be written for each rigid body in a 3D system. In addition, as many (thirdlaw) reaction force equations as are necessary will be written and the resulting set of equations solved simultaneously for the forces and moments. The number of secondlaw equations will be up to six times the number of individual parts in a threedimensional system (plus the reaction equations), meaning that even simple systems result in large sets of simultaneous equations. A computer is needed to solve these equations, though highend pocket calculators will solve large sets of simultaneous equations also. The reaction (thirdlaw) equations are often substituted into the secondlaw equations to reduce the total number of equations to be solved simultaneously. TwoDimensional Analysis All real machines exist in three dimensions but many threedimensional systems can be analyzed two dimensionally if their motions exist only in one plane or in parallel planes. Euler’s equations 3.17d show that if the rotational motions (w, a) and applied moments or couples exist about only one axis (say the z axis), then that set of three equations reduces to one equation,
∑ Mz = Izαz
(3.18a)
* This is a convenient choice for symmetric bodies but may be less convenient for other shapes. See F. P. Beer and E. R. Johnson, Vector Mechanics for Engineers, 3rd ed., 1977, McGrawHill, New York, Chap. 18, “Kinetics of Rigid Bodies in Three Dimensions.”
3
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because the w and a terms about the x and y axes are now zero. Equation 3.17b is reduced to
∑ Fx = max 3
∑ Fy = may
(3.18b)
Equations 3.18 can be written for all the connected bodies in a twodimensional system and the entire set solved simultaneously for forces and moments. The number of secondlaw equations will now be up to three times the number of elements in the system plus the necessary reaction equations at connecting points, again resulting in large systems of equations for even simple systems. Note that even though all motion is about one (z) axis in a 2D system, there may still be loading components in the z direction due to external forces or couples. Static Load Analysis The difference between a dynamic loading situation and a static one is the presence or absence of accelerations. If the accelerations in equations 3.17 and 3.18 are all zero, then for the threedimensional case these equations reduce to
∑ Fx = 0 ∑ Mx = 0
∑ Fy = 0 ∑ My = 0
∑ Fz = 0 ∑ Mz = 0
(3.19a)
∑ Fy = 0
∑ Mz = 0
(3.19b)
and for the twodimensional case,
∑ Fx = 0
Thus, we can see that the static loading situation is just a special case of the dynamic loading one, in which the accelerations happen to be zero. A solution approach based on the dynamic case will then also satisfy the static one with appropriate substitutions of zero values for the absent accelerations.
3.12
TWODIMENSIONAL, STATIC LOADING CASE STUDIES
This section presents a series of three case studies of increasing complexity, all limited to twodimensional static loading situations. A bicycle handbrake lever, a crimping tool, and a scissors jack are the systems analyzed. These case studies provide examples of the simplest form of force analysis, having no significant accelerations and having forces acting in only two dimensions.
C A S E
S T U D Y
1 A
Bicycle Brake Lever Loading Analysis Problem
Determine the forces on the elements of the bicycle brake lever assembly shown in Figure 326 during braking.
Chapter 3
KINEMATICS AND LOAD DETERMINATION
107
Fb 2 Fcable
Fsheath
3
brake lever
2
cable
pivot Px
Mh
3
1 Py
handlebar
handgrip Fb 1
FIGURE 326 Bicycle Brake Lever Assembly
Given
The geometry of each element is known. The average human’s hand can develop a grip force of about 267 N (60 lb) in the lever position shown.
Assumptions
The accelerations are negligible. All forces are coplanar and two dimensional. A Class 1 load model is appropriate and a static analysis is acceptable.
Solution
See Figures 326, 327, and Table 32, parts 1 and 2.
1 Figure 326 shows the handbrake lever assembly, which consists of three subassemblies: the handlebar (1), the lever (2), and the cable (3). The lever is pivoted to the handlebar and the cable is connected to the lever. The cable runs within a plasticlined sheath (for low friction) down to the brake caliper assembly at the bicycle’s wheel rim. The sheath provides a compressive force to balance the tension in the cable (Fsheath = –Fcable). The user’s hand applies equal and opposite forces at some points on the lever and handgrip. These forces are transformed to a larger force in the cable by the lever ratio of part 2. Figure 326 is a freebody diagram of the entire assembly since it shows all the forces and moments potentially acting on it except for its weight, which is small compared to the applied forces and is thus neglected for this analysis. The “broken away” portion of the handlebar can provide x and y force components and a moment if required for equilibrium. These reaction forces and moments are arbitrarily shown as positive in sign. Their actual signs will “come out in the wash” in the calculations. The known applied forces are shown acting in their actual directions and senses. 2 Figure 327 shows the three subassembly elements separated and drawn as freebody diagrams with all relevant forces and moments applied to each element, again neglecting the weights of the parts. The lever (part 2) has three forces on it, Fb2, F32, and F12. The twocharacter subscript notation used here should be read as, force of element 1 on 2 (F12) or force at B on 2 (Fb2), etc. This defines the source of the force (first subscript) and the element on which it acts (second subscript). This notation will be used consistently throughout this text for both forces and position vectors such as Rb2, R32, and R12 in Figure 327, which serve to locate the above three
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θ C
F32
Fb2
Sixth Edition
y
R32
F32

2
x
B
F12
3
A
F12
Vector diagram for brake lever
R12
3 F23
1 2°
F31 Fsheath Rd Mh
R31 F21
y A
D
R21
Px Rp Py
FIGURE 327
Rb2
2°
F13
φ
Fcabl e
Fb2
x
1 Rb1
Fb1
Bicycle Brake Lever FreeBody Diagrams *
Actually, for a simple static analysis such as the one in this example, any point (on or off the element) can be taken as the origin of the local coordinate system. However, in a dynamic force analysis it simplifies the analysis if the coordinate system is placed at the CG. So, for the sake of consistency, and to prepare for the more complicated dynamic analysis problems ahead, we will use the CG as the origin even in the static cases here. †
You may not have done this in your statics class but this approach makes the problem more amenable to a computer solution. Note that regardless of the direction shown for any unknown force on the FBD, we will assume its components to be positive in the equations. The angles of the known (given) forces (or the signs of their components) do have to be correctly input to the equations, however.
forces in a local, nonrotating coordinate system whose origin is at the center of gravity (CG) of the element or subassembly being analyzed.* On this brake lever, Fb2 is an applied force whose magnitude and direction are known. F32 is the force in the cable. Its direction is known but not its magnitude. Force F12 is provided by part 1 on part 2 at the pivot pin. Its magnitude and direction are both unknown. We can write equations 3.3b for this element to sum forces in the x and y directions and sum moments about the CG. Note that all unknown forces and moments are initially assumed positive in the equations. Their true signs will come out in the calculation.† However, all known or given forces must carry their proper signs:
∑ Fx = F12 x + Fb2 x + F32 x = 0 ∑ Fy = F12 y + Fb2 y + F32 y = 0 ∑ Mz = ( R12 × F12 ) + ( R b2 × Fb2 ) + ( R 32 × F32 ) = 0
(a)
The cross products in the moment equation represent the “turning forces” or moments created by the application of these forces at points remote from the CG of the element. Recall that these cross products can be expanded to
∑ M z = ( R12 x F12 y − R12 y F12 x ) + ( Rb2 x Fb2 y − Rb2 y Fb2 x ) + ( R32 x F32 y − R32 y F32 x ) = 0
(b)
Chapter 3
KINEMATICS AND LOAD DETERMINATION
109
We have three equations and four unknowns (F12x, F12y, F32x, F32y) at this point, so we need another equation. It is available from the fact that the direction of F32 is known. (The cable can pull only along its axis.) We can express one component of the cable force F32 in terms of its other component and the known angle, q of the cable: F32 y = F32 x tan θ
(c)
3 We could now solve the four unknowns for this element, but will wait to do so until the equations for the other two links are defined. 3 Part 3 in Figure 327 is the cable that passes through a hole in part 1. This hole is lined with a lowfriction material that allows us to assume no friction at the joint between parts 1 and 3. We will further assume that the three forces F13, F23, and Fcable form a concurrent system of forces acting through the CG and thus create no moment. With this assumption only a summation of forces is necessary for this element:
∑ Fx = Fcable ∑ Fy = Fcable
x
+ F13 x + F23 x = 0
y
+ F13 y + F23 y = 0
(d )
4 The assembly of elements labeled part 1 in Figure 327 can have both forces and moments on it (i.e., it is not a concurrent system), so the three equations 3.19b are needed:
∑ Fx = F21x + Fb1x + F31x + Px + Fsheath = 0 ∑ Fy = F21y + Fb1y + F31y + Py = 0 ∑ Mz = Mh + ( R 21 × F21 ) + ( R b1 × Fb1 ) + ( R 31 × F31 ) + ( R p × P ) + ( R d × Fsheath ) = 0 x
(e)
Expanding cross products in the moment equation gives the moment magnitude as
∑ M z = M h + ( R21x F21y − R21y F21x ) + ( Rb1x Fb1y − Rb1y Fb1x ) + ( R31x F31y − R31y F31x ) + ( RPx Py − RPy Px ) + ( Rdx Fsheath − Rdy Fsheath ) = 0 y
(f)
x
5 The total of unknowns at this point (including those listed in step 2 above) is 21: Fb1x, Fb1y, F12x, F12y, F21x, F21y, F32x, F32y, F23x, F23y, F13x, F13y, F31x, F31y, Fcablex, Fcabley, Fsheathx, Fsheathy,Px, Py, and Mh. We have only nine equations so far, three in equation set (a), one in set (c), two in set (d), and three in set (e). We need twelve more equations to solve this system. We can get seven of them from the Newton’s thirdlaw relationships between contacting elements: F23 x = − F32 x
F23 y = − F32 y
F21x = − F12 x
F21y = − F12 y
F31x = − F13 x
F31y = − F13 y
( g)
Fsheathx = − Fcablex
Two more equations come from the assumption (shown in Figure 326) that the two forces provided by the hand on the brake lever and handgrip are equal and opposite:* Fb1x = − Fb 2 x Fb1y = − Fb 2 y
(h)
*
But not necessarily colinear.
110
MACHINE DESIGN

An Integrated Approach

Sixth Edition
The remaining three equations come from the given geometry and the assumptions made about the system. The direction of the forces Fcable and Fsheath are known to be in the same direction as that end of the cable. In the figure it is seen to be horizontal, so we can set Fcabley = 0;
3
Fsheathy = 0
(i )
Because of our nofriction assumption, the force F31 can be assumed to be normal to the surface of contact between the cable and the hole in part 1. This surface is horizontal in this example, so F31 is vertical and F31x = 0
( j)
F13x
0.0
N
6 This completes the set of 21 equations (equation sets a, c, d, e, g, h, i, and j), and they can be solved for the 21 unknowns simultaneously “as is,” that is, all 21 equations could be put into matrix form and solved with a matrixreduction computer program. However, the problem can be simplified by manually substituting equations c, g, h, i, and j into the others to reduce them to a set of eight equations in eight unknowns. The known or given data are as shown in Table 32, part 1.
Fb2x
0.0
N
7 As a first step, for link 2, substitute equations b and c in equation a to get:
Fb2y
–267.0
N
184.0
deg
180.0
deg
Table 3–2—part 1 Case Study 1A Given Data Variable
θ φ Rb2x
Value
Unit
F12 x + Fb 2 x + F32 x = 0 F12 y + Fb 2 y + F32 x tan θ = 0
( R12 x F12 y − R12 y F12 x ) + ( Rb2 x Fb2 y − Rb2 y Fb2 x ) + ( R32 x F32 x tan θ − R32 y F32 x ) = 0
39.39
mm
Rb2y
2.07
mm
R32x
–50.91
mm
R32y
4.66
mm
R12x
–47.91
mm
Fcablex + F13 x − F32 x = 0
R12y
–7.34
mm
R21x
Fcable y + F13 y − F32 x tan θ = 0
7.0
mm
R21y
19.0
mm
Rb1x
47.5
mm
Rb1y
–14.0
mm
R31x
–27.0
mm
R31y
30.0
mm
Rpx
–27.0
mm
Rpy
0.0
mm
Rdx
–41.0
mm
Rdy
27.0
mm
(k )
8 Next, take equations d for link 3 and substitute equation c and also –F32x for F23x, and –F32y for F23y from equation g to eliminate those variables: (l )
9 For link 1, substitute equation f in e and replace F21x with –F12x, F21y with –F12y, F31x with –F13x , F31y with –F13y, and Fsheathx with –Fcablex from equation g, − F12 x + Fb1x − F13 x + Px − Fcablex = 0
(
) (
− F12 y + Fb1y − F32 x tan θ + Py = 0
M h + − R21x F12 y + R21y F12 x + Rb1x Fb1y − Rb1y Fb1x
(
) (
)
(m)
)
+ − R31x F13 y + R31y F13 x + RPx Py − RPy Px + Rdy Fcablex = 0
10 Finally, substitute equations h, i, and j into equations k, l, and m to yield the following set of eight simultaneous equations in the eight remaining unknowns: F12x, F12y, F32x, F13y, Fcablex, Px, Py, and Mh. Put them in the standard form, which has all unknown terms on the left and all known terms to the right of the equal signs.
Chapter 3
KINEMATICS AND LOAD DETERMINATION
111
F12 x + F32 x = − Fb 2 x F12 y + F32 x tan θ = − Fb 2 y Fcablex − F32 x = 0 F13 y − F32 x tan θ = 0
(n)
− F12 x + Px − Fcablex = Fb 2 x
(
3
− F12 y − F13 y + Py = Fb 2 y
)
R12 x F12 y − R12 y F12 x + R32 x tan θ − R32 y F32 x = − Rb 2 x Fb 2 y + Rb 2 y Fb 2 x M h − R21x F12 y + R21y F12 x − R31x F13 y + RPx Py − RPy Px + Rdy Fcablex = Rb1x Fb 2 y − Rb1y Fb 2 x
11 Form the matrices from equation n: 1 0 0 0 −1 0 − R12 y R21y
0 1 0 0 0 −1 R12 x − R21x
R32 x
1 tan θ −1 − tan θ 0 0 tan θ − R32 y 0
0 0 0 1 0 −1 0
0 0 1 0 −1 0 0
0 0 0 0 1 0 0
0 0 0 0 0 1 0
− R31x
Rdy
− RPy
RPx
F12 x F12 y F32 x F13 y × Fcablex Px Py 1 M h 0 0 0 0 0 0 0
= (o)
− Fb 2 x − Fb 2 y 0 0 Fb 2 x Fb 2 y −R F + R F b2x b2 y b2 y b2x Rb1x Fb 2 y − Rb1y Fb 2 x
12 Substitute the known data as shown in Table 32 part 1 (repeated opposite): 1 0 1 0 0 0 1 0.070 0 0 −1 0 0 1 0 0 −0.070 1 0 0 −1 0 0 0 −1 −1 −1 0 0 0 7.34 −47.91 −8.22 0 0 19 −7 0 27 27
0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 −27
0 0 0 0 0 0 0 1
F12 x F 12 y F32 x F13 y × F cablex Px Py M h
0 267 0 0 ( p) = 0 −267 10 517.13 −12 682.50
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MACHINE DESIGN
Table 3–2—part 2 Case Study 1A Calculated Data
3
Variable
Value
Unit
F32 x
–1909
N
F32 y
–133
N
F12 x
1909
N
F12 y
400
N
F23 x
1909
N
F23 y
133
N
F13 y
–133
N
Fcabl e x
–1909
N
Fcabl e y
0
N
Fb1x
0
N
Fb1y
267
N
F31 x
0
N
F31 y
133
N
F21 x
–1909
N
F21 y
–400
N
Px
0
N
Py
0
N
Mh
9
Nm
Fsheathx
1909
N

An Integrated Approach

Sixth Edition
13 The solution is shown in Table 32 part 2. This matrix equation can be solved with any one of a number of commercially available matrix solvers such as Mathcad, MATLAB, Maple, or Mathematica, as well as with many engineering pocket calculators. A customwritten program called Matrix is provided on the book’s website that can be used to solve a linear system of up to 16 equations. Equation p was solved with program Matrix to find the 8 unknowns listed in step 10. Those results were then substituted into the other equations to solve for the previously eliminated variables. 14 Table 32 part 2 shows the solution data for the data given in Figure 327 and Table 32 part 1. This assumes a 267N (60lb) force is applied by the person’s hand normal to the brake lever. The force generated in the cable (Fcable) is then 1909 N (429 lb) and the reaction force against the handlebar (F21) is 1951 N (439 lb) at –168°.
C A S E
S T U D Y
2 A
HandOperated CrimpingTool Loading Analysis Problem
Determine the forces on the elements of the crimping tool shown in Figure 328 during a crimp operation.
Given
The geometry is known and the tool develops a crimp force of 2000 lb (8896 N) at closure in the position shown.
Assumptions
The accelerations are negligible. All forces are coplanar and two dimensional. A Class 1 load model is appropriate and a static analysis acceptable.
Solution
See Figures 328 and 329, and Table 33, parts 1 and 2.
1 Figure 328 shows the tool in the closed position in the process of crimping a metal connector onto a wire. The user’s hand provides the input forces between links 1 and 2, shown as the reaction pair Fh. The user can grip the handle anywhere along its length, but we are assuming a nominal moment arm of Rh for the application of the resultant of the user’s grip force (see Figure 329). The high mechanical advantage of the tool transforms the grip force to a large force at the crimp. Figure 328 is a freebody diagram of the entire assembly, neglecting the weight of the tool, which is small compared to the crimp force. There are four elements, or links, in the assembly, all pinned together. Link 1 can be considered to be the “ground” link, with the other links moving with respect to it as the jaw is closed. The desired magnitude of the crimp force Fc is defined and its direction will be normal to the surfaces at the crimp. The third law relates the actionreaction pair acting on links 1 and 4: Fc1x = − Fc 4 x Fc1y = − Fc 4 y * Again, in a static analysis it is not necessary to take the CG as the coordinate system origin (any point can be used), but we do so to be consistent with the dynamic analysis approach in which it is quite useful to do so.
(a)
2 Figure 329 shows the elements of the crimpingtool assembly separated and drawn as freebody diagrams with all forces applied to each element, again neglecting their weights as being insignificant compared to the applied forces. The centers of gravity of the respective elements are used as the origins of the local, nonrotating coordinate systems in which the points of application of all forces on the elements are located.*
Chapter 3
KINEMATICS AND LOAD DETERMINATION
Fh
Table 3–3—part 1 Case Study 2A Given Data
1 2
Fh
113
3
4
force from hand
Fc crimp force
FIGURE 328 Wire Connector Crimping Tool
3 We will consider link 1 to be the ground plane and analyze the remaining moving links. Note that all unknown forces and moments are initially assumed positive. Link 2 has three forces acting on it: Fh is the unknown force from the hand, and F12 and F32 are the reaction forces from links 1 and 3, respectively. Force F12 is provided by part 1 on part 2 at the pivot pin and force F32 is provided by part 3 acting on part 2 at their pivot pin. The magnitudes and directions of both these pin forces are unknown. We can write equations 3.3b for this element to sum forces in the x and y directions and sum moments about the CG (with cross products expanded):
∑ Fx = F12 x + F32 x = 0 ∑ Fy = F12 y + F32 y + Fh = 0 ∑ M z = Fh Rh + ( R12 x F12 y − R12 y F12 x ) + ( R32 x F32 y − R32 y F32 x ) = 0
(b)
4 Link 3 has two forces on it, F23 and F43. Write equations 3.3b for this element:
∑ Fx = F23x + F43x = 0 ∑ Fy = F23 y + F43 y = 0 ∑ M z = ( R23x F23 y − R23 y F23x ) + ( R43x F43 y − R43 y F43x ) = 0
(c)
5 Link 4 has three forces acting on it: Fc4 is the known (desired) force at the crimp, and F14 and F34 are the reaction forces from links 1 and 3, respectively. The magnitudes and directions of both these pin forces are unknown. Write equations 3.19b for this element:
∑ Fx = F14 x + F34 x + Fc4 x = 0 ∑ Fy = F14 y + F34 y + Fc4 y = 0 ∑ M z = ( R14 x F14 y − R14 y F14 x ) + ( R34 x F34 y − R34 y F34 x ) + ( Rc 4 x Fc 4 y − Rc 4 y Fc 4 x ) = 0
(d )
6 The nine equations in sets b through d have 13 unknowns: F12x, F12y, F32x, F32y, F23x, F23y, F43x, F43y, F14x, F14y, F34x, F34y, and Fh. We can write the thirdlaw relationships between actionreaction pairs at each of the joints to obtain the four additional equations needed:
Variable
Value
Unit
Fc4x
–1956.30
lb
Fc4y
415.82
lb
R c4x
0.45
in
R c4y
0.34
in
R 12 x
1.40
in
R 12 y
0.05
in
R 32 x
2.20
in
R 32 y
0.08
in
Rh
–4.40
in
R 23 x
–0.60
in
R 23 y
0.13
in
R 43 x
0.60
in
R 43 y
–0.13
in
R 14 x
–0.16
in
R 14 y
–0.76
in
R 34 x
0.16
in
R 34 y
0.76
in
3
114
MACHINE DESIGN

An Integrated Approach

Sixth Edition
y
Fh
R21 Rc1
x
3 y
R23 F23
B
A
R43
all dimensions in inches
F21 Fc1
3
1
x 1.23
C
R41
F43
F41
D
1.0
y Rh
0.7
F34
F12 A 0.80
R12 R32
Fh
C
R34
B
x 2
y
Rc4
1.55
F32
Fc4
x F14
1.2 4 D
FIGURE 329
R14
FreeBody Diagrams of a Wire Connector Crimping Tool
F32 x = − F23 x ;
F34 x = − F43 x ;
F32 y = − F23 y ;
F34 y = − F43 y
(e)
7 The 13 equations b–e can be solved simultaneously by matrix reduction or by iteration with a rootfinding algorithm. For a matrix solution, the unknown terms are placed on the left and known terms on the right of the equal signs: F12 x + F32 x = 0 F12 y + F32 y + Fh = 0 Rh Fh + R12 x F12 y − R12 y F12 x + R32 x F32 y − R32 y F32 x = 0 F23 x + F43 x = 0 F23 y + F43 y = 0 R23 x F23 y − R23 y F23 x + R43 x F43 y − R43 y F43 x = 0 F14 x + F34 x = − Fc 4 x F14 y + F34 y = − Fc 4 y R14 x F14 y − R14 y F14 x + R34 x F34 y − R34 y F34 x = − Rc 4 x Fc 4 y + Rc 4 y Fc 4 x F32 x + F23 x = 0 F34 x + F43 x = 0 F32 y + F23 y = 0 F34 y + F43 y = 0
(f)
Chapter 3
KINEMATICS AND LOAD DETERMINATION
115
8 Substitute the given data from Table 33 part 1: F12 x + F32 x = 0 F12 y + F32 y + Fh = 0 −4.4 Fh + 1.4 F12 y − 0.05F12 x + 2.2 F32 y − 0.08 F32 x = 0 F23 x + F43 x = 0
3
F23 y + F43 y = 0 −0.6 F23 y − 0.13F23 x + 0.6 F43 y + 0.13F43 x = 0 F14 x + F34 x = 1 956.3 F14 y + F34 y = −415.82
( g)
(
)
−0.16 F14 y + 0.76 F14 x + 0.16 F34 y − 0.76 F34 x = −0.45 ( 415.82 ) − 0.34 1 956.3 = −852.26 F32 x + F23 x = 0 F34 x + F43 x = 0 F32 y + F23 y = 0 F34 y + F43 y = 0
9 Form the matrices for solution:
0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 1 0 0 1 0 1 0 0 0 0 0 0 0 0 −0.05 1.4 −0.08 2.2 −4.4 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 −0.13 −0.6 0.13 0.6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0.76 −0.16 −0.76 0.16 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1
F12 x F12 y 0 F32 x 0 0 F32 y 0 Fh 0 F23 x 0 F23 y = 1 956.30 (h) F43 x −415.82 −852.26 F43 y 0 F14 x 0 0 F14 y 0 F34 x F34 y
10 Table 33 part 2 shows the solution to this problem for the data given in Table 33 part 1, assuming a 2000lb (8896N) force applied at the crimp, normal to the crimp surface. Program Matrix (on the website) was used. The force generated in link 3 is 1547 lb (6888 N), the reaction force against link 1 by link 2 (F21) is 1561 lb (6 943 N) at 166°, the reaction force against link 1 by link 4 (F41) is 451 lb (2008 N) at 169°, and a –233.5 lbin (–26.6Nm) moment must be applied to the handles to generate the specified crimp force. This moment can be obtained with a 53.1lb (236N) force applied at midhandle. This force is within the physiological gripforce capacity of the average human.
116
MACHINE DESIGN
C A S E
Table 3–3—part 2 Case Study 2A Calculated Data Variable
3
Fh

An Integrated Approach
S T U D Y

Sixth Edition
3 A
Automobile ScissorsJack Loading Analysis
Value
Unit
53.1
lb
F12 x
1513.6
lb
F12 y
–381.0
lb
F32 x
–1513.6
lb
F32 y
327.9
lb
F43 x
–1513.6
lb
F43 y
327.9
lb
F23 x
1513.6
lb
F23 y
–327.9
lb
F34 x
1513.6
lb
F34 y
–327.9
lb
F14 x
442.7
lb
F14 y
–87.9
lb
F21 x
–1513.6
lb
F21 y
381.0
lb
F41 x
–442.7
lb
F41 y
87.9
lb
Problem
Determine the forces on the elements of the scissors jack in the position shown in Figure 330.
Given
The geometry is known and the jack supports a force of P = 1000 lb (4448 N) in the position shown.
Assumptions
The accelerations are negligible. The jack is on level ground. The angle of the elevated car chassis does not impart an overturning moment to the jack. All forces are coplanar and twodimensional. A Class 1 load model is appropriate and a static analysis is acceptable.
Solution
See Figures 330 through 333 and Table 34, parts 1 and 2.
1 Figure 330 shows a schematic of a simple scissors jack used to raise a car. It consists of six links, which are pivoted and/or geared together and a seventh link in the form of a lead screw, which is turned to raise the jack. While this is clearly a threedimensional device, it can be analyzed as a twodimensional one if we assume that the applied load (from the car) and the jack are exactly vertical (in the z direction). If so, all forces will be in the xy plane. This assumption is valid if the car is jacked from a level surface. If not, then there will be some forces in the yz and xz planes as well. The jack designer needs to consider the more general case, but for our simple example we will initially assume twodimensional loading. For the overall assembly as shown in Figure 330, we can solve for the reaction force Fg, given force P, by summing forces: Fg = –P. 2 Figure 331 shows a set of freebody diagrams for the entire jack. Each element or subassembly of interest has been separated from the others and the forces and moments shown acting on it (except for its weight, which is small compared to the applied forces and is thus neglected for this analysis). The forces and moments can be either internal reactions at interconnections with other elements or external loads from the “outside world.” The centers of gravity of the respective elements are used as the origins of the P dimensions in inches
6
3
ty p 2
4
1
y
7
2
x FIGURE 330 An Automobile Scissors Jack
5
6 Fg
30 ° typ
Chapter 3
KINEMATICS AND LOAD DETERMINATION
117
y P R43 Rp
R32
x
F32
y
M42
x
F12
3
y
F43
R23 2
F34
3
F23
R12
R34
x M24
4
R14
F41
F21
F14
1
F17
y
R17
x
7
y
F76 R67
y
M75
M57
x
F16
F61
F71
F67
F56 F65
x
6
R65
R56
R76 FIGURE 331
5
Fg
Rg
FreeBody Diagrams of the Complete Scissors Jack
local, nonrotating coordinate systems in which the points of application of all forces on the elements are located. In this design, stability is achieved by the mating of two pairs of crude (noninvolute) gear segments acting between links 2 and 4 and between links 5 and 7. These interactions are modeled as forces acting along a common normal shared by the two teeth. This common normal is perpendicular to the common tangent at the contact point. There are 3 secondlaw equations available for each of the seven elements, allowing 21 unknowns. An additional 10 thirdlaw equations will be needed for a total of 31. This is a cumbersome system to solve for such a simple device, but we can use its symmetry to advantage in order to simplify the problem. 3 Figure 332 shows the upper half of the jack assembly. Because of the mirror symmetry between the upper and lower portions, the lower half can be removed to simplify
R16
118
MACHINE DESIGN

An Integrated Approach

Sixth Edition
P 3
ty p 6"
3 A
y
x
FAy
2 1
4
B
FB y
FIGURE 332 FreeBody Diagram of the Symmetrical Upper Half of an Automobile Scissors Jack
the analysis. The forces calculated for this half will be duplicated in the other. If we wished, we could solve for the reaction forces at A and B using equations 3.3b from this freebody diagram of the halfjack assembly. 4 Figure 333a shows the freebody diagrams for the upper half of the jack assembly, which are essentially the same as those of Figure 331. We now have four elements but can consider the subassembly labeled 1 to be the “ground,” leaving three elements on which to apply equations 3.3. Note that all unknown forces and moments are initially assumed positive in the equations.
*
Note the similarity to equations (b) in Case Study 2A. Only the subscript for the reaction moment is different because a different link is providing it. The consistent notation of this force analysis method makes it easy to write the equations for any system.
5 Link 2 has three forces acting on it: F42 is the unknown force at the gear tooth contact with link 4; F12 and F32 are the unknown reaction forces from links 1 and 3 respectively. Force F12 is provided by part 1 on part 2 at the pivot pin, and force F32 is provided by part 3 acting on part 2 at their pivot pin. The magnitudes and directions of these pin forces and the magnitude of F42 are unknown. The direction of F42 is along the common normal shown in Figure 333b. Write equations 3.19b for this element to sum forces in the x and y directions and sum moments about the CG (with the cross products expanded):*
∑ Fx = F12 x + F32 x + F42 x = 0 ∑ Fy = F12 y + F32 y + F42 y = 0 ∑ M z = R12 x F12 y − R12 y F12 x + R32 x F32 y − R32 y F32 x + R42 x F42 y − R42 y F42 x = 0
(a)
6 Link 3 has three forces acting on it: the applied load P, F23, and F43. Only P is known. Writing equations 3.19b for this element gives
∑ Fx = F23x + F43x + Px = 0 ∑ Fy = F23 y + F43 y + Py = 0 ∑ M z = R23x F23 y − R23 y F23x + R43x F43 y − R43 y F43x + RPx Py − RPy Px = 0
(b)
Chapter 3
KINEMATICS AND LOAD DETERMINATION
P
common normal
y R43
Rp
R32
x
2
F12
3
F23
R12
FA x
F42
R23
F34
F24
4
F41
F21
( b) Gear tooth detail
x R24
1
4
B
R14
F14
FB x FB y
FAy (a) Freebody diagrams
FIGURE 333 FreeBody Diagrams of Elements of the Half Scissors Jack
7 Link 4 has three forces acting on it: F24 is the unknown force from link 2; F14 and F34 are the unknown reaction forces from links 1 and 3, respectively:
∑ Fx = F14 x + F24 x + F34 x = 0 ∑ Fy = F14 y + F24 y + F34 y = 0 ∑ M z = R14 x F14 y − R14 y F14 x + R24 x F24 y − R24 y F24 x + R34 x F34 y − R34 y F34 x = 0
(c)
8 The nine equations in sets a through c have 16 unknowns in them, F12x, F12y, F32x, F32y, F23x, F23y, F43x, F43y, F14x, F14y, F34x, F34y, F24x, F24y, F42x, and F42y . We can write the thirdlaw relationships between actionreaction pairs at each of the joints to obtain six of the seven additional equations needed: F32 x = − F23 x
F32 y = − F23 y
F34 x = − F43 x
F34 y = − F43 y
F42 x = − F24 x
F42 y = − F24 y
(d )
9 The last equation needed comes from the relationship between the x and y components of the force F24 (or F42) at the tooth/tooth contact point. Such a contact (or half) joint can transmit force (excepting friction force) only along the common normal[4] that is perpendicular to the joint’s common tangent as shown in Figure 333b. The common normal is also called the axis of transmission. The tangent of the angle of this common normal relates the two components of the force at the joint: F24 y = F24 x tan θ
θ
2
y
F43
R42 A
x
R34
x
F32
y
119
(e)
3
120
MACHINE DESIGN

An Integrated Approach

Sixth Edition
10 Equations (a) through (e) comprise a set of 16 simultaneous equations that can be solved either by matrix reduction or by iterative rootfinding methods. Putting them in standard form for a matrix solution gives:
3
Table 3–4—part 1 Case Study 3A Given Data Variable
Value
Unit
Px
0.00
lb
Py
–1000.00
lb
–0.50
in
R px R py
0.87
in
θ R 12 x
–45.00 –3.12
in
R 12 y
–1.80
in
R 32 x
2.08
in
R 32 y
1.20
in
R 42 x
2.71
in
R 42 y
1.00
in
R 23 x
–0.78
in
R 23 y
–0.78
in
R 43 x
0.78
in
R 43 y
–0.78
in
R 14 x
3.12
in
R 14 y
–1.80
in
R 24 x
–2.58
in
R 24 y
1.04
in
R 34 x
–2.08
in
R 34 y
1.20
in
deg
F12 x + F32 x + F42 x = 0 F12 y + F32 y + F42 y = 0 R12 x F12 y − R12 y F12 x + R32 x F32 y − R32 y F32 x + R42 x F42 y − R42 y F42 x = 0 F23 x + F43 x = − Px F23 y + F43 y = − Py R23 x F23 y − R23 y F23 x + R43 x F43 y − R43 y F43 x = − RPx Py + RPy Px F14 x + F24 x + F34 x = 0 F14 y + F24 y + F34 y = 0 R14 x F14 y − R14 y F14 x + R24 x F24 y − R24 y F24 x + R34 x F34 y − R34 y F34 x = 0 (f) F32 x + F23 x = 0 F32 y + F23 y = 0 F34 x + F43 x = 0 F34 y + F43 y = 0 F42 x + F24 x = 0 F42 y + F24 y = 0 F24 y − F24 x tan θ = 0
11 Substituting the given data from Table 34 part 1 gives: F12 x + F32 x + F42 x F12 y + F32 y + F42 y −3.12 F12 y + 1.80 F12 x + 2.08 F32 y − 1.20 F32 x + 2.71F42 y − 0.99 F42 x F23 x + F43 x F23 y + F43 y −0.78 F23 y + 0.78 F23 x + 0.78 F43 y + 0.78 F43 x F14 x + F24 x + F34 x F14 y + F24 y + F34 y 3.12 F14 y + 1.80 F14 x − 2.58 F24 y − 1.04 F24 x − 2.08 F34 y − 1.20 F34 x F32 x + F23 x F32 y + F23 y F34 x + F43 x F34 y + F43 y F42 x + F24 x F42 y + F24 y F24 y + 1.0 F24 x
=0 =0 =0 = 0.0 = 1 000 = − 500 =0 =0 =0 =0 =0 =0 =0 =0 =0 =0
( g)
Chapter 3
KINEMATICS AND LOAD DETERMINATION
121
12 Put these equations into matrix form.
1
0
1
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
1
0
1
0
0
0
0
0
0
0
0
0
0
1.80
3.12
1.20
2.08
1.00
2.71
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0`
1
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0.78
0.78
0.78
0.78
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
1
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
1
0
1
0
0
0
0
0
0
0
0
0
0
1.80
3.12
1.04
2.58
1.20
2.08
0
0
1
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
1
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
1
0
0
F12 x F12 y F32 x F32 y F42 x F42 y F23 x F23 y F43 x F43 y F14 x F14 y F24 x F24 y F34 x F34 y
0 0 0 0 1000 500 0 0 0 0 0 0 0 0 0 0
(h)
3
13 Table 34 part 2 shows the solution to this problem from program Matrix for the given data in Table 34 part 1, which assumes a vertical 1000lb (4448N) applied force P.
Table 3–4—part 2
14 The forces on link 1 can also be found from Newton’s third law.
Case Study 3A Calculated Data
FAx = − F21x = F12 x FAy = − F21y = F12 y FBx = − F41x = F14 x FBy = − F41y = F14 y
3.13
(i )
THREEDIMENSIONAL, STATIC LOADING CASE STUDY
This section presents a case study involving threedimensional static loading on a bicycle brake caliper assembly. The same techniques used for twodimensional load analysis also work for the threedimensional case. The third dimension requires more equations, which are available from the summation of forces in the z direction and the summation of moments about the x and y axes as defined in equations 3.17 and 3.19 for the dynamic and static cases, respectively. As an example, we will now analyze the bicycle brake arm that is actuated by the handbrake lever that was analyzed in Case Study 1A.
Variable
Value
Unit
F12x
877.8
lb
F12y
530.4
lb
F32x
–587.7
lb
F32y
–820.5
lb
F42x
–290.1
lb
F42y
290.1
lb
F23x
587.7
lb
F23y
820.5
lb
F43x
–587.7
lb
F43y
179.5
lb
F14x
–877.8
lb
F14y
469.6
lb
F24x
290.1
lb
F24y
–290.1
lb
F34x
587.7
lb
F34y
–179.5
lb
122
MACHINE DESIGN
C A S E

An Integrated Approach
S T U D Y

Sixth Edition
4 A
Bicycle Brake Arm Loading Analysis Problem
Determine the forces acting in three dimensions on the bicycle brake arm in its actuated position as shown in Figure 334. This brake arm has been failing in service and may need to be redesigned.
Given
The existing brake arm geometry is known and the arm is acted on by a cable force of 1046 N in the position shown. (See Case Study 1A.)
Assumptions
The accelerations are negligible. A Class 1 load model is appropriate and a static analysis is acceptable. The coefficient of friction between the brake pad and wheel rim has been measured and is 0.45 at room temperature and 0.40 at 150°F.
Solution
See Figures 334 and 335, and Table 35.
3
1 Figure 334 shows a centerpull brake arm assembly commonly used on bicycles. It consists of six elements or subassemblies, the frame and its pivot pins (1), the two brake arms (2 and 4), the cable spreader assembly (3), the brake pads (5), and the wheel rim (6). This is clearly a threedimensional device and must be analyzed as such. 2 The cable is the same one that is attached to the brake lever in Figure 326. The 267N (60lb) hand force is multiplied by the mechanical advantage of the hand lever and transmitted via this cable to the pair of brake arms as was calculated in Case Study 1A. We will assume no loss of force in the cable guides, thus the full 1046N (235lb) cable force is available at this end. 3 The direction of the normal force between the brake pad and the wheel rim is shown in Figure 334 to be at q = 172° with respect to the positive x axis, and the friction force is directed along the z axis. (See Figures 334 and 335 for the xyz axis orientations.) 4 Figure 335 shows freebody diagrams of the arm, frame, and cable spreader assembly. We are principally interested in the forces acting on the brake arm. However, we first need to analyze the effect of the cable spreader geometry on the force applied to the arm at A. This analysis can be two dimensional if we ignore the small zoffset between the two arms for simplicity. A more accurate analysis would require that the zdirected components of the cablespreader forces acting on the arms be included. Note that the cable subassembly (3) is a concurrent force system. Writing equations 3.19b in two dimensions for this subassembly, and noting the symmetry about point A, we can write from inspection of the FBD:
∑ Fx = F23x + F43x = 0 ∑ Fy = F23 y + F43 y + Fcable = 0
(a)
This equation set can be easily solved to yield Fcable 1046 =− = −523 N 2 2 F23 y −523 = = = −353 N tan ( 56° ) 1.483 = − F23 x = 353 N
F23 y = F43 y = − F23 x F43 x
(b)
Chapter 3
KINEMATICS AND LOAD DETERMINATION
123
Fcabl e y
y
cable
cable
3
3
3 x
z
brake arm
brake arm
2 4
4
2 frame
pad
1
pad wheel rim 6
5
θ 172 ° x N
5 wheel rim
FIGURE 334
6
V
ω
CenterPull Bicycle Brake Arm Assembly
Newton’s third law relates these forces to their reactions on the brake arm at point A: F32 x = − F23 x = 353 N F32 y = − F23 y = 523 N
(c)
F32 z = 0
5 We can now write equations 3.19a for the arm (link 2). For the forces:
∑ Fx = F12 x + F32 x + F52 x = 0; ∑ Fy = F12 y + F32 y + F52 y = 0; ∑ Fz = F12z + F32z + F52z = 0;
F12 x + F52 x = −353 F12 y + F52 y = −523
(d )
F12 z + F52 z = 0
For the moments:
∑ M x = M12 x + ( R12 y F12z − R12z F12 y ) + ( R32 y F32z − R32z F32 y ) + ( R52 y F52 z − R52 z F52 y ) = 0 ∑ M y = M12 y + ( R12z F12 x − R12 x F12z ) + ( R32z F32 x − R32 x F32z ) (
)
+ R52 z F52 x − R52 x F52 z = 0
∑ M z = ( R12 x F12 y − R12 y F12 x ) + ( R32 x F32 y − R32 y F32 x ) + ( R52 x F52 y − R52 y F52 x ) = 0
(e)
124
MACHINE DESIGN

An Integrated Approach

Fcabl e
frame
Sixth Edition
Fcabl e
1 cable
3
F21 z
F21 x
A
3
3
M21 y M12 y
F23 x
F32 y A
cable
F32 x
2
56 °
56 °
F23 y
F43 y
F43 x
R32
R32
2
x
R12 F12 y
brake arm B R52 F52 x
C
frame
z F12 z
F43 y
y
F32 z
y
F23 y
F32 y
F21 y
R12 R52
F12 x
M12 x
M21 x
F52 z
F52 y
F21 z 1
F52 y
FIGURE 335 Brake Arm FreeBody Diagrams
Note that all unknown forces and moments are initially assumed positive in the equations, regardless of their apparent directions on the FBDs. The moments M12x and M12y are due to the fact that there is a moment joint between the arm (2) and the pivot pin (1) about the x and y axes. We assume negligible friction about the z axis, thus allowing M12z to be zero. 6 The joint between the brake pad (5) and the wheel rim (6) transmits a force normal to the plane of contact. The friction force magnitude, Ff, in the contact plane is related to the normal force by the Coulomb friction equation, Ff = µN
(f)
where m is the coefficient of friction and N is the normal force. The velocity of the point on the rim below the center of the brake pad is in the z direction. The force components F52x and F52y are due entirely to the normal force being transmitted through the pad to the arm and are therefore related by Newton’s third law. F52 x = − N x = − N cos θ = − N cos172° = 0.990 N F52 y = − N y = − N sin θ = − N sin172° = −0.139 N
( g)
Chapter 3
KINEMATICS AND LOAD DETERMINATION
125
The direction of the friction force Ff must always oppose motion and thus it acts in the negative z direction on the wheel rim. Its reaction force on the arm has the opposite sense. F52 z = − Ff
(h)
7 We now have 10 equations (in the sets labeled d, e, f, g, and h) containing 10 unknowns: F12x, F12y, F12z, F52x, F52y, F52z, M12x, M12y, N, and Ff. Forces F32x, F32y, and F32z are known from equations c. These 10 equations can be solved simultaneously either by matrix reduction or by iterative rootfinding methods. First arrange the equations with all unknowns on the left and all known or assumed values on the right.
3
F12 x + F52 x = −353 F12 y + F52 y = −523 M12 x + R12 y F12 z − R12 z F12 y
Table 3–5—part 1
F12 z + F52 z = 0 + R52 y F52 z − R52 z F52 y = R32 z F32 y − R32 y F32 z
M12 y + R12 z F12 x − R12 x F12 z + R52 z F52 x − R52 x F52 z = R32 x F32 z − R32 z F32 x
(i )
R12 x F12 y − R12 y F12 x + R52 x F52 y − R52 y F52 x = R32 y F32 x − R32 x F32 y Ff − µN = 0 F52 x + N cos θ = 0 F52 y + N sin θ = 0 F52 z + Ff = 0
8 Substitute the known and assumed values from Table 35 part 1: F12 x + F52 x = −353 F12 y + F52 y = −523 F12 z + F52 z = 0 M12 x − 27.2 F12 z − 23.1F12 y − 69.7 F52 z − 0 F52 y = 0 ( 523) − 38.7 ( 0 ) = 0 M12 y + 23.1F12 x − 5.2 F12 z + 0 F52 x + 13F52 z = −75.4 ( 0 ) − 0 ( 353) = 0 5.2 F12 y + 27.2 F12 x − 13F52 y + 69.7 F52 x = 38.7 ( 353) + 75.4 ( 523) = 53 095 Ff − 0.4 N = 0 F52 x − 0.990 N = 0 F52 y + 0.139 N = 0 F52 z + Ff = 0
9 Form the matrices for solution.
( j)
Case Study 4A Given and Assumed Data Variable
Value
Unit
µ
0.4
none
θ
172.0
deg
R12x
5.2
mm
R12y
–27.2
mm
R12z
23.1
mm
R32x
–75.4
mm
R32y
38.7
mm
R32z
0.0
mm
R52x
–13.0
mm
R52y
–69.7
mm
R52z F32x
0.0
mm
353.0
N
F32y
523.0
N
F32z
0.0
N
M12z
0.0
N–m
126
MACHINE DESIGN
Table 3–5—part 2 Case Study 4A Calculated Data
3
Variable
Value
Unit
F12 x
–1805
N
F12 y
–319
N
F12 z
587
N
F52 x
1452
N
F52 y
–204
N
–587
N
F52 z M12 x
32 304
N–mm
M12 y
52 370
N–mm
N
1467
N
Ff
587
N

An Integrated Approach
1 0 0 0 1 0 0 1 0 0 0 1 0 1 1 0 0 0 0 −23.1 −27.2 0 0 −69.7 23.1 −5.2 13 0 0 0 0 69.7 −13 0 27.2 5.2 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 1 0 0
0 0 0 1 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 0 1 0 0 1

Sixth Edition
0 0 0 0 0 × 0 −0.4 −0.990 0.139 0
F12 x F12 y −353 −523 F12 z 0 F52 x 0 F52 y 0 = F52 z 53 095 0 M12 x 0 M12 y 0 Ff 0 N
( k )
10 Table 35 part 2 shows the solution to this problem from program Matrix for the given data in Table 35 part 1. This problem can be solved using any one of several commercial equationsolver programs such as Mathcad, MATLAB, Maple, Mathematica or with the program Matrix provided with the text.
3.14
DYNAMIC LOADING CASE STUDY
This section presents a case study involving twodimensional dynamic loading on a fourbar linkage designed as a dynamicload demonstration device. A photograph of this machine is shown in Figure 311 and a video is on the website (Fourbar Linkage). This machine can be analyzed in two dimensions since all elements move in parallel planes. The presence of significant accelerations on the moving elements in a system requires that a dynamic analysis be done with equations 3.1. The approach is identical to that used in the preceding static load analyses except for the need to include the mA and Ia terms in the equations.
C A S E † http://www.designofmachinery.com/MD/Fourbar_Machine. mp4
S T U D Y
5 A
Fourbar Linkage Loading Analysis View the Video
(35:38) †
Problem
Determine the theoretical rigidbody forces acting in two dimensions on the fourbar linkage shown in Figure 336.
Given
The linkage geometry, masses, and mass moments of inertia are known and the linkage is driven at up to 120 rpm by a speedcontrolled electric motor.
Assumptions
The accelerations are significant. A Class 4 load model is appropriate and a dynamic analysis is required. There are no external loads on the system; all loads are due to the accelerations of the links. The weight forces are insignificant compared to the inertial forces and will be neglected. The links are assumed to be ideal rigid bodies. Friction and the effects of clearances in the pin joints also will be ignored.
Chapter 3
KINEMATICS AND LOAD DETERMINATION
127
coupler rocker
crank
3
flywheel
counter weights gear motor
FIGURE 336
Copyright © 2018 Robert L. Norton: All Rights Reserved
Fourbar Linkage Dynamic Model (Photo by the author)
Solution
See Figures 336 through 338 and Table 36.
1 Figures 336 and 337 show the fourbar linkage demonstrator model. It consists of three moving elements (links 2, 3, and 4) plus the frame or ground link (1). The motor drives link 2 through a gearbox. The two fixed pivots are instrumented with piezoelectric force transducers to measure the dynamic forces acting in x and y directions on the ground plane. A pair of accelerometers is mounted to a point on the floating coupler (link 3) to measure its accelerations. 2 Figure 337 shows a schematic of the linkage. The links are designed with lightening holes to reduce their masses and mass moments of inertia. The input to link 2 can be an angular acceleration, a constant angular velocity, or an applied torque. Link 2 rotates fully about its fixed pivot at O2. Even though link 2 may have a zero angular acceleration a2, if run at constant angular velocity w2, there will still be timevarying angular accelerations on links 3 and 4 since they oscillate back and forth. In any case, the CGs of the links will experience timevarying linear accelerations as the linkage moves. These angular and linear accelerations will generate inertia forces and torques
3
O 2 O 4 = 18 in (457.2 mm) O 2 A = 6 in (152.4 mm) AB = 16 in (406.4 mm)
B
O4 2 B = 12 in (304.8 mm)
m 4 , IG 4
m 3 , IG 3
m 2 , IG 2 2 A
ω2, α2, T 2
4
O2 1
O4 1
FIGURE 337 Fourbar Linkage Schematic and Basic Dimensions (See Table 36 for more information)
128
MACHINE DESIGN
Case Study 5A Given and Assumed Data
3
Value
An Integrated Approach

Sixth Edition
as defined by Newton’s second law. Thus, even with no external forces or torques applied to the links, the inertial forces will create reaction forces at the pins. It is these forces that we wish to calculate.
Table 3–6—part 1
Variable

3 Figure 338 shows the freebody diagrams of the individual links. The local, nonrotating, coordinate system for each link is set up at its CG. The kinematic equations of motion must be solved to determine the linear accelerations of the CG of each link and the link’s angular acceleration for every position of interest during the cycle. (See equations 3.7) These accelerations, AGn and an, are shown acting on each of the n links. The forces at each pin connection are shown as xy pairs, numbered as before, and are initially assumed to be positive.
Unit
θ2
30.00
deg
ω2
120.00
rpm
mass2
0.525
kg
mass3
1.050
kg
mass4
1.050
kg
Icg2
0.057
kgm2
Icg3
0.011
kgm2
Icg4
0.455
kgm2
R12x
–46.9
R12y
–71.3
mm
R32x
85.1
mm
R32y
4.9
mm
R23x
–150.7
mm
R23y
–177.6
mm
R43x
185.5
mm
R43y
50.8
mm
R14x
–21.5
mm
R14y
–100.6
mm
R34x
–10.6
mm
R34y
204.0
mm
4 Equations 3.1 can be written for each moving link in the system. The masses and the mass moments of inertia of each link about its CG must be calculated for use in these equations. In this case study, a solid modeling CAD system was used to design the links’ geometries and to calculate their mass properties. 5 For link 2:
mm
∑ Fx = F12 x + F32 x = m2 aG 2 ∑ Fy = F12 y + F32 y = m2 aG 2 ∑ M z = T2 + ( R12 x F12 y − R12 y F12 x ) + ( R32 x F32 y − R32 y F32 x ) = IG 2α 2 x
(a)
y
6 For link 3:
∑ Fx = F23x + F43x = m3aG3 ∑ Fy = F23 y + F43 y = m3aG3 ∑ M z = ( R23x F23 y − R23 y F23x ) + ( R43x F43 y − R43 y F43x ) = IG 3α3 x
(b)
y
y α3 AG 3 F43
R43
y
R32
x
F32
F34
ω2, α2, T 2
x AG 2 F12
4 3
R12
2
α4
R34
F23
R14 AG 4
F21
FIGURE 338
y
R23
1
FreeBody Diagrams of Elements in a Fourbar Linkage
F41 1
F14
x
Chapter 3
KINEMATICS AND LOAD DETERMINATION
7 For link 4:
Table 3–6—part 2
∑ ∑ Fy = F14 y + F34 y = m4 aG 4 ∑ M z = ( R14 x F14 y − R14 y F14 x ) + ( R34 x F34 y − R34 y F34 x ) = IG 4 α 4
Case Study 5A Calculated Data
Fx = F14 x + F34 x = m4 aG 4 x
(c)
y
8 There are 13 unknowns in these nine equations: F12x, F12y, F32x, F32y, F23x, F23y, F43x, F43y, F14x, F14y, F34x, F34y, and T2. Four thirdlaw equations can be written to equate the actionreaction pairs at the joints. F32 x = − F23 x F32 y = − F23 y F34 x = − F43 x F34 y = − F43 y
(d )
9 The set of thirteen equations in a through d can be solved simultaneously to determine the forces and driving torque either by matrix reduction or by iterative rootfinding methods. This case study was solved by both techniques and files for both are on the website. Note that the masses and mass moments of inertia of the links are constant with time and position, but the accelerations are timevarying. Thus, a complete analysis requires that equations a–d be solved for all positions or time steps of interest. The models use lists or arrays to store the calculated values from equations a–d for 13 values of the input angle q2 of the driving link ( 0 to 360° by 30° increments). The model also calculates the kinematic accelerations of the links and their CGs, which are needed for the force calculations. The largest and smallest forces present on each link during the cycle can then be determined for use in later stress and deflection analyses. The given data and results of this force analysis for one crank position (q2 = 30°) are shown in Table 36, parts 1 and 2. Plots of the forces at the fixed pivots for one complete revolution of the crank are shown in Figure 339.
x and y forces on joint 12
0
force N
force N
200
y
–200
Variable
Value
Unit
F12x
–255.8
N
F12y
–178.1
N
F32x
252.0
N
F32y
172.2
N
F34x
–215.6
N
F34y
–163.9
N
F14x
201.0
N
F14y
167.0
N
F43x
215.6
N
F43y
163.9
N
F23x
–252.0
N
F23y
–172.2
N
T12
–3.55
α3
56.7
rad/sec2
α4
138.0
rad/sec2
Acg2x
–7.4
rad/sec2
Acg2y
–11.3
rad/sec2
Acg3x
–34.6
rad/sec2
Acg3y
–7.9
rad/sec2
Acg4x
–13.9
rad/sec2
Acg4y
2.9
rad/sec2
Nm
x and y forces on joint 14
100
–100
129
x
100 y 0
x –100
–300 0
60
120
180
240
300
360
crank angle  degrees FIGURE 339 Calculated RigidBody Dynamic Forces in the Fourbar Linkage of Case Study 5A
0
60
120
180
240
crank angle  degrees
300
360
3
130
MACHINE DESIGN
3.15
3

An Integrated Approach

Sixth Edition
VIBRATION LOADING
In systems that are dynamically loaded, there will usually be vibration loads superimposed on the theoretical loads predicted by the dynamic equations. These vibration loads can be due to a variety of causes. If the elements in the system were infinitely stiff, then vibrations would be eliminated. But all real elements, of any material, have elasticity and thus act as springs when subjected to forces. The resulting deflections can cause additional forces to be generated from the inertial forces associated with the vibratory movements of elements or, if clearances allow contact of mating parts, to generate impact (shock) loads (see below) during their vibrations. A complete discussion of vibration phenomena is beyond the scope of this text and will not be attempted here. References are provided in the bibliography at the end of this chapter for further study. The topic is introduced here mainly to alert the machine designer to the need to consider vibration as a source of loading. Often the only way to get an accurate measure of the effects of vibration on a system is to do testing of prototypes or production systems under service conditions. The discussion of safety factors in Section 1.7 mentioned that many industries (automotive, aircraft, etc.) engage in extensive test programs to develop realistic loading models of their equipment. This topic will be discussed further in Section 6.4 when fatigue loading is introduced. Modern finite element (FEA) and boundary element (BEA) analysis techniques also allow vibration effects on a system or structure to be modeled and calculated. It is still difficult to obtain a computer model of a complex system that is as accurate as a real, instrumented prototype. This is especially true when clearances (gaps) between moving parts allow impacts to occur in the joints when loads reverse. Impacts create nonlinearities that are very difficult to model mathematically. Natural Frequency When designing machinery, it is desirable to determine the natural frequencies of the assembly or subassemblies in order to predict and avoid resonance problems in operation. Any real system can have an infinite number of natural frequencies at which it will readily vibrate. The number of natural frequencies that are necessary or desirable to calculate will vary with the situation. The most complete approach to the task is to use Finite Element Analysis (FEA) to break the assembly into a large number of discrete elements. See Chapter 8 for more information on FEA. The stresses, deflections, and number of natural frequencies that can be calculated by this technique are mainly limited by time and the computer resources available. If not using FEA, we would like to determine, at a minimum, the system’s lowest, or fundamental, natural frequency, since this frequency will usually create the largest magnitude of vibrations. The undamped fundamental natural frequency wn, with units of rad/sec, or fn, with units of Hz, can be computed from the expressions: k m 1 ωn fn = 2π
ωn =
(3.20)
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131
where wn is the fundamental natural frequency, m is the moving mass of the system in true mass units (e.g., kg, g, blob, or slug, not lbm), and k is the effective spring constant of the system. (The period of the natural frequency is its reciprocal in seconds, Tn = 1 / fn.) Equation 3.20 is based on a singledegreeoffreedom, lumped model of the system. Figure 340 shows such a model of a simple camfollower system consisting of a cam, a sliding follower, and a return spring. The simplest lumped model consists of a mass connected to ground through a single spring and a single damper. All the moving mass in the system (follower, spring) is contained in m and all the “spring” including the physical spring and the springiness of all other parts is lumped in the effective spring constant k.
3
spriNg coNstaNt A spring constant k is an assumed linear relationship between the force, F, applied to an element and its resulting deflection d (see Figure 342): k=
F δ
(3.21a)
If an expression for the deflection of an element can be found or derived, it will provide this springconstant relationship. This topic is revisited in the next chapter. In the example of Figure 340, the spring deflection d is equal to the displacement y of the mass. k=
F y
(3.21b)
dampiNg All the damping, or frictional, losses are lumped in the damping coefficient d. For this simple model, damping is assumed to be inversely proportional to the velocity ydot of the mass. d=
F y
(3.22)
followe r spring
Fspring Fdamper spring k
roller
damper d
. ..
y,y,y
cam
m
m
. ..
mass
y,y,y
ω Fcam
( a) Actual system
FIGURE 340 Lumped Model of a CamFollower Dynamic System
( b ) Lumped model
Fcam
( c ) Freebody diagram
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Equation 3.20 simplifies this model even further by assuming the damping d to be zero. If damping is included, the expressions for the fundamental, damped natural frequency wd with units of radians/sec, or fd, with units of Hz, become k d 2 − m 2m 1 fd = ωd 2π
ωd =
3
(3.23)
This damped frequency wd will be slightly lower than the undamped frequency wn. eFFective vaLues Determining the effective mass for a lumped model is straightforward and requires only summing all the values of the connected, moving masses in appropriate mass units. Determining the values of the effective spring constant and the effective damping coefficient is more complicated and will not be addressed here. See reference 2 for an explanation. resoNaNce A condition called resonance can be experienced if the operating or forcing frequency applied to the system is the same as any one of its natural frequencies. That is, if the input angular velocity applied to a rotating system is the same as, or close to, wn, the vibratory response will be very large. This can create large forces and cause failure. Thus, it is necessary to avoid operation at or near the natural frequencies if possible. Dynamic Forces If we write equation 3.17 for the simple, oneDOF model of the dynamic system in Figure 340 and substitute equations 3.21 and 3.22, we get
∑ Fy = ma = my
Fcam − Fspring − Fdamper = my Fcam
(3.24)
= my + dy + ky
If the kinematic parameters of displacement, velocity, and acceleration are known for the system, this equation can be solved directly for the force on the cam as a function of time. If the cam force is known and the kinematic parameters are desired, then the wellknown solution to this linear, constantcoefficient differential equation can be applied. See reference 3 for a detailed derivation of that solution. Though the coordinate system used for a dynamic analysis can be arbitrarily chosen, it is important to note that both the kinematic parameters (displacement, velocity, and acceleration) and the forces in equation 3.24 must be defined in the same coordinate system. As an example of the effect of vibration on the dynamic forces of a system, we now revisit the fourbar linkage of Case Study 5A and see the results of actual measurements of dynamic forces under operating conditions.
Chapter 3
C A S E
S T U D Y
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133
5 B
Fourbar Linkage Dynamic Loading Measurement Problem
Determine the actual forces acting on the fixed pivots of the fourbar linkage in Figure 336 during one revolution of the input crank.
Given
The linkage is driven at 60 rpm by a speedcontrolled electric motor, and force transducers are placed between the fixed pivot bearings and the ground plane.
Assumptions
There are no applied external loads on the system; all loads are due to the accelerations of the links. The weight forces are insignificant compared to the inertial forces and will be neglected. The force transducers measure only dynamic forces.
Solution
See Figures 336, 337, and 341.
3
1 Figure 336 shows the fourbar linkage.* It consists of three moving elements (links 2, 3, and 4) plus the frame or ground link (1). The shockmounted motor drives link 2 through a gearbox and shaft coupling. The two fixed pivots are instrumented with piezoelectric force transducers to measure the dynamic forces acting in x and y directions on the ground plane. 2 Figure 341 shows an actual force and torque measured while the linkage was running at 60 rpm and compares them to the theoretical force and torque predicted by equations a–d in Case Study 5A.[5] Only the x component of force at the pivot between link 2 and the ground and the torque on link 2 are shown as examples. The other pin forces and components show similar deviations from their predicted theoretical values. Some of these deviations are due to variations in instantaneous angular velocity of the drive motor; others are due to backlash in the gearbox. The theoretical analysis assumes constant input shaft velocity. Vibrations and impacts account for other deviations.
This example of deviations from theoretical forces in a very simple dynamic system is presented to point out that our best calculations of forces (and thus the resulting stresses) on a system may be in error due to factors not included in a simplified force analysis. It is common for the theoretical predictions of forces in a dynamic system to understate reality, which is, of course, a nonconservative result. Wherever feasible, the testing of physical prototypes will give the most accurate and realistic results. The effects of vibration on a system can cause significant loadings and are difficult to predict without test data of the sort shown in Figure 341, where the actual loads are seen to be double their predicted values; this will obviously double the stresses. A traditional, and somewhat crude, approach used in machine design has been to apply overload factors to the theoretical calculated loads based on experience with the same or similar equipment. As an example, see Table 1217 in the chapter on spurgear design. This table lists industryrecommended overload factors for gears subjected to various types of shock loading. These sorts of factors should be used only if one cannot develop more accurate test data of the type shown in Figure 341.
*
A video of this linkage is in the Demos folder on the website as Fourbar Linkage.
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(a) Theoretical and actual dynamic force in x direction at crank pivot
10
Sixth Edition
actual
5 0 lbf
3

–5 –10
theoretical
–15 –20 (b) Theoretical and actual dynamic torque at crank pivot
400 300
actual theoretical
lbf – in
200 100 0
–100 –200 –300
FIGURE 341
Copyright © 2018 Robert L. Norton: All Rights Reserved
Theoretical and Measured Dynamic Forces and Torques in a Fourbar Linkage
3.16
IMPACT LOADING
The loading considered so far has either been static or, if timevarying, has been assumed to be gradually and smoothly applied, with all mating parts continually in contact. Many machines have elements that are subjected to sudden loads or impacts. One example is the crankslider mechanism, which forms the heart of an automobile engine. The piston head is subjected to an explosive rise in pressure every two crank revolutions when the cylinder fires, and the clearance between the circumference of the piston and the cylinder wall can allow an impact of these surfaces as the load is reversed each cycle. A more extreme example is a jackhammer, whose purpose is to impact pavement and break it up. The loads that result from impact can be much greater than those that would result from the same elements contacting gradually. Imagine trying to drive a nail by gently placing the hammer head on the nail rather than by striking it. What distinguishes impact loading from static loading is the time duration of the application of the load. If the load is applied slowly, it is considered static; if applied rapidly, then it is impact. One criterion used to distinguish the two is to compare the time of load application tl (defined as the time it takes the load to rise from zero to its peak value) to the period of the natural frequency Tn of the system. If tl is less than half Tn, it is considered to be impact. If tl is greater than three times Tn, it is considered static. Between those limits is a gray area in which either condition can exist. Two general cases of impact loading are considered to exist, though we will see that one is just a limiting case of the other. Burr[6] calls these two cases striking impact
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135
and force impact. Striking impact refers to an actual collision of two bodies, such as in hammering or the taking up of clearance between mating parts. Force impact refers to a suddenly applied load with no velocity of collision, as in a weight suddenly being taken up by a support. This condition is common in friction clutches and brakes (see Chapter 17). These cases can occur independently or in combination. 3
Severe collisions between moving objects can result in permanent deformation of the colliding bodies as in an automobile accident. In such cases the permanent deformation is desirable in order to absorb the large amount of energy of the collision and protect the occupants from more severe harm. We are concerned here only with impacts that do not cause permanent deformation; that is, the stresses will remain in the elastic region. This is necessary to allow continued use of the component after impact. If the mass of the striking object m is large compared to that of the struck object mb and if the striking object can be considered rigid, then the kinetic energy possessed by the striking object can be equated to the energy stored elastically in the struck object at its maximum deflection. This energy approach gives an approximate value for the impact loading. It is not exact because it assumes that the stresses throughout the impacted member reach peak values at the same time. However, waves of stress are set up in the struck body, travel through it at the speed of sound, and reflect from the boundaries. Calculating the effects of these longitudinal waves on the stresses in elastic media gives exact results and is necessary when the ratio of mass of the striking object to that of the struck object is small. The wave method will not be discussed here. The reader is directed to reference 6 for further information. Energy Method The kinetic energy of the striking body will be converted to stored potential energy in the struck body, assuming that no energy is lost to heat. If we assume that all particles of the combined bodies come to rest at the same instant, then just before rebound, the force, stress, and deflection in the struck body will be maximal. The elastic energy stored in the struck body will be equal to the area under the forcedeflection curve defined by its particular spring constant. A generalized forcedeflection curve for a linear spring element is shown in Figure 342. The elastic energy stored is the area under the curve between zero and any combination of force and deflection. Because of the linear relationship, this is the area of a triangle, A = 1/2bh. Thus, the energy stored at the point of peak impact deflection, di, is E=
1 Fi δ i 2
(3.25a)
E=
Fi 2 2k
(3.25b)
F k Fi
Substituting equation 3.21 gives
horizoNtaL impact Figure 343a shows a mass about to impact the end of a horizontal rod. This device is sometimes called a slide hammer and is used to remove dents from automobile sheet metal among other uses. At the point of impact, the portion of the kinetic energy of the moving mass that is imparted to the struck mass is
δi FIGURE 342 Energy Stored in a Spring
δ
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δi
l
3
m k m k
vi (a) Horizontal
h
( b) Vertical
δi
FIGURE 343 Axial Impact on a Slender Rod
1 E = η mvi 2 2
(3.26)
where m is its mass and vi its velocity at impact. We need to modify the kinetic energy term by a correction factor η to account for the energy dissipation associated with the particular type of elastic member being struck. If the dissipation is negligible, η will be 1. Assuming that all the kinetic energy transferred from the moving mass is converted to elastic energy stored in the struck member allows us to equate equations 3.25 and 3.26: Fi 2 mv 2 =η i 2k 2 Fi = vi ηmk
(3.27)
If the mass were allowed to statically load the struck member, the resulting static deflection would be dst = W / k where W = mg. Substituting these into equation 3.27 gives a ratio of dynamic force to static force or dynamic deflection to static deflection: Fi δ η = i = vi W δ st gδ st
(3.28)
The term on the right side of this equation is called the impact factor, which provides a ratio of impact to static force or deflection. Thus, if the static deflection can be calculated for the application of a force equal to the weight of the mass, an estimate of the dynamic force and dynamic deflection can be obtained. Note that equations 3.27 and 3.28 are valid for any case of horizontal impact, whether the object is loaded axially as shown here, in bending, or in torsion. Methods for calculating the deflections of various cases are addressed in the next chapter. The spring rate k for any object can be found by rearranging its deflection equation according to equation 3.21.
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137
verticaL impact For the case of a mass falling through a distance h onto a rod as shown in Figure 343b, equation 3.27 also applies with the impact velocity vi2 = 2gh. The potential energy for a drop through distance h is: E= η
mvi 2 = ηmgh = Wηh 2
(3.29a)
3
If the deflection at impact is small compared to the drop distance h, this equation is sufficient. But, if the deflection is significant compared to h, the energy of impact needs to include an amount due to the deflection through which the weight falls beyond h. The total potential energy given up by the mass on impact is then
(
E = Wηh + W δ i = W ηh +δ i
)
(3.29b)
Equate this potential energy to the elastic energy stored in the struck member and substitute equation 325b and the expression W = kdst Fi 2 = W ηh +δ i 2k W Fi 2 = 2 kW ηh +δ i = 2 W ηh +δ i δ st
(
(
)
)
(
)
2 2 ηh δ i 2 ηh Fi F + = + 2 i = W δ st δ st δ st W 2 Fi Fi 2 ηh =0 − 2 − δ st W W
(3.30 a)
giving a quadratic equation in Fi / W whose solution is: Fi δ 2 ηh = i = 1+ 1+ W δ st δ st
(3.30 b)
The right side expression is the impact ratio for the falling weight case. Equation 330b can be used for any impact case involving a falling weight. For example, if the weight is dropped on a beam, the static deflection of the beam at the point of impact is used. If the distance h to which the mass is raised is set to zero, equation 3.30b becomes equal to 2. This says that if the mass is held in contact with the “struck” member (with the weight of the mass separately supported) and then allowed to suddenly impart its weight to that member, the dynamic force will be twice the weight. This is the case of “force impact” described earlier, in which there is no actual collision between the objects. A more accurate analysis, using wave methods, predicts that the dynamic force will be more than doubled even in this noncollision case of sudden application of load.[6] Many designers use 3 or 4 as a more conservative estimate of this dynamic factor for the case of sudden load application. This is only a crude estimate, however, and, if possible, experimental measurements or a wavemethod analysis should be made to determine more suitable dynamic factors for any particular design. Burr derives the correction factors η for several impact cases in reference 6. Roark and Young provide factors for additional cases in reference 7. For the case of a mass axially impacting a rod as shown in Figure 344, the correction factor is[6]
η 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0
5
1 0 15 20 m mass ratio mb
FIGURE 344 Correction Factor η as a Function of Mass Ratio
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η=

Sixth Edition
1 m 1+ b 3m
(3.31)
where m is the mass of the striking object, and mb is the mass of the struck object. As the ratio of the striking mass to the struck mass increases, the correction factor η asymptotically approaches one. Figure 344 shows η as a function of mass ratio for three cases: an axial rod (equation 3.31) plotted in color, a simply supported beam struck at midspan (blacksolid), and a cantilever beam struck at the free end (blackdotted).[6] The correction factor η is always less than one (and > 0.9 for mass ratios > 5) , so assuming it to be one is conservative. However, be aware that this energy method gives approximate and nonconservative results in general, and it needs to be used with largerthanusual safety factors applied.
3
force ratio 7000 6000
EXAMPLE 33
5000 4000
Impact Loading on an Axial Rod
3000
Problem
The axial rod shown in Figure 343a is hit by a mass moving at 1 m/sec. a. Determine the sensitivity of the impact force to the length/diameter ratio of the rod for a constant 1kg moving mass. b. Determine the sensitivity of the impact force to the ratio of moving mass to rod mass for a constant length/diameter ratio of 10.
2000 1000 0 0
10
20
Given
The round rod is 100 mm long. The rod and moving mass are steel with E = 207 Gpa and a mass density of 7.86 g/cm3.
Assumptions
An approximate energy method will be acceptable. The correction factor for energy dissipation will be applied.
Solution
See Figures 343a, 345, and 346.
l/d ratio force in Newtons 70 000
1 Figure 343a shows the system. The moving mass strikes the flange on the end of the rod with the stated velocity of 1 m/sec.
60 000 50 000
2 For part (a), we will keep the moving mass constant at 1 kg, and the rod length constant at 100 mm and vary the rod diameter to obtain l / d ratios in the range of 1 to 20. In the equations that follow we will show only one calculation made for the l / d ratio of 10. We can also compute all the values for a list of l / d ratios. The static deflection that would result from application of the weight force of the mass is calculated from the expression for the deflection of a bar in tension. (See equation 4.8 in the next chapter for the derivation.)
40 000 30 000 20 000 10 000 0 0
10
20
δ st =
Wl AE
=
l/d ratio
(
9.81 N 100 mm 78.54 mm
2
)
(2.07 E5 N mm 2 )
= 0.06 µm
(a)
The correction factor η is calculated here for an assumed mass ratio of 16.2, FIGURE 345 Dynamic Force and Force Ratio as a Function of l/d Ratio for the System in Example 33
η=
1 1 = = 0.98 mb 0.0617 1+ 1+ 3 (1) 3m
(b)
Chapter 3
KINEMATICS AND LOAD DETERMINATION
These values (calculated for each different rod diameter d) are then substituted into equation 3.12 to find the force ratio Fi / W and the dynamic force Fi. For d = 10 mm Fi η m = vi =1 W gδ st s
(
)
0.98 = 1 285.9 m 9.81 2 0.00006 m s
(
Fi = 1 285.9 9.81N = 12 612 N
)
(c)
The variation in force ratio with changes in l / d ratio for a constant amount of moving mass and a constant impact velocity (i.e., constant input energy) is shown in Figure 345. As the l / d ratio is reduced, the rod becomes much stiffer and generates much larger dynamic forces from the same impact energy. This clearly shows that impact forces can be reduced by increasing the compliance of the impacted system. 3 For part (b), we will keep the l / d ratio constant at 10 and vary the ratio between the moving mass and the rod mass over the range of 1 to 20. The files EX0301B on the website calculate all values for a range of mass ratios. The results for a mass ratio of 16.2 are the same as in part (a) above. Figure 346a shows that the dynamic force ratio Fi / W varies inversely with the mass ratio. However, the value of the dynamic force is increasing with mass ratio as shown in Figure 346b, because the static force W is also increasing with mass ratio.
139
force ratio 5000 4500 4000 3500 3000 2500 2000 1500 1000 500 0
3
0
10
20
mass ratio force in ne wtons 16 000 14 000 12 000
3.17
BEAM LOADING
A beam is any element that carries loads transverse to its long axis and may carry loads in the axial direction as well. A beam supported on pins or narrow supports at each end is said to be simply supported, as shown in Figure 347a. A beam fixed at one end and unsupported at the other is a cantilever beam (Figure 347b). A simply supported beam that overhangs its supports at either end is an overhung beam (Figure 347c). If a beam has more supports than are necessary to provide kinematic stability (i.e., make the kinematic degree of freedom zero), then the beam is said to be overconstrained or indeterminate, as shown in Figure 347d. An indeterminate beam problem cannot be solved for its loads using only equations 3.19. Other techniques are necessary. This problem is addressed in the next chapter. Beams are typically analyzed as static devices, though vibrations and accelerations can cause dynamic loading. A beam may carry loads in three dimensions, in which case equations 3.19a apply. For the twodimensional case, equations 3.19b suffice. The review examples used here are limited to 2D cases for brevity. Shear and Moment A beam may be loaded with some combination of distributed and/or concentrated forces or moments as in Figure 347. The applied forces will create both shearing forces and bending moments in the beam. A load analysis must find the magnitudes and spatial distributions of these shear forces and bending moments on the beam. The shear forces V and the moment M in a beam are related to the loading function q(x) by
10 000 8 000 6 000 4 000 2 000 0 0
10
20
mass ratio FIGURE 346 Dynamic Force and Force Ratio as a Function of Mass Ratio for the System in Example 33
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l
y a

Sixth Edition
l
y a
w 〈x–a 〉0
3
〈x–a 〉–1
M1
x R1
F
x
R2
R1
( a) Simply supported beam with uniformly distributed loading
(b) Cantilever beam with concentrated loading
y
y
l
l
w 〈x–a〉1
a
b a
Mz 〈x–0〉–2
w 〈x–a 〉0
x
x
R2
R1
R1
( c) Overhung beam with moment and linearly distributed loading
R2
R3
( d) Statically indeterminate beam with uniformly distributed loading
FIGURE 347 Types of Beams and Beam Loadings
q(x) =
dV d 2 M = dx dx 2
(3.32a)
The loading function q(x) is typically known and the shear V and moment M distributions can be found by integrating equation 3.16a: VB
∫V V
V
dV =
A
xB
∫x
A
q dx = VB − VA
(3.32b)
Equation 3.16b shows that the difference in the shear forces between any two points, A and B, is equal to the area under the graph of the loading function, equation 3.32a. Integrating the relationship between shear and moment gives
positive shear
M
M
positive moment FIGURE 348 Beam Sign Convention
MB
∫M
A
dM =
xB
∫x
A
V dx = M B − M A
(3.32c)
showing that the difference in the moment between any two points, A and B, is equal to the area under the graph of the shear function, equation 3.32b. sigN coNveNtioN The usual (and arbitrary) sign convention used for beams is to consider a moment positive if it causes the beam to deflect concave downward (as if to collect water). This puts the top surface in compression and the bottom surface in tension. The shear force is considered positive if it causes a clockwise rotation of the section on which it acts. These conventions are shown in Figure 348 and result in positive moments being
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KINEMATICS AND LOAD DETERMINATION
141
created by negative applied loads. All the applied loads shown in Figure 347 are negative. For example, in Figure 347a, the distributed load magnitude from a to l is q = –w. equatioN soLutioN The solution of equations 3.19 and 3.32 for any beam problem may be carried out by one of several approaches. Sequential and graphical solutions are described in many textbooks on statics and mechanics of materials. One classical approach to these problems is to find the reactions on the beam using equations 3.19 and then draw the shear and moment diagrams using a graphical integration approach combined with calculations for the significant values of the functions. This approach has value from a pedagogical standpoint as it is easily followed, but it is cumbersome to implement. The approach most amenable to computer solution uses a class of mathematical functions called singularity functions to represent the loads on the beam. We present the classical approach as a pedagogical reference and also introduce the use of singularity functions that offer some computational advantages. While this approach may be new to some students, when compared to the methods usually learned in other courses, it has significant advantages in computerizing the solution. Singularity Functions Because the loads on beams typically consist of collections of discrete entities, such as point loads or segments of distributed loads that can be discontinuous over the beam length, it is difficult to represent these discrete functions with equations that are valid over the entire continuum of beam length. A special class of functions called singularity functions was invented to deal with these mathematical situations. Singularity functions are often denoted by a binomial in angled brackets as shown in equations 3.33. The first quantity in the brackets is the variable of interest, in our case x, the distance along the beam length. The second quantity a is a userdefined parameter that denotes where in x the singularity function either acts or begins to act. For example, for a point load, the quantity a represents the particular value of x at which the load acts (see Figure 347b). The definition of this singularity function, called the unit impulse or Dirac delta function, is given in equation 3.33d. Note that all singularity functions involve a conditional constraint. The unit impulse evaluates to ∞ if x = a and is 0 at any other value of x. The unit step function or Heaviside step function (Eq. 3.33c) evaluates to 0 for all values of x less than a and to 1 for all other x. Since these functions are defined to evaluate to unity, multiplying them by a coefficient creates any magnitude desired. Their application is shown in the following three examples and is explained in the most detail in Example 34B. If a loading function both starts and stops within the range of x desired, it needs two singularity functions to describe it. The first defines the value of a1 at which the function begins to act and has a positive or negative coefficient as appropriate to its direction. The second defines the value a2 at which the function ceases to act and has a coefficient of the same magnitude but opposite sign as the first. These two functions will cancel beyond a2, making the load zero. Such a case is shown in Example 46 in the next chapter. Quadratically distributed loads can be represented by a unit parabolic function, x−a
2
which is defined as 0 when x ≤ a, and equal to (x – a)2 when x > a.
(3.33a)
3
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Linearly distributed loads can be represented by a unit ramp function, x−a
1
(3.33b)
which is defined as 0 when x ≤ a, and equal to (x – a) when x > a.
3
A uniformly distributed load over a portion of a beam can be represented mathematically by a unit step function, x−a
0
(3.33c)
which is defined as 0 when x a, and is undefined at x = a. A concentrated force can be represented by the unit impulse function, x−a
−1
(3.33d )
which is defined as 0 when x a. Its integral evaluates to unity at a. A concentrated moment can be represented by the unit doublet function, x−a
−2
(3.33e)
which is defined as 0 when x a. It generates a unit couple moment at a. This process can be extended to obtain polynomial singularity functions of any order n to fit distributed loads of any shape. Four of the five singularity functions described here are shown in Figure 347, as applied to various beam types. A computer program is needed to evaluate these functions. Table 37 shows five of the singularity functions implemented in a “Basiclike” pseudocode. The For loops run the variable x from zero to the beam length l. The If test determines whether x has reached the value of a, which is the location of the start of the singularity function. Depending on this test, the value of y(x) is set either to zero or to the specified magnitude of the singularity function. This type of code can easily be implemented in any computer language (e.g., C+, Fortran, basic) or in an equation solver (e.g., Mathcad, MATLAB, TK Solver, EES). The integrals of these singularity functions have special definitions that, in some cases, defy common sense but nevertheless provide the desired mathematical results. For example, the unit impulse function (Eq. 3.17d) is defined in the limit as having zero width and infinite magnitude, yet its area (integral) is defined as equal to one (Eq. 3.18d). (See reference 8 for a more complete discussion of singularity functions.) The integrals of the singularity functions in equations 3.17 are defined as dλ =
x−a 3
3
dλ =
x−a 2
2
dλ = x − a
1
x
2
x
1
x
0
∫−∞ λ − a ∫−∞ λ − a ∫−∞ λ − a
(3.34 a)
(3.34 b)
(3.34c)
Chapter 3
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143
Pseudocode to Evaluate Singularity Functions
Pulse Singularity Function For x = 0 to l * Note: This routine does not generate the infinite value of the Dirac delta function. Rather, it generates the magnitude of a point load applied at location a for use in plotting a beam’s loading function.
If ABS (x – a) < 0.0001 Then y(x) = magnitude, Else y(x) = 0* Next x Step Singularity Function For x = 0 to l If x < a Then y(x) = 0, Else y(x) = magnitude Next x Ramp Singularity Function For x = 0 to l If x < = a Then y(x) = 0, Else y(x) = magnitude * (x – a) Next x Parabolic Singularity Function For x = 0 to l If x < = a Then y(x) = 0, Else y(x) = magnitude * (x – a)^2 Next x Cubic Singularity Function For x = 0 to l If x < = a Then y(x) = 0, Else y(x) = magnitude * (x – a)^3 Next x
x
−1
dλ = x − a
0
(3.34 d )
x
−2
dλ = x − a
−1
(3.34e)
∫−∞ λ − a ∫−∞ λ − a
where λ is just an integration variable running from –∞ to x. These expressions can be used to evaluate the shear and moment functions that result from any loading function that is expressed as a combination of singularity functions.
EXAMPLE 34A Shear and Moment Diagrams of a Simply Supported Beam Using a Graphical Method Problem
Determine and plot the shear and moment functions for the simply supported beam with uniformly distributed load shown in Figure 347a.
Given
Beam length l = 10 in, and load location a = 4 in. The magnitude of the uniform force distribution is w = 10 lb/in.
Assumptions
The weight of the beam is negligible compared to the applied load and so can be ignored.
3
144
MACHINE DESIGN
y
Solution
l a
An Integrated Approach

Sixth Edition
See Figures 347a and 349.
1 Solve for the reaction forces using equations 3.19. Summing moments about the right end and summing forces in the y direction gives
〈x–a 〉 0 w x
3 R1

R2
∑ M z = 0 = R1l −
w ( l − a )2 2
w ( l − a )2 10 (10 − 4 )2 = = 18 R1 = 2l 2 (10 )
Simply supported beam with uniformly distributed loading
FIGURE 347a
∑ Fy = 0 = R1 − w (l − a ) + R2
Repeated
R2 = w ( l − a ) − R1 = 10 (10 − 4 ) − 18 = 42 (a) Loading Diagram 40 q R2 30 20 R 1 10 0 w –10 x –20 0 5 10 (b) Shear Diagram 20 V 10 0 –10 –20 –30 –40 –50 0
–w 5.8 5
x 10
(c) Moment Diagram 100 M 88. 2 80
0 0
5
FIGURE 349 Example 34 Plots
3 If your reflexes are quick enough, you should try to catch the shear diagram (Figure 349b) as you fall, climb onto it, and repeat this backwardwalking trick across it to create the moment diagram, which is the integral of the shear diagram. Note in Figure 349c that from x = 0 to x = a this moment function is a straight line with slope = R1. Beyond point a, the shear diagram is triangular, and so integrates to a parabola. The peak moment will occur where the shear diagram crosses zero (i.e., zero slope on the moment diagram). The value of x at V = 0 can be found with a little trigonometry, noting that the slope of the triangle is –w: R1 18 =4+ = 5.8 w 10
(c)
Positive shear area adds to the moment value and negative area subtracts. So the value of the peak moment can be found by adding the areas of the rectangular and triangular portions of the shear diagram from x = 0 to the point of zero shear at x = 5.8:
40 5.8
(b)
2 The shape of the shear diagram can be sketched by graphically integrating the loading diagram shown in Figure 349a. As a “device” to visualize this graphical integration process, imagine that you walk backward across the loading diagram of the beam, starting from the left end and taking small steps of length dx. You will record on the shear diagram (Figure 349b) the area (force · dx) of the loading diagram that you can see as you take each step. As you take the first step backward from x = 0, the shear diagram rises immediately to the value of R1. As you walk from x = 0 to x = a, no change occurs, since you see no additional forces. As you step beyond x = a, you begin to see strips of area equal to –w · dx, which subtract from the value of R1 on the shear diagram. When you reach x = l, the total area w · (l – a) will have taken the value of the shear diagram to –R2. As you step backward off the beam’s loading diagram (Figure 349a) and plummet downward, you can now see the reaction force R2, which closes the shear diagram to zero. The largest value of the shear force in this case is then R2 at x = l.
[email protected] = 0 = a +
60 20
(a)
x
[email protected] x = 5.8 = R1 ( a ) + R1
1.8 1.8 = 18 ( 4 ) + 18 = 88.2 2 2
(d )
10
The above method gives the magnitudes and locations of the maximum shear and moment on the beam and is useful for a quick determination of those values. However,
Chapter 3
KINEMATICS AND LOAD DETERMINATION
145
all that walking and falling can become tiresome, and it would be useful to have a method that can be conveniently computerized to give accurate and complete information on the shear and moment diagrams of any beamloading case. Such a method will also allow us to obtain the beam’s deflection curve with little additional work. The simple method shown previously is not as useful for determining deflection curves, as will be seen in the next chapter. We will now repeat this example using singularity functions to determine the loading, shear, and moment diagrams.
3
EXAMPLE 34B Shear and Moment Diagrams of a Simply Supported Beam Using Singularity Functions Problem
Determine and plot the shear and moment functions for the simply supported beam with uniformly distributed load shown in Figure 347a.
Given
Beam length l = 10 in, and load location a = 4 in. The magnitude of the uniform force distribution is w = 10 lb/in.
Assumptions
The weight of the beam is negligible compared to the applied load and so can be ignored.
Solution
See Figures 347a and 349.
∫
−1
−w x−a
V = qdx = R1 x − 0
∫
0
0
+ R2 x − l
−1
− w x − a 1 + R2 x − l
M = Vdx = R1 x − 0 1 −
w x−a 2
l a
〈x–a 〉 0 w x
1 Write equations for the load function in terms of equations 3.33 and integrate the resulting function twice using equations 3.34 to obtain the shear and moment functions. For the beam in Figure 347a, q = R1 x − 0
y
2
(a) 0
+ C1
+ R2 x − l 1 + C1 x + C2
(b) (c)
There are two reaction forces and two constants of integration to be found. We are integrating along a hypothetical infinite beam from –∞ to x. The variable x can take on values both before and beyond the end of the beam. If we consider the conditions at a point infinitesimally to the left of x = 0 (denoted as x = 0–), the shear and moment will both be zero there. The same conditions apply at a point infinitesimally to the right of x = l (denoted as x = l +). These observations provide the four boundary conditions needed to evaluate the four constants C1, C2, R1, R2: when x = 0–, V = 0, M = 0; when x = l+, V = 0, M = 0. 2 The constants C1 and C2 are found by substituting the boundary conditions x = 0–, V = 0, and x = 0–, M = 0 in equations (b) and (c), respectively:
R1
R2
Simply supported beam with uniformly distributed loading
FIGURE 347a Repeated
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MACHINE DESIGN

An Integrated Approach
( )
V 0 − = 0 = R1 0 − − 0
0

Sixth Edition
− w 0− − a
1
+ R2 0 − − l
0
w − 0 −a 2
2
+ R2 0 − − l
1
+ C1
C1 = 0
( )
M 0 − = 0 = R1 0 − − 0
3
1
−
C2 = 0
( )
(d )
+ C1 0 − + C2
Note that in general, the constants C1 and C2 will always be zero if the reaction forces and moments acting on the beam are included in the loading function, because the shear and moment diagrams must close to zero at each end of the beam. 3 The reaction forces R1 and R2 can be calculated from equations (c) and (b) respectively by substituting the boundary conditions x = l+, V = 0, M = 0. Note that we can substitute l for l+ to evaluate it since their difference is vanishingly small. (a) Loading Diagram 40 q R2 30 20 R 1 10 0 w –10 x –20 0 5 10 (b) Shear Diagram 20 V 10 0 –10 –20 –30 –40 –50 0
l −0 +
0 = R1l − R1 =
( )
(
1
+
2l +
−w
(
w l+ − a
w l+ − a
)
V l + = R1 l + − 0
l+ − a
)
2
+ R2 l + − l
2
1
=0
2
(e)
2 2
=
0
w ( l − a )2 10 (10 − 4 )2 = = 18 2l 2 (10 )
− w l+ − a
1
+ R2 l + − l
0 = R1 − w ( l − a ) + R2
0
=0 (f)
R2 = w ( l − a ) − R1 = 10 (10 − 4 ) − 18 = 42
Since w, l, and a are known from the given data, equation (e) can be solved for R1, and this result substituted in equation (f) to find R2. Note that equation (f) is just ΣF = 0, and equation (e) is the sum of moments taken about point l and set to 0.
–w 5.8 5
x 10
(c) Moment Diagram 100 M 88. 2 80 60 40 20
5.8
0 0
( ) = R1
M l
+
5
x 10
F I G U R E 3  4 9 Repeated Example 34 Plots
4 To generate the shear and moment functions over the length of the beam, equations (b) and (c) must be evaluated for a range of values of x from 0 to l, after substituting the above values of C1, C2, R1, and R2 in them. The independent variable x was run from 0 to l = 10 at 0.1 increments. The reactions, loading function, shearforce function, and moment function were calculated from equations (a) through (f) above and are plotted in Figure 349. The files EX0304 that generate these plots are on the website. 5 The largest absolute values of the shear and moment functions are of interest for the calculation of stresses in the beam. The plots show that the shear force is largest at x = l and the moment has a maximum Mmax near the center. The value of x at Mmax can be found by setting V to 0 in equation (b) and solving for x. (The shear function is the derivative of the moment function and so must be zero at each of its minima and maxima.) This gives x = 5.8 at Mmax. The function values at these points of maxima or minima can then be calculated from equations b and c respectively by substituting the appropriate values of x and evaluating the singularity functions. For the maximum absolute value of shear force at x = l,
Chapter 3
KINEMATICS AND LOAD DETERMINATION
Vmax = [email protected] x = l − = R1 l − − 0
0
(
− w l− − a
1
+ R2 l − − l
147
0
)
= R1 − w l − − a + 0
( g)
= 18 − 10 (10 − 4 ) + 0 = −42
3
Note that the first singularity term evaluates to 1 since l – > 0 (see Eq. 3.33c), the second singularity term evaluates to (l – a) because l – > a in this problem (see Eq. 3.33b), and the third singularity term evaluates to 0 as defined in equation 3.33c. The maximum moment is found in similar fashion: M max = [email protected] x = 5.8 = R1 5.8 − 0 1 − w = R1 5.8 1 − w = 18 ( 5.8 ) − 10
5.8 − a 2
5.8 − 4 2
2
(5.8 − 4 )2 2
2
+ R2 5.8 − l
1
+ R2 5.8 − 10
1
(h)
+ 0 = 88.2
The third singularity term evaluates to 0 because 5.8 10, making it other than a short column. Calculate the slenderness ratio of the tangent point between the Johnson and Euler lines of Figure 442:
( Sr )D =
π
2 ( 30 E 6 ) 2E =π = 84.5 83E 3 Sy
(c)
The slenderness ratio of this column is less than that of the tangent point between the Johnson and Euler lines shown in Figure 442. It is thus an intermediatecolumn and the Johnsoncolumn formula (Eq. 4.43) should be used to find the critical load: 2 1 S y Sr Pcr = A S y − E 2 π
(d )
2 1 83 000 ( 34 ) = 4765 lb = 0.125 (.5) 83 000 − 30 E 6 2π
Fh 1 2
Fh
force from hand
3
4
Fc crimp force
F I G U R E 3  2 8 Repeated Wire Connector Crimping Tool
243
* Even a small amount of clearance in the holes will prevent the pins from acting as a moment joint along their axes, thus creating an effective pinnedpinned connection in two dimensions unless and until the clearance is taken up by deflection and a pin is jammed in its hole.
MACHINE DESIGN
244

An Integrated Approach 
Sixth Edition
curved beam approximation
A Fh
1560
52
2000
1
y
1.23
1548
0.6 r P
B
x
3 Fh
0.7
D
452
1548
C
52
1560
1.0 A
B
1548
0.75
C
0.35
1548 2
0.80
All dimensions in inches and forces in pounds. See Table 33 for complete force information.
1.55
a
2000 1.2
452 D
4
FIGURE 449 FreeBody Diagrams, Dimensions, and Force Magnitudes for a Wire Connector Crimping Tool
The critical load is 3.1 times larger than the applied load. It is safe against buckling. Link 2 is a shorter, wider column than link 3 and has lower axial forces so can be assumed to be safe against buckling based on the link 3 calculations. 2 Since it does not buckle, the deflection of link 3 in axial compression is (Eq. 4.7): x=
1548 (1.23) Pl = = 0.001 in AE 0.0625 ( 30 E 6 )
(e)
3 Any of the links could also fail in bearing in the 0.25india holes. The largest force on any pin is 1560 lb. This worstcase bearing stress (Eqs. 4.7 and 4.10) is then σb =
P P 1560 = = = 49 920 psi Abearing length ( dia ) 0.125 ( 0.25)
(f)
There is no danger of tearout failure in links 2 or 3, since the loading is toward the center of the part. Link 1 has ample material around the holes to prevent tearout. 4 The 0.25india pins are in single shear. The worstcase direct shear stress from equation 4.9 is: τ=
P 1560 = = 31 780 psi Ashear π ( 0.25)2 4
( g)
Chapter 4
STRESS, STRAIN, AND DEFLECTION
5 Link 4 is a 1.55inlong beam, simply supported at the pins and loaded with the 2 000lb crimp force at 0.35 in from point C. Write the equations for the load, shear, moment, slope, and deflection using singularity functions, noting that the integration constants C1 and C2 will be zero: q = R1 x − 0 V = R1 x − 0
−1 0
−F x−a
−F x−a
1
0
−1
+ R2 x − l 0
+ R2 x − l
1
M = R1 x − 0 − F x − a + R2 x − l
−1
(h)
1
1 R1 x−0 EI 2
2
−
F x−a 2
2
+
R2 x−l 2
2
+ C3
1 R1 y= x−0 EI 6
3
F − x−a 6
3
R + 2 x−l 6
3
+ C3 x + C4
θ=
(i )
6 The reaction forces can be found from ΣM = 0 and ΣF = 0 (see Appendix B): R1 =
F ( l − a ) 2000 (1.55 − 0.35) = = 1548 lb 1.55 l
( j)
R2 =
Fa 2000 ( 0.35) = = 452 lb 1.55 l
(k )
The maximum moment is 1548(0.35) = 541.8 lbin at the applied load. The shear and moment diagrams for link 4 are shown in Figure 450. 7 The beam depth at the point of maximum moment is 0.75 in and the thickness is 0.187. The bending stress is then 0.75 541.8 2 Mc σ= = = 30 905 psi I 0.187 ( 0.75)3 12
(l )
8 The beam slope and deflection functions require calculation of the integration constants C3 and C4, which are found by substituting the boundary conditions x = 0, y = 0 and x = l, y = 0 in the deflection equation. 1 R1 0−0 EI 6 C4 = 0
3
1 R1 l−0 EI 6
3
0=
0=
−
F 0−a 6
3
+
R2 0−l 6
3
+ C3 ( 0 ) + C4 (m )
−
F l−a 6
3
+
R2 l−l 6
3
+ C3 ( l )
1 1 C3 = F ( l − a )3 − R1l 3 = 2000 (1.55 − 0.35)3 − 1548 (1.55)3 = −248.4 6l 6 (1.55)
( n)
9 The deflection equation is found by combining equations i, j, k, m, and n: y=
F 6lEI
{( )(
)
l − a x 3 + ( l − a )2 − l 2 x − l x − a
3
+a x−l
3
}
(o)
245
MACHINE DESIGN
246
Loading Diagram (lb) 1500
An Integrated Approach 
Sixth Edition
and the maximum deflection at x = 0.68 in is ymax =
0

Fa ( l − a ) 2 a + ( l − a )2 − l 2 6l EI
(
)
2000 ( 0.35)(1.55 − 0.35) = 0.352 + (1.55 − 0.35)2 − 1.552 = 0.0005 in 6 (1.55)( 30 E 6 )( 0.0066 )
–1000
( p)
–2000 0
0.8
1.6
Shear Diagram (lb) 1500 1000 500 0 –500 0
0.8
1.6
Moment Diagram (lbin) 600 400
Only a very small deflection is allowed here to guarantee the proper crimp stroke, and this amount is acceptable. The slope and deflection diagrams are shown in Figure 450. Also see the file CASE2B1. 10 Link 1 is relatively massive compared to the others and the only area of concern is the jaw, which is loaded by the 2000lb crimp force and has a hole in the cross section at its root. While the shape of this element is not exactly that of a curved beam with concentric inside and outside radii, this assumption will be acceptably conservative if we use an outer radius equal to the smallest section dimension as shown in Figure 449. This makes its inside radius 0.6 in and its approximate outside radius 1.6 in. The eccentricity e of the curved beam’s neutral axis versus the beam’s centroidal axis rc is found from equation 4.12a,while accounting for the section’s hole in the integration: e = rc −
200 0 0
0.8
1.6
Slope Diagram (rad) 0.0010 0.0005 0 –0.0005 –0.0010 –0.0015 0 0.8 1.6 Deflection Diagram (in) 0 –0.0001 –0.0002 –0.0003 –0.0004 –0.0005 0 0.8 1.6 FIGURE 450 Link 4 Plots—Case 2B
0.313 (1 − 0.25) A = 0.103 = 1.1 − 1.600 dr dA 0.975 dr + 0.313 0 r 0.600 r 1.225 r
∫
ro
∫
∫
(q)
The radius to the neutral axis (rn) and the distances (ci and co) from the inner and outer fiber radii (ri and ro) to the neutral axis are then (see Figure 416) rn = rc − e = 1.10 − 0.103 = 0.997 ci = rn − ri = 0.997 − 0.600 = 0.397
(r )
co = ro − rn = 1.600 − 0.997 = 0.603
11 The applied bending moment on the curved beam section is taken as the applied load times its distance to the beam’s centroidal axis: M = Fl = 2000 ( 0.7 − 0.6 + 1.1) = 2400 lbin
(s )
12 Find stresses at the inner and outer fibers from equations 4.12b and 4.12c. Reduce the beam cross sectional area by the hole area: σi = +
M ci 2400 0.397 = = 65 kpsi eA ri 0.103 (1.0 − 0.25)( 0.313) 0.60
(t )
M co 2400 0.603 =− σo = − = −37 kpsi eA ro 0.103 (1.0 − 0.25)( 0.313) 1.60
13 There is also a direct axial tensile stress, which adds to the bending stress in the inner fiber at point P: σa =
F 2000 = = 8.5 kpsi A (1.0 − 0.25)( 0.313)
σ max = σ a + σ i = 65 + 8.5 = 74 kpsi
(u)
Chapter 4
STRESS, STRAIN, AND DEFLECTION
(a)
(b) FIGURE 451
Copyright © 2018 Robert L. Norton: All Rights Reserved
Finite Element Analysis of the Stresses in the Crimping Tool of Case Study 2B
This is the principal stress for point P, since there is no applied shear or other normal stress at this edge point. The maximum shear stress at point P is half this principal stress or 37 kpsi. The bending stress at the outer fiber is compressive and thus subtracts from the axial tensile stress for a net of –37 + 8.5 = –28.5 kpsi. 14 There is significant stress concentration at the hole. The theoretical stressconcentration factor for the case of a circular hole in an infinite plate is Kt = 3 as defined in equation 4.32a and Figure 435. For a circular hole in a finite plate, Kt is a function of the ratio of the hole diameter to the plate width. Peterson gives a chart of stressconcentration factors for a round hole in a flat plate under tension[5] from which we find that Kt = 2.42 for a dia / width ratio = 1/4. The local axial tensile stress at the hole is then 2.42(8.5) = 20.5 kpsi, which is less than the tensile stress at the inner fiber. 15 While this is far from a complete stress and deflection analysis of these parts, the calculations done address the areas judged to be most likely to fail or to have problem deflections. The stresses and deflections in link 1 were also computed using the ABAQUS finite element analysis program, which gave an estimated maximum principal stress at point P of 81 kpsi compared to our estimate of 74 kpsi. The FEA mesh and stress distribution calculated by the FEA model is shown in Figure 451. Case Study 2D in Chapter 8 presents the complete FEA analysis of this assembly. Our analysis simplified the part geometry in order to allow the use of a known closedform model (the curved beam) whereas the FEA model included all the material in the actual part but discretized its geometry. Both analyses should be recognized as only estimates of the stress states in the parts, not exact solutions. 16 Redesign may be needed to reduce these stresses and deflections, based on a failure analysis. This case study will be revisited in the next chapter after various failure theories are presented. You may examine the models for this case study by opening the files CASE2B1, CASE2B2, and CASE2B3 in the program of your choice. A stress analysis of this case study is done in Chapter 8 using Finite Element Analysis (FEA).
247
248
MACHINE DESIGN
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An Integrated Approach 
S T U D Y
Sixth Edition
3 B
Automobile ScissorsJack Stress and Deflection Analysis Problem
Determine the stresses and deflections at critical points in the scissorsjack assembly shown in Figures 35 (repeated here) and 452.
Given
The geometry and loading are known from Case Study 3A. The design load is 2000 lb total or 1000 lb per side. The width of the links is 1.032 in and their thickness is 0.15 in. The screw is a 1/213 UNC thread with root dia = 0.406 in. The material of all parts is ductile steel with E = 30E6 psi and Sy = 60 000 psi.
Assumptions
The most likely failure points are the links as columns, the holes in bearing where the pins insert, the connecting pins in shear, the gear teeth in bending, and the screw in tension. There are two sets of links, one set on each side. Assume the two sides share the load equally. The jack is typically used for very few cycles over its lifetime, so a static analysis is appropriate.
Solution
See Figures 452 to 453 and the files CASE3B1 and CASE3B2.
1 The forces on this jack assembly for the position shown were calculated in the previous installment of this case study (3A) in Chapter 3. Please see that section and Table 34 for additional force data. 2 The force in the jack screw is four times the 878lb component F21x at point A because that force is from the upper half of the jack in one plane only. The lower half exerts an equal force on the screw, and the back side doubles their sum. These forces put the screw in axial tension. The tensile stress is found from equation 4.7 using the 0.406 in root diameter of the thread to calculate the crosssectional area. This is a conservative assumption, as we shall see when analyzing threaded fasteners in Chapter 15: σx =
4 (878 ) 3512 P = = = 27 128 psi A π ( 0.406 )2 0.129 4
(a)
The axial deflection of the screw is found from equation 4.8: x=
Pl 4 (878 )(12.55) = = 0.011 in 0.129 ( 30 E 6 ) AE
(b)
3 Link 2 is the most heavily loaded of the links due to the applied load P being slightly offset to the left of center, so we will calculate its stresses and deflections. This link is loaded as a beamcolumn with both an axial compressive force P between points C and D and a bending couple applied between D and E. Note that the force F12 is virtually colinear with the link axis. The axial load is equal to F12 cos(1°) = 1026 lb and the bending couple created by F42 acting about point D is M = 412(0.9) = 371 inlb. This couple is equivalent to the axial load being eccentric at point D by distance e = M / P = 371 / 1026 = 0.36 in. The secantcolumn formula (Eq. 4.46c) can be used with this effective eccentricity e accounting for the applied couple in the plane of bending; c is 1/2 of the 1.032in width of the link. Since it is a pinnedpinned column, leff = l from Table 44. The radius of gyration k is taken in the xy plane of bending for this calculation (Eq. 4.34):
Chapter 4
STRESS, STRAIN, AND DEFLECTION
249
P dimensions in inches
6
3
ty p
2
4
1
y
2
7
x
5
30 ° typ
6 Fg
F I G U R E 3  3 0 Repeated An Automobile Scissors Jack
k=
0.15 (1.032 )3 I bh3 = = = 0.298 12bh 12 ( 0.15)(1.032 ) A
(c)
The slenderness ratio is leff / k = 20.13. The secant formula can now be applied and iterated for the value of P (see Figure 453):
common normal
P = 1000 per side
y
x x
F32 = 1009 D
6"
3 F23
E
4
F43
(b) Gear tooth detail
F34 = 613 2 C
0.9
F42 = 412
4
30°
F41 = 996
F21 = 1026
F14 = 996
1
F12 = 1026 @ 31° FAx = –F21x = 878
A
FAy = –F21y = 530 FIGURE 452
θ
2
F24 = 412
B (a) Freebody diagrams
FreeBody Diagrams, Dimensions, and Forces for Elements of the Scissors Jack
FBx = –F41x = –878
FBy = –F41y = 470
MACHINE DESIGN
250

An Integrated Approach 
P = A
S yc leff ec 1 + 2 sec k k
(
P 4 EA
Sixth Edition
= 18 975 psi (d )
)
Pcrit = 0.155 18 975 = 2937 lb
The column also has to be checked for concentriccolumn buckling in the weaker (z) direction with c = 0.15 / 2. The radius of gyration in the z direction is found from k=
1.032 ( 0.15)3 I bh3 = = = 0.043 12bh 12 (1.032 )( 0.15) A
(e)
The slenderness ratio in the z direction is Sr =
leff k
=
6 = 138.6 0.043
(f)
This needs to be compared to the slenderness ratio (Sr)D at the tangency between the Euler and Johnson lines to determine which buckling equation to use for this column:
( Sr )D = π
2 ( 30 E 6 ) 2E =π = 99.3 60 000 Sy
( g)
The Sr for this column is greater than (Sr)D, making it an Euler column (see Figure 453). The critical Euler load is then found from equation 4.38a: Pcr =
π 2 EI l
=
2
π 2 ( 30 E 6 )(1.032 )( 0.15)3 12 ( 6 )2
= 2387 lb
(h)
Thus, it is more likely to buckle in the weaker z direction than in the plane of the applied moment. Its safety factor against buckling is 2.3. unit load psi x 103 70 short 60 50 Euler 40 Johnson 30 fails 20 10 secant 0 20 0 50 100 150 200 slenderness ratio FIGURE 453 Solution to Eccentric Column in Case Study 3B
4 The pins are all 0.437in dia. The bearing stress in the most heavily loaded hole at C is σ bearing =
P Abearing
=
1026 = 15 652 psi 0.15 ( 0.437 )
(i)
The pins are in single shear and their worst shear stress is τ=
1026 P = = 6841 psi Ashear π ( 0.437 )2 4
( j)
5 The gear tooth on link 2 is subjected to a force of 412 lb applied at a point 0.22 in from the root of the cantilevered tooth. The tooth is 0.44 in deep at the root and 0.15 in thick. The bending moment is 412(0.22) = 91 inlb and the bending stress at the root is σ=
91( 0.22 ) Mc = = 18 727 psi I 0.15 ( 0.44 )3 12
(k )
Chapter 4
STRESS, STRAIN, AND DEFLECTION
6 This analysis could be continued, looking at other points in the assembly and, more importantly, at stresses when the jack is in different positions. We have used an arbitrary position for this case study but, as the jack moves to the lowered position, the link and pin forces will increase due to poorer transmission angles. A complete stress analysis should be done for multiple positions. This case study will be revisited in the next chapter for the purpose of failure analysis. You may examine the models for this case study by opening the files CASE3B1 and CASE3B2 in the program of your choice.
C A S E
S T U D Y
4 B
Bicycle Brake Arm Stress Analysis Problem
Determine the stresses at critical points in the bicycle brake arm shown in Figures 39 (repeated here) and 454.
Given
The geometry and loading are known from Case Study 4A and are shown in Table 35. The castaluminum arm is a teesection curved beam whose dimensions are shown in Figure 454. The pivot pin is ductile steel. The loading is threedimensional.
Assumptions
The most likely failure points are the arm as a doublecantilever beam (one end of which is curved), the hole in bearing, and the connecting pin in bending as a cantilever beam. Since this is a marginally ductile cast material (5% elongation to fracture), we can ignore the stress concentration on the basis that local yielding will relieve it.
Solution
See Figures 454 to 456.
1 Each brake arm is a doublecantilever beam. Each end of an arm can be treated separately. The curvedbeam portion has a teeshaped cross section as shown in Section XX of Figure 454. The neutral axis of a curved beam shifts from the centroidal axis toward the center of curvature a distance e as described in Section 49 and in equation 4.12a. To find e requires an integration of the beam cross section and knowledge of its centroidal radius. Figure 455 shows the teesection broken into two rectangular segments, flange and web. The radius of the centroid of the tee is found by summing moments of area for each segment about the center of curvature:
∑ M = A1rc
1
rct = rct =
+ A2 rc2 = At rct
A1rc1 + A2 rc2 At
=
A1 ( ri + y1 ) + A2 ( ri + y2 ) A1 + A2
( a)
( 20 )( 7.5)(58 + 3.75) + (10 )( 7.5)(58 + 11.25) = 64.25 mm ( 20 )( 7.5) + (10 )( 7.5)
See Figures 453 and 454 for dimensions and variable names. The integral dA/r for equation 4.12a can be found in this case by adding the integrals for web and flange: ro
∫0
dA A1 A2 ( 20 )( 7.5) (10 )( 7.5) = + = + = 3.51 mm r rc1 rc2 58 + 3.75 58 + 11.25
(b)
251
252
MACHINE DESIGN

An Integrated Approach 
Sixth Edition
Fcabl e y
y
cable
cable
3
3 x
z
brake arm
brake arm
2 4
4
2 frame
pad
1
pad
θ
wheel rim 6
172 ° x N
5
5 wheel rim
F I G U R E 3  3 4 Repeated
6
V
ω
CenterPull Bicycle Brake Arm Assembly
The radius of the neutral axis and the distance e are then rn =
At 225 = = 64.06 mm dA 3.51 0 r
∫
ro
e = rc − rn = 64.25 − 64.06 = 0.187 mm
(c)
The magnitude of the bending moment acting on the curved section of the beam at Section XX can be approximated by taking the cross product of the force F32 and its position vector RAB referenced to the pivot at B in Figure 454: M AB = RABx F32 y − RAB y F32 x = −80.6 ( 523) − 66 ( 353) = 65 452 Nmm
(d )
The stresses at the inner and outer fibers can now be found using equations 4.12b and 4.12c (with lengths in mm and moments in Nmm for proper unit balance): σi = +
65 452 ( 6.063) M ci = = 162 MPa eA ri ( 0.1873)( 225)( 58 )
65 452 (8.937 ) M co = = −190 MPa σo = − eA ro ( 0.1873)( 225)( 73)
(e)
2 The hub cross section, shown as Section BB in Figure 454, is a possible location of failure, since there is a combination of bending and axial tension stresses here and the pin hole removes substantial material. The bending stress is due to the maximum mo
Chapter 4
80.6
F32y +523 A
STRESS, STRAIN, AND DEFLECTION
F32y +523
15
F32x
7.5
brake arm
Section XX
+353 X
F12x
B –1805
F52x
frame
20
11 d
X
58 r
42.5
29
10
2
66
Section BB
B
F52y –204
+1452
2
F12z
–319
F52z
y
–587
13 18.2
x
FIGURE 454
pin
F52y –204
M21x –32 304
y
z
ment acting on the curved beam at its root and the tensile stress is due to the y component of the force at A. There is also a shear stress due to transverse loading, but this will be zero at the outer fiber where the sum of the bending and axial stresses is maximum. The area and area moment of inertia of the hub cross section are needed: Ahub = length ( d out − din ) = 28.5 ( 25 − 11) = 399 mm 2
(
12
) = 28.5 (253 − 113 ) = 33 948 mm 4
(f)
12
The stress on the left half of Section BB is the sum of the bending and axial stresses: σ hub =
Mc F32 y 65 452 (12.5) 523 + = + = 25.4 MPa 33 948 399 I hub Ahub
( g)
The stress on the right half of Section BB is lower, because the compression due to bending is reduced by the axial tension. 3 The straight portion of the brake arm is a cantilever beam loaded in two directions, in the xy plane and in the yz plane. The section moduli and moments are different in these bending directions. The z moment in the xy plane is equal and opposite to the moment on the curved section. The cross section at the root of the cantilever is a rectangle of 23 by 12 mm as shown in Figure 454. The bending stress at the outer fiber of the 23mm side due to this moment is
σ y1
Mc = = I
12 65 452 2 23 (12 )3 12
= 118.6 MPa
F21y +319
1
F21x = +1805 M21y = – 52 370
BrakeArm FreeBody Diagrams, Forces in N, Moments in Nmm, and Dimensions in mm
3 3 − din length d out
F21z –587
+587
23 by 12 mm 28.5
F12y
C
I hub =
253
(h)
MACHINE DESIGN
254
A1
y1

An Integrated Approach 
The x moment is due to force F52z acting at the 42.5 radius, bending the link in the z direction. The bending stress at the surface of the 12mm side is
σ y2
flange
Mc = = I
23 589 ⋅ 42.5 2 12 ( 23)3 12
ri rc1
(a) Centroid of flange
web y2 rc2
A2
ri
Sixth Edition
= 23.7 MPa
(i )
These two ydirection normal stresses add at the corners of the two faces to give σ y = σ y1 + σ y2 = 118.6 + 23.7 = 142.2 MPa
( j)
4 Another possible failure point is the slot in the cantilever arm. Though the moment is zero there, the shear force is present and can cause tearout in the z direction. The tearout area is the shear area between the slot and edge: Atearout = thickness ( width ) = 8 ( 4 ) = 32 mm 2 τ=
F52z Atearout
(k )
589 = = 18.4 MPa 32
(b) Centroid of web
yt
5 The pivot pin is subjected to force F21, which has both x and y components and to a couple M21 due to the forces F12z and F52z. The force F21 creates a bending moment having components F21xl and F21yl in the yz and xz planes, respectively, where l = 29 mm is the length of the pin:
tee
ri
At
rct
(c) Centroid of tee
yt
e
M pin =
(M
21x
− F12 y ⋅ l
) + (− M 2
21y
+ F12 x ⋅ l
)
2
=
( −32 304 + 319 ⋅ 29)2 + (52 370 − 1805 ⋅ 29)2
=
( −32 304 + 9 251)2 + (52 370 − 52 345)2
=
( −23 053)2 + ( 25)2 = 23 053 Nmm
25 θ M pin = tan −1 ≅ 0° −23 053 rn
(l )
(m )
rct
Figure 456a shows the moment of the couple M21 and Figure 456b shows the moment of the force F21. Their combination is shown in Figure 456c.
(d) Neutral axis of tee
It is this combined moment that creates the largest bending stresses in the pin at 0° and 180° around its circumference. The maximum bending stress in the pin (with lengths in mm and moments in Nmm for unit balance) is
At
ri
FIGURE 455 Finding Neutral Axis of a TeeSection Curved Beam—Case Study 4B
σ pin =
M pin c pin I pin
=
11 23 053 2 π (11) 64
4
=
23 053 ( 5.5) 718.7
= 176 MPa
(n)
Chapter 4
STRESS, STRAIN, AND DEFLECTION
255
6 A more complete analysis could be done using finite element methods to determine the stresses and deformations at many other locations on the part. You may examine the model for this case study by opening the file CASE4B in the program of your choice. A stress analysis of this case study is also done in Chapter 8 using Finite Element Analysis (FEA).
y
4.19
SUMMARY
The equations used for stress analysis are relatively few and are fairly easy to remember. (See the equation summary later in this section.) The major source of confusion among students seems to be in understanding when to use which stress equation and how to determine where in the part’s continuum to calculate the stresses, since they vary over the part’s internal geometry. There are two types of applied stresses of interest, normal stress σ and shear stress τ. Each may be present on the same stress element and they will combine to create a set of principal normal stresses and maximum shear stress, as evidenced on the Mohr’s circle plane. It is ultimately these principal stresses that we need to find in order to determine the safety of the design. So, regardless of the source of loading or type of stress that may be applied to the part, you should always determine the principal stresses and maximum shear stress that result from their combination. (See Sections 4.3 and 4.5.) There are only a few types of loading that commonly occur on machine parts, but they may occur in combination on the same part. Loading types that create applied normal stresses are bending loads, axial loads, and bearing loads. Bending loads will always create both tensile and compressive normal stresses at different locations within the part. Beams provide the most common example of bending loads. (See Section 4.9.) Axial loads create normal stresses that can be either tensile or compressive (but not both at once), depending on whether the axial load is in a tensile or compressive direction. (See Section 4.7.) Fasteners such as bolts often have significant axialtension loads. If the axial load is compressive, then there may be a danger of column buckling, and the equations of Section 4.16 must also be applied. Bearing loads create compressive normal stresses in the shaft and bushing (bearing). Loading types that create applied shear stresses are torsional loads, direct shear loads, and bending loads. Torsional loads involve the twisting of a part around its long axis by application of a torque. A transmission shaft is a typical example of a torsionally loaded part. (Chapter 10 deals with the design of transmission shafts.) Direct shear can be caused by loads that tend to slice the part transversely. Fasteners such as rivets or pins sometimes experience direct shear loads. A pin trying to tear its way out of its hole also causes direct shear on the tearout area. (See Section 4.8.) Bending loads also cause transverse shear stresses on the cross section of the beam. (See Section 4.9.) Stresses can vary continuously over the internal continuum of a part’s geometry and are calculated as acting at an infinitesimally small point within that continuum. To do a complete analysis of the stresses at all the infinite number of potential sites within the part would require infinite time, which we obviously do not have. So, we must intelligently select a few sites for our calculations such that they represent the worstcase situations.
x
pin
32.3 Nm
–52.4 Nm (a) Moment of couple F12z  F52z
52.4 Nm
pin
9.3 Nm
(b) Moment of force F21
pin
23.0 Nm
(c) Combined moments
FIGURE 456 Bending Moments on Pivot Pin  Case Study 4B
MACHINE DESIGN
256
l a
M1
F 〈x—a〉 – 1 x
R1
FIGURE 457 Cantilever Beam with Concentrated Load
Loading Diagram 60 40 20 0 –20 –40 –60 0 5 10 Shear Diagram 40 30 20 10 0 5
10
Moment Diagram 50 0 –50 –100 –150 –200 0
5
FIGURE 458 Load Distributions
An Integrated Approach 
Sixth Edition
The student needs to understand how the various stresses are distributed within the continuum of a loaded part. There are two aspects to determining appropriate locations on a given part at which to make the stress calculations. The first aspect concerns the load distribution over the part’s geometry and the second concerns the stress distribution within the part’s cross section. For example, consider a straight cantilever beam loaded with a single force at some point along its length as shown in Figure 457. The first aspect requires some knowledge of the way the loads on the beam are distributed in response to the applied force. This comes from an analysis of the beam’s shear and moment diagrams as shown in Figure 458, which indicate that in this case, the section with greatest load is at the wall. We would then concentrate our attention on a vanishingly thin “bologna slice” taken from this beam at the wall. Note that the presence of stress concentrations at other locations having lower nominal stresses would require their investigation too. The second aspect is then to determine where in this crosssectional “bologna slice” the stresses will be greatest. Figures in the relevant sections of this chapter show the stress distributions across sections for various types of loadings. These stressdistribution diagrams are collected in Figure 459, which also shows the relevant stress equations for each case. Since the loading in this beam example creates bending stresses, we must understand that there will be a normal stress that is maximally compressive at one extreme fiber and maximally tensile at the other extreme fiber as shown in Figures 415 and 459c. Thus, we would take a stress element at the outer fiber of this slice of the beam to calculate the worstcase bending normal stress from equation 4.11b. Bending loads also cause shear stress, but its distribution is maximum at the neutral plane and zero at the outer fiber as shown in Figures 419 and 459d. So, a different stress element is taken at the neutral plane of the crosssectional slice for calculation of the shear stress due to transverse loading, using the appropriate equation such as equation 4.14b for a rectangular cross section. Each of these two stress elements will have its own set of principal stresses and maximum shear stress, which can be calculated from equation 4.6a for this 2D case.
50
0

10
More complicated loadings on more complicated geometries may have multiple stresses applied to the same infinitesimal stress element. It is very common in machine parts to have loadings that create both bending and torsion on the same part. Example 49 deals with such a case and should be studied carefully. Stress is only one consideration in design. The deflections of parts must also be controlled for proper function. Often, a requirement for small deflections will dominate the design and require thicker sections than would be necessary to guard against excessive stress. The deflections of a designed part, as well as its stresses, should always be checked. Equations for deflection under various loadings are given in the relevant sections and are also collected in Appendix B for beams of various types and loadings. Figure 460 shows a flow chart depicting a set of steps that can be followed to analyze stresses and deflections under static loading.
Chapter 4
STRESS, STRAIN, AND DEFLECTION
257
Static Loading Stress Analysis Materials are assumed to be Homogeneous and Isotropic
y σ (a) Uniaxial tension, stress distribution across section
z
x
σ=
P A
Eq. 4.7
y τ (b) Direct shear, averagestress distribution across section
z
x
P Ashear
τ=
Eq. 4.9
y σ (c) Bending, normalstress distribution across section
z
x
σ=
My I
Eq. 4.11a
y
τ
z
τ
x
y
τ
z τ
VQ Ib
Eq. 4.13d
τ=
Tr J
Eq. 4.23b
τ τ
FIGURE 459
τ=
Determine the stress distributions within the cross sections of interest and identify locations of the highest applied and combined stresses.
Calculate the applied stresses acting on each face of every element, and then calculate the principal stresses and maximum shear stress resulting therefrom.
τ
τ (e) Torsion, shearstress distribution across section
Based on the load distributions over the part’s geometry, determine what cross sections of the part are most heavily loaded.
Draw a 3D stress element for each of the selected points of interest within the section and identify the stresses acting on it.
σ
(d) Bending, shearstress distribution across section
Find all applied forces, moments, torques, etc. and draw the freebody diagrams to show them applied to the part’s geometry.
τ
Distribution of Stresses Across a Cross Section Under Various Types of Loading
Calculate critical deflections of the parts.
FIGURE 460 Flow Chart for Static Stress Analysis
258
MACHINE DESIGN

An Integrated Approach 
Sixth Edition
Important Equations Used in This Chapter See the referenced sections for information on the proper use of these equations. The Stress Cubic—its roots are the 3D principal stresses (Section 4.3):
σ 3 − C2 σ 2 − C1σ − C0 = 0
(4.4c)
where C2 = σ x + σ y + σ z C1 = τ 2xy + τ 2yz + τ 2zx − σ x σ y − σ y σ z − σ z σ x C0 = σ x σ y σ z + 2τ xy τ yz τ zx − σ x τ 2yz − σ y τ 2zx − σ z τ 2xy Maximum Shear Stresses (Section 4.3):
τ13 =
σ1 − σ 3 2
σ 2 − σ1
τ 21 =
;
2
;
τ32 =
σ3 − σ 2 2
(4.5)
TwoDimensional Principal Stresses (Section 4.3):
σ a ,σ b =
σx + σy 2
2 σx − σy 2 ± + τ xy 2
σ c =0 σ1 − σ 3
τ max = τ13 =
2
(4.6a)
(4.6b)
Axial Tension Stress (Section 4.7):
σx =
P A
(4.7)
s=
Pl AE
(4.8)
P Ashear
(4.9)
Axial Deflection (Section 4.7):
Direct Shear Stress (Section 4.8):
τ xy = Direct Bearing Area (Section 4.8):
Abearing =
π ld 4
(4.10 b)
Maximum Bending Stress—Straight Beams (Section 4.9):
σ max =
Mc I
(4.11b)
Maximum Bending Stress—Curved Beams (Section 4.9):
σi = +
M ci eA ri
(4.12b)
Chapter 4
STRESS, STRAIN, AND DEFLECTION
Transverse Shear Stress in Beams—General Formula (Section 4.9):
τ xy =
V bI
c
∫y
y dA
(4.13 f )
1
Maximum Transverse Shear Stress—Rectangular Beam (Section 4.9):
3V 2 A
τ max =
(4.14 b)
Maximum Transverse Shear Stress—Round Beam (Section 4.9):
τ max =
4V 3 A
(4.15c)
Maximum Transverse Shear Stress—IBeam (Section 4.9):
τ max ≅
V Aweb
(4.16)
The General Beam Equations (Section 4.9):
q d4y = 4 EI dx
(4.18a)
V d3y = 3 EI dx
(4.18b)
M d2y = EI dx 2
(4.18c)
θ=
dy dx
y = f (x)
(4.18d ) (4.18e)
Maximum Torsional Shear Stress—Round Section (Section 4.12):
τ max =
Tr J
(4.23b)
Maximum Torsional Deflection—Round Section (Section 4.12):
θ=
Tl JG
(4.24)
Maximum Torsional Shear Stress—Nonround Section (Section 4.12):
τ max =
T Q
(4.26a)
Maximum Torsional Deflection—Nonround Section (Section 4.12):
θ=
Tl KG
(4.26b)
259
260
MACHINE DESIGN

An Integrated Approach 
Sixth Edition
Spring Rate or Spring Constant—linear (a), angular (b) (Section 4.14):
k=
F y
(4.27a)
k=
T θ
(4.27b)
Stress with Stress Concentration (Section 4.15):
σ max = K t σ nom
(4.31)
τ max = K ts τ nom Column Radius of Gyration (Section 4.16):
I A
(4.34)
l k
(4.33)
k= Column Slenderness Ratio (Section 4.16):
Sr =
Column Critical Unit Load—Euler Formula (Section 4.16):
Pcr π 2 E = 2 A Sr
(4.38c)
Column Critical Unit Load—Johnson Formula (Section 4.16):
2E S yc
( Sr )D = π
(4.42)
2 Pcr 1 S yc Sr = S yc − A E 2 π
(4.43)
Column Critical Unit Load—Secant Formula (Section 4.16):
P = A
S yc leff ec 1 + 2 sec k k
P 4 EA
(4.46c)
Pressurized Cylinder (Section 4.17)
σt =
σr =
σa =
pi ri2 − po ro2 ro2 − ri2
pi ri2 − po ro2 ro2 − ri2
pi ri2 − po ro2 ro2 − ri2
+
−
ri2 ro2 ( pi − po )
(
r 2 ro2 − ri2
)
ri2 ro2 ( pi − po )
(
r 2 ro2 − ri2
)
(4.47a)
(4.47b)
(4.47c)
Chapter 4
4.20
STRESS, STRAIN, AND DEFLECTION
REFERENCES
261
Table P40†
1 I. H. Shames and C. L. Dym, Energy and Finite Element Methods in Structural Mechanics. Hemisphere Publishing: New York, Sect. 1.6, 1985.
Topic/Problem Matrix
2 I. H. Shames and F. A. Cossarelli, Elastic and Inelastic Stress Analysis. PrenticeHall: Englewood Cliffs, N.J., pp. 4650, 1991.
Sect. 4.1, 4.2 Stress, Strain
3 R. E. Peterson, Stress Concentration Factors. John Wiley & Sons: New York, 1974.
Sect. 4.5 Mohr’s Circles
4 R. J. Roark and W. C. Young, Formulas for Stress and Strain. 6th ed. McGrawHill: New York, 1989.
41, 479, 493, 4100
5 R. E. Peterson, Stress Concentration Factors. John Wiley & Sons: New York, p. 150, 1974.
42, 418, 461, 474a, 495
6 W. D. Pilkey, Peterson’s Stress Concentration Factors, John Wiley & Sons: New York, 1997. 7 H. T. Grandin and J. J. Rencis, Mechanics of Materials, John Wiley & Sons: New York, pp. 176177, 2006. 8 N. Troyani, C. Gomes, and G. Sterlacci, “Theoretical Stress Concentration Factors for Short Rectangular Plates With Centered Circular Holes.” ASME J. Mech. Design, V. 124, pp. 126128, 2002. 9 N. Troyani, et al., “Theoretical Stress Concentration Factors for Short Shouldered Plates Subjected to Uniform Tensions.” IMechE J. Strain Analysis, V. 38, pp. 103113, 2003. 10 N. Troyani, G. Sterlacci, and C. Gomes, “Simultaneous Considerations of Length and Boundary Conditions on Theoretical Stress Concentration Factors.” Int. J. Fatigue, V. 25, pp. 353355, 2003.
4.21
BIBLIOGRAPHY For general information on stress and deflection analysis, see:
F. P. Beer and E. R. Johnston, Mechanics of Materials. 2nd ed. McGrawHill: New York, 1992. J. P. D. Hartog, Strength of Materials. Dover: New York, 1961.
Sect. 4.8 Direct Shear, Bearing, and Tearout 44, 45, 46, 47, 49, 415, 419, 420, 422, 447, 459, 460, 474f, 487 Sect. 4.9 Straight Beams 410, 411, 412, 413, 414, 427, 440, 443a, 464, 465, 466, 467, 468, 474b,474 c, 474g, 491 Sect. 4.9 Curved Beams 417, 437, 462, 463, 469 to 472, 473, 474e Sect. 4.10 Deflection 48, 416, 423, 424, 425, 426, 428, 443b, 444, 448, 498 Sect. 4.11 Castigliano’s Method 484, 485, 486, 496, 497 Sect. 4.12 Torsion 421, 434, 446, 474d, 474h, 481, 482, 489, 490 Sect. 4.13 Combined Stresses
I. H. Shames, Introduction to Solid Mechanics. PrenticeHall: Englewood Cliffs, N.J., 1989.
Sect. 4.14 Spring Rates
S. Timoshenko and D. H. Young, Elements of Strength of Materials. 5th ed. Van Nostrand: New York, 1968.
*41
Sect. 4.7 Axial Tension
R. J. Roark and W. C. Young, Formulas for Stress and Strain. 6th ed. McGrawHill: New York, 1989.
I. H. Shames and F. A. Cossarelli, Elastic and Inelastic Stress Analysis. PrenticeHall: Englewood Cliffs, N.J., 1991.
4.22
455, 456, 457, 458, 494
PROBLEMS A differential stress element has a set of applied stresses on it as indicated in each row of Table P41. For the row(s) assigned, draw the stress element showing the applied
43, 434, 446, 488, 492
429, 430, 431, 432, 435, 438, 439 Sect. 4.15 Stress Conc. 436, 475 to 478 Sect. 4.16 Columns 445, 449, 450, 451, 452, 453, 454 Sect. 4.17 Cylinders 441, 442, 480, 483, 499 * Answers
to these problems are provided in Appendix D.
MACHINE DESIGN
262
Table P41

An Integrated Approach 
Sixth Edition
Data for Problem 41 (in psi) Rows a–g and k–m are twodimensional, others are 3D problems
60 mm
170 mm
F
T
Row
σx
a
1000
0
0
500
0
0
b
–1000
0
0
750
0
0
σy
σz
τ xy
τ yz
τ zx
c
500
–500
0
1000
0
0
d
0
–1500
0
750
0
0
e
750
250
0
500
0
0
f
–500
1000
0
750
0
0
g
1000
0
–750
0
0
250
h
750
500
250
500
0
0
i
1000
–250
–750
250
500
750
j
–500
750
250
100
250
1000
FIGURE P41
k
1000
0
0
0
0
0
Problem 43 (A Solidworks model
l
1000
0
0
0
500
0
m
1000
0
0
0
0
500
n
1000
1000
1000
500
0
0
o
1000
1000
1000
500
500
0
p
1000
1000
1000
500
500
500
of this is on the book’s website)
40 mm
stresses, find the principal stresses and maximum shear stress analytically, and check the results by drawing Mohr’s circles for that stress state. 42 A 400lb chandelier is to be hung from two 10ftlong solid steel cables in tension. Choose a suitable diameter of cable which will not exceed an allowable stress of 5000 psi. What will be the deflection of the cables? State all assumptions. FIGURE P42
†43
For the bicycle pedal arm assembly in Figure P41 with a riderapplied force of 1500 N at the pedal, determine the maximum principal stress in the pedal arm if its cross section is 15 mm in dia. The pedal attaches to the pedal arm with a 12mm screw thread. What is the stress in the pedal screw?
†*44
The trailer hitch shown in Figure P42 and Figure 11 has loads applied as defined in Problem 34. The 100kgf (980 N) tongue weight acts downward and the pull force of 4905 N acts horizontally. Using the dimensions of the ball bracket shown in Figure 15, determine:
Problems 44, 45, 46
(A Solidworks model of this is on the book’s website)
* Answers
to these problems are provided in Appendix D.
(a) (b) (c) (d) (e)
Problem numbers in boldface are extended from similar problems in earlier chapters with the same dash number. †
Problems with numbers in italics are design problems.
The principal stresses in the shank of the ball where it joins the ball bracket. The bearing stress in the ball bracket hole. The tearout stress in the ball bracket. The normal and shear stresses in the attachment bolts if they are 19mm dia. The principal stresses in the ball bracket as a cantilever.
†45
Repeat Problem 44 for the loading conditions of Problem 35.
†*46
Repeat Problem 44 for the loading conditions of Problem 36.
†*47
Design the wrist pin of Problem 37 for a maximum allowable principal stress of 20 kpsi if the pin is hollow and loaded in double shear.
†*48
A paper mill processes rolls of paper having a density of 984 kg/m3. The paper roll is 1.50 m outside dia (OD) X 220 mm inside dia (ID) X 3.23m long and is on a simply supported, hollow, steel shaft. Find the shaft ID needed to obtain a maximum deflection
Chapter 4
STRESS, STRAIN, AND DEFLECTION
263
P F
P
F 0.5cm grid
FIGURE P43 Problem 49 (A Solidworks model of this is on the website)
at the center of 3 mm if the shaft OD is 220 mm. Assume the shaft to be the same length between supports as the length of the paper roll. 49 For the ViseGrip® plierwrench drawn to scale in Figure P43, and for which the forces were analyzed in Problem 39, find the stresses in each pin for an assumed clamping force of P = 4000 N in the position shown. The pins are 8mm dia and are all in double shear. *410
*411
The overhung diving board of Problem 310 is shown in Figure P44a. Assume crosssection dimensions of 305 mm X 32 mm. The material has E = 10.3 GPa. Find the largest principal stress at any location in the board when a 980 N person is standing in the center of the width of the board at the free end. What is the maximum deflection? Repeat Problem 410 using the loading conditions of Problem 311. Assume the board weighs 284.4 N and deflects 131 mm statically when the person stands on it. Find the largest principal stress anywhere in the board when the 980 N person in Problem 410 jumps up 250 mm and lands back on the board. Find the maximum deflection.
* Answers
to these problems are provided in Appendix D. Problem numbers in boldface are extended from similar problems in earlier chapters with the same dash number. †
Problems with numbers in italics are design problems.
412 Repeat Problem 410 using the cantilevered diving board design in Figure P44b. 413 Repeat Problem 411 using the diving board design shown in Figure P44b. Assume the board weighs 186.3 N and deflects 85 mm statically when the person stands on it. † 414
Figure P45 shows a child’s toy called a pogo stick. The child stands on the pads, applying half her weight on each side. She jumps up off the ground, holding the pads up 2m
2m 0.7 m
( a ) Overhung diving board
FIGURE P44 Problems 410 through 413
P
0.7 m
( b) Cantilevered diving board
P
MACHINE DESIGN
264
W/2

An Integrated Approach 
Sixth Edition
W/2 FIGURE P46 Problem 416
against her feet, and bounces along with the spring cushioning the impact and storing energy to help each rebound. Assume a 60lb (266.9 N) child and a spring constant of 100 lb/in (17.5 N/mm). The pogo stick weighs 5 lb (22.2 N). Design the aluminum cantilever beam sections on which she stands to survive jumping 2 in (50.8 mm) off the ground. Allowable stress = 20 kpsi (138 MPa). Size the beam and define its shape. * † 415
P FIGURE P45 Problem 414
F F
A
Design a shear pin for the propeller shaft of an outboard motor if the shaft through which the pin is placed is 25mm diameter, the propeller is 200mm diameter, and the pin must fail when a force > 400 N is applied to the propeller tip. Assume an ultimate shear strength for the pin material of 100 MPa.
416 A track to guide bowling balls is designed with two round rods as shown in Figure P46. The rods are not parallel to one another but have a small angle between them. The balls roll on the rods until they fall between them and drop onto another track. The angle between the rods is varied to cause the ball to drop at different locations. Each rod’s unsupported length is 30 in and the angle between them is 3.2°. The balls are 4.5in dia and weigh 2.5 lb. The center distance between the 1india rods is 4.2 in at the narrow end. Find the maximum stress and deflection in the rods. (a) Assume the rods are simply supported at each end. (b) Assume the rods are fixed at each end.
r
417 A pair of ice tongs is shown in Figure P47. The ice weighs 50 lb and is 10 in wide across the tongs. The distance between the handles is 4 in, and the mean radius r of a tong is 6 in. The rectangular crosssectional dimensions are 0.75 in deep X 0.312 in wide. Find the stress in the tongs.
W
*418
FIGURE P47 Problem 417 * Answers
to these problems are provided in Appendix D. Problem numbers in boldface are extended from similar problems in earlier chapters with the same dash number. †
Problems with numbers in italics are design problems.
* † 419
A set of steel reinforcing rods is to be stretched axially in tension to create a tensile stress of 30 kpsi prior to being cast in concrete to form a beam. Determine how much force will be required to stretch them the required amount and how much deflection is required. There are 10 rods; each is 0.75in dia and 30 ft long. The clamping fixture used to pull the rods in Problem 418 is connected to the hydraulic ram by a clevis like that shown in Figure P48. Determine the size of the clevis pin needed to withstand the applied force. Assume an allowable shear stress of 20 kpsi and an allowable normal stress of 40 kpsi. Determine the required outside radius of the clevis end to not exceed the above allowable stresses in either tearout or bearing if the clevis flanges are each 0.8 in thick.
Chapter 4
† 420
STRESS, STRAIN, AND DEFLECTION
265
Repeat Problem 419 for 12 rods each 10mm dia and 10 m long. The desired rod stress is 200 MPa. The allowable normal stress in the clevis and pin is 280 MPa and their allowable shear stress is 140 MPa. Each clevis flange is 20 mm wide.
P
421 Figure P49 shows an automobile wheel with two common styles of lug wrench being used to tighten the wheel nuts, a singleended wrench in (a), and a doubleended wrench in (b). In each case two hands are required to provide forces, respectively, at A and B as shown. The distance between points A and B is 1 ft in both cases and the handle diameter is 0.625 in. The wheel nuts require a torque of 70 ftlb. Find the maximum principal stress and maximum deflection in each wrench design. *422
An inline “rollerblade” skate is shown in Figure P410. The polyurethane wheels are 72mm dia and spaced on 104mm centers. The skatebootfoot combination weighs 2 kg. The effective “spring rate” of the personskate system is 6000 N/m. The axles are 10mmdia steel pins in double shear. Find the stress in the pins for a 100kg person landing a 0.5m jump on one foot. (a) Assume that all four wheels land simultaneously. (b) Assume that one wheel absorbs all the landing force.
*423
A beam is supported and loaded as shown in Figure P411a. Find the reactions, maximum shear, maximum moment, maximum slope, maximum bending stress, and maximum deflection for the data given in the assigned row(s) in Table P42.
*424
A beam is supported and loaded as shown in Figure P411b. Find the reactions, the maximum shear, maximum moment, maximum slope, maximum bending stress, and maximum deflection for the data given in the assigned row(s) in Table P42.
*425
A beam is supported and loaded as shown in Figure P411c. Find the reactions, the maximum shear, maximum moment, maximum slope, maximum bending stress, and maximum deflection for the data given in the assigned row(s) in Table P42.
*426
A beam is supported and loaded as shown in Figure P411d. Find the reactions, maximum shear, maximum moment, maximum slope, maximum bending stress, and maximum deflection for the data given in the assigned row(s) in Table P42.
† 427
A storage rack is to be designed to hold the paper roll of Problem 48 as shown in Figure P412. Determine suitable values for dimensions a and b in the figure. Consider bending, shear, and bearing stresses. Assume an allowable tensile/compressive stress of 100 axle
axle
3 in
3 in A
B
lug wrench
A
lug wrench B F
F
F
FIGURE P49 Problem 421
(a)
tire
F
tire (b)
P FIGURE P48 Problems 419 and 420
* Answers
to these problems are provided in Appendix D. Problem numbers in boldface are extended from similar problems in earlier chapters with the same dash number. †
Problems with numbers in italics are design problems.
266
MACHINE DESIGN

An Integrated Approach 
Sixth Edition
l
l b
a
a
F
F w x
R1
w
M1
x
R2
R1
( a)
( b)
l
l b
b
F
a
a
F w
w x R1
R2
x R1
R2
(c)
R3
(d)
FIGURE P411 Beam Loadings for Problems 423 to 426, 429 to 432 and 496—See Table P42 for Data
MPa and an allowable shear stress of 50 MPa for both stanchion and mandrel, which are steel. The mandrel is solid and inserts halfway into the paper roll. Balance the design to use all of the material strength. Calculate the deflection at the end of the roll.
†
Problems with numbers in italics are design problems. Problem numbers in boldface are extended from similar problems in earlier chapters with the same dash number. Problems in succeeding chapters may also continue and extend these problems.
†428
Figure P413 shows a forklift truck negotiating a 15° ramp to drive onto a 4fthigh loading platform. The truck weighs 5000 lb and has a 42in wheelbase. Design two (one for each side) 1ftwide ramps of steel to have no more than 1in deflection in the worst case of loading as the truck travels up them. Minimize the weight of the ramps by using a sensible crosssectional geometry.
*429
Find the spring rate of the beam in Problem 423 at the applied concentrated load for the row(s) assigned in Table P42.
*430
Find the spring rate of the beam in Problem 424 at the applied concentrated load for the row(s) assigned in Table P42.
* Answers
to these problems are provided in Appendix D.
b stanchion paper roll
a mandrel FIGURE P412 Problem 427
base
Chapter 4
Table P42
*
STRESS, STRAIN, AND DEFLECTION
267
Data for Problems 423 through 426, 429 through 432, and 496
Use only data relevant to the particular problem. Lengths in m, forces in N, I in m4.
Row
l
a
b
w*
F
I
c
E
a
1.00
0.40
0.60
200
500
2.85E–08
2.00E–02
steel
b
0.70
0.20
0.40
80
850
1.70E–08
1.00E–02
steel
c
0.30
0.10
0.20
500
450
4.70E–09
1.25E–02
steel
d
0.80
0.50
0.60
65
250
4.90E–09
1.10E–02
steel
e
0.85
0.35
0.50
96
750
1.80E–08
9.00E–03
steel
f
0.50
0.18
0.40
450
950
1.17E–08
1.00E–02
steel
g
0.60
0.28
0.50
250
250
3.20E–09
7.50E–03
steel
h
0.20
0.10
0.13
400
500
4.00E–09
5.00E–03
alum
i
0.40
0.15
0.30
50
200
2.75E–09
5.00E–03
alum
j
0.20
0.10
0.15
150
80
6.50E–10
5.50E–03
alum
k
0.40
0.16
0.30
70
880
4.30E–08
1.45E–02
alum
l
0.90
0.25
0.80
90
600
4.20E–08
7.50E–03
alum
m
0.70
0.10
0.60
80
500
2.10E–08
6.50E–03
alum
n
0.85
0.15
0.70
60
120
7.90E–09
1.00E–02
alum
Note that w is a unit force of N/m
*431
Find the spring rate of the beam in Problem 425 at the applied concentrated load for the row(s) assigned in Table P42.
*432
Find the spring rate of the beam in Problem 426 at the applied concentrated load for the row(s) assigned in Table P42.
*433
For the bracket shown in Figure P414 and the data in the row(s) assigned from Table P43, determine the bending stress at point A and the shear stress due to transverse loading at point B. Also find the torsional shear stress at both points. Then determine the principal stresses at points A and B.
*434
For the bracket shown in Figure P414 and the data in the row(s) assigned from Table P43, determine the deflection at load F.
*435
For the bracket shown in Figure P414 and the data in the row(s) assigned from Table P43, determine the spring rate of the tube in bending, the spring rate of the arm in bend
ramp
FIGURE P413 Problem 428
1ft grid
* Answers
to these problems are provided in Appendix D.
MACHINE DESIGN
268

An Integrated Approach 
Sixth Edition
F
l a A B
t
y tube
wall
arm
z
h
x od
id
FIGURE P414 Problems 433 to 436 (A Solidworks model of this is on the book’s website)
ing, and the spring rate of the tube in torsion. Combine these into an overall spring rate in terms of the force F and the linear deflection at force F. 436 For the bracket shown in Figure P414 and the data in the row(s) assigned from Table P43, redo Problem 433 considering the stress concentration at points A and B. Assume a stressconcentration factor of 2.5 in both bending and torsion. * Answers
to these problems are provided in Appendix D.
*437
A semicircular curved beam as shown in Figure P415 has OD = 150 mm, ID = 100 mm, and t = 25 mm. For a load pair F = 14 kN applied along the diameter, find the eccentricity of the neutral axis and the stress at the inner and outer fibers.
† 438
Design a solid, straight, steel torsion bar to have a spring rate of 10 000 inlb per radian per foot of length. Compare designs of solid round and solid square cross sections. Which is more efficient in terms of material use?
† 439
Design a 1ftlong steel, endloaded cantilever spring for a spring rate of 10 000 lb/in at the load. Compare designs of solid round and solid square cross sections. Which is more efficient in terms of material use?
† 440
Redesign the roll support of Problem 48 to be like that shown in Figure P416. The stub mandrels insert to 10% of the roll length at each end. Choose appropriate dimensions a and b to fully utilize the material’s strength, which is the same as in Problem 427. See Problem 48 for additional data.
*441
A 10mm ID steel tube carries liquid at 7 MPa. Determine the principal stresses in the wall if its thickness is: (a) 1 mm, (b) 5 mm.
†
Problems with numbers in italics are design problems.
t
b typ OD ID
F
stanchion paper roll
F
a typ
mandrel
FIGURE P415 Problem 437 (A Solidworks model is on the book’s website)
FIGURE P416
base
Problem 440 (A Solidworks model of this is on the book’s website)
Chapter 4
Table P43
STRESS, STRAIN, AND DEFLECTION
Data for Problems 433–436, 449–452, and 489–492 Incl. Use only data that are relevant to the particular problem. Lengths in mm, forces in N.
Row
l
a
t
h
F
OD
ID
E
a
100
400
10
20
50
20
14
steel
b
70
200
6
80
85
20
6
steel
c
300
100
4
50
95
25
17
steel
d
800
500
6
65
160
46
22
alum
e
85
350
5
96
900
55
24
alum
f
50
180
4
45
950
50
30
alum
g
160
280
5
25
850
45
19
steel
h
200
100
2
10
800
40
24
steel
i
400
150
3
50
950
65
37
steel
j
200
100
3
10
600
45
32
alum
k
120
180
3
70
880
60
47
alum
l
150
250
8
90
750
52
28
alum
m
70
100
6
80
500
36
30
steel
n
85
150
7
60
820
40
15
steel
442 A cylindrical tank with hemispherical ends is required to hold 150 psi of pressurized air at room temperature. Find the principal stresses in the 1mmthick wall if the tank diameter is 0.5 m and its length is 1 m. 443 Figure P417 shows an offloading station at the end of a paper rolling machine. The finished paper rolls are 0.9m OD by 0.22m ID by 3.23m long and have a density of 984 kg/m3. The rolls are transferred from the machine conveyor (not shown) to the forklift truck by the Vlinkage of the offload station, which is rotated through 90° by an air cylinder. The paper then rolls onto the waiting forks of the truck. The forks are 38mm thick by 100mm wide by 1.2m long and are tipped at a 3° angle from the horizontal.
Vlinks
1m crank arm
paper rolling machine forks A rod FIGURE P417 Problems 443 to 447
offloading station
air cylinder
fork lift truck
269
270
MACHINE DESIGN

An Integrated Approach 
Sixth Edition
rollers FIGURE P418 Problem 448
Find the stresses in the two forks on the truck when the paper rolls onto it under two different conditions (state all assumptions):
†
Problems with numbers in italics are design problems.
(a) The two forks are unsupported at their free end. (b) The two forks are contacting the table at point A. † 444
Determine a suitable thickness for the Vlinks of the offloading station of Figure P417 to limit their deflections at the tips to 10 mm in any position during their rotation. Assume that there are two Vlinks supporting the roll, arranged at the 1/4 and 3/4 points along the roll’s length, and that each of the V arms is 10 cm wide by 1 m long. The V arms are welded to a steel tube that is rotated by the air cylinder. See Problem 443 for more information.
445 Determine the critical load on the aircylinder rod in Figure P417 if the crank arm that it rotates is 0.3 m long and the rod has a maximum extension of 0.5 m. The 25mmdia rod is solid steel with a yield strength of 400 MPa. State all assumptions. 446 The Vlinks of Figure P417 are rotated by the crank arm through a shaft that is 60mm dia by 3.23 m long. Determine the maximum torque applied to this shaft during the motion of the Vlinkage and find the maximum stress and deflection for the shaft. See Problem 443 for more information. 447 Determine the maximum forces on the pins at each end of the air cylinder of Figure P417. Determine the stress in these pins if they are 30mm dia and in single shear. † 448
A 100kg wheelchair marathon racer wants an exerciser that will allow indoor practicing in any weather. The design shown in Figure P418 is proposed. Two freeturning rollers on bearings support the rear wheels. A platform supports the front wheels. Design the 1mlong rollers as hollow tubes of aluminum to minimize the height of the platform and also limit the roller deflections to 1 mm in the worst case. The wheelchair has 65cmdia drive wheels separated by a 70cm track width. The flanges shown on the rollers limit the lateral movement of the chair while exercising and thus the wheels can be anywhere between those flanges. Specify suitably sized steel axles to support the tubes on bearings. Calculate all significant stresses.
*449
A hollow, square column has a length l and material E, as shown in the row(s) assigned in Table P43. Its crosssectional dimensions are 4 mm outside and 3 mm inside. Use Sy = 150 MPa for aluminum and 300 MPa for steel. Determine if it is a Johnson or an Euler column and find the critical load:
* Answers
to these problems are provided in Appendix D.
(a) (b) (c) (d) *450
If its boundary conditions are pinnedpinned. If its boundary conditions are fixedpinned. If its boundary conditions are fixedfixed. If its boundary conditions are fixedfree.
A hollow, round column has a length of 1.5 m, material E, and crosssectional dimensions OD and ID as shown in the row(s) assigned in Table P43. Use Sy = 150 MPa for
Chapter 4
STRESS, STRAIN, AND DEFLECTION
aluminum and 300 MPa for steel. Determine if it is a Johnson or an Euler column and find the critical load: (a) (b) (c) (d) *451
to these problems are provided in Appendix D.
If its boundary conditions are pinnedpinned. If its boundary conditions are fixedpinned. If its boundary conditions are fixedfixed. If its boundary conditions are fixedfree.
P
A
If its boundary conditions are pinnedpinned. If its boundary conditions are fixedpinned. If its boundary conditions are fixedfixed. If its boundary conditions are fixedfree.
B
A solid, circular column has a length l, material E, diameter OD, and an eccentricity t as shown in the row(s) assigned in Table P43. Use Sy = 150 MPa for aluminum and 300 MPa for steel. Determine if it is a Johnson or an Euler column and find the critical load: (a) (b) (c) (d)
†453
* Answers
A solid, rectangular column has a length l, material E, and crosssectional dimensions h and t as shown in the row(s) assigned in Table P43. Use Sy = 150 MPa for aluminum and 300 MPa for steel. Determine if it is a Johnson or an Euler column and find the critical load: (a) (b) (c) (d)
*452
271
If its boundary conditions are pinnedpinned. If its boundary conditions are fixedpinned. If its boundary conditions are fixedfixed. If its boundary conditions are fixedfree.
Design an aluminum, hollow, circular column for the following data: 3 m long, 5 mm wall thickness, 900 N concentric load, material yield strength of 150 MPa, and safety factor of 3: (a) If its boundary conditions are pinnedpinned. (b) If its boundary conditions are fixedfree.
454 Three round, 1.25india bars are made of SAE 1030 hotrolled steel but are of different lengths, 5 in, 30 in, and 60 in, respectively. They are loaded axially in compression. Compare the loadsupporting capability of the three bars if the ends are assumed to be: (a) (b) (c) (d)
Pinnedpinned. Fixedpinned. Fixedfixed. Fixedfree.
455 Figure P419 shows a 1.5india, 30inlong steel rod subjected to tensile loads P = 10 000 lb applied at each end of the rod, acting along its longitudinal Y axis and through the centroid of its circular cross section. Point A is 12 in below the upper end and point B is 8 in below A. For this bar with this loading find: (a) All components of the stress tensor matrix (equation 4.1a) for a point midway between A and B. (b) The displacement of point B relative to point A. (c) The elastic strain in the section between A and B. (d) The total strain in the section between A and B. 456 The rod in Figure P419, with the loading of Problem 455, is subjected to a reduction in temperature from 80°F to 20°F after the load is applied. The coefficient of thermal expansion for steel is approximately 6 µin/in/°F. Find: (a) All components of the stress tensor matrix (equation 4.1a) for a point midway between A and B. (b) The displacement of point B relative to point A. (c) The elastic strain in the section between A and B. (d) The total strain in the section between A and B. 457 Figure P420 shows a steel bar fastened to a rigid ground plane with two 0.25india hardened steel dowel pins. For P = 1500 lb, find:
P Copyright © 2018 Robert L. Norton: All Rights Reserved
FIGURE P419 Problems 455, 456 and 494 †
Problems with numbers in italics are design problems.
P h 0.25"
P
bar 4"
pin
2" pin ground Copyright © 2018 Robert L. Norton: All Rights Reserved
FIGURE P420 Problems 457 and 458
(A Solidworks model of this is on the book’s website)
MACHINE DESIGN
272

An Integrated Approach 
P
Sixth Edition
A C 2.5"
40
B
A 10"
B
3"
7"
10
5"
3"
D 1.5" 8" FIGURE P422 35
12" P
3"
0.5"
Copyright © 2018 Robert L. Norton: All Rights Reserved
P
Problems 460 and 461 (A Solidworks model of this is on the book’s website)
P dimensions in mm Copyright © 2018 Robert L. Norton: All Rights Reserved
FIGURE P421 Problem 459 (A Solidworks model is on the book’s website)
P
20"
20"
(a) The shear stress in each pin. (b) The direct bearing stress in each pin and hole. (c) The minimum value of dimension h to prevent tearout failure if the steel bar has a shear strength of 32.5 kpsi. 458 Repeat Problem 457 for P = 2200 lb. 459 Figure P421 shows a rectangular section aluminum bar subjected to offcenter forces P = 4000 N applied as shown: (a) Solve for the maximum normal stress in the midregion of the bar well away from the eyes where the loads are applied. (b) Plot the normal stress distribution across the cross section at this midregion of the bar. (c) Sketch a “reasonable” plot of the normal stress distribution across the cross section at the ends, close to the applied loads. 460 Figure P422 shows a bracket machined from 0.5inthick steel flat stock. It is rigidly attached to a support and loaded with P = 5000 lb at point D. Find: (a) The magnitude, location, and the plane orientation of the maximum normal stress at section AA. (b) The magnitude, location, and the plane orientation of the maximum shear stress at section AA. (c) The magnitude, location, and the plane orientation of the maximum normal stress at section BB. (d) The magnitude, location, and the plane orientation of the maximum shear stress at section BB. 461 For the bracket of Problem 460, solve for the deflection and slope of point C. 462 Figure P423 shows a 1india steel bar supported and subjected to the applied load P = 500 lb. Solve for the deflection at the load and the slope at the roller support.
Copyright © 2018 Robert L. Norton: All Rights Reserved
FIGURE P423 Problem 462
463 Figure P424 shows a 1.25india solid steel shaft with several twisting couples applied in the directions shown. For TA = 10 000 lbin, TB = 20 000 lbin, TC = 30 000 lbin, find: (a) The magnitude and location of the maximum torsional shear stress in the shaft. (b) The corresponding principal stresses for the location determined in part (a). (c) The magnitude and location of the maximum shear strain in the shaft.
Chapter 4
TA
STRESS, STRAIN, AND DEFLECTION
TB
TC
273
TD
l l 2 l 4
B
A 18"
C 12"
F
D 10"
B FIGURE P424
Copyright © 2018 Robert L. Norton: All Rights Reserved
Problems 463 and 464
d A
464 If the shaft of Problem 463 were rigidly attached to fixed supports at each end (A and D) and loaded only by the applied couples TB, and TC, then find: (a) The reactions TA and TD at each end of the shaft. (b) The rotation of section B with respect to section C. (c) The magnitude and location of the maximum shear strain. †465
Copyright © 2018 Robert L. Norton: All Rights Reserved
Figure P425 shows a pivot pin that is a press fit in part A and a slip fit in part B. If F = 100 lb and l = 1.5 in, what pin diameter is needed to limit the maximum stress in the pin to 50 kpsi?
†466
Figure P425 shows a pivot pin that is a press fit in part A and a slip fit in part B. If F = 100 N and l = 64 mm, what pin diameter is needed to limit the maximum stress in the pin to 250 MPa?
†467
Figure P425 shows a pivot pin that is a press fit in part A and a slip fit in part B. Determine the l/d ratio that will make the pin equally strong in shear and bending if the shear strength is equal to onehalf the bending strength.
75.4
FIGURE P425 Problems 465 to 467
†
Problem numbers in italics are design problems.
6.4
28.4 (b)
(c)
31.8
3.2 typ.
63.5 A 9.6 (a)
31.8
9.6 (d)
(e)
all dimensions in mm
31.8 3.2
A FIGURE P426 Problems 469 to 473 (A Solidworks model of this is on the book’s website)
Sections AA Copyright © 2018 Robert L. Norton: All Rights Reserved
274
†
Problems with numbers in italics are design problems.
MACHINE DESIGN

An Integrated Approach 
Sixth Edition
†468
Write a computer program in any language, or use an equationsolver program or spreadsheet to calculate and plot the variation in crosssectional area, area moment of inertia, radius of gyration, slenderness ratio, and critical load with respect to the inside diameter of both a Euler and a Johnson column having roundhollow cross sections. Assume that the outside diameter of each column is 1 in. The effective length of the Euler column is 50 in. The effective length of the Johnson column is 10 in. Both are made of steel with Sy = 36 000 psi. Let the inside diameter vary from 10% to 90% of the outside diameter for this parametric study. Comment on the advantages of hollowround columns over solidround columns of each type (Euler and Johnson) that have the same outer diameters, respective lengths, and materials.
*469
Figure P426a shows a Cclamp with an elliptical body dimensioned as shown. The clamp has a Tsection with a uniform thickness of 3.2 mm at the throat as shown in Figure P426b. Find the bending stress at the inner and outer fibers of the throat if the clamp force is 2.7 kN.
470 A Cclamp as in Figure P426a has a rectangular cross section as in Figure P426c. Find the bending stress at the inner and outer fibers of the throat if the clamp force is 1.6 kN. * Answers
to these problems are provided in Appendix D. Problem numbers in italics are design problems.
471 A Cclamp as in Figure P426a has an elliptical cross section as in Figure P426d. Dimensions of the major and minor axes of the ellipse are given. Determine the bending stress at the inner and outer fibers of the throat if the clamping force is 1.6 kN. 472 A Cclamp as in Figure P426a has a trapezoidal cross section as in Figure P426e. Determine the bending stress at the inner and outer fibers of the throat if the clamping force is 1.6 kN. †473
We want to design a Cclamp with a Tsection similar to the one shown in Figures P426a and b. The depth of the section will be 31.8 mm as shown but the width of the flange (shown as 28.4 mm) is to be determined. Assuming a uniform thickness of 3.2 mm and a factor of safety against static yielding of 2, determine a suitable value for the width of the flange if the Cclamp is to be made from 604018 ductile iron and the maximum design load is 1.6 kN.
474 A round steel bar is 10 in long and has a diameter of 1 in. (a) Calculate the stress in the bar when it is subjected to a 1000lb force in tension. (b) Calculate the bending stress in the bar if it is fixed at one end (as a cantilever beam) and has a 1000lb transverse load at the other end. (c) Calculate the transverse shear stress in the bar of part (b). (d) Determine how short the bar must be when loaded as a cantilever beam for its maximum flexural bending stress and its maximum transverse shear stress to provide equal tendency to failure. Find the length as a fraction of the diameter if the failure stress in shear is half the failure stress in bending. (e) Calculate the torsional shear stress when a 10 000 inlb couple is applied about the centerline (axis) of the cantilever beam at its free end. (f) If the force on the cantilever beam in (b) is eccentric, inducing torsional as well as bending stress, what fraction of the diameter would the eccentricity need to be in order to give a torsional stress equal to the transverse shear stress? (g) Calculate the direct shear stress that would result in the bar of (a) if it were the pin in a pinandclevis connection as in Figure P48 that is subjected to a 1000lb pull. (h) Calculate the direct bearing stress that would result on the bar of (a) if it were the pin in a pinandclevis connection as in Figure P48 that is subjected to a 1000lb pull if the center part (the eye or tongue) is 1in wide. (i) Calculate the maximum bending stress in the bar if it is formed into a semicircle with a centroidal radius of 10/π in and 1000lb opposing forces are applied at the ends and in the plane of the ends similar to Figure P415. Assume that there is no distortion of the cross section during bending.
Chapter 4
Table P44
STRESS, STRAIN, AND DEFLECTION
275
Data for Problems 475 through 478 Use only data that are relevant to the particular problem Lengths are in mm, forces in N, and moments in Nm
Row
D
d
h
M
P
a
40
20
4
10
80
8000
b
26
20
1
12
100
9500
c
36
30
1.5
8
60
6500
d
33
30
1
8
75
7200
e
21
20
1
10
50
5500
1.5
7
80
8000
5
8
400
15 000
f
51
50
g
101
100
r
Copyright © 2018 Robert L. Norton: All Rights Reserved
*475
For a filleted flat bar loaded in tension similar to that shown in Figure C9 (Appendix C) and the data from the row(s) assigned from Table P44, determine the nominal and maximum axial stress in the bar.
476 For a filleted flat bar loaded in bending similar to that shown in Figure C10 (Appendix C) and the data from the row(s) assigned from Table P44, determine the nominal and maximum bending stress in the bar. 477 For a shaft, with a shoulder fillet, loaded in tension similar to that shown in Figure C1 (Appendix C) and the data from the row(s) assigned from Table P44, determine the nominal and maximum axial stress in the shaft. 478 For a shaft, with a shoulder fillet, loaded in bending similar to that shown in Figure C2 (Appendix C) and the data from the row(s) assigned from Table P44, determine the nominal and maximum bending stress in the shaft. 479 A differential stress element has a set of applied stresses on it as shown in Figure 41. For σx = 850, σy = –200, σz = 300, τxy = 450, τyz = –300, and τzx = 0; find the principal stresses and maximum shear stress and draw the Mohr’s circle diagram for this threedimensional stress state. 480 Write expressions for the normalized (stress/pressure) tangential stress as a function of the normalized wall thickness (wall thickness/outside radius) at the inside wall of a thickwall cylinder and for a thinwall cylinder, both with internal pressure only. Plot the percent difference between these two expressions and determine the range of the wall thickness to outside radiusratio for which the stress predicted by the thinwall expression is at least 5% greater than that predicted by the thickwall expression. 481 A hollow square torsion bar such as that shown in Table 43 has dimensions a = 25 mm, t = 3 mm, and l = 300 mm. If it is made of steel with a modulus of rigidity of G = 80.8 GPa, determine the maximum shear stress in the bar and the angular deflection under a torsional load of 500 Nm. 482 Design a hollow rectangular torsion bar such as that shown in Table 43 that has dimensions a = 45 mm, b = 20 mm, and l = 500 mm. It is made of steel with a shear yield strength of 90 MPa and has an applied torsional load of 135 Nm. Use a factor of safety against yielding of 2. 483 A pressure vessel with closed ends has the following dimensions: outside diameter, OD = 450 mm, and wall thickness, t = 6 mm. If the internal pressure is 690 kPa, find the
* Answers
to these problems are provided in Appendix D.
MACHINE DESIGN
276

An Integrated Approach 
l
Sixth Edition
l
a
a w
w x
R1
R2 (a)
FIGURE P427
R3
x R1
R2
R3 ( b)
Copyright © 2018 Robert L. Norton: All Rights Reserved
Beams and Beam Loadings for Problem 485 to 486—see Table P42 for Data
principal stresses on the inside surface away from the ends. What is the maximum shear stress at the point analyzed? 484 A simply supported steel beam of length l with a concentrated load, F, acting at midspan has a rectangular crosssection of width, b, and depth, h. If the strain energy due to transverse shear loading is Us and that due to bending loading is Ub, derive an expression for the ratio Us / Ub and plot it as a function of h / l over the range 0.0 to 0.10. 485 A beam is supported and loaded as shown in Figure P427a. Find the reactions for the data given in row a from Table P42. 486 A beam is supported and loaded as shown in Figure P427b. Find the reactions for the data given in row a from Table P42. a
30°
w A leg t wheel assembly t = 20 mm w = 45 mm a = 75 mm P Copyright © 2018 Robert L. Norton: All Rights Reserved
FIGURE P428 Problems 487 and 488
487 Figure P428 shows a portion of the landing gear on a lightweight, experimental airplane. The landing gear attaches to the fuselage at A. Find the maximum shear stress on the crosssection of the leg at A if P = 3.5 kN. 488 Figure P428 shows a portion of the landing gear on a lightweight, experimental airplane. The landing gear attaches to the fuselage at A. Find the principal stresses on the inner and outer edges of the crosssection of the leg at A if P = 5 kN. 489 For the footpedal and pushrod linkage shown in Figure P429 and the data in the row(s) assigned from Table P43, determine the angular deflection of the tube between sections B and C. 490 For the footpedal and pushrod linkage shown in Figure P429 and the data in the row(s) assigned from Table P43, determine the maximum torsional shear stress in the tube. 491 For the footpedal and pushrod linkage shown in Figure P429 and the data in the row(s) assigned from Table P43, determine the maximum bending stress in the tube. 492 For the footpedal and pushrod linkage shown in Figure P429 and the data in the row(s) assigned from Table P43, determine the maximum applied and principal stresses in the tube. 493 A differential stress element has a set of applied stresses on it as shown in Figure 41. For σx = −2100, σy = 1575, σz = 0, τxy = −1100, τyz = 650, and τzx = 800; find the principal stresses and maximum shear stress and draw the Mohr’s circle diagram for this threedimensional stress state.
Chapter 4
A
a
B
STRESS, STRAIN, AND DEFLECTION
C
l
a
277
D
Z foot pedal
foot pedal
F
10 h
y tube
bearing
bearing z
x
OD
ID link
link push rod
P Z
push rod Section ZZ
FIGURE P429
Copyright © 2018 Robert L. Norton: All Rights Reserved
Problem 489 to 492—see Table P43 for Data
494 Figure P419 shows a 12mmdia, 250mmlong steel rod subjected to tensile loads applied at each end of the rod, acting along its longitudinal Y axis and through the centroid of its circular cross section. Before the tensile load was applied the points A and B were 100mm apart. After the load was applied they were 100.03 mm apart. For this bar with its loading, find: (a) All components of the strain tensor matrix (equation 4.3a) for a point midway between A and B. (b) All components of the stress tensor matrix (equation 4.1a) for a point midway between A and B. (c) The magnitude of the tensile load. 495 A 15mlong, vertical pole supports a 500N antenna at the top. Four steel guy wires are attached to the pole at an elevation of 12 m. The angle between the guy wires and the pole is 30° and they oppose each other in pairs at the point of attachment. The total load on the pole must not exceed 20 kN. What is the maximum tension allowed in each guy wire? If they have a 12mmdia, what is the stress in a crosssection far from either end and how much must each wire stretch to achieve the required load? 496 A beam is supported and loaded as shown in Figure P411a. Using Castigliano’s method find the deflection under the load F for the data given in the assigned row(s) from Table P42. 497 A beam is supported and loaded as shown in Figure P411b. Using Castigliano’s method find the deflection under the load F for the data given in the assigned row(s) from Table P42. 498 A beam is supported and loaded as shown in Figure P411b. Using the Beam Tables in Appendix B find the reactions, maximum shear, maximum moment, and maximum deflection for the data given in the assigned row(s) from Table P42. 499 A pressure vessel with closed ends has the following dimensions: outside diameter, OD = 4 in, and wall thickness, t = 0.5 in. If the internal pressure is 10 000 psi, find the principal stresses on the inside surface away from the ends. What is the maximum shear stress at the point analyzed?
10 t
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An Integrated Approach 
Sixth Edition
4100 A differential stress element has a set of applied stresses on it as shown in Figure 41. For σx = 1880, σy = −1255, σz = 240, τxy = 0, τyz = 1230, and τzx = 940; find the principal stresses and maximum shear stress and draw the Mohr’s circle diagram for this threedimensional stress state.
STATIC FAILURE THEORIES The whole of science is nothing more than a refinement of everyday thinking. Albert einstein
5.0
INTRODUCTION
Why do parts fail? This is a question that has occupied scientists and engineers for centuries. Much more is understood about various failure mechanisms today than was known even a few decades ago, largely due to improved testing and measuring techniques. If you were asked to respond to the above question based on what you have learned so far, you would probably say something like “Parts fail because their stresses exceed their strength,” and you would be right up to a point. The followup question is the critical one; what kind of stresses cause the failure: Tensile? Compressive? Shear? The answer to this one is the classic, “It depends.” It depends on the material in question and its relative strengths in compression, tension, and shear. It also depends on the character of the loading (whether static or dynamic) and on the presence or absence of cracks in the material. Table 50 shows the variables used in this chapter and references the equations or sections in which they are used. At the end of the chapter, a summary section is provided that also groups all the significant equations from this chapter for easy reference and identifies the chapter section in which their discussion can be found. Two master lecture videos on this chapter’s topics and a demonstration video are linked. Lecture 5 covers sections 5.1, and Lecture 6 covers sections 5.2–5.5. The demonstration video, Failure Modes, shows the effects of the failure theories in this chapter on test specimens. Figure 51a shows the Mohr’s circle for the stress state in a tensile test specimen. The tensile test (see Section 21) slowly applies a pure tensile loading to the part and causes a tensile, normal stress. However, the Mohr’s circle shows that a shear stress is also present, which happens to be exactly half as large as the normal stress. Which stress failed the part, the normal stress or the shear stress? 279
5
280
MACHINE DESIGN
Opening page public domain photograph courtesy of the NASALewis Research Center
Table 50 Symbol

An Integrated Approach

Sixth Edition
Variables Used in This Chapter Variable
ips units
SI units
See
a
halfwidth of crack
in
m
Sect. 5.3
b
halfwidth of cracked plate
in
m
Sect. 5.3
E
Young's modulus
psi
Pa
Sect. 5.1
K
stress intensity
kpsiin0.5
MPam0.5
Sect. 5.3
Kc
fracture toughness
kpsiin0.5
MPam0.5
Sect. 5.3
N NFM
safety factor
none
none
Sect. 5.1
safety factor for fracturemechanics failure
none
none
Sect. 5.3
Suc
ultimate compressive strength
psi
Pa
Sect. 5.2
Sut
ultimate tensile strength
psi
Pa
Sect. 5.2
Sy
tensile yield strength
psi
Pa
Eq. 5.8a, 5.9b
Sys
shear yield strength
psi
Pa
Eq. 5.9b, 5.10
U Ud
total strain energy
inlb
Joules
Eq. 5.1
distortion strain energy
inlb
Joules
Eq. 5.2
Uh
hydrostatic strain energy
inlb
Joules
Eq. 5.2
β
stressintensity geometry factor
none
none
Eq. 5.14c
ε
strain
none
none
Sect. 5.1
ν σ1
Poisson's ratio
none
none
Sect. 5.1
principal stress
psi
Pa
Sect. 5.1
τ
σ2
principal stress
psi
Pa
Sect. 5.1
σ3
principal stress
psi
Pa
Sect. 5.1
σ
σ
ModifiedMohr effective stress
psi
Pa
Eq. 5.12
σ′
von Mises effective stress
psi
Pa
Eq. 5.7
3
τ
σx
σ σ
σ
Figure 51b shows the Mohr’s circle for the stress state in a torsion test specimen. The torsion test (see Section 21) slowly applies a pure torsion loading to the part and causes a shear stress. However, the Mohr’s circle shows that a normal stress is also present, which happens to be exactly equal to the shear stress. Which stress failed the part, the normal stress or the shear stress?
(a)
τ τxy σ
σ τyx
(b)
FIGURE 51 Mohr's Circles for Unidirectional Tensile Stress (a) and Pure Torsion (b)
σ
σ
In general, ductile, isotropic materials in static tensile loading are limited by their shear strengths while brittle materials are limited by their tensile strengths (though there are exceptions to this rule when ductile materials can behave as if brittle). This situation requires that we have different failure theories for the two classes of materials, ductile and brittle. Recall from Chapter 2 that ductility can be defined in several ways, the most common being a material’s percent elongation to fracture, which, if >5%, is considered ductile. Most ductile metals have elongations to fracture >10%. Most importantly, we must carefully define what we mean by failure. A part may fail if it yields and distorts sufficiently to not function properly. Also, a part may fail by fracturing and separating. Either of these conditions is a failure, but the mechanisms causing them can be very different. Only ductile materials may yield significantly before fracturing. Brittle materials proceed to fracture without significant shape change. The
Chapter 5
Stress σ
STATIC FAILURE THEORIES
true
Sut y
Sy
u
f
281
Stress σ
true u
eng'g
y el pl
pl el (a)
f
(b)
eng'g
3 offset line E
E Elastic Range
Plastic Range
Strain ε
Strain ε
0.002
F I G U R E 2  2 Repeated Engineering and True StressStrain Curves for Ductile Materials: (a) LowCarbon Steel (b) Annealed HighCarbon Steel
Stress σ
stressstrain curves of each type of material reflect this difference, as shown in Figures 22 and 24, which are reproduced here for your convenience. Note that if cracks are present in a ductile material, it can suddenly fracture at nominal stress levels well below the yield strength, even under static loads. Another significant factor in failure is the character of the loading, whether it is static or dynamic. Static loads are slowly applied and remain essentially constant with time. Dynamic loads are either suddenly applied (impact loads) or repeatedly varied with time (fatigue loads), or both. The failure mechanisms are quite different in either case. Table 31 defined four classes of loading based on the motion of the loaded parts and the time dependence of the loading. By that definition, only Class 1 loading is static. The other three classes are dynamic to a greater or lesser degree. When the loading is dynamic, the distinction between ductile and brittle materials’ failure behavior blurs, and ductile materials fail in a “brittle” manner. Because of the significant differences in failure mechanisms under static and dynamic loading, we will consider them each separately, discussing failures due to static loading in this chapter and failures due to dynamic loading in the next chapter. For the static loading case (Class 1), we will consider the theories of failure separately for each type of material, ductile and brittle. 5.1
FAILURE OF DUCTILE MATERIALS UNDER STATIC LOADING View the Lecture 5 Video (46:02)* View a Demonstration Video (09:41)†
While ductile materials will fracture if statically stressed beyond their ultimate tensile strength, their failure in machine parts is generally considered to occur when they yield under static loading. The yield strength of a ductile material is appreciably less than its ultimate strength. Historically, several theories have been formulated to explain this failure: the maximum normalstress theory, the maximum normalstrain theory, the total strainenergy theory, the distortionenergy (von MisesHencky) theory, and the maximum shearstress theory. Of these only the last two agree closely with experimental data for this case, and of those, the von MisesHencky theory is the most accurate. We will discuss only the last two in detail, starting with the most accurate (and preferred) approach.
Sut
f
Sy
y offset line E
0.002
Strain ε
F I G U R E 2  4 Repeated StressStrain Curve of a Brittle Material
*
http://www.designofmachinery. com/MD/05_Ductile_Failure_ Theory.mp4 †
http://www.designofmachinery. com/MD/Failure_Modes.mp4
282
MACHINE DESIGN
An Integrated Approach

Sixth Edition
σ
The von MisesHencky or DistortionEnergy Theory
σi
The microscopic yielding mechanism is now understood to be due to relative sliding of the material’s atoms within their lattice structure. This sliding is caused by shear stress and is accompanied by distortion of the shape of the part. The energy stored in the part from this distortion is an indicator of the magnitude of the shear stress present.
E
3

U εi
ε
FIGURE 52
ToTal STrain EnErgy It was once thought that the total strain energy stored in the material was the cause of yield failure, but experimental evidence did not bear this out. The strain energy U in a unit volume (strain energy density) associated with any stress is the area under the stressstrain curve up to the point of the applied stress, as shown in Figure 52 for a unidirectional stress state. Assuming that the stressstrain curve is essentially linear up to the yield point, then we can express the total strain energy in a unit volume at any point in that range as
Internal Strain Energy Density in a Deflected Part
U=
1 σε 2
(5.1a)
Extending this to a threedimensional stress state gives U=
1 (σ1ε1 + σ 2 ε2 + σ 3 ε3 ) 2
(5.1b)
using the principal stresses and principal strains that act on planes of zero shear stress. This expression can be put in terms of principal stresses alone by substituting the relationships 1 (σ1 − νσ 2 − νσ 3 ) E 1 ε 2 = ( σ 2 − νσ1 − νσ 3 ) E 1 ε3 = ( σ 3 − νσ1 − νσ 2 ) E ε1 =
(5.1c)
where ν is Poisson’s ratio, giving U=
1 σ12 + σ 22 + σ 32 − 2 ν ( σ1σ 2 + σ 2 σ 3 + σ1σ 3 ) 2E
(5.1d )
HydroSTaTic loading Very large amounts of strain energy can be stored in materials without failure if they are hydrostatically loaded to create stresses that are uniform in all directions. This can be done in compression very easily by placing the specimen in a pressure chamber. Many experiments have shown that materials can be hydrostatically stressed to levels well beyond their ultimate strengths in compression without failure, as this just reduces the volume of the specimen without changing its shape. P. W. Bridgman subjected water ice to 1 Mpsi hydrostatic compression with no failure. The explanation is that uniform stresses in all directions, while creating volume change and potentially large strain energies, cause no distortion of the part and thus no shear stress. Consider the Mohr’s circle for a specimen subjected to σx = σy = σz = 1 Mpsi compressive stress. The Mohr’s “circle” is a point on the σ axis at –1 Mpsi and σ1 = σ2 = σ3. The shear stress is zero, so there is no distortion and no failure. This is true for ductile or brittle materials when the principal stresses are identical in magnitude and sign.
Chapter 5
STATIC FAILURE THEORIES
Den Hartog[1] describes the condition of rocks at great depth in the earth’s crust where they withstand uniform, hydrostatic compressive stresses of 5500 psi/mile of depth due to the weight of the rock above. This is in excess of their typical 3000 psi ultimate compressive strength as measured in a compression test. While it is much more difficult to create hydrostatic tension, Den Hartog[1] also describes such an experiment done by the Russian scientist Joffe in which he slowly cooled a glass marble in liquid air, allowed it to equilibrate to a stressfree state at the low temperature, then removed it to a warm room. As the marble warmed from the outside in, the temperature differential versus its cold core created uniform tensile stresses calculated to be well in excess of the material’s tensile strength, but it did not crack. Thus, it appears that distortion is the culprit in tensile failure as well. componEnTS of STrain EnErgy The total strain energy in a loaded part (Eq. 5.1d) can be considered to consist of two components—one due to hydrostatic loading, which changes its volume, and one due to distortion, which changes its shape. If we separate the two components, the distortionenergy portion will give a measure of the shear stress present. Let Uh represent the hydrostatic or volumetric component and Ud the distortionenergy component, then U = Uh + Ud
(5.2)
We can also express each of the principal stresses in terms of a hydrostatic (or volumetric) component σh that is common to each face and a distortion component σιd that is unique to each face, where the subscript i represents the principal stress direction, 1, 2, or 3: σ1 = σ h + σ1d σ 2 = σ h + σ 2d
(5.3a)
σ 3 = σ h + σ 3d
Adding the three principal stresses in equation 5.3a gives: σ1 + σ 2 + σ 3 = σ h + σ1d + σ h + σ 2d + σ h + σ 3d
(
σ1 + σ 2 + σ 3 = 3σ h + σ1d + σ 2d + σ 3d
(
)
3σ h = σ1 + σ 2 + σ 3 − σ1d + σ 2d + σ 3d
(5.3b)
)
For a volumetric change with no distortion, the term in parentheses in equation 5.3b must be zero, giving an expression for the volumetric or hydrostatic component of stress σh: σh =
σ1 + σ 2 + σ 3 3
(5.3c)
which you will note is merely the average of the three principal stresses. Now, the strain energy Uh associated with the hydrostatic volume change can be found by replacing each principal stress in equation 5.1d with σh: 1 2 σ h + σ 2h + σ 2h − 2 ν ( σ h σ h + σ h σ h + σ h σ h ) 2E 1 2 = 3σ h − 2 ν 3σ 2h 2E − ν 1 2 3( ) 2 Uh = σh E 2 Uh =
( )
(5.4 a)
283
3
284
MACHINE DESIGN

An Integrated Approach

Sixth Edition
and substituting equation 5.3c: 2 3 (1 − 2 ν ) σ1 + σ 2 + σ 3 E 2 3 1 − 2ν 2 = σ1 + σ 22 + σ 32 + 2 ( σ1σ 2 + σ 2 σ 3 + σ1σ 3 ) 6E
Uh =
3
(5.4 b)
diSTorTion EnErgy The distortion energy Ud is now found by subtracting equation 5.4b from 5.1d in accordance with equation 5.2: Ud = U − Uh
{
}
1 σ12 + σ 22 + σ 32 − 2 ν ( σ1σ 2 + σ 2 σ 3 + σ1σ 3 ) 2E 1 − 2ν 2 − σ1 + σ 22 + σ 32 + 2 ( σ1σ 2 + σ 2 σ 3 + σ1σ 3 ) 6E 1+ ν 2 σ1 + σ 22 + σ 32 − σ1σ 2 − σ 2 σ 3 − σ1σ 3 Ud = 3E =
{
} (5.5)
To obtain a failure criterion, we will compare the distortion energy per unit volume given by equation 5.5 to the distortion energy per unit volume present in a tensile test specimen at failure, because the tensile test is our principal source of materialstrength data. The failure stress of interest here is the yield strength Sy. The tensile test is a uniaxial stress state where, at yield, σ1 = Sy and σ2 = σ3 = 0. The distortion energy associated with yielding in the tensile test is found by substituting these values in equation 5.5: Ud =
1+ ν 2 Sy 3E
(5.6a)
and the failure criterion is obtained by equating the general expression 5.5 with the specific failure expression 5.6a to get 1+ ν 2 1+ ν 2 σ1 + σ 22 + σ 32 − σ1σ 2 − σ 2 σ 3 − σ1σ 3 S y = Ud = 3E 3E S y2 = σ12 + σ 22 + σ 32 − σ1σ 2 − σ 2 σ 3 − σ1σ 3
(5.6b)
S y = σ12 + σ 22 + σ 32 − σ1σ 2 − σ 2 σ 3 − σ1σ 3
which applies to the threedimensional stress state. * Note that this assumption will be consistent with the conventional ordering of principal stresses in the 3D case (σ1 > σ2 > σ3) only if σ3 σ3, σ2 = 0, then only the first and fourth quadrants of Figure 512 need to be drawn, as shown in Figure 513, which plots the stresses normalized by N / Sut where N is the safety factor. Figure 513 also depicts three planestress conditions labeled A, B, and C. Point A represents any stress state in which the two nonzero principal stresses, σ1, σ3 are positive. Failure will occur when the load line OA crosses the failure envelope at A’. The safety factor for this situation can be expressed as N=
Sut σ1
(5.12a)
Chapter 5
STATIC FAILURE THEORIES
297
3
FIGURE 512 Biaxial Fracture Data of Gray Cast Iron Compared to Various Failure Criteria (From Fig 7.13, p. 255, in Mechanical Behavior of Materials by N. E. Dowling, PrenticeHall, Englewood Cliffs, NJ, 1993.)
If the two nonzero principal stresses have opposite sign, then two possibilities exist for failure, as depicted by points B and C in Figure 513. The only difference between these two points is the relative values of their two stress components σ1, σ3. The load line OB exits the failure envelope at B’ above the point Sut, –Sut, and the safety factor for this case is given by equation 5.12a above. If the stress state is as depicted by point C, then the intersection of the load line OC and the failure envelope occurs at C’ below point Sut, –Sut. The safety factor can be found by solving for the intersection between the load line OC and the failure line. Write the equations for these lines and solve simultaneously to get the modified Mohr equation: N=
Sut Suc
Suc σ1 − Sut ( σ1 + σ 3 )
(5.12b)
If the stress state is in the fourth quadrant, both equations 5.12a and 5.12b should be checked and the smaller resulting safety factor used. Compare equation 5.12b to the lessaccurate equation for the unmodified CoulombMohr theory (which is not recommended for use): N=
Sut Suc Suc σ1 − Sut σ 3
To use the preferred modified Mohr theory of equation 5.12b, it would be convenient to have expressions for an effective stress that would account for all the applied stresses and allow direct comparison to a materialstrength property, as was done for ductile materials with the von Mises stress. Dowling[5] develops a set of expressions for this effective stress involving the three principal stresses:*
σ3
N ⋅ stress / Sut Sut, Sut
1.0 0
A’
A
O
σ1
B
–1.0
C
–2.0
B’ Sut, —Sut C’
–3.0 0 —Suc
–4.0 0
1.0
N ⋅ stress / Sut FIGURE 513 ModifiedMohr Failure Theory for Brittle Material
* See reference 5 for a complete derivation for both two and threedimensional CoulombMohr and modifiedMohr theories and the effective stress.
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MACHINE DESIGN

An Integrated Approach

C1 =
2 Sut − Suc 1 σ1 + σ 2 ) σ1 − σ 2 + ( 2 − Suc
C2 =
2 Sut − Suc 1 σ 2 + σ 3 ) σ 2 − σ3 + ( 2 − Suc
C3 =
2 Sut − Suc 1 σ 3 − σ1 + ( σ 3 + σ1 ) 2 − Suc
3
Sixth Edition
(5.12c)
The largest of the set of six values (C1, C2, C3, plus the three principal stresses) is the desired effective stress as suggested by Dowling: = MAX (C1 , C2 , C3 , σ1 , σ 2 , σ 3 ) σ = 0 if MAX < 0 σ
(5.12d )
where the signed function max denotes the algebraically largest of the six supplied arguments. If all of the arguments are negative, then the effective stress is zero. This modifiedMohr effective stress can now be compared to the ultimate tensile strength of the material to determine a safety factor: N=
Sut σ
(5.12e)
This approach allows easy computerization of the process.
EXAMPLE 52 Failure of Brittle Materials Under Static Loading Problem
Determine the safety factors for the bracket rod shown in Figure 59 (repeated on next page) based on the modifiedMohr theory.
Given
The material is class 50 gray cast iron with Sut = 52 500 psi and Suc = –164 000 psi. The rod length l = 6 in and arm a = 8 in. The rod outside diameter d = 1.5 in. Load F = 1000 lb.
Assumptions
The load is static and the assembly is at room temperature. Consider shear due to transverse loading as well as other stresses.
Solution
See Figures 59 and 433 and Examples 49 and 51.
1 The rod of Figure 59 is loaded both in bending (as a cantilever beam) and in torsion. The largest tensile bending stress will be in the top outer fiber at point A. The largest torsional shear stress will be all around the outer circumference of the rod. First take a differential element at point A where both of these stresses combine. Find the normal bending stress and torsional shear stress on point A using equations 4.11b and 4.23b: σx =
Mc 1000 ( 6 )( 0.75) = = 18 108 psi I 0.249
(a)
Chapter 5
STATIC FAILURE THEORIES
299
y l A
wall
d
rod
B
z
arm
3 x
F
A y
a
F I G U R E 5  9 Repeated
z
Bracket for Examples 51 and 52
B
τ xz =
Tr 1000 (8 )( 0.75) = = 12 072 psi J 0.497
T
(b)
2 Find the maximum shear stress and principal stresses that result from this combination of applied stresses using equations 4.6:
(a) Two points of interest for stress calculations
2 2 18 108 − 0 σ − σz τ max = x + τ 2xz = + 12 0722 = 15 090 psi 2 2
σ1 =
σx + σz
σ2 = 0
2
+ τ max =
18 108 2
F
y
+ 15 090 = 24 144 psi
x z
(c)
τxz
Tr/J
18 108 σ + σz σ3 = x − τ max = − 15 090 = −6036 psi 2 2
3 The principal stresses for point A can now be plotted on a modifiedMohr diagram as shown in Figure 514a. This shows that the load line crosses the failure envelope above the Sut,–Sut point, making equation 5.12a appropriate for the safetyfactor calculation: N=
Sut 52 400 = = 2.2 σ1 24 144
=
y x z
2 Sut − Suc 1 σ1 + σ 2 ) σ1 − σ 2 + ( 2 − Suc
(
)
V A
2 52 500 − 164 000 1 24 144 − 0 + 24 144 + 0 −164 000 2
(
) = 16 415 psi
(e)
)
2 52 500 − 164 000 1 = 0 − ( −6036 ) + ( 0 − 6036) = 1932 psi −164 000 2
Tr/J (c) Stress element at point B Repeated
2 Sut − Suc 1 C2 = σ 2 − σ 3 + (σ 2 + σ 3 ) 2 − Suc
(
Mc/I
(b) Stress element at point A
(d )
4 An alternative approach that does not require drawing the modifiedMohr diagram is to find the Dowling factors C1, C2, C3 using equation 5.12c: C1 =
σx
τzx
Note that these stresses are identical to those of Example 51.
FIGURE 433 (f)
Stress Elements at Points A and B within Cross Section of Rod for Example 410
300
MACHINE DESIGN
C3 = σ3
3
52.5 kpsi
= σ1
0

An Integrated Approach

Sixth Edition
2 Sut − Suc 1 σ 3 + σ1 ) σ 3 − σ1 + ( 2 − Suc
(
)
2 52 500 − 164 000 1 24 144 − ( −6036 ) + 24 144 − 6036 −164 000 2
(
) = 18 348
( g)
5 Then find the largest of the six stresses C1, C2, C3, σ1, σ2, σ3: = MAX (C1 , C2 , C3 , σ1 , σ 2 , σ 3 ) σ
A
(
(h)
)
= MAX 16 415, 1932, 18 348, 24 144, 0, − 6036 = 24 144 σ –52.5
which is the modifiedMohr effective stress. 6 The safety factor for point A can now be found using equation 5.12e: N=
Sut 52 500 = = 2.2 σ 24 144
(i )
which is the same as was found in step 3. –164 0
7 Since the rod is a short beam, we need to check the shear stress due to transverse loading at point B on the neutral axis. The maximum transverse shear stress at the neutral axis of a solid round rod was given as equation 4.15c:
24.1 52.5 kpsi
(a) Stresses at point A
τ bending =
σ3 52.5 kpsi σ1 B
–12.8
4V 4 (1000 ) = = 755 psi 3A 1.767
( j)
Point B is in pure shear. The total shear stress at point B is the algebraic sum of the transverse shear stress and the torsional shear stress, which both act on the same planes of the differential element and, in this case, act in the same direction as shown in Figure 433b. τ max = τtorsion + τ bending = 12 072 + 755 = 12 827 psi
(k )
8 Find the principal stresses for this pure shear loading: –52.5
σ1 = τ max = 12 827 psi σ2 = 0
(l )
σ 3 = −τ max = −12 827 psi
9 These principal stresses for point B can now be plotted on a modifiedMohr diagram as shown in Figure 514b. Because this is a pure shear loading, the load line crosses the failure envelope at the Sut’–Sut point, making equation 5.12a appropriate for the safetyfactor calculation:
–164 0 12.8
(b) Stresses at point B
FIGURE 514 Example 52
N=
52.5 kpsi
Sut 52 400 = = 4.1 σ1 12 827
(m)
10 To avoid drawing the modifiedMohr diagram, find the Dowling factors C1, C2, C3 using equations 5.12c: C1 = 8721 psi;
C2 = 4106 psi;
C3 = 12 827 psi;
(n)
Chapter 5
STATIC FAILURE THEORIES
301
11 And find the largest of the six stresses C1, C2, C3, σ1, σ2, σ3: = 12 827 psi σ
P
(o)
which is the modifiedMohr effective stress. 12 The safety factor for point B can now be found using equation 5.12e: 52 500 S N = ut = = 4.1 σ 12 827
( p)
and it is the same as was found in step 9.
c
13 The files EX0502 are on the book’s website.
5.3
P
FRACTURE MECHANICS
The static failure theories discussed so far have all assumed that the material is perfectly homogeneous and isotropic, and thus free of any defects such as cracks, voids, or inclusions, which could serve as stressraisers. This is seldom true for real materials. Actually, all materials are considered to contain microcracks too small to be seen with the naked eye. Dolan[6] says that “... every structure contains small flaws whose size and distribution are dependent upon the material and its processing. These may vary from nonmetallic inclusions and microvoids to weld defects, grinding cracks, quench cracks, surface laps, etc.” Scratches or gouges in the surface due to mishandling can also serve as incipient cracks. Functional geometric contours that are designed into the part may raise local stresses in predictable ways and can be taken into account in the stress calculations as was discussed in Chapter 4 (and will be further discussed in the next chapter). Cracks that occur spontaneously in service, due to damage or material flaws, are more difficult to predict and account for. The presence of a sharp crack in a stress field creates stress concentrations that theoretically approach infinity. See Figure 435 and equation 4.32a, which are repeated here for your convenience: a Kt = 1 + 2 c
3
a
(4.32a)
Kt
10 9 8 7 6 5 4 3 2 1 0 0 2 4
6 8 10 c/a
F I G U R E 4  3 5 Repeated Stress Concentration at the Edge of an Elliptical Hole in a Plate
Note that when the value of c approaches zero, the stress concentration, and thus the stress, approaches infinity. Since no material can sustain such high stresses, local yielding (for ductile materials), local microfracture (for brittle materials), or local crazing (for polymers) will occur at the crack tip[7]. If stresses are high enough at the tip of a crack of sufficient size, a sudden, “brittlelike” failure can result, even in ductile materials under static loads. The science of fracture mechanics has been developed to explain and predict this suddenfailure phenomenon. Cracks commonly occur in welded structures, bridges, ships, aircraft, land vehicles, pressure vessels, etc. Many catastrophic failures of tankers and Liberty Ships built during World War II occurred.* [8], [9] Twelve of these failures occurred shortly after the ships were launched and before they had sailed anywhere. They simply split in half while tied to the pier. One such ship is shown in Figure 515. The hull material was welded,
* “Nearly 80 ships broke in two, and almost 1000 were found to have deck plates with long brittle fractures.” D. A. Canonico, “Adjusting the Boiler Code,” Mechanical Engineering, Feb. 2000, p.56.
302
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Sixth Edition
3
FIGURE 515 WW II Tanker Cracked in Two While Berthed Prior to Being Placed in Service, Portland, Oregon, January 16, 1943 (Public Domain—Courtesy of the Ship Structures Committee, U.S. Government)
ductile steel and the ship had not yet been dynamically loaded to any significant degree. The nominal stresses were well below the material’s yield strength. Other examples of sudden failure at stresses below yield strength have occurred in this century, such as the Boston molassestank rupture in January 1919, which drowned 21 people and many horses under 2.3 million gallons of the sticky liquid.[10] A more recent example is the failure of a 22ftdia rocketmotor case while being pressure tested by the manufacturer. Figure 516 shows the pieces of the rocket case after failure. It “… was designed to stand proof pressures of 960 psi (but) failed … at 542 psi.”[11] These and other sudden “brittlelike” failures of ductile materials under static loading led researchers to seek better failure theories, since the ones then available could not adequately explain the observed phenomena. Where human life is at risk, as in bridges, aircraft, etc., periodic structuralsafety inspections for cracks are required by law or government regulation. These inspections can be by Xray, ultrasonic energy, or just be visual. When cracks are discovered, an engineering judgment must be made whether to repair or replace the flawed part, retire the assembly, or to continue it in service for a further time subject to more frequent inspection. (Many commercial aircraft currently flying contain structural cracks.) These decisions can now be made sensibly through the use of fracturemechanics theory. FractureMechanics Theory Fracture mechanics presumes the presence of a crack. The stress state in the region of the crack may be one of plane strain or plane stress (see Section 4.4). If the zone of yielding around the crack is small compared to the dimensions of the part, then linearelastic fracturemechanics (LEFM) theory is applicable. LEFM assumes that the bulk of the material is behaving according to Hooke’s law. However, if a significant portion of the
Chapter 5
STATIC FAILURE THEORIES
303
3
FIGURE 516 Failed RocketMotor Case (Public Domain—Courtesy of NASALewis Research Center)
bulk material is in the plastic region of its stressstrain behavior, then a more complicated approach is required than that described here. For the following discussion, we will assume that LEFM applies. modES of crack diSplacEmEnT Depending on the orientation of the loading versus the crack, the applied loads may tend to pull the crack open in tension (Mode I), shear the crack inplane (Mode II), or shear (tear) it outofplane (Mode III) as shown in Figure 517. Most of the fracturemechanics research and testing has been devoted to the tensile loading case (Mode I), and we will limit our discussion to it. STrESS inTEnSiTy facTor k Figure 518a shows a plate (not to scale) of width 2b under tension with a through crack of width 2a in the center. The crack is assumed to be sharp at its ends, and b is much larger than a. The crack’s cross section is in the xy plane. An rθ polar coordinate system is also set up in the xy plane with its origin at the crack tip as shown in Figure 518b. From the theory of linear elasticity, for b >> a the stresses around the crack tip, expressed as a function of the polar coordinates, are σx =
3θ θ θ K cos 1 − sin sin + 2 2 2 2 πr
(a) Mode I
(b) Mode II
(5.13a) (c) Mode III
σy =
τ xy =
3θ θ θ K cos 1 + sin sin + 2 2 2 2 πr
(5.13b)
θ θ 3θ K sin cos cos + 2 2 2 2 πr
(5.13c)
Copyright © 2018 Robert L. Norton All Rights Reserved
FIGURE 517 Three Modes of Crack Displacement
304
MACHINE DESIGN
P
or

An Integrated Approach
σz = 0

Sixth Edition
for plane stress
(
σz = ν σx + σ y
)
τ yz = τ zx = 0 2a
3
(5.13e)
with higherorder terms of small value omitted. Note that, when the radius r is zero, the xy stresses are infinite, which is consistent with equation 4.32b and Figure 436. The stresses diminish rapidly as r increases. The angle θ defines the geometric distribution of the stresses around the crack tip at any radius. The quantity K is called the stress intensity factor. (A subscript can be added to designate the mode I, II, III of loading as in KI, KII, KIII. Since we are dealing only with mode I loading, we will eliminate the subscript and let K = KI.)
a
b 2b
If we take the planestress case and compute the von Mises stress σ' from the x, y, and shear components (Eqs. 5.13a, b, c), we can plot the distribution of σ' versus θ for any chosen r as shown in Figure 519a for r = 1E–5 in and K = 1. The maximum occurs at 71°. If we set θ to that angle and compute the distribution of σ' as a function of r, it looks like Figure 519b, which plots r from 1E–5 to 1 in on a log scale.
P (a)
y
The high stresses near the crack tip cause local yielding and create a plastic zone of radius ry as shown (not to scale) in Figure 519c. For any radius r and angle θ, the stress state in this plastic zone at the crack tip is directly proportional to the stress intensity factor K. If b >> a, then K can be defined for the centercracked plate as
r θ
crack tip
x
(b)
FIGURE 518 A ThroughCrack in a Plate in Tension *
(5.13d )
for plane strain
The nominal stress for a fracturemechanics analysis is calculated based on the gross crosssectional area, without any reduction for the crack area. Note that this is different from the procedure used for calculation of nominal stress when using stressconcentration factors in a regular stress analysis. Then, the net cross section is used to find the nominal stress.
K = σ nom π a
a 0.025b
(b) Channel
b
2
h
A95 = 0.05bh
6
Sixth Edition
nonrotating
1
A95 = 0.076 6 d 2
d

2
nonrotating A951−1 = 0.10 bt
t
A952 − 2 = 0.05bh, t > 0.025b
1
(c) Solid rectangular
(d) I beam
FIGURE 625
Formulas for 95% Stressed Areas of Various Sections Loaded in Bending (Data from: Fig. 715, p. 294, Shigley and Mitchell, Mechanical Engineering Design, 4th ed., McGrawHill, New York, 1983)
Brinell hardness (HB) 120
160
200
240
280
320
1.0
360
400
440
520
240
260
mirrorpolished
0.9
fineground or commercially polished
0.8 0.7
mac
hine
0.6 surface factor, Csurf
480
d
0.5
hot
0.4
rolle
d
asfo
0.3
rged
0.2 corroded in tap water
0.1
corroded in salt water 0 60
80
FIGURE 626
100
120
140
160
180
200
220
tensile strength, Sut (kpsi)
Surface Factors for Various Finishes on Steel (Data from R. C. Juvinall, Stress, Strain, and Strength,
McGrawHill, New York, 1967, p. 234)
Chapter 6
Csurf
FATIGUE FAILURE THEORIES
1.0 4 2 1 0.9 16 8 32 63 0.8 125 250 0.7 500 1000 2000 0.6 surface finish Ra (µin) 0.5 0.4 40 60 80 100 120 140 160 180 200 220
369
240
Sut (kpsi) FIGURE 627 Surface Factor as a Function of Surface Roughness and Ultimate Tensile Strength (Data from R. C. Johnson, Machine Design, vol. 45, no. 11, 1967, p. 108, Penton Publishing, Cleveland, Ohio)
6
materials are more sensitive to the stress concentrations introduced by surface irregularities. In Figure 626 corrosive environments are seen to drastically reduce strength. These surface factors have been developed for steels and should only be applied to aluminum alloys and other ductile metals with the caution that testing of actual parts under realistic loading conditions be done in critical applications. Cast irons can be assigned a Csurf = 1, since their internal discontinuities dwarf the effects of a rough surface. R. C. Johnson[25] provides the data shown in Figure 627 that gives more detail for machined and ground surfaces by relating Csurf to tensile strength based on the measured surfaceaverageroughness Ra in microinches.* If Ra is known for a machined or ground part, Figure 627 can be used to determine a suitable surface factor Csurf. The surfacefactor curves in Figure 626 for hotrolled, forged, and corroded surfaces should still be used, as they account for decarburization and pitting effects as well as surface roughness. Shigley and Mischke[39] suggest using an exponential equation of the form Csurf ≅ A ( Sut )
b
if Csurf > 1.0, set Csurf = 1.0
(6.7e)
to approximate the surface factor with Sut in either kpsi or MPa. The coefficients A and exponents b for various finishes were determined from data similar to those in Figure 626 and are shown in Table 63. This approach has the advantage of being computer programmable and eliminates the need to refer to charts such as Figures 626 and 627. Table 63
Coefficients for SurfaceFactor Equation 6.7e Some data taken from Shigley, Mischke and Budynas, Mechanical Engineering Design, 7th ed., McGrawHill, New York, 2004, p. 329 For Sut in MPa, use
Surface Finish
A
b
For Sut in psi, use
A
b
Ground
1.58
–0.085
2.411
–0.085
Machined or coldrolled
4.51
–0.265
16.841
–0.265
Hotrolled Asforged
57.7 272
–0.718
2052.9
–0.718
–0.995
38 545.0
–0.995
*
There are many parameters used to characterize surface roughness, all of which are typically measured by passing a sharp, conical diamond stylus over the surface with controlled force and velocity. The stylus follows and encodes the microscopic contours and stores the surface profile in a computer. A number of statistical analyses are then done on the profile, such as finding the largest peaktopeak distance (Rt), the average of the 5 highest peaks (Rpm), etc. The most commonly used parameter is Ra (or Aa), which is the arithmetic average of the absolute values of the peak heights and valley depths. It is this parameter that is used in Figure 627. See Section 7.1 for more information on surface roughness.
370
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Sixth Edition
150 120 alternating stress, 90 σa (kpsi)
unplated
60
plated
30 0
100
101
102
103
104
105
106
107
108
cycles to failure, N FIGURE 628
Copyright © 2018 Robert L. Norton: All Rights Reserved
The Effect of Chrome Plating on Fatigue Strength of Steel (Data from C. C. Osgood, Fatigue Design, Pergamon Press, London, 1982)
Note that slightly different values of Csurf will be obtained from equation 6.7e than from the chart in Figure 626 due to the different data used to develop equation 6.7e and its factors in Table 63. Surface treatments such as electroplating with certain metals can severely reduce the fatigue strength, as shown in Figure 628 for chrome plating because electroplating typically creates residual tensile stress in the surface. Plating with soft metals such as cadmium, copper, zinc, lead, and tin appears not to severely compromise fatigue strength. Electroplating with chrome and nickel is generally not recommended for parts stressed in fatigue unless additional surface treatments such as shot peening (see below) are also applied. An exception may be when the part is in a corrosive environment and the corrosion protection afforded by the plating outweighs its strength reduction. The strength lost to plating can be recovered by shotpeening the surface after plating to introduce compressive stresses as shown in Figure 629, which shows the advantage of plating before shot peening. Choosing that order of application increases endurance strength above 70 alternating stress, σa (kpsi)
6
nickel plated, then peened
60
unplated, unpeened
50 40
peened, then nickel plated
30 nickel plated, unpeened 20 104
105
106
107
108
cycles to failure, N FIGURE 629
Copyright © 2018 Robert L. Norton: All Rights Reserved
The Effect of Nickel Plating and Shot Peening on Fatigue Strength of Steel (Data from Almen and Black, Residual Stress and Fatigue in Metals, McGrawHill, 1963, p. 148)
Chapter 6
FATIGUE FAILURE THEORIES
the unpeened material, but peening before plating has the opposite effect. Shot peening and other means of creating residual stresses are addressed in Section 6.8. temperature Fatigue tests are most commonly done at room temperature. The fracture toughness decreases at low temperatures and increases at moderately high temperatures (up to about 350°C). But, the endurancelimit knee in the SN diagram disappears at high temperatures, making the fatigue strength continue to decline with number of cycles, N. Also, the yield strength declines continuously with temperatures above room ambient and, in some cases, this can cause yielding before fatigue failure. At temperatures above about 50% of the material’s absolute melting temperature, creep becomes a significant factor and the stresslife approach is no longer valid. The strainlife approach can account for the combination of creep and fatigue under hightemperature conditions and should be used in those situations. Several approximate formulas have been proposed to account for the reduction in endurance limit at moderately high temperatures. A temperature factor Ctemp can be defined. Shigley and Mitchell[26] suggest the following: for T ≤ 450°C (840°F ) :
Ctemp = 1
for 450°C < T ≤ 550°C:
Ctemp = 1 − 0.0058 ( T − 450 )
for 840°F < T ≤ 1 020°F:
Ctemp = 1 − 0.0032 ( T − 840 )
(6.7 f )
Note that these criteria are based on data for steels and should not be used for other metals such as Al, Mg, and Cu alloys. reLiaBiLity Many of the reported strength data are mean values. There is considerable scatter in multiple tests of the same material under the same test conditions. Haugen and Wirsching[27] report that the standard deviations of endurance strengths of steels seldom exceed 8% of their mean values. Table 64 shows reliability factors for an assumed 8% standard deviation. Note that a 50% reliability has a factor of 1 and the factor reduces as you choose higher reliability. For example, if you wish to have 99.99% probability that your samples meet or exceed the assumed strength, multiply the mean
vacuum alternating stress, σa
air presoak corrosion fatigue 103
104
105 106 cycles to failure, N
FIGURE 630
107
108
Copyright © 2018 Robert L. Norton: All Rights Reserved
The Effect of Environment on Fatigue Strength of Steel (Data from Fuchs and Stephens, Metal Fatigue in Engineering, New York, 1980, p. 220)
371
Table 64 Reliability Factors for Sd = 0.08 µ Reliability % Creliab 50
1.000
90
0.897
95
0.868
99
0.814
99.9
0.753
99.99
0.702
99.999
0.659
99.9999
0.620
6
MACHINE DESIGN
endurance limit, Se (kpsi)
372

An Integrated Approach

Sixth Edition
100 80
in air (all steels)
60 (chromium steels)
40
in fresh water
20 0
(carbon and lowalloy steels) 0
40
80
120
160
200
240
tensile strength, Sut (kpsi) FIGURE 631
Copyright © 2018 Robert L. Norton: All Rights Reserved
The Effect of Fresh Water on Fatigue Strength of Steel (Data from P. G. Forrest, Fatigue of Metals, Pergamon Press, Oxford, 1962, p. 212)
6
strength value by 0.702. The values in Table 64 provide strengthreduction factors Creliab for chosen reliability levels. environment The environment can have significant effects on fatigue strength as is evidenced by the curves for corroded surfaces in Figure 626. Figure 630 shows schematically the relative effects of various environments on fatigue strength. Note that even room air reduces strength compared to vacuum. The higher the relative humidity and temperature, the larger will be the reduction of strength in air. The presoak line represents parts soaked in a corrosive environment (water or seawater) and then tested in room air. The increased roughness of the corroded surface is thought to be the reason for the loss of strength. The corrosion fatigue line shows a drastic reduction of strength and elimination of the endurancelimit knee. The corrosionfatigue phenomenon is not yet fully understood but empirical data such as those in Figures 630 and 631 depict its severity. Figure 631 shows the effect of operation in fresh water on the SN curves of carbon and lowalloy steels. The relationship between Se’ and Sut becomes constant at about 15 kpsi. So lowstrength carbon 250
σm = 0
air 3% NaCl solution
200 σa (MPa)
150
30
20
100 Al  7.5
50
σa (kpsi)
10
Zn  2.5 Mg
0
0 104
105
106
107
108
cycles to failure, N FIGURE 632
Copyright © 2018 Robert L. Norton: All Rights Reserved
The Effect of Saltwater on the Fatigue Strength of Aluminum (Data from Stubbington and Forsyth,
“Some CorrosionFatigue Observations on a HighPurity AluminumZincMagnesium Alloy and Commercial D.T.D. 683 Alloy," J. of the Inst. of Metals, London, U.K., vol. 90, 1961–1962, pp. 347–354)
Chapter 6
FATIGUE FAILURE THEORIES
373
steel is as good as highstrength carbon steel in this environment. The only steels that retain some strength in water are chromium steels (including stainless steels), since that alloying element provides some corrosion protection. Figure 632 shows the effects of saltwater on the fatigue strength of one alloy of aluminum. Only limited data are available for material strengths in severe environments. Thus, it is difficult to define any universal strengthreduction factors for environmental conditions. The best approach is to extensively test all designs and materials in the environment that they will experience. (This is difficult to do for situations in which the longterm effects of lowfrequency loading are desired because it will take an unreasonable time to obtain the data.) Based on Figure 631, for carbon and lowalloy steels in fresh water, the relationship between Se’ and Sut in equation 6.5a should be modified to Se ' ≅ 15 kpsi (100 MPa )
for carbon steel in fresh water
(6.8)
Presumably, a saltwater environment would be even worse.
6
Calculating the Corrected Fatigue Strength Sf or Corrected Endurance Limit Se The strengthreduction factors can now be applied to the uncorrected endurance limit S e’ or to the uncorrected fatigue strength Sf’ using equation 6.6 to obtain corrected values for design purposes. Creating Estimated SN Diagrams Equations 6.6 provide information about the material’s strength in the highcycle region of the SN diagram. With similar information for the lowcycle region, we can construct an SN diagram for the particular material and application as shown in Figure 633. The bandwidth of interest is the HCF regime from 103 to 106 cycles and beyond. Let the material strength at 103 cycles be called Sm. Test data indicate that the following estimates of Sm are reasonable:[28] bending:
Sm = 0.9 Sut
axial loading:
Sm = 0.75Sut
(6.9)
The estimated SN diagram can now be drawn on loglog axes as shown in Figure 633. The x axis runs from N = 103 to N = 109 cycles or beyond. The appropriate Sm Sn
Sn
(a)
Sm
(b)
Sm Se
Sf N
103 104 105 106 107 108 109 N1 N2
N 103 104 105 106 107 108 109 N1 N2
FIGURE 633 Estimated SN Curves for (a) Materials with Knee, (b) Materials Without Knee
374
MACHINE DESIGN

An Integrated Approach

Sixth Edition
from equation 6.9 for the type of loading is plotted at N = 103. Note that the correction factors from equation 6.6 are not applied to Sm. If the material exhibits a knee, then the corrected Se from equation 6.6 is plotted at Ne = 106 cycles and a straight line is drawn between Sm and Se. The curve is continued horizontally beyond that point. If the material does not exhibit a knee, then the corrected Sf from equation 6.6 is plotted at the number of cycles for which it was generated (shown at Nf = 5 x 108) and a straight line is drawn between Sm and Sf. This line may be extrapolated beyond that point, but its accuracy is questionable, though probably conservative (see Figure 610). The equation of the line from Sm to Se or Sf can be written as S (N ) = a N b
(6.10 a)
log S ( N ) = log a + b log N
(6.10 b)
or
6
where S(N) is the fatigue strength at any N and a, b are constants defined by the boundary conditions. For all cases, the y intercept is S(N) = Sm at N = N1 =103. For the endurancelimit case (Figure 633), S(N) = Se at N = N2 = 106. For a material that does not exhibit an endurancelimit knee, the fatigue strength is taken at some number of cycles: S(N) = Sf at N = N2 (Figure 633b). Substitute the boundary conditions in equation 6.10b and solve simultaneously for a and b: S 1 b = log m z Se
where
z = log N1 − log N 2
log ( a ) = log ( Sm ) − b log ( N1 ) = log ( Sm ) − 3b
(6.10c)
Note that N1 is always 1000 cycles and its log10 is 3. For a material with a knee at N2 = 106, z = (3 – 6) = –3 as shown in Table 65. This curve is valid only to the knee, beyond which S(N) = Se as shown in Figure 633a. For a material with no knee and S(N) = Sf at N = N2 (Figure 633b), the values of z corresponding to various values of N2 are easily calculated. For example, the curve in Figure 633b shows Sf at N2 = 5E8 cycles. The value for z is then
Table 65 zfactors for Eq. 610c
1.0E6 5.0E6 1.0E7 5.0E7 1.0E8 5.0E8 1.0E9 5.0E9
–3.000 –3.699 –4.000 –4.699 –5.000 –5.699 –6.000 –6.699
[email protected] E 8 = log(1000) − log(5E 8) = 3 − 8.699 = −5.699 S 1 [email protected] E 8 =– log m 5.699 Sf
(6.10 d )
for S f @ N 2 = 5E 8 cycles
The constants for any other boundary conditions are determined in the same way. Some values of z for a range of values of N2 with N1 set to 103 are shown in Table 65. These equations for the SN curve allow the estimated finite life N to be found for any fully reversed fatigue strength S(N), or the estimated fatigue strength S(N) can be found for any N.
Chapter 6
FATIGUE FAILURE THEORIES
375
EXAMPLE 61 Determining Estimated SN Diagrams for Ferrous Materials Problem
Create an estimated SN diagram for a steel bar and define its equations. How many cycles of life can be expected if the alternating stress is 100 MPa?
Given
The Sut has been tested at 600 MPa. The bar is 150 mm square and has a hotrolled finish. The operating temperature is 500°C maximum. The loading will be fully reversed bending.
Assumptions
Infinite life is required and is obtainable since this ductile steel will have an endurance limit. A reliability factor of 99.9% will be used.
Solution
6
1 Since no endurancelimit or fatigue strength information is given, we will estimate Se’ based on the ultimate strength using equation 6.5a. Se ' ≅ 0.5 Sut = 0.5 ( 600 ) = 300 MPa
(a)
2 The loading is bending so the load factor from equation 6.7a is Cload = 1.0
(b)
3 The part is larger than the test specimen and is not round, so an equivalent diameter based on its 95% stressed area (A95) must be determined and used to find the size factor. For a rectangular section in nonrotating bending loading, the A95 area is defined in Figure 625c and the equivalent diameter is found from equation 6.7d: A95 = 0.05bh = 0.05 (150 )(150 ) = 1125 mm 2 dequiv =
1125 mm 2 A95 = = 121.2 mm 0.0766 0.0766
(c)
and the size factor is found for this equivalent diameter from equation 6.7b: Csize = 1.189 (121.2 )−0.097 = 0.747
(d )
4 The surface factor is found from equation 6.7e and the data in Table 63 for the specified hotrolled finish. b Csurf = A Sut = 57.7 ( 600 )−0.718 = 0.584
(e)
5 The temperature factor is found from equation 6.7f: Ctemp = 1 − 0.0058 ( T − 450 ) = 1 − 0.0058 ( 500 − 450 ) = 0.71
(f)
6 The reliability factor is taken from Table 64 for the desired 99.9% and is Creliab = 0.753
7 The corrected endurance limit Se can now be calculated from equation 6.6:
( g)
376
MACHINE DESIGN
1000 500 300 200
σ (MPa)
100

An Integrated Approach
Sut

Sixth Edition
Sm
failure point
σa
σa Se
50 30 20 10
100
101
102
103
104
105
106
107
108
109
number of cycles, N FIGURE 634
6
SN Diagram and Alternating Stress Line Showing Failure Point for Example 61
Se = Cload Csize Csurf Ctemp Creliab Se ' = 1.0 ( 0.747 )( 0.584 )( 0.71)( 0.753)( 300 )
(h)
Se =70 MPa
8 To create the SN diagram, we also need a number for the estimated strength Sm at 103 cycles based on equation 6.9 for bending loading: Sm = 0.90 Sut = 0.90 ( 600 ) = 540 MPa
(i )
9 The estimated SN diagram is shown in Figure 634 with the above values of Sm and Se. The expressions of the two lines are found from equations 6.10a through 6.10c assuming that Se begins at 106 cycles: S 1 1 540 b = – log m = – log = −0.295 765 70 3 3 Se
(
)
log ( a ) = log ( Sm ) − 3b = log [ 540 ] − 3 −0.295 765 :
( j)
a = 4165.707
S ( N ) = a N b = 4165.707 N −0.295 765 MPa
103 ≤ N ≤ 106
S ( N ) = Se = 70 MPa
N > 106
(k )
10 The number of cycles of life for any alternating stress level can now be found from equations (k). For the stated stress level of 100 MPa, we get 100 = 4165.707 N −0.295 765
103 ≤ N ≤ 106
log100 = log 4165.707 − 0.295 765 log N 2 = 3.619 689 − 0.295 765 log N log N =
2 − 3.619 689 −0.295 765
(l )
= 5.476 270
N = 105.476 270 = 3.0 E 5 cycles
Figure 634 shows the intersection of the alternating stress line with the failure line at N = 3E5 cycles. 11 The files EX0601 are on the website.
Chapter 6
FATIGUE FAILURE THEORIES
377
EXAMPLE 62 Determining Estimated SN Diagrams for Nonferrous Materials Problem
Create an estimated SN diagram for an aluminum bar and define its equations. What is the corrected fatigue strength at 2E7 cycles?
Given
The Sut for this 6061T6 aluminum has been tested at 45 000 psi. The forged bar is 1.5 in round. The maximum operating temperature is 300°F. The loading is fully reversed torsion.
Assumptions
A reliability factor of 99.0% will be used. The uncorrected fatigue strength will be taken at 5E8 cycles.
Solution
1 Since no fatiguestrength information is given, we will estimate Sf ’ based on the ultimate strength using equation 6.5c: S f ' ≅ 0.4 Sut
(
)
for Sut < 48 kpsi
(a)
S f ' ≅ 0.4 45 000 = 18 000 psi
This value is at N = 5E8 cycles. There is no knee in an aluminum SN curve. 2 The loading is pure torsion, so the load factor from equation 6.7a is Cload = 1.0
(b)
because the applied torsional stress will be converted to an equivalent von Mises normal stress for comparison to the SN strength. 3 The part size is greater than the test specimen and it is round, so the size factor can be estimated with equation 6.7b, noting that this relationship is based on steel data:
(
Csize = 0.869 dequiv
)−0.097 = 0.869 (1.5)−0.097 = 0.835
(c)
4 The surface factor is found from equation 6.7e using the data in Table 63 for the specified forged finish, again with the caveat that these relationships were developed for steels and may be less accurate for aluminum: b Csurf = A Sut = 39.9 ( 45)−0.995 = 0.904
(d )
5 Equation 6.7f is only for steel so we will assume: Ctemp = 1
(e)
6 The reliability factor is taken from Table 64 for the desired 99.0% and is Creliab = 0.814
(f)
7 The corrected fatigue strength Sf at N = 5E8 can be calculated from equation 6.6: S f = Cload Csize Csurf Ctemp Creliab S f '
(
)
= 1.0 ( 0.835)( 0.904 )(1.0 )( 0.814 ) 18 000 = 11 063 psi
( g)
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378
MACHINE DESIGN

An Integrated Approach

Sixth Edition
8 To create the SN diagram, we also need a number for the estimated strength Sm at 103 cycles based on equation 6.9. Note that the bending value is used for torsion:
(
)
Sm = 0.90 Sut = 0.90 45 000 = 40 500 psi
(h)
9 The coefficient and exponent of the corrected SN line and its equation are found using equations 6.10a through 6.10c. The value of z is taken from Table 65 for Sf at 5E8 cycles: b=–
S 40 500 1 1 log m = – log = −0.0989 5.699 5.699 11 063 Sf
log ( a ) = log ( Sm ) − 3b = log 40 500 − 3 ( −0.0989 ) :
6
(i )
a = 80 193
10 The fatigue strength at the desired life of N = 2E7 cycles can now be found from the equation for the corrected SN line: S ( N ) = aN b = 80 193 N −0.0989 = 80 193 ( 2e 7 )−0.0989 = 15 209 psi
( j)
S(N) is larger than Sf because it is at a shorter life than the published fatigue strength. 11 Note the order of operations. We first found an uncorrected fatigue strength Sf’ at some “standard” cycle life (N = 5E8), then corrected it for the appropriate factors from equations 6.7. Only then did we create equation 6.10a for the SN line so that it passes through the corrected Sf at N = 5E8. If we had created equation 6.10a using the uncorrected Sf’, solved it for the desired cycle life (N = 2E7), and then applied the correction factors, we would get a different and incorrect result. Because these are exponential functions, superposition does not hold. 12 The files EX0602 are on the website.
6.7
NOTCHES AND STRESS CONCENTRATIONS
Notch is a generic term in this context and refers to any geometric contour that disrupts the “force flow” through the part as described in Section 4.15. A notch can be a hole, a groove, a fillet, an abrupt change in cross section, or any disruption to the smooth contours of a part. The notches of concern here are those that are deliberately introduced to obtain engineering features such as Oring grooves, fillets on shaft steps, fastener holes, etc. It is assumed that the engineer will follow good design practice and keep the radii of these notches as large as possible and reduce stress concentrations as described in Section 4.15. Notches of extremely small radii are poor design practice and, if present, should be treated as cracks and the principles of fracture mechanics used to predict failure (see Sections 5.3 and 6.57). A notch creates a stress concentration that raises the stresses locally and may even cause local yielding. In the discussion of stress concentration in Chapters 4 and 5 where only static loads were being considered, the effects of stress concentrations were only of concern for brittle materials. It was assumed that ductile materials would yield at the local stress concentration and lower the stress to acceptable levels. With dynamic loads, the situation is different, since ductile materials behave as if brittle in fatigue failures. The geometric (theoretical) stressconcentration factors Kt for normal stress and Kts for shear stress were defined and discussed in Section 4.15. (We will refer to them both here as Kt.) These factors give an indication of the degree of stress concentration
Chapter 6
FATIGUE FAILURE THEORIES
379
at a notch having a particular contour and are used as a multiplier on the nominal stress present in the cross section containing the notch (see equation 4.31). Many of these geometric or theoretical stressconcentration factors have been determined for various loadings and part geometries and are published in various references.[30], [31] For dynamic loading, we need to modify the theoretical stressconcentration factor based on the notch sensitivity of the material to obtain a fatigue stressconcentration factor, Kf, which can be applied to the nominal dynamic stresses. Notch Sensitivity Materials have different sensitivity to stress concentrations, which is referred to as the notch sensitivity of the material. In general, the more ductile the material, the less notch sensitive it is. Brittle materials are more notch sensitive. Since ductility and brittleness in metals are roughly related to strength and hardness, lowstrength, soft materials tend to be less notch sensitive than highstrength, hard ones. Notch sensitivity is also dependent on the notch radius (which is a measure of notch sharpness). As notch radii approach zero, the notch sensitivity of materials decreases and also approaches zero. This is quite serendipitous, since you will recall from Section 4.15 that the theoretical stressconcentration factor Kt approaches infinity as the notch radius goes to zero. If not for the reduced notch sensitivity of materials at radii approaching zero (i.e., cracks), engineers would be at a loss to design parts able to withstand any nominal stress level when notches are present. Neuber[32] made the first thorough study of notch effects and published an equation for the fatigue stressconcentration factor in 1937. Kuhn[33] later revised Neuber’s equation and experimentally developed data for the Neuber constant (a material property) needed in his equation. Peterson[30] subsequently refined the approach and developed the concept of notch sensitivity q defined as q=
K f −1 Kt − 1
(6.11a)
where Kt is the theoretical (static) stressconcentration factor for the particular geometry and Kf is the fatigue (dynamic) stressconcentration factor. The notch sensitivity q varies between 0 and 1. This equation can be rewritten to solve for Kf : K f = 1 + q ( K t − 1)
(6.11b)
The procedure is to first determine the theoretical stress concentration Kt for the particular geometry and loading, then establish the appropriate notch sensitivity for the chosen material and use them in equation 6.11b to find the dynamic stressconcentration factor Kf. The nominal dynamic stress for any situation is then increased by the factor Kf for tensile stress (Kfs for shear stress) in the same manner as was done for the static case: σ = K f σ nom τ = K fs τ nom
(6.12)
Note in equation 6.11 that when q = 0, Kf = 1 and it does not increase the nominal stress in equation 6.12. When q = 1, Kf = Kt and the entire effect of the geometric stressconcentration factor is felt in equation 6.12.
6
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An Integrated Approach

Sixth Edition
Neuber Constant,
a ( in0.5 )
0.5
0.4
Wrought Aluminums heat treated (T )
0.3
annealed or strainhardened (O and H )
0.2 Lowalloy Steels loaded in torsion 0.1
loaded in tension
6 0 0
40
80
120
160
200
240
Ultimate Tensile Strength, Sut (kpsi) FIGURE 635
Copyright © 2018 Robert L. Norton: All Rights Reserved
Neuber Constants for Steel and Aluminum (Data from P. Kuhn, el al,"NACA Technical Note 2805, 1952 and NACA Technical Note D1259, 1962)
The notch sensitivity q can also be defined from the KuhnHardrath formula[33] in terms of Neuber’s constant a and the notch radius r, both expressed in inches. Sut kpsi (MPa)
NotchSensitivity Factors for Steels (mm)
1.0
0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
200 160 140 120 100 80 70 60 50
5.0
0.9 0.8 0.7 q
0.6
Note: For torsional loading, use a curve for an Sut that is 20 kpsi higher than that of the material selected
0.5 0.4 0.3 0.2 0.1
(in)
0
1379 1103 965 827 689 552 483 414 345
0
FIGURE 636
0.02 Part 1
0.04
0.06
0.08
0.10
0.12
0.14
0.16
0.18
0.20
notch radius, r Copyright © 2018 Robert L. Norton: All Rights Reserved
NotchSensitivity Curves for Steels Calculated from Equation 6.13 Using NACA Data from Figure 635 and Table 66
Chapter 6
FATIGUE FAILURE THEORIES
1
q= 1+
381
(6.13)
a r
NotchSensitivity Factors for HeatTreated Aluminum (T) (mm)
1.0
0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
kpsi (MPa)
0.9 0.8 0.7 q
Sut
90
621
60
414
40
276
0.6
30
207
0.5
20
138
0.4 0.3 0.2 0.1 (in)
0
0
0.02
0.04
0.06
0.08
0.10
0.12
0.14
0.16
0.18
0.20
notch radius, r
(mm)
1.0
0
NotchSensitivity Factors for Annealed and StrainHardened Aluminum (O & H) 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0
kpsi (MPa)
0.9
45
310
35
241
0.7
25
172
0.6
20
138
0.5
15
103
0.8
q
Sut
0.4 0.3 0.2 0.1 (in)
0
0
0.02
FIGURE 636
Part 2
0.04
0.06
0.08 0.10 0.12 notch radius, r
0.14
0.16
0.18
0.20
Copyright © 2018 Robert L. Norton: All Rights Reserved
NotchSensitivity Curves for Aluminums Calculated from Equation 6.13 Using NACA Data from Figure 635 and Tables 67 and 68
6
382
MACHINE DESIGN
Table 66 Neuber's Constant for Steels Sut (ksi)
6
a (in0.5)
50 55 60 70 80 90 100 110 120 130 140
0.130 0.118 0.108 0.093 0.080 0.070 0.062 0.055 0.049 0.044 0.039
160 180 200 220 240
0.031 0.024 0.018 0.013 0.009
Table 67 Neuber's Constant for Annealed Aluminum Sut (kpsi)
a (in0.5)
10 15 20 25
0.500 0.341 0.264 0.217
30 35 40 45
0.180 0.152 0.126 0.111
Table 68 Neuber's Constant for Hardened Aluminum Sut (kpsi)
a (in0.5)
15 20 30
0.475 0.380 0.278
40 50 60 70 80
0.219 0.186 0.162 0.144 0.131
90
0.122

An Integrated Approach

Sixth Edition
Note that the Neuber constant is defined as the square root of a, not as a, so it is directly substituted in equation 6.13, while the value of r must have its square root taken. A plot of the Neuber constant √a for three types of materials is shown in Figure 635, and Tables 66 to 68 show data taken from that figure. Note in Figure 635 that, for torsional loads on steel, the value of √a should be read for Sut 20 kpsi higher than that of the material. Figure 636, parts 1 and 2 show sets of notch sensitivity curves for steels and aluminums (respectively) generated with equation 6.13 using the data in Figure 635 and Tables 66 through 68. These curves are for notches whose depth is less than four times the root radius and should be used with caution for deeper notches. The full value of the fatigue stressconcentration factor Kf applies only to the high end of the HCF regime (N = 106 – 109 cycles). Some authors [10], [30], [34] recommend applying a reduced portion of Kf, designated Kf’, to the alternating stress at N = 103 cycles. For highstrength or brittle materials Kf’ is nearly equal to Kf, but for lowstrength, ductile materials, Kf’ approaches 1. Others [35] recommend applying the full value of Kf even at 103 cycles. The latter approach is more conservative since data indicate that the effects of stress concentration are less pronounced at lower N. We will adopt the conservative approach and apply Kf uniformly across the HCF range, since the uncertainties surrounding the estimates of fatigue strength and its collection of modifying factors encourage conservatism.
EXAMPLE 63 Determining Fatigue StressConcentration Factors Problem
A rectangular, stepped bar similar to that shown in Figure 436 is to be loaded in bending. Determine the fatigue stressconcentration factor for the given dimensions.
Given
Using the nomenclature in Figure 436, D = 2, d = 1.8, and r = 0.25. The material has Sut = 100 kpsi.
Solution
1 The geometric stressconcentration factor Kt is found from the equation in Figure 436: b r Kt = A d
(a)
where A and b are given in the same figure as a function of the D / d ratio, which is 2 / 1.8 = 1.111. For this ratio, A = 1.014 7 and b = –0.217 9, giving 0.25 K t = 1.014 7 1.8
−0.217 9
= 1.56
(b)
2 The notch sensitivity q of the material can be found by using the Neuber factor √a from Figure 635 and Tables 66 to 68 in combination with equation 6.13, or by reading q directly from Figure 636. We will do the former. The Neuber factor from Table 66 for Sut = 100 kpsi is 0.062. Note that this is the square root of a:
Chapter 6
q=
1 a 1+ r
=
FATIGUE FAILURE THEORIES
1 = 0.89 0.062 1+ 0.25
383
(c)
3 The fatigue stressconcentration factor can now be found from equation 6.11b: K f = 1 + q ( K t − 1) = 1 + 0.89 (1.56 − 1) = 1.50
(d )
4 The files EX0603 are on the website.
6.8
RESIDUAL STRESSES
Residual stress refers to stresses that are “builtin” to an unloaded part. Most parts will have some residual stresses from their manufacturing processes. Any procedure such as forming or heat treatment that creates localized strains above the yield point will leave stresses behind when the strain is removed. Good design requires that the engineer try to tailor the residual stresses to, at a minimum, not create negative effects on the strength and preferably to create positive effects. Fatigue failure is a tensilestress phenomenon. Figures 617 and 618 show the beneficial effects of mean compressive stresses on fatigue strength. While the designer has little or no control over the presence or absence of a mean compressive stress in the loading pattern to which the part will be subjected, there are techniques that allow the introduction of compressive residual stresses in a part prior to its being placed in service. Properly done, these compressive residual stresses can make significant improvements in fatigue life. There are several methods for introducing compressive residual stresses: thermal treatments, surface treatments, and mechanical prestressing treatments. Most of them create biaxial compressive stresses at the surface, triaxial compressive stresses below the surface, and triaxial tensile stresses in the core. Since the part is in equilibrium, the compressive stresses near the surface have to be balanced by tensile stresses in the core. If the treatment is overdone, the increased tensile core stresses can cause failure, so a balance must be struck. These treatments have the greatest value when the applied stress distribution due to loading is nonuniform and is maximally tensile at the surface, as in reversed bending. Bending in one direction will have peak tensile stress on one side only and the treatment can then be applied just to that side. Axialtension loading is uniform across the section and so will not benefit from a nonuniform residual stress pattern, unless there are notches at the surface to cause local increases in tensile stress. Then compressive residual stresses at the surface are very helpful. In fact, regardless of loading, the net tensile stresses at notches will be reduced by adding residual compressive stresses in those locations. Since designed notches are usually at the surface, a treatment can be applied to them. The deliberate introduction of residual compressive stresses is most effective on parts made of highyieldstrength materials. If the yield strength of the material is low, then the residual stresses may not stay long in the part due to later yielding from high applied stresses in service. Faires [36] found that steels with Sy 1.0, set Csurf = 1.0
for T ≤ 450°C (840°F ) :
Ctemp = 1
for 450°C < T ≤ 550°C:
Ctemp = 1 − 0.0058 ( T − 450 )
for 840°F < T ≤ 1 020°F:
Ctemp = 1 − 0.0032 ( T − 840 )
437
(6.7e)
(6.7 f )
Corrected Fatigue Strength Estimates (Section 6.6):
Se = Cload Csize Csurf Ctemp Creliab Se ' S f = Cload Csize Csurf Ctemp Creliab S '
(6.6)
f
Approximate Strength at 1000 Cycles (Section 6.6):
bending:
Sm = 0.9 Sut
axial loading:
Sm = 0.75Sut
(6.9)
SN Diagram (Section 6.6):
log S ( N ) = log a + b log N S 1 b = log m z Se
(6.10 b)
z = log N1 − log N 2
where
(6.10c)
log ( a ) = log ( Sm ) − b log ( N1 ) = log ( Sm ) − 3b Notch Sensitivity (Section 6.7):
1
q= 1+
(6.13)
a r
Fatigue StressConcentration Factors (Sections 6.7 and 6.11):
K f = 1 + q ( K t − 1)
(6.11b)
if K f σ maxnom < S y then:
K fm = K f
if K f σ maxnom > S y then:
K fm =
if K f σ maxnom − σ minnom > 2 S y then:
K fm = 0
S y − K f σ anom σ mnom
(6.17)
Safety Factor  Fully Reversed Stresses (Section 6.10):
Nf =
Sn σ'
(6.14)
ModifiedGoodman Diagram (Section 6.11):
−
σ 'm σ 'a + =1 S yc S yc
(6.16a)
6
438
MACHINE DESIGN

An Integrated Approach

Sixth Edition
σ 'a = S f
(6.16b)
σ 'm σ 'a + =1 Sut Sf
(6.16c)
σ 'm σ 'a + =1 Sy Sy
(6.16d )
σ' Sy
Safety Factor  Fluctuating σ 'Stresses − a S y 6.11): m @Q = 1 (Section
Case 1:
Sy σ 'a = 1 − σ σ Sy ' ' [email protected] σZ'm m = 1 − Sf S σ 'm @ Q
Nf = σ 'a @ P
6
ut
Case 2:
(
Case 4:
(6.18b)
)
Sut S f 2 − S f σ 'a + Sut σ 'm 2 S f Sut σ' 2 N f = Smf @+R S=ut ' ' σ σ S [email protected] Z a ut + σ 'm S f Sf =− σ 'm @ S ) + S f ( Sut
σ 'm @ S = σ 'a @ S
σ 'm 1 − S ut
Sf σ 'a @ P = σ 'a @ Z σ 'a
Nf = Case 3:
(6.18a)
(6.18e)
( σ ' m − σ ' m @ S )2 + ( σ ' a − σ ' a @ S )2
ZS =
(6.18 f )
( σ ' a )2 + ( σ ' m )2
OZ =
OZ + ZS OZ
Nf =
(6.18 g)
Sines Method for Multiaxial Stresses in Fatigue  3D (Section 6.12):
σ 'a =
(σ
xa
− σ ya
) + (σ 2
ya
− σ za
) + (σ 2
− σ xa
za
) + 6( τ 2
2 xya
+ τ 2yza + τ 2zxa
)
2
(6.21a)
σ 'm = σ xm + σ ym + σ zm Sines Method for Multiaxial Stresses in Fatigue  2D (Section 6.12):
σ 'a = σ 2xa + σ 2ya − σ xa σ ya + 3τ 2xya
(6.21b)
σ 'm = σ xm + σ ym General Multiaxial Stresses in Fatigue  3D (Section 6.12):
σ 'a =
σ 'm =
(σ
xa
− σ ya
) + (σ 2
ya
− σ za
) + (σ 2
za
− σ xa
) + 6( τ 2
2 xya
+ τ 2yza + τ 2zxa
)
2
(σ
xm
− σ ym
) + (σ 2
ym
− σ zm
) + (σ 2
zm
2
− σ xm
) + 6( 2
τ 2xym
+ τ 2yzm
+ τ 2zxm
)
(6.22a)
Chapter 6
FATIGUE FAILURE THEORIES
439
General Multiaxial Stresses in Fatigue  2D (Section 6.12):
σ 'a = σ 2xa + σ 2ya − σ xa σ ya + 3τ 2xya
(6.22b)
σ 'm = σ 2xm + σ 2ym − σ xm σ ym + 3τ 2xym SEQA Method for Complex Multiaxial Stresses in Fatigue (Section 6.12): 1
σ SEQA = 2
3 2 3 2 9 4 2 1 + Q + 1 + Q cos 2φ + Q 2 16 4
(6.23)
Fracture Mechanics in Fatigue (Section 6.5):
K = βσ max πa − βσ min πa = β πa ( σ max − σ min ) da = A (∆K )n dN
6.16
(6.3b)
(6.4 a)
REFERENCES
1 R. P. Reed, J. H. Smith, and B. W. Christ, The Economic Effects of Fracture in the United States: Part I, Special Pub. 6471, U. S. Dept. of Commerce, National Bureau of Standards, Washington, D.C., 1983. 2 N. E. Dowling, Mechanical Behavior of Materials. PrenticeHall: Englewood Cliffs, N.J., p. 340, 1993. 3 J. W. Fischer and B. T. Yen, Design, Structural Details, and Discontinuities in Steel, Safety and Reliability of Metal Structures, ASCE, Nov. 2, 1972. 4 N. E. Dowling, Mechanical Behavior of Materials. PrenticeHall: Englewood Cliffs, N.J., p. 355, 1993. 5 D. Broek, The Practical Use of Fracture Mechanics. Kluwer Academic Publishers: Dordrecht, The Netherlands, p. 10, 1988. 6 N. E. Dowling, Mechanical Behavior of Materials. PrenticeHall: Englewood Cliffs, N.J., p. 347, 1993. 7 R. C. Juvinall, Engineering Considerations of Stress, Strain and Strength. McGrawHill: New York, p. 280, 1967. 8 J. E. Shigley and C. R. Mischke, Mechanical Engineering Design. 5th ed. McGrawHill: New York, p. 278, 1989. 9 A. F. Madayag, Metal Fatigue: Theory and Design. John Wiley & Sons: New York, p. 117, 1969. 10 N. E. Dowling, Mechanical Behavior of Materials. PrenticeHall: Englewood Cliffs, N.J., p. 418, 1993. 11 R. C. Juvinall, Engineering Considerations of Stress, Strain and Strength. McGrawHill: New York, p. 231, 1967.
6
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MACHINE DESIGN

An Integrated Approach

Sixth Edition
12 J. A. Bannantine, J. J. Comer, and J. L. Handrock, Fundamentals of Metal Fatigue. PrenticeHall: Englewood Cliffs, N. J., p. 13, 1990. 13 P. C. Paris and F. Erdogan, A Critical Analysis of Crack Propagation Laws. Trans. ASME, J. Basic Eng., 85(4): p. 528, 1963. 14 J. M. Barsom, FatigueCrack Propagation in Steels of Various Yield Strengths. Trans. ASME, J. Eng. Ind., Series B(4): p. 1190, 1971. 15 H. O. Fuchs and R. I. Stephens, Metal Fatigue in Engineering. John Wiley & Sons: New York, p. 88, 1980. 16 J. A. Bannantine, J. J. Comer, and J. L. Handrock, Fundamentals of Metal Fatigue. PrenticeHall: Englewood Cliffs, N. J., p. 106, 1990. 17 J. M. Barsom and S. T. Rolfe, Fracture and Fatigue Control in Structures, 2nd ed. PrenticeHall: Englewood Cliffs, N. J., p. 256, 1987.
6
18 H. O. Fuchs and R. I. Stephens, Metal Fatigue in Engineering. John Wiley & Sons: New York, p. 89, 1980. 19 J. M. Barsom and S. T. Rolfe, Fracture and Fatigue Control in Structures. 2nd ed. PrenticeHall: Englewood Cliffs, N. J., p. 285, 1987. 20 P. G. Forrest, Fatigue of Metals. Pergamon Press: London, 1962. 21 J. E. Shigley and L. D. Mitchell, Mechanical Engineering Design. 4th ed. McGrawHill: New York, p. 293, 1983. 22 R. Kuguel, A Relation Between Theoretical StressConcentration Factor and Fatigue Notch Factor Deduced from the Concept of Highly Stressed Volume. Proc. ASTM, 61: pp. 732748, 1961. 23 R. C. Juvinall, Engineering Considerations of Stress, Strain and Strength. McGrawHill: New York, p. 233, 1967. 24 Ibid., p. 234. 25 R. C. Johnson, Machine Design, vol. 45, p. 108, 1973. 26 J. E. Shigley and L. D. Mitchell, Mechanical Engineering Design. 4th ed. McGrawHill: New York, p. 300, 1983. 27 E. B. Haugen and P. H. Wirsching, “Probabilistic Design.” Machine Design, vol. 47, pp. 1014, 1975. 28 R. C. Juvinall and K. M. Marshek, Fundamentals of Machine Component Design. 2nd ed. John Wiley & Sons: New York, p. 270, 1967. 29 Ibid., p. 267. 30 R. E. Peterson, StressConcentration Factors. John Wiley & Sons: New York, 1974. 31 R. J. Roark and W. C. Young, Formulas for Stress and Strain. 5th ed. McGrawHill: New York, 1975. 32 H. Neuber, Theory of Notch Stresses. J. W. Edwards Publisher Inc.: Ann Arbor Mich., 1946. 33 P. Kuhn and H. F. Hardrath, An Engineering Method for Estimating Notchsize Effect in Fatigue Tests on Steel. Technical Note 2805, NACA, Washington, D.C., Oct. 1952.
Chapter 6
FATIGUE FAILURE THEORIES
441
34 R. B. Heywood, Designing Against Fatigue. Chapman & Hall Ltd.: London, 1962. 35 R. C. Juvinall, Engineering Considerations of Stress, Strain and Strength. McGrawHill: New York, p. 280, 1967. 36 V. M. Faires, Design of Machine Elements. 4th ed. Macmillan: London, p. 162, 1965. 37 R. B. Heywood, Designing Against Fatigue of Metals. Reinhold: New York, p. 272, 1962. 38 H. O. Fuchs and R. I. Stephens, Metal Fatigue in Engineering. John Wiley & Sons: New York, p. 130, 1980. 39 J. E. Shigley and C. R. Mischke, Mechanical Engineering Design. 5th ed. McGrawHill: New York, p. 283, 1989. 40 N. E. Dowling, Mechanical Behavior of Materials. PrenticeHall: Englewood Cliffs, N J., p. 416, 1993. 41 R. C. Juvinall, Engineering Considerations of Stress, Strain and Strength. McGrawHill: New York, p. 280, 1967. 42 G. Sines, Failure of Materials under Combined Repeated Stresses Superimposed with Static Stresses, Technical Note 3495, NACA, 1955. 43 F. S. Kelly, A General Fatigue Evaluation Method, Paper 79PVP77, ASME, New York, 1979. 44 Y. S. Garud, A New Approach to the Evaluation of Fatigue Under Multiaxial Loadings, in Methods for Predicting Material Life in Fatigue, W. J. Ostergren and J. R. Whitehead, ed. ASME: New York. pp. 249263, 1979. 45 M. W. Brown and K. J. Miller, A Theory for Fatigue Failure under Multiaxial StressStrain Conditions. Proc. Inst. Mech. Eng., 187(65): pp. 745755, 1973. 46 J. O. Smith, The Effect of Range of Stress on the Fatigue Strength of Metals. Univ. of Ill., Eng. Exp. Sta. Bull., (334), 1942. 47 J. E. Shigley and L. D. Mitchell, Mechanical Engineering Design. 4th ed. McGrawHill: New York, p. 333, 1983. 48 B. F. Langer, Design of Vessels Involving Fatigue, in Pressure Vessel Engineering, R. W. Nichols, ed. Elsevier: Amsterdam. pp. 59100, 1971. 49 J. A. Collins, Failure of Materials in Mechanical Design. 2nd ed. J. Wiley & Sons: New York, pp. 238254, 1993. 50 S. M. Tipton and D. V. Nelson, Fatigue Life Predictions for a Notched Shaft in Combined Bending and Torsion, in Multiaxial Fatigue, K. J. Miller and M. W. Brown, Editors. ASTM: Philadelphia, PA. pp. 514550, 1985. 51 R. C. Rice, ed. Fatigue Design Handbook. 2nd ed. Soc. of Automotive Engineers: Warrendale, PA. p. 260, 1988. 52 T. Nishihara and M. Kawamoto, The Strength of Metals under Combined Alternating Bending and Torsion with Phase Difference. Memoirs College of Engineering, Kyoto Univ., Japan, 11(85), 1945. 53 Shot Peening of Gears. American Gear Manufacturers Association AGMA 938A05, 2005.
6
442
MACHINE DESIGN
Table P60† Topic/Problem Matrix Sect. 6.4 Fatigue Loads 61 Sect. 6.5 Fracture Mechanics 651, 652, 653, 672, 673, 674, 682, 683 Sect. 6.6 S/N Diagrams 62, 64, 65, 654, 655, 656, 657, 675, 684, 688, 692
6
Sect. 6.7 Stress Conc. 615, 658, 659, 660, 663 to 666, 676, 677, 678, 685, 689, 693 Sect. 6.10 Fully Reversed 66, 67, 68, 620, 626, 629, 633, 635, 637, 646, 648, 649, 650, 686, 687, 690 Sect. 6.11 Fluctuating 63, 69, 610, 611, 612, 613, 614, 616, 617, 618, 619, 621, 622, 623, 624, 625, 627, 628, 630, 631, 632, 634, 636, 638, 640, 643, 644, 645, 670, 679, 680, 681, 691 Sect. 6.12 Multiaxial 639, 641, 642, 661, 662, 667, 668, 669 †
Problem numbers in boldface are extended from similar problems in earlier chapters with the same dash number. Problem numbers in italics are design problems.
6.17

An Integrated Approach

Sixth Edition
BIBLIOGRAPHY
For more information on fatigue design, see:
J. A. Bannantine, J. J. Comer, and J. L. Handrock, Fundamentals of Metal Fatigue. PrenticeHall: Englewood Cliffs, N.J., 1990. H. E. Boyer, ed., Atlas of Fatigue Curves. Amer. Soc. for Metals: Metals Park, Ohio. 1986. H. O. Fuchs and R. I. Stephens, Metal Fatigue in Engineering. John Wiley & Sons: New York, 1980. R. C. Juvinall, Engineering Considerations of Stress, Strain and Strength. A. J. McEvily, ed. Atlas of StressCorrosion and Corrosion Fatigue Curves. Amer. Soc. for Metals: Metals Park, Ohio. 1990. McGrawHill: New York, 1967. For more information on the strainlife approach to lowcycle fatigue, see:
N. E. Dowling, Mechanical Behavior of Materials. PrenticeHall: Englewood Cliffs, N.J., 1993. R. C. Rice, ed. Fatigue Design Handbook. 2nd ed. Soc. of Automotive Engineers: Warrendale, PA. 1988. For more information on the fracture mechanics approach to fatigue, see:
J. M. Barsom and S. T. Rolfe, Fracture and Fatigue Control in Structures. 2nd ed. PrenticeHall: Englewood Cliffs, N.J., 1987. D. Broek, The Practical Use of Fracture Mechanics. Kluwer Academic Publishers: Dordrecht, The Netherlands, 1988. For more information on residual stresses, see:
J. O. Almen and P. H. Black, Residual Stresses and Fatigue in Metals. McGrawHill: New York, 1963. For more information on multiaxial stresses in fatigue, see:
A. Fatemi and D. F. Socie, A Critical Plane Approach to Multiaxial Fatigue Damage Including OutofPhase Loading. Fatigue and Fracture of Engineering Materials and Structures, 11(3): pp. 149165, 1988. Y. S. Garud, Multiaxial Fatigue: A Survey of the State of the Art. J. Test. Eval., 9(3), 1981. K. F. Kussmaul, D. L. McDiarmid and D. F. Socie, eds. Fatigue Under Biaxial and Multiaxial Loading. Mechanical Engineering Publications Ltd.: London. 1991. G. E. Leese and D. Socie, eds. Multiaxial Fatigue: Analysis and Experiments. Soc. of Automotive Engineers: Warrendale, Pa., 1989. K. J. Miller and M. W. Brown, eds. Multiaxial Fatigue. Vol. STP 853. ASTM: Philadelphia, Pa., 1985. G. Sines, Behavior of Metals Under Complex Static and Alternating Stresses, in Metal Fatigue, G. Sines and J. L. Waisman, eds. McGrawHill: New York. 1959. R. M. Wetzel, ed., Fatigue Under Complex Loading: Analyses and Experiments. SAE Pub. No. AE6, Soc. of Automotive Engineers: Warrendale, Pa., 1977.
Chapter 6
6.18 *61
FATIGUE FAILURE THEORIES
PROBLEMS For the data in the row(s) assigned in Table P61, find the stress range, alternating stress component, mean stress component, stress ratio, and amplitude ratio.
62 For the steelmaterial strength data in the row(s) assigned in Table P62, calculate the uncorrected endurance limit and draw a strengthlife (SN) diagram. *63
For the bicycle pedal arm assembly in Figure P61 assume a riderapplied force that ranges from 0 to 1 500 N at the pedal each cycle. Determine the fluctuating stresses in the 15mmdia pedal arm. Find the fatigue safety factor if Sut = 500 MPa.
64 For the aluminummaterial strength data in the row(s) assigned in Table P62, calculate the uncorrected fatigue strength at 5E8 cycles and draw a strengthlife (SN) diagram for the material.
*67 *68
to these problems are provided in Appendix D. Problem numbers in italics are design problems. Problem numbers in boldface are extended from similar problems in earlier chapters with the same dash number. Problems in succeeding chapters may also continue and extend these problems.
Table P61 Data for Problem 61
σ max
σ min
a
1000
0
b
1000
–1000
c
1500
500
d
1500
–500
e
500
–1000
Design the wrist pin of Problem 37 for infinite life with a safety factor of 1.5 if the 2500g acceleration is fully reversed and Sut = 130 kpsi.
f
2500
–1200
g
0
–4500
A paper machine processes rolls of paper having a density of 984 kg/m3.
h
2500
1500
For the trailer hitch from Problem 36 (also see Figure P62 and Figure 15), find the infinitelife fatigue safety factors for all modes of failure, assuming that the horizontal impact force of the trailer on the ball is fully reversed. Use steel with Sut = 600 MPa and Sy = 450 MPa.
The paper roll is 1.50m outside dia (OD) x 0.22m inside dia (ID) x 3.23 m long and is on a simply supported, hollow, steel shaft with Sut = 400 MPa. Find the shaft ID needed to obtain a dynamic safety factor of 2 for a 10year life running three 8hour shifts per day if the shaft OD is 22 cm and the roll turns at 50 rpm.
69 For the ViseGrip® plierwrench drawn to scale in Figure P63, and for which the forces were analyzed in Problem 39 and stresses in Problem 49, find the safety factors for
Table P63 Row
* Answers
Row
65 For the data in the row(s) assigned in Table P63, find the corrected endurance strength (or limit), create equations for the SN line, and draw the SN diagram. *66
443
Data for Problem 65
Material
Sut kpsi
Shape
Size inches
Surface Finish
Loading
Temp °F
Reliability
a
steel
110
round
2
ground
torsion
b
steel
90
square
4
mach.
axial
c
steel
80
Ibeam
16 x 18*
hot rolled
bending
d
steel
200
round
5
forged
torsion
e
steel
150
square
7
cold rolled
axial
f
aluminum
70
round
9
mach.
bending
ground
torsion
room
99.9
cold rolled
axial
room
99.0
900
50
room
90
round
4
ground
bending
room
99.99
40
square
6
forged
torsion
room
99.999
k
ductile iron
70
round
5
cast
axial
room
50
room
90
50
90
bronze
80
square
6
cast
axial
alum.
alum.
60
n
steel
25 000
35 000
99.999
aluminum
torsion
150 000
e
h
–50
aluminum
bending
d
alum.
j
forged
steel
99.99
i
cast
steel
120 000
room
9
9
250 000
c
alum.
24 x 36*
7
b
40 000
Ibeam
round
steel
70 000
square
square
90 000
g
85
60
a
f
50
90
mat'l
99.0
aluminum
bronze
sut (psi)
99.9
aluminum
ductile iron
Row
600
g
l
Data for Problems 62, 64
room
h
m
Table P62
212
99.999
6
444
MACHINE DESIGN

An Integrated Approach

Sixth Edition
60 mm
P F
170 mm
F
P F
6
0.5cm grid
FIGURE P63
T
Problem 69 (A Solidworks model of this is on the website)
FIGURE P61 Problem 63 (A Solidworks
each pin for an assumed clamping force of P = 4000 N in the position shown. The steel pins are 8mm dia with Sy = 400 MPa, Sut = 520 MPa, and are all in double shear. Assume a desired finite life of 5E4 cycles.
model is on the book’s website)
40 mm *610
An overhung diving board is shown in Figure P64a. A 100kg person is standing on the free end. Assume crosssectional dimensions of 305 mm x 32 mm. What is the fatigue safety factor for finite life if the material is brittle fiberglass with Sf = 39 MPa @ N = 5E8 cycles and Sut = 130 MPa in the longitudinal direction?
*611
Repeat Problem 610 assuming the 100kg person in Problem 610 jumps up 25 cm and lands back on the board. Assume the board weighs 29 kg and deflects 13.1 cm statically when the person stands on it. What is the fatigue safety factor for finite life if the material is brittle fiberglass with Sf = 39 MPa @ N = 5E8 cycles and Sut = 100 MPa in the longitudinal direction?
FIGURE P62 Problem 66 (A Solidworks model is on the book’s website)
612 Repeat Problem 610 using the cantilevered diving board design in Figure P64b. 613 Repeat Problem 611 using the diving board design shown in Figure P64b. Assume the board weighs 19 kg and deflects 8.5 cm statically when the person stands on it.
* Answers
to these problems are provided in Appendix D. Problem numbers in italics are design problems. Problem numbers in boldface are extended from similar problems in earlier chapters with the same dash number. Problems in succeeding chapters may also continue and extend these problems.
614 Figure P65 shows a child’s toy called a pogo stick. The child stands on the pads, applying half her weight on each side. She jumps up off the ground, holding the pads up against her feet, and bounces along with the spring cushioning the impact and storing energy to help each rebound. Assume a 60lb child and a spring constant of 100 lb/in. The pogo stick weighs 5 lb. Design the aluminum cantilever beam sections on which she stands to survive jumping 2 in off the ground with a dynamic safety factor of 2 for a finite life of 5E4 cycles. Use 2000 series aluminum. Define the beam shape and size. 2m
0.7 m
(a) Overhung diving board
FIGURE P64 Problems 610 through 613
2m P
0.7 m
(b) Cantilevered diving board
P
Chapter 6
Table P64
FATIGUE FAILURE THEORIES
445
Data For Problem 615 (kpsi)
Row
(in)
Material
Loading
a
100
3.3
0.25
steel
bending
b
90
2.5
0.55
steel
torsion
c
150
1.8
0.75
steel
bending
d
200
2.8
1.22
steel
torsion
e
20
3.1
0.25
soft aluminum
bending
f
35
2.5
0.28
soft aluminum
bending
g
45
1.8
0.50
soft aluminum
bending
h
50
2.8
0.75
hard aluminum
bending
i
30
3.5
1.25
hard aluminum
bending
j
90
2.5
0.25
hard aluminum
bending
6 W/2
*615
W/2
For a notched part having notch dimension r, geometric stressconcentration factor Kt, material strength Sut, and loading as shown in the assigned row(s) of Table P64, find the Neuber factor a, the material’s notch sensitivity q, and the fatigue stressconcentration factor Kf.
616 A track to guide bowling balls is designed with two round rods as shown in Figure P66. The rods have a small angle between them. The balls roll on the rods until they fall between them and drop onto another track. Each rod’s unsupported length is 30 in and the angle between them is 3.2°. The balls are 4.5india and weigh 2.5 lb. The center distance between the 1india rods is 4.2 in at the narrow end. Find the infinitelife safety factor for the 1india SAE 1010 coldrolled steel rods. (a) Assume rods are simply supported at each end. (b) Assume rods are fixed at each end. *617
A pair of ice tongs is shown in Figure P67. The ice weighs 50 lb and is 10in wide across the tongs. The distance between the handles is 4 in, and the mean radius r of a steel tong is 6 in. The rectangular crosssectional dimensions are 0.750 in x 0.312 in. Find the safety factor for the tongs for 5E5 cycles if their Sut = 50 kpsi.
618 Repeat Problem 617 with the tongs made of Class 40 gray cast iron.
FIGURE P66 Problem 616
P FIGURE P65 Problem 614 * Answers
to these problems are provided in Appendix D. Problem numbers in italics are design problems. Problem numbers in boldface are extended from similar problems in earlier chapters with the same dash number. Problems in succeeding chapters may also continue and extend these problems.
446
MACHINE DESIGN
F F
A
An Integrated Approach

Sixth Edition
Determine the size of the clevis pin shown in Figure P68 needed to withstand an applied repeated force of 0 to 130 000 lb for infinite life. Also determine the required outside radius of the clevis end to not fail in either tearout or bearing if the clevis flanges each are 2.5 in thick. Use a safety factor of 3. Assume Sut = 140 kpsi for the pin and Sut = 80 kpsi for the clevis.
620 A ±100 Nm torque is applied to a 1mlong, solid, round steel shaft. Design it to limit its angular deflection to 2° and select a steel alloy to have a fatigue safety factor of 2 for infinite life.
r
W
6
*619

FIGURE P67
621 Figure P69 shows an automobile wheel with two styles of lug wrench, a singleended wrench in (a), and a doubleended wrench in (b). The distance between points A and B is 1 ft in both cases and the handle diameter is 0.625 in. How many cycles of tightening can be expected before a fatigue failure if the average tightening torque is 100 ftlb and the material Sut = 60 kpsi? *622
Problem 617
P
An inline “rollerblade” skate is shown in Figure P610. The polyurethane wheels are 72mm dia and spaced on 104mm centers. The skatebootfoot combination weighs 2 kgf. The effective “spring rate” of the personskate system is 6000 N/m. The 10mmdia axle pins are in double shear and of steel with Sut = 550 MPa. Find the infinitelife fatigue safety factor for the pins when a 100kg person lands a 0.5m jump on one foot. (a) Assume that all four wheels land simultaneously. (b) Assume that one wheel absorbs all the landing force.
P FIGURE P68
*623
The beam in Figure P611a is subjected to a sinusoidal force time function with Fmax = F and Fmin = –F / 2, where F and the beam’s other data are given in the row(s) assigned from Table P65. Find the stress state in the beam due to this loading and choose a material specification that will give a safety factor of 3 for N = 5E8 cycles.
*624
The beam in Figure P611b is subjected to a sinusoidal force time function with Fmax = F N and Fmin = F / 2, where F and the beam’s other data are given in the row(s) assigned from Table P65. Find the stress state in the beam due to this loading and choose a material specification that will give a safety factor of 1.5 for N = 5E8 cycles.
*625
The beam in Figure P611c is subjected to a sinusoidal force time function with Fmax = F and Fmin = 0, where F and the beam’s other data are given in the row(s) assigned from
Problem 619
axle
axle
3 in
3 in A
B
lug wrench
A
lug wrench B F
F FIGURE P610 Problem 622 * Answers
to these problems are provided in Appendix D. Problem numbers in italics are design problems. Problem numbers in boldface are extended from similar problems in earlier chapters with the same dash number. Problems in succeeding chapters may also continue and extend these problems.
F
FIGURE P69 Problem 621
(a)
tire
F
tire (b)
Chapter 6
Table P65
FATIGUE FAILURE THEORIES
447
l
Data for Problems 623 through 626
Use Only Data Relevant to a Particular Problem. Lengths in m, Forces in N, I in m4
b
F
Row a
1.00
0.40
0.60
500
2.85E–08
2.00E–02
steel
b
0.70
0.20
0.40
850
1.70E–08
1.00E–02
steel
c
0.30
0.10
0.20
450
4.70E–09
1.25E–02
steel
d
0.80
0.50
0.60
250
4.90E–09
1.10E–02
steel
e
0.85
0.35
0.50
750
1.80E–08
9.00E–03
steel
f
0.50
0.18
0.40
950
1.17E–08
1.00E–02
steel
g
0.60
0.28
0.50
250
3.20E–09
7.50E–03
steel
h
0.20
0.10
0.13
500
4.00E–09
5.00E–03
alum.
i
0.40
0.15
0.30
200
2.75E–09
5.00E–03
alum.
j
0.20
0.10
0.15
80
6.50E–10
5.50E–03
alum.
k
0.40
0.16
0.30
880
4.30E–08
1.45E–02
alum.
l
0.90
0.25
0.80
600
4.20E–08
7.50E–03
alum.
m
0.70
0.10
0.60
500
2.10E–08
6.50E–03
alum.
n
0.85
0.15
0.70
120
7.90E–09
1.00E–02
alum.
x R1
R2 (a )
l F
M1
x R1
l b
Table P65. Find the stress state in the beam due to this loading and choose a material specification that will give a safety factor of 2.5 for N = 5E8 cycles. *626
*627
The beam in Figure P611d is subjected to a sinusoidal force time function with Fmax = F lb and Fmin = –F, where F and the beam’s other data are given in the row(s) assigned from Table P65. Find the stress state in the beam due to this loading and choose a material specification that will give a safety factor of 6 for N = 5E8 cycles.
F
x R2
R1 (c)
A storage rack is to be designed to hold the paper roll of Problem 68 as shown in Figure P612. Determine a suitable value for dimension a in the figure for an infinitelife fatigue safety factor of 2. Assume dimension b = 100 mm and that the mandrel is solid and inserts halfway into the paper roll: (more overleaf)
l b
(a) If the beam is a ductile material with Sut = 600 MPa. (b) If the beam is a castbrittle material with Sut = 300 MPa.
a
628 Figure P613 shows a forklift truck negotiating a 15° ramp to drive onto a 4fthigh loading platform. The truck weighs 5000 lb and has a 42in wheelbase. Design two (one for each side) 1ftwide ramps of steel to have a safety factor of 2 for infinite life in the worst case of loading as the truck travels up them. Minimize the weight of the ramps
(b )
F x
R1
R2
R3
(d )
b stanchion
FIGURE P611
paper roll
Beams and Beam Loadings for Problems 623 to 626: See Table P65 for Data
a mandrel FIGURE P612 Problem 627
* Answers
base
to these problems are provided in Appendix D. Problem numbers in italics are design problems. Problem numbers in boldface are extended from similar problems in earlier chapters with the same dash number. Problems in succeeding chapters may also continue and extend these problems.
6
448
MACHINE DESIGN

An Integrated Approach

Sixth Edition
ramp 1ft grid
FIGURE P613 Problem 628
by using a sensible crosssectional geometry. Choose an appropriate steel or aluminum alloy.
6 * Answers
to these problems are provided in Appendix D. Problem numbers in italics are design problems. Problem numbers in boldface are extended from similar problems in earlier chapters with the same dash number. Problems in succeeding chapters may also continue and extend these problems.
*629
A bar 22 mm x 30 mm in cross section is loaded axially in tension with F(t) = ±8 kN. A 10mm hole passes through the center of the 30mm side. Find the safety factor for infinite life if the material has Sut = 500 MPa.
630 Repeat Problem 6.29 with Fmin = 0, Fmax = 16 kN. *631
Repeat Problem 6.29 with Fmin = 8 kN, Fmax = 24 kN.
632 Repeat Problem 6.29 with Fmin = –4 kN, Fmax = 12 kN. *633
The bracket in Figure P614 is subjected to a sinusoidal force time function with Fmax = F and Fmin = –F, where F and the beam’s other data are given in the row(s) assigned from Table P66. Find the stress states at points A and B due to this fully reversed loading and choose a ductile steel or aluminum material specification that will give a safety factor of 2 for infinite life if steel or N = 5E8 cycles if aluminum. Assume a geometric stressconcentration factor of 2.5 in bending and 2.8 in torsion.
*634
The bracket in Figure P614 is subjected to a sinusoidal force time function with Fmax = F and Fmin = 0, where F and the beam’s other data are given in the row(s) assigned from Table P66. Find the stress states at points A and B due to this repeated loading and choose a ductile steel or aluminum material specification that will give a safety factor of 2 for infinite life if steel or N = 5E8 cycles if aluminum. Assume a geometric stressconcentration factor of 2.8 in bending and 3.2 in torsion.
635 Repeat Problem 633 using a cast iron material. 636 Repeat Problem 634 using a cast iron material.
F
l a A
tube
y
B wall
t arm
z x od
id
FIGURE P614 Problems 633 to 636 (A Solidworks model of this is on the book’s website)
h
Chapter 6
Table P66
FATIGUE FAILURE THEORIES
449
Data for Problems 633 through 636 Use only data that are relevant to the particular problem. Lengths in mm, forces in N.
Row
l
a
t
h
F
od
id
E
a
100
400
10
20
50
20
14
steel
b
70
200
6
80
85
20
6
steel
c
300
100
4
50
95
25
17
steel
d
800
500
6
65
160
46
22
alum.
e
85
350
5
96
900
55
24
alum.
f
50
180
4
45
950
50
30
alum.
g
160
280
5
25
850
45
19
steel
h
200
100
2
10
800
40
24
steel
i
400
150
3
50
950
65
37
steel
j
200
100
3
10
600
45
32
alum.
k
120
180
3
70
880
60
47
alum.
l
150
250
8
90
750
52
28
alum.
m
70
100
6
80
500
36
30
steel
n
85
150
7
60
820
40
15
steel
*637
A semicircular curved beam as shown in Figure P615 has od = 150 mm, id = 100 mm and t = 25 mm. For a load pair F = ±3 kN applied along the diameter, find the fatigue safety factor at the inner and outer fibers: (a) If the beam is steel with Sut = 700 MPa. (b) If the beam is cast iron with Sut = 420 MPa.
638 A 42mmdia steel shaft with a 19mm transverse hole is subjected to a sinusoidal combined loading of σ = ±100 MPa bending stress and a steady torsion of 110 MPa. Find its safety factor for infinite life if Sut = 1 GPa. *639
A 42mmdia steel shaft with a 19mm transverse hole is subjected to a combined loading of σ = ±100 MPa bending stress and an alternating torsion of ±110 MPa, which are 90° outofphase. Find its safety factor for infinite life if Sut = 1 GPa.
640 Redesign the roll support of Problem 68 to be like Figure P616. The mandrels insert to 10% of the roll length. Design dimensions a and b for an infinitelife safety factor of 2. (a) If the beam is steel with Sut = 600 MPa. (b) If the beam is cast iron with Sut = 300 MPa. *641
A 10mmID steel tube carries liquid at 7 MPa. The pressure varies periodically from zero to maximum. The steel has Sut = 400 MPa. Determine the infinitelife fatigue safety factor for the wall if its thickness is: (a) 1 mm. (b) 5 mm.
642 A cylindrical tank with hemispherical ends is required to hold 150 psi of pressurized air at room temperature. The pressure cycles from zero to maximum. The steel has Sut = 500 MPa. Determine the infinitelife fatigue safety factor if the tank diameter is 0.5 m, the wall thickness is 1 mm, and its length is 1 m.
t
OD ID
F F
6 FIGURE P615 Problem 637 (A Solidworks
model is on the book’s website)
* Answers
to these problems are provided in Appendix D. Problem numbers in italics are design problems. Problem numbers in boldface are extended from similar problems in earlier chapters with the same dash number. Problems in succeeding chapters may also continue and extend these problems.
450
MACHINE DESIGN

An Integrated Approach

Sixth Edition
b typ stanchion paper roll
a typ
mandrel base
FIGURE P616
Problem 640 (A Solidworks model of this is on the book’s website)
643 The paper rolls in Figure P617 are 0.9mOD x 0.22mID x 3.23 m long and have a density of 984 kg/m3. The rolls are transferred from the machine conveyor (not shown) to the forklift truck by the Vlinkage of the offload station, which is rotated through 90° by an air cylinder. The paper then rolls onto the waiting forks of the truck. The forks are 38mmthick by 100mmwide by 1.2 m long and are tipped at a 3° angle from the horizontal and have Sut = 600 MPa. Find the infinitelife fatigue safety factor for the two forks on the truck when the paper rolls onto it under two different conditions (state all assumptions):
6
(a) The two forks are unsupported at their free end. (b) The two forks are contacting the table at point A. *
Problem numbers in italics are design problems. Problem numbers in boldface are extended from similar problems in earlier chapters with the same dash number. Problems in succeeding chapters may also continue and extend these problems.
644 Determine a suitable thickness for the Vlinks of the offloading station of Figure P617 to limit their deflections at the tips to 10 mm in any position during their rotation. Two Vlinks support the roll, at the 1/4 and 3/4 points along the roll’s length, and each of the V arms is 10 cm wide by 1 m long. What is their infinitelife fatigue safety factor when designed to limit deflection as above? Sut = 600 MPa. See Problem 443 for more information. 645 Determine the infinitelife fatigue safety factor based on the tension load on the aircylinder rod in Figure P617. The tension load cycles from zero to maximum (compression
Vlinks
1m crank arm
paper rolling machine forks A rod FIGURE P617 Problems 643 to 647
offloading station
air cylinder
forklift truck
Chapter 6
FATIGUE FAILURE THEORIES
451
rollers FIGURE P618 Problem 648
loads below the critical buckling load will not affect fatigue life). The crank arm that it rotates is 0.3 m long and the rod has a maximum extension of 0.5 m. The 25mmdia rod is solid steel with Sut = 600 MPa. State all assumptions.
6
646 The Vlinks of Figure P617 are rotated by the crank arm through a shaft that is 60mm dia by 3.23m long. Determine the maximum torque applied to this shaft during the motion of the Vlinkage and find the infinitelife fatigue safety factor for the shaft if its Sut = 600 MPa. See Problem 643 for more information. *647
Determine the maximum forces on the pins at each end of the air cylinder of Figure P617. Determine the infinitelife fatigue safety factor for these pins if they are 30mm dia and in single shear. Sut = 600 MPa. See Problem 643 for more information.
648 Figure P618 shows an exerciser for a 100kg wheelchair racer. The wheelchair has 65cmdia drive wheels separated by a 70cm track width. Two freeturning rollers on bearings support the rear wheels. The lateral movement of the chair is limited by the flanges. Design the 1mlong rollers as hollow tubes of aluminum (select alloy) to minimize the height of the platform and also limit the roller deflections to 1 mm in the worst case. Specify suitably sized steel axles to support the tubes on bearings. Calculate the fatigue safety factors at a life of N = 5E8 cycles.
* Answers
to these problems are provided in Appendix D. Problem numbers in italics are design problems. Problem numbers in boldface are extended from similar problems in earlier chapters with the same dash number. Problems in succeeding chapters may also continue and extend these problems.
649 Figure P619 shows a machined pivot pin that is a pressfit into part A and is a slip fit in part B. If F = 100 lb, l = 2 in, and d = 0.5 in, what is the pin’s safety factor against fatigue when made of SAE 1020 coldrolled steel? The loading is fully reversed and a reliability of 90% is desired. There is a bending stress concentration factor Kt = 1.8 at the section where the pin leaves part A on the righthand side.
l l 2 l 4
650 Figure P619 shows a machined pivot pin that is a pressfit into part A and is a slip fit in part B. If F = 100 N, l = 50 mm, and d = 16 mm, what is the pin’s safety factor against fatigue when made of class 50 cast iron? The loading is fully reversed and a reliability of 90% is desired. There is a bending stress concentration factor Kt = 1.8 at the section where the pin leaves part A on the righthand side. 651 A component in the shape of a large sheet is to be fabricated from 7075T651 aluminum, which has fracture toughness Kc = 24.2 MPam0.5 and a tensile yield strength of 495 MPa. Determine the number of loading cycles that can be endured if the nominal stress varies from 0 to onehalf the yield strength and the initial crack had a total length of 1.2 mm. The values of the coefficient and exponent in equation 6.4 for this material n n are A = 5 x 10–11 ( mm/cycles) MPa ( m ) and n = 4. *652
A component in the shape of a large sheet is to be fabricated from SAE 4340 steel, which has a fracture toughness Kc = 98.9 MPam0.5. The sheets are inspected for crack flaws after fabrication, but the inspection device cannot detect flaws smaller than 5 mm. Determine the minimum thickness required for the sheet to have a minimum cycle life
F
B d A
Copyright © 2018 Robert L. Norton: All Rights Reserved
FIGURE P619 Problems 649 and 650
452
MACHINE DESIGN

An Integrated Approach

Sixth Edition
of 106 cycles (using fracturemechanics criteria) if its width is 400 mm and the load normal to the crack varies from 20 to 170 kN. The values of the coefficient and exponent in n n equation 6.4 for this material are A = 4 x 10–9 ( mm/cycles) MPa ( m ) and n = 3. 653 A closed, thinwall cylinder is made from an aluminum alloy that has a fracture toughness of 38 MPam0.5 and has the following dimensions: length = 200 mm, OD = 84 mm, and ID = 70 mm. A 2.8mmdeep semicircular crack is discovered on the inner diameter away from the ends, oriented along a line parallel to the cylinder axis. If the cylinder is repeatedly pressurized from 0 to 75 MPa, how many pressure cycles can it withstand? The values of the coefficient and exponent in equation 6.4 for this material are A = 5 x n n 10–12 ( mm/cycles) MPa ( m ) and n = 4. (Hint: The value of the geometry factor for a semicircular surface flaw is β = 2 / p and the crack grows in the radial direction.) 654 A nonrotating, hotrolled, steel beam has a channel section with h = 64 mm and b = 127 mm. It is loaded in repeated bending with the neutral axis through the web. Determine its corrected fatigue strength with 90% reliability if it is used in an environment that has a temperature that is below 450°C and has an ultimate tensile strength of 320 MPa.
6
655 A nonrotating, machined, steel rod has a round section with d = 50 mm. It is loaded with a fluctuating axial force. Determine its corrected fatigue strength with 99% reliability if it is used in an environment that has a temperature below 450°C and has an ultimate tensile strength of 480 MPa. 656 A nonrotating, colddrawn, steel rod has a round section with d = 76 mm. It is loaded in repeated torsion. Determine its corrected fatigue strength with 99% reliability if it is used in an environment that has a temperature of 500°C and has an ultimate tensile strength of 855 MPa. 657 A nonrotating, ground, steel rod has a rectangular section with h = 60 mm and b = 40 mm. It is loaded in repeated bending. Determine its corrected fatigue strength with 99.9% reliability if it is used in an environment that has a temperature that is below 450°C and has an ultimate tensile strength of 1550 MPa. 658 A grooved steel shaft similar to that shown in Figure C5 (Appendix C) is to be loaded in bending. Its dimensions are: D = 57 mm, d = 38 mm, r = 3 mm. Determine the fatigue stressconcentration factor if the material Sut = 1130 MPa. 659 A steel shaft with a transverse hole similar to that shown in Figure C8 (Appendix C) is to be loaded in torsion. Its dimensions are: D = 32 mm, d = 3 mm. Determine the fatigue stressconcentration factor if the material Sut = 808 MPa. 660 A hardened aluminum filleted flat bar similar to that shown in Figure C9 (Appendix C) is to be loaded axially. Its dimensions are: D = 1.20 in, d = 1.00 in, r = 0.10 in. Determine the fatigue stressconcentration factor if the material Sut = 76 kpsi.
D
d
r Copyright © 2018 Robert L. Norton: All Rights Reserved
FIGURE P620 Problem 661, 686, 690
661 A rotating shaft with a shoulder fillet seated in the inner race of a rolling contact bearing with the shoulder against the edge of the bearing is shown in Figure P620. The bearing has a slight eccentricity that induces a fully reversed bending moment in the shaft as it rotates. Measurements indicate that the resulting alternating stress amplitude due to bending is σa = 57 MPa. The torque on the shaft fluctuates from a high of 90 Nm to a low of 12 Nm and is in phase with the bending stress. The shaft is ground and its dimensions are D = 23 mm, d = 19 mm, and r = 1.6 mm. The shaft material is SAE 1040 coldrolled steel. Determine the infinitelife fatigue safety factor for the shaft for a reliability of 99%. 662 A tension member in a machine is filleted as shown in Figure P621. The member has a manufacturing defect that causes the fluctuating tension load to be applied eccentrically resulting in a fluctuating bending load as well. Measurements indicate that the
Chapter 6
FATIGUE FAILURE THEORIES
maximum bending stress is 16.4 MPa and the minimum is 4.1 MPa. The tensile load fluctuates from a high of 3.6 kN to a low of 0.90 kN and is in phase with the bending stress. The member is machined and its dimensions are D = 33 mm, d = 25 mm, h = 3 mm and r = 3 mm. The material is SAE 1020 coldrolled steel. Determine the infinitelife fatigue safety factor for the member for 90% reliability. 663 For a filleted flat bar in tension similar to that shown in Figure C9 (Appendix C) and the data from the row(s) assigned from Table P67, determine the alternating and mean axial stresses as modified by the appropriate stress concentration factors in the bar. 664 For a filleted flat bar in bending similar to that shown in Figure C10 (Appendix C) and the data from the row(s) assigned from Table P67, determine the alternating and mean bending stresses as modified by the appropriate stress concentration factors in the bar. 665 For a shaft, with a shoulder fillet, in tension similar to that shown in Figure C1 (Appendix C) and the data from the row(s) assigned from Table P67, determine the alternating and mean axial stresses as modified by the appropriate stress concentration factors in the shaft. 666 For a shaft, with a shoulder fillet, in bending similar to that shown in Figure C2 (Appendix C) and the data from the row(s) assigned from Table P67, determine the alternating and mean bending stresses as modified by the appropriate stress concentration factors in the shaft. 667 A machine part is subjected to fluctuating, simple, multiaxial stresses. The fully corrected nonzero stress ranges are: σxmin = 50 MPa, σxmax = 200 MPa, σymin = 80 MPa, σymax = 320 MPa, τxzmin = 120 MPa, τxymax = 480 MPa. The material properties are: Se = 525 MPa and Sut = 1200 MPa. Using a Case 3 load line, calculate and compare the infinitelife safety factors given by the Sines and von Mises methods. 668 A cylindrical tank with hemispherical ends has been built. It was made from hot rolled steel that has Sut = 380 MPa. The tank outside diameter is 300 mm with 20 mm wall thickness. The pressure may fluctuate from 0 to an unknown maximum. For an infinitelife fatigue safety factor of 4 with 99.99% reliability, what is the maximum pressure to which the tank may be subjected? 669 A rotating shaft has been designed and fabricated from SAE 1040 HR steel. It is made from tubing that has an outside diameter of 60 mm and a wall thickness of 5 mm. Strain gage measurements indicate that there is a fully reversed axial stress of 68 MPa and a torsional stress that fluctuates from 12 MPa to 52 MPa in phase with the axial stress at the critical point on the shaft. Determine the infinitelife fatigue safety factor for the shaft for a reliability of 99%.
Table P67
Data for Problems 663 through 666 Use only data that are relevant to the particular problem. Lengths in mm, forces in N, and moments in Nm.
Row
D
d
a
40
20
r
h
Mmin
Mmax
Pmin
4
10
80
320
12
100
500
9500
47 500
SAE 1040 CR
8
60
180
6500
19 500
SAE 1020 CR
8000
Pmax
Material
32 000
SAE 1020 CR
b
26
20
1
c
36
30
1.5
d
33
30
1
8
75
300
7200
28 800
SAE 1040 CR
e
21
20
1
10
50
150
5500
16 500
SAE 1050 CR
f
51
50
1.5
7
80
320
8000
32 000
SAE 1020 CR
g
101
100
5
8
400
800
15 000
60 000
SAE 1040 CR
Copyright © 2018 Robert L. Norton: All Rights Reserved
453
F d r h
D F Copyright © 2018 Robert L. Norton: All Rights Reserved
FIGURE P621 Problem 662
6
454
MACHINE DESIGN
Data for Problem 670
6
σm’
An Integrated Approach

Sixth Edition
670 For the mean and alternating stress values (in MPa) in the row(s) assigned in Table P68, find the safety factor for each of the four loading variation cases based on the ModifiedGoodman diagram if Se = 100, Sy = 150, and Sut = 200 MPa.
Table P68
Row

σa’
a
50
30
b
70
30
c
100
10
d
20
60
e
80
40
f
40
40
g
120
50
h
80
80
Copyright © 2018 Robert L. Norton: All Rights Reserved
671 A rotating shaft with a shoulder fillet, in torsion similar to that shown in Appendix C Figure C3 is made from SAE 1020 CR steel and has dimensions D = 40 mm, d = 20 mm, and r = 4 mm. The shaft is ground and is subjected to a fully reversed torque of +/– 80 Nm. Determine the shaft’s infinitelife safety factor for 99.9% reliability. 672 A 1.6mmlong central crack is found on a fuselage panel of a commuter aircraft during routine inspection. The material has the same properties as the panel in Problem 651. The plane takes off and lands 32 times each day (creating a pressure cycle each time), 260 days per year. On the ground the panel is unstressed. In the air the maximum tensile stress is 300 MPa. If the next inspection will be done six months later, estimate the expected crack length at that time. 673 Steel sheets 400 mm wide by 2000 mm long by 2mm thick are fabricated from the same material given in Problem 652 with the inspection device used in that problem. Using fracturemechanics criteria, estimate the minimum potential cycle life of these sheets if they will be subjected to a lengthwise tensile load that continuously varies from 50 to 500 kN. 674 Repeat problem 653 for a cylinder with dimensions: length = 180 mm, OD = 80 mm, and ID = 72 mm; a 2.4mmdeep semicircular crack; and a pressure that varies from 0 to 70 MPa. 675 A nonrotating, hotrolled, steel beam has an Ibeam section with h = 200 mm, b = 200 mm, and t = 25 mm. It is loaded in repeated bending with the neutral axis through the web. Determine its corrected fatigue strength with 95% reliability if it is used in an environment that has a temperature that is below 450C and has an ultimate tensile strength of 365 MPa. 676 A steel, shaft with shoulder fillet similar to that shown in Figure C3 (Appendix C) is to be loaded in torsion. Its dimensions are: D = 52 mm, d = 39 mm, r = 4 mm. Determine the fatigue stressconcentration factor if the material Sut = 1330 MPa. 677 A steel, grooved shaft similar to that shown in Figure C4 (Appendix C) is to be loaded in axial tension. Its dimensions are: D = 60 mm, d = 40 mm, r = 5 mm. Determine the fatigue stressconcentration factor if the material Sut = 1200 MPa. 678 A steel, grooved shaft similar to that shown in Figure C6 (Appendix C) is to be loaded in torsion. Its dimensions are: D = 48 mm, d = 40 mm, r = 2 mm. Determine the fatigue stressconcentration factor if the material Sut = 1450 MPa. 679 A cylindrical tank with hemispherical ends is made from hot rolled steel that has Sut = 450 MPa. The tank outside diameter is 350 mm with 25 mm wall thickness. The pressure fluctuates from 0 to 15 MPa. Determine the infinitelife fatigue safety factor for 99.9% reliability. 680 A cylindrical tank with hemispherical ends is to be fabricated from coldrolled steel that has Sut = 550 MPa. The tank outside diameter is to be 250 mm. The pressure will fluctuate from 0 to 15 MPa. For an infinitelife fatigue safety factor of 2 with 99% reliability, determine the minimum wall thickness that may be used. 681 A ground steel shaft with a shoulder fillet has dimensions: D = 30 mm, d = 25 mm, and r = 2 mm. It is subjected to a combined nominal alternating loading of σσ = 50 to 180 MPa bending stress and torsional stress of στ = 10 to 50 MPa, which are in phase. Find the safety factor for infinite life if Sut = 965 MPa and the reliability is 95%.
Chapter 6
FATIGUE FAILURE THEORIES
455
682 A component in the shape of a large sheet is to be fabricated from 2024T351 aluminum. Estimate the number of loading cycles that can be endured if the nominal stress varies from 0 to six tenths of the yield strength and the initial crack had a total length of 1 mm. The values of the coefficient and exponent in equation 6.4 for this material are A = 4.8 x 1011 (mm/cycle) and n = 3.5. 683 A machine part is made from 2024T351 aluminum plate. For each cycle of the machine the part undergoes a tensile stress that varies from 0 to 1/2 of the material’s tensile yield strength. The machine cycles 20 times per minute, 24 hours per day, 365 days per year. At the last inspection of the machine the part was found to have a central crack 1.7 mm long. The machine is inspected every six months. Estimate the expected crack length at next inspection. The values of the coefficient and exponent in equation 6.4 for this material are A = 4.8 x 1011 (mm/cycle) and n = 3.5. 684 A non rotating, hotrolled, SAE 1010 steel beam has a rectangular section with h = 80 mm and b = 40 mm. It is loaded in repeated bending with the neutral axis through the long dimension. Determine its corrected fatigue strength with 95% reliability if it is used in an environment that has a temperature that is below 450C. 685 A hardened aluminum, notched, flat bar similar to that shown in Figure C11 (Appendix C) is to be loaded axially. Its dimensions are: D = 1.625 in, d = 1.250 in, r = 0.125 in. Determine the fatigue stressconcentration factor if the material Sut = 55 kpsi. 686 A rotating shaft with a shoulder fillet seated in the inner race of a rolling contact bearing with the shoulder against the edge of the bearing is shown in Figure P620. The bearing has a slight eccentricity that induces a fully reversed bending moment in the shaft as it rotates. Measurements indicate that the resulting fully reversed bending moment at the shaft shoulder (in Nm) is (4.3x103)d where d is in mm . Design a steel shaft for this application that has a dynamic factor of safety of 2 for an infinite life. Bearings are available with IDs of 10 mm to 100 mm in increments of 2 mm. To seat the bearing properly the maximum fillet radius on the shaft is rmax = d x 0.05 and the maximum shoulder diameter is Dmax = d + (4 x rmax). 687 A rotating, steel shaft with a shoulder fillet similar to that shown in Figure C2 (Appendix C) is to be loaded in bending. The dimensions of the fillet radius and shoulder diameter are related to the small diameter as r = d x 0.05 and D = d + (4 x r). For a fully reversed bending moment at the fillet of 880d lbin, where d is in in, a material with Sut = 100 kpsi, and an infinite life calculate and plot the dynamic safety factor as a function of the diameter d. 688 A nonrotating, coldrolled, SAE 1040 steel rod has a square section with h = 50 mm. It is loaded with a fluctuating axial force. Determine its corrected fatigue strength with 99% reliability if it is used in an environment that has a temperature below 450C. 689 An annealed aluminum, flat bar with a transverse hole similar to that shown in Figure C14 (Appendix C) is to be loaded in bending. Its dimensions are: W = 1.625 in, d = 0.125 in, h = 0.125 in. Determine the fatigue stressconcentration factor if the material Sut = 34 kpsi. 690 A rotating shaft with a shoulder fillet seated in the inner race of a rolling contact bearing with the shoulder against the edge of the bearing is shown in Figure P620. The bearing has a slight eccentricity that induces a fully reversed bending moment in the shaft as it rotates. Measurements indicate that the resulting fully reversed bending moment at the shaft shoulder (lbin) is 880d where d is in inches. Design a steel shaft for this application that has a dynamic factor of safety of 2 for an infinite life. Bearings are available with IDs of 0.5 in to 5.0 in in increments of 0.125 in. To seat the bearing properly the maximum fillet radius on the shaft is rmax = d x 0.05 and the maximum shoulder diameter is Dmax = d + (4 x rmax).
6
456
MACHINE DESIGN

An Integrated Approach

Sixth Edition
691 A 1.625india SAE 1050 HR steel shaft with a 1.250india groove is subjected to a combined loading of σσ = ±8 kpsi bending stress and an alternating torsional stress that varies from −2 kpsi to +6 kpsi, which are 45 deg out of phase. Find the safety factor for infinite life for a reliability of 99% if the shaft is ground and operates at room temperature. The groove radius is 0.125 in. 692 For the part in Problem 654, find the corrected endurance strength (or limit), create equations for the SN line, draw the SN diagram, and find the fatigue strength at 105 cycles. 693 A steel, flat bar with a transverse hole similar to that shown in Figure C13 (Appendix C) is to be loaded in bending. Its dimensions are: W = 1.625 in and d = 0.125 in. Determine the fatigue stressconcentration factor if the material Sut = 127 kpsi.
6
SURFACE FAILURE
7
Use it up, wear it out; Make it do, or do without. New eNglaNd MaxiM
7.0
INTRODUCTION
There are only three ways in which parts or systems can “fail”: obsolescence, breakage, or wearing out. My old computer still works well but is obsolete and no longer of any use to me. My wife’s favorite vase is now in pieces since I dropped it on the floor, and it is irrecoverable. However, my 123 000mile automobile is still quite serviceable and useful despite showing some signs of wear. Most systems are subject to all three types of possible failure. Failure by obsolescence is somewhat arbitrary. (My grandchild is now getting good use of the old computer.) Failure by breakage is often sudden and may be permanent. Failure by “wearing out” is generally a gradual process and is sometimes repairable. Ultimately, any system that does not fall victim to one of the other two modes of failure will inevitably wear out if kept in service long enough. Wear is the final mode of failure, which nothing escapes. Thus, we should realize that we cannot design to avoid all types of wear completely, only to postpone them. The previous chapters have dealt with failure of parts by distortion (yielding) and breakage (fracture). Wear is a broad term that encompasses many types of failures, all of which involve changes to the surface of the part. Some of these socalled wear mechanisms are still not completely understood, and rival theories exist in some cases. Most experts describe five general categories of wear: adhesive wear, abrasive wear, erosion, corrosion wear, and surface fatigue. The following sections discuss these topics in detail. In addition, there are other types of surface failure that do not fit neatly into one of the five categories or that can fit into more than one. Corrosion fatigue has aspects of the last two categories as does fretting corrosion. For simplicity, we will discuss these hybrids in concert with one of the five main categories listed above. Failure from wear usually involves the loss of some material from the surfaces of solid parts in the system. The wear motions of interest are sliding, rolling, or some 457
457
458
MACHINE DESIGN
Opening page photograph courtesy of V. S. Makhijani.
Table 70 Symbol
a Aa
7
An Integrated Approach

Sixth Edition
Variables Used in This Chapter Variable
ips units
SI units
See
halfwidth of contact patch
in
m
Sect. 7.8–7.10
apparent area of contact
in2
m2
Sect. 7.2
Ar
real area of contact
in2
m2
Eq. 7.1
B b d E F, P f f max
geometry factor
1/in
1/m
Eq. 7.9b
halflength of contact patch
in
m
Sect. 7.8–7.10
depth of wear
in
m
Eq. 7.7
Young's modulus
psi
Pa
all
force or load
lb
N
all
friction force
lb
N
Eq. 7.2
maximum tangential force
lb
N
Eq. 7.22f
K l L m1, m2
wear coefficient
none
none
Eq. 7.7
length of linear contact
in
m
Eq. 7.7
length of cylindrical contact
in
m
Eq. 7.14
material constants
1/psi
m2/N
Eq. 7.9a
penetration hardness
psi
kg/mm2
Eq. 7.7
number of cycles
none
none
Eq. 7.26
H N Nf
* A 1977 study sponsored by the ASME estimated that the energy cost to the U.S. economy associated with the replacement of equipment that failed from wear accounted for 1.3% of total U.S. energy consumption. This was then equivalent to about 160 million barrels of oil per year. See O. Pinkus and D. F. Wilcock, Strategy for Energy Conservation through Tribology, ASME, New York, 1977, p. 93.

safety factor in surface fatigue
none
none
Example 75
p p avg
pressure in contact patch
psi
N/m2
Sect. 7.8–7.10
avg pressure in contact patch
psi
N/m2
Sect. 7.8–7.10
p max
max pressure in contact patch
psi
N/m2
Sect. 7.8–7.10
R,R
radii of curvature
in
m
Eq. 7.9b
Sus
ultimate shear strength
psi
Pa
Sect. 7.3
Sut
ultimate tensile strength
psi
Pa
Sect. 7.3
Sy
yield strength
psi
Pa
Sect. 7.3
Syc
yield strength in compression
psi
Pa
Sect. 7.3
V x, y, z µ ν σ σ1
volume
in3
m3
Eq. 7.7
generalized length variables
in
m
all
coefficient of friction
none
none
Eq. 7.2–7.6
Poisson's ratio
none
none
all
normal stress
psi
Pa
all
principal stress
psi
Pa
Sect. 7.11
σ2
principal stress
psi
Pa
Sect. 7.11
σ3
principal stress
psi
Pa
Sect. 7.11
τ τ 13
shear stress
psi
Pa
all
maximum shear stress
psi
Pa
Sect. 7.11
τ 21
principal shear stress
psi
Pa
Sect. 7.11
τ 32
principal shear stress
psi
Pa
Sect. 7.11
combination of both. Wear is a serious cost to the national economy.* It only requires the loss of a very small volume of material to render the entire system nonfunctional. Rabinowicz[1] estimated that a 4000lb automobile, when completely “worn out,” will
Chapter 7
SURFACE FAILURE
459
have lost only a few ounces of metal from its working surfaces. Moreover, these damaged surfaces will not be visible without extensive disassembly, so it is often difficult to monitor and anticipate the effects of wear before failure occurs. Table 70 shows the variables used in this chapter and references the equations, tables, or sections in which they are used. At the end of the chapter, a summary section is provided that also groups all the significant equations from this chapter for easy reference and identifies the chapter section in which their discussion can be found. A master lecture video on the topics in sections 7.1–7.13 is linked as Lecture 12.
7.1
SURFACE GEOMETRY View a Video of Lecture 12 (52:25)*
*
Before discussing the types of wear mechanisms in detail it will be useful to define the characteristics of an engineering surface that are relevant to these processes. (Material strength and hardness will also be factors in wear.) Most solid surfaces that are subject to wear in machinery will be either machined or ground, though some will be ascast or asforged. In any case, the surface will have some degree of roughness that is concomitant with its finishing process. Its degree of roughness or smoothness will have an effect on both the type and degree of wear that it will experience. Even an apparently smooth surface will have microscopic irregularities. These can be measured by any of several methods. A profilometer passes a lightly loaded, hard (e.g., diamond) stylus over the surface at controlled (low) velocity and records its undulations. The stylus has a very small (about 0.5 µm) radius tip that acts, in effect, as a lowpass filter, since contours smaller than its radius are not sensed. Nevertheless, it gives a reasonably accurate profile of the surface with a resolution of 0.125 µm or better. Figure 71 shows the profiles and SEM* photographs (100x) of both ground (a) and machined
(a)
FIGURE 71
http://www.designofmachinery. com/MD/12_Wear_and_Surface_Fatigue.mp4
7
*
Scanning Electron Microscope
(b) Copyright © 2018 Robert L. Norton: All Rights Reserved
Scanning Electron Microscope SurfaceReplica Photographs (100x) and Profiles of Ground (a) and Milled (b) Cam Surfaces
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MACHINE DESIGN

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(b) surfaces of hardened steel cams. The profiles were measured with a Hommel T20 profilometer that digitizes 8000 data points over the sample length (here 2.5 mm). The microscopic “mountain peaks” on the surfaces are called asperities. From these profiles a number of statistical measures may be calculated. ISO defines at least 19 such parameters. Some of them are shown in Figure 72 along with their mathematical definitions. Perhaps the most commonly used parameters are Ra, which is the average of the absolute values of the measured points, or Rq, which is their rms average. These are very similar in both value and meaning. Unfortunately, many engineers specify only one of these two parameters, neither of which tells enough about the surface. For example, the two surfaces shown in Figures 73a and b have the same Ra and Rq values but are clearly different in nature. One has predominantly positive, and the other predominantly negative, features. These two surfaces will react quite differently to sliding or rolling against another surface.
7
FIGURE 72 DIN and ISO Surface Roughness, Waviness, and Skewness Parameter Definitions
Chapter 7
Ra or Rq
SURFACE FAILURE
461
Ra or Rq
(a )
( b)
FIGURE 73 Different Surface Contours Can Have the Same Ra or Rq Values
In order to differentiate these surfaces that have identical Ra or Rq values, other parameters should be calculated. Skewness Sk is a measure of the average of the first derivative of the surface contour. A negative value of Sk indicates that the surface has a predominance of valleys (Figure 73a) and a positive Sk defines a predominance of peaks (Figure 73b). Many other parameters can be computed (see Figure 72). For example, Rt defines the largest peaktovalley dimension in the sample length, Rp the largest peak height above the mean line, and Rpm the average of the 5 largest peak heights. All the roughness measurements are calculated from an electronically filtered measurement that zeros out any slowchanging waves in the surface. An average line is computed from which all peak/valley measurements are then made. In addition to these roughness measurements (denoted by R), the waviness Wt of the surface can also be computed. The Wt computation filters out the highfrequency contours and preserves the longperiod undulations of the raw surface measurement. If you want to completely characterize the surfacefinish condition, note that using only Ra or Rq is not sufficient.
7.2
MATING SURFACES
When two surfaces are pressed together under load, their apparent area of contact Aa is easily calculated from geometry but their real area of contact Ar is affected by the asperities present on their surfaces and is more difficult to accurately determine. Figure 74 shows two parts in contact. The tops of the asperities will initially contact the mating part and the initial area of contact will be extremely small. The resulting stresses in the asperities will be very high and can easily exceed the compressive yield strength of the material. As the mating force is increased, the asperity tips will yield and spread until their combined area is sufficient to reduce the average stress to a sustainable level, i.e., some compressive penetration strength of the weaker material. We can get a measure of a material’s compressive penetration strength from conventional hardness tests (Brinell, Rockwell, etc.), which force a very smooth stylus into
FIGURE 74 The Actual Contact Between Two Surfaces Is Only at the Asperity Tips
7
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the material and deform (yield) the material to the stylus’ shape. (See Section 2.4.) The penetration strength Sp is easily calculated from these test data and tends to be of the order of 3 times the compressive yield strength Syc of most materials.[2] The real area of contact can then be estimated from Ar ≅
F F ≅ S p 3S yc
(7.1)
where F is the force applied normal to the surface and the strengths are as defined in the above paragraph, taken for the weaker of the two materials. Note that the contact area for a material of particular strength under a given load will be the same regardless of the apparent area of the mating surfaces.
7.3
FRICTION
Note that the real area of contact Ar (equation 7.1) is independent of the apparent area Aa that is defined by the geometry of the mating parts. This is the reason that Coulomb friction between two solids is also independent of the apparent area of contact Aa. The equation for Coulomb sliding friction is
7
f = µF
(7.2a)
where f is the force of friction, µ is the coefficient of dynamic friction, and F is the normal force. The normal force presses the two surfaces together and creates elastic deformations and adhesions (see next section) at the asperities’ tips. We can define the dynamic Coulomb friction force f as being the force necessary to shear the adhered and elastically interlocked asperities in order to allow a sliding motion. This shearing force is equal to the product of the shear strength of the weaker material and the actual contact area Ar plus a “plough force” P: f = Sus Ar + P
* This
will be true only if the two surfaces have about the same hardness. If one surface is harder and rougher than the other, there could be a significant plough force.
(7.2b)
The plough force P is due to loose particles digging into the surfaces and is negligible compared to the shear force,* so can be ignored. Recalling equation 7.1 gives Ar ≅
F 3S yc
(7.2c)
Substituting equation 7.2c in 7.2b (ignoring P) gives f ≅F
Sus 3S yc
(7.2d )
Combining equations 7.2a and 7.2d gives µ=
S f ≅ us F 3S yc
(7.3)
which indicates that the coefficient of friction µ is a function only of a ratio of material strengths of the weaker of the two materials in contact.
Chapter 7
SURFACE FAILURE
463
The ultimate shear strength can be estimated based on the ultimate tensile strength of the material. steels:
Sus ≅ 0.8 Sut
other ductile metals:
Sus ≅ 0.75Sut
(7.4)
The yield strength in compression as a fraction of the ultimate tensile strength varies with the material and alloy over a fairly broad range, perhaps 0.5Sut < S yc < 0.9 Sut
(7.5)
Substituting equations 7.4 and 7.5 in equation 7.3 gives 0.75Sut 0.8 Sut Fi as shown in Figure 1522e). The additional deflection ∆δ creates a new load situation in both the bolt and material as shown in Figure 1524b. The load in the material is reduced by Pm and moves down the material stiffness line to point D with a new value Fm. The load in the bolt is increased by Pb and moves up the bolt stiffness line to point C with a new value Fb. Note that the applied load P is split into two components, one (Pm) taken by the material and one (Pb) taken by the bolt. P = Pm + Pb
(15.12a)
The compressive load Fm in the material is now Fm ≥ 0*
Fm = Fi − Pm :
(15.12b)
and the tensile load Fb in the bolt becomes Fb = Fi + Pb
* If Fm has a negative value, set Fm = 0, because the material cannot support a tensile force—it will separate.
(15.12c)
Note what has happened as a result of the preload force Fi. The “spring” of the material was “wound up” under preloading. Any applied loads are partially supported by the “unwinding” of this spring. If the relative stiffness of bolt and material are as shown in Figure 1524 (i.e., material stiffer than bolt), the material supports the majority
F A
B
A
Fi km δm
kb
∆δ δb
(a) Preload force and initial deflections
δ
Pb
F Fb Fi D
Pm
C B
P
Fm
δm
15
∆δ δb
(b) Load deflection and resulting forces
FIGURE 1524 Effects on Bolt and Material from Preload: (a) Preload and (b) Applied Load
δ
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of the applied load and the bolt feels little additional load above that of the initial preload. This is one aspect of the justification for the earlier statement that “if the bolt doesn’t fail when preloaded, it probably won’t fail in service.” There is also another reason for this being true, which will be discussed in a later section. Note, however, that if the applied load P is large enough to cause the component Pm to exceed the preload force Fi, then the joint will separate, Fm will be zero, and the bolt will feel the full value of the applied load P. The material can no longer contribute to supporting the load if the joint is separated. This is one reason for the very large recommended preloads as a percentage of bolt proof strength. In order to get the full benefit of material load sharing, the preload should be high. We can summarize the information in Figure 1524 in the following way. The common change in deflection ∆δ due to the applied load P is δ=
Pb Pm = kb km
(15.13a)
kb Pm km
(15.13b)
Pb =
or:
Substitute equation 15.12a to get Pb = or
kb P km + kb
Pb = CP
where C =
kb km + kb
(15.13c)
The term C is called the joint’s stiffness constant or just the joint constant. Note that C is typically < 1, and if kb is small compared to km, C will be a small fraction. This confirms that the bolt will see only a portion of the applied load P. In like fashion, Pm =
km P = (1 − C ) P kb + km
(15.13d )
These expressions for Pb and Pm can be substituted into equations 15.12b and 15.12c to get expressions for the bolt and material loads in terms of the applied load P:
15
Fm = Fi − (1 − C ) P
(15.14 a)
Fb = Fi + CP
(15.14 b)
Equation 15.14b can be solved for the preload Fi needed for any given combination of applied load P and maximum allowable bolt (proof) load Fb, provided that the joint constant C is known. The load P0 required to separate the joint can be found from equation 15.14a by setting Fm to zero. P0 =
Fi (1 − C )
(15.14c)
Chapter 15
SCREWS AND FASTENERS
931
A safety factor against joint separation can be found from N separation =
P0 Fi = P P (1 − C )
(15.14 d )
EXAMPLE 152 Preloaded Fasteners in Static Loading Problem
Determine a suitable bolt size and preload for the joint shown in Figure 1523 (repeated here). Find its safety factor against yielding and separation. Determine the optimum preload as a percentage of proof strength to maximize the safety factors.
Given
The joint dimensions are D = 1 in and l = 2 in. The applied load P = 2000 lb.
Assumptions
Both of the clamped parts are steel. The effects of the flanges on the joint stiffness will be ignored. A preload of 90% of the bolt’s proof strength will be applied as a first trial.
Solution
See Figure 1525.
1 As with most design problems, there are too many unknown variables to solve the necessary equations in one pass. Trial values must be chosen for various parameters and iteration used to find a good solution. We actually made several iterations to solve this problem, but will only present two in the interest of brevity. Thus, the trial values used here have already been massaged to reasonable values. 2 The bolt diameter is the principal trial value to be chosen along with a thread series and a bolt class to define the proof strength. We choose a 5/1618 UNC2A steel bolt of SAE class 5.2. (This was actually our third trial choice.) For a clamp length of 2 in, assume a bolt length of 2.5 in to allow sufficient protrusion for the nut. The preload is taken at 90% of proof strength as assumed above.
d P/2
P/2
3 Table 156 shows the proof strength of this bolt to be 85 kpsi. The tensilestress area from equation 151a is 0.052 431 in2. The preload is then
(
)(
)
Fi = 0.9 S p At = 0.9 85 000 0.052 431 = 4011 lb
(a)
4 Find the lengths of thread lthd and shank ls of the bolt as shown in Figure 1521: lthd = 2d + 0.25 = 2 ( 0.3125) + 0.25 = 0.875 in ls = lbolt − lthd = 2.5 − 0.875 = 1.625 in
l1 l l2
(b)
15
from which we can find the length of thread lt that is in the clamp zone: lt = l − ls = 2.0 − 1.625 = 0.375 in
(c)
P/2 D
5 Find the stiffness of the bolt from equation 15.11a: l l 1.625 ( 4 ) 1 0.375 = t + s = + kb At E Ab E 0.052 431( 30 E 6 ) π ( 0.3125)2 ( 30 E 6 ) k b = 1.059 E 6 lb/in
P/2
FIGURE 1523 (d )
Repeated
A Preloaded Bolt Compressing a Cylinder to Which External Loads Are Applied
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6 The calculation for the stiffness of the clamped material is simplified in this example by its relatively small diameter. We can assume in this case that the entire cylinder of material is compressed by the bolt force. (We will soon address the problem of finding the clamped area in a continuum of material.) The material stiffness from equation 15.11d is km =
(
)
(
π D2 − d 2 E π 1.0 2 − 0.3122 m = 4 4 l
) (30 E6) = 1.063E 7 lb/in 2.0
(e)
7 The joint stiffness factor from equation 15.13c is C=
kb 1.059 E 6 = = 0.090 56 km + kb 1.063E 7 + 1.059 E 6
(f)
8 The portions of the applied load P felt by the bolt and the material can now be found from equations 15.13: Pb = CP = 0.090 56(2000) = 181 lb
(
)
Pm = (1 − C ) P = 1 − 0.090 56 (2000) = 1819 lb
( g)
9 Find the resulting loads in bolt and material after the load P is applied: Fb = Fi + Pb = 4011 + 181 = 4192 lb Fm = Fi − Pm = 4011 − 1 819 = 2192 lb
(h)
Note how little the applied load adds to the preload in the bolt. 10 The maximum tensile stress in the bolt is σb =
Fb 4192 = = 79 953 psi At 0.052 431
(i )
Note that no stressconcentration factor is used because it is static loading. 11 This is a uniaxial stress situation, so the principal stress and von Mises stress are identical to the applied tensile stress. The safety factor against yielding is then Ny =
Sy
=
σb
92 000 79 953
= 1.15
( j)
The yield strength is found from Tables 156 and 157. 12 The load required to separate the joint and the safety factor against joint separation are found from equations 15.14c and 15.14d:
15
P0 =
Fi
(1 − C )
=
(
4011
1 − 0.090 56
N separation =
)
= 4410 lb
P0 4410 = = 2.2 P 2000
(k )
(l )
13 The safety factor against separation is acceptable. The yielding safety factor is low but this is to be expected, since the bolt is deliberately preloaded to a level close to its yield strength.
Chapter 15
SCREWS AND FASTENERS
933
5.0 Ny
4.5 4.0 safety factor
3.5 3.0
Nseparation
2.5
B
2.0
A
1.5
B’
1.0 .05 0 0
0.1
0.2
0.3 0.4 0.5 0.6 0.7 0.8 preload as decimal % of proof strength
FIGURE 1525
0.9
1.0
Copyright © 2018 Robert L. Norton: All Rights Reserved
Safety Factors Versus Preload for the Statically Loaded Bolt in Example 152
14 The model was solved for the range of possible preloads from zero to 100% of proof strength and the safety factors plotted versus preload percentage. The results are shown in Figure 1525. The separation safety factor rises linearly with increasing preload, but is S y then:
K fm =
S y − K f σ anom σ mnom
K f σ maxnom = K f σ anom + σ mnom = 5.9 864 + 77 364 = 462 548 psi
15
462 548 psi > S y = 92 000 psi K fm =
S y − K f σ anom σ mnom
=
92 000 − 5.9 (864 ) 77 364
= 1.12
8 The local mean and alternating stresses in the bolt are then:
(e)
Chapter 15
F
SCREWS AND FASTENERS
Pb
B
force (lb X 103)
5.0
Fb Pm
4.0
937
Fi A
Fm 3.0 2.0 1.0
km
kb δ
0 –0.0005
0
0.001
0.002
0.003
0.004
deflection (in)
FIGURE 1527
Dynamic Forces in Bolt and Material for Example 153, Drawn to Scale
σ a = K f σ anom = 5.9 (864 ) = 5107 psi
(
(f)
)
σ m = K fm σ mnom = 1.12 77 364 = 86 893 psi
9 The stress at the initial preload is σ i = K fm
Fi 4011 = 1.12 = 85 923 psi 0.052 431 At
( g)
10 An endurance strength must be found for this material. Using the methods of Section 6.6 we find
(
)
Se' = 0.5Sut = 0.5 120 000 = 60 000 psi Se = Cload Csize Csurf Ctemp Creliab Se'
(
(h)
)
= 0.70 ( 0.995)( 0.76 )(1)( 0.81) 60 000 = 25 726 psi
(i )
where the strength reduction factors are taken from the tables and formulas in Section 6.6 for, respectively, axial loading, the bolt size, a machined finish, room temperature, and 99% reliability. 11 The corrected endurance strength and the ultimate tensile strength are used in equation 15.16 to find the safety factor from the Goodman line: Nf = =
Se ( Sut − σ i )
Se ( σ m − σ i ) + Sut σ a
(
(
25 837 120 000 − 85 923
)
)
25 726 86 893 − 85 923 + 120 000 ( 5107 )
= 1.38
The modifiedGoodman diagram for this stress state is shown in Figure 1528.
( j)
15
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Sixth Edition
30 alternating stress (kpsi)
S 25
to
Goo
dma
20
n li
ne
load line for 90% preload
15 10 σa 5
0
10
20
30
40
50
70
60
80
mean stress (kpsi)
FIGURE 1528
Su
S
σi
0
90 σm
100
11
120
ModifiedGoodman Diagram for Example 153
12 The static bolt stress after initial local yielding and the yielding safety factor are: σs =
Fbolt 4102 = = 78 227 psi 0.045 36 At
Ny =
Sy σb
=
92 000 78 227
= 1.18
(k )
13 The preload required to obtain these safety factors is
(
)(
)
Fi = 0.90 S p At = 0.90 85 000 0.052 431 ≅ 4011 lb
(l )
14 The safety factor against joint separation is found from equation 15.14d: N separation =
Fi 4011 = = 4.4 P (1 − C ) 1000 1 − 0.090 56
(
)
(m )
15 The fatigue and separation safety factors are acceptable. The yielding safety factor is low but is nevertheless acceptable, since the bolt is being deliberately preloaded to a level close to its yield strength.
15
16 The model was solved for the range of possible preloads from zero to 100% of proof strength and the safety factors plotted versus percent preload. The results are shown in Figure 1529. The fatigue and separation safety factors are 2x the screw diameter (10 threads), which is the minimum recommended length for a steel screw in aluminum threads. For U.S. standard bolts up to 6in long, the thread length is 2d + 0.25 = 0.875 in of this bolt’s length,[2] which allows the desired penetration. The trial clamp length for the stiffness calculations is then 1.25 in, since the entire cap screw is engaged.
132
1000
p psig
Fg lb 0
180
360
crank angle (deg) F I G U R E 9  2 Repeated Pressure and Force Within Cylinder During One Cycle
15
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Sixth Edition
5 Try an SAE grade 7 screw preloaded to 70% of its proof strength. Table 156 shows the proof strength of this screw to be 105 kpsi. The tensilestress area from equation 151 is 0.052 431 in2. The required preload is then
(
)(
)
Fi = 0.7 S p At = 0.7 105 000 0.052 431 = 3853.66 lb
(c)
6 Find the joint aspect ratio and plate to screw modulus from equations 15.18a and b: j=
d 0.313 = = 0.25 1.25 l
E 10.4 E 6 = 0.347 r = material = 30 E 6 Ebolt
(d )
7 Since this joint has the same material (aluminum) throughout, equation 15.19 is all that is needed to calculate the joint constant Cng for the metal without the gasket. The parameters needed for the equation are found by interpolating for j = 0.25 in Table 158. They are: p0 = 0.653, p1 = –1.207, p2 = 1.103, and p3 = –0.383: Cng = p3r 3 + p2 r 2 + p1r + p0 = –0.383 ( 0.347 )3 + 1.103 ( 0.347 )2 − 1.207 ( 0.347 ) + 0.653 = 0.351
(e)
8 We can approximate the stiffness of the screw kb’ with equation 15.17, then estimate material stiffness km with no gasket by using equation 15.13c, given kb’ and Cng : length of thread:
lthd = 2d + 0.25 = 2 ( 0.313) + 0.25 = 0.875 in
length of shank:
ls = lbolt − lthd = 1.25 − 0.875 = 0.375 in
length of clamped thread:
lt = l − ls = 1.25 − 0.375 = 0.875 in
d −1 At Ab Eb kb ′ ≅ 1 + l Ab lt + At ls
(f)
0.052 ( 0.077 ) 0.313 −1 kb ′ ≅ 1 + 30 E 6 = 1.112 E 6 lb/in 1.25 0.077 ( 0.875) + 0.052 ( 0.375) 1 − Cng kb ′ 1 − 0.351 Cng = km1 = kb ′ = 1.112 E 6 = 2.060 E 6 lb/in 0.351 km + kb ′ Cng 1
9 Now consider the gasket. The area of the unconfined gasket subjected to the clamp force can be assumed to be that of one “bolt’s worth” of the total clamped area which extends from the outside diameter of the cylinder head to the bore: Atotal =
15
(
) (
)
π 2 π Do − D12 = 5.6252 − 3.1252 ≅ 17.18 in 2 4 4
( g)
Dividing by the number of bolts and subtracting the screw hole area gives the area of the clamped gasket around any one screw as: Ag =
Atotal π 2 17.18 π − d = − 0.3132 = 2.071 in 2 4 8 4 nb
(h)
10 The stiffness of this piece of gasket is then km2 =
Ag Eg lg
=
2.071(13.5E 6 ) ≅ 4.659 E 8 lb/in 0.06
(i )
Chapter 15
SCREWS AND FASTENERS
959
The modulus of elasticity of the gasket material is found in Table 1510. 11 The material and gasket stiffness combine according to equation 14.2b. The combined stiffness of the gasketed joint is then 1
k mg =
1 km1
+
1
=
km2
1 1 1 + 2.060 E 6 4.659 E 8
kmg = 2.051E 6 lb/in
( j)
Note that the combined stiffness in this case is dominated by the aluminum because the copperasbestos gasket is stiffer. 12 The joint constant with the unconfined gasket is C=
kb ' 1.112 E 6 = ≅ 0.352 kmg + kb ' 2.051E 6 + 1.112 E 6
(1 − C ) = 0.648
and
(k )
13 The 1000lb load is assumed to be divided equally among the 8 bolts at 125 lb each. The portions of the applied load felt by each screw and material (Eq. 15.13) are: Pb = CP = 0.352(125) ≅ 43.95 lb Pm = (1 − C ) P = 0.614(125) ≅ 81.05 lb
(l )
14 The resulting peak loads in screw and material are Fb = Fi + Pb = 3853.66 + 43.95 = 3897.61 lb Fm = Fi − Pm = 3853.66 − 81.05 = 3772.61 lb
(m)
15 The alternating and mean components of force on the screw are Falt = Fmean =
Fb − Fi 3897.61 − 3853.66 = ≅ 21.98 lb 2 2 Fb + Fi 3897.61 + 3853.66 = ≅ 3875.63 lb 2 2
(n)
16 The nominal mean and alternating stresses in the screw are σ anom =
Falt 21.98 = ≅ 419.2 psi At 0.052 431
σ mnom =
Fmean 3875.63 = ≅ 73 919 psi At 0.052 431
(o)
17 The fatigue stressconcentration factor for this diameter thread is found from equation 15.15c and the mean stressconcentration factor Kfm is found from equation 6.17.
15
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MACHINE DESIGN

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Sixth Edition
K f = 5.7 + 0.6812d = 5.7 + 0.6812 ( 0.313) = 5.9 if K f σ maxnom > S y then:
K fm =
S y − K f σ anom σ mnom
K f σ maxnom = K f σ anom + σ mnom = 5.9 460.2 + 73 960 = 440 039 psi 440 039 psi > S y = 115 000 psi K fm =
S y − K f σ anom σ mnom
=
115 000 − 5.9 ( 460.2 ) 73 960
= 1.52
( p)
18 The local mean and alternating stresses in the screw are Falt 21.98 = 5.9 ≅ 2478 psi At 0.052 431
σa = K f
σ m = K fm
(q)
Fmean 3875.63 = 1.52 ≅ 112 522 psi At 0.052 431
19 The stresses at the initial preload and at the maximum screw force are σ i = K fm
Fi 3853.66 = 1.52 ≅ 111 883 psi At 0.052 431
σ b = K fm
Fb 3897.61 = 1.52 = 113 160 psi At 0.052 431
(r )
20 The endurance strength for this material is found with the methods of Section 6.6:
(
)
Se' = 0.5Sut = 0.5 133 000 = 66 500 psi Se = Cload Csize Csurf Ctemp Creliab Se'
(
(s )
)
= 0.70 (.995)( 0.739 )(1)( 0.753) 66 500 = 25 778 psi
(t )
where the strengthreduction factors are taken from the tables and formulas in Section 6.6 for, respectively, axial loading, the screw size, a machined finish, room temperature, and 99.9% reliability. 21 The corrected endurance strength and the ultimate tensile strength are used in equation 15.16 to find the fatigue safety factor from the Goodman line: Nf =
15
=
Se ( Sut − σ i )
Se ( σ m − σ i ) + Sut σ a
(
(
25 778 133 000 − 111 883
)
)
25 778 112 522 − 111 883 + 133 000 ( 2478 )
≅ 1.6
(u)
22 The static screw stress after initial local yielding and the yielding safety factor are: σs =
Fbolt 3897.61 = = 74 338 psi 0.052 431 At
Ny =
Sy σb
=
115 000 74 338
≅ 1.5
23 The safety factor against joint separation is found from equation 15.14d:
(v)
Chapter 15
N separation =
SCREWS AND FASTENERS
Fi 3853.66 = ≅ 47 P (1 − C ) 125 (1 − 0.352 )
961
( w)
24 The joint will leak unless the clamping forces are sufficient to create more pressure at the gasket than exists in the cylinder. The minimum clamping pressure can be found from the total area of the gasketed joint and the minimum clamping force Fm : pavg =
4 Fm Fm 4 ( 3772.6 ) = = ≅ 228 psi A j π Do2 − Di2 − nb Ab π 5.6252 − 3.1252 − 8(0.077)
(
)
(
N leak =
pavg pcyl
)
(x)
228 = ≅ 1.7 130
This ratio of clamp pressure vs. cylinder pressure makes the screw spacing acceptable. 25 The torque required to obtain the preload of 3853.66 lb found in step 5 is: Ti ≅ 0.21Fi d = 0.21( 3853.66 )( 0.313) ≅ 253 lbin
( y)
26 This design uses eight 5/1618 UNC2A, grade 7 hexhead cap screws, 1.25 in long, preloaded to 70% of proof strength and equispaced on a 4.375india bolt circle. This design has a safety factor against leakage of 1.7, a fatigue safety factor of 1.6, and can stand 47 times the operating pressure before joint separation will occur. These safety factors are acceptable. The files CASE8D can be found on the book’s website.
15.12 SUMMARY This chapter has dealt with only a small sample of commercially available fasteners. An extremely varied collection of fasteners is available. The “right” fastener can usually be found for any application, and if not (and the required quantity is high enough), some vendor will make a new one for you. Many standards exist that define the configurations, sizes, strengths, and tolerances of fasteners. Threaded fasteners are made to one or another of these standards, which provides good interchangeability. Unfortunately, metric and English threads are not interchangeable, and both are in wide use in the United States. Power screws are threaded devices used primarily to move loads or accurately position objects. They have low efficiency due to their large friction losses unless the ballscrew variety is used, which lowers the friction significantly. However, lowfriction screws give up one of their advantages, which is selflocking or the ability to hold a load in place with no input of energy (as in a jack). Backdrivable screws are the opposite of selflocking and can be used as a lineartorotary motion converter. Threaded fasteners (bolts, nuts, and screws) are the standard means of holding machinery together. These fasteners are capable of supporting very large loads, especially if they are preloaded. Preloading tightens the fastener to a high level of axial tension before any working loads are applied. The tension in the fastener causes compression in the clamped parts. This compression has several salutary effects. It keeps the joint tightly together and thus able to contain fluid pressure and resist shear loads with its interfacial friction. The compressive forces in the clamped material also serve to protect the fastener from fluctuating fatigue loads by absorbing most of the applied load oscillations. The
15
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MACHINE DESIGN

An Integrated Approach

Sixth Edition
high clamping forces also guard against vibratory loosening of the fastener by creating high friction forces in the threads. Threaded fasteners are also capable of resisting shear loads and are used extensively in that manner in structural applications. In machine design it is more common to rely on closefitted dowel pins to take shear loads and let the threaded fasteners provide the tension to hold the joint together. The interested reader is referred to the publications listed in the bibliography of this chapter for more information on the diverse and fascinating world of fasteners. Important Equations Used in This Chapter See the referenced sections for information on the proper use of these equations. Torque Required to Raise the Load with a Power Screw (Section 15.2):
Tu = Tsu + Tc =
( ) + µ P dc ( π d p cos α − µL ) c 2
Pd p µπ d p + L cos α 2
(15.5a)
SelfLocking of a Power Screw Will Occur if (Section 15.2):
µ≥
L cos α πd p
µ ≥ tan λ cos α
or
(15.6a)
Efficiency of a Power Screw (Section 15.2):
e=
Wout PL = 2 πT Win
(15.7c)
Spring Constant of a Threaded Fastener (Section 15.7):
l l − lt l l 1 = t + = t + s kb At Eb Ab Eb At Eb Ab Eb
∴
kb =
At Ab Eb Ab lt + At ls
d −1 kb ′ = 1 + kb l
(15.11a)
(15.17alt )
Spring Constant of the Clamped Material if Am is known (Section 15.7):
km =
15
Am Em l
(15.11c)
Load Taken by the Preloaded Material (Section 15.7):
Pm =
km P = (1 − C ) P kb + km
(15.13d )
Load Taken by a Preloaded Bolt and the Joint Constant C (Section 15.7):
Pb = or
kb P km + kb
Pb = CP
kb where C = km + kb
(15.13c)
Chapter 15
SCREWS AND FASTENERS
963
Minimum Load in Material and Maximum Load in Bolt (Section 15.7):
Fm = Fi − (1 − C ) P
(15.14 a)
Fb = Fi + CP
(15.14 b)
Load Required to Separate a Preloaded Joint (Section 15.7):
Fi
P0 =
(15.14c)
(1 − C )
Mean and Alternating Loads Felt by a Preloaded Bolt (Section 15.7):
Falt =
Fb − Fi , 2
Fmean =
Fb + Fi 2
(15.15a)
Mean and Alternating Stresses in a Preloaded Bolt (Section 15.7):
σa = K f
Falt , At
σ m = K fm
Fmean At
(15.15b)
Fatigue Stress Concentration Factor in Threads:
or
K f = 5.7 + 0.6812d
d in inches
K f = 5.7 + 0.026 82d
d in mm
(15.15c)
Preload Stress in a Bolt (Section 15.7):
σ i = K fm
Fi At
(15.15d )
Fatigue Safety Factor for a Preloaded Bolt (Section 15.7):
Nf =
Se ( Sut − σ i )
(15.16)
Se ( σ m − σ i ) + Sut σ a
Approximate Torque Needed to Preload a Bolt (Section 15.9):
Ti ≅ 0.21Fi d
(15.23d )
Centroid of a Group of Fasteners (Section 15.10):
∑ 1 Ai xi , x = n ∑ 1 Ai
∑ 1 Ai yi y = n ∑ 1 Ai
n
n
(15.24)
15
Forces on Fasteners Eccentrically Loaded in Shear (Section 15.10):
F1i = F2i =
P n
Mri
∑
n
r2 j =1 j
(15.25a) =
Plri
∑ j =1 rj2 n
(15.25b)
964
Table P150†

An Integrated Approach

Sixth Edition
15.13 REFERENCES
Topic/Problem Matrix
1 Product Engineering, vol. 41, p. 9, Apr. 13, 1970.
15.2 Power Screws
2 H. L. Horton, ed. Machinery’s Handbook, 21st ed. Industrial Press, Inc.: New York. p. 1256, 1974.
151, 152, 153, 1537, 1538 15.3 Stresses in Threads
3 ANSI/ASME Standard B1.11989, American National Standards Institute, New York, 1989.
1515, 1516, 1539, 1547, 1553
4 ANSI/ASME Standard B1.131983 (R1989), American National Standards Institute, New York, 1989.
15.6 Strengths of Bolts
5 T. H. Lambert, Effects of Variations in the Screw Thread Coefficient of Friction on Clamping Force of Bolted Connections, J. Mech Eng. Sci., 4: p. 401, 1962.
1528, 1529, 1540, 1541, 1548 15.8 Joint Stiffness Material Only 1517, 1518, 1519, 1531, 1532, 1533, 1542, 1543 No GasketStatic Load 157, 158, 1523, 1530, 1549 No GasketDynamic Load 159, 1510, 1524, 1525, 1527 Unconfined Gasket 1526, 1542, 1543, 1544, 1545, 1546 Different Materials in Joint 1542, 1543, 1544, 1545, 1546 15.9 Controlling Preload Torque Only
6 R. E. Peterson, StressConcentration Factors. John Wiley & Sons: New York, p. 253, 1974. 7 H. H. Gould and B. B. Mikic, Areas of Contact and Pressure Distribution in Bolted Joints, Trans ASME, J. Eng. for Industry, 94: pp. 864–869, 1972. 8 N. Nabil, Determination of Joint Stiffness in Bolted Connections, Trans. ASME, J. Eng. for Industry, 98: pp. 858–861, 1976. 9 Y. Ito, J. Toyoda and S. Nagata, Interface Pressure Distribution in a BoltFlange Assembly, Trans. ASME, J. Mech. Design, 101: pp. 330–337, 1979. 10 J. F. Macklin and J. B. Raymond, Determination of Joint Stiffness in Bolted Connections using FEA, Major Qualifying Project, Worcester Polytechnic Institute, Worcester, Mass., Dec. 31, 1994. 11 B. Houle, An Axisymmetric FEA Model Using Gap Elements to Determine Joint Stiffness, Major Qualifying Project, Worcester Polytechnic Institute, Worcester, Mass., Dec. 31, 1995. 12 J. E. Shigley and C. H. Mischke, Mechanical Engineering Design, 5th ed. McGrawHill: New York, p. 354, 1989. 13 J. Wileman, M. Choudhury, and I. Green, “Computation of Member Stiffness in Bolted Connections,” Trans. ASME, J. Mech. Design, 113, pp. 432–437, 1991.
1511, 1512, 1513, 1514, 1554
14 T. F. Lehnhoff, K. I. Ko, and M. L. McKay, “Member Stiffness and Contact Pressure Distribution of Bolted Joints,” Trans. ASME, J. Mech. Design, 116, pp. 550557, 1994.
No GasketStatic Load
15 R. E. Cornwell, “Computation of Load Factors in Bolted Connections,” Proc. IMechE, J. Mech. Eng. Sci., 223c, pp. 795808, 2009.
154, 155, 156, 1550, 1556, 1559 No GasketDynamic Load 1522, 1551, 1552, 1557, 1560, 1562
15
MACHINE DESIGN
16 A. R. Kephart, “Fatigue Acceptance Test Limit Criterion for Larger Diameter Rolled Thread Fasteners,” ASTM Workshop on Fatigue and Fracture of Fasteners, May 6, 1997, St. Louis, MO. 17 C. Crispell, “New Data on Fastener Fatigue,” Machine Design, pp. 7174, April 22, 1982.
Unconfined Gasket 1520, 1521, 1555, 1558, 1561
15.14 BIBLIOGRAPHY
15.10 Fasteners in Shear
American Institute of Steel Construction Handbook. AISI: New York.
1534, 1535, 1536
Helpful Hints for Fastener Design and Application. Russell, Burdsall & Ward Corp.: Mentor, Ohio, 1976.
Problem numbers in italics are design problems. Problem numbers in boldface are extended from similar problems in earlier chapters with the same dash no.
†
SAE Handbook. Soc. of Automotive Engineers: Warrendale, Pa., 1982. “Fastening and Joining Reference Issue,” Machine Design, vol. 55, Nov. 17, 1983.
Chapter 15
SCREWS AND FASTENERS
965
J. H. Bickford, An Introduction to the Design and Behavior of Bolted Joints, 2nd ed. Marcel Dekker, New York, 1990. H. L. Horton, ed., Machinery’s Handbook, 21st ed. Industrial Press: New York, 1974.
40 mm
R. O. Parmley, ed. Standard Handbook of Fastening and Joining. McGrawHill: New York. 1977. H. A. Rothbart, ed., Mechanical Design and Systems Handbook. McGrawHill: New York, Sections 20, 21, 26, 1964.
15.15 PROBLEMS 151 Compare the tensile load capacity of a 5/1618 UNC thread and a 5/1624 UNF thread made of the same material. Which is stronger? Make the same comparison for M8 x 1.25 and M8 x 1 ISO threads. Compare them all to the strength of a 5/1614 Acme thread. *152
A 3/46 Acme thread screw is used to lift a 2 kN load. The mean collar diameter is 4 cm. Find the torque to lift and to lower the load using a ballbearing thrust washer. What are the efficiencies? Is it selflocking?
FIGURE P151 Problems 154 to 156
* Answers
to these problems are provided in Appendix D.
153 A 1 3/84 Acme thread screw is used to lift a 1ton load. The mean collar diameter is 2 in. Find the torque to lift and to lower the load using a ballbearing thrust washer. What are the efficiencies? Is it selflocking? *†154
The trailer hitch from Figure 11 (p. 11) has loads applied as shown in Figure P151. The tongue weight of 100 kgf acts downward and the pull force of 4905 N acts horizontally. Using the dimensions of the ball bracket in Figure 15 (p. 14), draw a freebody diagram of the ball bracket and find the tensile and shear loads applied to the two bolts that attach the bracket to the channel in Figure 11. Size and specify the bolts and their preload for a safety factor of at least 1.7.
155 For the trailer hitch of Problem 34, determine the horizontal force that will result on the ball from accelerating a 2000kg trailer to 60 m/s in 20 sec. Assume a constant acceleration. Size and specify the bolts and their preload for a safety factor of at least 1.7.‡ *156
For the trailer hitch of Problem 34, determine the horizontal force that will result on the ball from an impact between the ball and the tongue of the 2000kg trailer if the hitch deflects 1 mm on impact. The tractor weighs 1000 kg. The velocity at impact is 0.3 m/s. Size and specify the bolts and their preload for a safety factor of at least 1.7.‡
*157
A 1/2india UNC, class 7 bolt with rolled threads is preloaded to 80% of its proof strength when clamping a 3inthick sandwich of solid steel. Find the safety factors against static yielding and joint separation when a static 1000lb external load is applied.‡
158 An M14 x 2, class 8.8 bolt with rolled threads is preloaded to 75% of its proof strength when clamping a 3cmthick sandwich of solid aluminum. Find the safety factors against static yielding and joint separation when a static 5kN external load is applied.‡ *159
A 7/16india UNC, class 7 bolt with rolled threads is preloaded to 70% of its proof strength when clamping a 2.75inthick sandwich of solid steel. Find the safety factors against fatigue failure, yielding, and joint separation when a 1000lb (peak) fluctuating external load is applied.‡
1510 An M12 x 1.25, class 9.8 bolt with rolled threads is preloaded to 85% of its proof strength when clamping a 5cmthick sandwich of aluminum. Find the safety factors
†
Problem numbers in italics are design problems. Problem numbers in boldface are extended from similar problems in earlier chapters with the same dash no.
‡
For these problems, assume that the nut and washer, together, have a thickness equal to the bolt diameter, and assume that bolts are available in length increments of either 0.25 in or 5 mm.
15
966
MACHINE DESIGN

An Integrated Approach

Sixth Edition
against fatigue failure, yielding, and joint separation when a 2.5kN (peak) fluctuating external load is applied.‡ to these problems are provided in Appendix D.
*1511
†
*1513
* Answers
Problem numbers in italics are design problems. Problem numbers in boldface are extended from similar problems in earlier chapters with the same dash number.
Find the tightening torque required for the bolt in Problem 157.
1512 Find the tightening torque required for the bolt in Problem 158. Find the tightening torque required for the bolt in Problem 159.
1514 Find the tightening torque required for the bolt in Problem 1510. 1515 An automobile manufacturer would like a feasibility study of the concept of buildingin electricmotorpowered screw jacks at each end of the car to automatically jack the car wheels off the ground for service. Assuming a 2ton vehicle with a 60/40 front/ rear weight distribution, design a selflocking screw jack capable of lifting either end of the car. The jack body will be attached to the car frame and the screw will extend downward to engage the ground. Assume a minimum installed clearance of 8 in under the retracted screw in the up position. It must lift the car frame at least 8 additional inches. Use rollingelement thrust bearings. Determine a minimum screw size safe against column buckling. Determine its required lifting torque and efficiency and the power required to lift it to full height in 45 sec. What is your recommendation as to the feasibility of this idea? 1516 Design a manual screw jack similar to that shown in Figure 154 for a 20ton lift capacity and a 10cm lift stroke. Assume that the operator can apply a 400N force at the tip of its bar handle to turn either the screw or nut depending on your design. Design the cylindrical bar handle to fail in bending at the design load before the jack screw fails, so that one cannot lift an overload and fail the screw. Use rollingelement thrust bearings. Seek a safety factor of 3 for thread or column failure. State all assumptions. *1517
(a) (b) (c) (d) (e)
Table P151 Data for Problem 1519 Row
Member Mat'l (mm)
m8 x 1
steel
30
b
m8 x 1
alum
40
c
m14 x 2
steel
38
d
m14 x 2
alum
45
e
m24 x 3
steel
75
f
m24 x 3
alum
90
a
15
Bolt Thd
Determine the effective spring constant of the following sandwiches of materials under compressive load. All are uniformly loaded over their 10cm2 area. The first and thirdlisted materials are each 10 mm thick and the middle one is 1 mm thick, together making a 21mmthick sandwich. aluminum, copperasbestos, steel steel, copper, steel steel, rubber, steel steel, rubber, aluminum steel, aluminum, steel
In each case determine which material dominates the calculation. *1518
Determine the effective spring constant of the following sandwiches of materials under compressive load. All are uniformly loaded over their 1.5in2 area. The first and thirdlisted materials are each 0.4 in thick and the middle one is 0.04 in thick, together making a 0.84inthick sandwich. (a) (b) (c) (d) (e)
Copyright © 2018 Robert L. Norton: All Rights Reserved
aluminum, copperasbestos, steel steel, copper, steel steel, rubber, steel steel, rubber, aluminum steel, aluminum, steel
In each case determine which material dominates the calculation. ‡
For these problems, assume that the nut and washer, together, have a thickness equal to the bolt diameter, and assume that bolts are available in length increments of either 0.25 in or 5 mm.
1519 A preloaded steel bolt similar to that shown in Figure 1521 clamps two flanges of total thickness l. Using the data given in the rows assigned in Table P151, find the joint stiffness constant.‡ *1520
A singlecylinder aircompressor head sees forces that range from 0 to 18.5 kN each cycle. The head is 80mmthick aluminum, the unconfined gasket is 1mmthick Teflon,
Chapter 15
SCREWS AND FASTENERS
967
and the block is aluminum. The effective clamp length of the cap screw is 120 mm. The piston is 75mm dia and the cylinder is 140mm outside dia. Specify a suitable number, class, preload, and bolt circle for the cylinderhead cap screws to give a minimum safety factor of 1.2 for any possible failure mode. 1521 A singlecylinder engine head sees explosive forces that range from 0 to 4000 lb each cycle. The head is 2.5inthick cast iron, the unconfined gasket is 0.125inthick copperasbestos, and the block is cast iron. The effective clamp length of the cap screw is 3.125 in. The piston is 3in dia and the cylinder is 5.5in outside dia. Specify a suitable number, class, preload, and bolt circle for the cylinderhead cap screws to give a minimum safety factor of 1.2 for any possible failure mode. †1522
The forgedsteel connecting rod for the engine of Problem 1521 is split around the 38mmdia crankpin and fastened with two bolts and nuts that hold its two halves together. The total load on the two bolts varies from 0 to 8.5 kN each cycle. Design these bolts for infinite life. Specify their size, class, and preload.‡
*1523
(See also Problem 433.) The bracket in Figure P152 is fastened to the wall by 4 cap screws equispaced on a 10cmdia bolt circle and arranged as shown. The wall is the same material as the bracket. The bracket is subjected to a static force F, where F and the beam’s other data are given in the row(s) assigned from Table P152. Find the forces acting on each of the 4 cap screws due to this loading and choose a suitable cap screw diameter, length, and preload that will give a minimum safety factor of 2 for any possible mode of failure.‡
*1524
(See also Problem 633.) The bracket in Figure P152 is fastened to the wall by 4 cap screws equispaced on a 10cmdia bolt circle and arranged as shown. The wall is the same material as the bracket. The bracket is subjected to a sinusoidal forcetime function with Fmax = F and Fmin = –F, where F and the beam’s other data are given in the row(s) assigned from Table P152. Find the forces acting on each of the 4 cap screws due to this fully reversed loading and choose a suitable cap screw diameter, length, and preload that will give a minimum safety factor of 1.5 for any possible mode of failure for N = 5E8 cycles.‡
*1525
(See also Problem 634.) The bracket in Figure P152 is fastened to the wall by 4 cap screws equispaced on a 10cmdia bolt circle and arranged as shown. The bracket is subjected to a sinusoidal forcetime function with Fmax = F and Fmin = 0, where F and the beam’s other data are given in the row(s) assigned from Table P152. Find the forces acting on each of the 4 cap screws due to this fluctuating loading and choose a suitable cap screw diameter, length, and preload that will give a minimum safety factor of 1.5 for any possible mode of failure for N = 5E8 cycles.‡
†1526
(See also Problem 642.) A cylindrical, steel tank with hemispherical ends is required to hold 425 kPa of pressurized air at room temperature. The pressure cycles from zero
y
wall
t
z x FIGURE P152 Problems 1523 through 1525
* Answers
to these problems are provided in Appendix D.
‡
For these problems, assume that the nut and washer, together, have a thickness equal to the bolt diameter, and assume that bolts are available in length increments of either 0.25 in or 5 mm.
15
a
od
Problem numbers in italics are design problems. Problem numbers in boldface are extended from similar problems in earlier chapters with the same dash number.
F
l
t
†
id
arm
h
968
MACHINE DESIGN

An Integrated Approach

Sixth Edition
Table P152 Data for Problems 1523 through 1525 Use Only Data Relevant to the Particular Problem—Lengths in mm, Forces in N Row
†
Problem numbers in italics are design problems. Problem numbers in boldface are extended from similar problems in earlier chapters with the same dash number.
a
100
400
10
20
50
10
4
steel
b
70
200
6
80
85
12
6
steel
c
300
100
4
50
95
15
7
steel
d
800
500
6
65
250
25
15
alum
e
85
350
5
96
900
40
30
alum
f
50
180
4
45
950
30
25
alum
g
160
280
5
25
850
45
40
steel
h
200
100
2
10
800
40
35
steel
i
400
150
3
50
950
45
38
steel
j
200
100
3
10
600
30
20
alum
k
120
180
3
70
880
60
55
alum
l
150
250
8
90
750
45
30
alum
m
70
100
6
80
500
20
12
steel
n
85
150
7
60
820
25
15
steel
to maximum. The tank diameter is 0.5 m and its length is 1 m. The hemispherical ends are attached by some number of bolts through mating flanges on each part of the tank. A 0.5mmthick, compressed asbestos, unconfined gasket is used between the 10mmthick steel flanges. Determine a suitable number, class, preload for, and size of bolts to fasten the ends to the tank. Specify the bolt circle and outside diameter of the flange needed to prevent leakage. A minimum safety factor of 2 is desired against leakage and a safety factor of 1.5 against bolt failure for infinite life.‡ 1527 Repeat Problem 1526 using a confined Oring gasket.‡ 1528 Calculate the proof load (load that causes a tensile stress equal to the proof strength) for 1/213 UNC bolts in each SAE grade listed in Table 156. 1529 Calculate the proof load (load that causes a tensile stress equal to the proof strength) for M20 x 2.50 bolts in each class listed in Table 157.
‡
For these problems, assume that the nut and washer, together, have a thickness equal to the bolt diameter, and assume that bolts are available in length increments of either 0.25 in or 5 mm.
15
1530 Determine the joint stiffness constant for the bolt and members in Problem 157.‡ 1531 Determine the joint stiffness constant for the bolt and members in Problem 158.‡ 1532 Determine the joint stiffness constant for the bolt and members in Problem 159.‡ 1533 Determine the joint stiffness constant for the bolt and members in Problem 1510.‡ 1534 Figure P153 shows a bolted and doweled joint eccentrically loaded in shear. The shear loads are taken by the dowel pins, the number and size of which are given in Table P153. Though the figure shows 5 dowel pins, that is not the case for every row in the table. For a = 4 in, b = 4 in, l = 10 in, P = 2500 lb, and the data in the row(s) assigned from Table P153, find the magnitude and direction of the total shear force acting on each dowel. 1535 Figure P153 shows a bolted and doweled joint eccentrically loaded in shear. The shear loads are taken by the dowel pins, the number and size of which are given in Table P153. Though the figure shows 5 dowel pins, that is not the case for every row in the
Chapter 15
SCREWS AND FASTENERS
969
l A
B b/2
C
D
b E
a/2 P
a FIGURE P153
Copyright © 2018 Robert L. Norton: All Rights Reserved
Problems 1534 through 1536
table. For a = 4 in, b = 4 in, l = 10 in, P = 2500 lb, and the data in the row(s) assigned from Table P153, find the total shear stress in each dowel. 1536 Figure P153 shows a bolted and doweled joint eccentrically loaded in shear. The shear loads are taken by the alloy steel dowel pins, the number and size of which are given in Table P153. Though the figure shows 5 dowel pins, that is not the case for every row in the table. For a = 4 in, b = 4 in, l = 10 in, P = 2500 lb, and the data in the row(s) assigned from Table P153, find the factor of safety against yielding in shear for each dowel pin. See Table 1512 for strength data. 1537 The coefficient of friction for an oillubricated, singlestart, power screw threadnut combination is 0.10. Which of the American Standard Acme Threads in Table 153 will be selflocking for this threadnut combination? Which will be the least likely to back down in the presence of dynamic loading? Which will be the most likely to back down in the presence of a dynamic load? 1538 The coefficient of friction for an oillubricated, singlestart, power screw threadnut combination is 0.20. Which of the American Standard Acme Threads in Table 153 will have the greatest efficiency (neglecting thrustcollar friction)?
Table P153 Data for Problems 1534 through 1536 Dowels with Diameters of Zero Are Not Present Row
Number of Dowels, n
a
5
0.250
0.250
0.250
0.250
0.250
b
4
0.250
0.250
0
0.250
0.250
c
5
0.375
0.250
0.250
0.375
0.250 0.250
d
5
0.375
0.375
0.250
0.250
e
3
0.375
0.375
0
0.375
0
f
3
0.375
0
0
0.375
0.375
Copyright © 2018 Robert L. Norton: All Rights Reserved
15
970
MACHINE DESIGN
* Answers
*1539
to these problems are provided in Appendix D.

An Integrated Approach

Sixth Edition
Determine the number of engaged screw threads required to make the total strippingshear area of the engaged threads equal to twice the tensile stress area for each of the fine pitch screw threads in Table 152.
1540 Calculate the ultimate tensile loads (load that causes a tensile stress equal to the tensile strength) for 1/213 UNC bolts in each SAE grade listed in Table 156. *1541
Calculate the ultimate tensile loads (load that causes a tensile stress equal to the tensile strength) for M20 x 2.50 bolts in each class listed in Table 157.
1542 3/816 UNC bolts and nuts clamp 0.75inthick aluminum cover plate to a 0.50inthick steel flange. Determine the stiffness factor of the joint at each bolt. 1543 M14 x 2.0 bolts and nuts clamp a 16mmthick aluminum cover plate to a 12mmthick steel flange. Determine the stiffness factor of the joint at each bolt. 1544 M16 x 2.0 bolts and nuts clamp a 50mmthick aluminum cover plate to a 30mmthick steel flange. Determine the stiffness factor of the joint at each bolt. 1545 5/1618 UNC bolts and nuts clamp a 1.625inthick castiron cover plate to a 1.5inthick steel flange. Determine the stiffness factor of the joint at each bolt. 1546 M16 x 1.5 bolts and nuts clamp an 8mmthick titanium cover plate to a 8mmthick stainless steel flange. Determine the stiffness factor of the joint at each bolt. 1547 An M12 X 1.25 soft steel nut is assembled with a hardened steel bolt. The nut is 11 mm thick and has a shear yieldstrength of 120 MPa. Determine the axial force that will cause stripping failure of the nut if the nut threads fail before the bolt fails. 1548 Compare the yield loads (load that causes a tensile stress equal to the yield strength) to the proof loads (load that causes a tensile stress equal to the proof strength) for M12 x 1.25 bolts in each class listed in Table 147. 1549 An M16 x 1.50, class 4.8 bolt with cut threads is preloaded to 85% of its proof strength when clamping a 20mmthick sandwich of solid steel. Find the safety factors against static yielding and joint separation when a static 3kN external load is applied. 1550 A 15mmthick steel cap is to be fastened to a 15mmthick steel flange with 6 bolts and nuts. The external load on the cap is 30 kN. Size and specify the bolts for a safety factor of at least 1.5 and specify the torque required on each bolt to obtain the preload if the threads are lubricated. 1551 Repeat Problem 1550 with a total external load on the six bolts that varies from 0 to 30 kN per cycle. Design these bolts for infinite life with a factor of safety of at least 1.5. Specify their size, class, preload, and tightening torque.
15
1552 A 20mmthick aluminum cap is to be fastened to a 20mmthick aluminum flange with 8 bolts and nuts. The external load on the cap varies from 0 to 40 kN per cycle. Size and specify the bolts for infinite life and a safety factor of at least 1.5 and specify the torque required on each bolt to obtain the preload if the threads are lubricated. 1553 An M8 X 1.00 soft steel nut is assembled with a hardened steel bolt. The nut is 8 mm thick and has a shear yieldstrength of 140 MPa. Determine the axial force that will cause stripping failure of the nut if the nut threads fail before the bolt fails. 1554 An M8 x 1.00, class 9.8 bolt is preloaded to 80% of its proof strength. Find the tightening torque required to obtain this condition. 1555 A cylinder with an unconfined gasket between head and body applies a static load of 15.5 kN to its head. The head is aluminum, the unconfined gasket is 2mmthick plain rubber, and the cylinder flange is aluminum. The pressure diameter is 80 mm and the outside
Chapter 15
SCREWS AND FASTENERS
971
flange diameter is 130 mm. The effective clamp length of a cap screw is 80 mm. Determine the safety factor against static failure of the cap screws if there are 12 M8 x 1.25, class 8.8 screws preloaded to 90% of proof strength on a 110mmdiameter bolt circle. 1556 A 0.600inthick steel cap is to be fastened to a 0.600inthick steel flange with 5 bolts and nuts. The external load on the cap is 5500 lb. Size and specify the bolts for a safety factor of at least 1.5 and specify the torque required on each bolt to obtain the preload if the threads are lubricated. 1557 A 3/4inthick aluminum cap is to be fastened to a 3/4inthick aluminum flange with 6 bolts and nuts. The external load on the cap varies from 0 to 6750 lb per cycle. Size and specify the bolts for infinite life and a safety factor of at least 1.5 and specify the torque required on each bolt to obtain the preload if the threads are lubricated. 1558 A cylinder with an unconfined gasket between head and body applies a static load of 14 kN to its head. The head is steel, the copperasbestos, unconfined gasket is 3mmthick and the cylinder flange is steel. The pressure diameter is 80 mm and the outside flange diameter is 140 mm. The effective clamp length of a cap screw is 80 mm. Specify a suitable number, class, preload, and bolt circle for the cylinder head cap screws to give a minimum safety factor of 1.5 for any possible failure mode and specify the torque required on each bolt to obtain the preload if the threads are lubricated. 1559 Experiments have shown that, with a certain lubricant, the thread coefficient of friction is µ = 0.12 and the clamping coefficient of friction is µc = 0.10 for UNF2 threads in the range of 1/4 to 1/2 in. Calculate the torque coefficient for the standard bolt sizes in this range when this lubricant is used. 1560 Two steel machine members will be held together by a single bolt and nut. The total clamped length is 2.75 in. The bolt will be subjected to a fluctuating peak load of 2000 lb. Size and specify the bolt for infinite life and a fatigue safety factor of at least 1.4 and specify the torque required on the bolt to obtain the preload if the threads are lubricated. 1561 A cylinder with an unconfined gasket between head and body applies a static load of 3150 lb to its head. The head is steel, the copperasbestos, unconfined gasket is 0.125inthick and the cylinder flange is steel. The pressure diameter is 3 in and the outside flange diameter is 5.5 in. The effective clamp length of a cap screw is 3 in. Specify a suitable number, class, preload, and bolt circle for the cylinder head cap screws to give a minimum safety factor of 1.5 for any possible failure mode and specify the torque required on each bolt to obtain the preload if the threads are lubricated. 1562 Two steel machine members will be held together by a single bolt and nut. The total clamped length is 70 mm. The bolt will be subjected to a fluctuating peak load of 9 kN. Size and specify the bolt for infinite life and a fatigue safety factor of at least 1.3 and specify the torque required on the bolt to obtain the preload if the threads are lubricated.
15
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An Integrated Approach
NOTES
15

Sixth Edition
WELDMENTS To invent, you need a good imagination and a pile of junk. Thomas a. Edison
16.0
INTRODUCTION
Weldments (welded assemblies) are used in many applications such as machine frames, machine parts, building structures, bridges, ships, vehicles, construction equipment, and many other systems. Our focus here will be on their use in machine design rather than as structural elements in buildings and bridges, though the principles of weldment design are similar across applications. We also do not address welded pressure vessels that see elevated temperatures and can have corrosion issues. ASME publishes detailed codes for these applications. A halfcentury ago, machine frames were commonly made as greyiron castings, which have good damping. Now it is as common to see machines with welded steel frames. One reason for the change is the superior stiffness of steel over grey cast iron (30E6 psi versus 15E6 psi). A steel frame can then be lighter than a castiron casting and have equivalent stiffness or be the same weight and much stiffer. Unlike structural steel applications such as buildings, which have relatively loose tolerances on their dimensions, machine frames and parts usually have to be made to tight dimensional tolerances. But it is difficult to hold tight tolerances when welding an assembly. This requires (as it does for castings) that the weldment be machined as an assembly once it is welded into the desired configuration on any surfaces whose dimensions are critical. Noncritical surfaces can be left in the aswelded condition. Most metals can be welded though some are easier to weld than others. Low carbon steel is one of the easiest to weld. High carbon and alloy steels are more difficult to weld and, if parts are hardened or coldrolled before welding to improve strength, the high localized heat of welding will tend to anneal it locally, reducing its strength. For these reasons, it is generally recommended to limit steel weldments to low carbon and lowalloy steels. Aluminum can be welded but requires proper attention to choice of process and This chapter is Copyright © 2018 Robert L. Norton: All Rights Reserved
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16
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MACHINE DESIGN
Opening page photograph A man gas metal arc welding by William M. Plate Jr., USAF (photograph is in public domain).
Table 160 Symbol

An Integrated Approach

Sixth Edition
Variables Used in This Chapter Variable
ips units
SI units
See
A
amplitude ratio
none
none
Eq. 6.1d
Ashear
shear area of weld
in2
mm2
Eq. 16.4
Aw
shear area per unit length of weld
in
mm
Eq. 16.4
Cf
Coefficient for Sfr equation
none
none
Table 164
Exx
min. ultimate tensile strength of electrode
kpsi
—
fb
bending load per unit length of weld
lb/in
N/mm
Eq. 164
fn
normal load per unit length of weld
lb/in
N/mm
Eq. 164
fs
shear load per unit length of weld
lb/in
N/mm
Eq. 164
ft
torsional load per unit length of weld
lb/in
N/mm
Eq. 164
Fr
torsional load per unit length of weld
lb/in
N/mm
Eq. 164
Jw
polar 2nd moment per unit length of weld
in3
mm3
Eq. 16.4
N Nf
number of cycles
none
none
Ex. 162
fatigue stress safety factor
none
none
Ex. 164
Nfr
fatigue stressrange safety factor
none
none
Eq. 16.3
Nyield
static yieldstress safety factor
none
none
Eq. 16.3
R Ser
stress ratio
none
none
Eq. 6.1d
tensile stressrange endurance strength
psi
MPa
Table 164
Sers
shear stressrange endurance strength
psi
MPa
Ex. 162
Sfr
tensile stressrange fatigue strength
psi
MPa
Eqs. 16.2
Sf r s
shear stressrange fatigue strength
psi
MPa
Ex. 164
Sw
section modulus per unit length of weld
in2
mm2
Eq. 16.4
t
weld throat dimension
in
mm
Fig. 164
w
weld leg dimension
in
mm
Fig. 164
Z
section modulus
in3
mm3
Eq. 4.11d
∆σ σx σ1,2,3 σa σm σ' σmax σmin ∆τ τxy τallow
Table 164
stress range
psi
MPa
Eq. 6.1a
normal stress
psi
MPa
Eq. 4.7
principal stresses
psi
MPa
Ex. 164
alternating normal stress
psi
MPa
Eq. 6.1b
mean normal stress
psi
MPa
Eq. 6.1c
von Mises effective stress
psi
MPa
Ex. 164
maximum applied normal stress
psi
MPa
Eq. 6.1a
minimum applied normal stress
psi
MPa
Eq. 6.1a
shear stress range
psi
MPa
Ex. 162
shear stress
psi
MPa
Eq. 4.9
allowable shear stress
psi
MPa
Ex. 161, 2
technique. Machine frames are usually made of steel for strength and stiffness. Their large mass is not of concern since they are static. Machine parts that move, such as links, may be better made from aluminum if they experience large accelerations. Many links will nevertheless need to be of steel for strength, stiffness, or wear issues.
Chapter 16
WELDMENTS
The reader should consider this chapter to be but a brief introduction into what is actually a quite fascinating and complicated technology. This short treatment will not make you an expert weldment designer by any means. But it will get you started and point you to the extensive literature and code publications available on the topic.* Several organizations do research in welding and publish recommendations and requirements. Some of these are: American Association of State Highway and Transportation Officials (AASHTO)— http://www.transportation.org/
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*
The Lincoln Electric Company (www.lincolnelectric.com) offers a short course called “Blodgett’s Welding Design,” which is an excellent introduction to welding processes and weldment design for any engineer.
American Institute of Steel Construction (AISC)—http://www.aisc.org/ American Petroleum Institute (API)—http://www.api.org/ American Society of Mechanical Engineers (ASME)—http://www.asme.org/ American Welding Society (AWS)—http://www.aws.org/ James F. Lincoln Arc Welding Foundation—http://www.jflf.org/ Welding Research Council—http://www.forengineers.org
See the references at the end of this chapter for some of their relevant publications.
16.1
WELDING PROCESSES
Arc welding of metals requires the localized application of sufficient heat to melt the base material while adding compatible filler material to join the two parts. A properly applied weld can be as strong as the material adjacent to it but if improperly done can leave the assembly severely weakened. The heat is typically supplied by bringing an electrode close to or in contact with the surface, causing an arc to jump between electrode and workpiece. The “arcwelding” machine supplies either DC or AC current at sufficient voltage to create the arc which has a temperature of 6000–8000 °F, well above the melting point of steel.† The weld filler metal is supplied either as the electrode itself or as a separate wire fed into the arc and are consumed in the process. A good weld requires fusion of the metal on both sides of the joint with the added weld metal and fusion requires atomic cleanliness. The oxygen in air will contaminate the surface with metal oxide rapidly at these temperatures. The nitrogen in air also can compromise weld quality and trap bubbles in the molten metal as it cools causing porosity that weakens the weld. Moisture in the air or on the metal will cause hydrogen embrittlement and weaken the weld. To prevent contamination of the heated metal, either a flux material is supplied that covers the molten pool of metal with slag while it cools, or a stream of inert gas, such as argon or helium, is used to displace the air. If slag is present, it is chipped off when the weld cools. A good weld also requires that the molten mass penetrate into the parent metal, making the final weld metal a combination of the filler material and the parent material. There will also be a heat affected zone or HAZ formed at the edges of the weld as shown in Figure 161. This HAZ can be weaker than the parent material in higher strength steel (above 50 kpsi tensile strength), or stronger and harder than the parent in lowstrength steel, which promotes crack formation. Aluminum strength is reduced up to 50% in the HAZ. Figure 161 shows typical weld terminology. The toes are at the interface between the weld and the base material on the weld face. The root is at the base of the weld.
†
Oxyacetylene gas welding is not generally used where highstrength welds are required. It is slow and the gas flame causes too much oxidization that contaminates and weakens the weld. It is sometimes used for field repairs but not generally for new work.
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reinforcement
face

An Integrated Approach
throat
heat affected zone (HAZ) backing
Sixth Edition
reinforcement
toe
throat

weld outline toe penetration
root root opening (a) General weld terminology
note flaw (crack) in throat
HAZ (b) A fillet weld cross section showing the HAZ and a flaw
FIGURE 161
Copyright © 2018 Robert L. Norton: All Rights Reserved
Weld Cross Section and Terminology Photo Courtesy of Lincoln Electric Co, Cleveland, OH
Preparation of the parts for welding may involve contouring the edges and leaving a root opening to allow the heat and weld metal to penetrate fully. A root opening may require a backing strip be used to keep the molten puddle in place until it solidifies. The backing can be the same or different material. If the same, it will become welded to the joint. It can be left in place or removed by grinding. The latter is usually recommended if the joint is dynamically loaded as the backing creates stress concentrations. The reinforcement is the amount of weld material that protrudes above the surface of the base metal. This can be left in place for statically loaded joints but should be ground flush to remove the stress concentration at the toes if dynamically loaded. The material in the reinforcement is not considered to contribute to the strength of the weld regardless of loading. The throat dimension, which is used to determine the stress area excludes any weld material outside the part thickness or weld outline. Types of Welding in Common Use Shielded Metal Arc Welding (SMAW), also called “stick welding,” uses discrete lengths of electrode (sticks) that are coated with flux on the outside. As the arc melts the electrode, liquefied flux flows onto the pool to cover and protect it from air contact. This method is commonly used outdoors or for field repairs as it has no gas stream that could be blown away by wind. Flux Cored Arc Welding (FCAW) uses hollow electrode wire with flux trapped in its hollow core. This allows it to be spooled in long lengths. The welding machine has a wire feeder that drives the wire through the welding head at a rate controlled by the welder, making it a continuous and more efficient process. FCAW wire electrodes are available for use either with or without an inert gas stream. The gas stream makes it easier to use indoors, but with the right electrode it can also be used without gas.
16
Gas Metal Arc Welding (GMAW), also called MIG (Metal Inert Gas) welding, uses a wire electrode with no flux. An inert gas is directed onto the weld to displace the air. This makes cleaner welds because of the absence of slag and its necessary cleanup but cannot be used outside if the wind blows over 5 m.p.h. Gas Tungsten Arc Welding (GTAW), also called TIG (Tungsten Inert Gas) or Heliarc welding, uses a nonconsumable tungsten wire electrode. A separate metal filler rod or wire is fed into the molten pool. An inert gas, such as argon, is directed onto the weld to protect it. The original gas used was helium, which gave it the Heliarc name. This method is often used on aluminum, titanium, and magnesium, and for fine repairs. It gives a clean weld but has the same wind limitation outside as does GMAW.
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Submerged Arc Welding (SAW) uses powdered flux that is applied in such quantity as to bury or “submerge” the entire weld in a blanket so thick that the arc cannot be seen. The operator needs only untinted eye protection. The flux is delivered to the arc as a powder flowing through a tube adjacent to or concentric with the electrode. After the weld cools, the unmelted powder is brushed or vacuumed away and can be reused. The melted slag is chipped away to expose the weld. This process is limited to welds done on a top surface and is best suited to production welds in a shop where the motion and path of the electrode can be controlled automatically by a robot or semiautomatically by guides. It gives a good appearing weld that is free of spatter. Resistance welds are created in thin sheet metal by a similar electrical process. Electrodes pinch the metal sandwich and a high current is passed through, fusing the two materials together in a “spot.” If the electrodes are moved along the parts with the current on, it will create a “seam” weld. A laser can be used to create the needed heat instead of electrodes. No filler material is added in these welds. They are commonly used to weld automotive bodies together as well as sheet metal enclosures and other thinwalled structures but are not used for thicker sections. Why Should a Designer Be Concerned with the Welding Process? It is useful for the design engineer to have a basic understanding of these welding processes and their limitations, just as one needs to understand how a part can (and cannot) be machined on a lathe or milling machine. But most design engineers are not also machinists nor are they often certified welders. So just as engineers don’t try to tell an expert machinist how to make the part, they should leave the detailed decisions on welding process to the expert welder. But they nevertheless must be familiar with its limitations. The design engineer’s task is to design the weldment according to good and accepted engineering practice, size the welds based on the techniques to be explained here (and in the references of this chapter) so that they are safe against failure in expected use, choose the needed strength of weld material, and specify this information on the drawing.
16.2
WELD JOINTS AND WELD TYPES
There are five types of weld joints as shown in Figure 162, butt, tee, corner, lap, and edge. The choice of joint type will to some degree be dictated by the desired geometry of the weldment and a given weldment may have several types within it. There are three general types of welds that can be used with one or more of the five joint types: groove welds, fillet welds, and plug/slot welds as shown in Figure 163. Groove welds break into two subcategories, having either complete or partial penetration. It is generally recommended that plug and slot welds be avoided as they tend to be weaker than the others. We will focus here on the two subtypes of groove welds and fillet welds. Groove welds are suitable for butt joints, outside corner joints, and edge joints in materials with sufficient thickness. Fillet welds are suitable for tee joints, lap joints, and inside corner joints. Groove welds can have complete joint penetration (CJP) or partial joint penetration (PJP) as shown in Figures 163 and 164. A CJP groove butt weld statically loaded in tension will be as strong as the thinner of the two materials joined. The strength of a PJP joint depends on the depth of its “throat” as defined in Figure 164. PJP welds
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MACHINE DESIGN

An Integrated Approach

Sixth Edition
butt tee
CJP Groove edge
PJP Groove
corner lap FIGURE 162
Fillets
Copyright © 2018 Robert L. Norton: All Rights Reserved
Types of Welded Joints
are typically used on both sides of thick sections where a CJP weld would be larger than necessary. Note that any “reinforcement” in the bead that protrudes outside the surface of the welded parts is not included in the throat measurement. In fatigue loaded applications, unless the stress range is sufficiently low, the reinforcement may need to be ground flush with the material to eliminate stress concentrations in the rippled surface of the weld and at the toes. The total weld throat area is the throat dimension times the length of weld. The fusion area is the area of the junction between the weld and the base material as shown in Figure 164.
Plug
Slot Copyright © 2018 Robert L. Norton: All Rights Reserved
FIGURE 163 Types of Welds
Fillet welds are defined by their leg dimension w, but the weld strength is limited by the throat dimension t as shown in Figure 164. Fillet welds are typically at 45° between two orthogonal parts but can join parts at any angle. If the pieces joined are orthogonal and the fillet at 45°, then the throat width t is 0.707 times the leg width w. Any reinforcement is ignored. The total weld area is the throat width t times the length of weld but the fusion area that determines whether the weld tears away from the base metal is the leg width w times the length of weld on each leg for a fillet weld. The stress on each leg’s fusion area may be the same or different depending on part loading.
reinforcement
throat (t)
CJP
reinforcement fusion area
fusion area leg (w)
16
face
Fillet
reinforcement
fusion area
fusion areas
FIGURE 164 Throat Dimensions of Welded Joints Data from Lincoln Arc Welding Co., Cleveland, OH
(t)
root
at
ro
th
throat (t)
PJP
toe
leg (w) Copyright © 2018 Robert L. Norton: All Rights Reserved
Chapter 16
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979
Joint Preparation Better welds will result if the joint is properly prepared to allow the heat and weld metal to reach and fuse to all portions of the joint area. Unless the sections are thin, the weld joint should be prepared by removing metal from one or both sides of the joint. Several joint shapes are recommended, U, J, and V, as shown in Figure 165. The J or U joint detail leaves a small amount of base material in the bottom of the slot to prevent the molten weld metal from running out but is thin enough to allow good penetration. A V groove is easier to machine but may need to have a gap at the bottom to get good penetration. This gap can be closed with a backing strip of metal or ceramic to contain the weld metal until it cools. If the backing strip is the same material as the parts joined, it too will become welded to the joint. It can be removed, but if left in place it should be continuous over the entire length of the joint. If any splices are allowed in the backing strip and if they are not fully welded, high stress concentrations will be created that could cause failure. In like fashion, one can choose to weld a long seam only intermittently rather than put down a continuous bead over its length. But, it is often better to use a continuous weld because every interruption of the bead causes a stress concentration, undesirable especially with dynamic loading as is common in machine parts. Moreover, for the same strength a bead that runs only half the length of a joint must have twice the throat dimension, which quadruples its crosssectional area. A longer, smaller crosssection bead will use less weld material and be more economical. The designer needs to select and specify the size of the groove to be machined into the parts before welding. Recommended sizes based on part thickness can be found in welding handbooks and specifications such as references 1–4. It is highly recommended that these handbooks and codes be used when designing weldments. If one is designing buildings or bridges, then the AISC and AWS codes must be adhered to. They contain very specific rules and procedures that must be followed when designing structures whose failure can put people’s lives in danger. The machine designer is usually not required to follow these codes but would be wise to do so anyway to be sure of good results.
Square edge
Single bevel
Single V
Single U
Single J
Copyright © 2018 Robert L. Norton: All Rights Reserved
FIGURE 165 Joint Preparation
Weld Specification Welds and their joint preparation are specified on a drawing using a standard form of welding symbol as shown in Figure 166a. At a minimum, it has a reference line and an arrow. There can also be an optional tail at the end opposite the arrow. The arrow points to the joint and the weld symbol defines the type of weld (fillet, CJP, PJP, etc.). Weld symbols above the reference line refer to the side opposite the arrow and those below the line refer to the arrow side. The arrow may point either up or down. The symbols are always read right to left regardless of which end the arrow is on. Figure 166b shows some of the possible weld symbols. Refer to AWS A2.4[3] for details.
16.3
PRINCIPLES OF WELDMENT DESIGN
It is just as important to pay attention to the geometry of the weldment as it is to consider the sizes of the welds if one wants the design to succeed. It is also very important to arrange the welds so that a sensible and safe load path exists to take the applied loads to their reaction points. The “forceflow” concept introduced in Chapter 4 applies here.
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An Integrated Approach

Sixth Edition
(a)
(b)
FIGURE 166
Copyright © 2018 Robert L. Norton: All Rights Reserved
Standard Welding Symbols
The following rules have been developed by welding design experts over many years and more detail on them can be found in reference 1: 1 Provide a path for applied forces to enter into the section(s) of the weldment that lie parallel to the direction of the applied force. 2 Forces will follow the stiffest path to ground so it is better to have relatively uniform stiffness to distribute reaction loads evenly in the weldment.
16
3 There are no secondary members in weldments. It is essentially one piece. But the welds can be either primary or secondary. A primary weld carries the entire load directly and its failure causes the weldment to fail. Secondary welds just hold the parts together and have low forces on them. 4 Do not put welds in bending if possible. If bending moments exist, try to place welds near locations of zero or low moment, or arrange the welds to handle shear or axial tension/compression. 5 Where possible, do not apply tensile loads across a transverse thickness of the parent
Chapter 16
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981
material (i.e., “across the grain”). Wrought metals do have a grain in the direction of rolling and are slightly weaker across the grain than with the grain. (See Section 14.3 for a discussion of grain in metals.) Welds loaded in tension on a transverse cross section (see Figure 1610) may cause “lamellar tearing” of the material under the weld. Shear loads applied to such a surface through a weld are preferable. 6 Where section size changes across a joint, taper the material around the joint to improve force flow and reduce stress concentration.
Figure 167 shows alternate designs for a hangar welded to a beam. Part a shows an inferior arrangement in which the hangar is turned 90° and so loads the flange of the beam. Since the stiffest path will carry the bulk of the load, the small length of weld that spans the central web will handle most of the force. If the entire length of the weld is sized to have acceptable stress levels, it may fail because the nonuniform stress distribution across the flange overloads the center section of the weld. The flange compliance near its ends reduces its ability to transfer load to the central web. Note the force flow arrows for each design. Part b shows an acceptable arrangement wherein the load from the hangar is transferred directly into the web of the Ibeam through the weld. Figure 168 shows a similar assembly that uses a rectangular section beam. In part a, the solution that worked in Figure 167 is now inferior because the load applied to the center of the bottom surface of the section must travel across it to reach the side webs that carry the forces into the beam. This weld experiences bending. A better arrangement is shown in part b where the hangar is redesigned to weld directly to the side webs, put the welds in shear, and carry the forces into the sections that lie parallel to the load through the hangar, which is stiffer in bending than the section’s web. Note the force flow arrows for each design. Figure 169 shows parts of differing width or thickness welded together. The AWS welding code D1.1 calls for a taper of at least 2.5:1 (≤ 22°) be provided at all such junctions across nonuniformsize butt joints when cyclically loaded.
(a) Poor design—load taken through flange into web and the center of weld takes the brunt of load
(b) Good design—load taken directly into web and whole weld shares the load Copyright © 2018 Robert L. Norton: All Rights Reserved
FIGURE 167 Examples of Good and Poor Welded Joints Data from
Lincoln Arc Welding Co., Cleveland, OH
16.4
STATIC LOADING OF WELDS
Compared to the calculation of stresses in general machine parts of complex geometry, calculation of weld stresses is fairly straightforward. If we can avoid loading welds in bending, the loading will typically be direct tension/compression or direct shear. In either
16 (a) Poor design—load taken through bottom into side webs (b) Good design—load taken directly into side webs
FIGURE 168
Copyright © 2018 Robert L. Norton: All Rights Reserved
Examples of Good and Poor Welded Joints Data from Lincoln Arc Welding Co., Cleveland, OH
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2.5
An Integrated Approach
22°

Sixth Edition
2.5
1.0
(a) Tapered width dimension
22° 1.0
(b) Tapered thickness dimension
FIGURE 169
Copyright © 2018 Robert L. Norton: All Rights Reserved
Mismatched Joints Must Be Tapered Data from Lincoln Electric Company, Cleveland, OH
case the equation for stress is simple. For axial tension or compression the normal stress was defined in equation 4.7 as: σx =
P A
(4.7)
Direct shear, which is common in welds, is defined in equation 4.9 as: τ xy =
P Ashear
(4.9)
In both cases P is the applied load and A or Ashear is the area of the weld. For a butt joint with a CJP weld in tension or compression, the weld throat area is equal to the cross section of the smaller part joined. For a PJP weld in tension or compression or a fillet weld in tension, compression, or shear, the area is simply the throat dimension t times the length of the weld. Figure 164 defines throat size for a variety of welds. Note that whether a filletwelded tee joint is loaded in tension or shear, the weld will experience shear stress and may also have tensile stress. The fusion areas between weld and base metal may have one or both types of stresses applied depending on loading.
16.5
16
* The yield strength of steel electrode material is usually taken as 75% of Sut.
STATIC STRENGTH OF WELDS
In Chapter 5 we discussed static failure theories in depth and concluded that static failures were due to shear stress and that the distortion energy theory best defines a safe stress level for a ductile material using any desired safety factor. The metals used in welding and the weld filler materials are ductile, so this theory should apply, and it does. However, extensive testing done on welded assemblies over the last halfcentury by the welding industry has developed good data on allowable strengths of welds and weldments. An electrode’s part number appears as E followed by 4 or 5 digits, the first 2 or 3 of which define its minimum ultimate tensile strength in kpsi and the rest indicate the position in which it can be used and its coating. We note its strength generally as Exx. For example, an E70 electrode has a minimum Sut = 70 kpsi and E110 has a minimum Sut = 110 kpsi.* Note that a given sample may actually exceed the stated value. It is recommended that the strength of the selected electrode be approximately matched to that of the base metal to be welded, and this is a requirement with CJP welds loaded in tension. It can be undermatched (weld metal weaker than base metal) in some cases, particularly if higher strength steels are welded or better crack resistance is
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lo ng itu di na l
needed. Undermatching is often done with fillet welds. Overmatching (weld metal stronger than base metal) is generally not recommended. Residual Stresses in Welds Welds will always have large residual tensile stresses in them. This is because weld metal expands to about six times its elongation at yield when melted. When it cools, it shrinks by that same factor. No solid metal has 600% elongation to yield, so this means that the weld material is guaranteed to yield in tension as it shrinks, since the solid base metal adjacent to the weld cannot move with it. There will be a balancing compressive stress formed in the adjacent base metal and the residual stress distribution will look like Figure 1610 when cooled. Using an undermatched (i.e., weaker) weld metal will reduce the residual stress due to its lower yield strength. It is not recommended that a significantly overmatched weld metal be used in any case as this could result in a nonconservative design. Direction of Loading
transverse (a) CJP buttwelded plates
tensile stress
compressive stress
The direction of applied load versus the weld axis has a significant effect on fillet weld strength. Tests show that fillet welds loaded orthogonal (transverse) to the long axis of the weld, have 50% greater strength than the same weld loaded along (longitudinal to) the weld axis (see Figure 1610a). This is due in part to the effective throat of a transversely loaded fillet weld between two orthogonal parts being at 67.5° versus 45° when loaded longitudinally. The 67.5° plane has 30% more shear area. Nevertheless, the AWS D1.1 code specifies that the effective throat area, defined by the shortest distance from root to weld face, be used for any direction of applied load.[9] However, longitudinal welds have the advantage of allowing more deformation before yield than transverse welds do.[9] Nevertheless, welds are designed against failure by rupture rather than yield because, even though ductile, the weld volume is so small compared to the entire part, the magnitude of weld deformation between yield and fracture is too small to provide any warning of incipient failure.
tensile stress
base metal
weld compressive stress (c) Longitudinal residual stress Copyright © 2018 Robert L. Norton: All Rights Reserved
Allowable Shear Stress for Statically Loaded Fillet and PJP Welds For static loading, the AWS[3] recommends that shear stresses in fillet or PJP welds be limited to 30% of the electrode tensile strength, Exx : τ allow = 0.30 Exx
(b) Transverse residual stress
FIGURE 1610 Residual Stresses in Welds
(16.1)
This value has a builtin minimum safety factor against fracture ranging from 2.21 to 4.06 for various weld loadings using electrode strengths from E60xx to E110xx in extensive testing.* Table 161 shows more detail on these safety factors. These safety factors and equation 16.1 are based on extensive testing done on welded assemblies by AISC[6] and assume that the weld material strength approximately matches that of the base metal in each case.† Equation 16.1 is included in the AWS D1.1 Structural Welding Code.[3] The nominal safety factor for equation 16.1 is usually stated as 2.5, which is within the range of longitudinal weld safety factor values in Table 161. Equation 16.1 is a bit unusual in that it compares applied shear stress to tensile ultimate strength as a reference value. It also appears to give a safety factor of 3.33 (the
* Note that E60XX electrodes are now considered essentially obsolete and the lowest strength electrode now in common use is E70XX. † Equation 16.1 has been verified by extensive FEA modeling that shows good correlation with this estimate.[8]
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Table 161

An Integrated Approach

Sixth Edition
Safety Factors Against Static Failure When Using Equation 16.1
As reported by Testing Engineers, Inc., 1968 (Data from Reference 16) Factor of Safety Using Stress on Throat Area Equal to 0.3 Tensile Strength of Electrode
Table 162
Minimum Fillet Weld Sizes* Base Metal Thickness (T)
Minimum Weld Size
sizes in inches
T ≤ 1/4
1/8
1/4 < T ≤ 1/2 1/2 < T ≤ 3/4 3/4 < T
3/16 1/4 5/16
sizes in mm
T≤6
3
6 < T ≤ 12
5
12 < T ≤ 20
6
20 < T
8
* Data from AWS D1.1 Copyright © 2018 Robert L. Norton: All Rights Reserved
Table 163 Minimum Strengths of Some ASTM Structural Steels
16
ASTM Number
Sy kpsi (MPa)
Sut kpsi (MPa)
A36
36 5880 (250) (400500)
A572 Gr42
42 (290)
60 (415)
A572 Gr50
50 (345)
65 (450)
A514
100 (690)
120 (828)
Copyright © 2018 Robert L. Norton: All Rights Reserved
Longitudinal Welds
Transverse Welds
Base Metal
Electrode Class
Average
Minimum
Average
Minimum
A36
E60xx
2.88
2.67
—
—
A441
E70xx
2.95
2.67
4.62
4.06
A514
E110xx
2.41
2.21
3.48
3.30
Copyright © 2018 Robert L. Norton: All Rights Reserved
reciprocal of 0.30) rather than the stated 2.5. This anomaly is explained by the mixing of shear stress and tensile strength in the equation. Equation 2.5b gives an approximate ratio between ultimate shear strength and ultimate tensile strength of ductile metals of 0.75 to 0.8. Multiplying 0.75 by 3.33 gives 2.5 as a factor against shear rupture. But, it is more convenient to use the stated minimum tensile strength of the electrode and adjust the factor in the equation from 1 / 2.5 to 1 / 3.33 = 0.30. The AWS D1.1 Structural Welding Code defines minimum sizes for welds based on thickness of the material being welded. Some of these data are shown in Table 162. The minimum weld size is to ensure that sufficient heat is applied to achieve good fusion.
EXAMPLE 161 Design of a StaticallyLoaded Fillet Weld Problem
A tee section of 0.5inthick by 4in wide ASTM 36 hotrolled steel on both legs as shown in Figure 1611 is to be fillet welded on both sides. Determine the required weld throat size t.
Given
The tee is statically loaded in tension on the center web with P = 16 800 lb applied at the 1india hole. Strengths of ASTM steels are shown in Table 163.
Assumptions
Use matching strength electrode material and run the welds the full width of the parts. The weld is directly loaded and will fail in shear along a 45° plane in its throat.
Solution
1 Table 163 gives the tensile strength of this material as 5880 kpsi. Select an electrode with approximately the same strength as the average of the material strength. Electrodes come in increments of 10 kpsi and the closest available is an E70 with 70 kpsi tensile strength. (E60xx electrodes are now considered obsolete.) 2 Determine the allowable strength based on 30% of the Exx value for this electrode using equation 16.1:
Chapter 16
WELDMENTS
τ allow = 0.30 Exx = 0.30 ( 70 ) = 21 kpsi
985
(a)
3 Determine the shear area in the throat needed to limit the stress to this value: 16 800 P = Ashear Ashear
τ xy = τ allow = 21 kpsi = Ashear =
16 800 21 000
= 0.8 in 2
(b) 0.5"
4 Determine the throat dimension of two full length fillet welds (one on each side of the joint) that will give the required area:
L = 4" P
Ashear = 2 Lt = 2(4)t = 0.8 in 2 t = 0.1 in
(c)
5 Convert this throat dimension t to a leg width w assuming an equalleg fillet in a 90° tee joint: w=
t
( )
cos 45
=
0.1 = 0.141 in 0.707
(d )
6 Check this against the recommended minimum weld size for this thickness part. Table 162 indicates that a 0.5inthick part needs at least a 3/16in weld leg width, so increase the weld’s leg width to 0.187 in. 7 Now check whether the part will fail in the fused base metal. There are two areas of concern, the areas between the welds and the base, labeled areas ’A’, which are in tension, and the areas between welds and web, labeled areas ’B’,which are in shear. Since both are the same total area and the shear strength of the base metal is about half that of its tensile strength, we need only check the shear areas ’B’ against failure. The material’s minimum tensile yield strength from Table 163 is 36 kpsi: τ xy =
16 800 P P = = = 11 230 psi A fusion 2 Lw 2 ( 4 ) (0.187)
N yield =
Ss y τ xy
=
36 000 ( 0.577 ) 11 230
=
20 772 11 230
A
0.5"
= 1.85
P/2
P/2
A
areas 'B' web
t
fillet weld
throat base areas 'A' w
leg width
View AA (e)
This is acceptable, especially since the yield strength is a guaranteed minimum value.
Copyright © 2018 Robert L. Norton: All Rights Reserved
FIGURE 1611 Examples 161 and 162
8 Check the part’s strength against tensile failure across the section at the centerline of the 1india hole: P 16 800 = = 11 200 psi A 3(0.5) S y 36 000 N yield = = = 3.2 σ x 11 200
σx =
(f)
The part is safe against yielding in tension at the hole and against failure in the weld, which limits the design. To complete the design, it needs to be checked for tearout and bearing failure at the hole, and preloaded bolts need to be designed to clamp the base of the tee. These tasks are left to the reader.
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16.6

An Integrated Approach

Sixth Edition
DYNAMIC LOADING OF WELDS
As described in Chapter 6, parts loaded dynamically fail at much lower stress levels than parts loaded statically, In that chapter we learned about fully reversed, repeated, and fluctuating stresses (see Figure 66) and also learned that the presence of a mean stress component in addition to an alternating stress component made the situation worse and required that a Goodman, Gerber, or Soderbergline analysis be done (Figures 642, 643, and 644). The mean and alternating stress components σm and σa, the stress range ∆σ, the stress ratio R, and the amplitude ratio A, were defined in equations 6.1a – 6.1d, repeated here for your convenience:
R=
σ min σ max
σ = σ max − σ min
(6.1a)
σa =
σ max − σ min 2
(6.1b)
σm =
σ max + σ min 2
(6.1c) A=
σa σm
(6.1d )
Effect of Mean Stress on Weldment Fatigue Strength
* The 2005 AISC specification[2] states on p. 16.1400: Extensive test programs using fullsize specimens, substantiated by theoretical stress calculations have confirmed the following general conclusions ... : (1) Stress range and notch severity are the dominant stress variables for welded details and beams; (2) Other variables such as minimum stress, mean stress, and maximum stresses are not significant for design purposes; and (3) Structural steels with yield points of 36 to 100 kpsi (250 to 690 MPa) do not exhibit significantly different fatigue strengths for given welded details fabricated in the same manner.
16
†
Fatigue crack propagation and fatigue life of most weldments is proportional to the third power of the stress range[10]
Weldments loaded dynamically behave in a surprisingly different way than nonwelded parts, which makes the mean stress irrelevant to their potential fatigue failure.* Figure 1612 shows fatigue test data for both unwelded and welded test specimens. The unwelded specimens were rectangular crosssection, hotrolled steel bars loaded in axial tension/compression. The welded specimens were pieces of the same bar cut, then joined with transverse CJP butt welds, making their overall geometry, material, and finish the same as the unwelded ones. Samples were tested with axial loading at stress ratios R of 1/4 (fluctuating), 0 (repeated), and –1 (fully reversed). The first two have nonzero mean stress and the last has zero mean stress. Note that nonzero mean stress reduces the strength of unwelded parts as we saw in Figure 616. But, the data for welded parts show no significant difference between fully reversed data and data with mean stress added. The stress range (equation 6.1a) is the single determining factor in failure of dynamically loaded weldments. This makes the calculation for dynamically loaded weldments simpler than for nonwelded machine parts. A Goodmanline analysis is not needed here. Instead, the stress range that a weldment sees in the cycle is compared to an acceptable stressrange fatigue strength Sfr obtained from test data.† Are Correction Factors Needed For Weldment Fatigue Strength? Recall from Chapter 6 that the fatigue strength Sf of a steel part is never more than 50% of its ultimate tensile stress Sut and is usually much less due to a number of factors that involve its surface finish, size, type of loading, and other factors (equation 6.6). The data that give the uncorrected endurance limit Se’ = 0.5 Sut come from testing smalldiameter, rotatingbeam specimens with polished surfaces and these are reported as average values. Thus, the uncorrected value Se’ must be reduced by the factors noted to account for differences in size, surface finish, etc. between the test specimen and your part and by a statistical reliability factor to obtain a corrected endurance strength Se.
WELDMENTS
987
∆σ
∆σ
Chapter 16
FIGURE 1612 FatigueStrength Data for Unwelded and Welded Parts at Various Stress Ratios[4]
Fatigue and endurance strength data for weldments are not obtained from polished laboratory specimens but rather from actual welded assemblies in a variety of configurations. Also, these test specimens are large (think building and bridgesized parts) and are made from hotrolled steel with rough surfaces, residual stresses from the rolling process, and real welds containing stress concentrations and tensile residual stress. So, we don’t have to factor these fatigue test data down to match our parts on the basis of size, surface finish, etc., as our parts are similar to the test specimens in those respects. Effect of Weldment Configuration on Fatigue Strength The fatigue resistance of a weldment also will vary according to the presence or absence of interruptions in the geometry of the assembly and of weld beads, both of which create stress concentrations.* So, the test specimens were made with deliberate stress concentrations in the form of interrupted welds, added stiffeners, and all varieties of weld joints and welds listed above. AISC defined and tested many different welded configurations, and based on the fatigue strength of the base metal, grouped them into eight categories labeled A, B, B’, C, D, E, E’, and F in order of decreasing resistance to dynamic loading. Category A is most resistant and E’ least resistant to fatigue. Note that F is the shear strength of the weld metal itself, and the others are for tensile strength of the fusion area between the weld and base material. Sketches of these configurations and their associated letter categories and fatigue strengths for various numbers of cycles for each letter category are all published in reference 2. The chart is too large to reproduce here in its entirety, but a few selected examples are shown in Figure 1613.
*
According to Barsom and Rolfe:[10] One of the most sensitive weld details is a fillet weld termination oriented perpendicular (transverse) to the applied cyclic stress field. In this case, fatigue cracking initiates from the toe of the fillet weld and propagates through the adjacent base metal. In fact, the majority of weldrelated fatigue failures initiate at the surface, generally at the weld toe. Note that several examples in this chapter have fillet welds loaded in this manner.
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Figure 1613a shows a category A part. Note that it has no welds. This is the strongest category and is in effect a reference against which the other categories can be compared. The relatively low fatigue strength of this specimen (24 kpsi) compared to a rotating beam specimen of the same material (about 30 kpsi) has to do with its much larger size, rough surface, residual stresses from the hotrolling process, and flame cut edges. Figure 1613b has uninterrupted welds running the full length or width of the part and is a category B. Figure 1613c is similar to 1613b but has attachments added. In the direction of stress, the attachments are short (just the thickness of the attached member), but some stress will flow from the main member into the attachment, creating
Category B Category A
(b)
(a)
Category C
Category B Category B'
Table 164
Category C
Reliability Factors Versus 95% for Sd = 0.08 µ
(c )
(d)
Reliability % Creliab
16
50
1.152
90
1.033
95
1.000
99
0.938
99.9
0.868
99.99
0.809
99.999
0.759
Copyright © 2018 Robert L. Norton: All Rights Reserved
Category C
(e)
FIGURE 1613
Copyright © 2018 Robert L. Norton: All Rights Reserved
AISC Strength Categories for Welded Parts Subject to Fatigue Loading Adapted from [2]
Chapter 16
WELDMENTS
989
a stress concentration and the stress category is therefore reduced to category C. Figure 1613d can be any one of three categories depending on the base material and whether or not the weld reinforcement is ground flush. Figure 1613e has very different geometry from 1613c but is also a category C with a Note C attached that applies if the web size is larger than 0.5 in. Note C is also shown in Figure 1613e. Note the level of detail in these test specimen types—there are many more in the AISC Specification.* The weld metal itself is nearly always category F except in the case of a transverse CJP weld, which can be a category B or C depending on reinforcement removal. Extensive testing of weldments in each of these categories was done by the Highway Research Board in the 1960s.[11] Testing of 374 beams of 10foot span and depths of about 15 inches having various welded details were done at two separate university laboratories. The data from both labs had close statistical correlation. Figure 1614a shows test data for category A beams and Figure 1614b shows data for category E beams. The three sets of data in each plot are from loadings with different values of minimum stress but with the same stress range. All cluster together showing that the only stress parameter that had any effect was stress range. The maximum, mean, and minimum stresses were not factors in the failures.
Table 165a[2] U.S. Coefficient Cf and Ser for Equation 16.2
Regression lines are fitted to the data on loglog axes. The equations are shown on the plots. One has a slope of 3.372 and the other a slope of 2.877. The lines above and below the mean line represent ±2 standard deviations, which includes 95% of the population. The lower band line is taken as the fatigue strength line. Exponential relationships for weldment fatigue strength as a function of number of cycles were developed from the data and are shown below. The average slope of all the data for categories A to E’ was rounded to 1/3 for the design equation 16.2a below. Figure 1614c shows the SN diagrams for each of the AISC weldment categories based on these test data. These are different from the SN diagrams in Chapter 6 in that the ordinate shows stress range ∆σ (labelled as fatigue strength range Sfr) rather than alternating stress σa (denoted as fatigue strength Sf) They also differ in that they do not show average fatigue strength values, but rather use values that are two standard deviations below the average. While not truly minimum values, they are close to minimum because 95% of the population will be within plus or minus two standard deviations from the average, meaning only 2.5% lie below these values. So it is not necessary to apply reliability factors to reduce them further unless one wants to have more than a 95% confidence level in the data, then use Table 164.
S fr = Creliab ≥ Ser N
Cf ips
Ser kpsi
A
250 E08 120 E08 61 E08 44 E08 22 E08 11 E08 3.9 E08 150 E10
24.0 16.0 12.0 10.0 10.0 4.5 2.6 8.0
B B' C D E E' F
Copyright © 2018 Robert L. Norton: All Rights Reserved
Table 165b[2] SI Coefficient Cf and Ser for Equation 16.2 AISC Category
Cf
Ser
SI
MPa
A
170 E10 83 E09 42 E09 30 E09 15 E09 7.6 E09 2.7 E09 10 E12
165 110 82 69 48 31 18 55
All categories except F have the same slope of 1/3 and their intercepts decrease with each higher category letter. Category F, which is for the weld metal rather than the base metal near the weld, has a shallower slope of 1/6 and a low intercept. Note that the knees where infinite life begins also vary with the category from 2E6 to over 1E7 cycles. These curves are linear on a loglog plot up to the knee so exponential equations can be fitted to them. For all categories except F the allowable fatigue stress range Sfr is 1 Cf 3
AISC Category
B B' C D E E' F
(16.2a)
where N is the required number of stress cycles. CF and Ser (the stress range endurance strength) are shown in Tables 165a and b for U.S. and SI units, respectively. These values are for hotrolled steels with tensile yield strengths from 36 to 110 kpsi.
Copyright © 2018 Robert L. Norton: All Rights Reserved
*
The AISC Specification For Structural Steel Buildings is a free download from www.aisc.org.
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For category F, which is for shear stress in the weld metal, it becomes: 1
S frs * The Welding Research Council[12] recommends that the exponent in equation 16.2b be 1/5 rather than 1/6.
Cf 6 = Creliab ≥ Sers N
and CF and Sers for category F are also shown in Tables 165a and b. Creliab is given in Table 164.*
(b) Experimental data for Category E
Stressrange fatigue strength Sfr or Sfrs (kpsi)
(a) Experimental data for Category A
16
(16.2b)
60 50
weldment tensile strength range Sfr weld metal shear strength range Sfrs Category
30
A
20
B B' C
10
D
5
F
E
3
E' near horizontal lines are stressrange endurance strength Ser or Sers
2
1 5
10
6
10
10
7
4.0 E7
cycles to failure, N (c) Regression lines fitted to experimental fatigue strength data for all categories
FIGURE 1614
Copyright © 2018 Robert L. Norton: All Rights Reserved
Experimental FatigueStrength Data for Welded Parts in the AISC Categories and for Weld Metal Some data from Ref. 11
Chapter 16
WELDMENTS
991
The values in Tables 165 can be reduced by a factor of three for use with aluminum. To use these strength data, calculate your applied stress range ∆σ or ∆τ and determine the safety factor for the fusion area in tension or weld metal in shear, respectively, as: N fr =
S fr σ
or
N frs =
S frs τ
(16.3)
Is There an Endurance Limit for Weldments? Until recently it was assumed that once the knee of a curve in Figure 1614c was reached, the curve remained horizontal out to infinite cycles for steel and a few other materials such as titanium. This allowed one to use that value as an endurance limit for infinite life as was done in Chapter 6 for unwelded parts. More recent research[12] indicates that the fatigue strength of weldments continues to decline beyond the knee. Figure 1615 shows sample stressrange fatigue curves for both steel and aluminum from reference 12. These data result from extensive testing done by the Welding Research Council (WRC) on weldment assemblies of various geometries. The WRC has defined categories similar to those of the AISC shown in Figure 1613 except that WRC has many more categories that are denoted by numbers rather than letters. See reference 12 for their definitions. The numbers assigned to the symbols in Figure 1615 refer to the strength of that number WRC welding category taken at 2 million cycles. The knee for steel and aluminum in tension is taken at 1E7 cycles, but shear fatigue data (not shown) have the knee at 1E8 cycles. These figures show the tensilestressrange fatigue strength of steel and aluminum weldments declining with stress cycles beyond the curve’s knee but with a much lower slope. The exponent of the curve’s equation becomes 1/22 beyond the knee.[12] Figure 1614 also reflects this fact by showing those lines with a small negative slope beyond the knee. We also avoid calling the value at the knee an endurance limit, instead referring to Sers as an endurance strength.
16 (a) Stressrange curves for steel weldments
(b) Stressrange curves for aluminum weldments
FIGURE 1615 Experimental TensileStressRange FatigueStrength Data for Welded Parts in Various Welding Research Council Categories [12]
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MACHINE DESIGN

An Integrated Approach

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lo ng itu di na l
Fatigue Failure in Compression Loading?
transverse (a) CJP buttwelded plates
tensile stress
compressive stress (b) Transverse residual stress
tensile stress
base metal
weld compressive stress
Another difference between unwelded and welded parts in fatigue has to do with the tensile residual stress in the welds. Recall from Chapter 6 that fatigue failures were stated to be due only to oscillating tensile stress. Compressive stress oscillations could be safely ignored. In fact, we saw in Chapter 15 how to use residual compressive stress in the clamp zone of preloaded bolts to partially “hide” the applied oscillating tensile stress from the tensionloaded bolt. We regard compressive stress as a friend in dynamically loaded, unwelded parts. When welds are introduced to a part this is no longer true. Oscillating compressive stress can also cause fatigue cracks. How can this be? The answer is residual tensile stress. As described earlier and as depicted in Figure 1610, a weld will always have residual tensile stress at the material’s yield point. Consider two different loading cases on a given welded part whose yield strength is 50 kpsi. In the first case, an oscillating tensile stress that varies between zero and 10 kpsi is applied in the region of the weld. On the first cycle, the stress in the weld area will go above the yield strength. The material will yield locally relieving some of the residual stress, about 10 kpsi of it. When the load cycles back to zero it has only 40 kpsi of residual stress there. The next and all successive cycles will oscillate from 40 to 50 kpsi tensile stress at that point with a stress range of 10 kpsi. Now get a fresh specimen with the same 50 kpsi residual stress in its weld and change the applied load to a compressive oscillation from zero to negative 10 kpsi. The local stress in the weld now goes from 50 to 40 kpsi tensile stress each cycle. Since the phase of the stress oscillation does not matter, this is the same tensile stress oscillation in both cases. Applied compression loading stresses that location with a varying tensile stress and so can develop cracks in the weld. These cracks only grow in the residual tensile stress zone and don’t propagate into the base metal but they still weaken the weld and can cause failure there. Chapter 6 warned against allowing residual tensile stress in fatigueloaded parts. Unfortunately, the nature of welds guarantees high tensile residual stress. This residual stress can be reduced by hammerpeening as discussed in Section 6.8, but shot peening is less effective on welds.
(c) Longitudinal residual stress Copyright © 2018 Robert L. Norton: All Rights Reserved
FIGURE 1610
Repeated
Residual Stresses in Welds
16
EXAMPLE 162 Design of a DynamicallyLoaded Fillet Weld Problem
The welded tee section of Figure 1611 (repeated here) is going to be loaded dynamically with repeated stress ranging from zero to a maximum. Determine the largest repeated tensile load that the welds can safely withstand for infinite life with a fatigue safety factor Nfrs = 1.5.
Given
The tee is 0.5inthick by 4in wide ASTM A36 hotrolled steel on both legs, and has 3/16in fillet welds full length on both sides. This matches the minimum weld size specified in Table 162.
Assumptions
Matching strength electrode material was used. Shear stress in the weld throat governs the design (see Example 161). The load is evenly distributed along the weld length.
Solution
1 A repeated stress will range from zero to a maximum value each cycle. Adapting equation 6.1a, the shear stress range ∆τ is τ = τ max − τ min = τ max
(a)
Chapter 16
WELDMENTS
993
2 The geometry of this weldment is similar to that of Figure 1613e. Note C in Figure 1613e indicates that this part is a category C based on its thickness. However, this design has the welds taking the load directly. In that case we must use the fatigue strength for category F, the weld metal itself, in the weld throat area. From Table 165a, the tensile stress range endurance strength Sers for category F and infinite life is 8000 psi. 3 Calculate an allowable shear stress range based on Sers and the desired safety factor: τ allow
S 8000 = ers = = 5333 psi 1.5 N frs
(b)
P
4 The part as designed has two 4inlong 3/16inleg fillet welds. We need the throat area for this calculation which is 0.707 as wide as the leg. The load to create the allowable shear stress from equation 4.9 is then: Pmax = τ allow Ashear = 5333 ( 2 )( 4 )( 0.187 )( 0.707 ) = 5641 lb
Pmax 5641 = = 3770 psi A fusion 2 ( 4 )( 0.187 )
A
0.5" (c)
5 Check that the fusion area of weld to base metal is safe. The most likely point of failure is at the weld toe. The weld leg area rather than throat area is used here: σ toe =
0.5" L = 4"
P/2 (d )
A category C weldment has Ser = 10 kpsi from Table 165a, giving a safety factor of 10 000 / 3770 = 2.65. The weld throat limits the design, as is typical in fillet welds. 6 We still need to check that this repeated load will not fail the part statically or in fatigue at locations away from the welds. For example, a Goodman line analysis (Chapter 6) should be done for this repeated load applied to the smallest tensile area across the centerline of the 1india hole and for the tearout shear areas of the same hole. Any fasteners used to attach the base will need to be checked for fatigue under their applied preload (Chapter 15). These tasks are left to the reader.
P/2
A
areas 'B' t
web fillet weld
throat base areas 'A' w
leg width
View AA Copyright © 2018 Robert L. Norton: All Rights Reserved
16.7
TREATING A WELD AS A LINE
The designer is usually trying to determine the size of weld (crosssection and length) needed to withstand the applied loads. One approach is to assume a weld size, calculate the safety factor and if inadequate, change the weld size assumption, recalculate and iterate this process until satisfied with the result. A more direct and somewhat simpler approach as defined by Blodgett[1] is to treat the weld as a line. Instead of stress, this calculation will give a load/length (lb/in or N/m) number that is easily converted to a required weld cross section area using the allowable stress for the weld from either equation 16.1 for static loads or equations 16.2 for dynamic loads. The weld area and the load are essentially normalized to one unit of weld throat length. Allowable unit load factors relating fillet weld leg width w to electrode strength for static loads are shown in Table 166. These are calculated as 0.707(0.30)Exx[5] and so are for 90° corners. Loading on a given weld location is typically one or a combination of direct tension or compression, direct shear, bending, and torsion. The stresses associated with each of these loading conditions were defined in earlier chapters. The stress equations can be converted into fx = load per length of weld throat t as follows:
F I G U R E 1 6  1 1 Repeated Examples 161 and 162
Table 166
Data from [5]
Allowable Static Unit Force on Fillet Welds as a Function of Weld Leg Length w. Electrode Allowable Unit Number Force lb/in E60 E70 E80 E90 E100 E110 E120
12 730 w 14 850 w 16 970 w 19 090 w 21 210 w 23 330 w 25 450 w
Copyright © 2018 Robert L. Norton: All Rights Reserved
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P Aw V fs = Aw
fn =
direct tension or compression direct shear
M fb = Sw Tc ft = Jw
bending torsion
(16.4)
The units of Aw are area / length = length. Sw is the section modulus Z / length = length2 and Jw is the polar area moment of inertia J / length = length3. This makes fx = load per length of weld throat in all cases. Figure 1616 shows nine configurations of weldments and equations for calculating the factors Aw, Sw, and Jw. Their application will be shown in an example.
EXAMPLE 163 Design of a Statically Loaded Weldment Assembly Problem
The weldment assembly in Figure 1617 has fillet welds all around between the pipe and each end plate. Determine the weld size needed to withstand a static load P = 2700 lb.
Given
The material is ASTM A36 structural steel and an E70xx welding electrode is used. The Schedule 40 pipe is 4.5in OD by 0.24in wall. Dimension a = 15 in and r = 10 in.
Assumptions
The weld takes the load directly. Stress in the weld will limit the design since the fusion areas for fillet welds are larger than their throat areas. Ignore weight of arm and pipe.
Solution
See Figure 1617.
1 The offset load puts the pipe section and weld in combined bending, torsion, and direct shear at the root of the cantilever beam where the moment and torque are both maximum. The torsion and direct shear are assumed to be uniformly distributed along the weld. The location of highest stress will be on the top of the pipe at the weld toe (labeled A) where the bending stress is maximum tensile. We first need to calculate the unit loads on the weld due to each mode of loading and then find their vector sum. 2 This is Case 9 in Figure 1616, which shows the direct shear component factor Aw = πd. Use this to find the unit load fs at point A due to direct shear:
16
fs =
P P 2700 = = = 191.0 lb/in Aw πd 4.5π
(a)
3 Find the unit load fb at point A due to the bending moment using Sw from Case 9: fb =
2700 (15) M Pa = = = 2546.5 lb/in Sw π d 2 4 π 4.52 4
(
)
(b)
Chapter 16
WELDMENTS
995
16
FIGURE 1616 Geometry Factors to Analyze a Weld as a Line (reprinted from [7])
996
MACHINE DESIGN

An Integrated Approach

Sixth Edition
y pipe
A
ft
fb
fb
fs
x
wall
z FR
ft
FR
y
y r
a A wall z
fb
x
arm
fs FR P
P FIGURE 1617
Copyright © 2018 Robert L. Norton: All Rights Reserved
Examples 163 and 164
4 Find the unit load ft at point A due to the torsional moment using Jw from Case 9: ft =
(
)
Tc Pr d 2 2700 (10 ) 4.5 2 = = = 848.8 lb/in Jw π d 3 4 π 4.53 4
(
)
(c)
5 Find the magnitude of the resultant force at point A (the maximum weld load): FR =
fs2 + fb2 + ft2 = 2691 lb/in
(d )
6 This is the load per inch of weld. The throat area of one linear inch of weld is equal to the throat dimension. So, if we set the throat stress equal to the allowable value from equation 16.1, use this unit load and calculate the area needed to achieve that allowable stress, we will define the required throat dimension. From equation 16.1, an E70 electrode has an allowable stress of 0.30(70 000) = 21 000 psi: t=
FR τ allow
=
2691 lb/in 21 000 lb/in 2
= 0.128 in 2 /in
(e)
7 This is the throat dimension, but fillet welds are specified by their leg dimension. Assuming an equalleg fillet in a 90° joint, the leg dimension will be: w = 1.414t = 1.414 ( 0.128 ) = 0.181 in
16
(f)
8 Specify a 3/16” fillet weld. This meets the minimum weld size specified in Table 162 and will have a safety factor of approximately 2.5 based on equation 16.1. 9 The weld between the pipe and arm is stressed at a lower level than the weld at A because the bending moment is zero at the end of the cantilever. It sees only direct shear and torsional shear which are 32% of the stress at A. Use a 3/16” weld here also for consistency of fabrication. It is also the minimum for this wall thickness.
Chapter 16
WELDMENTS
997
EXAMPLE 164 Design of a Dynamically Loaded Weldment Assembly Problem
The weldment assembly in Figure 1617 has fillet welds all around between the pipe and each end plate. Determine the weld size needed to withstand a dynamic load that varies between Fmin = –80 lb and Fmax = 600 lb for infinite life with a fatigue safety factor Nfr = 1.5.
Given
The material is ASTM A36 structural steel and an E70xx welding electrode is used. The Schedule 80 pipe is 4.5in OD by 0.337in wall (3.83in ID). Dimension a = 15 in and r = 10 in.
Assumptions
The weldment is a Category C but the weld may limit as Category F.
Solution
See Figure 1617.
1 Failure due to dynamic loading of weldments has been shown to depend only on the stress range or oscillation between the minimum and maximum values of stress seen during the cycle.[2] The force range is Fmax – Fmin = 600 –(–80) = 680 lb. 2 This is Case 9 in Figure 1616, which shows the direct shear component factor Aw = πd. Use this to find the unit load fs at point A due to direct shear. fs =
680 P P = = = 48.1 lb/in Aw πd 4.5π
(a)
3 Find the unit load fb at point A due to the bending moment using Sw from Case 9: fb =
680 (15) M Pa = = = 641.3 lb/in Sw π d 2 4 π 4.52 4
(
)
(b)
4 Find the unit load ft at point A due to the torsional moment using Jw from Case 9: ft =
(
)
Tc Pr d 2 680 (10 ) 4.5 2 = = = 213.8 lb/in Jw π d 3 4 π 4.53 4
(
)
(c)
5 Find the magnitude of the resultant unit force at point A: FR =
fs2 + fb2 + ft2 = 678 lb/in
(d )
This is the load per inch of weld length. 6 From Table 16.5a, a Category F joint has a shear stressrange endurance strength Sers = 8000 psi. Apply the safety factor to this strength to get an allowable stress: τ allow =
Sers 8000 = = 5333 psi 1.5 N fr
(e)
7 The throat area of one linear inch of weld length is equal to the throat dimension. So, if we set the throat stress equal to the allowable value from equation (e), use this unit load and calculate the unit area needed to achieve that allowable stress, we will define the required throat dimension.
16
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MACHINE DESIGN

An Integrated Approach
t=
FR
=
τ allow
678 lb/in 5333 lb/in 2

Sixth Edition
= 0.127 in 2 /in
(f)
8 This is the throat dimension (in), but fillet welds are specified by their leg dimension. Assuming a 45° fillet, the leg dimension will be: w = 1.414t = 1.414 ( 0.127 ) = 0.180 in
( g)
9 Specify a 3/16” (0.187”) fillet weld to obtain the specified safety factor of 1.5. This meets the minimum weld size specified in Table 162. 10 The weld between the pipe and arm has lower stress than the weld at A because the bending moment is zero at the cantilever end. That weld sees only direct shear and torsional shear which are 32% of the stress at A. Table 162 gives a minimum 3/16” weld for this wall thickness—the same size as at A, which simplifies fabrication. 11 We also need to check that the stress in the fusion area between weld and pipe will not fail. Remember that only the stress range is of concern with weldments and it is due to the range of force oscillation, here 680 lb from step 1. Find the normal bending stress range and torsional shear stress range on point A using equations 4.11b and 4.23b, respectively: σx =
680 (15)( 2.25) Mc ( Pa )c = = = 2388 psi I I π 4.54 − 3.834 64
τ xz =
Pr c 680 (10 )( 2.25) Tr = = = 796 psi J J π 4.54 − 3.834 32
(
( )
)
(
)
(h)
(i )
12 Find the maximum shear stress, principal stresses, and von Mises stress that result from this stress combination using equations 4.6 and 5.7c: 2 2 σ − σz 2388 − 0 τ max = x + τ 2xz = + 7962 = 1435 psi 2 2
σ1 =
σx + σz 2
+ τ max =
2388 + 1435 = 2629 psi 2
σ2 = 0 σ + σz 2388 σ3 = x − τ max = − 1435 = −241 psi 2 2
σ ' = σ12 − σ1σ 3 + σ 32 = 2 6292 − 2 629 ( −24 ) + ( −24 )2 = 2 757 psi
( j)
(k )
13 Assume this assembly is an AISC Category C. Table 165a shows the infinitelife fatigue strength for a Category C weldment as 10 000 psi. The safety factor is then:
16
Nf =
Sf σ'
=
The weld material limits this design.
10 000 2757
= 3.62
(l )
Chapter 16
16.8
WELDMENTS
999
ECCENTRICALLY LOADED WELD PATTERNS
Weldments are often used to support offset or eccentric loads as in Examples 163 and 164. There are many other common arrangements as shown in Figure 1616. Some of these such as the one shown in Figure 1618 require that the centroid of the weld pattern be found. This is the same procedure as was described in Section 1510 and Example 156 where a pattern of bolts and dowels was used to support an eccentric load and its centroid was needed. The arrangement of Figure 1618 is defined in part 5 of Figure 1616. An equation for the location of the weld pattern’s centroid is defined there as well as expressions for the unit loads due to direct shear and either bending or torsional loading depending on the location of the applied force being either in or outofplane. This application will be shown in an example.
EXAMPLE 165 Design of an Eccentrically Loaded Weldment Assembly Problem
The weldment assembly in Figure 1618 has fillet welds on three sides of the joint. Determine the weld size needed to withstand a static load of P = 4000 lb. Check for yielding in the beam also.
Given
The material is ASTM A36 structural steel and an E70xx welding electrode is used. The plates are 1/2” thick. Dimension a = 12”, b = 3”, and d = 6”.
Assumptions
Consider the weld as a line. The weld pattern matches that of part 5 of Figure 1616. The weld will limit the design as Category F.
Solution
See Figure 1618.
1 Find the centroid of the weld pattern with the equation given in Figure 1616: x=
32 9 b2 = = = 0.75 in 2b + d 2 ( 3) + 6 12
(a)
2 Find the radius from applied load to centroid: r = a + b − x = 12 + 3 − 0.75 = 14.25 in
(b)
3 The weld joint is loaded in direct shear and also has a torsional moment about its centroid. Find the unit load in the weld due to direct shear: fs =
4000 P P = = = 333.3 lb/in Aw 2b + d 6 + 2 ( 3)
(c)
4 As far as the weld is concerned, the torsional moment acts about its centroid which is at radius r from the applied load P: T = Pr = 4000 (14.25) = 57 000 lbin
(d )
5 The torsional geometry factor for this weld versus the centroid from Figure 1616 is: Jw =
( 2b + d )3 12
−
b 2 ( b + d )2 (12 )3 9 ( 9 )2 = − = 83.25 in3 2b + d 12 12
(e)
16
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MACHINE DESIGN

An Integrated Approach
b
Sixth Edition
P
A
x

f th
A centroid of weld pattern
d
ftv FR fs
a r
(b) Vector sum of unit forces at point A
(a) Weld geometry
FIGURE 1618
Copyright © 2018 Robert L. Norton: All Rights Reserved
Example 165
6 The highest stress in the weld due to the torsional moment will occur at the point furthest from the centroid, labelled A in Figure 1618. It will be simplest if we compute the horizontal and vertical components at point A and then resolve them in combination with the vertical direct shear component into a resultant unit force. For the horizontal component, the radius from the centroid is d / 2. For the vertical component, the radius from the centroid is r – a: fth = ftv = FR =
Td 2 Jw
=
( ) = 2054.1 lb/in
57 000 6 2 83.25
T (r − a ) = Jw
57 000 (14.2512 ) 83.25
(
ft2h + ftv + fs
)
2
= 1540.5 lb/in
(f)
= 2054.12 + (1540.5 + 333.3)2 = 2780 lb/in
7 The unit area of the throat will then be t=
FR τ allow
=
2780 lb/in
(
)
0.30 70 000 lb/in 2
= 0.132 in 2 /in
( g)
using the Sut of E70 electrode and equation 16.1 for τallow. The leg dimension is w = 1.414t = 1.414 ( 0.132 ) = 0.187 in
(h)
8 The minimum recommended weld size for 1/2in plate is 3/16”, which matches this number. Use a 3/16” fillet weld.
16.9 16
DESIGN CONSIDERATIONS FOR WELDMENTS IN MACHINES
Weldments can be a practical choice for oddshaped subassemblies that locate and support machine parts. They may not prove to be an economic choice if not properly designed, however. A significant fraction of their cost is in fixturing and setup of the pieces to hold them in position while welded. In some cases, it can be less expensive to machine a complex part from solid billet with modern numerical control (CNC) machines, which can machine complex shapes defined with solidmodeling CAD systems rapidly and relatively inexpensively and can run unattended after being programmed, creating parts while you sleep. The CNC tool path information generated from the CAD model can
Chapter 16
WELDMENTS
1001
be sent directly to the machining center electronically and the part made in one or a few setups with very little operator interaction. Welded assemblies, on the other hand, often are laborintensive. It may be worth seeking quotes on direct machining of parts from billet that were initially designed as weldments. That said, what can be done during its design to reduce the cost of a weldment? It may be possible to design the weldment assembly to be fully or partially “selffixturing.” The degree that the individual pieces to be welded are designed to fit together easily in the correct orientation and alignment for welding as opposed to requiring external fixtures to be made to hold them will reduce cost. Obviously, minimizing the number and size of welds will also reduce cost. Be cautious about using weldments for parts subject to dynamic loading. Try to place welds in locations of lower stress when possible. The residual tensile stress in all welds is a danger when oscillating stress is applied to those regions. If that situation cannot be designed around, then weld reinforcements and any backing strips should be removed and weld surfaces ground smooth to reduce stress concentration at the toes and on the weld surface. Hammer peening the welds will reduce tensile residual stress and improve fatigue resistance, but this can add significant cost. Parts machined from solid billet will often prove less expensive than a weldment in these cases and should be investigated as an alternative. If quantities permit, forging will give parts with better fatigue resistance, but high tooling cost eliminates this option for small batches that are typical with custom machine parts. If a weldment requires postweld machining, it will probably be necessary to thermally stress relieve the weldment before machining it. This will relieve the residual stresses globally and improve part life. Otherwise the part will likely distort when metal removal during machining releases the residual stress locally where machining is done.
16.10 SUMMARY This brief introduction to the design of weldments probably raises as many questions in the mind of the designer as it provides answers. Welding is a complex subject and is based on a large body of research and experimental data. It has advanced tremendously since its application to ship construction in World War II. Some of the problems encountered then are described in Chapters 5 and 6. If the reader needs to design weldments, they are strongly encouraged to delve more deeply into the subject than can be attempted here. See the references noted. One of the most interesting results of extensive weldment fatigue testing is the independence of failure to the presence of mean stress, which is a serious issue in fatigue failure of unwelded parts. This simplifies the weldment designer’s task since it eliminates the need for a Goodman line analysis. Treating the weld as a line greatly simplifies the task of determining appropriate weld sizes for various loading situations. The existence of design codes both simplifies and complicates the designer’s task. It simplifies by providing rules and guidelines for design. But it also complicates the situation because those rules are far from simple and require much effort to fully understand and use properly. The AISC provides design guidelines and failure stress data based on extensive experiment that should be used for weldment design. The AISC and AWS specifications and codes should be studied by the serious designer.
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MACHINE DESIGN

An Integrated Approach

Sixth Edition
Weldments work best when made of lowcarbon hotrolled steel and statically loaded, which is typical of machine frames and structures. Parts that are highly stressed or loaded dynamically may not be not good candidates for weldment design. One does not find engine crankshafts and connecting rods made as weldments, for example. Those highly loaded, dynamically stressed parts are usually hot forged. Important Equations Used in This Chapter Allowable Shear Stress in Statically Loaded Weldments (Section 16.5):
τ allow = 0.30 Exx
(16.1)
Tensile StressRange Fatigue Strength—Cat. AE (Section 16.6): 1
S fr
Cf 3 = Creliab ≥ Ser N
(16.2a)
Shear StressRange Fatigue Strength—Cat. F (Section 16.6): 1
S frs
Cf 6 = Creliab ≥ Sers N
(16.2b)
See Table 165 for values of Cf and Ser, Sers. Safety Factor in Fatigue for Weldments
N fr =
S fr σ
or
N frs =
S frs τ
(16.3)
16.11 REFERENCES 1 O. Blodgett, Design of Weldments, J. F. Lincoln Foundation, Cleveland, OH, 1963.
Table P160† Topic/Problem Matrix 16.5 Static Strength of Welds
3 American Welding Society, Miami, FL.
161, 162, 163, 1613, 1617
4 SAE Fatigue Design Handbook AE22 3ed., p. 95, Society of Automotive Engineers, Warrandale. PA, 1997.
16.6 Dynamic Strength of Welds
5 O. Blodgett, “Stress Allowables Affect Welding Design,” J. F. Lincoln Foundation, Cleveland, OH, 1998.
165, 167, 169, 1614, 1618
6 T. R. Higgins and F. R. Preece, “Proposed Working Stresses for Fillet Welds in Building Construction,” AISC Engineering Journal, Vol. 6, No. 1, pp. 1620, 1969.
16.7 Weld as a Line
16
2 ANSI/AISC 36005, Specification for Structural Steel Buildings, p. 161–400, American Institute of Steel Construction, March 9, 2005.
166, 167, 168, 169, 1619 16.8 Eccentrically Loaded Weld Patterns 164, 165, 1615, 1616, 1620 Problem numbers in italics are design problems.
†
7 R. L. Mott, Machine Elements in Mechanical Design, 4ed., p. 786, PrenticeHall, Upper Saddle Brook, NJ, 2004. 8 M. A. Weaver, “Determination of Weld Loads and Throat Requirements Using Finite Element Analysis with Shell Element Models—A Comparison with Classical Analysis,” Welding Research Supplement, pp. 1s11s, 1999. 9 D. K. Miller, “Consider Direction of Loading When Sizing Fillet Welds,” Welding Innovation, Vol. XV. No. 2, 1998.
Chapter 16
WELDMENTS
1003
10 J. M. Barsom and S. T. Rolfe, Fracture and Fatigue Control in Structures, 3ed., pp. 238, 269, PrenticeHall, Upper Saddle Brook, NJ, 1999. 11 J. W. Fisher et al. , Effect of Weldments on the Fatigue Strength of Steel beams, National Cooperative Highway Research Program Report 102, Highway Research Board, National Research Council, 1970.
P 2"
P
0.25"
12 Recommendations for Fatigue Design of Welded Joints and Components, WRC Bulletin 520, The Welding Research Council Inc., 2009. bar
6"
16.12 PROBLEMS *161
A submergedarc complete joint penetration (CJP) butt weld was made between two sections of A36 hotrolled steel plate. The plate is 10in wide by 1/2inthick. E70 electrode was used. How much tensile load can the assembly withstand across the weld without yielding in either base metal or weld?
3"
162 The plate of problem 161 has partial joint penetration welds applied from each side. Each weld throat is 1/4 in. What is the maximum allowable tensile load across the weld. *†163
A tee bracket similar to Figure 1611 has 1/2inthick A572 Grade 42 steel welded with a 3/16 in fillet weld along both inside corners using an E70 electrode. It will be subjected to a 20 kip tensile load on the leg of the tee. Determine the minimum required length L of the bracket based on fulllength welds.
164 Figure P161 shows a bar welded to a base with 3/16 in fillet welds on three sides using an E70 electrode. Material is A572 Grade 50 hotrolled steel. What is your recommended maximum static load P that can safely be applied? *165
*†166
*†167
Figure P161 shows a bar welded to a base with 3/16 in fillet welds on three sides using an E70 electrode. Material is A572 Grade 50 hotrolled steel. What is your recommended maximum dynamic repeated load, zero to Pmax that can be applied for 10E8 cycles with a safety factor of 1.6? Figure P162 shows a bracket welded to a wall with a fillet weld using an E70 electrode. For the row(s) assigned in Table P161, determine the fillet weld size needed between the tube and wall for a static load F and h = 1.2OD, a = 2OD, and l = 2.5OD. The pipe and wall material are A36 steel.
base Copyright © 2018 Robert L. Norton: All Rights Reserved
FIGURE P161 Problems 164 and 165
Table P161 Data for Problems 166 to 167 Lengths in inches, forces in kip. Row F
Figure P162 shows a bracket welded to a wall with a fillet weld using an E70 electrode. For the row(s) assigned in Table P161, determine the fillet weld size needed between the tube and wall for a dynamic load that varies from – 0.1F to + 0.2F. The pipe and wall material are A36 steel, h = 1.2OD, a = 2OD, and l = 2.5OD. Use a safety factor of 1.5.
x
id
wall
FIGURE P162 Problems 166 to 167
arm
b
a
2.5
3.500
3.068 1/2
b
3.8
4.500
4.026 1/2
c
4.8
5.563
5.047 1/2
d
8.0
6.625
6.065 1/2
e
9.0
8.625
7.981 3/4
f 11.0 10.750 10.020 3/4
to these problems are provided in Appendix D.
z
pipe
b
* Answers
a od
ID
Copyright © 2018 Robert L. Norton: All Rights Reserved
y
F
l
OD
h
Copyright © 2018 Robert L. Norton: All Rights Reserved
†
Problem numbers in italics are design problems.
16
1004
MACHINE DESIGN

An Integrated Approach

Sixth Edition
A
75
support
75
425
75
B 200
300
b
P FIGURE P163
P Copyright © 2018 Robert L. Norton: All Rights Reserved
Problems 168 and 169 * Answers
to these problems are provided in Appendix D.
†168
Figure P163 shows a bracket machined from 12mmthick A572 Grade 50 hotrolled steel flat stock. It is welded to a support with a fillet weld all around using an E80 electrode. Determine the fillet weld size needed between the bracket and the support for a static load of P = 12 kN.
*†169
Figure P163 shows a bracket machined from 12mmthick A572 Grade 50 hotrolled steel flat stock. It is welded to a support with a fillet weld all around using an E90 electrode. Determine the fillet weld size needed between the bracket and the support for a dynamic load that varies from 0 to +3 kN using a safety factor of 1.8.
†
Problem numbers in italics are design problems.
1610 Two 8mmthick by 50mmwide A572 Grade 42 steel straps are welded together with fillet welds in a lap joint using E70 electrode. The tensile load of 45 kN on the straps is orthogonal (transverse) to the welds. What is your recommended weld size for the two fulllength welds? 1611 Two 8mmthick by 50mmwide A572 Grade 42 steel straps are welded together with fillet welds in a lap joint using E70 electrode. Determine the fillet weld size needed for a dynamic load that varies from 0 to +12 kN using a safety factor of 1.5 for infinite life of the two fulllength welds. 1612 Two 12mmthick by 50mmwide weldable aluminum straps are welded together with fillet welds in a lap joint using an aluminum electrode. Determine the fillet weld size needed for a dynamic load that varies from 0 to +5 kN using a safety factor of 2.0 for infinite life of the two fulllength welds. 1613 Two 10mmthick by 60mmwide, hot rolled SAE 1050 steel plates are to be buttwelded together with a CJP weld. What electrode part number would you recommend if the weldment is loaded in tension? 1614 Calculate and plot the 106cycle stressrange fatigue strength for AISC Categories A through E’ at a reliability of 99%. Use SI units.
16
1615 Figure P161 shows a bar welded to a base on three sides with 4mm fillet welds using an E80 electrode. The material is SAE 1035 hotrolled steel. Determine the safety factor against static failure if the load P = 5 kN. 1616 Figure P161 shows a bar welded to a base on three sides with 4mm fillet welds using an E80 electrode. The material is SAE 1035 hotrolled steel. Determine the safety factor against dynamic failure for 10E8 cycles if the repeated load varies from zero to 2.5 kN
Chapter 16
WELDMENTS
1005
1617 A tee bracket similar to that shown in Figure 1611 is to have 1/2inthick A572 Grade 50 steel welded with fillet welds along both inside corners. The length of the bracket is L = 6 in. If it will be subjected to a 30kip tensile load on the leg of the tee, specify a suitable electrode and weld size for a minimum factor of safety of 1.4 against any failure mode in the weld. 1618 Figure P163 shows a bracket machined from 16mmthick A572 Grade 42 hotrolled steel flat stock. It is welded to a support with a fillet weld all around. Determine the fillet weld size needed between the bracket and the support for a dynamic load that varies from 0 to +700 lb using a safety factor of 1.8. 1619 Case 6 of Figure 1616 shows a channel section welded to a wall with a fillet weld using an E70 electrode. Determine the fillet weld size needed between the channel and the wall for a static bending load P = 3200 lb. The channel and wall material are A36 steel and a = 6 in, b = 3 in, and d = 1.410 in. 1620 Case 6 of Figure 1616 shows a channel section welded to a wall with a fillet weld using an E70 electrode. Determine the fillet weld size needed between the channel and the wall for a static torsion load P = 3200 lb at a = 16 in. The channel and wall material are A36 steel and b = 3 in and d = 1.410 in.
16
1006
MACHINE DESIGN

An Integrated Approach
NOTES
16

Sixth Edition
CLUTCHES AND BRAKES A big book is a big nuisance. CallimaChas, 260 bC
17.0
INTRODUCTION
Clutches and brakes are essentially the same device. Each provides a frictional, magnetic, hydraulic, or mechanical connection between two elements. If both connected elements can rotate, then it is called a clutch. If one element rotates and the other is fixed, it is called a brake. A clutch thus provides an interruptible connection between two rotating shafts as, for example, the crankshaft of an automobile engine and the input shaft of its transmission. A brake provides an interruptible connection between one rotating element and a nonrotating ground plane as, for example, the wheel of an automobile and its chassis. The same device may be used as either clutch or brake by fixing its output element to a rotatable shaft or by fixing it to ground. Brakes and clutches are used extensively in production machinery of all types, not just in vehicle applications where they are needed to stop motion and allow the internalcombustion engine to continue turning (idling) when the vehicle is stopped. Clutches also allow a highinertia load to be started with a smaller electric motor than would be required if it were directly connected. Clutches are commonly used to maintain a constant torque on a shaft for tensioning of webs or filaments. A clutch may be used as an emergency disconnect device to decouple the shaft from the motor in the event of a machine jam. In such cases, a brake will also be fitted to bring the shaft (and machine) to a rapid stop in an emergency. To minimize injuries, many U.S. manufacturers require their production machinery to stop within one or fewer revolutions of the main driveshaft if a worker hits the “panic bar” that typically spans the length of the machine. This can be a difficult specification to achieve on large machines (10 to 100 feet long) driven by multihorsepower electric motors. Manufacturers provide clutchbrake combinations in the same package for such applications. Applying power disengages the brake and engages the clutch, making it failsafe. Failsafe brakes are engaged (typically by inter1007
17
1008
MACHINE DESIGN
Title page photograph courtesy of the Logan Clutch Corporation, Cleveland, Ohio.
Table 170

An Integrated Approach

Sixth Edition
Variables Used in this Chapter
Symbol
Variable
ips units
SI units
See
a
length
in
m
Sect. 17.6
b
length
in
m
Sect. 17.6
c
length
in
m
Sect. 17.6
d
diameter
in
m
various
F
force
lb
N
various
Fa
applied force
lb
N
Sect. 17.6
Ff
friction force
lb
N
Sect. 17.6
Fn
normal force
lb
N
Sect. 17.6
Rx
reaction force
lb
N
Sect. 17.6
Ry
reaction force
lb
N
Sect. 17.6
K
arbitrary constant
none
none
various
l
length
in
m
various
M
moment
lbin
Nm
various
N
number of friction surfaces
none
none
Eq. 17.2
P
power
hp
Watts
Ex. 171
p
pressure
psi
N/m2
Sect. 17.4
pmax
maximum pressure
psi
N/m2
Sect. 17.4
r
radius
in
m
various
ri
inside radius of disk lining
in
m
various
ro
outside radius of disk lining
in
m
various
T
torque
lbin
Nm
various
V
linear velocity
in/sec
m/sec
Eq. 17.4
W
wear rate
psiin/sec
Pam/sec
Eq. 17.4
w
width
in
m
various
θ
angular position
rad (deg)
rad (deg)
Sect. 17.6
µ
coefficient of friction
none
none
Eq. 17.2
ω
angular velocity
rad/sec
rad/sec
Ex. 171
nal springs) unless power is applied to disengage them. Thus, they “fail safe” and stop the load if the power fails. Highway truck and railroadcar air brakes are of this type. Air pressure releases the brake, which is normally engaged. If the railroad car or truck trailer breaks loose and severs its airhose connection to the engine or tractor, the brakes automatically engage.
17
This chapter will describe a number of types of commercially available clutches and brakes and their typical applications, and will also discuss the theory and design of a few particular types of friction clutches/brakes. Table 170 lists the variables used in this chapter and indicates the equation or section in which they can be found.
Chapter 17
17.1
CLUTCHES AND BRAKES
1009
TYPES OF BRAKES AND CLUTCHES
Brakes and clutches can be classified in a number of ways, by the means of actuation, the means of energy transfer between the elements, and the character of the engagement. Figure 171 shows a flow chart depicting these characteristics. The means of actuation may be mechanical, as in the pushing of an automobile’s clutch pedal, pneumatic or hydraulic, in which fluid pressure drives a piston to mechanically engage or disengage as with vehicle brakes, electrical, which is typically used to excite a magnetic coil, or automatic, as in an antirunaway brake that engages by relative motion between the elements. Positive ContaCt ClutChes The means of energy transfer may be a positive mechanical contact, as in a toothed or serrated clutch, which engages by mechanical interference as shown in Figure 172. The character of engagement is mechanical interference obtained with jaws of square or sawtoothed shape, or with teeth of various shapes. These devices are not as useful for brakes (except as holding devices) because they cannot dissipate large amounts of energy as can a friction brake, and as clutches they can be engaged only at low relative velocities (about 60 rpm max for jaw clutches and 300 rpm max for toothed clutches). Their advantage is positive engagement, and, once
17 FIGURE 171 Classification of Clutches and Brakes Source: Z. Hinhede, Machine Design Fundamentals: A Practical Approach, PrenticeHall, 1983.
1010
MACHINE DESIGN

An Integrated Approach
FIGURE 172

Sixth Edition
Copyright © 2018 Robert L. Norton: All Rights Reserved
A Positive Contact Clutch Photo by author
* Automotive transmissions typically use helical gears, for quiet operation, as was noted in Chapter 13. The helical gears cannot be easily shifted in and out of engagement in manual transmissions because of their helix angle. So, they are all kept in constant mesh and clutched/ declutched from the transmission shaft to engage a particular ratio. Each gear has a synchromesh clutch connecting it to its shaft. This clutch actually consists of conical friction surfaces that drag the two elements (shaft and gear) into near zero relative velocity before the teeth of its companion positivecontact clutch engage. The shift lever moved by the driver is shifting these synchromesh clutches into and out of engagement, rather than moving gears around in the transmission.
coupled, they can transmit large torque with no slip. They are sometimes combined with a frictiontype clutch, which drags the two elements to nearly the same velocity before the jaws or teeth engage. This is the principle of a synchromesh clutch in a manual automotive transmission.* FriCtion ClutChes and Brakes are the most common types used. Two or more surfaces are pressed together with a normal force to create a friction torque. The friction surfaces may be flat and perpendicular to the axis of rotation, in which case the normal force is axial (disk brake or clutch), as shown in Figure 173, or they may be cylindrical with the normal force in a radial direction (drum brake or clutch), as shown in Figures 179 and 1710, or conical (cone brake or clutch). Cone clutches can tend to grab or refuse to release and are not now used very much in the United States, but are popular in Europe.[1]
inlet port
piston
needle thrust bearing
cylinder cylinder bearing
17
pressure plate FIGURE 173 A Multiplate Disk Clutch Actuated by Fluid Pressure Courtesy of Logan Clutch Corporation, Cleveland, Ohio
Chapter 17
CLUTCHES AND BRAKES
1011
At least one of the friction surfaces is typically metal (cast iron or steel) and the other is usually a highfriction material, referred to as the lining. If there are only two elements, there will be either one or two friction surfaces to transmit the torque. A cylindrical arrangement (drum brake or clutch) has one friction surface, and an axial arrangement (disk brake or clutch) has one or two friction surfaces, depending on whether the disk is sandwiched between two surfaces of the other element or not. For higher torque capacity, disk clutches and brakes are often made with multiple disks to increase the number of friction surfaces (see Figure 173). A clutch or brake’s ability to transfer the heat of friction generated can become the limiting factor on its capacity. Multidisk clutches are more difficult to cool so are appropriate for highload, lowspeed applications. For highspeed dynamic loads, fewer friction surfaces are better.[1] Friction clutches may be operated either dry or wet, the latter running in an oil bath. While the oil severely reduces the coefficient of friction, it greatly increases the heat transfer. Friction coefficients of clutch/brake material combinations typically range from 0.05 in oil to 0.60 dry. Wet clutches often use multiple disks to make up for the lower friction coefficient. Automatic transmissions for automobiles and trucks contain many wet clutches and brakes operating in oil that is circulated out of the transmission for cooling. Manual transmissions for offroad vehicles, such as motorcycles, use sealed, oilfilled, multidisk wet clutches to protect the friction surfaces from dust, water, and dirt. Manualtransmission automobiles and trucks typically use singledisk, dry clutches. overrunning ClutChes (also called oneway clutches) operate automatically based on the relative velocity of the two elements. They act on the circumference and allow relative rotation in only one direction. If the rotation attempts to reverse, the internal geometry of the clutch mechanism grabs the shaft and locks up. These backstop clutches can be used on a hoist to prevent the load from falling back if power to the shaft is interrupted, for example. These clutches are also used as indexing mechanisms. The input shaft can oscillate back and forth but the output turns intermittently in only one direction. Another common application of an overrunning clutch is in the rear hub of a bicycle to allow freewheeling when the wheel speed exceeds that of the drive sprocket. Several different mechanisms are used in oneway clutches. Figure 174a shows a sprag clutch, which has an inner and outer race like a ball bearing. But, instead of balls, the space between the races is filled with oddshaped sprags, which allow motion in one direction but jam and lock the races together in the other, transmitting a oneway torque. A similar result is obtained with balls or rollers captured in wedgeshaped chambers between the races, then called a roller clutch. Figure 174b shows another type of oneway or overrunning clutch called a spring clutch, which uses a spring wrapped tightly around the shaft. Rotation in one direction wraps the spring tighter on the shaft and transmits torque. Counterrotation unwinds the spring slightly, allowing it to slip. CentriFugal ClutChes engage automatically when the shaft speed exceeds a certain magnitude. Friction elements are thrown radially outward against the inside of a cylindrical drum to engage the clutch. Centrifugal clutches are sometimes used to couple an internalcombustion engine to the drive train. The engine can idle decoupled from the wheels and when the throttle is opened, its increased speed automatically engages the clutch. These are common on gokarts. Used on chain saws for the same purpose, they also serve as an overload release that slips to allow the engine to continue running when the chain jams in the wood.
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MACHINE DESIGN
outer race
sprag

An Integrated Approach

Sixth Edition
energizing spring
output hub wrap spring input hub inner race
spring retainer (a)
( b)
FIGURE 174 Overrunning Clutches (a) Sprag Clutch (b) WrapSpring Clutch Courtesy of Warner Electric, South Beloit, Ill. 61080
MagnetiC ClutChes and Brakes are made in several types. Friction clutches are commonly electromagnetically operated, as shown in Figure 175. These have many advantages, such as rapid response times, ease of control, and smooth starts and stops, and are available powered on or powered off (failsafe). Both clutch and brake versions are supplied as well as combined clutchbrake modules. Magneticparticle clutches and brakes (not shown) have no direct frictional contact between the clutch disk and the housing and no friction material to wear. The gap between the surfaces is filled with a fine ferrous powder. When the coil is energized, the powder particles form chains along the magnetic field’s flux lines and couple the disk to the housing with no slip. The torque can be controlled by varying the current to the coil and the device will slip when the applied torque exceeds the value set by coil current, providing a constant tension. Magnetic hysteresis clutches and brakes have no mechanical contact between the rotating elements and so have zero friction when disengaged. The rotor, also called the drag cup, is dragged along (or braked) by the magnetic field set up by the field coil (or permanent magnet). These devices are used to control torque on shafts in applications such as winding machines, where a constant force must be applied to a web or filament of material as it is wound up. The torque on a hysteresis clutch is controllable independent of speed. These devices are extremely smooth, quiet, and longlived, as there is no mechanical contact within the clutch except in its bearings.
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Eddycurrent clutches (not shown) are similar in construction to hysteresis devices in that they have no mechanical contact between rotor and pole. The coil sets up eddy currents, which magnetically couple the clutch together. There will always be some slip in this type of clutch, as there has to be relative motion between rotor and pole to generate
Chapter 17
armature
CLUTCHES AND BRAKES
1013
coil
magnet or field bushings
armature spline
friction material
FIGURE 175 Magnetically Operated Friction Clutch Courtesy of Warner Electric, South Beloit, Ill. 61080,
the eddy currents that supply the coupling force; so an eddycurrent brake cannot hold a load stationary, only slow it from one speed to another. These have similar advantages to the hysteresis devices and are used for similar applications, such as coil or filament winders, etc. Fluid CouPlings transmit torque through a fluid, typically an oil. An impeller having a set of blades is turned by the input shaft and imparts angular momentum to the oil that surrounds it. A turbine (or runner) with similar blades is attached to the output shaft and is turned by the moving oil impinging on it. The principle of operation is similar to placing two electric fans face to face and turning only one on. The airflow from the powered fan blades will cause the facing, unpowered blades to windmill, passing power without any mechanical contact. Using incompressible oil in a confined volume is much more efficient than two openair fans, especially when the impeller and turbine blades are optimally shaped to pump the oil. A fluid coupling provides extremely smooth starts and absorbs shocks, as the fluid simply shears when there is a speed differential, then gradually accelerates (or decelerates) the output turbine to nearly match the speed of the impeller. There will always be some slip,* meaning the turbine can never reach 100% of the impeller’s speed (0% slip), but it can operate at 100% slip when the turbine is stalled. All the input energy will then be converted to heat in shearing the oil. Heat transfer is an important consideration when designing a fluid coupling. The outside case is often finned to improve heat transfer. A fluid coupling transmits the input torque to the output shaft at any speed including stall, so it can never be totally decoupled like a friction clutch. The output must be braked to hold it stationary when the input shaft is turning.† The horsepower rating of a fluid coupling varies as the fifth power of its diameter. A 15% increase in diameter doubles its power capacity. If used as a brake, a fluid coupling can only provide a drag to slow a device, as in a dynamometer, but cannot hold a load stationary.
*
This slip is the reason that autombiles with fluidcoupled, automatic transmissions get worse mileage per gallon than a comparable standard tranmission auto. To counter this, many automatic transmissions are now fitted with a mechanical lockup clutch that engages above some road speed (like 30 mph) to lock the impeller and output turbine together. This clutch is released automatically when the car slows below the set speed. †
This is why you must keep your foot on the brake when stopped at a traffic light in an automobile that has an automatic transmission if the engine is running and the transmission is in “drive.” The fluid coupling between the engine and transmission is always transmitting torque and the car will creep forward at idle unless the brakes are applied.
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MACHINE DESIGN

An Integrated Approach
flywheel
oil
turbine attached to transmission input shaft

Sixth Edition
impeller or pump attached to flywheel oil
oil oil
stator
direction of oil flow oneway clutch between stator and transmission case
engine crankshaft
transmission input shaft pilot bearing
oil oil oil
FIGURE 176
transmission case oil stator is stationary at stall and during acceleration but rotates in same direction as pump and turbine at steady state Copyright © 2018 Robert L. Norton: All Rights Reserved
Schematic of a Torque Converter
If a third, stationary, element with a set of curved blades, called a reactor or stator, is placed between impeller and turbine, additional angular momentum is imparted to the fluid and the device is then called a torque converter as shown in Figure 176. Torque converters are used in vehicles to couple the engine to an automatic transmission. The engine can idle with the vehicle stopped (turbine stalled—100% slip). At stall, the impeller and reactor blades create about a 2:1 torque multiplication, which is available to help accelerate the vehicle when the brakes are released and the engine speed increased. As the vehicle (and thus turbine) speed increases, the turbine will approach the impeller speed and the torque multiplication will decrease to essentially zero at a slip of a few percent. If more torque is needed momentarily (as in passing), the slip between impeller and turbine will automatically increase when the engine is sped up to provide more torque and power to accelerate the vehicle.
17.2
17
CLUTCH/BRAKE SELECTION AND SPECIFICATION
Manufacturers of specialty clutches and brakes such as those described above provide extensive information on the torque and power capacities of their various models in catalogs, many of which are as informative as a textbook on the particular subject. They also define procedures for selection and specification, usually based on the anticipated torque and power for the application plus suggested service factors that attempt to accommodate different loading, installation, or environmental factors than those under which the products were tested. For example, the manufacturer’s standard rating for a clutch model may be based on a smooth driver, such as an electric motor. If the particular application uses an internalcombustion engine of the same power, there will be impulse loads, and
Chapter 17
CLUTCHES AND BRAKES
1015
a clutch or brake of larger capacity than dictated by the average power will need to be selected. This is sometimes referred to as derating the clutch (or brake), meaning that its actual capacity under the anticipated conditions is considered to be less than the rated capacity of the chosen device. serviCe FaCtors According to many clutch manufacturers, a common cause of clutch trouble is the designer’s failure to properly apply adequate service factors to account for the particular conditions of the application.[1] This may, in part, be due to confusion engendered by a lack of standardization for definitions of service factors. One manufacturer may recommend a service factor of 1.5 for a particular condition while another manufacturer recommends 3.0 for the same condition. Both will be correct for their particular clutch designs because, in one case, the manufacturer may have already builtin a safety factor while the other applies it in the service factor. The wise designer will carefully follow each manufacturer’s recommended selection procedures for their products, realizing that they are typically based on extensive (and expensive) testing programs as well as on fieldservice experience with that particular product. A clutch even slightly too small for the applied loading will slip and overheat. A clutch too large for the load is also bad, as it adds unnecessary inertia and may overload the motor that must accelerate it. Most manufacturers of machine parts are generous in providing engineering help to properly size and specify their products for any application. The machine designer’s principal concern should be to accurately define the loading and environmental conditions that the device must accommodate. This may require extensive and tedious calculations of such things as the moments of inertia for all the elements in the drive train actuated by the clutch or brake. The loadanalysis methods of Chapter 3 are applicable to such a task. ClutCh loCation When a machine has both high and lowspeed shafts (as it will whenever a speed reducer is used, such as in Case Studies 7 and 8), and a clutch is needed in the system, the question immediately arises, should the clutch be placed on the low or highspeed side of the gear reducer? Sometimes the answer is dictated by function. For example, it would make little sense to put an automotive clutch on the output shaft of the transmission instead of the input side, since in this instance the principal purpose of the clutch is to interrupt the connection between engine and transmission, ergo, it must go on the highspeed side. In other cases, function does not dictate the clutch location, as, for example, in Case Study 8, where the coupling on either shaft could be replaced by a clutch if it were desired to decouple the compressor from the engine. (See Figure 91.) The choice is less clear in these situations, and there are two competing schools of thought. The torque (and any shock load) is larger on the lowspeed shaft than on the highspeed shaft by the gear ratio. The power is essentially the same at both locations (neglecting losses in the gear train), but the kinetic energy at the highspeed shaft is larger by the square of the gear ratio. A clutch on the lowspeed side must be larger (and thus more expensive) in order to handle the larger torque. However, a smaller and cheaper clutch on the highspeed side must dissipate the greater kinetic energy at that location and thus may more readily overheat. Some manufacturers recommend always using the highspeed side for the clutch location if function allows, opting for its better initial economy. Other clutch manufacturers suggest that the higher initial cost of the larger, lowspeed clutch will be counterbalanced by lower maintenance cost in the long run. The balance of expert opinion seems to tilt toward the highspeed location with the caveat that each situation should be individually evaluated on its own merits.[1]
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MACHINE DESIGN
17.3

An Integrated Approach

Sixth Edition
CLUTCH AND BRAKE MATERIALS
Materials for the structural parts of brakes and clutches, such as the disks or drums, are typically made of gray cast iron or steel. The friction surfaces are usually lined with a material having a good coefficient of friction and sufficient compressive strength and temperature resistance for the application. Asbestos fiber was once the most common ingredient in brake or clutch linings, but is no longer used in many applications because of its danger as a carcinogen. Linings may be molded, woven, sintered, or of solid material. Molded linings typically use polymeric resins to bind a variety of powdered fillers or fibrous materials. Brass or zinc chips are sometimes added to improve heat conduction and wear resistance and reduce scoring of drums and disks. Woven materials typically used long asbestos fibers. Sintered metals provide highertemperature resistance and compressive strength than molded or woven materials. Materials such as cork, wood, and cast iron are sometimes used as linings as well. Table 171 shows some frictional, thermal, and mechanical properties of a few frictionlining materials.
17.4
DISK CLUTCHES
The simplest disk clutch consists of two disks, one lined with a highfriction material, pressed together axially with a normal force to generate the friction force needed to transmit torque, as shown in Figure 177. The normal force can be supplied mechanically, pneumatically, hydraulically, or electromagnetically and is typically quite large. The pressure between the clutch surfaces can approach a uniform distribution over the surface if the disks are flexible enough. In such cases, the wear will be greater at larger diameters because wear is proportional to pressure times velocity (pV) and the velocity increases linearly with radius. However, as the disks wear preferentially toward the outside, the loss of material will change the pressure distribution to a nonuniform one and the clutch will approach a uniform wear condition of pV = constant. Thus, the two extremes are a uniformpressure and a uniformwear condition. A flexible clutch may be close to a uniformpressure condition when new, but will tend toward a uniformwear condition with use. A rigid clutch will more rapidly approach the uniformwear condition with use. The calculations for each condition are different, and the uniformwear assumption gives a more conservative clutch rating, so is favored by some designers. A
lining
dr
ro F
r
F ri
17
A FIGURE 177 A SingleSurface Axial Disk Clutch
Section AA
Chapter 17
Table 171
CLUTCHES AND BRAKES
1017
Properties of Common Clutch/Brake Lining Materials
Friction Material Against Steel or CI
Dynamic Coefficient of Friction
Maximum Pressure psi
Maximum Temperature °F
°C
400–500
204–260
400–500
204–260
dry
in oil
kPa
Molded
0.25–0.45
0.06–0.09
Woven
0.25–0.45
0.08–0.10
Sintered metal
0.15–0.45
0.05–0.08
150–300 1 030–2 070
450–1 250 232–677
Cast iron or hard steel
0.15–0.25
0.03–0.06
100–250
500
150–300 1 030–2 070 50–100
345–690
690–720
260
Uniform Pressure Consider an elemental ring of area on the clutch face of width dr as shown in Figure 177. The differential force acting on this ring is dF = pθr dr
(17.1a)
where r is the radius, θ is the angle of the ring in radians and p is the uniform pressure on the clutch face. For a full circumference clutch as shown in Figure 177, θ will be 2π. The total axial force F on the clutch is found by integrating this expression between the limits ri and ro: F=
ro
∫r
pθr dr =
i
(
pθ 2 ro − ri2 2
)
(17.1b)
The friction torque on the differential ring element is dT = pθµr 2 dr
(17.2a)
where µ is the coefficient of friction. The total torque for one clutch disk is T=
ro
∫r
pθµr 2 dr =
i
(
pθµ 3 3 ro − ri 3
)
(17.2b)
For a multipledisk clutch with N friction faces: T=
(
)
pθµ 3 3 ro − ri N 3
(17.2c)
Equations 17.1b and 17.2c can be combined to give an expression for torque as a function of axial force: T = NµF
( (
3 3 2 ro − ri 3 ro2 − ri2
) )
(17.3)
Uniform Wear The constantwear rate W is assumed to be proportional to the product of pressure p and velocity V:
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MACHINE DESIGN

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Sixth Edition
W = pV = constant
(17.4 a)
And the velocity at any point on the face of the clutch is V = rω
(17.4 b)
Combine these equations, assuming a constant angular velocity ω: pr = cons tan t = K
(17.4c)
The largest pressure pmax must then occur at the smallest radius ri : K = pmax ri
(17.4 d )
Combining equations 17.4c and 17.4d gives an expression for pressure as a function of the radius r: p = pmax
ri r
(17.4e)
where the maximum allowable pressure pmax will vary with the lining material used. Table 171 shows recommended values of pmax and coefficients of friction for various clutch/brake lining materials. The axial force F is found by integrating equation 17.1a for the differential force on the ring element of Figure 177 with equation 17.4e substituted for p: F=
ro
∫r
pθr dr =
i
ro
∫r
i
r θ pmax i r dr = ri θpmax ( ro − ri ) r
(17.5a)
The torque is found by integrating equation 17.2a with the same substitution: T=
ro
∫r
pθµr 2 dr =
i
(
θ µri pmax ro2 − ri2 2
)
(17.5b)
Combine equations 17.2a and b for an expression relating torque to axial force for the uniformwear case: T = NµF
(ro + ri ) 2
(17.6)
where N is the number of friction surfaces in the clutch. From equations 17.5a and 17.5b, it can be shown that the maximum torque for any outside radius ro will be obtained when the inside radius is: ri = 1 3ro = 0.577ro
17
(17.7)
Note that the uniformwear assumption gives a lower torque capacity for the clutch than does the uniformpressure assumption. The higher initial wear at the larger radii shifts the center of pressure radially inward, giving a smaller moment arm for the resultant friction force. Clutches are usually designed based on uniform wear. They will have a greater capacity when new, but will end up close to the predicted design capacity after they are worn in.
Chapter 17
CLUTCHES AND BRAKES
1019
EXAMPLE 171 Design of a Disk Clutch Problem
Determine a suitable size and required force for an axial disk clutch.
Given
The clutch must pass 7.5 hp at 1725 rpm with a service factor of 2.
Assumptions
Use a uniformwear model. Assume a single dry disk with a molded lining.
Solution
1 The service factor of 2 requires derating the clutch by that factor, so we will design for 15 hp instead of 7.5. Find the torque required for that power at the design rpm: inlb sec 15 hp 6600 hp P = 548.05 lbin T= = ω 2π rad sec 1725 rpm 60 rpm
(a)
2 Find the coefficient of friction and maximum recommended pressure for a dry, molded material from Table 171. Use the average of the ranges of values shown: pmax = 225 psi and µ = 0.35. 3 Substitute equation 17.7 relating ri to ro for maximum torque into equation 17.5b to get
(
)
1 T = πµri pmax ro2 − ri2 = πµ ( 0.577ro ) pmax ro2 − ro2 = 0.3849ro3 πµpmax 3 1
T ro = 0.3849 πµp
1
3 3 548.05 = 0.3849 π ( 0.35)( 225) = 1.79 in max
( b)
4 From equation 17.7: ri = 0.577ro = 0.577 (1.79 ) = 1.03 in
(c)
5 The axial force needed (from equation 17.5a) is: F = 2 πri pmax ( ro − ri ) = 2 π (1.034 )( 225)(1.792 − 1.034 ) = 1108 lb
(d )
6 The clutch specification is then a single disk with 3.6in outside diameter and 2in inside diameter, a molded lining with µdry ≥ 0.35, and an actuating force ≥ 1 108 lb. 7 The files EX1701 can be found on the book’s website.
17.5
DISK BRAKES
The disk clutch equations also apply to disk brakes. However, disk brakes are seldom made with linings covering the entire circumference of the face, because they would then overheat. Use the appropriate value of θ for the included angle of the brake pads in equations 171 through 175 to calculate the force and torque in a disk brake. Note that
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MACHINE DESIGN

An Integrated Approach
cable

Sixth Edition
cable
brake arm
brake arm frame
pad 1
pad 2 wheel rim
FIGURE 178
wheel rim θ
Bicycle DiskType Brake
N will be at least 2 for a disk brake as it has opposing pads in pairs as shown in Figure 178. While clutches are often used with light duty cycles (engagement time a small fraction of total time), brakes often must absorb large amounts of energy in repeated applications. Caliper disk brakes, such as those used on automobiles, use friction pads applied against only a small fraction of the disk circumference, leaving the rest exposed for cooling. The disk is sometimes ventilated with internal air passages to promote cooling. The caliper typically straddles the disk and contains one or more pairs of pads, each pair rubbing both sides of the disk. This cancels the axial force and reduces axial loads on the bearings. The common bicycle caliper brake as shown in Figure 178 is another example in which the wheel rim is the disk and the calipers pinch against only a small fraction of the circumference. Disk brakes are now commonly used on automobiles, particularly on the front wheels, which provide more than half the stopping force. Some advantages of disk over drum brakes are their good controllability and linearity (braking torque directly proportional to applied axial force).
17.6
17
DRUM BRAKES
Drum brakes (or clutches) apply the friction material to the circumference of a cylinder, either externally, internally, or both. These devices are more often used as brakes than as clutches. The part to which the friction material is riveted or bonded with adhesive is called the brake shoe, and the part against which it rubs, the brake drum. The shoe is forced against the drum to create the friction torque. The simplest configuration of a drum brake is the band brake, in which a flexible shoe is wrapped around a majority of the outer circumference of the drum and squeezed against it. Alternatively a relatively rigid, lined shoe (or shoes) can be pivoted against the outer or inner circumference (or both) of the drum. If the shoe contacts only a small angular portion of the drum, it is a
Chapter 17
CLUTCHES AND BRAKES
1021
shortshoe brake, otherwise a longshoe brake. The geometry of the short versus long shoe contact requires a different analytical treatment in each case. We will examine the cases of external shortshoe and external longshoe drum brakes to illustrate their differences and features, particularly in contrast to disk brakes. The principles are the same for internalshoe brakes as well. ShortShoe External Drum Brakes Figure 179a shows a schematic of a shortshoe external drum brake. If the angle θ subtended by the arc of contact between shoe and drum is small (< about 45°), then we can consider the distributed force between shoe and drum to be uniform, and it can be replaced by the concentrated force Fn in the center of the contact area, as shown in Figure 179b. For any maximum allowable lining pressure pmax (Table 171) the force Fn can be estimated as Fn = r θ w pmax
(17.8)
where w is the width of the brake shoe in the z direction and θ is the subtended angle in radians. The frictional force Ff is Ff = µFn
(17.9)
where µ is the coefficient of friction for the brake lining material (Table 171). The torque on the brake drum is then T = Ff r = µFn r
(17.10)
Summing moments about point O on the freebody diagram of Figure 179b and substituting equation 17.9 gives
∑ M = 0 = aFa − bFn + cFf
Fa =
bFn − cFf a
=
(17.11a)
bFn − µcFn b − µc = Fn a a
Fa
Fa
shoe θ
(17.11b)
a
r
Ff
Ry
c
O
Rx
O
Fn drum
ω b (a) Brake assembly
FIGURE 179 Geometry and Forces for a ShortShoe External Drum Brake
(b) Freebody diagram
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An Integrated Approach

Sixth Edition
The reaction forces at the pivot are found from a summation of forces: Rx = − Ff Ry = Fa − Fn
(17.12)
selFenergizing Note in Figure 179b that with the direction of drum rotation shown, the friction moment cFf adds to the actuating moment aFa. This is referred to as selfenergizing. Once any application force Fa is applied, the friction generated at the shoe acts to increase the braking torque. However, if the brakedrum rotation is reversed from that shown in Figure 179a, the sign of the frictionmoment term cFf of equation 17.11a becomes negative, and the brake is then selfdeenergizing. This selfenergizing feature of drum brakes is a potential advantage, as it reduces the required application force compared to a disk brake of the same capacity. Drum brakes typically have two shoes, one of which can be made selfenergizing in each direction, or both in one direction. The latter arrangement is commonly used in automobile brakes to aid in stopping forward motion at the expense of stopping backward motion, which is normally at lower speeds. selFloCking Note in equation 17.11 that if the brake is selfenergizing and the product µc ≥ b, the force Fa needed to actuate the brake becomes zero or negative. The brake is then said to be selflocking. If the shoe touches the drum, it will grab and lock. This is usually not a desired condition except in socalled backstopping applications, as were described under overrunning clutches above. In effect, a selflocking brake can function as an overrunning clutch to backstop a load and prevent it from running away if power is lost. Such brakes are sometimes used in hoists for that purpose.
EXAMPLE 172 Design of a ShortShoe Drum Brake
* Files EX1702 are on the book’s website.
Problem
For the drumbrake arrangement shown in Figure 179, determine the ratio c / r that will give a selfenergizing ratio Fn / Fa of 2. Also find the c / r ratio that will cause selflocking.
Given
The dimensions are a = b = 6, r = 5.
Assumptions
Coefficient of friction µ = 0.35.
Solution*
See Figure 179.
1 Rearrange equation 17.11 to form the desired ratio: Fn a = Fa b − µc
(a)
2 Substitute the desired selfenergizing ratio, the given dimensions, and solve for c:
17
Fn 6 =2= Fa 6 − 0.35c −3 = 8.571 c= −0.35
(b)
Chapter 17
CLUTCHES AND BRAKES
1023
3 Form the c / r ratio for a selfenergizing ratio of 2 with the given brake geometry: c 8.571 = = 1.71 5 r
(c)
4 For selflocking to begin, Fa becomes zero, making Fn / Fa = ∞ and Fa / Fn = 0. The second of these ratios will need to be used to avoid division by zero. Rearrange equation 17.11 to form the desired ratio and solve for c: Fa b − µc 6 − 0.35c = = =0 Fn a 6 c = 17.143
(d )
5 Form the c / r ratio for selflocking with the given brake geometry: c 17.143 = = 3.43 5 r
(e)
6 Note that these ratios are specific to the dimensions of the brake. The length a was set equal to b in this example in order to eliminate the effect of the lever arm ratio a / b, which further reduces the application force Fa required for any normal force Fn.
LongShoe External Drum Brakes If the angle of contact θ between shoe and drum in Figure 179 exceeds about 45°, then the assumption of uniform pressure distribution over the shoe surface will be inaccurate. Most drum brakes have contact angles of 90° or more, so a more accurate analysis than the shortshoe assumption is needed. Since any real brake shoe will not be infinitely rigid, its deflections will affect the pressure distribution. An analysis that accounts for deflection effects is very complicated and is not really warranted here. As the shoe wears, it will pivot about point O in Figure 1710, and point B will travel farther than point A because of its greater distance from O. The pressure at any point on the shoe will also
a Fa
Fa shoe B
dFf
dθ
drum
dFn C A
y
θ
ω
r – b cos θ
b cos θ
θ2 θ1 x
B
dFn
C A
y θ
dFf
(a) Brake assembly
FIGURE 1710 Geometry and Forces for a LongShoe External Drum Brake
x Rx
O r
Ry
b sin θ
O
b ( b) Freebody diagram and dimensions
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An Integrated Approach

Sixth Edition
vary in proportion to its distance from O. Assume that the drum turns at constant velocity and that wear is proportional to the friction work done, i.e., the product pV. Then, at any arbitrary point on the shoe, such as C in Figure 1710, the normal pressure p will be proportional to its distance from point O: p ∝ b sin θ ∝ sin θ
(17.13a)
Since the distance b is constant, the normal pressure at any point is just proportional to sin θ. Call its constant of proportionality K: p = K sin θ
(17.13b)
If the maximum allowable pressure for the lining material is pmax (Table 171) then the constant K can be defined as K=
pmax p = sin θ sin θmax
(17.13c)
where θmax is the smaller of θ2 and 90°. Then p=
pmax sin θ sin θmax
(17.13d )
Equation 17.13d defines the normal pressure at any point on the shoe and it varies as sin θ, since pmax and θ2 are constant for any particular brake. Thus, the friction force is small at small θ, is optimum at θ = 90°, and diminishes at angles larger than 90°. Little is gained by using θ1 120°. To obtain the total force on the shoe, the pressure function must be integrated over the angular range of the shoe. Consider the differential element dθ shown in Figure 179. Two differential forces act on it, dFn and dFf. They have respective moment arms about point O of b sin θ and r – b cos θ, as shown in Figure 1710b. Integrating to create their moments about O for the entire surface gives, for the moment due to the normal force: θ2
M Fn =
∫θ
=
∫θ
pwr d θ b sin θ =
1
θ2
∫θ
wrb p sin θ d θ
1
θ2
wrb
1
M Fn = wrb
pmax sin 2 θ d θ sin θmax
pmax sin θmax
1 1 2 ( θ2 − θ1 ) − 4 ( sin 2θ2 − sin 2θ1 )
(17.14 a)
where w is the drum width in the z direction and the other variables are as defined in Figure 1710. For the moment due to the frictional force: θ2
M Ff =
∫θ
=
∫θ
1
θ2 1
17
M Ff = µwr
µpwr d θ ( r − b cos θ ) µwr
pmax sin θ ( r − b cos θ ) d θ sin θmax
pmax sin θmax
(
)
b 2 2 − r ( cos θ2 − cos θ1 ) − 2 sin θ2 − sin θ1
(17.14 b)
Chapter 17
CLUTCHES AND BRAKES
1025
Summing moments about point O gives Fa =
M Fn M Ff
(17.14c)
a
where the upper sign is for a selfenergizing brake and the lower one for a selfdeenergizing brake. Selflocking can occur only if the brake is selfenergizing and MFf > MFn. The torque for the brake is found by integrating the expression for the product of the friction force Ff and drum radius r : T=
θ2
∫θ
µpwr d θ r
1
=
θ2
∫θ
µwr 2
1
T = µwr 2
pmax sin θ d θ sin θmax
pmax ( cos θ1 − cos θ2 ) sin θmax
(17.15)
The reaction forces Rx and Ry are found by summing forces in the x and y directions (see Figure 1710b):
∫ θ =∫ θ
∫
Rx = cos θ dFn + sin θ dFf 2
wrp cos θ dθ + µ
1
=
θ2 1
∫ θ =∫ θ
wrp sin θ dθ
1
∫θ
wr
pmax sin θ cos θ dθ + µ sin θmax
pmax sin θmax
Rx = wr
θ2
∫θ
θ2
∫θ
1
wr
pmax sin 2 θ dθ sin θmax
sin 2 θ sin 2 θ1 2 − − 2 2 1 1 +µ 2 ( θ2 − θ1 ) − 4 ( sin 2θ2 − sin 2θ1 )
∫
(17.16a )
Ry = cos θ dFf + sin θ dFn − Fa 2
1
Ry = wr
µwr
pmax sin θ cos θ dθ + sin θmax
pmax sin θmax
θ2
∫θ
1
µwr
pmax sin 2 θ dθ − Fa sin θmax
sin 2 θ sin 2 θ1 2 − −µ 2 2 − Fa 1 1 + ( θ2 − θ1 ) − ( sin 2θ2 − sin 2θ1 ) 4 2
(17.16b)
17
1026
MACHINE DESIGN

An Integrated Approach

Sixth Edition
EXAMPLE 173 Design of a LongShoe Drum Brake Problem
For the drumbrake arrangement shown in Figure 1710, determine the friction torque T, application force Fa, and reaction forces Rx, Ry.
Given
The dimensions are a = 180 mm, b = 90 mm, r = 100 mm, w = 30 mm, θ1 = 30°, θ2 = 120°, θmax = 90°.
A