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Design for Strength and Endurance

understanding and predicting the static strength,

fatigue life and limit load of mechanical parts

A multimedia self-instructional educational and

knowledge content reference package covering

• Two & three dimensional stress and strain analysis

• Static strength and ductility properties of materials

• Effects of processing on material properties

• Ductile and brittle failure theories under static load

• Nominal and statistical factor of safety application

• Cyclic load effects on material properties

• Fatigue life prediction under fluctuating loads

• Effect of part geometry on limit loading

Carl F. Zorowski Professor Emeritus N.C. State University

Copyright 2002

Copyright © 2002 by Carl F. Zorowski All rights reserved

ISBN: 0-9713126-1-3


_____________________________________________________________________________________ C.F. Zorowski 2002

To Louise

for her patience and support

and

in respectful recognition and admiration of the motivation and dedication of all life-long self-learners

"Every day you may make progress. Every step may be fruitful. Yet there will stretch out before you an ever-lengthening, ever-ascending, ever-improving path. You know you will never get to the end of the journey. But this, so far from discouraging, only adds to the joy and glory of the climb." Winston Churchill


_____________________________________________________________________________________ C.F. Zorowski 2002


C.F. Zorowski 2002


Table of Contents Page

Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1 Course Objective and Overview, Content and Format, Proposed Study Guide, Installation and Operation of CD.

Chapter 1 – Review of Basic Concepts . . . . . . . . . . . . . . . . . . . . . . .11 Strength and Endurance Factors, Normal Stress, Normal Strain, Shear Stress, Shear Strain, Hooke’s Law, Axial Elongation, Internal Beam Loading, Bending and Shear Characteristics, Bending Stress Distribution, Shear Stress in Bending, Maximum Bending Shear Stress, Shaft in Torsion, Shear Stress Distribution, Angular Twist.

Chapter 2 – Two Dimensional Stress Analysis . . . . . . . . . . . . . . . . .29

Definition of Stress Components, Rectilinear Stresses, Rotated Stress System, Double Angle Formulation, Average Stress, Graphical Interpretation, Principle Stresses, Mohr’s Circle for Stress, Orientation of Principle Stresses – 2D, Principal Stresses – 3D, Mohr’s Circle – 3D.

Chapter 3 – Two Dimensional Strain Analysis . . . . . . . . . . . . . . . . . 53

Rectilinear Strain Components, Rotated Strain Components, Total Rotated Normal Strain, Rotated Shear Strain Components, Total Rotated Shear Strain, Graphical Interpretation, Moh’rs Circle for Strain, 45o Strain Rosette, Hooke’s Law, Poisson’s Ratio, Relation between E and G.

Chapter 4 – Static Material Properties . . . . . . . . . . . . . . . . . . . . . . . . 77

Mechanical Tests, Tensile Test, Elastic Behavior, Moduli and Poisson’s Ratio, Plastic Behavior, Nominal Stress-Strain Diagram, Necking, True Stress and Strain, True Stress-Strain Diagram, Offset Yield, Typical Tensile Properties, Brinell and Rockwell Hardness Tests, Hardness Scale Comparison.

Chapter 5 – Effects of Material Processing . . . . . . . . . . . . . . . . . . . . 97

Overview, Casting – Sand, Investment, Shell Mold, & Die, Hot Working – Rolling, Extrusion, & Forging, Cold Working – Rolling, Drawing, Heading, Roll Treading, & Stamping, Hot/Cold Property Comparison, Heat Treatment – Annealing, Normalizing, Quenching, Tempering, Case Hardening, Effects of Heat Treatment.


C.F. Zorowski 2002

Chapter 6 – Theories of Static Failure . . . . . . . . . . . . . . . . . . . . . . . . 121 Design Issues, Designer’s Dilemma, Max Normal Stress Theory, Max Shear Stress Theory, Distortion Energy Theory, Mohr’s Theory, Coulomb Mohr Theory, Ductile Material Failure – Applicable Ductile Theories, Brittle Failure, Modified Coulomb Mohr Theory, Applicable Brittle Theories.

Chapter 7 – Factors of Safety . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

Purpose and Definition, Issues Effecting FS, Example Variation Scenarios, Modified Theories of Failure, Generic Factors for Safety, Actual and Load Capability Distributions, Interpretation of Load Differences, Application of Gaussian Distributions, Statistical Determination of Percent Failures, No Failure Scenario

Chapter 8 – Fatigue Strength and Endurance. . . . . . . . . . . . . . . . . . .169

Fatigue Failure Process, Fatigue Testing and S-N Diagram, Fatigue Strength Equation, Endurance Limit Behavior and Approximations, Modifying Factors including Surface, Size, Load, Temperature, Stress Concentration and Miscellaneous Effects

Chapter 9 – Fluctuating Load Analysis . . . . . . . . . . . . . . . . . . . . . . . .189

Fluctuating Stresses, Stress-Time Relations, Modified Goodman Diagram, Mean/Fluctuating Stress Diagram, Failure Theories – Soderberg, Goodman, & Gerber, Torsional Fatigue, Combined Loading Modes, Cumulative Fatigue Damage, Palmgren-Minor Theory, Mason’s Modification

Chapter 10 – Limit Loading. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223

Basis of Limit Load Analysis, Linear and Non-Linear Beam Bending, Plastic Moment, Plastic Hinge Behavior, Determinant and Indeterminant Beam Ultimate Load Examples, Ring Collapse Load

Appendix. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245

Chapter Quizzes, Chapter Quiz Solutions, Off Line Exercise Solutions, Final Course Project, Index

Design for Strength and Endurance- Introduction

_____________________________________________________________________________________Introduction - 1 - C.F. Zorowski 2002

Introduction

Screen Titles

Course Objective and Overview Content and Format Proposed Study Guide Installation and Operation



Course Objective and Overview

The Design for Strength and Endurance instructional package was created to meet the needs of engineering life-long learners who desire through self-study to continue their professional development. It was designed with the instructional objective to provide a learning environment that overcomes the frustrations and difficulties often encountered by the self-learner in attempting to decipher and understand commonly used text books on subjects of interest. A clarification is also in order at this point to establish the manner in which the activity of “design” will be dealt with in this unique instructional approach. The preferred definition of the word “design”, according to Webster, is “a mental process or scheme in which the means to an end are laid down”. An acceptable definition of “design” in the context of engineering might well be “an activity under taken by engineers that results in the creation of the description and specification of a device, system or process that meets some recognized or identified need of society based on existing technology while making use of available natural resources within some specified set of constraints applicable to the circumstance in question.” Although the engineering definition is lengthy and possesses greater specificity there can be no disagreement of its consistency with the more generic Webster definition. However, it is precisely the greater specificity of design as an engineering process that requires further consideration of the subject content of this educational module and its manner of presentation. The word design in the title Design for Strength and Endurance is not meant to refer to the process of creating a description of the size, form, shape or operational characteristics of some physical device or system. The purpose of this educational module is to develop and describe the application of analytical predictive processes used to determine whether a proposed physical design will satisfy the criteria

and constraints that will insure its mechanical strength and life endurance requirements. Throughout the instructional treatment that follows application of these predictive processes will be limited to what might be classified generically as physical “parts” in some mechanical device. More specifically, the size, shape, loadings and materials of the bars, beams, shafts and other similar components treated will always be specified. The subject of Design for Strength and Endurance is dealt with in this fashion to more readily achieve the instructional objective of providing the student with the knowledge and practice to determine the static strength, fatigue life and ultimate strength of mechanical components as they are influenced by their geometry, external loading and material properties. Having the student develop effective levels of competence in the following specific topics will fulfill this educational objective:

1. Introductory definitions and concepts. 2. Analysis of two dimensional stress states. 3. Two dimensional strain states and

Hooke’s law. 4. Static strength behavior of materials. 5. Effect of processing on material

properties, 6. Theories for predicting static ductile and

brittle failure. 7. Nominal and statistical factors of safety. 8. Cyclic reversed loading effects on material

properties. 9. Fatigue life prediction under general

fluctuating loads. 10. Effect of part geometry on limit loading.

Knowledge prerequisites for this study

are the basic concepts provided in an introductory engineering course in strength of materials. Some repetition of these earlier materials is included to provide their thorough understanding to insure complete familiarity and confidence in their application as the entire subject is developed.



Content and Format

The complete Design for Strength and Endurance instructional package consists of two items, a CD disk and a printed Instructional Supplement. The disk contains a multimedia audio enhanced graphical presentation of the subject content. This covers the knowledge required to both understand and analytically predict the static strength, fatigue life and limit load capacity of mechanical parts subjected to static and fluctuating external loading. The complete CD presentation consists of ten self-contained chapters that can be played independently using installed software together with the disk. The ten chapters contain some 300 graphically illustrated and animated screens accompanied by complete audio explanations of the presented material. Navigation aids are provided on each screen. This permits the viewer to stop the presentation, exit the chapter, replay the audio or select other screens in the chapter or any other chapter. Capability to view the text of the audio in a pop up window is also available. Each chapter includes both example and interactive exercise problems that illustrate the application of the material covered. Hyperlinks to solutions of the interactive problem are included to allow the viewer to verify their own results. At the end of each chapter Review Exercises with immediate feedback are provided for the viewer to check their general knowledge of the material covered. Also included is an off line exercise for practice in the application of the chapter concepts beyond the presentation. Instructions for the use of the software program on the disk are covered in the section on Installation and Operation. The Instructional Supplement is provided as an additional learning aid and ready hard copy reference to all materials presented on the CD disk. Its contents include grayscale copies of all instructional screens on the CD together with the printed script used for the audio enhancement of each screen. The printed supplement can be used in a variety of ways to enhance the instructional experience to best suit the preferred learning style of the student.

Only two screens appear on each page. This provides space for the viewer to take notes or make other relevant comments while listening to the disk presentation. With the full text of the audio provided important phrases and concepts in the text as well as important equations on the graphics can be highlighted as desired by the student. The interactive problem solutions are separated from the main chapter content to reduce the temptation to look up the answer before attempting to solve the problem. Quick visual reference from one screen to another in any given chapter is readily possible as might be helpful in the solution of the Off Line Exercise. Also, quick reference to equations and physical concepts is readily available with use of the index in the Appendix of the supplement. Imaginative serious students of the subject will undoubtedly employ other complementary uses of the CD disk together with the Instructional Supplement. At the end of each chapter are solutions to the interactive practice problems included in the chapter. Also included with each chapter is a short quiz that students can use to further evaluate their comprehension of the chapter content. These are most effectively taken as timed ten-minute exercises without reference to the supplement or disk. Solutions to the quizzes are provided in the Appendix. Solutions to the Off Line Exercises are provided in the Appendix along with a topical Index to provide a ready reference to specific concepts and subjects included through out the Instructional Supplement. Finally, the Appendix includes instructions for the student to undertake and carry out an extended formal project representative of a real practical problem. This more extensive final exercise provides the learner with an experience that synthesizes the application of all knowledge content included in the educational module.



Suggested Study Guide Although individual learning styles may differ widely the following guide is suggested as a study guide to achieve both effective coverage and self-evaluation of the understanding and proper application of the knowledge content of Design for Strength and Endurance. First, it is useful to recognize that the ten chapters of the module can be subdivided into the following areas of emphasis:

1. Review of the fundamental applicable concept of strength of materials. (Chapter 1)

2. Analytical and graphical analysis of general states of stress and strain and their relationship as dictated by Hooke’s Law. (Chapter 2 &3)

3. Typical mechanical properties of metals under static loading and the effects of mechanical and thermal processing. (Chapters 4&5)

4. Theories for predicting the static ductile and brittle failure of mechanical parts together with nominal and statistical factors of safety. (Chapter 6 & 7)

5. Effects of reversed loading on material properties and fatigue failure analysis under cyclic loading. (Chapters 8 & 9)

6. Support geometry effects on the limit loading of beam elements. (Chapter 10)

Area 1 is preparation for the remainder of the subject. Areas 2 and 3 cover the fundamental required for consideration of Area 4. Area 5 extends the static analysis to application of fluctuating loads and Area 6 introduces the concept of ultimate loading. The chapters should be covered in the order listed for continuity. It is recommended that the study of each chapter be started with the audio enhanced presentation provided on the CD. As suggested on the chapter index page proceed through all pages in a sequential pattern. As the audio proceeds do not hesitate to interrupt the presentation (stop button) and start it over (replay button) if you do not understand the

explanation. The page may be replayed as many times as necessary. The manual may be used during this process for reference to the audio text, highlighting important items or making relevant notes. The audio text is also available on the pages of the CD presentation by clicking on the text button after the audio is complete or has been stopped. Complete interactive exercises as they are encountered to evaluate your understanding of the most recent material covered. Check your results by clicking on the solution button on the page. If correct return to the presentation and proceed to the next page, otherwise satisfy your self that you understand where you went wrong. You may return to the chapter index page (index button) to review any page that may be useful. Use the Review Exercises at the end of the chapter for an immediately evaluation of your general knowledge of the important concepts covered. Make use of the “Hot words” (links to other screens) as appropriate. At this point undertake the solution of the Off Line Exercise. The supplement may again be useful at this point to quickly review or reference important equations and concepts. When completed check your solution against the solution listed in the appendix. Finally, take the quiz at the end of the chapter without reference to either the CD or the supplement and then check your answers against those provided in the Appendix. If you are satisfied with your comprehension of the content of the chapter proceed on. When you have completed all ten chapters it is recommended that you undertake the project presented in the Appendix. This activity is more extensive than the preceding exercises. It provides an experience in synthesizing all the material covered with the added requirement of requiring assumptions concerning aspects of the situation presented that are not defined. The value gained from this activity will be in proportion to the time and effort committed.



Installation and Operation

The Design for Strength and Endurance CD instructional module is a self-contained PowerAudioTM instructional product created in Asymetrix Toolbook II AssistantTM. Be sure to read the following installation and operating instructions before proceeding. Hardware/Software Requirements: PC running Windows 95/98/2000 or NT CD 2x or higher drive (the faster the better!) Sound card with speakers (capable of playing wave and midi files). MS Windows media player To Install CD Software Program (First, close all running applications before installing Design for Strength and Endurance)

1. Insert the Design for Strength and Endurance disc in the CD ROM drive. 2. From the Start menu choose Run in

Windows 95/98/2000 or NT. 3. In the Run window, type location from

which program is to be installed (for example D:\). Then type setup.exe. (see note below)

4. Click OK and follow instructions on screen in Install Wizard.

5. Choose either “typical” or “custom” installation. “Typical” places only the program operating system on hard drive with all remaining files left on CD. “Custom” permits installation of educational module files on hard drive. This can require significantly more storage space. Advantage of installation on hard drive is more rapid transfer between chapter and audio files.

6. Restart computer. (Note: Following step 3 above a window may appear with icons of files on the CD disc. If this occurs double click on setup.exe file and proceed with step 4)

Alternative:

1. Insert the Design for Strength and Endurance disc in the CD drive.

2. Open Windows file Explorer application.

3. Double click on CD drive to list files. 4. Select and double click on setup.exe

file. 5. Install Wizard will appear. 6. Follow steps 5 and 6 as previously.

If “Typical” installation is chosen (or all files are not placed on hard drive) CD must remain or be placed in drive to play complete educational module presentation. To Run Design for Strength and Endurance :

1. From the Start menu in Windows 95/98/2000 or NT point to Programs then Design for Strength and Endurance shortcut.

2. Click on Begin Module file. 3. Program will begin and detailed

navigation instructions will be provided on the Main Menu page

Or 1. From the Start menu in Windows

95/98/2000 or NT point to Programs, then to Design for Strength and Endurance.

2. Click on any Chap. _ desired. 3. Program will open at Chapter selected.

To Exit Design for Strength and Endurance:

1. From any page in any chapter click on the Stop button until Replay button appears. 2. Click on Exit button. 3. The Exit page will appear and allow two choices: to exit the module or return to the Main Menu.



Navigation, Audio and Special Features:

Six navigation and operation buttons always appear in the bottom border of each page in the module. Their functions are as described below: Stops audio and forwards to next page. Stops audio and returns to previous page. (Note: Above navigation buttons are only active if pointers are highlighted. If active they will respond even if audio is playing.) Stops audio and slide animation. On repeated clicks will bring up Replay button and Click Here button.

Transfers to exit page to permit Exiting module or return to Main Menu page

Shows “popup” window in lower right of page with text of audio for that page. Returns to Chapter Index Page for selection of a specific page Four additional operational buttons that appear intermittently when the audio finishes are the “Replay”, “Click Here”, “Return” and “Main Menu” buttons.

Replays page from the beginning with audio and all slide animation

Appears when audio is finished and Next Page button should be clicked.

Returns reader from problem solution page to next page in chapter.

Returns reader to main Chapter Menu to permit selection of another chapter or exit module

Some Special Notes

1. Older sound cards may not be equipped to play midi (synthesizer music) files. If no music occurs when second page of Begin Module appears wait 30 seconds and next page with audio will come up automatically.

2. If audio volume requires adjustment refer to your operating systems manual (Windows 95/98/2000 or NT) for instructions on changing volume of wave files.

3. Pages of Design for Strength and Endurance education module should fill about 2/3 of monitor screen. Size of presentation window can be adjusted. Refer to your operating systems manual (Windows 95/98/2000 or NT) for instructions on changing monitor settings.

4. Pop up windows are closed by clicking on X in top right corner.

5. If content, video and text chapter files remain on CD rather than loaded on hard drive transfer time of hyperlinks (special buttons) may slow down depending on speed of computer. Click once to activate hyperlink and apply patience! Hour Glass icon will appear indicating transfer is underway,

Exit

Stop

Click Here

Text

Index

Replay

Return

Main Menu

Design for Strength and Endurance – Chapter 1

_____________________________________________________________________________________Review of Basic Concepts - 11 - C.F. Zorowski 2002

Chapter 1 Review of Basic Concepts

Screen Titles

Strength and Endurance Factors Review of Basic Concepts Normal Stress Definition Normal Strain Definition Shear Stress Definition Shear Strain Definition Hooke’s Law Axial Elongation Internal Beam Loading Beam Loading Problem Bending and Shear Characteristics Bending Stress Distribution Shear Stress IN Bending Qmax for Rectangle Maximum Shear Stress Shaft in Torsion Shear Stress Distribution Angular Twist Review Exercises Off Line Exercises



1. Title page Chapter one provides an introductory overview of the factors that influence the strength and endurance of mechanical components as well as a review of some basic concepts that underlie a detailed study of this subject. The factors discussed include applied loading, geometric property characteristics of the mechanical components, material properties and failure criteria. The basic concepts reviewed cover the definition of normal and shear stress and strain, elongation and shear deformation, Hooke’s law and the analysis of simple beams and shafts.

2. Page Index Listed on this page are all the individual pages in Chapter 1. Each title is hyperlinked to its specific page and can be accessed by clicking on the title. It is suggested, however, that the reader first proceed through all pages sequentially Clicking on the text button at the bottom of the page will provide a pop up window with the text for that page. Clicking on the x in the top right corner closes the text page. Clicking on the index button returns the presentation to the chapter page index.



3. Strength and Endurance Factors There are three generic factors that directly influence the strength and endurance of mechanical components. One factor is the loading applied to the component. More specifically, three important characteristics of the loading are its magnitude, its location and direction and whether the load is static or dynamic. A second factor deals with the mechanical component itself. In this case the important characteristics influencing strength and endurance are the shape and size of the component. The third factor is the material from which the component has been made. Here the important characteristics are the material’s stress and strain behavior under both static and dynamic loading. A fourth consideration that also needs to be considered is the failure criteria that define the strength and endurance characteristics. Failure may not simply be designated by fracture of the component. There may also be design considerations dealing with excessive deformation or permanent distortion that will dictate the definition of strength and endurance.

4. Review of Basic Concepts The brief review that follows deals primarily with some of the more important definitions, material behavior assumptions and specific component behavior related to the topic called strength of materials. The first concept reviewed is that of normal stress and shear stress within a loaded body resulting from external loading. This is followed by the definition of elongation and shear strain that accompany these stresses. Hooke’s law is introduced to relate stress to strain in terms of the moduli of the material from which the component is made. Finally these concepts are applied to the determination of normal and shear stress distributions in transverse loaded beams and twisted shafts. All these topics should indeed be a review for the reader but in no way does this material represent a complete review of basic strength of materials.



5. Normal Stress Definition Normal stress is most easily visualized and defined in terms of the behavior of a bar in simple tension or compression as shown in the figure. Consider cutting the bar in tension transversely and treat the left portion as a free body diagram. This portion of the bar must be kept in equilibrium by the distribution of internal force elements across the cut cross-section that balances the applied external load P. These internal force elements are given the units of load per unit area or a kind of pressure and are called the normal stress. Assuming these force element distribution is uniform across the section the axial normal stress in a bar subjected to an axial load is the magnitude of the load divided by the cross sectional area of the bar. Hence, the units of normal stress are lbs. per square in. or PSI. For a bar under compression the normal stress is negative since the applied load that is compressive is considered as being a negative force.

6. Normal Strain Definition The axial bar from the previous page will lengthen under the action of a tensile load and shorten when compressed. The change in length of the bar under axial load is used to define a generic parameter that represents this axial deformation. The increase in length divided by the original length of the bar is defined as the positive axial strain, epsilon. In order words, epsilon is simply delta L divided by L. Note that axial strain is dimensionless since both delta L and L have the units of length. For a bar shortened by compression the strain will be negative since the shortened length delta L is considered negative.



7. Shear Stress Definition Consider a block element that is fixed along its bottom edge and subjected to a horizontal force V applied parallel to the top edge acting to the right as shown in the figure on the left. To keep this block in horizontal equilibrium there must exist distributed force elements along its bottom surface directed to the left to balance the magnitude of the applied force V. Assuming that these force elements are uniform over the fixed surface of the block this leads to the concept of shear stress acting parallel to the bottom surface as the magnitude of the force V divided by the surface area of the block where it is fixed. In other words, shear stress is defined as the shear force V divided by the area over which it acts. Note that the dimensions of shear stress are pressure or force per unit area expressed as PSI similar to that of normal stress. The basic difference between normal and shear stress is that the first acts normal to a surface while the second act parallel to a surface.

8. Shear Strain Definition The application of the shear force V on the previous page produces an angular distortion of the block as indicated by the dotted parallelogram on the figure. When this occurs the right angles of the corners of the original block are distorted. The change in angle gamma of the bottom left corner is defined as the shear strain associated with the shear stress resulting from the shear force V. Since this is a change in angle and is measured in radians the units of shear strain are dimensionless just like the units of normal strain.



9. Hooke’s Law Robert Hooke, an English experimental physicist is credited with discovering the first law of elasticity in the late 17th century. This law simply states that stress and strain are linearly related. That is normal stress is proportional to normal strain and shear stress is proportional to shear strain. The constant of proportionality for normal stress is designated E and is called the modulus of elasticity or Young’s modulus. The constant of proportionality for shear strain is designated G and is called the shear modulus of elasticity. Both of these physical quantities must be determined experimentally for a given specific material. Most solid materials and in particular metals behave in accordance with this law at least for small values of strain. More detail on this behavior will be discussed in a later chapter.

10. Axial Elongation Using the definitions of normal stress and strain together with Hook’s law it is possible to express the elongation of a bar in tension in terms of the applied load P. First begin with the relationship expressing Hooke’s law, that is the normal stress in the bar, sigma, is equal to E times epsilon. Now replace sigma with its equivalent which is the load P divided by the cross sectional area of the bar A. Also replace the strain epsilon with its definition which is the change in length divided by the original bar length L. Now solve the resulting relationship for the bar elongation delta L. This gives the final expression PL over AE.



11. Internal Beam Loading Transversely loaded simple beams will now be reviewed. Before proceeding to normal and shear stresses in bending it is first necessary to determine the internal bending moment and shear force distribution due to the effect of external transverse loadings. The internal shear force and bending moment at any point along a beam is determined as the force and moment which must exist at the location of a cut cross section of beam to satisfy equilibrium treating the remaining beam portion either to the right or left of that location as a free body diagram. Applying the principle that the sum of all vertical forces acting on the free body beam portion must be zero will result in an equation from which the magnitude and direction of the shear force at the cut cross section can be determined. Setting to zero the sum of all moments of all external forces acting on that portion of the beam about the cut cross section location together with the internal bending moment will result in an equation defining the direction and magnitude of the internal bending moment. The sign conventions for bending moment and shear force are given below the figure. The distributions of these internal reactions along the length of the beam are referred to as the bending moment and shear force diagrams.

12. Beam Loading Problem Before going on consider the application of the concept and method of analysis described on the previous page to the beam loaded and supported as shown in the figure. Assume that the internal and left end supports can only sustain vertical reaction forces. For the dimensions and loading given calculate expressions for the shear force and bending moment distributions in the beam. Plot the shear and bending moment diagrams along the length of the beam and determine the maximum bending moment. When you have completed you solution to the problem click on the solution button to check your results against mine.

(Solution on Pages 27 and 28)



13. Bending and Shear Characteristics Listed on this page are some characteristics of shear force and bending moment diagram behavior that are useful aids in developing these characteristics of internal loading in beam. The slope of the shear diagram will be equal to the magnitude of any distributed load on the beam at each point. That is if the loading is 200 lb/ft then the shear will change at the rate of 200 lbs per foot. It also follows that between concentrated external loads the slope of the shear diagram will be zero. The slope of the bending moment diagram is always equal to the magnitude of the shear force at each point. Thus if the shear force is constant over a length section of the beam the slope of the bending moment diagram over that same section will be a constant. Applied external forces cause vertical jumps in the shear force diagram equal to the magnitude of the applied load. In a similar fashion applied external moments cause jumps in the moment diagram but do not effect the shear force distribution. To see how some of these characteristics apply you may wish to revisit the problem solution on the previous page.

14. Bending Stress Distribution A positive bending moment applied to a beam will cause it to assume a concave upward deformation. This in turn means that the top fibers of the beam will be in compression while the bottom fibers of the beam will be put in tension. The strain distribution is linear and passes through zero at the centroid of the section. This location is referred to as the neutral axis of the beam. The accompanying internal normal stress distribution is also linear and given by the expression sigma equal to the bending moment M times the distance measured from the neutral axis y divided by the area moment of inertia I of the cross section through its centroid about an axis perpendicular to the plane of the external loading. Thus the normal bending stress is zero at the neutral axis and is maximum at the top and bottom of the beam cross section. For a rectangular cross section the moment of inertia I is given as 1/12 the base b or width of the section times the height h cubed. It is important to remember that this stress formula is only valid for a beam loaded in a plane of symmetry of the cross section.



15. Shear Stress in Bending If a section of beam is in pure bending it will experience no shearing stresses at an internal cross section over this section. This is because the internal shear force is zero since the moment is a constant. However, if an internal shear force exists then there will be corresponding shear stress that accompanies the normal bending stress. The magnitude of the shear stress tau is given by the expression of the product of the shear force V times the first moment of the area of the cross section above the point of the stress about the an axis through the centroid perpendicular to the plane of external loading divided by the area moment of inertia I of the cross section and the width of the cross section b. For a rectangular cross section this gives rise to a parabolic shear distribution as shown in which the shear stress is maximum at the neutral axis and zero at the top and bottom of the cross section.

16. Qmax for Rectangle To determine the maximum shear stress for a rectangular cross section it is first necessary to determine the first moment Q of the top half of the cross section since the maximum shear stress will be at the neutral axis through the centroid of the cross section. This is computed by multiplying the area of the top half of the cross section by the square of the distance from the centroid of the top half to the centroid of the total cross section. Hence Q max becomes the product of half the height h/2 times the width b times the square of h/4. The final result for Q max is given by 1/8 bh2.



17. Maximum Shear Stress in Bending The expression VQ over I b is now used to calculate the magnitude of the maximum shear stress in a rectangular cross section. Substituting Q equal to 1/8 bh2 and I equal to 1/12 bh3 results in a maximum shear stress of 3/2 times the shear force divided by the area of the cross section. In other words the maximum shear stress in bending in a rectangular cross section is 150 % higher than what would be expected if the it were assumed that the shear stress was uniform over the section and just equal to V divided by A. This multiplier for the maximum shear stress as compared to the average shear stress of V/A is of course a function of the shape of the cross section and must be determine d for other geometries.

18. Shaft in Torsion A circular shaft in torsion will experience an internal shear stress distribution on cross section perpendicular to the axis of the shaft. The formula for its calculation is similar in appearance and form to that for normal bending stress except that M is replaced by the torque T acting at the section, the distance y is replaced by the radius r measured from the center to the point of stress and I is replaced by J which is the polar moment of inertia of the circular cross section. Since the torque is in the units of force times distance or length and J is in the units of length to the fourth power and r has the units of length the shear stress has the units of force per length squared or PSI as expected.



19. Shear Stress Distribution Since the magnitude of the shear stress in a twisted shaft is proportional to r or the distance measured from the center of the section then the shear stress distribution is liner and maximum at the outer surface of the shaft. It is important that this analysis is only applicable to shafts of circular cross section, however they need not be solid. Hence the analysis is also applicable to hollow shafts.

20. Angular Twist The angular twist, theta, of one end of the shaft relative to the other requires the introduction of the shear Modulus G and is given by the expression TL over JG. In this expression T is the Torque carried by the shaft, L is the length of the shaft, J is the polar moment of inertia of the circular cross section and G is the shear modulus. Satisfy yourself that if G is given in the units of force per unit area that the units of theta the angle of twist are dimensionless. Thus the magnitude of the twist that will be calculated will be in radians.



21. Review Exercises At the end of each chapter one or more review exercises are provided to help readers satisfy themselves that they under stand the material contents presented. In this exercise the items in the list on the left are to be matched with the mathematical relationships on the right. Place the cursor over an item on the left and hold down the left button. A pencil will appear that can be dragged to one of the green dots on the right. If the right choice is made the arrow will remain. If the selection is incorrect the arrow will disappear. After the exercise is completed proceed to the next page.

22. Off Line Exercises Each chapter includes one or more off-line exercises. These are to be completed by the reader and submitted at required by the calendar provided by the course instructor. When you have finished with this page click on the main menu button. This will return the program to the chapter index page to select another chapter or exit the module.

(Solution in Appendix)



Chapter 1 Review of Basic Concepts

Problem Solutions

Screen Titles

Support Reactions Shear Force Distribution Bending Moment Distribution Shear and Moment Diagrams



1. Support Reactions The reaction forces on the beam at the support points A and B must be determined first using equilibrium. The reactions are represented by vertical forces F1 and F2 since A and B are simple supports. Summing moments about point A permits F1 to be determined directly as 3750 lbs in the direction assumed since it is positive. Summing forces vertically with F1 now known permits F2 to be calculated as minus 1250 lbs. The negative sign indicates that it acts down at A rather than up as initially assumed.

2. Shear Force Distribution Assume the internal shear force V at section 1 is positive. Equilibrium of the beam from section 1 to point C indicates that the shear force is a constant value of 2500 lbs. from point C to B since section 1 is a general location. Assume again that the shear force V is positive at section 2. Equilibrium applied to the beam from section 2 to point C gives a value for V of –1250 lbs. The negative sign indicates that the positive shear assumption was incorrect. Since section 2 is a general location between point A and B the shear in this section of –1250 N is again constant.



3. Bending Moment Distribution The bending moment at section 1 is now determined by applying moment equilibrium to the beam section from 1 to C. It is necessary to define the location of section 1 by the dimension x since moment arms need to be established. Setting clockwise moments equal to zero on the length x to C on the beam gives the moment M at section 1 as minus 2500x. Hence the moment varies linearly from C to B and is negative. Summing clockwise moments ay section two for the beam from 2 to C and setting them equal to zero gives a second linear equation in x for the moment. Note that the moment value at B is 12500 ft lbs whether it is calculated from the equation for the portion C to B or for the portion B to A. Also note that the moment at location A from the second equation is zero.

4. Shear and Moment Diagrams The shear diagram has a constant value of 2500 lbs. from point C to B as calculated previously. At B it changes its value to –1250 lbs. due to the effect of the reaction at B and continues constant over the beam length from B to A. The bending moment diagram is plotted directly below the shear diagram to see the effects of the shear on the bending moment. From the equations on the previous slide the bending moment starts at zero at point C and increases negatively in linear fashion until it reaches the value of 12500 ft. lbs. at point B. At this point the second equation for the moment becomes valid which decreases its negative value from 12500 ft. lbs. at B to zero at point A. Placing the shear force diagram immediately above the bending moment diagram leads to a very useful observation. The bending moment diagram is the integral of the shear diagram beginning from the left end of the beam. Although no general proof of this is given here it is applicable to all beam bending problems. Another interesting observation is that the maximum bending moment occurs where the shear force passes through zero. This is a useful way to quickly determine where the largest values of bending moment will occur.


_____________________________________________________________________________________Two Dimensional Stresses - 29 - C.F. Zorowski 2002

Chapter 2 Two Dimensional Stress Analysis

Screen Titles

Definitions – Stress Components Rectilinear Components Rotated Stress System Normal Stress – X’ direction Shear Stress – Y’ direction Normal Stress – Y’ direction Double Angle Formulation Average Stress (introduction) Average Stress (continuation) Graphical Interpretation Principal Stresses Mohr’s Circle Construction Orientation – Principal Stresses General Stress State Principal Stresses – 3D Orientation Principal Stress Equation – 3D Mohr’s Circle – 3D Stress State Review Exercise Off line Exercise



1. Title page Chapter two covers the definition and analysis of a two dimensional rectilinear state of stress involving both normal and shear components. The topics discussed include the equilibrium of cross shear stresses, development of the equations defining stress components with respect to a rotated axis system, graphical interpretation of rotated axis system stress equations, principal stress components, Mohr’s circle construction and use together with a brief introduction to the analysis of a generalized three dimensional state of stress. Several sample problems demonstrating the application of the theory presented are also included.

2. Page Index Listed on this page are all the individual pages in Chapter 2 with the exception of the sample problems. Each title is hyperlinked to its specific page and can be accessed by clicking on the title. It is suggested that the reader first proceed through all pages sequentially. Clicking on the text button at the bottom of the page provides a pop up window with the text for that page. The text page is closed by clicking on the x in the top right corner of the frame. Clicking on the index button returns the presentation to the chapter page index.



3. Definitions – Stress Components The figure illustrates some general solid body in equilibrium under the action of a number of external forces. If a plane is passed through the body there must exist a distribution of force elements delta F on incremental area elements delta A of the cut surface to keep the bottom portion of the body in equilibrium. The incremental force element delta F can be broken down into two components, one that is normal to the surface of delta A designated delta Fn and a second component parallel to the surface element designated delta Ft. The limit of the ratio of delta Fn to delta A as delta A goes to zero is mathematically defined as the normal stress sigma at that point. The limit of the ratio of delta Ft to delta A as delta A approaches zero is defined as the shear stress at that point. Both the normal stress and the shear stress may vary over the surface depending on the shape of the body and the location and direction of the externally applied loads.

4. Rectilinear Stress Components Consider a cubical element of material with dimensions dx, dy and dz within a loaded body with the dx and dy edges parallel to an xy axes system as shown. The third dz edge is parallel to a z-axis, which is normal to the page. In a two dimensional state of stress the right face perpendicular to the x axis is acted on by a normal component of stress sigma x and a positive shear stress component tau xy pointing in the y direction. To satisfy both horizontal and vertical equilibrium the stress components on the left face perpendicular to the x-axis must be as shown. In the most general two-dimensional case the upper face normal to the y-axis will also be acted on by a normal stress component sigma y and a positive shear component tau yx pointing in the x direction. To again satisfy horizontal and vertical equilibrium the normal and shear components on the lower face perpendicular to the y axis must contain normal and shear components as shown. To satisfy moment equilibrium about the center of the element the force couples created by tau xy and tau yx stresses must be equal to each other. Assuming the dz length of the cube is unity equality of these couples results in tau xy equal to tau yx. This is commonly referred to as equilibrium of cross shears. Hence, in the most general case a two dimensional stress state in rectangular coordinates consists of two normal stresses, sigma x and sigma y together with only a single shear stress, tau xy.



5. Rotated Stress System Given a two dimensional state of stress with respect to a given xy axis system the question then is what is the state of stress with respect to some x’y’ axis system that makes an angle theta with the original axis system as shown in the figure. Do the values of the normal stresses sigma x’ and sigma y’ and shear stress tau xy’ measured relative to planes perpendicular to the x’y’ axes change and how are they related to the given state of stress. To answer this question it is necessary to perform an equilibrium analysis on elements that share planes perpendicular to the x’y’ axes and the xy axes. This is carried out on the next two pages.

6. Normal Stress – x’ direction To determine the normal stress in the x’ direction the triangular element on the left is employed. The plane normal to x’ is acted on by the stresses sigma x’ and tau xy’ while the original stress components sigma x, sigma y and tau xy act on the two faces normal to the x and y-axes. Force equilibrium is now applied in the x’ direction. In this development the element is assumed to be of unit depth in the z direction and the length of the inclined edge is also assumed to be of unit length. Thus the edge parallel to the x-axis is sin theta and the edge parallel to the y-axis is cos theta. The force due to sigma x’ is simply the stress times the area it acts over which is one by one. The force contribution of sigma x in the x’ direction is the stress times the area, cos theta times one, times cos theta to give the component in the x’ direction. In a similar fashion the force contribution of tau xy on the x face is the stress times the area, cos theta times 1, times sin theta to give the component in the x’ direction. The force contribution of sigma y in the x’ direction is the stress times the area, sin theta times one, times sin theta to give the component in the x’ direction. Like wise the force contribution of tau xy on the y face is the stress times the area, sin theta times 1, times cos theta to give the component in the x’ direction. All contribution due to the original stress state are in the negative x’ direction. Combining trigonometric terms gives the final equation at the bottom of the page. It is seen that all components of the original stress state contribute to the magnitude of sigma x’.



7. Shear Stress – y’ direction The triangular element from the previous page can also be used to determine an equation for tau xy’ in terms of the original stress state by applying the condition for force equilibrium in the y’ direction. The force due to the tau xy’ stress is simply tau xy’ times the area it acts on which is one by one. The force contribution of sigma x will be positive and is given by the stress times its area, cos theta times 1, times sin theta to give the component in the y’ direction. The contribution of tau xy on the x face which will be negative is the stress times the area, cos theta times 1, times cos theta for its component in the y’ direction. On the y face the contribution of sigma y will be the stress times the area, sin theta times 1, times cos theta for its component in the y’ direction. Finally the positive force contribution of tau xy on the y face will be the stress times the area sin theta times 1 times sin theta to give the component in the y’ direction. Appropriately combining terms results in the equation for tau xy’ at the bottom of the page. Again all components of the original stress state contribute to the magnitude of tau xy’.

8. Normal Stress – y’ direction To satisfy yourself that you really understand the developments on the previous two pages undertake the exercise of generating the equation for the normal stress sigma y’ in the y’ direction in terms of the original xy axis stress state and the angle theta. This will require a different triangular element, one that possesses a plane perpendicular to the y’ direction as well as planes perpendicular to the x and y axes. The correct result for sigma y’ is given by the equation at the bottom of the page. If needed you can review the solution to this exercise by clicking on the solution button, or when you complete the development go on to the next page.

(Solution on Page 47)



9. Double Angle Formulation For reasons that will soon become apparent it is convenient to write the equations developed for sigma x’, sigma y’ and tau xy’ in terms of the double angle of axis rotation, two theta. In particular the identities which prove useful for this purpose are sin theta cos theta equal to sin two theta, sin squared theta equal to ½ - ½ cos two theta and cos squared theta equal to ½ + ½ cos two theta. This permits the earlier equations written in terms of theta to be expressed in terms of half the sum of sigma x and sigma y, half the difference between sigma x and sigma y, tau xy and the sin and cos of two theta as shown on the page. It is of interest to note that these are periodic functions all of which have maximum and minimum values for some prescribed values of theta. To show that this is true and to determine what these extreme values are a further rearrangement of these equations will be undertaken on the next two pages.

10. Average Stress - Introduction This further development is begun by defining sigma average as one half the sum of sigma x plus sigma y. This term is then subtracted from the equation for sigma x’ and the result is squared. The gives an equation with the rather extended right side as shown on the page. A further manipulation of this result is continued on the next page.



11. Average Stress (continued) The equation for tau xy’ is now squared and added to the equation of the previous page. This significantly simplifies the right side of the resulting equation to give the expression shown in the middle of the page. By defining R squared as the sum of the quantity sigma x minus sigma y all over two squared plus tau xy squared the even simpler form of the equation shown at the bottom of the page is obtained. This should be recognized as the equation of a circle in the sigma x’, tau xy’ axes system whose center is on the sigma x’ axis at the location sigma avg with a radius of R. A graphical representation is shown on the next page.

12. Graphical Interpretation A graphical representation of the circle represented by the last equation on the previous page is shown on the left plotted on a sigma x’, tau xy’ axis system. The center is located at sigma avg on the sigma x’ axis and the circle is drawn with a radius R. The expressions for the circle and the definitions of sigma avg and the radius R in terms of the original stress components are given to the right. Any point on the circle represents the sigma and tau stress components on some rotated plane from that of the original defined stress state. It is observed that the normal stress will have both a maximum and minimum stress where the circle cuts the sigma x’ axis. At this location the accompanying value of the shear stress will be zero. These extreme values of the normal stress are called the principal stress of the stress state. It is also observed that the shear stress attains its maximum value at the highest point on the circle and is just equal to the radius. On the plane orientation where the shear stress is maximum the normal stress is just equal to sigma avg or the average of the normal stresses in the original stress state.



13. Principal Stresses Using the graphical representation from the previous page it is now a simple matter to write the equations for the maximum and minimum normal stresses and maximum shear stress in terms of the xy axis stress components. The maximum normal stress is equal to the average normal stress plus the radius of the stress circle. This become the quantity sigma x plus sigma y over two plus the square root of the quantity sigma x minus sigma y over two squared plus tau xy squared. The minimum normal stress from the circle becomes the average stress minus the radius of the circle. Hence sigma min is made up of the two terms that make up sigma max with the exception that the terms are subtracted rather than added. This makes the two equations easy to remember. Finally the maximum shear stress is simply given by the radius of the circle, which is the square root term in the expressions for the two principal stresses.

14. Sample Problem 1 This sample problem is included to help you understand the application of the equations for the principal stresses and maximum shear stress for a given initial stress state. Solve this problem before proceeding further. Click on the solution button to check your answer and/or review the solution process then go on to the next page.




15. Mohr circle construction The circular representation of the general two dimensional stress equations can also be used to determine the stress state on any set of rotated axes relative to the axis system of the original stress state. This construction is well known as Mohr’s circle, named after its developer. The question to be answered is: given a set of two dimensional stresses relative to an xy axis system as shown on the left what will be the normal and shear stress on a plane perpendicular to an x’ axis rotated through an angle theta in the counter clockwise direction. The graphical construction will take place on the sigma’ - tau’ coordinate system shown on the right. A point whose coordinates are sigma x and tau xy is plotted together with a second point whose coordinates are sigma y and minus tau xy in this axis field. A straight line is then drawn connecting these two points. This line is the diameter of the stress circle. Where it crosses the sigma ‘ axis is the center of the

circle. A circle defined in this fashion is now drawn. Next, an angle two theta is measured counterclockwise from the radius defined by point sigma x, tau xy. At this angular position another diameter is drawn on the circle. The point of intersection of the diameter with the circle represents the coordinates sigma x’ and tau xy’ the two desired stress on the plane perpendicular to the x’ direction. The point where the far end of this second diameter intersects the circle represents the coordinates sigma y’ and minus tau xy’ acting on a plane perpendicular to the y’ direction. It is further observed that the angle two theta between the planes on which the principal stress act and the maximum shear stresses act is 90 degrees. This means the physical planes on which these two sets of stresses act are at 45 degrees with respect to each other.



16. Orientation – Principal Stresses It will be helpful to graphically visualize the orientation of the planes of principal stress to the planes of maximum shear stress discussed on the previous page. On the left is shown an element on which the principal stress sigma max and sigma min are assumed to be acting relative to axes that are horizontal and vertical. The planes on which the maximum shear stress act will be perpendicular to an axis system that is rotated 45 degrees counter clockwise as shown on the figure on the right. Satisfy yourself that the normal stresses on these planes of maximum shear stress are just equal to sigma average or sigma max plus sigma min divided by two. Also determine what the value of the maximum shear stress will be if sigma max is equal to minus sigma min. This is referred to as a state of pure shear such as occurs in the simply torsion of a shaft.

17. Sample problem -2 This sample problem will help you understand the application of Mohr’s circle. For the same initial stress state as in Sample Problem –1 determine the direction of the principal stress planes. That is, calculate the value and the direction of the orientation of the axis for the perpendicular plane on which sigma max acts. Solve this problem before proceeding further. Click on the solution button to check your answer and/or review the process of solution then go on to the next page.




18. General Stress State The discussions and developments so far have dealt with two dimensional stress states. Before leaving this topic it is worthwhile including some introductory material on three dimensional stress states. First consider the type and number of components associated with a general three-dimensional state of stress. This requires considering what stress components can exist on a plane perpendicular to the z-axis and the effect of these components on the two dimension stresses relative to an xy axis system. In general three stresses will exist on the z-axis plane. These will consist of a normal stress sigma z and two components of shear stress tau xz and tau yz. The two shear stresses require cross shears to exist on the planes perpendicular to the xy axes. By the principal of equal cross shears they will have the magnitude of the shears on the z face and the directions shown in the figure. Hence, in a general three-dimensional stress state there exist three normal components of stress, sigma x, sigma y and sigma z. There are also three shear stress components tau xy, tau xz and tau yz

19. Principal Planes – 3 D Orientation Just as in a two dimensional state of stress it is possible to show that there exist a set of rotated coordinates axes in three dimensions for which there are only normal stresses on a cubical element whose planes are perpendicular to this axis system. This is illustrated in the graphic where no shear stresses are shown on the rotated planes. Hence, the normal stresses measured relative to the rotated axis system are the principal stresses for the given three-dimensional stress state. The analysis that leads to the equations for the directions of these rotated axes and the magnitudes of the associated normal stress are beyond the scope of this study. In addition these specific results are seldom of use in real problems. However, there are some special instances in which the general knowledge of this behavior is important.



20. Principal Stress Equation in 3D Should an instance occur where the principal stresses for a general three-dimensional state of stress are required they can be determined by solving the cubic equation listed on this page. Note that the coefficients of this third order equation involve a variety of combinations of all the stress state components relative to the xyz axis system. Satisfy yourself before proceeding on that if all shear stresses in this equation are set equal to zero substituting either sigma x, sigma y or sigma z into the resulting simplification will indeed satisfy the equation.

21. Mohr Circles – 3D Stress State Each set of planes associated with the principal axes in a three dimensional stress state will possess a set of principal normal stress. One such set can be expressed as sigma 1 and sigma 2 relative to the x’y’ axis system. A second set would be sigma 2 and sigma 3 relative to the y’z’ axes and finally sigma 1 and sigma 3 relative to the x’z’ axes. Each of these sets of principle stresses can be represented graphically as a Mohr circle as shown on the right. Note that all three circles are tangent to one another as they pass through the points sigma 1, sigma 2 and sigma 3 on the horizontal axis. This is because each pair shares common stresses with each other. Also these stress values lie on the horizontal sigma axis since the shear stresses on planes of principal normal stress are all zero. The radius of each of the circles corresponds to the maximum shear stress associated with that particular primed axis system. Thus the maximum shear stress for the total state of stress is tau 3, the radius of the circle of largest diameter. Being able to determine this maximum shear stress is sometimes an important requirement and can be missed in dealing with two dimensional stress states. This is illustrated in Sample Problem – 3 on the next page.



22. Sample Problem – 3 In this sample problem you are asked to determine the maximum shear stress in the axial portion of a thin walled pressure vessel subjected to an internal pressure of 1300 psi with the dimensions given in the figure. The correct value can only be calculated when the three dimensional stress considerations of the previous slide are properly taken into account. Solve this problem before proceeding further. Click on the solution button to check your answer and/or to review the process of solution then go on to the next page.

(Solution on Pages 48,49 and 50)

23. Review Exercise In this review exercise select the answer to the question by clicking on the appropriate button in the group of choices below the question. An immediate feedback will be provided indicating whether the selection was correct or what material should be reviewed to obtain the right result. You can also click on the hot word in the question to pop up the relevant page from the chapter. To remove the feedback click the mouse and proceed to the next question. When all questions have been answered correctly proceed to the next page.



24. Off Line Exercise This off line exercise requires the application of the two dimensional stress analysis theory and equations developed in this chapter. If you can correctly answer all four questions for the four example stress states listed on the next page you will have a good understanding of two dimensional stress analysis. This ability is fundamental to the topics to be taken up in subsequent chapters.

25. Off Line Exercise 9 (continued) When finished with the information on this page click on the exit button to leave the chapter or the index button to select any specific page to be revisited.




Chapter 2 Two Dimensional Stress Analysis

Problem Solutions

Screen Titles

Normal Stress - Y’ direction Maximum Normal Stress Minimum Normal Stress / Shear Stress Generic Principal Axis Formula Principal Axis Orientation Axial Stress Hoop Stress Stress Calculations Complete Stress State Maximum Shear Stress



1. Normal Stress - y’ direction The element shown on the left is chosen to determine the normal stress in the y’ direction since it contains planes perpendicular to the x, y and y’ axes. Again force equilibrium is applied in the y’ direction taking into account the contributions of all the stress components acting on the element. First recognize that if the length of the inclined edge of the element is taken to be unity then the length of the edge parallel to the x axis is cos theta and the edge parallel to the y axis is sin theta. It is also assumed that the element is of unit depth in the z direction. To apply equilibrium all stress components must be multiplied by the area over which they act before being multiplied by the appropriate trigonometric function of theta to give their contibution in the y’ direction. The force contribution of sigma y’ is simply the stress times the area one time one. On the x face sigma x has a negative contribution equal to the stress times the area sin theta times one times the sin of theta. The positive force contribution of tau xy on the x face is the stress times the area times cos theta. On the y face the negative force contribution of sigma y in the y’ direction is the stress times the area cos theta time one times cos theta. Tau xy on the y face contributes a positive force contribution in the y’ direction equal to the stress times the area times sin theta. Carrying through the multiplication, combining the tau xy terms and solving for sigma y’ gives the equation at the bottom of the page.

2. Maximum Normal Stress Begin with the equation for the maximum normal stress, sigma 1, that is the sum of sigma average, the quantity sigma x plus sigma y divided by two, plus the radius of the stress circle, the square root of the quantity sigma x minus sigma y over 2 squared plus tau xy squared. Next substitute the values of the given stress state for the parameters into the equation. These are sigma x equal to 11,000 psi, sigma y equal to zero and tau xy equal to 6,900 psi. Carrying out the indicated mathematical operations gives a final answer for sigma 1, the maximum normal stress, of 14,320 psi. tension.



3. Min Normal Stress/ Max Shear Stress Now begin with the equation for sigma 2, which is the difference between the value of the sigma average and the radius of the stress circle. Substituting the value of the stress components sigma x, sigma y and tau xy, into this equation gives the difference of the two numbers used to determine sigma 1. This results in sigma 2 having a final value of - 3,320 psi compression. The value of the maximum shear stress is just equal to the radius of the stress circle, which is the square root term in the sigma 1 and sigma 2 equations. The result is a maximum shear stress of 8,820 psi.

4. Generic Principal Axis Formula To determine the direction of the principal axes attention is directed to the generic Mohr circle shown on the left. The highlighted red triangle can be used to graphically define the tangent of two theta that represents the rotation of the initial stress state circle diameter defined by sigma x and tau xy to the horizontal sigma axis that defines the principal stress orientation. From the geometry of the figure it is seen that the height of the triangle is simply tau xy while the base is equal to sigma x minus sigma average, that is sigma x plus sigma y over 2, giving the result sigma x minus sigma y over 2. Thus the tangent of two theta that defines the orientation of the principal axes becomes 2 tau xy divided by the quantity sigma x minus sigma y.



5. Principal Axis Orientation The values of the given stress state are now substituted into the formula from the previous page for the tangent of two theta. This gives a value of two theta equal to 51.4 degrees. Recall that this is double the actual physical angle of rotation. Hence, the orientation of the principal axes is 25.7 degrees counter clockwise from the x-axis since the physical rotation is opposite from that on the circle. The figure on the right illustrates graphically the orientation of the principal axes relative to the xy axes system.

6. Axial Stress If the cylinder is cut by a plane perpendicular to its central axis then the resulting free body must exhibit an axial stress distribution on the cut plane to balance the force of the pressure acting on the interior surface of the spherical end cap. Thus the pressure load must be equal to the axial stress load resultant to satisfy equilibrium. The pressure load is just the pressure multiplied by the projected area of the surface it is acting on. In this case the projected area is just Pi times R squared. The axial stress resultant is equal to the stress sigma a times the area it acts over which is the circumference 2 Pi times R times t. Setting the pressure load equal to the stress reaction force and solving for sigma a gives simply PR over 2t. As the length L of the cylinder of the free body is shrunk to zero it is seen that the axial stress on the spherical end cap is also the same as the axial stress in the cylinder.



7. Hoop Stress Next consider a free body portion of the cylinder that is one unit in length along the axis of the cylinder and cut by a horizontal plane that contains the axis of the cylinder. In the figure shown the radial effect of the pressure acting in the interior of this free body must be balanced by hoop stress reactions acting vertically down on the cut surfaces of the thickness. Again the effective pressure force acting upward must equal the hoop stress reaction forces acting down to satisfy equilibrium. The effective pressure load is the pressure times the effective area which is just 2R times the unit depth of the free body. The stress reaction force on one side is the hoop stress times the area it acts on which is t times the unit depth. Hence the equilibrium condition becomes P times 2R is equal to 2 sigma h times t. Solving for sigma h gives PR over t.

8. Stress Calculations Employing the formulas for the axial stress, sigma a, and the hoop stress, sigma h, developed on the previous two pages the values of these stresses can now be calculated using the numerical parameters given in this problem. Note that both stresses are tensile and the hoop stress is twice the axial stress. Carrying out these calculations gives an axial stress of 31,200 psi and a hoop stress of 62,400 psi.



9. Complete Stress State Now consider a cubical element at the inside surface of the cylinder with its edges oriented in the axial, circumferential and radial directions as shown. Acting on the axial face is the normal tensile stress, sigma a. Acting on the face perpendicular to the circumferential direction is the tensile hoop stress, sigma h. Finally, acting on the face perpendicular to the radial direction on the inside surface of the tank is the internal pressure P which can be represented by a compressive normal stress, sigma p, equal to -1300 psi. Since there are no shear stresses acting on any of the cubical element faces the three normal stresses represent a three-dimensional principal stress state.

10. Maximum Shear Stress The maximum shear stress can now be determined by drawing Mohr circles for the principal stresses from the previous page. Its numeric value will be equal to the radius of the largest circle. The graphic on the left illustrates the three Mohr circles for this stress state. It is easily seen that the largest circle is the one corresponding to the axis system associated with the circumferential and radial directions. Thus the maximum shear stress is given by the hoop stress minus the pressure stress divided by two. For this problem the numerical value is 31,850 psi. It should be noted that if the problem had been treated as a two-dimensional stress state involving only the hoop and axial stresses an incorrect


_____________________________________________________________________________________Two Dimensional Strains - 53 - C.F. Zorowski 2002

Chapter 3 Two Dimensional Strain Analysis

And Hooke’s Law

Screen Titles

Rectilinear Strain Components Rotated Strain Component - #1 Rotated Strain Component - #2 Rotated Strain Component - #3 Total Strain – Direction A Total Strain – Direction B Rotated Shear Strain Component - #1 Rotated Shear Strain Component - #2 Rotated Shear Strain Component - #3 Total Rotated Shear Strain Total Rotated Strain Equations Graphic Interpretation Principal Strain Equations Mohr’s Circle for Strain 45o Strain Rosette Hooke’s Law Poisson’s Ratio Strain from 2D State Pure Shear Stress and Strain Relation between E and G Review Exercise Off Line Exercises



1. Title page Chapter three covers the definition and analysis of a two dimensional rectilinear state of strain involving both normal and shear components. The topics discussed include development of the equations defining strain components with respect to a rotated axis system, graphical interpretation of rotated axis system strain equations, principal strain and maximum shear strain components, Mohr’s circle construction for two dimensional strain states, Hooke’s law, Poisson’s ratio, two dimensional stress-strain equations, and the relation between elastic and shear moduli. Several sample problems demonstrating the application of the theory are also included.

2. Page Index – Chapter 2 Listed on this page are all the individual pages in Chapter 3 with the exception of the sample problems. Each title is hyperlinked to its specific page and can be accessed by clicking on the title. It is suggested that the reader first proceed through all pages sequentially. Clicking on the text button at the bottom of the page provides a pop up window with the text for that page. The text page is closed by clicking on the x in the top right corner of the frame. Clicking on the index button returns the presentation to the chapter page index.



3. Rectilinear Strain Components The definitions introduced for normal and shear strain in Chapter 1 are now used to define a two-dimensional state of strain in a body. Consider the deformation of an incremental rectangular element of length lx and height ly. The extension of the edge lx is designated delta x while the extension of the edge ly is designated delta y. These extensions divided by the length of the edges are designated epsilon x and epsilon y, the normal strains in the x any y directions. Extensional strains are considered positive. In addition to these incremental extensions the element can also undergo a shear deformation that changes the original right angle at the bottom left corner of the rectangular element. The change in this angle is designated the shear strain, gamma xy. If the angle decreases the shear strain is taken as positive. The two normal strains together with the shear strain define the two-dimensional strain state in the body.

4. Rotated Strain Component - #1 Given a specific two-dimensional strain state relative to an xy axis system it is desirable to determine the state of strain with respect to a set of axes rotated through some angle theta with respect to the original axis system. Each component of the original strain state will contribute to the strains relative to the rotated axis system. Consider the contribution of epsilon x to epsilon a measured in the a direction in the figure. A rectangular element dx dy is chosen such that its diagonal is in the direction a. The element is then given an incremental positive strain epsilon x resulting in the elongation of dx by epsilon x dx. This gives rise to an extension of the diagonal ds in the direction a. The magnitude of this extension of ds due to the epsilon x extension is determined using the small right triangle in the upper right corner of the extended element dx dy. The hypotenuse of the triangle is epsilon x dx so the extension of the diagonal ds becomes epsilon x dx times cosine theta. The strain component in the a direction due to epsilon x then becomes epsilon x dx cosine theta divided by ds. However, dx over ds is just cosine theta so that the final expression for epsilon a due to epsilon x is epsilon x cosine squared theta.



5. Rotated Strain Component - #2 The dx dy element from the previous page is now subjected to an incremental positive strain in the y direction, epsilon y. This extends the dy dimension by epsilon y dy which again increases the length of the diagonal ds that still lies in the a direction. Again the small right triangle in the upper right hand corner is used to determine the amount by which ds is extended. In this instance the hypotenuse of the triangle is epsilon y dy so that the extension of a is given by epsilon y dy times sine theta. The strain component in the direction of a due to epsilon y can then be written as epsilon y dy sine theta divided by ds. But, dy divided by ds is just equal to sine theta. The final contribution to epsilon a from epsilon y then becomes epsilon y sine squared theta.

Rotated Strain Component - #3 The third contribution to epsilon a comes from gamma xy, the shear strain relative to the xy axes. The development begins again with the element dx dy with diagonal ds in the a direction. An incremental positive shear strain gamma xy is applied to create a rotation of the dx edge while dy remains as it was originally. For a small angle of rotation the hypotenuse of the right triangle in the upper right can be expressed as gamma x dx. Thus the extension of ds becomes gamma xy dx times sine theta. The strain component in the direction of a due to gamma xy can then be written as gamma xy dx times sine theta divided by ds. Since dx over ds is just cosine theta the final expression for the contribution of gamma xy to epsilon a becomes gamma xy times sine theta cosine theta.



7. Total Rotated Strain (direction a) The three xy strain state contributions from the previous pages are now added together to give the normal stain epsilon a in the a direction. The equation for epsilon a becomes epsilon x cosine squared theta plus epsilon y sine squared theta plus gamma xy sine theta cosine theta. It is convenient to rewrite this equation into double angle form using the trig identities indicated. This gives the final expression epsilon a equal to the quantity epsilon x plus epsilon y divided by 2 plus the quantity epsilon x minus epsilon y divided by two times cos two theta plus the quantity gamma xy divided by 2 times sine two theta. Does this look familiar? It should, recall the development of a similar equation for a two-dimensional state of stress.

8. Total Rotated Strain (direction b) The magnitude of the normal strain in direction b, the second strain component for the rotated rectilinear axis system shown in the figure, is now desired. This can be obtained from the epsilon a equation by substituting theta equal to the quantity theta plus pi over 2 into the equation for epsilon a. Do this as an exercise to show that the equation for epsilon b at the bottom of the page is obtained. Click on the solution button to check the procedure for getting this result or go on to the next page.




9. Rotated Shear Stain Component - #1 A similar development will now be undertaken to obtain an expression for the shear strain gamma ab relative to the ab axis system in terms of the strain state specified with respect to the xy axis system. For this purpose an elemental rectangle of dx’ dy is added to the element dx dy. The dimension dx’ is chosen such that its diagonal ds’ is perpendicular to ds the diagonal of the element dx dy. The combined element is subjected to a incremental normal strain epsilon x. The right side of the rectangular element extends epsilon x dx while the left side extends epsilon x dx’. This results in extensions of both ds and ds’ giving rise to two parts to the shear strain component gamma ab due to epsilon x. These components are represented by the change in angle of ds and ds’. The change in the angle of ds is the distance A A’ divided by ds. This is positive since the rotation tend to make the angle between ds and ds’ smaller. The change in angle contribution due to the rotation of ds’ is the distance B B’ divided by ds’. This contribution is negative as the change tends to increase the angle between ds and ds’. From the triangle in the upper right the distance A A’ is just epsilon x dx cosine theta while the distance B B’ from the right triangle in the upper left is epsilon x dx’ sine theta. Since dx over ds is cosine theta and dx’ over ds’ is sine theta the final expression for gamma ab due to epsilon x is gamma xy times the quantity cosine squared theta minus sine squared theta.

10. Rotated Shear Strain Component - #2 To develop the contribution to gamma ab from a normal strain in the y direction an incremental strain epsilon y is applied to the elemental rectangle used on the previous page. This extends the upper edge by an amount epsilon y dy which again results in small rotations of the diagonal ds and ds’. The change in the angle of ds is again positive and is given by the distance A A’ divided by ds. The change in the angle of ds’ is also positive and can be expressed as B B' divided by ds’. The distance A A’ from the small upper right triangle is epsilon y dy cosine theta while B B’ from the upper left right triangle is epsilon y dy sine theta. With dy over ds equal to sine theta and dy over ds’ equal to cos theta the final expression for gamma ab due to epsilon y is given by 2 epsilon y sine theta cosine theta.



11. Rotated Shear Strain Component - #3 An incremental shear strain gamma xy is now applied to the rectangular element to develop its contribution to gamma ab. The lower edge of the rectangle is rotated about the origin by the angle gamma xy while the vertical dimension remains parallel to the y-axis. This gives rise to a positive change in the angle of ds and a negative change in the angle of ds’. The change in angle of ds is given by plus A A’ divided by ds while the angle change in ds’ can be written as B B’ divided by ds’. From the upper right triangle A A’ is equal to gamma xy dx cosine theta and B B’ from the upper left triangle is given by gamma xy dx sine theta. Again recognizing that dx over ds is cosine theta and dx over ds’ is sine theta the final expression for the gamma ab contribution from gamma xy is gamma xy times the quantity cosine squared theta minus sine squared theta.

12. Total Rotated Shear Strain The three xy strain state contributions from the previous pages are now added together to give the total shear stain gamma ab relative to the rotated ab axis system. The equation for gamma ab becomes two times the quantity epsilon y minus epsilon x times sine theta cosine theta plus gamma xy times the quantity cosine squared theta minus sine squared theta. It is again convenient to rewrite this equation into double angle form using the trig identities indicated. This gives the final expression gamma ab equal to the quantity epsilon y minus epsilon x times sine two theta Plus gamma xy cos two. This should again look familiar recalling a similar looking equation for shear stress relative to a rotated axis system in the study of two dimensional stresses.



13. Total Rotated Strain Equations Listed below are the three equations for determining all components of a two dimensional state of strain measured relative to an ab rotated axis system in terms of the specified two dimensional components of strain relative to some xy axis system and the angle of rotation to the new ab directions. These three equations are in effect exactly the same as the equations for a two dimensional stress state with respect to a rotated axis system except for one small difference. What is that difference? Check on the compare button to see if your answer is correct.

(Comparison on Page 71)

14. Graphic Interpretation Since the rotated axis strain equations are effectively the same as the rotated axis stress equations they can be given a similar graphical interpretation and representation. By designating epsilon avg as the quantity epsilon x plus epsilon y over 2 and eliminating the angle theta between two equations on the previous page an equation is obtained which describes a circle in the epsilon and gamma over two axis system whose center is at epsilon average on the epsilon axis as shown on the left. As indicated on the right the radius of the circle is given by R which is equal to the quantity epsilon x minus epsilon y over 2 quantity squared plus gamma xy over two squared. From the circle it is seen that at some rotated axis position the normal strains will have maximum and minimum values together with a shear strain of zero. There also exist a maximum value of shear strain relative to a set of axes for which the normal strains are equal.



15. Principal Strain Equations From the circle representation on the previous page equations can now be written for the maximum and minimum values of normal strain and the maximum shear strain for any given two-dimensional strain state. Note that the maximum and minimum normal strains only differ by the sign of the second term. Also the value of the maximum shear strain is just equal to the second term in the max and min normal strain equations.

16. Mohr Circle for Strain A Mohr circle for a two dimensional strain state can be constructed and used in a similar manner to that presented for a two-dimensional stress state in chapter 2. The construction is begun by establishing an epsilon and gamma over two coordinate system. In this coordinate field two sets of points are plotted. The first is epsilon x and gamma xy over two. The second set is epsilon y and minus gamma xy over two. A line connecting these two sets of coordinates is the diameter of the circle. All states of strain relative to rotated axes now lie as coordinates at the end of some diameter of this circle. To establish this diameter for the rotated axis system x’y’ an angle of two theta is measured from the original diameter in the direction opposite to the rotation theta for the real axes. Where this new diameter intersects the circle are the coordinates epsilon x’ and gamma x’y’ over two at one end and epsilon y’, minus gamma x’y’ over two at the other end.



17. Sample Problem - 1 In this sample problem the values of a two-dimensional strain state are given for an xy axis system. In part a determine the values of epsilon a, epsilon b and gamma ab for an axis system rotated 60 degrees counterclockwise from the x axis. In part b determine the values of the principal strains and maximum shear strain for the given two-dimensional state of strain. After working out results for both parts check your process of solution and results by clicking on the solution button.


18. Sample Problem – 2 For the same given strain state as in sample problem 1 determine for part a the orientation of the axes of principal strain relative to the xy axis system. In part b determine the orientation of the axes of maximum shear strain relative to the xy axis system. Click on the solution button to check your results and solution process.




19. 45 Degree Strain Rosette If the normal strains are known in three directions, which are separated by angles of 45 degrees this, is sufficient information to establish the Mohr strain circle. This is useful when a 45-degree strain gauge is used to measure stain in an actual loaded member. The construction is begun by setting up an epsilon, gamma over two coordinate system. Epsilon 1, epsilon 2 and epsilon 3 are marked off along the epsilon axis. The center of the circle is then located at epsilon 1 plus epsilon 3 over two. The distance between the center and the epsilon two value which is the quantity epsilon 1 plus epsilon 3 minus two epsilon 2 all over 2 is determined and measured vertically up at the location of epsilon 1. This vertical distance corresponds to the value of gamma 13 over 2. Hence this defines one end of the diameter of the circle. The other end is of course epsilon 3, minus gamma 13 over 2. With this diameter the circle can be drawn. A line drawn from where epsilon 2 intersects the circle and it center represents on half of the diameter for the axis system defined by the direction of epsilon 2 and an axis perpendicular to it. Note that the construction results in angles of two theta of 90 degrees between the diameter for axes 13 and the direction of epsilon three, which agrees with the fact that on the rosette axes 1,2 and 3 are separated by 45 degrees.

20. Hooke’s Law The page on Hooke’s law from Chapter 1 is repeated here to introduce a brief overview of how two-dimensional strain and stress states are related through material properties. This law of elasticity simply states that for most materials stress and strain are linearly related for small values of strain. The constant of proportionality for normal stress and strain is the modulus of elasticity, E, sometimes called Young’s modulus. For shear stress and strain the constant of proportionality is designated G and is called the shear modulus. To relate two dimensional stress and strain states a third material property covered on the next page is required.



21. Poisson’s Ratio Consider the physical deformation behavior of an incremental element dx dy of a real material subjected to an extensional strain epsilon x. The x dimension of the element elongates by an amount epsilon x dx. At the same time the y dimension of the element contracts by an amount that is proportional to the x direction elongation and is expressed mathematically by the expression nu epsilon x dx. The parameter nu is a constant of proportionality called Poisson’s ratio after a French mathematician and is a material property like the two moduli E and G. It is defined as the ratio of unit lateral contraction to unit axial elongation. Thus epsilon y for this loading and deformation state is just nu times epsilon x. The value of nu for real materials always lies between zero and one half.

22. Strains from 2 D Stress State Now consider the same incremental element subjected to both sigma x and sigma y stresses at the same time. The epsilon x strain will now consists of two terms. The first is the direct elongation due to the sigma x stress or simply sigma x divided by E. The second term will be a contraction in the x direction due the Poisson’s ratio effect of the applied sigma y stress. It is expressed as minus nu times sigma y divided by E. Added together they give the total strain in the x direction as indicated. In a similar fashion the strain epsilon y in the y direction is given by the same equation with sigma x and sigma y reversed. Now write out the three equations for epsilon x, y and z for a cubical element subjected to the three normal stresses sigma x, y and z. Also for the two dimensional state of strain develop equations for sigma x and sigma y in terms of epsilon x, epsilon y and nu. Check your answers by clicking on the solution button.




23. Sample Problem – 3 The purpose of this problem is to demonstrate why the value of nu cannot exceed one half for real materials. The small strain assumption means that squared and cubed strain terms can be neglected relative to first order strain terms. Click on the solution button to check your result for the change in volume.


24. Pure Shear Stress and Strain States The next two pages will be used to develop a relationship between E, G and nu based on Hook’s law of deformation behavior. This development is begun by considering a state of pure shear stress, Tau, in two dimensions. The Mohr circle for this stress state is shown directly below the element on which the shear stresses act. It is observed that the principle stress values, sigma 1 and sigma 2, are simply plus and minus Tau. From the Mohr strain circle below the element on which the principal stresses act it is observed the value of maximum shear strain is just twice epsilon 1 or minus epsilon 2.



25. Relation Between E and G Epsilon 1 is now written in terms of sigma 1 and sigma 2 both of which can then be replaced by Tau with consideration of sign. A second equation for epsilon 1 is written in terms of gamma max over two, which can be replaced by Tau over 2G. The two equations for epsilon 1 are now set equal to one another and the common Tau term is divided out. This is then rearranged to give the expression G equal to E divided by 2 times the quantity one minus two nu. Hence G can be calculated if E and Nu are known. Also since nu is limited between zero and one half this in turn means that G must lie between one third and one half E. Note how this works for steel that has an E of 30 million psi and a G of 12 million psi.

26. Review Exercises In this exercise the items in the list on the left are to be matched with the most appropriate phrase on the right. Place the cursor over an item on the left and hold down the left button. A pencil will appear that can be dragged to one of the green dots on the right. If the right choice is made the arrow will remain. If the selection is incorrect the arrow will disappear. After the exercise is completed proceed to the next page.




27. Off Line Exercises This off line exercise is a practical problem that requires using materials covered both in Chapter 3 and Chapter 2. In addition to determining the specific numerical values requested it is suggested that the appropriate Mohr circles be drawn to check your calculated values. If you can successfully complete this problem you will have a good understanding of two-dimensional stress and strain states and their relation to one another. Good luck.



Chapter 3 Two Dimensional Strain Analysis

Problem Solutions

Screen Titles

Total Rotated Strain (direction b) Rotated Stress Equations Rotated Normal Strains Rotated Shear Strain Principal Strains Maximum Shear Strain Generic Principal Axis Direction Principal Direction Calculation 3D Strain Equations Stress in terms of strain Volume Calculations Simplifying Results



1.Total Rotated Strain (direction b Begin with the equation for epsilon a with theta plus Pi over 2 substituted for theta. Now cosine of two times the quantity theta plus pi over two is equal to minus cos two theta and sine of two times the quantity theta plus pi over 2 is equal to minus sine two theta. Thus the equation for epsilon b contains the same three terms as the equation for epsilon a with the exception that the second and third terms are now negative. Click on the return button to go back to the next page in chapter 3.

2. Rotated Stress Equations The three listed rotated axis stress equations will look exactly like the rotated axis strain equations if the normal stresses are replaced by normal strains and the shear stresses are replaced by shear strains divided by two. In other words sigma x’, y’ become epsilon a’, b’ and sigma x, y become epsilon x, y together with tau ab’ replaced by gamma xy’ over two and tau xy replaced by gamma xy over two. Click on the return button to go back to the next page in chapter 3.



3. Rotated Normal Strains Substituting the given values of epsilon x, epsilon y and gamma xy into the equation for epsilon a together with two theta equal to a positive 120 degrees and carrying out the indicated mathematical manipulations gives a final value for epsilon a of 386 micro inches per inch. Epsilon b is similarly determined by simply applying the appropriate sign changes for the last two terms in the epsilon a equation. This gives a final value of 254 micro inches per inch.

4. Rotated Shear Strain Again the given strain state values along with two theta equal to 120 degrees are substituted into the equation for gamma ab and the required mathematical manipulations are carried out. This gives a final value for gamma ab of minus 492 micro inches per inch. Shown to the right, approximately to scale, is the Mohr Strain circle with the original and rotated axis strain values. Note that the negative rotation on the circle of 120 degree results in a negative value of gamma ab associated with epsilon a as specified by the numerical determination. Click on the return button to go back to the next page in chapter 3.



5. Principal Strains The given xy strain state from the previous problem are now substituted into the equation for the maximum normal strain. Carrying out the indicated mathematical manipulations results in epsilon max equal to 569 micro inches per inch. Since the minimum principal strain only changes by the sign of the second term in epsilon max the numerical result simply becomes 71 micro inches per inch.

6. Maximum Shear Strain The maximum shear strain is determined using the equation for gamma max over two. This is just the last term in the principal stain equations, which gives a final value of 498 micro inches per inch. Again the Mohr circle for strain from the previous problem is shown on the right with the principal strains and maximum shear strain indicated. Note that they also seem to be correct values as indicated by the approximate scale of the circle. Now click on the return button to go back to the next page in chapter 3.



7. Generic Principal Axis Direction To determine the direction of the principal axes attention is directed to the generic Mohr circle for strain shown on the left. The highlighted red triangle can be used to graphically define the tangent of two theta that represents the rotation of the initial strain state circle diameter defined by epsilon x and gamma xy over two to the horizontal epsilon axis that defines the principal strain orientation. From the geometry of the figure it is seen that the height of the triangle is simply gamma xy over two while the base is equal to epsilon x minus epsilon a average, that is epsilon x plus epsilon y over 2, giving the result epsilon x minus epsilon y over 2. Thus the tangent of two theta that defines the orientation of the principal strain axes becomes gamma xy divided by the quantity epsilon x minus epsilon y.

8. Principal Direction Calculation The appropriate xy axis strain state values are substituted into the equation for tangent two theta. Since tangent of two theta is one then two theta is 45 degrees which makes theta just 22.5 degrees. Since two theta is clockwise on the Mohr circle for strain in the previous problem the rotation of the principal axes will be counterclockwise in the real physical system shown as indicated in the figure on the right. The axis relative to which the maximum shear strain acts is shown in red at 45 degrees counterclockwise from the epsilon max direction. Click the return button to go to the next page in chapter 3.



9. 3D Strain Equations The Poisson’s ratio effect due to the sigma z stress must be added to the two-dimensional strain equation for epsilon x. The added term is given by minus nu times sigma z divided by E assuming that sigma z is positive. This same term must also be added to the epsilon y strain equation in two dimensions to include the Poisson’s ratio effect of the sigma z stress. Using the form of the two expanded strain equations for epsilon x and epsilon y a similar equation can be written for the strain in the z direction. It too will consist of three terms. The first is the direct effect of the sigma z stress while the second two terms are the Poisson’s ratio effect of the sigma x and sigma y stresses.

10. Stress in terms of strain An equations for the sigma y stress in terms of normal strains in two dimension is obtained by eliminating the sigma x stress between the two strain equations. This is easily accomplished by multiplying the epsilon x equation by nu and adding it to the equation for epsilon y. The result is an equation that only contains the stress sigma y and the two normal strain equations. Solving for sigma y gives E over the quantity one minus nu squared multiplied by the quantity epsilon y plus nu times epsilon x. A similar equation can be developed for sigma x in terms of epsilon x and epsilon y. Click the return button to go to the next page in chapter 3.



11. Volume Calculations The volume of the unloaded cylindrical rod is simply pi R squared times the length, L. Applying an axial normal stress with a corresponding axial strain of epsilon to the rod results in a new length expressed as L times the quantity one plus epsilon. At the same time the radius of the deformed rod becomes R times the quantity one minus nu times epsilon due the Poisson’s ratio contraction from the effect of the axial stress. Thus the deformed volume becomes Pi R squared times the quantity one minus nu epsilon squared times L times the quantity one plus epsilon. The change in volume defined by the deformed volume in the original volume can then be expressed by the equation at the bottom of the page.

12. Simplifying Results The bracketed terms on the right side of the delta volume divided by Pi R squared L equation are now multiplied out. Canceling out the ones and neglecting all higher order epsilon terms due to the small value of the strain gives the final result that the change in volume term is just equal to epsilon times the quantity one minus two nu. If nu become greater than one half the change in volume of the rod will become negative which is not possible under the action of an axial elongation. Hence the value of Poisson’s ratio is limited physically to one half.


_____________________________________________________________________________________ Static Material Properties - 77 - C.F. Zo0rowski 2002

Chapter 4 Static Material Properties

Screen Titles

Common Mechanical Tests Tensile Test Elastic Behavior Moduli & Poisson’s Ratio Plastic Behavior Nominal Stress Strain Diagram Necking Behavior True Strain True Stress True Stress Strain Diagram Offset Yield Stress Typical Tensile Properties Plain Carbon Steel Hardness Tests Brinell Hardness Test Rockwell Hardness Test Hardness Scale Comparison Review Exercise Off Line Exercises


_____________________________________________________________________________________ Static Material Properties - 78 - C.F. Zo0rowski 2002


Static Material Properties - - C.F. Zorowski 2002 79

1. Title page Chapter 4 deals with the subject of the static tensile strength and ductility properties of both ferrous and non-ferrous metals. The topics covered include: tensile testing, elastic and plastic behavior, comparative modulus properties, yield, ultimate and fracture stress, work hardening, elongation and area reduction at fracture, ductility, necking phenomena, true stress and strain, off set yield stress, typical metal tensile properties, effect of carbon content on steel tensile properties, hardness testing, comparison of Brinell and Rockwell hardness scales and their relationship to tensile strength. Sample exercises that demonstrate the application of the subject content are also included.

2. Page Index Listed on this page are all the individual pages in Chapter 4 with the exception of the sample exercises. Each title is hyperlinked to its specific page and can be accessed by clicking on the title. It is suggested that the reader first proceed through all pages sequentially. Clicking on the text button at the bottom of the page provides a pop up window with the text for that page. The text page is closed by clicking on the x in the top right corner of the frame. Clicking on the index button returns the presentation to the page index of chapter 4



3. Common Mechanical Tests. The three most commonly used mechanical tests for determining the static strength properties of both metals and non-metals are the tensile test, the compression test and hardness testing. Standard strength and ductility characteristics used particularly in the design of components that undergo elongation and shear loading come principally from tensile testing in which cylindrical test specimens are elongated to fracture. Compression testing is used principally for the testing of brittle materials like cast iron, concrete and stone or other such materials that are customarily use to carry compressive rather than tensile loading. This topic together with compressive properties will not be covered in this chapter. Hardness testing is a means of determining strength characteristics non-destructively by measuring the material surface indentation properties produced by standard indenters under specific magnitudes and forms of loading.

4. Tensile Test In a standard tensile test a cylindrical specimen of specified standard length is elongated slowly until fracture of the specimen takes place. The elongation of the specimen places the material under an axial tensile load P. Dividing this load by the original cross sectional area of the specimen results in an axial tensile stress sigma. This is referred to as the engineering stress that the specimen experiences under the elongation delta L. The state of elongation is defined in terms of the ratio of delta L to L which is defined as the engineering strain, epsilon. As epsilon increases there will be a corresponding increase in sigma.



5. Elastic Behavior The generic behavior of a ductile material subjected to a tensile test will now be discussed in terms the various characteristics of its elongation and load behavior. As elongation proceeds producing an increase in the applied value of strain it is observed that for small values of strain the resulting internal tensile stress will increase in direct proportion to the strain. These observations and the measurement of the magnitudes of both the engineering stress and strain are made on special machines designed specifically for conducting this type of test. This linear behavior of the material as depicted on the graphic follows Hooke’s law and is referred to as elastic behavior of the material. The slope of the linear portion of the stress strain curve is the modulus of elasticity, E, of the material. This property is also referred to as Young’s modulus or simply the modulus of the material. Its units are the same as stress since the units of strain are dimensionless.

6. Common Moduli & Poisson’s Ratio Listed in the table are actual measured average values of the elastic modulus E for a variety of ferrous and non-ferrous materials along with corresponding shear modulus G and Poisson’s ratio values. Note that the order of magnitude of the modulus E is ten to the sixth pounds per inch. This is indicative that the strains associated with elastic behavior are indeed very small. As an example a stress of thirty thousand psi in steel that behaves elastically corresponds to an associated strain of only one times ten to the minus three inches per inch or a thousandth of an inch per inch of length. Also note that the ratio of E for steel compared to lead is only a factor of about six greater. The corresponding values of the shear modulus G are also of magnitude of ten to the sixth pounds per inch. This is not surprising in that in chapter three it was shown theoretically that G should be between one third and one half E. Also note that all of the Poisson’s ratio values lie between the theoretical values of zero and one half and seem to cluster around 0.3 to 0.4.



7. Exercise - 1 The purpose of this exercise is to check the elastic property relationship developed theoretically in chapter 2. Using the relationship from Chapter 2 between the shear modulus, G, elastic modulus E and Poisson’s ratio apply the E and Nu values from the previous page to calculate G for copper and carbon steel. How do they compare with the average measured values given in the table? When you have completed your analysis click on the solution button to check your results or go on to the next page.


8. Plastic Behavior The response of the material specimen to the tension test now enters a second phase of behavior. At some value of applied strain the material will deviate from the straight-line elastic behavior and the strain will begin to increase more rapidly than the induced stress. The stress level at which this deviation from linear behavior takes place is designated the yield stress in tension, sigma y. This second phase of response beyond the yield stress is called plastic behavior indicating some form of permanent unrecoverable deformation is taking place. If at some stress level greater than the yield stress the load is removed from the specimen the stress strain curve will unload along a line parallel to the initial elastic behavior until the load and stress goes to zero. At this point there is an unrecoverable residual strain in the specimen producing a permanent set which can be observed as an increase in length over the original length. If the specimen is reloaded the stress will increase along the unloading curve until it reaches the stress level from which it was unloaded. It will then continue to behave plastically again under additional elongation. Thus the yield stress of the material has been increased beyond what it was originally. This phenomenon is referred to as work hardening and represents a mechanism for increasing the yield stress of the material. This is what occurs when wire is drawn down to a smaller diameter through a die.



9. Nominal Stress Strain Diagram. Plastic behavior continues in the specimen with increased elongation until a third phase of extension behavior is observed. The induced stress continues to increase but with significant increases in the applied strain until the stress reaches a maximum value referred to as the ultimate stress or the tensile strength of the material, sigma u. At this point the value of the load and engineering stress begins to decrease and continues to do so until sufficient elongation takes place to produce physical fracture of the specimen into two separate pieces. In this third phase the deformation behavior of the specimen changes radically as the diameter of the specimen begins to suddenly reduce at a localized position along the length of the specimen. This phenomenon is referred to as necking and is very observable in a material like low carbon steel. The extent of the total strain at which fracture takes place is a measure of the ductility of the material. Ductility can be thought of as the degree of plastic behavior the material can undergo before failure takes place. The greater the fracture strain the more ductile the material is or the greater the amount of plastic behavior and deformation it can withstand before fracture occurs. is gamma xy times the quantity cosine squared theta minus sine squared theta.

10. Necking Behavior As the tensile specimen is stretched first through the elastic and the initial plastic range of behavior it elongates uniformly along its length while its diameter and subsequently the cross sectional area also decrease uniformly along its length as represented by the graphic on the left. When the ultimate stress is reached the deformation behavior undergoes a radical change. The rate at which work is being done on the material can no longer be distributed uniformly at a fast enough rate throughout the entire specimen. It becomes locally concentrated which results in the diameter of the bar decreasing more rapidly at that location than anywhere else along its length. This can be physically observed as a local necking down of the bar as indicated in the middle graphic. Further extension of the specimen results in a decrease in its load carrying capacity with a continuing decrease in the necked down diameter until fracture takes place and the specimen separates into two parts as depicted in the third graphic at the fracture stress. The behavior of the material beyond the ultimate stress level is really a state of instability. Materials that exhibit high degrees of ductility are more apt to undergo this necking process. It is very apparent in low carbon steels.



11. Exercise - 2 The purpose of this exercise is to apply to a physical problem the concepts and observations of nominal stress strain behavior for a ductile material undergoing tensile elongation and the effects of work hardening on the yield stress of the material. When you have completed your solution you can click on the solution button to check your results or go on to the next page.

(Solution on Page 93 and 94)

12. True Strain The definition of engineering strain as the change in length divided by the original length is sufficiently accurate for very small levels of strain but is not a true measure of strain under elongation into the plastic behavior of the material. A more accurate description of the strain is the incremental change in length dy associated with that specific length y integrated from the original length Lo to the length L at which the strain is desired. Carrying out this integration results in the true strain being given by the natural logarithm of the ratio of L to Lo. However if L is expressed as delta L plus L then this logarithmic term can be written as the log of the quantity one plus delta L over L. This second term is just the engineering strain. Since epsilon will always be smaller than one the natural log of the term one plus epsilon can be expressed as the power series epsilon times the quantity one minus one half epsilon plus one-third epsilon squared plus etc. Again considering that epsilon is a small quantity the higher order terms of the power series can be neglected giving finally an approximation for the true strain as the engineering strain times the quantity one minus the engineering strain dived by two. Thus it is seen that the true strain of the tensile specimen will always be somewhat smaller than the engineering strain but not by a great deal.



13. True Stress The definition of engineering stress is the applied load P divided by the original cross sectional area of the tensile specimen A. As the specimen elongates the cross sectional area will decrease, therefore the true tensile stress in the material is more accurately given by the applied load P divided by the actual cross sectional area Ao at that load value P. Thus the true stress can be expressed in terms of the engineering stress simply as the engineering stress multiplied by the original area A divided by the actual area at load P which is Ao. Since the actual cross sectional area under load will always be less than the original cross sectional area the true stress will always be greater than the engineering stress. In the elastic range of the material the reduced cross sectional area can be expressed in terms of the Poisson’s ratio effect. Again making use of a power series expansion and neglecting higher order strain terms the true stress is given approximately by the engineering stress times the quantity 1 plus two times the Poisson’s ratio, nu. In the plastic region but below sigma ultimate and assuming a constant volume process the true stress can be expressed as the engineering stress times the quantity one plus the engineering strain. This makes it somewhat higher than in the elastic region. After necking takes place it is difficult to express the true stress analytically in terms of the engineering stress. However, at fracture it is not uncommon to observe in very ductile materials that the cross sectional area can be reduced to 40 % of its original value which results in a true stress that can be more than double the fracture stress measured in engineering terms

14. True Stress Strain Diagram A generic true stress strain diagram for the tensile behavior of a ductile material is shown in the graphic compared to the engineering stress strain diagram for the same material. Up to the value of the ultimate stress it is observed that the true stress is somewhat higher than the engineering stress but that it follows the same form of behavior relative to the strain. Beyond the ultimate stress the true stress curve departs radically from the engineering stress curve. This is the consequence of the necking behavior, which decreases the minimum cross sectional area of the specimen dramatically. Consequently the true stress in this last deformation phase it actually observed to increase even though the load on the specimen is decreasing. As indicated on the previous page the true fracture stress can in some instances be greater than double the fracture stress in engineering terms. One might conclude from this that in a physical design application a material could be expected to carry stress loads higher than the ultimate stress defined by the peak on the engineering stress strain curve. This is an incorrect conclusion since the behavior beyond the ultimate stress is unstable. Remember that the load begins decreasing after this point even though the strain increases. Hence the ultimate stress represents a true realistic maximum design value for the material and is therefore referred to as the tensile strength of the material.



15. Offset Yield Stress All engineering materials do not exhibit a well-defined yield stress as represented on the previous page. This is particularly true of non-ferrous materials like copper, aluminum, magnesium and their alloys. Hence, an arbitrary standard procedure has been established to define a yield stress for such materials. The process for doing this is to draw a line parallel to the initial slope of the stress strain curve of the material at some prescribed value of strain on the epsilon axis as shown in the figure. The value of stress associated with where this parallel line intersects the stress strain curve is defined as the off set yield stress for that material. In most instances the value used for the magnitude of the off set stain is 0.2 %. Within this range beginning at zero strain the material is considered to behave elastically for design considerations.

16. Typical Tensile Properties The three most common tensile properties published for materials are the yield stress, the ultimate stress or tensile strength and the percent elongation associated with a specified gage length used in the test. The table presented here lists approximate ranges of these property values for a number of different materials. The ranges are a consequence of alloying and possible work hardening. Note that the stresses are given in the units of kpsi so that the values in the table must be multiplied by 1000 to obtain psi. Cast iron is not normally considered an effective tensile material for practical applications hence information is scarce. It should be observed that as the yield stress and tensile strength increase the ductility of carbon steel and stainless decreases significantly as indicated by the decrease in percent elongation. It is also observed that the highest yield and tensile strengths are achieved in the ferrous materials with a particularly wide range for stainless steel. Material property values that account for specific alloying and processing should be obtained from a reliable test source before use in design applications.



17. Plain Carbon Steel The effects of alloying and processing on material properties can be dramatic and are well demonstrated by considering plain carbon steel. The chart on this page shows how the yield point, tensile strength and ductility change with an increase in the percent carbon content. By increasing the carbon content from 0.1 % to 0.5 % the yield stress and ultimate stress (tensile strength) of hot rolled (HR) steel are both doubled in value. However this improvement in strength come with a significant decrease in ductility from 27 % elongation to 17% elongation. Even more dramatic changes occur from the hot rolled steels to cold drawn steels (CD) of similar alloy composition. Here it is observed that the work hardening effect dramatically improves the yield stress with less but still positive increase in the tensile strength also. For 0.5 % carbon the yield increases from 50 kpsi to 83 psi while the tensile strength improves from 89kpsi to 100 kpsi. However, the effect of cold work processing decreases the ductility for the same composition steel from 17 % to about 10% over a 2 in gage length. Not all material undergo as dramatic changes as carbon steel but it is important to know what the specific composition and properties of a given material are for a given design application.

18. Hardness Testing In the introduction to this chapter hardness testing was cited as one of the more common static material properties tests employed. Hardness is generally referred to as the resistance of the material to some form of surface indentation. These tests are useful from several points of view. To begin with the more common types of test can be related to one another quantitatively and correlated to the tensile strength of the material. Secondly they are non destructive and virtually have no effect on the strength of finished mechanical parts and as such can be used to control the quality and uniformity of their manufacture. The two most common forms of hardness testing that will be discussed here are the Brinell and Rockwell tests.



19. Brinell Hardness Test The Brinell hardness test makes use of a loading device that indents the surface of the material with a 10 mm diameter ball with a load of 3000 kg for ferrous materials. For non-ferrous materials the indenting load is reduced to 500 kg. The load is removed and the diameter of the indentation is then measured with a microscope. The pressure in kg per mm squared is then determined from the surface area of the impression. This pressure is designated the Brinell Hardness number or Bhn. For steels this number is generally range from 100 to 500. The formula for determining the Brinell hardness number in terms of the load in kg, the indenter ball diameter in mm and the indention diameter in mm is as given at the bottom of the page.

20. Rockwell Hardness Test. Rockwell harness testing also indents the surface of the material but its process and measurements are quite different from the Brinell test. In the Rockwell test the surface is first indented under a standard load. The load is then increased a specified amount and reduced back to the initial load and the increase in the depth of the indentation between the two loads is measured. This increment of additional indentation provides the Rockwell harness number. The initial load applied is 10 kg. The additional load may vary. Several scales of Rockwell hardness exist. The Rockwell C scale uses a sphero-conical indenter with a maximum load of 150 kg. The Rockwell B scale uses a 1/16 in. diameter ball with a maximum load of 100 kg. Typical numbers for medium carbon steel are Rockwell C-20 or B-100 while for very hard steel it would be Rockwell C-100.



21. Hardness Scale Comparisons The table on this page lists corresponding values of Brinell hardness number, Rockwell C scale and B scale numbers together with tensile strength in kpsi for steel. Note that Rockwell C scale measurements are discontinued at 15 because lower values are difficult to obtain and would be inaccurate. Similarly Rockwell B values are discontinued above about 100 for the same reason. Also observe that the Brinell numbers range from about 100 to 500 for steel as indicated earlier. Hardness comparison scales between Brinell and Rockwell numbers are available in a variety of sources such as the Machinist’s Handbook. Good sources for the correlation of these hardness numbers to tensile strength for other materials are much more difficult to come by. This simply indicates that such a correlation may need to be determined for a specific material design application by conducting both the tensile tests and hardness test on that specific material.

22. Exercise -3 This exercise deals with establishing a correlation between Brinell hardness numbers and tensile strength for steel from the table on the previous page. The hint is provided to help guide the solution process. When you have completed the development of this relationship check your result by clicking on the solution button and then proceed on to the next page.




23. Review Exercises This review exercise consists of a series of statements taken from the subject content of the chapter that are either true or false. Indicate your response by clicking on either the true or false button at the end of the statement. An immediate visual feedback will be provided. To remove the feedback and go on to the next statement hit the tab key or click the mouse. Clicking on the hot word, in red, in the statement will popup the page on which the subject of the statement is covered. After completing all questions go on to the next page

24. Off Line Exercises This off line exercise consists of two practical problems that require using material covered in this and earlier chapters. The first requires an understanding of a ductile material undergoing tensile test behavior and the difference and significance of true stress and strain as related to engineering stress and strain. The second problem deals with determining both Brinell and Rockwell hardness numbers given the results of a specific test. After reviewing these two problem statements as long as necessary and if you are through with this chapter click on the main menu button to exit or go to another chapter.




Chapter 4 Static Material Properties

Problem Solutions

Screen Titles

Calculation of Shear Modulus Mode of Deformation Elongation due to Applied Load Hardness Exercise Hardness Exercise (cont.)


_____________________________________________________________________________________Static Material Properties - 92 - C.F. Zorowski 2002



1. Calculation of Shear Modulus The modulus E for Copper from the table is 17.2 times ten to the sixth psi while its Poisson’s ration is 0.325. Substituting these values into the theoretical relationship for the shear modulus gives a value of 6.49 times ten to the sixth. This is exactly the value for G as given in the table. The modulus E for carbon steel from the table is 30 times ten to the sixth psi and its Poisson’s ratio is given as o.292. Substituting these values into the theoretical relationship for the shear modulus gives a calculated value of 11.6 times ten to the sixth psi. This compares with 11.58 times ten to the sixth given in the table. Thus the calculated shear modulus numbers are essentially the same as given in the property table. Click on the return button to return to the next page in chapter four.

2. Mode of Deformation Determine the stress by dividing the load by the cross sectional area of the bar. With the dimensions of ¼ in. by 5/8 in. the cross sectional area becomes 0.156 inches squared. Thus the stress is 4800 lbs. divided by 0.156 inches squared giving 30,770 psi. Since the stress level is greater than the yield stress the material has gone beyond its elastic range. Also since the calculated stress is below the ultimate stress the material is not yet loaded sufficiently high enough to cause it to begin necking. Thus it is appropriate to conclude that the material is in its initial plastic deformation mode.



3. Elongation due to Applied Load When the load is removed and reapplied the material will be work hardened and the yield stress will be increased to 30,700 psi. Thus the deformation of the reapplied load will deform the material elastically and the modulus E can be used to calculate the magnitude of the resulting strain as the ratio of the stress to the modulus. Substituting the values of these two parameters gives a strain of 1.025 times ten to the minus three inches per inch. Multiplying the strain by the length of the bar, 50 inches, gives a total elongation of .051 inches. Click on the return button to go back to the next page in chapter four.

4. Hardness Exercise Applying the hint assume that the tensile strength is a constant times the Brinell hardness number. Use the values as given in the table to determine the constant C for all combinations of Brinell hardness number and tensile strength. These values for C are presented in a table on the next page. From the 20th Edition of Machinery Handbook the value of the constant C is given as 515 for Bhn values less than 175 and 495 for Bhn values greater than 175.



5. Hardness Exercise (cont.) The table lists the Brinell hardness number in the first column, the tensile strength in the second column and the corresponding value of the constant C as the ration of tensile strength to Brinell hardness number in the third column. It is observed that the values of C are close to 500. The percentage variation between the calculated values of C and the values of C as given in Machinery Handbook are listed in column four. It is observed that the C values as given in Machinery Handbook predict the tensile strength to less that 2.5% error in all cases. Hence the assumption of a proportional relationship appears to be quite appropriate. Click on the return button to go back to the next page in chapter four.


_____________________________________________________________________________________Material Processing - - C.F. Zorowski 2002 97

Chapter 5 Effects of Material Processing

Screen Titles

Processing Overview Sand Casting Investment Casting Shell Molding Die Casting Extrusion Forging Cold Working Hot/Cold Property Comparison Cold Forming Processes Heat Treatment Annealing Normalizing Quenching T-T-T Diagram Tempering Effects of Tempering Case Hardening Review Exercise Off Line Exercises



1. Title page Chapter 5 presents an quick overview of some of the more common mechanical and thermal processes used to form and treat ferrous and non-ferrous metals together with effects of these processes on material properties. A variety of common casting, hot working and cold working processes are described and illustrated. The thermal processes described include annealing, normalizing, quenching, tempering and case hardening particularly as applied to the preparation and conditioning of carbon steels.

2. Page Index Listed on this page are all the individual pages in Chapter 5. Each title is hyperlinked to its specific page and can be accessed by clicking on the title. It is suggested that the reader first proceed through all pages sequentially. Clicking on the text button at the bottom of the page provides a pop up window with the text for that page. The text page is closed by clicking on the x in the top right corner of the frame. Clicking on the index button returns the presentation to the page index of chapter 5.



3. Processing Overview All ferrous and non-ferrous materials undergo some form of processing in becoming a finished part of some final product. Those processes that can have the greatest impact on their strength, ductility and hardness properties are dealt with in this chapter. These fall into the two major categories of mechanical processing that changes the intrinsic material properties by physical deformation or changes of shape and thermal processes that produce material transformations in grain size and crystal formation. The three mechanical processes that cover the majority of means by which materials are reshaped can be included in casting, the solidification of a molten metal in a mold, hot working which results in large plastic deformations and changes of shape and cold working used primarily to achieve dimensional control and surface finish quality. Thermal processing consists of raising the temperature of the material to elevated levels and then controlling both it’s time and rate of cooling. This includes processes such as annealing, quenching, tempering and case hardening which play very special roles in achieving specific strength and ductility properties in carbon steels and steel alloys. The effects and role of metal removal as exemplified by cutting, machining, drilling, grinding, etc. are not covered in this chapter. Mechanical and thermal processing could by itself serve as the subject of a very extensive study and exploration. This chapter only highlights the very surface of this topic and is only intended to provide a quick overview to develop an appreciation for the reader of it potential impact and complexity.

4. Casting Casting can be very simply defined as the process of pouring a molten metal into a mold and allowing the liquid metal to solidify. A variety of currently used casting processes are probably best categorized by the material from which the mold is made, how it formed and how it is used. The oldest of these processes and probably the best know is sand-casting. The mold is formed by packing moist sand around a pattern of the shape to be made. Since the pattern must be removed to create a cavity for the molten metal the mold is made of an upper and lower half, the cope and drag. This requires the pattern, which may be wood, plaster or metal to be made in two halves also. If the cast shape contains an internal cavity a separate core is placed inside the mold before pouring takes place. Vents and risers from the outside to the mold cavity allow gases to be vented from the molten metal. The metal is poured into the mold through a channel called the sprue. After solidification takes place the mold halves are separated, the casting is removed and the sand is reconditioned and used again to repeat the process. Casting dimensions are nominal and the surface of the part is rough. Sand casting is relatively simple and low cost but not a high number part production process. Commonly used for large parts produced in small numbers. Click on the graphics button below to review illustrations of a generic sand casting mold.

(Graphic on Page 113)



5. Investment casting Used to produce identical modest size cast parts, of the order of thousands, to close to finished dimensions with a better finish than sand casting. Often referred to as the “lost wax process”. Mold sections of the part pattern are first machined in metal. This mold is then used to cast multiple numbers of part shapes in wax. These wax parts are then joined by runners to a central tree. The tree and runners become the channels through which the molten metal will flow in the mold when the final casting is made. The tree with the attached wax part shapes is then coated with a slurry of plaster or fine silica. The entire coated assembly is then baked in an oven during which the wax is driven off leaving a mold of the part shapes, runners and tree. The molten metal is then poured into the mold creating multiple cast parts. After solidification the mold is destroyed when it is stripped off the casting. The parts are then separated from the tree. The process is then repeated starting with the casting of the wax patterns of the shape desired. Click on the two buttons below to view generic illustration of the six most important parts of the process.

(Graphics on Pages 114 and 115)

6. Shell Molding This process takes its name after the type of mold used to produce the final casting. A hot metal pattern of machined aluminum, brass or cast iron is immersed in a mixture of sand and thermosetting resin. The heat of the pattern melts the plastic that combines with the sand to form a 3/8 to ½ inch shell around the pattern. The system is cured at about 600 degrees F. After cooling the shell is cut in half and stripped from the pattern. At this point the shell resembles the two halves of a sand casting mold. The shell halves are then fastened together to form the final mold. The mold is placed in a container of back up material like lead shot to provide support. When the molten metal is poured in the mold its heat burns away the plastic bond allowing the gases to escape and the casting to be air-cooled. Close dimensional tolerances and smooth finishes can be achieved by this process.



7.Die Casting In die casting molten metal is forced under pressure into a metal die that has been machined to be the mold of the part shape desired. This process can be used to produce large numbers, ten of thousands, of small to moderate sized parts due to the long life of the metal die. Both hot and cold chamber versions of the process are employed. In the hot chamber version the pressurizing system is immersed in the molten metal. Potential chemical reaction between the plunger and molten metal restricts this process to zinc, lead and tin alloys. Aluminum and copper alloys casting are produced in the cold chamber version in which the pressurizing system is separate from the molten medal. The melting point of the cast materials must lie significantly below that of the steel plunger used for pressurization. After solidification the die is separated and the part ejected. Injection pressures normally range from 2500 to 4000 psi but systems achieving pressures of 20,000 psi are possible.

8. Hot Working Processes All hot working processes take advantage of the material being deformed at high temperature allowing very large plastic deformations due to the softened state of the metal. The first of these processes that virtually every metal undergoes is the hot rolling of large ingots or slabs. Rolling is the process of passing a red-hot ingot or slab between large parallel rolls which reduces it’s thickness and increases it length. This can occur in multiple passes through a single mill to produce long thin sheets stored in coils. Hot rolling is performed on steel, aluminum, copper and magnesium alloys. By using progressive shaped roll passes this process is also used to produce round and square rods as well as structural shapes. High pressure seamless tubing is produced by passing round rods through inclined rolls that force the rod over a mandrel to form the inside diameter of the tube. Pipe is formed by rolling a flat sheet into a circular shaped section with a lengthwise seam that is then welded together. Click on the graphics example button below to view illustrations of hot rolling and seamless tube piercing.




9. Extrusion Extrusion is a hot working process in which a heated billet is placed in a confining cylinder and forced under high pressure by a plunger behind the material to flow through a die to produce long bars with a specific cross sectional shape. The process works best on lower melting point metals like aluminum, copper, magnesium and their alloys that flow easier than steel at comparable elevated temperatures. Product can be formed with high dimensional accuracy into very complex solid and hollow cross sectional shapes. It is more costly than rolling because of initial die expense coupled with short life resulting from high wear rates. The process has many variations such as indirect extrusion where the die moves instead of the plunger, employing hydrostatic pressure around the billet to facilitate metal flow and the use of lubricants to reduce die wear. Click on the graphic example button to view a generic illustration of direct extrusion and a photo of different cross sectional shapes of actual product.


10. Forging In the process of forging large plastic deformations are produced by compressive stresses created by shaped or flat dies using large hammers or presses. The simplest example of this process was practiced by the village black smith with his strong arm and heavy hammer on iron bars heated in an open charcoal furnace augmented by hand operated bellows. This process produces a refined grain structure resulting in higher yield strength and greater ductility than is produced by casting. The die can be open or closed and the hammer or press can be pneumatic or hydraulic driven. It is the first part of the forming process for large items like railroad axles and turbine rotors. Drop forging is a closed die process in which the hammer is actuated by gravity. Closed die forging generally produces a dimensionally more accurate and stronger part than casting. Click on the graphic example button to view an illustration of three generic steps in the production of a large forged ring.

(Graphic on 118)



11. Cold-Working Processes Cold working processes deform metals plastically at or slightly above room temperature. Since the material is less plastic at these temperatures compared to hot working the rates of size reduction are significantly smaller. The grain size of the metal is not appreciably changed however it does become quite distorted. Cold working essentially work hardens the material producing both higher yield stress and ultimate strength but at the expense of significant reduction in ductility. The product has a much smoother finish, is more accurate dimensionally and requires less machining or other mechanical finishing. Yield strengths are higher than comparable cast products.

12. Cold and Hot Working Comparison The figure compares the engineering stress strain diagrams for a generic 0.3 % carbon steel that has been hot rolled and then subsequently cold worked. It is first observed that the yield stress has essentially been doubled by the effects of the cold working. The ultimate strength is also increased but percentage wise by a smaller amount. The comparative yield and ultimate stresses for the hot rolled steel are 40 kpsi and 60 kpsi whereas for the cold worked material these same property figures are 80 kpsi and 90 kpsi. It is further observed that the strain at failure, a measure of ductility, is decreased by a factor of two from the hot rolled material to that which has been cold worked. These changes and their general comparative magnitudes are experienced by the entire family of conventional carbon steels as they go from being hot rolled to cold worked.



13. Cold Working Processes Some of the more common cold working processes with their generic description and product results are presented on this page. Cold rolling is used to reduce hot rolled strip into thin sheet product by passing the material through multiple sets of parallel rolls. Each pass reduces the thickness by some small amount until the final desired thickness is achieved. Cold drawing is used to reduce the cross sectional dimensions of hot rolled bars. Cleaned bars are pulled through dies that can reduce the cross section by up to 10%. The product is normally referred to as cold finished bars. Even though the product has a smooth finish no metal has been removed only deformed. Heading is a cold working process in which a portion of the product is upset or flattening out. Examples of cold-headed products are bolts and rivets and other similar shaped parts. Roll threading is a cold working process used to form threads on round stock by squeezing and rolling the material between thread shaped dies. Stamping is a generic cold working term that covers press operations that include blanking, coining and forming or bending. All portions of the material that undergo deformation by cold working will exhibit the change in yield stress, ultimate strength and ductility already described.

14. Heat Treatment The heat treatment of materials refers to some form of thermal processing, that is heating and cooling, the interrupts or changes the transformation process of a metal or its alloy as prescribed by its equilibrium phase diagram. These changes can have major effects on the mechanical properties of the final material product. In the discussion that follows thermal processing of steels will be emphasized. The more important processes considered together with their effect on strength, ductility and hardness include annealing, quenching, tempering, and case hardening. The principal differences between these processes are the temperatures to which the material is elevated, how long it may be held at a specified temperature and the rate at which it is cooled.



15. Annealing Both hot and cold working of ferrous materials introduces residual stress into the product as well as increasing its hardness or the resistance of its surface to indentation. Annealing is a process that permits the material to slowly transform in accordance with its equilibrium diagram to relieve these two work effects. The temperature of the material is raised to about 100 oF above its critical or equilibrium crystallization temperature and held at that point to allow the carbon in the steel to dissolve. The material is then allowed to cool very slowly usually in the furnace in which it was heated and held. The process can take hours or days. This softens the material, reduces its hardness, relieves the residual stresses and refines the grain structure.

16. Normalizing Normalizing is considered as being included in the annealing thermal processing category. The temperature is raised slightly higher than for a full anneal. The material is then cooled in still air at room temperature. This results in a faster temperature decrease than in a full anneal with subsequently less time for the material to reach its normal equilibrium state. The result is material with a slightly coarser grain size that is easier to machine. This is particularly true in the low carbon steel family, 0.1% to 1.7%-1.8% carbon content. The material is also slightly harder than fully annealed steel. Normalizing is often used as the final processing treatment for steels.



17. Quenching Quenching, as the term implies, refers to the rapid cooling of a heated material in some liquid that has the capacity to quickly conduct thermal energy from the metal. The rate of thermal extraction is a function of the quenching media. Two common liquids used to quench steel are oil and water. The resulting structure of the material becomes a function of the carbon content of the steel, the temperature to which the material is heated, the time it is held at the temperature and the rate at which it is then cooled. This time and temperature dictate the degree and type of transformation and crystal structure that will result from the process. If steel is cooled very rapidly the transformation that takes place produces a structure called martensite. This is a very hard and brittle steel that contains high levels of residual stress.

18. T-T-T Diagram The T-T-T or time temperature transformation diagram dictates the phase combinations and crystalline structure of steel that has been heated and then cooled. The two red knee shaped curves indicate when transformation begins and is completed at a given temperature from austenite to pearlite or bainite depending on the isothermal hold time. Four specific transformation and cooling curves are shown. Curve 1 illustrates austenite transforming completely to pearlite at a little less than 600 oC after about 103 seconds. Curve 2 at about 525oC shows austenite transforming partially to pearlite in about 10 second after which rapid cooling produces martensite in the remaining material. Curve 3 shows austenite at 375oC being transformed totally into martensite. Curve 4 illustrates austenite being held at about 280 oC for about 103 seconds being transformed partially into bainite that followed by rapid cooling produces martensite in the remaining material. These examples are provided simply to make the reader aware of the complex nature of the transformation that a steel can undergo by thermal processing depending on its carbon content. The reader is directed to references on material science for greater detail and explanation of the structures cited.



19. Tempering Steels that have been fully hardened by quenching are very brittle and contain significant residual stresses. A thermal process called tempering can relieve these undesirable conditions. It is sometimes referred to as stress relieving and/or softening. The material is reheated to below the critical temperature and allowed to cool in air. The key element here is that that critical temperature is not exceeded. The temperature is dependent on the composition of the steel and degree of hardness desired. The process releases carbon in the martensite that forms carbide crystals. The resulting structure is referred to as tempered martensite.

20. Effects of Tempering The graph depicted here illustrates the effect of tempering temperature on the more important mechanical properties of medium carbon steel. It is observed that the yield stress, ultimate strength and Brinell hardness are all reduced by about a factor of two as the tempering temperature is increased from 400 oF to 1200 oF. This upper temperature incidentally is about 150 oF below the critical temperature where transformation of the structure begins to take place. At the same time the reduction in area and the elongation increases with tempering temperature by about the same proportion. Hence the effect of tempering is to decrease the strength and hardness of the material while increasing its ductility without severely changing its internal structure.



21. Case Hardening. Case hardening is a process used to preserve the ductility and toughness of the core of a steel part while producing a very hard thin surface finish. This is done by simply increasing the carbon content of the surface. To accomplish this the steel part is placed in a carburizing media, which can be either a gas, solid, or liquid at some specified temperature for a prescribed period of time. The time and temperature depend on the composition of the steel and the depth of hardness desired. The surface simply absorbs the excess carbon of its environment. The steel is then quenching and tempered. Some of the more common practices include pack carburizing, gas carburizing, and nitriding as well as induction hardening and flame hardening.

22. Review Exercise In this exercise the mechanical and thermal process on the left are to be matched with the appropriate relevant characteristic listed on the right. Place the cursor over an item on the left and hold down the left button. A pencil will appear that can be dragged to one of the green dots on the right. If the right choice is made the arrow will remain. If the selection is incorrect the arrow will disappear. After the exercise is completed proceed to the next page.



23. Off Line Exercise This exercise deals with the analysis and interpretation of both elastic and plastic tensile test data. A newly developed steel alloy test specimen, a half-inch in diameter, is pulled in tension to failure. Its elongation over a two-inch gauge length is recorded for every 1000 pounds up to eleven thousand pounds. Beginning at 9000 pounds its actual area is also determined and recorded for each additional thousand pounds until failure takes place. This information is listed on the next page. Use this data this data to plot the engineering stress-strain diagram (to an appropriate scale) for both the elastic and plastic behavior. Calculate the modulus of the material, determine its 0.2 % yield stress, estimate its ultimate stress and calculate the percentage of area reduction at the maximum strain. Click on the button below for a useful hint on how to calculate the strain in the plastic region.

(Hint on Page 119) (Solution in Appendix)


_____________________________________________________________________________________Material Processing - 111 - C.F. Zorowski 2002

Chapter 5 Effects of Material Processing

Supplementary Graphics

Screen Titles

Sand Casting Investment Casting - 1 Investment Casting - 2 Hot Rolling Extrusion Forging Hint for Chapter 5 Off line Exercise



1. Sand Casting Shown on the left is an illustration of the part to be cast. The center drawing shows a lengthwise section of the complete mold for the part with the molten metal depicted in red. Note how the core used to create the central cavity in the part is held in place by the drag and the cope of the mold. The sprue is placed at one end of the casting cavity while the riser and vent is located at the other end to insure good flow and distribution of the molten material. The final illustration on the right is the section AA through the mold again illustrating the function of the core to produce the internal cavity on the part.



2. Investment Casting - 1 The illustration on the left is a section through the machined metal mold in which the wax facsimile of the part, shown in black, is cast. In the center illustration multiple wax facsimiles of the part are joined together by wax runners to a central wax tree that supports the system of wax parts. The final illustration on the right depicts the slurry coating of plaster or fine silica that adheres to the surfaces of the multiple part tree. This coating is actually quite thin compared to the how it appears on the illustration.



3. Investment Casting - 2 The figure on the left illustrated the mold as it is baked and the wax is driven off leaving a blank cavity, shown in white. In the center figure the molten material depicted in red is shown having filled the mold. In the final illustration on the right the mold has been stripped away permitting the individual parts to be separated from the tree.



4. Hot Rolling This graphic illustrated two distinct hot rolling processes. On the left is a conventional rolling operation in which the metal slab passing between the rolls is reduced from an initial thickness H to a final thickness h. From simple conservation of volume the slab lengthens as it passes through the rolls. The material is pulled through the rolls by the surface friction in the arc of contact produced by the torque that drives the rolls. The figure on the right illustrates the process of seamless tube piercing. A circular billet is drawn and rotated between two-inclined barrel shaped rolls from right to left as shown. This rolling action cause a fracture to initiate at the center of the billet which then passes over a fixed mandrel to size the inside diameter of the tube.



5. Extrusion The drawing on the left illustrates a generic direct hot extrusion process. A hot billet is placed in a fixed container. Force is applied to a plunger behind the billet causing the material to flow through a die machined to produce the desired cross section of the final product. An extension can be placed on the front end of the plunger illustrated by dotted lines to produce a hollow product using the same process. On the right is a photograph of a number of typical extruded products showing the variety of complex cross sectional shapes that can be produced.



6. Forging This graphic illustrates three generic forging steps that might be used in creating a large ring. On the left the first step is to upset a circular billet into a thick flat disk shape using a large press. In the second step the disk is pierced by a punch shaped die again using a large press. On the right the ring is shown being hammered into its final thickness and diameter between a flat die on the hammer and a circular fixed mandrel that supports the ring. The ring must be moved in a circular manner as this process is carried out. In forging large parts the material being processed will normally require being reheated between the individual steps and possible even during the final hammer shaping shown as the last step.



Hint for Chapter 5 Off line Exercise


_____________________________________________________________________________________ Theories of Static Failure - - C.F. Zorowski 2002 121

Chapter 6 Theories of Static Failure

Screen Titles

Introduction Design Scenario Issues Designer’s Dilemma Theories of Failure Max Normal Stress Theory –1, 2 Max Shear Stress Theory – 1, 2 Distortion Energy Theory – 1, 2 Mohr’s Theory Coulomb Mohr Theory – 1, 2 Ductile Material Failure Applicable Theories Brittle Failure Modified Coulomb – Mohr Applicable Theories Example Problem Review Exercise Off Line Exercises


____________________________________________________________________________________Theories of Static Failure - - C.F. Zorowski 2002 122



1. Title page Chapter 6 presents a number of theories that have been developed to predict when both ductile and brittle materials will yield or fail when subjected to static loading that creates a state of combined stress within the material. Each of the theories investigated represent a different criteria for predicting yielding or failure under a state of combined stress based on the material properties measured in a simple tension or compression test. In addition to developing these theories their applicability and accuracy of predicting yielding or failure is also discussed and established. The chapter concludes with two sets of recommendations as to which theories represent the best behavior predictors for both ductile and brittle materials under static loading.

2. Page Index Listed on this page are all the individual pages in Chapter 6. Each page title is hyperlinked to its specific page and can be accessed by clicking on the title. It is suggested that the reader first proceed through all pages sequentially. Clicking on the text button at the bottom of the page provides a pop up window with the text for that page. The text page is closed by clicking on the x in the top right corner of the frame. Clicking on the index button returns the presentation to the page index of chapter 6.



3. Introduction Ideally, a part designer should have available results from a great many tests for the material chosen for the part to insure that the design decisions made will provide the strength and endurance desired for its final application. It would be best if these tests could be performed on materials possessing the identical properties of alloying, processing and heat treatment to be used in the part as well as being loaded in a similar manner as the part being designed, that is in tension, bending, torsion and/or combined loads applied statically or dynamically as specified by the design requirements. In other words actually testing the part itself. Unfortunately, this is seldom possible. Whether this is practical, possible or feasible is a consequence of how the design fits into an overall design scenario defined by issues involving varying degrees of safety, production volume, cost and available time.

4. Design Scenario Issues The issues that impact a design are many and varied with different influences on the testing that takes place on the materials that go into the design prior to or during the design process. Material testing may cover a wide spectrum that includes a simple tensile test at one end to complex life tests of prototype parts at the other. Listed here are five design scenario issues that vary in their effect on material testing that may take place. The issue of public safety is the most prominent. Any likelihood of endangerment of life or significant physical harm from material failure dictates extensive testing that provides the best material properties or part behavior available. The question then becomes when has enough testing been performed. Production volume impacts material testing in that the cost of testing will impact the final cost of the part. Where the production volume is high an extensive testing program may represent only a small portion of the final part cost. The greater the importance of operational reliability in the final product the more justifiable an appropriate material testing program irrespective of its cost. The space program is a good example. There are also instances in which the time available for an extensive testing program simply isn’t available. It is hoped that this never has to be the case but instances may arise. When parts fail with undesirable effects there is always the question of whether sufficient testing was undertaken as well as were the results properly interpreted and applied.



5. Designer’s Dilemma It is unfortunately clear that mechanical property results from extensive and complex individual part or material tests will not always be available to the designer. In fact it will normally be the exception when the designer is provided with such complete information. In many instances the designer is faced with only being able to access published values of yield strength, ultimate strength and percentage area reduction or elongation at failure from tensile tests particularly in the early stages of design. However, even with this minimal material information the designer must be able to estimate and predict static strength or failure under loading conditions that result in complex stress states. The question of how best to approach and deal with this problem is dealt with in this chapter for both ductile and brittle materials subjected to static loading.

6. Theories of Failure Five so-called theories of failure that predict yielding or fracture under combined states of stress created by static loading will be developed and studied. The first three listed, the maximum normal stress theory, the maximum shear stress theory and the distortion energy theory are based on criteria implied by their names. The first two obviously deal with the maximum normal stress and maximum shear stress of the combined state of stress to which the material is subjected. The third theory is based on a portion of the total strain energy possessed by the material due to the combined stress state. These three theories are of greatest importance in predicting the behavior of ductile materials. The fourth and fifth theories are based on an interpretation of the results of a simple tension test, a simple torsion test and a simple compression test. These have greater applicability to predicting the behavior of brittle materials subjected to combined state of stress created by static loading.



7. Maximum Normal Stress Theory - 1 In all the theory developments that follow their application will be limited to a general two-dimensional state of stress. This makes understanding the development and application of the theory easier. In addition most loading cases can be reduced to creating a two-dimensional state of stress anyway. The maximum normal stress theory simply proposes that yielding will occur in a material when the maximum normal stress in a combined stress state in that material is equal to the maximum normal stress in a simple tension test. Any general two-dimensional state of stress can be defined in terms its two principal stresses sigma 1 and sigma 2. Assume that these are ordered as listed. i.e. sigma 1 is greater than zero (tensile) and sigma two is negative (compressive). This theory predicts that yielding will take place when sigma 1 is equal to the yield stress, sigma y in tension, or when sigma 2 is equal to minus sigma y in compression whichever occurs first. This can best be visualized in terms of a graphical representation of these limiting criteria as depicted on the next page.

8. Maximum Normal Stress Theory - 2 The maximum normal stress theory in two dimensions is best visualized by plotting the limiting conditions imposed by the theory on a two dimensional axis system defined by sigma 1 and sigma 2 coordinates as illustrated. If sigma 1 and sigma 2 are both positive, that is the stress state lies in the first quadrant then the material will not yield unless either sigma 1 or sigma 2 are equal to the yield stress sigma y. The points sigma y super t on the two axes represent states of simple tension. Since the yield stress in simple compression might be numerically higher than the yield stress in tension the diagram can be shown as extended somewhat in the second, third and fourth quadrants to the limiting values of sigma y super c for simple compression on the negative sigma 1 and negative sigma 2 axes. Then if sigma 1 is positive and sigma 2 is negative, that is the fourth quadrant, then yielding is defined by the theory if either sigma 1 has a value of sigma y super t or sigma 2 becomes equal to minus sigma y super c. In a similar fashion the yielding boundaries in the second and third quadrants can be defined in a similar fashion as prescribed by the theory. Any state of two dimensional stress that lies inside the rectangular figure denotes a condition that yielding will not occur whereas a state of combined stress on any boundary of the figure indicates that yielding will occur. This theory is a pretty good predictor in the first and third quadrants but not in the second and fourth.



9. Maximum Shear Stress Theory - 1 The maximum shear stress theory proposes that yielding will occur in a material subjected to a combined state of stress when the maximum shear stress is equal to the maximum shear stress experienced by material in a simple tension test. For simplicity it will be assumed that the yield stress in tension and the yield stress in compression for a ductile material will be equal. This is not unreasonable. The maximum shear stress in a simple tension test is just equal to one half the tensile yield stress or sigma y over 2. For a two-dimensional stress state in which sigma 1 and sigma 2 are the principal stresses and sigma 1 is greater than sigma 2 the maximum shear stress is given by the quantity sigma 1 minus sigma 2 divided by 2. Hence the maximum shear stress theory stated analytically is that sigma 1 minus sigma 2 is equal to sigma yield. If sigma 2 is greater than sigma 1 then the expression of the maximum shear stress theory is that sigma 2 minus sigma 1 is equal to sigma yield. Again this can be best visualized graphically as presented on the next page.

10. Maximum Shear Stress Theory - 2 The equation sigma 1 minus sigma 2 equal to sigma y is a 45-degree line that passes through the points sigma y on the sigma 1 axis and minus sigma y on the negative sigma 2 axis. In a similar manner the equation sigma 2 minus sigma 1 equal to sigma y is also a 45-degree line that passes through sigma y on the positive sigma 2 axis and minus sigma y on the negative sigma 1 axis. These two lines define the limiting condition for yielding as predicted by the maximum shear stress theory. Any state of stress that lies between these two lines will not produce yielding while any point on either line will result in yielding of the material. Note that this implies that very large, in fact infinite values of sigma 1 equal to sigma 2 will not predict yielding. This of course is not reasonable. However, it has already been stated that the maximum normal stress theory is a pretty good predictor in quadrants one and three. Hence a combination of the maximum normal stress theory in quadrants one and three together with the maximum shear stress theory in quadrants two and four might well be a good combination. This will be seen later to be quite acceptable for ductile materials.



11. Yielding in Pure Shear Before going on determine the value of the maximum shear stress to produce the onset of yielding in a load state of pure torsion, that is a circular shaft subjected to a twisting torque about its central axis based on the maximum shear stress theory. After solving the problem click on the solution button to check your answer.


12. Distortion Energy Theory - 1 The distortion energy theory is based on using a portion of the total strain energy contained in a material subjected to a combined stress state as its criteria for establishing the onset of yielding. More specifically it recognizes that the work done on a material under load can be expressed as energy due to volume change and energy due to distortion of shape. The theory states that yielding will take place when the energy per unit volume due to distortion under combined stress is equal to the distortion energy per unit volume at yielding in a simple tension test. The distortion energy is determined by subtracting the energy due to volume change from the total strain energy. This is done for a material under combined stress and then set equal to this same quantity for a simple tension test. The detailed development follows on the next several pages.



13. Distortion Energy Theory - 2 In this development it is necessary to begin with a general three-dimensional state of stress and then reduce the final result to two dimensions for comparison to the maximum normal stress and maximum shear stress theories. The total strain energy per unit volume in terms of the principal stresses in three-dimensions is given by the first equation. Note that it contains both the modulus of the material and Poisson’s ratio since it will be dependent on the specific material being loaded. To review the development of this expression for total strain energy click on the review button and return here when the review is completed. The stain energy per unit volume due to volume change is determined by first defining an average principal stress as one-third the sum of the three principal stresses in the general combined state and substituting this average into the expression for the total strain energy. Click on the second review button for an explanation of how a general three dimensional stress state can be separated into that component that creates just a change in volume leaving a remaining component associated with distortion of shape. The equation for the volume change energy is then expanded back into the three principal stresses as shown in the last equation on the page. Note that the stress part of the equation looks similar to that of the total strain equation with the exception of a change in sign and the elimination of Poisson’s ratio as a multiplier of the product stress terms.

(Reviews on Pages 141 and 142)

14. Distortion Energy - 3 The distortion energy per unit volume is now obtained by subtracting the energy due to change in volume from the total strain energy which gives the equation in which the differences of the principal stresses squared appear for the three possible sets in three dimensions. The expression for the distortion energy in a simple tension test is obtained by simply substituting sigma y the yield stress for sigma 1 in the general equation recognizing that sigma 2 and sigma three are both zero in this instance. Setting sigma three equal to zero in the general equation for distortion energy and equating the result to the distortion energy in a simple tension test gives the equation governing the relationship between the principal stresses in a two dimensional stress state and the yield stress in tension as predicted by the distortion energy theory. Again this is best interpreted graphically as illustrated and described on the next page.



15. Distortion Energy - 4 The equation that defines the distortion energy theory of yielding for a two dimensional state of stress is an ellipse whose major axis lies at a 45 degree angle through the origin of the sigma 1, sigma 2 axis system. If sigma 1 or sigma 2 is set equal to zero it is observed that the ellipse passes through the same points of simple tension and simple compression just like the combined maximum normal stress and maximum shear stress theories. Thus the distortion energy theories will predict the onset of yielding to occur due to a combined state of stress in a similar manner to that of the previous two combined theories but not quite as conservatively. The differences between these various criteria are illustrated by the sample problem on the following page.

16. Yielding in Pure Shear Based on the distortion energy theory of yielding determine the value of the maximum shear stress in a state of pure two-dimensional torsion. That is, a shaft subjected to a twisting torque about its central axis. Compare this the maximum shear stress for the same loading previously calculated using the maximum shear stress theory. After solving this problem click on the solution button to check your answer.

(Solution on Page143)



17. The Mohr Theory A different approach to predicting the onset of yielding is presented in what is called the “Mohr Theory”. This is based on test results obtained from a simple tensile test, a pure torsion test and a simple compression test. The yield stress from each of these three tests are used to plot three separate Mohr circles as shown in the diagram. An upper and lower envelope represented by the dotted lines is then drawn tangent to the three circles through points A, B and C as indicated. The theory proposes that any Mohr circle associated with a two dimensional sate of stress that is just tangent to these dotted lines will produce the onset of yielding in the material. Or conversely if the Mohr circle of any two dimensional state of stress lies inside of this envelope yielding will not occur. Analytical application of this theory presents a problem since the equations of the dotted lines are undefined and are dependent on the relationship of the three yield stresses measured. A resolution of this difficulty is presented on the next page.

18. Coulomb-Mohr Theory - 1 The Coulomb Mohr theory is proposed as a means of resolving the problem of analytically expressing the criteria of the Mohr theory. This modification of the Mohr theory simply proposes that the points A and B which are the tangents of the simple compression and tension circles lie on a straight line. In essence this neglects the effect of the result of the yield stress measured in the pure torsion test. Recognizing that the yield in compression may be greater than in tension particularly for a brittle material this results in the straight-line equation to the right of the figure. An interesting exercise is to develop this equation from the geometry of the figure.



19. Coulomb Mohr Theory - 2 The equation for the Coulomb Mohr theory shown gives the diagonal line in the fourth quadrant. Reversing sigma one and sigma 2 gives the diagonal line in the second quadrant. The figure is completed by combining these new criteria with the maximum normal stress theory in the first and third quadrants accounting for a higher yield stress in compression than in tension. The maximum shear stress theory diagram in which the yield in tension and compression were assumed equal is shown for comparison. It is observed that the Coulomb Mohr theory is essentially an expansion of the maximum shear stress theory taking into account that the compression yield can be higher than the tensile yield, which is most often the case in a brittle material. The question that now needs to be answered is how well do these theories actually predict the onset of yielding in both ductile and brittle materials.

20. Ductile Material Failure In the figure illustrated the results of a number of typical test results, indicated by the red pluses, for yielding under states of combined stress for ductile materials are plotted on top of the combined maximum normal stress and shear stress theories as well as the distortion energy theory in the first and fourth quadrants. Note that the coordinate axis parameters sigma 1 and sigma 2 have been normalized by dividing each by the yield stress. It is observed that the results from actual tests fall very closely to the yield criteria lines defined by these two theories. The criteria of the combined maximum stress and shear stress theory appears to be somewhat conservative while the distortion energy theory appears to be the more accurate. This is a good rule of thumb to remember in their application. Test results in the second and third quadrants would be the reverse of what is presented here since it really would only represent reversing sigm1 and sigma 2. Again it should be noted that the test results illustrated here are generic for ductile materials.



21. Applicable Theories Presented in the table are the equations from the maximum normal stress, maximum shear stress and distortion energy theories that are applicable for predicting the onset of yielding for ductile materials depending on the relationship of the principal stresses that represent the combined state of stress under consideration. All combined stress state of importance are covered by the behavior in quadrants one and four. A red yes in the table denoted the most conservative applicable approach while a green yes represents that more accurate prediction. An NA indicates that the particular theory is not applicable. These same equations and their application have been also shown to be equally applicable to predicting the ultimate strength of the material under combined stress if the yield stress is simply replaced by the ultimate strength measured in a tensile test.

22. Brittle failure In the figure presented here are characteristic test results of brittle materials under combined two-dimensional stress states that produce fracture plotted again as red pluses on top of the Coulomb-Mohr Theory in the first and fourth quadrants. It is observed that again the maximum normal stress theory appears quite applicable in the first quadrant where the limiting stress is the ultimate strength in a simple tensile test. In the fourth quadrant the Coulomb-Mohr Theory appears to be overly conservative for the combined states of stress for which sigma 2 is less than one half the ultimate stress in tension. However, a more acceptable fit with the data is obtained if the maximum normal stress theory is extended into the fourth quadrant to the point where sigma 2 is equal to minus one half the ultimate strength in tension and then a diagonal from that point is extended to the point where sigma 2 is equal to minus the ultimate strength in compression. This modification of the Coulomb-Mohr diagram is shown as a dotted yellow line in the fourth quadrant. This dotted portion of the figure is referred to as the Modified Coulomb-Mohr Theory.



23. Modified Coulomb-Mohr Equation In this exercise show that the equation listed on this page satisfies the Modified Coulomb-Mohr Theory relationship for the principal stresses sigma 1 and sigma 2 in the lower half of the fourth quadrant as depicted on the previous page. Note that the ultimate stress in tension is replaced by sigma T and the ultimate stress in compression is replaced by sigma C. This is because either the ultimate strength or the yield stress can be used in this relationship. If ultimate stresses are used the prediction is fracture. If yield stresses are used the prediction is the onset of yielding. Of course yielding is usually of less consequence in a brittle material than in a ductile material. When satisfied that this equation is valid click on the solution button to check your result or proceed on.


24. Applicable Theories Presented in the table are the equations from the Maximum Normal Stress, Coulomb Mohr and Modified Coulomb Mohr Theories that are applicable for predicting the onset of yielding and fracture for brittle materials depending on the relationship of the principal stresses that represent the combined state of stress under consideration. All combined stress state of importance are covered by the behavior in quadrants one and four. A red yes in the table denoted the most conservative applicable approach while a green yes represents that more accurate prediction. An NA indicates that the particular theory is not applicable. These equations and their application can be used equally well for predicting the onset of yielding or ultimate strength. It is only necessary to substitute the appropriate values for sigma T or sigma C in the equations. However with brittle material the possibility of fracture is the more important event to consider.



25. Example Problem –1 The theories developed and discussed in this chapter will now be applied in the following example problem. A stationary solid circular shaft is subjected to both a bending moment M and twisting torque T as shown in the figure. Using a conservative yield stress theory for ductile materials determine a general relationship between the moment M and the torque T that will result in the onset of yielding. If the yield stress in tension of the steel is 35,000 psi, the shaft has a diameter of 1 inch and the magnitude of the applied moment M is twice the torque T calculate the magnitude of the torque in inch pounds that will initiate yielding.

26. Example Problem –2 The first step is to determine the components of the two-dimensional stress state created by the loading system. The bending load will introduce a sigma x stress that can be expressed as M y over I. The shear stress tau xy is given by T r over J from the twist on the shaft. There is no load to create a sigma y stress so it is zero. For a circular shaft y is given by r the radius of the shaft and I the moment of inertia of the cross section is just equal to J the polar moment of inertia divided by 2. Thus sigma x becomes two M times r over J while tau xy remains T times r divided by J. These expressions are now used to define the principal stresses sigma 1 and sigma 2. The first term in sigma 1 is the sum of sigma x and sigma y over 2. This becomes simply M r over J. The second term in sigma 1 is the square root of the quantity sigma x minus sigma y over 2 squared plus tau xy squared. With sigma y zero this second term becomes the square root of M r over J squared plus T r over J squared. Sigma 2 is made up of the same two terms as sigma 1 but with a negative sign in front of the second term. It is observed from these two relationships for sigma 1 and sigma 2 that sigma 2 will be negative.



27. Example Problem –3 With sigma 1 positive and sigma 2 negative the most conservative predictor of yielding will be the maximum shear stress theory. Hence sigma 1 and sigma 2 must now be substituted into the expression sigma 1 minus sigma 2 is equal to sigma yield. Performing this substitution results in the equation that 2 times r over J times the square root of M squared plus T squared is equal to sigma y. Squaring both sides of the equation and rearranging gives the final result of M squared plus T squared is equal to sigma y squared over 4 times the quantity J over r quantity squared.

28 Example Problem –4 Now calculate T for a yield stress of 35,000 psi, a shaft radius of 0.5 inches and a load condition that M is equal to 2T. First J is determined to be .098 inches fourth. Then substituting M equal to 2T in the equation relating these loads the final expression for T becomes T equal to sigma y over the square root of 20 all times the quantity J over r. Substituting the appropriate numerical values into this equation and carrying out the mathematical manipulations gives a final answer for T of 1534 inch pounds of torque. The only question now is this really the right answer?



29. Example Problem –5 To check the answer for the torque the magnitudes of sigma x and sigma y as well as sigma 1 and sigma 2 will be calculated to see if they are the right order of magnitude. Substituting the appropriate numbers into the expressions for sigma x and tau xy gives values of 31,306 and 7,826 psi respectively for these stresses. These values are then used to calculate sigma 1 and sigma 2 which turn out to be 33,140 psi and –1,847 psi. Note that sigma 2 is negative but is not very large numerically. However with sigma 2 negative and sigma 1 positive the maximum shear stress theory for yielding was the appropriate choice and since sigma 1 is close to the yield stress it would not be expected that sigma 2 would be very large numerically.

30. Example Problem - 6 The fourth quadrant of the maximum shear stress theory diagram is shown here with the coordinate point defined by the magnitude of the principal stresses for this problem indicated. It is observed that this point does lie on the yield line defined by the theory indicating that the answer determined earlier is correct. A surprising result in this example is how close the actual point lies to the sigma one axis. There was no way to anticipate this at the beginning of the solution process.



31. Review Exercise In this exercise the items in the list on the left are to be matched with the mathematical relationships on the right. Place the cursor over an item on the left and hold down the left button. A pencil will appear that can be dragged to one of the green dots on the right. If the right choice is made the arrow will remain. If the selection is incorrect the arrow will disappear. After the exercise is completed proceed to the next page.

32. Off Line Exercise A portion of a cast metallic part has a rectangular cross section (1.25 in. x 0.25 in.) and is subjected to a normal compressive load of P equal to 4000 lbs. This section of the part is also required to carry a twisting torque T as shown in the diagram. If the metal has an ultimate strength in tension of 20,000 psi. and an ultimate compressive strength of 35,000 psi. determine the maximum value of torque that can be carried by the part as predicted by the appropriate theory of failure. When finished with this statement click on the exit button or the return to main menu button.




Chapter 6 Theories of Static Failure

Problem Solutions

Screen Titles

Work Due to Internal Stresses Total Strain Energy Yielding in Pure Shear Tri Axial Stress Breakdown Yielding in Pure Shear Modified Coulomb Mohr Theory


_____________________________________________________________________________________Theories of Static Failure - 140 - C.F. Zorowski 2002



1. Work Due to Internal Stresses The work done by the stresses acting on a material as it deforms is stored in the material as strain energy. Within the elastic limit of the material this energy is recoverable as in a compressed spring. To determine an analytical expression for the strain energy begin with the definition of work done by a force as the integral of the force over the distance the force acts. Internal to a stressed body in tension the force is the stress times the area over which it acts and the distance the force acts is the elongation which can be expressed as the strain times the length of the body. Substituting this into the general expression for work the strain energy per unit volume becomes the integral of the stress with respect to the strain. However, since stress and strain are linearly related by the modulus in the elastic region the strain energy per unit volume becomes half the modulus times the strain squared or one over 2E multiplied by the stress squared or one half the stress times the strain. This can be seen to be just the area under the stress strain curve in the elastic region of the material.

2. Total Strain Energy The total strain energy stored in a material subjected to a three dimensional state of stress can be represented as the sum of the unit strain energy components due to each of the three principal stresses. Thus it can be written as one half the sum of the products of the stress times the strain in the three principal directions. By introducing the generalized Hooke’s law each strain components can be expressed in terms of the stress in that same direction together with the strain component due to the Poisson ratio effect of the stresses in the other two directions. This leads to a final equation in terms of the square of the principal stress and a term modified by Poisson’s ratio times the sum of the products of the principal stresses.



3. Yielding in Pure Shear In a simple tension test at yielding sigma 1 is equal to the yield stress and sigma 2 is zero. For a state of pure shear sigma 1 is equal to minus sigma 2. Therefore the maximum shear stress, which is sigma I minus sigma 2 over two is just equal to sigma 1. From the maximum shear stress theory sigma 1 minus sigma 2 is equal to the yield stress. Therefore for pure shear sigma 1 is equal to the yield stress over two which makes the maximum shear stress in pure torsion equal to the yield stress over two.

4. Tri-Axial Stress Breakdown Consider an element of material subjected to a general three-dimensional state of stress represented by sigma 1, sigma 2 and sigma 3. This can be represented by an equivalent combination of two sets of effects. The first of this equivalent set is represented by a state of hydrostatic tension or compression in which the normal stress on each face is equal and is the average of the three original stresses. The second element in the equivalent set is subjected to three orthogonal stress represented by the original stress on each face minus the average of the three stresses. The element under the state of hydrostatic tension or compression will under go a simple change in volume in which each of its three perpendicular dimensions change the same amount. The second element of the equivalent set then under goes what is described as simple distortional changes in shape. The sum of all changes in this equivalent set are equal to the changes experienced by the element subjected to the original state of stress.



5. Yielding in Pure Shear In a simple tension test at yielding sigma 1 is equal to the yield stress and sigma 2 is zero. For a state of pure shear sigma 1 is equal to minus sigma 2. Therefore the maximum shear stress which is sigma I minus sigma 2 over two is just equal to sigma 1. From the distortion energy theory sigma 1 squared plus sigma 2 squared minus the product of sigma 1 and sigma 2 is equal to the yield stress squared. Therefore for pure shear sigma 1 squared is equal to the yield stress squared over three. Thus the maximum shear stress in pure torsion is equal to the yield stress divided by the square root of three or .707 times the yield stress. Hence the value predicted by the distortion energy theory required to initiate yielding in pure torsion is slightly higher than that predicted by the maximum shear stress theory.

6. Modified Coulomb-Mohr Theory At sigma 1 equal to sigma T sigma 2 should be equal to minus sigma T to satisfy the equation for the Modified Coulomb-Mohr Theory. Substituting these values into the equation shows that the equation is satisfied. A second check point is that if sigma 2 is equal to minus sigma c then sigma 1 should be zero. Substituting these values into the Modified Coulomb-Mohr equation shows that it is again satisfied. Since the equation satisfies two points on the straight line it must be the correct equation for the line.


_____________________________________________________________________________________Factors of Safety - - C.F. Zorowski 2002 145

Chapter 7 Factors of Safety

Screen Titles

Purpose and Definition Items Effecting Factor of Safety Variation Scenario – 1 Variation Scenario – 2 Modified Theories of Failure Generic Factors of Safety Actual Load Distribution Load Capability Distribution Difference of Distribution Gaussian (Normal) Distribution Transformation of Variables Transformation of Normal Distribution Application to Lc-L Distribution Determination of tf

Sample Problem Problem Solution Alternate form for n No Failure Scenario

Review Exercise Off Line Exercises Off Line Exercises



1. Title page Chapter 7 deals with the subject of the factor of safety in design from both a classical and statistical point of view. The presentation begins with a consideration of the purpose of introducing this concept and its definition in a classical sense. This is followed by looking at factors that influence its impact on a design and how it is included in the established theories of static failure. Some recommendations for factors of safety based on generically defined design scenarios are presented and discussed. The chapter content then takes up how a statistical treatment can be used in very large part populations to predict required factors of safety in terms of the properties of the sample’s statistical representation. A number of exercise problems are included to demonstrate the application of the concepts and principles discussed.

2. Page Index Listed on this page are all the individual pages in Chapter 7 with the exception of the exercise problems. Each page title is hyperlinked to its specific page and can be accessed by clicking on the title. It is suggested that the reader first proceed through all pages sequentially. Clicking on the text button at the bottom of the page provides a pop up window with the text for that page. The text page is closed by clicking on the x in the top right corner of the frame. Clicking on the index button returns the presentation to the page index of chapter 7.



3. Purpose and Definition The term “factor of safety” is ubiquitous in the practice of design. Everyone seems to know and appreciate what it means generically but similarly find it difficult to specify what it should be in a given situation. It has been called the “factor of ignorance” or the “factor of uncertainty” by many. The second designation is probably the more accurate. Another sense of its importance is summarized in the couplet “If in doubt make it stout”. One of the better ways to begin its study is to look at what might be considered its principal purpose in design. This can be simply stated as “the concept of factor of safety is introduced into design for the purpose of minimizing the risk of potential part failure”. This risk of potential failure is a direct consequence of the approximate analytical techniques used to determine load and strength levels and the uncertainty and variability in the values used in the numerical calculations involved. Classically this is incorporated in a simple analytical definition that states that the factor of safety is equal to the material strength of a part divided by the working stress that the part is designed to carry. Thus, if the factor of safety is two it simply means that it is anticipated that the part can be subjected to effectively twice the load it was designed for before the design criteria of either yielding or fracture, whichever has been chosen, will be exceeded in a static application.

4. Items Effecting Factor of Safety The natural variations and the level of confidence in understanding and knowing the extent of these variations in four specific areas give rise to the need and utility of a factor of safety to meet the purpose cited on the previous page. The first is the mathematical model with its included assumptions used to calculate the working stress in the part. For example, is simple bending theory based on elastic behavior appropriate to determine the bending stress in the part under consideration? The second area is the geometry of the part. This is characterized by the variability in the dimensions of the part as dictated by how it was physically processed into its final shape. In other words how close are the final dimensions to those used in calculating the working stress? The third area is the degree of certainty of the magnitude of the applied loads. Is there some question about the level of overloading the part may be subjected to or the direction in which the load may be applied? Finally, there is the area of the strength properties of the material from which the part will be made. How accurately are the material properties known and what is the degree of their variability?



5. Variation Scenario - 1 To determine the effect of variations of parameters associated with the areas discussed on the previous page consider the following numerical example. A bending moment of 10,000 in. lbs. plus or minus 500 in. lbs. acts on a rectangular cross section beam with nominal dimensions of 2 in. by 1 in. The width is only known to within plus or minus 1/16 inch and the height may vary by as much as plus or minus 1/8 inch. Using the nominal load and dimensions the nominal bending stress at the top of the section is calculated to be 14,992 psi. However, using the maximum possible bending moment together with the minimum possible dimensions results in a comparative maximum working stress in bending of 19,114 psi at the top of the section. This is 1.27 times the nominal working stress due to the uncertainty of the loading and the geometry.

6. Variation Scenario - 2 Now assume that the yield stress for the beam material is 15,000 psi plus or minus 10% or 1500 psi as established from a series of tensile tests. Therefore the minimum yield stress will be 13,500 psi or 9/10ths of sigma y nominal. Applying the criteria of the maximum normal stress theory can be stated as the maximum working stress must be less than or equal to the minimum yield stress. Substituting 1.27 times the nominal working stress for the maximum working stress and .9 nominal yield stress for the minimum yield stress gives the inequality that 1.27 sigma working nominal must be equal to or less than .9 sigma yield nominal. The concept of the factor of safety is now introduced by the equation that the nominal working stress should be equal to the nominal yield stress divided by the factor of safety. Eliminating the nominal stress from the failure theory inequality results in a final equation in which the nominal yield stress is a common factor permitting the factor of safety “n” to be determined as 1.4. This is the value of “n” that should be applied in the nominal design calculations to account for the uncertainty of the parameter values of load, geometry and material property to insure that yielding will not take place.



7. Modified Theories of Failure - 1 Incorporating the concept of a factor of safety into the equations for the static theories of failure for both ductile and brittle materials is a simple modification of the expressions developed in Chapter 6. It is only necessary to divide the strength property representation of the material designated by either the yield or ultimate stress that normally appears on the right side of the various ductile theory equations by the factor of safety “n”. In the equations shown here the material strength parameter in the first three theory relations has been placed in the denominator of the left side so that the right side simply becomes one over the factor of safety “n”. It should be noted that in the modified Coulomb Mohr theory where the material properties in tension and compression already appear on the left side of the equation that the right side is again simply one over the factor of safety “n”. This implies that the same factor of safety is being applied to both the tensile and compressive properties.

8. Exercise Problem In this first interactive exercise problem you are to determine the factor of safety for a shaft in pure torsion applying the maximum shear stress theory where the twisting torque may vary by plus 30%, the diameter is known to within 2% and the yield stress in tension is assumed to be within plus or minus 7% of the nominal value listed in a table of properties. When you have completed your calculation click on the solution button to check your answer. Then continue on by use of the return button.




9. Generic Factors of Safety Unfortunately, variations in the parameters representing the loading, geometry and material properties are seldom known precisely at the beginning of the design process. This makes it virtually impossible to establish apriori what the specific magnitude of the factor of safety should be. To circumvent this difficulty accumulated design experience and judgment over many years has lead to suggested recommendations for possible factors of safety to be used based on generic classification of such general characteristics as material properties, loading and application environment. One such table of recommendations is shown on this page taken from the 20th edition of Machinery’s Handbook. Note how broad and generic the classifications are. Their precise interpretation is of course left up to the judgment of the designer. It is to be observed that in the two numerical examples already covered both would seem to fit into the category of the materials properties being know to a level of high reliability and the load and environment as being not severe, whatever that means, and that the weight of the part has significant importance. You can be sure that these kinds of recommendations are very strongly conservative irrespective of how they are interpreted.

10. Exercise Problem –2 In this exercise the parameters of the specified part design are expressed in nominal terms for the loading, geometry and the yield property of the material. Your task is to determine whether the part will yield where the stresses are maximum and if yielding does not take place what minimum factor of safety is provided by the design. Assume that the part is made of a ductile material and that the load is constant. After you have completed your analysis click on the solution button to check your conclusions. When you are satisfied with the solution click on the return button to continue to the next page of the chapter.




11. Actual Load Distribution When a part is designed and fabricated in very large numbers for the same application it is of value to apply statistical considerations to the concept of a factor of safety. The remainder of this chapter will be devoted to this subject. Begin by considering that the loads to be carried by the total population of parts will have some probability distribution around a mean value of the load designated by L bar. Any specific load L in this distribution will fall between the limits of L bar plus or minus delta L representing the maximum variation of the load from the mean value. In general the natural probability distribution of L will look something like the bell shaped curve in the figure. That is, it is expected that the probability of L will be highest around the mean value and will tail off in both the positive and negative directions from L bar approaching zero at L bar plus or minus delta L.

12. Load capability In a similar fashion the load capability of the part, that is, the ability of the part to carry a load L will also possesses a probability distribution due to variances in its geometric and material property values. Again the distribution is defined in terms of a mean load capability Lc bar and variance of plus or minus delta Lc as depicted in the figure. The height and breadth of the distribution may be different from that of the actual load distribution but its appearance will again may be expected to be bell shaped for a very large number of parts.



13. Comparison of Distributions Now consider the interpretation of these two distributions plotted beside each other on the same axis system. It is expected that the mean value of the load capability Lc bar will be greater than the actual mean load L bar to prevent part failure. However, if any portion of the two distributions overlap as shown in the figure it means that there are parts that will be subjected to loads that are greater than the available load capability of some parts. Hence, the region of overlap represents potential failure. Whether such a region exists in a specific instance is a function of the mean values of the actual load and the load capability and the variances in both these parameters. If the mean actual load and load capability are interpreted as nominal load and load capability then it is appropriate to define a factor of safety as the ratio of Lc bar to L bar as was done earlier in the chapter.

14. Difference of Distributions It is now convenient to consider the interpretation of the difference of these two distributions. This difference will also be a bell shaped probability distribution as depicted in the figure. The distance from the origin to the peak value of the distribution will just be equal to the difference in the mean values of the separate distributions. If a portion of the distribution extends over into the negative side off the origin the area under that portion of the curve represents the portion of parts subject to failure from the total population, which is represented by the total area under the total combined distribution curve.



15. Gaussian (Normal) Distribution) The assumption is now made that the bell shaped probability distribution of the difference between the load capability and the actual load curve can be represented as a Gaussian or normal distribution. This permits some of the mathematics of statistics to be introduced into this analysis to create an analytic model for predicting numerical factors of safety where large populations are involved. The normal probability density distribution y of a variable x is defined by the exponential function adjacent to the curve in the figure. In this exponential function x bar is the arithmatic mean of the variable x as N approaches infinity as indicated by the accompanying equation. The symbol Dx is called the standard deviation and is defined as the square root of one over N minus 1 times the sum over N of the quantity the difference of the variable x and its mean value squared. The most important property of the standard deviation in this development is that the area under the probability density curve within three standard deviations on either side of the mean value of x represents 99.73 percent of the total population under the entire probability curve from minus to plus infinity.

16. Transformation of Variables A transformation of variables is now introduced to put this statistical model into a form in which the results of integrating the normal distribution over some portion of its variable can be obtained from available standard tables of numerical results. The importance of this will be clarified shortly. The transformation introduced is a new variable t that is equal to x minus x bar divided by the standard deviation Dx. It is demonstrated on this page that the mean value t bar of the variable t is zero and that its standard deviation Dt is simply 1. Hence, the transformed probability distribution of t, which is also normal, is uniformly distributed around the origin of the axis system. The transformed normal distribution of the variable t is depicted in the figure. It is seen as centered on the origin since its mean value is zero. The area under the portion of the total distribution represented by the horizontal coordinates plus and minus 3 constitutes 99.73 percent of the total population. In this instance the total area under the curve is unity. The shaded area that corresponds to the integral from minus infinite to minus tf represents the decimal percentage of all points in the population with values less than minus tf. The importance of this shaded area as it relates to factor of safety will be made clear on the next page.



17. Transformed Normal Distribution The transformed probability distribution is now expressed more simply as y equal to one over the square root of two Pi multiplied by e raised to the power of minus t squared over two. The curve is now observed centered with respect to the origin with its mean value at t equal to zero. With a standard deviation of one the area under the portion of the curve between t equal to –3 to t equal to +3 represents 99.73 % of the total area under the curve from minus infinity to plus infinity. The total area under this transformed distribution is simply unity. The shaded area under the portion of the curve from minus infinity to –tf represents the decimal % of all points with values less than –tf. The parameter Tf plays a very important role in the statistical definition of factor of safety as will be demonstrated on the next two slides.

18. Application to Lc-L distribution To relate these properties of the unit normal distribution to factor of safety the variable t is now written in terms of the parameters of the difference of the load capability and actual load probability distribution assuming this distribution is also normal. That is, t is now defined as the quantity Lc minus L minus the quantity Lc bar minus L bar all divided by the standard deviation of the Lc minus L distribution. This standard deviation can itself be written as the square root of the sum of the squares of the standard deviations of the individual Lc and L distributions. For proof of this refer to any standard text on statistics. Now set t equal to -tf for which Lc minus L is equal to zero. Click on the recall button to pop the previous graph of Lc minus L to see that this is true. Having satisfied yourself that this is the case substitute these conditions into the general expression for tf, solve the result for Lc bar and divide both sides of the equation by L bar. Thus the left side of the resulting equation is simply the factor of safety “n” defined as the mean load capability divided by the mean actual load and the right side of the equation contains terms which define the characteristics of the Lc and L distributions multiplied by tf. Appropriate application of this equation permits the determination of a factor of safety for a large population of parts whose load characteristics approximate normal probability distributions.



19. Determination of tf To apply the equation from the previous page to calculate a factor of safety it is first necessary to determine the appropriate value of tf. Recall that tf is the negative upper limit of the integral of the area from minus infinity of the unit normal distribution that now represents the decimal percentage of failure due to Lc being less than L. Unfortunately this integration cannot be carried out in closed form. Hence, it is necessary to refer to numerical tables that list the value of the integral for specific values of tf. A portion of such a table is presented on this page in which values of the integral are given for values of tf from 2.00 to 3.08 in intervals of .02. As an example if a specific design instance will permit a failure potential of 2% or two parts per hundred the value to be used for the integral of the area representing this failure level is 0.02. The closest value of tf from the table corresponding to this integral value is approximately equal to 2.04. This is the value of tf that would be used in the equation for determining the associated factor of safety depending of course on the values of other distribution parameters in the equation. by the direction of epsilon 2 and an axis perpendicular to it. Note that the construction results in angles of two theta of 90 degrees between the diameter for axes 13 and the direction of epsilon three, which agrees with the fact that on the rosette axes 1,2 and 3 are separated by 45 degrees.

20. Approximate values of tf For low values of acceptable failure rates a convenient approximation for tf can be employed. First define Fd as the percent decimal failure. That is, Fd is equal to the integral of the unity distribution function from minus infinity to minus tf. Then for Fd between .001 and .015 that represents failure rates from .1% to 1.5% tf can be approximated by the equation tf equal to 1.29 divided by Fd raised to the 0.128 power. This equation for the limits previously specified is accurate to within 2% error. This approximation is then substituted into the factor of safety equation resulting in a relationship that includes the decimal percent failure rate directly along with the other parameters describing the actual load and load capability distributions.



21. Sample Problem This statistical approach for determining a factor of safety will now be applied to a specific problem. Assume that a bar in tension is to carry a load of 1000 lbs. with a possible variation of +/- 25 %. Thus the maximum load capability must be at least 1250 lbs. However, the load capability can only be known to within +/- 10 % variation. Assume that a sufficient number of parts are to be produced such that these variables can be considered normally distributed. Calculate the necessary factor of safety to be applied to the mean values of load and load capability if the failure rate is not to exceed 2 %.

22. Problem Solution From the conditions of the problem L bar is1000 lbs. and delta L is 250 lbs. The value of Lc bar should be at least 1250 lbs. To be a little conservative it will be assumed to be 1300 lbs. Its variance will then be 130 lbs. The standard deviation for the actual load will be delta L over three or 83.3 lbs. while the standard deviation for the load capacity will be delta Lc over three or 43.3 lbs. A value of tf is now needed for a failure rate of 2%. Click on the % failure button to pop up the numerical chart from which it is seen that tf is just about 2.04. Substituting all these values into the equation for the factor of safety and carrying out the indicated mathematical manipulations gives a final answer of about 1.2. This value would be applied to design calculations based on the nominal values of actual load and load capability.



23. Alternate form for n An alternate formulation for n in terms of just the variations in actual load and load capability can be developed by manipulating the form of the equation used in the previous problem. Starting with the equation on the first line on this page the standard deviations are first replaced with delta Lc over 3 and delta L over 3. The L bar from the denominator of the second term is taken inside the square root term. It is then recognized the term delta Lc over L bar can be rewritten as delta Lc over Lc bar times the factor of safety n. This permits the starting equation for n to be written in the form given at the bottom of the page.

24. Alternate form for n (continued) The last equation from the previous page is rearranged so that only the square root term remains on the right. This expression is now squared and the common terms of n and its powers are combined. This results in the quadratic equation for n at the bottom of the page in which the coefficients are only functions of the variations of the actual load and the load capability along with a measure of the failure rate in the parameter tf. One further modification that could be made to this equation would be to replace tf by it equivalent representation in terms of the decimal percentage failure Fd provided Fd is limited to a failure rate between .1 to 1.5 %. This is left for the reader to do if desired.



25. Exercise Problem - 3 This exercise deals with determining the factor of safety “n” for the sample problem just solved using the alternate form for “n” as a quadratic equation. Recall that the mean applied load was 1000 lbs. with a variation of 25% and the mean load capability was taken to be 1300 lbs. with a variation of +/- 10 %. The acceptable failure rate was 2% for which tf was 2.04. Compare the result with the solution for “n” obtained in the sample problem. When you have calculated your answer click on the solution button to check your result. When finished with the solution click on the return button to continue on to the next page.


26. “No Failure” Scenario A relationship will now be developed for “n” for the case in which the failure rate is to be zero. This condition is defined by the inequality that L bar plus delta L must be less than or equal to Lc bar minus delta Lc. In other words the maximum actual load must be less than the minimum load capability. This inequality is first solved for Lc bar. Then both sides of the equation are divided by L bar. The last term of delta Lc over L bar is rewritten as delta Lc over Lc bar times n the factor of safety. The resulting expression is then solved for n giving the final relation that “n” is greater than or equal to the ratio of one plus delta L over L bar to one minus delta Lc over Lc bar. Note that this equation only involves the percentage variation of the actual load and the percentage variation of the load capability.



27. Example Problem –4 In this final chapter 7 exercise determine the factor of safety for the previous problem assuming the “no failure scenario”. Recall again that the mean applied load is 1000 lbs. with a variation of +/- 25% and the mean load capability was taken to be 1300 lbs. with a variation of +/- 10 %. Compare this “no failure” factor of safety with the value calculated for a 2 % failure rate. When you have calculated your answer click on the solution button to check your result. When finished with the solution click on the return button to continue on to the next page.

28. Review Exercise In this exercise the items in the list on the left are to be matched with the symbols and mathematical relationships on the right. Place the cursor over an item on the left and hold down the left button. A pencil will appear that can be dragged to one of the green dots on the right. If the right choice is made the arrow will remain. If the selection is incorrect the arrow will disappear. After the exercise is completed proceed to the next page.



29. Off Line Exercise An established design procedure calls for the use of a factor of safety of 2 for a class of specific non-moving parts used in an automotive suspension system. If the material properties variation is expected to be 10% determine the allowable projected load variation for zero failure potential based on the “no failure” scenario. How does this load variation change if 1 part per 100 or 1 part per 1000 is the prescribed failure rate? When you have finished with this statement click on the exit or main menu button to leave the chapter.




Chapter 7 Factors of Safety

Screen Titles

Problem 1 - Solution Problem 1 – Solution (cont.) Problem 2 - Solution Problem 2 – Solution (cont.) Problem 2 – Solution (cont.) Problem 2 – Solution (cont.) Problem 3 - Solution Problem 3 – Solution (cont.) Problem 4 - Solution


_____________________________________________________________________________________Factors of Safety - 164 - C.F. Zorowski 2002



1. Problem 1 - Solution For a shaft in pure torsion the nominal principal stress sigma one is equal to the maximum shear stress defined by TR over J with J equal to Pi R fourth over two. This give sigma one nominal equal to two T over Pi times R cubed. It then follows that sigma one max is equal to two T max divided by Pi times R min cubed. Substituting T max and R min into this expression gives sigma one max equal to 1.34 times sigma one nominal. The problem also states that sigma yield min is equal to .9 sigma yield nominal.

2. Problem 1 Solution (cont.) From the maximum shear stress theory of failure sigma one max must be less than or equal to sigma yield min over two. The previous expressions for sigma one max and sigma yield min are now substituted into this inequality. The factor of safety is introduced with the equation that sigma one nominal is given by sigma yield nominal divided by 2n to satisfy the maximum shear stress theory of failure. Using this equation to eliminate sigma one nominal from the previous inequality allows n to be determined as the ratio of 1.34 to .9 for a final result of 1.49. When finished with this solution click on the return button to go to the next page in Chapter 7.



3. Problem 2- Solution The location of the maximum bending stress in the bar is where the bar and rod are welded together. This stress is equal to Mc over I where I is the moment of inertia of the rectangular cross section given by b times h cubed over 12. The magnitude of the moment M is the force, 30 lbs. times the length of the bar, two inches, giving 60 in. lbs. With the given dimensions of the cross section the I is calculated to be 1.1 times ten to the minus three inches to the fourth power. Taking c to be half the height of the section the maximum bending stress becomes 10,230 psi. Since this is the only normal stress in the bar the applicable theory of failure is the normal stress theory. With the yield stress in tension for the material in the bar being 50,000 psi the factor of safety for this portion of the part is determined to be 4.9

4. Problem 2 – Solution (cont.) Now consider the stresses at the fixed end of the rod. These will consist of a shear stress due to the twisting torque created by the force F and a bending stress resulting from the bending moment due to the force F. Both the twisting torque T and the bending moment M have the same numerical value of 60 in. lbs. Since the rod is a quarter inch in diameter its polar moment of inertia J is calculated to be .35 time ten to the minus three inches to the fourth. The moment of inertia needed for the bending stress calculation is just half the value of J since the cross section is a solid circle. Substituting these values into the Mc over I and Tr over J equations for the stress gives a shear stress of 21,420 psi and a bending stress of 42,840 psi.



5. Problem 2 – Solution (cont.) Next the principal stresses must be calculated before applying an appropriate theory of failure to determine the factor of safety for this portion of the part. Since there is no normal stress perpendicular to the bending stress the equation for the maximum principal stress is given by the equation at the top of the page. Substituting the appropriate values for the normal stress and the shear stress into this equation gives sigma one equal to 51,700 psi. In a similar fashion the minimum principal stress is calculated to be –8,900 psi. With the two principal stresses being opposite in sign the appropriate theory of failure to apply is the maximum shear stress theory. Substituting the numerical values of sigma 1, sigma 2 and the yield stress of 75,000 psi into the equation for the maximum shear stress with the factor of safety included results in a numerical value for the factor of safety of 1.24. This is less than the value calculated for the bar section and thus is the minimum factor of safety for the part. When you have finished with this solution click on the return button to go to the next page in Chapter 7.

6. Problem 3 – Solution The solution is begun by recalling the equation for n in quadratic form. The fractional variations for the load capability of .10 corresponding to 10% and for the actual load of .25 representing 25 % are substituted into this equation along with the value of tf of 2.04 from the sample problem. This permits the coefficients of the n terms in the quadratic equation to be numerically determined. The resulting equation is then divided through by the coefficient of the n squared term and the quadratic formula is applied to determine a final value for n of 1.19 or approximately 1.2. It is observed that this is the same result as determined in the sample problem without an assumption being required as to the magnitude of the mean value of the load capability. When you have finished with this page click on the return button to go to the next page in Chapter 7.



7. Problem 4 – Solution To calculate the factor of safety for the sample problem for the condition of a zero failure rate requires the application of the equation for the “No failure “ scenario. The only parameters needed to carry out this application numerically are the fractional variations of the actual load and the load capability. Respectively these values are .25 and .10. Substituting these values into the “No failure” scenario relationship gives a numerical value for the factor of safety of 1.39. Compared with the value of 1.2 calculated for n with a 2 % acceptable failure rate this represents a 16 % increase in the factor of safety. When you have finished with this page click on the return button to go to the next page in Chapter 7.


_____________________________________________________________________________________Fatigue Strength and Endurance - - C.F. Zorowski 2002 169

Chapter 8 Fatigue Strength and Endurance

Screen Titles

Fatigue Failure Process – 1, 2 Fatigue Strength Testing S-N Diagram Fatigue Strength Equation – 1, 2 Endurance Limit Behavior Endurance Limit Equations Modifying Factors Surface Factor Surface Factor Impact Size Factor Size Factor Impact Load Factor Temperature Factor Miscellaneous Effects Stress Concentration Example Problem Problem Solution Review Exercise Off Line Exercise



1. Title page Chapter 8 begins the study of the behavior of mechanical elements subjected to fluctuating stress states that can produce fatigue failure. This chapter deals with the fatigue strength and endurance of materials. The subject is introduced by a generic description of the fatigue process. The subject of fatigue testing, cycle life behavior and endurance limit is considered in conjunction with estimating the endurance limit from the tensile strength of a material. Consideration is then given to factors that modify the endurance limit like surface condition, part size, load types, operating temperature and other miscellaneous effects. Several exercise problems and an extended sample problem is included to demonstrate the application of the subject content.




3. Fatigue Failure Process Fatigue behavior in mechanical components is associated with the application of reversed or fluctuating stress states as contrasted to static loading behavior already studied. The number of reversed cycles of loading must be at least 103 repetitions or greater for this behavior to become important. The associated failure mechanism in fatigue begins with the initiation of a small crack on the surface of the part. This may be due to some surface imperfection from processing particularly at sudden changes in geometry. The crack itself produces a high stress concentration that leads to continued growth of the crack.

4. Fatigue Failure Process (continued) As the crack enlarges the remaining stress-bearing area of the part reduces in size and the applied stress levels increase. With additional cyclic loading this conditions worsens and failure of the part occurs. This failure is sudden and catastrophic. The broken surface has two appearances. Where the crack initiated and grew the surface appears shiny and almost hammered. The surface where the final fracture takes place displays the appearance of a brittle material that has failed in tension. The stress level at which fracture take place is generally significantly below the yield stress of the material. Hence fatigue is truly a different mechanism of failure from what is experienced by material pulled in tension through yielding on to the ultimate strength and final fracture.



5. Fatigue Strength Testing To determine the fatigue strength of a material it is necessary to test the material to failure in a very different way from that used to determine its ultimate strength in tension. The technique employed makes use of a test specimen that is rotated at high speed subject to a pure reversed bending load. The dimensions and geometry of a common standard fatigue test specimen is shown in the figure with the loading forces in red. Tests to failure are conducted with such specimens on a Moore high-speed rotating beam machine at a number of increasing loads. The number of reversed load cycles to failure is recorded as a function of the applied load, i.e. maximum applied pure bending stress. Several tests are required at each load level to account for the statistical nature of the failure mechanism.

6. S-N Diagram The results of the rotating beam specimen for steels behave generically as illustrated in the figure where fatigue strength in psi is plotted as a function of cycles to failure. This is referred to as an S-N diagram where S corresponds to stress and N corresponds to the number of cycles to failure. At one cycle or N equal to 100 the fatigue strength is taken as equal to the ultimate strength in tension of the material. As the loading cycles increase the maximum stress level at which failure occurs decreases. Over the region of 1 to 103 or 1000 cycles, referred to as low cycle fatigue, the reduction in the strength figure is very little. For N in the range of 103 (one thousand) to 106 (one million) cycles the reduction in the fatigue strength is much more significant. Repeated loading above 103 cycles is considered high cycle fatigue. At some high cyclic level, usually around 106 or 107 load repetitions the S-N curve exhibits a distinct knee, becomes horizontal and the strength no longer decreases. This knee in the curve corresponds to the fact that a specimen at this stress level will effectively exhibit infinite life. The level of fatigue strength associated with this phenomenon is referred to as the endurance limit of the material and is measured as a stress in psi. Effectively all ferrous material and alloys exhibit this endurance limit behavior but non-ferrous materials do not. Their fatigue strength continues to decrease beyond 106 cycles but normally at a reduced rate.



7. Exercise Problem -1 It is somewhat difficult to conceptualize the true difference between 103 and 106 events, which have great significance in fatigue, without attaching some physical significance to such events in terms of every day experiences. The two exercises presented on this page will help accomplish this. Work out both of them and check your answer by clicking on the solution button before going on.

(Solution 0n Page 187)

8. Fatigue Strength Equation To carry out numerical analyses and predictions of fatigue behavior it is necessary to adopt some mathematical model to represent the generic behavior depicted on the S-N diagram previously described. For 103 to 106 cycles of load repetition, which is considered the most important region in high cycle finite fatigue life, the relation of fatigue strength to cycles to failure is assumed to be a straight line on a log-log plot. That is, the fatigue strength, sigma f, is represented as a constant “a” times N, the cycles to failure, raised to some power “b”. The numerical values of “a” and “b” are dependent on the specific properties of the material in question. To determine “a” and “b” two conditions are assumed. The first is that at 103 cycles the fatigue strength, sigma f, is assumed to be equal to 90% of the ultimate tensile strength in tension of the material. This is consistent with the S-N diagram behavior where the decrease in fatigue strength from the ultimate strength is quite small in the low cycle region. The second condition is that at 106 cycles the fatigue strength will be equal to the endurance limit of the material designated as sigma e. The log of both sides of the assumed fatigue behavior equation are now taken and the conditions representing the two ends of this line are substituted in to give equations 1 and 2 at the bottom of the page.



9. Fatigue Strength Equation (cont.) Equation 2 is now subtracted from two times equation 1. This eliminates b from the resulting equation permitting “a” to be determined. It is seen that the final result for “a” is dependent on the tensile strength and the endurance limit of the material in question. To determine “b” equation 2 is again subtracted directly from equation 1 to eliminate “a”. Solving for “b” gives the result shown that is again dependent on the tensile strength and the endurance limit of the material. The final result at the bottom of the page can be used to determine sigma f for any N between 103 and 106 cycles or conversely if given a specific fatigue stress to which the material is subjected it finite life in cycles can be determined. It should be noted here that the minimum ultimate strength in tension sigma u should always be used in this equation for fatigue strength.

10. Endurance Limit Behavior Since endurance limit values are not easily obtainable it is both desirable and convenient to have some means of relating the endurance limit to the ultimate tensile strength of the material since data on ultimate strength is usually more readily available. It might be anticipated or at least hoped for that the endurance limit might be somehow directly related to ultimate strength. As the ultimate tensile strength increases it seems reasonable that endurance limit might increase also. Consider a plot of endurance limit test results versus tensile strength for a variety of ferrous materials and alloys. The graph shown here depicts three possible linear relationships where the ratio of tensile strength to endurance limit is 0.4, 0.5 and 0.6. Tensile and rotary beam tests on a variety of wrought iron specimens gives the blue cluster of results as shown on the graph. Similarly, test results from a variety of carbon steels gives the two red clusters depicted on the graph. Finally, tests on a variety of alloy steels gives the two green clusters of results on the graph. It is observed that up to a tensile strength of 200 kpsi the blue, red and green clusters seem to average out close to the sigma e over sigma u line of 0.5. Above a tensile strength of 200 kpsi the results from both the carbon and alloy steels appear to flatten out quite appreciably. Based on these results a conservative model has been adopted and is generally used for estimating the endurance limit of a ferrous material based on its ultimate tensile strength.



11. Endurance Limit Equations From statistical considerations of a large number of test results of ferrous materials it has been determined that for materials with ultimate tensile strengths below 200 kpsi a reasonable assumption for estimating their endurance limit is given by the equation sigma e is equal to .504 times sigma u. In this relationship the endurance limit sigma e is what would be anticipated from a rotating beam test as described earlier. For materials with tensile strengths in excess of 200 kpsi it is generally assumed that the endurance limit remains constant at 100 kpsi. This method of estimating the endurance limit is of course approximate but does provide the means for carrying out analytical fatigue analysis of designs where exact material properties in fatigue may be questionable or unavailable.

12. Exercise Problem – 2 This exercise provides an opportunity to apply the material and models just covered to estimating endurance limit, the fatigue strength for finite life and the life associated with a given application of fatigue loading. When finished with your analysis click on the solution button to check your numerical results. Then proceed on to the next page in the chapter.




13. Modifying Factors The endurance limit discussed so far has been the test value either obtained directly from rotating beam tests or estimated as such from the tensile strength of the material. This value of endurance limit is in almost every instance higher than that of an actual part due to a number of factors not included in the beam test conducted under ideal standard conditions. This is represented analytically by multiplying the rotating beam endurance limit by a number of k factors to give the endurance limit of the part. To begin with a rotating beam specimen is polished longitudinally to reduce any surface imperfection that might initiate a crack. The rougher condition of processed real part surfaces will always reduce the value of the endurance limit. Other factors that affect the actual endurance limit include the size of the part, the type of loading it is subjected to, the temperature at which it operates and other miscellaneous conditions including stress concentration factors due to changes in geometry. Each of these factors will now be examined and treated separately.

14. Surface Factor The surface condition factor is represented analytically by the equation ka is equal to a constant “a” times the tensile strength raised to the “b” power. Values of “a” and “b” for different processed part surface conditions are indicated in the included table. As might be expected the value of ka is always less than one and decreases in value proceeding from a smooth ground surface finish to a rough forged part surface. The condition of the part surface has one of the greatest impacts on the reduction of the ideal endurance limit value.



15. Surface Factor Impact To illustrate the effect of the surface factor consider the value of ka for a part with a machined surface and a tensile strength of 65 kpsi. Applying the formula and values of “a” and “b” from the previous page gives a value of ka of 0.894 or an 11 % reduction in the endurance limit. If the same part were forged the ka value would drop to 0.627 or a 37 % reduction in sigma e. From the values in the table it is seen that increasing the tensile strength of the material reduces ka even further to the point that a forged part with a tensile strength of 125 kpsi reduces the endurance limit of the material by more than 65 %. As mentioned previously surface condition has one of the largest effects on endurance limit. Why do you think this is the case?

16. Size factor Somewhat surprising is that the over all size of the part can also effect its endurance limit. For a solid rotating shaft up to 2 inches in diameter the size factor Kb is given by the expression the quantity the diameter divided by 0.3 raised to the –0.1133 power. Again the kb factor will be less than one. For solid-rotating shafts greater than 2 inches in diameter a value of kb between 0.6 and 0.7 should be used with the smaller value employed for a more conservative design. For hollow or non-rotating shafts an effective diameter equal to 0.370 of the outside diameter should be used in the previous equation. For a rectangular cross section the effective diameter is given by 0.808 times the base multiplied by the height with the product raised to the 0.5 power. For axial loads there is no size factor effect so that Kb is one.



17. Size Factor Impact The numerical effect of the size factor is demonstrated by calculating kb for a rotating solid shaft 2 inches in diameter subjected to a bending load. Substituting 2 inches into the kb formula gives a result of 0.81 or a 19 % reduction in endurance limit by itself. However, if coupled with a machined surface part for which Ka is 0.89 the combined effect is a multiplier of 0.72 or a 28% reduction in endurance limit. Now compare the value of kb for the solid shaft with a hollow shaft. The effective diameter is .74 giving a value of Kb of 0.90, which is higher than for the solid shaft. Of course the stress for a given bending load would also be higher in the hollow shaft. Finally consider a rotating solid shaft with a diameter of .5 inches. In this case the size factor calculates out to be 0.94 or only a 6 % reduction.

18. Load Factor This factor has one of the least effects on the endurance limit of the part except for the case of pure torsion. The value of the factor for axial load is dependent on the tensile strength of the material. For materials with tensile strengths below 220 kpsi the value of kc is taken to be 0.923. For materials with tensile strengths above 220kpsi the value of kc is unity. For bending loads Kc is taken as unity for all values of tensile strength. For torsion or direct shear the kc value is given as 0.577 for all tensile strengths. Does this number look familiar? How about the distortion energy theory?



19. Temperature Factor The behavior of endurance limit with operating temperature of the part is assumed to be similar to that of the tensile strength of the material. Contrary to what might be expected tensile strength actually increases slightly with operating temperature compared to room values up to between 400 and 500 degrees F. The value of kd for the temperature effect on endurance limit is taken to be equal to the ratio of tensile strength at operating temperature to tensile strength at room temperature as given in the table on this page. Note that the reduction impact of this factor does not become very significant until operating temperatures are above 600 degree F.

20. Miscellaneous Effects Factor Within this category are grouped effects that include residual stress, corrosion, electrolytic plating, metal spraying, cyclic frequency, fretting corrosion and stress concentrations. Compressive residual stresses increase fatigue strength. This accounts for why shot peening, surface hammering and cold working of surfaces are employed. Corrosion tends to reduce fatigue strength as it leads to a rougher surface on which cracks can more easily initiate. Plating can reduce fatigue life by as much as 50% while the reduction effect of metal spraying is more like 15 %. The effect of frequency changes or the rate of loading is usually considered minimal but may take on more importance at elevated temperatures and with corrosion. Frettage corrosion due to microscopic movement at tight joints like press fits or bolted connections can have an appreciable effect but is extremely hard to quantify. Finally stress concentrations introduced by sudden changes in part geometry and shape will also reduce fatigue strength and can be handled more quantitatively than the other miscellaneous factors. This will be treated in more detail on the next page.



21. Stress Concentrations For low cycle fatigue or static loads, stress concentration effects can generally be neglected unless the part material is brittle. For cyclic loading above 106 cycles the factor ke is expressed as the reciprocal of the geometric stress concentration factor Kf. The appropriate values of Kf must be obtained from references dealing specifically with the topic of stress risers due to changes in part geometry and the type of applied loading. This will not be dealt with here. For cyclic loading between 103 and 106 cycles the Kf prime factor is given by “a” times the tensile strength raised to the “b” power where “a” is one over Kf and “b” is minus one third of the log of one over Kf. This is similar to the model used for calculating finite life fatigue strength between a thousand and a million cycles of loading.

22. Example Problem All of the previous subject content will now be applied to a realistic example problem. A ¾ inch diameter shaft that will be operate at 1800 rpm under a constant bending load is machined from steel with the properties given in Exercise Problem 2, that is the tensile strength is 97 kpsi and the yield strength is 75 kpsi. The shaft will operate at room temperature. It contains a radial hole that introduces a stress riser of 1.5. Estimate the values of the modifying factors and determine the endurance limit for the part for the condition specified. This problem will be worked out numerically on the next three pages.



23. Problem Solution - 1 The solution is begun by estimating the rotating beam specimen endurance limit sigma e. Since the tensile strength is less than 100 kpsi the value of sigma e is obtained by multiplying the tensile strength by .504 to give 48.8 kpsi. Next the modifying factors are each calculated for the given problem conditions. With the surface of the shaft being machined the surface factor ka is given by 2.7 times the tensile strength 97 raised to the –0.265 power. This gives a ka value of 0.802.

24. Problem Solution - 2 With the problem specifying that the shaft is rotating and subject to a bending load the size factor kb is given by the shaft diameter of .75 inches divided by 0.3 with the quotient raised to the –0.113 power. This gives kb a value of .902. Since the loading is bending and the operation is at room temp both the loading factor, kc, and the temperature factor kd are equal to 1.



25. Problem Solution - 3 The stress concentration factor ke is given by the reciprocal of the geometric stress concentration factor of 1.5 due to the hole through the shaft. This gives a value for Ke of .667. The final value for the modified endurance limit is then the product of all the modifying factors times the estimated endurance limit of the rotating beam specimen. Carrying out these multiplications gives a value of 23.7 kpsi for the endurance limit of the problem shaft. Note that this is a reduction of more than 50 % of the estimated test specimen material property. This illustrates quite clearly how the geometric, loading and operating conditions of the part can significantly impact its finished fatigue strength.

26. Review Exercise In this review exercise you are to type in answers from the keyboard into the blank spaces. Clicking on the tab key will provide immediate feed. If the answer is incorrect another response can be entered. A subsequent click on the tab key will move the curser to the next blank. Hot words are provided in each question to revisit the page in the chapter on which the correct answer can be found. After completing the exercise click on the next page button.



27. Off Line Exercise A rotating cantilevered beam stepped shaft supported by two bearing carries a load of 175 lbs. as indicated. The shaft surface is machined and ground from a steel with a yield stress of 95 kpsi and a tensile strength of 105 kpsi. The step in the shaft is radiused introducing a stress concentration factor of 1.30 and the part will operate at room temperature. Estimate the life of the part. When you have finished with this problem statement click on the exit or main menu button to leave the chapter.



_____________________________________________________________________________________Fatigue Strength and Endurance - 185 - C.F. Zorowski 2002

Chapter 8 Fatigue Strength and Endurance

Problem Solutions

Screen Titles

Problem 1 Solution Problem 1 Solution (cont.) Problem 2 Solution Problem 2 Solution (cont.)



1. Problem 1 - Solution To determine the distance traveled by a car in miles per million revolutions of its engine begin with the average speed of the car, 30 mph. Then multiply by the reciprocal of the minutes per hour and the reciprocal of the revolutions per min. This gives the units of miles per revolution on the right side of the equation. Multiplying both sides by ten to the sixth revolutions gives a final answer of 200 miles. A rather significant distance at this average speed. Contrast this with the distance associated with a thousand cycles that gives a distance of .2 of mile or just slightly more than a thousand feet. Similarly the time required to travel the 200 miles is obtained by dividing the distance by the average speed of 30 mph giving 6.7 hours while the time required to travel .2 miles is less than half a minute. Not surprising since this corresponds to 1000 cycles of an engine turning over at 2500 rpm.

2. Problem 1 Solution (cont.) For a clock ticking one cycle a second the time required for a million ticks is obtained by multiplying one cycle per second times the reciprocal of 60 sec per min times the reciprocal of 60 min per hour times the reciprocal of 24 hr per day all multiplied by ten to the sixth clicks. This gives the result of 11.57 days. In contrast the time required for a thousand ticks is just over one quarter of an hour. The number of ticks in a year is calculated to be over 30 million. Hopefully, these examples provide a better concept of how small 103 events is compared to 106 repeated occurrences particularly as it applies to design for fatigue strength and endurance. When you have finished with this solution click on the return button to go to the next page in Chapter 8.



3. Problem 2- Solution Since the tensile strength is less than 200 kpsi then for part a the estimate of the endurance limit is simply given by .504 times the tensile strength. This gives an endurance limit of 48.8 kpsi. To calculate the fatigue strength at 105 cycles the factors “a” and “b” must first be determined using the given tensile strength and estimated endurance limit. Carrying out the required calculations gives “a” equal to 156 kpsi and “b” equal to minus .084. Substituting these values into the equation for fatigue strength at a finite life gives a final value of 59.3 kpsi for sigma f. Note that this value lies between sigma e and sigma u as it should.

4. Problem 2 – Solution (cont.) The finite life fatigue strength equation is again used to determine the life in cycles to failure for an applied reversed stress loading of 63 kpsi. The values of “a” and ”b” remain the same as on the previous page. Solving for N with sigma f specified results in a finite life value of 48 time 104 cycles. When you have finished with this solution click on the return button to go to the next page in Chapter 8.


_____________________________________________________________________________________Fluctuating Load Analysis - - C.F. Zorowski 2002 189

Chapter 9 Fluctuating Load Analysis

Screen Titles

Fluctuating Stresses Generic Stress-Time Behavior Stress-Time Relations Modified Goodman Diagram Mean/Fluctuating Stress Diagram Soderberg Failure Theory Goodman Failure Theory Gerber Failure Theory Sample Problem – 1 Problem – 1 Solution Torsional Fatigue Combined Loading Modes Sample Problem – 2 Problem – 2 Solution Cumulative Fatigue Damage Palmgren-Minor Theory Sample Problem – 3 Problem – 3 Solution Mason’s Modification Review Exercise Off Line Exercise



1. Title page Chapter 9 continues the study of the fatigue analysis and behavior of mechanical parts. Attention is now directed to elements subjected to general fluctuating loads as contrasted to complete reversed stress states treated in Chapter 8. The presentation begins with defining generic parameters to represent fluctuating loads. General fatigue failure under fluctuating normal stress loading is then discussed and three specific failure theories are introduced together with fatigue under fluctuating torsional stresses. A method of fatigue analysis for combined stress states is also covered. The chapter concludes with two methods of determining the effects of cumulative fatigue damage due to multiple loads applied for different durations of cyclic application. Several exercise problems and extended sample problems are included to demonstrate the application of the subject content.




3. Fluctuating Stresses Chapter 8 considered the fatigue behavior of mechanical elements subjected to complete reversed stress states in which the stress level varied from some specific negative value to the same positive value and then repeated this behavior continuously. This chapter will deal with repeated stress states that may vary from zero to some specific value and then back to zero continuously or even more generally from some minimum value to a different numeric positive value and then back to the minimum value repeatedly. In this instance the variation will possess some mean value of stress that will lie between the minimum and maximum value of the applied stress state.

4. Generic Stress Time Behavior Depicted in the figure is an idealized general fluctuating stress versus time behavior curve. Although this is represented as a smooth sinusoidal function the parameters that will be used to generalize its description will be the same irrespective of the specific shape of the curve provided it is periodic in behavior. In other words the shape could be saw tooth or a square wave, it would still be characterized for fatigue analysis in the same way. The four parameters used to define the curves characteristics for the fatigue discussions and analyses to follow are its maximum value, sigma max, its minimum value, sigma min, its mean value, sigma m and its alternating value about the mean, sigma a.



5. Stress –Time Relations Of primary interest in the developments to follow are the mean stress and the alternating stress components. The mean stress is of course just the average of the two extreme values or one half the sum of the maximum and minimum values. The alternating stress, which is sometimes referred to as the stress amplitude, is simply one half the difference between the maximum and minimum values. Of occasional value is the stress ratio, R, defined as the minimum stress divided by the maximum stress and the amplitude ratio, A, which is the alternating component divided by the mean component.



6. Modified Goodman Diagram Fatigue failure behavior from experiments conducted under different combinations of fluctuating normal stress loading states are conveniently presented and described on a “Modified Goodman Diagram”. In this representation the mean stress is plotted on the horizontal axis and all other stresses are plotted vertically with tension to the right and up. A 45-degree line from the origin in the first quadrant represents the mean stress of an applied stress state. When the mean stress is zero fatigue failure is represented by plus and minus the endurance stress on the vertical axis. This is indicated by the red dot at the origin and the green dot on the vertical axis. A green dot on the negative vertical axis is not shown due to space limitations. Now consider the case where the mean stress plus the stress amplitude are equal to the yield stress of the material. This is illustrated by the second set of red and green dots. Assuming that yielding is considered as undesirable, that is equivalent to “failure”, then lines drawn from the second set of green dots to the positive and negative endurance limits on the vertical axes corresponds to a proposed criteria for defining fatigue failure for a generalized fluctuating stress loading. The failure boundary is completed by extending the line from the upper green dot on the maximum stress line to the value of the yield stress on the mean stress line and from that point to the lower green dot on the minimum stress line. Note that when the mean stress is equal to the yield stress no alternating stress is necessary to produce yielding. However, as the mean stress decreases the corresponding value of alternating stress required to define fatigue

failure increases as indicated by the third set of red and green dots. Thus the area enclosed by the red boundary in the first quadrant proposes states of stress for which fatigue failure will not occur. Experimental results have verified that this is a good approximation for defining fatigue behavior under a general fluctuating state of stress.



7. Mean/Fluctuating Stress Diagram A second way of depicting fatigue failure under the action of a fluctuating stress state is obtained by plotting the mean stress horizontally and the alternating stress vertically. The yield stress of the material is first plotted on both the vertical axis and the positive and negative axes. Then diagonal lines are used to join these three points as illustrated on the graphic. Next the endurance limit is plotted on the vertical axis together with the ultimate tensile strength on the horizontal axis. In the first quadrant a line is drawn from the endurance limit to the tensile strength. Any combination of alternating stress and mean stress that falls on the lower of the fatigue line or yield line, indicated by the heavy red boundary, defines a condition of failure either by fatigue or yielding. It has been determined experimentally that compressive mean stresses have virtually no effect on reducing the fatigue strength of a mechanical element. Hence in the second quadrant a horizontal line is extended from the endurance limit until it intersects with the yield line. Any combination of alternating stress and compressive mean stress that falls on the lower of this fatigue line or the yield line again indicated by the heavy red boundary defines a condition of failure either by fatigue or yielding. Since this is just another way of presenting the proposed behavior depicted by the modified Goodman diagram experimental test results indicates that it is a reasonable approximation of fatigue behavior under application of a general fluctuating stress state.

8. Exercise Problem 1 Application of the behavior depicted on the previous page to a design analysis process requires a mathematical model. This exercise provides an introduction to the development of this type of model. Following the exercise the presentation will introduce and discuss three theories of fatigue failure based on this type of development. After completing the exercise click on the solution button to check your result.




9. Soderberg Failure Theory Three distinct proposed failure theories for fatigue design application will now be presented and discussed. The first is the Soderberg Theory. Using the graphic on the previous page this theory proposes that designs for fluctuating normal stress states should be based on a limiting condition defined by a straight line drawn from the endurance limit on the vertical axis to the yield point on the horizontal axis in the first quadrant. This is defined analytically by the equation that the ratio of the alternating stress, sigma a, to the endurance limit, sigma e, plus the ratio of the mean stress, sigma m, to the yield stress, sigma y, is equal to 1. A factor of safety, n, can be introduced into this equation by dividing the right side of the equation by n. This can be seen to be a fairly conservative design approach.

10. Goodman Failure Theory The second proposed failure theory for fatigue design application under a general fluctuating normal stress loading is the Goodman Theory. It proposes a failure line that extends from the endurance limit on the vertical axis to the tensile strength on the horizontal axis. In effect it discounts yielding as a failure condition and is less conservative than the Soderberg theory particularly for mean stress values in excess of the yield strength. Analytically it is defined by the equation that the ratio of the alternating stress, sigma a, to the endurance limit, sigma e, plus the ratio of the mean stress, sigma m, to the tensile stress, sigma u, is equal to one. Again a factor of safety, n, can be introduced by dividing the right side of the equation by n.



11. Gerber Failure Theory.) The Gerber Failure Theory differs from the Soderberg and the Goodman theories in that it represents the failure line as a quadratic curve that passes through the endurance limit and the tensile stress. Of the three theories it is the least conservative and is considered by many to be the more accurate of the true behavior and impact of fluctuating loads on fatigue strength. Analytically it is represented by the equation the ratio of the alternate stress, sigma a, to the endurance limit, sigma e, plus the square of the ratio of the mean stress, sigma m, to the tensile strength, sigma u, is equal to one. To introduce a factor of safety into this express n is added to the numerator of the two stress ratios on the left side of the equation. Of the three theories presented preference for design applications and analysis is generally accorded to the Goodman theory. As such it will be used throughout the remainder of this chapter.

12. Exercise Problem - 2 In this exercise you are asked to determine numerically the magnitude of the difference between the Goodman and Gerber theories of failure over the range of sigma m over sigma u from zero to one. Since both theories pass through the endurance limit and the tensile strength there of course is no difference between them at these two points. When you have completed your analysis click on the solution button to check your answer.




13. Sample Problem 1 A sample problem will now be solved to illustrate a typical application of the Goodman theory of failure analysis. In this instance a flat cantilevered spring is subjected to a fluctuating end load. The dimensions of the element are obtained by clicking on the hot word in red at the beginning of the problem statement. The tensile strength of the spring material is specified as well as a stress concentration at the end where it is clamped. With these specifications and the load variation it is desired to determine the factor of safety for three specific conditions. The first requires that the mean stress be held constant, the second requires that the alternating stress be held constant and the third specifies that the amplitude ratio of the alternating to the mean stress be held constant.

(Linked Figure on Page 219)

14. Prob. – 1 Solution – Page 1 Since only the tensile strength of the material is given the first task is to obtain an estimate for the endurance limit of the steel. A value for the uncorrected or test specimen endurance limit is obtained by multiplying the tensile strength, sigma u, by 0.504 to give a value of 42.8 kpsi. Next the correction factors need to be determined to reduce the uncorrected endurance limit to what it will be for the mechanical element. Using the appropriate constants for a ground surface finish the surface factor, ka, is calculated to be 0.918.



15. Prob. – 1 Solution – Page 2 The size factor is determined by first substituting the cross section dimensions of the spring into the expression for the effective diameter. This gives a value of 0.1429 inches. This effective diameter is then substituted into the size factor equation for shafts less than 2 inches in diameter. The result of the calculation is a value for kb of 1.08. Since size factors are normally considered to be less than one kb will be taken as one to be conservative. With the spring subjected to a bending load the load factor, kc , will also be one as will the temperature factor, kd, since the spring will operate at room temperature. Finally the factor ke due to the stress concentration at the clamp, where the stress loading will be the greatest is one over 1.2 giving a value of 0.835.

16. Prob. – 1 Solution – Page 3 The part endurance limit is now obtained by multiplying sigma e by the product of all the correction factors. This results in a value of 32.73 kpsi for the corrected endurance limit sigma e prime. Next the loads and stresses need to be calculated. The mean load of the variation from 6 to 14 lbs is just 10 lbs while the alternating load component is 4 lbs. The bending stresses due to these loads will be a maximum at the clamp and will be given by the expression Mc over I. I is determined from the cross sectional dimensions and the expression bh cubed over twelve as 1.02 times ten to the minus fifth power. Substituting the loads, beam length, I and c as 1 /32 inch into the stress formula gives a mean stress of 30.7 kpsi and an alternating stress of 12.3 kpsi.



17. Prob. – 1 Solution – Page 4 Now consider the construction of a Goodman theory diagram with the numerical values calculated so far. Sigma e prime equal to 32.73 kpsi is plotted on the vertical axis while the tensile strength sigma u equal to 85 kpsi is plotted on the horizontal axis. A straight line is then drawn between these two points depicting the Goodman theory of failure line. Next the coordinates sigma m of 30.7 kpsi and sigma a of 12.3 kpsi define the point marked by the light gray dot. The solution to the first part of the problem is obtained by holding sigma m constant and allowing sigma a to increase to the value of sigma a super a at the red dot and then dividing sigma a super a by sigma a to obtain a factor of safety. A similar procedure is used to solve the second part of the problem with the exception that sigma a is held constant and sigma m is permitted to increase to sigma m super b at the green dot. For the third part the ratio of sigma a to sigma m is held constant and both values are permitted to increase along the diagonal to the blue dot at point c.

18. Prob. – 1 Solution – Page 5 The ratios of the sides of similar triangles will now be used to determine the unknown mean and alternating stress values for failure in the three scenarios described on the previous page. For part a the two triangles to be used are shown on the Goodman diagram by clicking on the graphic button. An equation to solve for sigma a super a is obtained by equating the vertical to the base on the green triangle to the vertical to the base on the red triangle. Now close the popup window. By substituting the appropriate numbers for all the known quantities sigma a super a can be determined to be 20.91 kpsi. Dividing this value by sigma a from the original loading gives a factor of safety of 1.70. Another way of interpreting this result is that the alternating value of the load can be increased by 70 % while keeping the mean load constant before failure due to fatigue will take place. For part b click on the graphic button to see that the ration of the vertical side of the green triangle to its base is similar to the vertical side of the red triangle to its base. This results in an equation in which the only unknown is sigma m super b. Now close the pop up window. Substituting the known parameters into the equation of side ratios permits sigma m super b to be determined as 53.06 kpsi. Dividing this value by sigma m equal to 30.7 gives a factor of safety of 1.73 if sigma a is held constant.



19. Prob. –1 Solution – Page 6 For part b the figure shows that the ratio of the vertical side of the green triangle to its base is similar to the vertical side of the red triangle to its base. This results in an equation in which the only unknown is sigma m super b. Substituting the known parameters into the equation of side ratios permits sigma m super b to be determined as 53.06 kpsi. Dividing this value by sigma m equal to 30.7 gives a factor of safety of 1.73 if sigma a is held constant.

20. Prob. –1 Solution – Page 7 In part c the equation involving the unknown stress parameters that will bring on fatigue failure is obtained using the similar triangles in the figure obtained by again clicking on the graphic button. Once more the equation becomes the ratio of the vertical side of the green triangle to its base equated to the ratio of the vertical side to the base of the red triangle. Now go back to the presentation by closing the pop up window. It is observed that the equation involves two unknowns, sigma a super c and sigma m super c. However, the ratio of sigma a to sigma m is to be held constant in this part of the solution so sigma a super c can be replaced by 0.401 times sigma m super c. Solving for sigma m super c gives a value of 41.64 kpsi. The factor of safety is again determined by dividing sigma m super c by sigma m to give value of 1.36. Why is this factor of safety significantly less than that calculated for parts a and b?



21.Exercise -3 This exercise will give you practice with solving a problem similar to the sample just concluded. Note that the applied tensile load consists of two components, one that is constant and a second component that is fluctuating. Also note that the ratio of the endurance stress to the tensile strength is not 0.504. When you obtain your result click on the solution button to check your answer before going on to the rest of the presentation.

(Linked Sketch on Page 219) (Solution on Pages 216, 217 and 218)

22. Fluctuating Torsional Fatigue All the subject material covered to this point in Chapter 9 has dealt with fatigue due to a fluctuating normal stress loading condition. The question is now raised as to how to treat this type of problem when the applied stress is due to a torsional loading. A Goodman diagram is again employed with the endurance limit replaced by a corrected torsional endurance limit and the tensile strength is replaced by an ultimate shear strength. The process of analysis remains the same after the alternating and mean shear stress components are determined from the torsional loading. In determining the corrected shear endurance limit from the endurance limit the load factor kc is always taken to be 0.577 for torsion. The ultimate strength in shear is estimated as 0.67 times the tensile strength after the work of Joerres at Associated Spring.



23. Combined Loading Mode The next problem of interest is how to perform a fatigue analysis when the loading system results in both normal and shear fluctuating stress in a combined mode of application. On this and the following page are the series of steps generally accepted as appropriate to analyze this type of problem. The first step is to determine the mean and alternating stress components for all applied loads. The second step is to apply stress concentration factors to the alternating components of all resultant stresses. The third step is to multiply any alternating axial stress components by 1.083. With all stress components known, modified and combined appropriately use Mohr’s circle to determine the principal mean stresses from the mean stress component system and the principal alternating stresses from the alternating stress component system.

24. Combined Loading Mode (continued) With the mean and alternating principal stresses determined effective values of mean stress and alternating stress are calculated using the equation in step 5. This effective stress is called the von Misses stress and is a consequence of a distortion energy consideration. Step 6 specifies that the material properties to be used are the endurance limit in bending and the tensile strength. However, the endurance limit should not be corrected for stress concentration effects, as these were included in step 2. Finally apply the Goodman fatigue analysis using the effective alternating stress, sigma a super e, and the effective mean stress, sigma m super e.



25. Sample Problem - 2 A sample problem involving combined loading will now be solved using the process outlined on the last two pages. A rotating shaft transmits a constant torque that generates a shear stress of 8 kpsi. The shaft is also subjected to an axial load that produces a tension of 10 kpsi. In addition the shaft carries a bending load that results in a maximum alternating bending stress of +/- 23 kpsi. Determine the factor of safety for this shaft it its material has a tensile strength of 75 kpsi and its corrected endurance limit is 37 kpsi.

26. Prob. 2 Solution Page 1 All corrections for axial load and stress concentrations are neglected in this problem for simplicity. Begin by considering the stressed element on the upper left of the page. It is subjected to a fluctuating bending stress and constant axial stress as well as a constant shear. The alternating normal stress component is one half the difference between the maximum and minimum normal stresses combing the components from bending and axial tension. This gives a value for sigma a equal to 23 kpsi. The mean normal stress is the average of the maximum and minimum normal stress which gives a value of 10 kpsi. These results could have been seen by inspection since the axial tension is constant and the bending contribution is completely reversed. The mean shear stress, tau m is 8 kpsi since there is no alternating component which in turn means that tau a is zero. The principal mean and alternating components are now calculated by substituting the mean and alternating stress components into the equations that define principal stress. This gives values of 14.43 kpsi for sigma one super m and –4.43 for sigma 2 super m.



27. Prob. 2 Solution Page 2 The principal alternating stresses are simple to calculate since there is no alternating shear stress component. Hence sigma one super a is just 23 kpsi and sigma two super a is zero. Next the effective mean and alternating von Mises stresses are calculated using the formula that comes from distortion energy considerations. Substituting the values of sigma one super m and sigma two super m into the equation gives a value of 17.7 kpsi for the effective mean stress, sigma e super m.

28. Prob. 2 Solution Page 3 The effective alternating stresses are again easy to calculate since there is no alternating shear stress component. Consequently sigma e super a is just 23 kpsi. The Goodman diagram with all the appropriate stress parameters is depicted on the lower portion of the page. Again the ratios of the sides of similar triangles are used to establish an equation relating sigma e super m prime to sigma e super a prime that designates a failure point on the Goodman diagram assuming the ratio of sigma e super a to sigma e super m remains constant.



29. Prob. 2 Solution Page 4 The final step in the solution is to solve the equation for the failure condition from the previous slide for one of the unknown primed effective stresses. This is accomplished by first recognizing that holding constant the ratio of the applied fluctuating effective stress to the applied effective mean stress permits sigma e super a prime to be replaced by 1.35 times sigma e super m prime. The resulting equation is then solved for sigma e super m prime to give a value of 20.1 kpsi. The factor of safety is then given by the ratio of sigma e super m prime divided by sigma e super m. The final factor of safety is 1.17. Although this sample was somewhat simplified to shorten the required calculations the fatigue analysis process used to include the effect of a combined stress state has been demonstrated.

30. Cumulative Fatigue Damage If a reversed stress in excess of the endurance limit is applied for a finite number of cycles less than the fatigue life associated with that stress level the material will undergo fatigue damage. That is, the endurance limit for the material will be lowered as a consequence of the damage so that further application of another level of reversed stress will reduce its fatigue life from what it could be if applied initially. A method of analytically treating this problem called the Palmgren-Minor Summation Theory will now be discussed. Basically this theory is expressed analytically by the equation that the sum of the ratios of the number of cycles of applied load to the original fatigue life at that stress level is equal to a constant. That is n1 over cap N1 plus n2 over cap N2 plus additional like terms is equal to a constant C. Experimentally the value of this constant appears to vary from 0.7 to 2.2. For simplicity and since it will in general be conservative the constant is taken to be unity for fatigue design analysis purposes. The theory is then expressed as the sum of the ratios is simply equal to one.



31. Palmgren-Minor Theory Application The application of the Palmgren-Minor theory can be demonstrated graphically on the S-N diagram shown on this page. Both of the axes are log scales so that the fatigue life versus cycles of reversed load application can be represented as a straight line over the range of 103 to 106 cycles starting at 0.9 of the tensile strength, sigma u, and decreasing to the endurance limit of the original material, sigma e super 0, before it becomes a horizontal line. Now consider the application of a reversed stress of magnitude sigma 1 applied for n1 cycles as indicated. At this stress level the normal fatigue life would be cap N 1cycles as designated by the point sigma f super 0. The difference between cap N 1and n 1 is plotted as a point at the sigma 1 stress level as shown by the point sigma f super 1. To apply the Palmgren-Minor theory a straight line parallel to the original fatigue life line is drawn through the (cap N 1 –n 1) point to 106 cycles before it becomes horizontal. This is indicated by the dotted red curve. It is observed that the intersection of this line with a vertical from the 106 cycle point on the horizontal line is at a stress level less than the original endurance limit, sigma e super 0. The reduced endurance limit due to the fatigue damage of the material

sustained from the application of sigma one for n 1 cycles is designated sigma e super 1. Although it won’t be shown here the dotted red line can be used to derive the generic Palmgren-Minor summation equation equal to 1. Note that n 2 on the graph which is the reduced life at the original endurance limit due to the damage can be solved for directly with the equation that n 2 is equal to the 1 minus the ratio of n 1 to cap N 1 with the entire quantity multiplied by cap N 2. This will all be demonstrated in a numerical answer that follows immediately.



32. Sample Problem 3 To demonstrate the numerical application of the Palmgren-Minor theory consider the following example problem. A part with a tensile strength of 90 kpsi and an endurance limit of 40 kpsi is subjected to a reversed normal stress of 65 kpsi for 3000 cycles. For these conditions find: a. the remaining life of the part if the stress is maintained at 65 kpsi., b. the remaining life if the stress is reduced to 40 kpsi from 65 kpsi. and c. the endurance limit after being subjected to 65 kpsi for 3000 cycles.

33. Prob. – 3 Solution – page 1 To determine the remaining fatigue life at 65 kpsi it is first necessary to find the total fatigue life at this stress level. Again the ratio of the sides of similar triangles on the figure on the previous page will be used for this purpose. Click on the graphic button to pop up a figure illustrating these similar triangles. The first equation on this page in which cap N 1 is the only unknown is developed using the green and red triangles depicted. Keep in mind that the equation is written in terms of the logs of the parameters since the axes on the figure are to a log scale both horizontally and vertically. Substituting the parameters values into the equation and taking logs as indicated leads to a numerical value for the log of cap N 1. Taking the anti log of this number gives the fatigue life at 65 kpsi as 8729 cycles. Hence the remaining life after 3000 cycles at 65 kpsi is 5729 cycles. This is the solution to part a of the problem.




34. Prob. – 3 Solution – page 2 For part b the Palmgren-Minor summation equation can be used to solve for n 2 the number of cycles the damaged material will under go after the stress level is reduced to 40 kpsi which was the endurance limit of the undamaged material. This is simply a matter of substituting the appropriate values into the equation for n 2 given by one minus the ratio of n 1 to cap N1 with the entire quantity multiplied by cap N 2 which would be 106 cycles at the original endurance limit. Carrying out this calculation indicates that the fatigue life is reduced to a finite value of 656 thousand cycles from an initial fatigue life of 1 million cycles or more.

35. Prob. – 3 Solution – page 3 To determine the reduced endurance limit of the damaged material as requested in part c again the ratio of the sides of similar triangles will be used to obtain an equation in which sigma e super 1 is the only unknown. Click on the graphic button to uncover a pop up window that shows the triangles used to obtain this equation. Again note that both the vertical and horizontals axes are prescribed in log scale divisions. The green triangle is used for the left side of the equation and the red triangle for the right side. Once again the known parameters are inserted in the equation log are taken as required and the log of sigma e super 1 is determined to be 4.582. Taking the anti log gives a value of 38.2 kpsi for the endurance limit of the damaged material. Note that his represents a 4.5 %decrease from the original value of 40 kpsi.




36. Mason’s Modification. It should be recognized that the Palmgren-Minor theory in proposing a parallel line to the original fatigue life line to represent the reduced fatigue life of the damaged material introduces an error at 103 cycles since it also reduces the value of the ultimate strength used to establish the starting point of the fatigue life line. Mason has proposed a modification to correct this error. In this modification the reduced fatigue life line of the damaged material still passes through the point sigma f super 1 at cap N 1minus n 1 cycles as already calculated but the line starts at 0.9 sigma u as on the original fatigue life line. This is illustrated by the graphic on this page. The consequence of this is that the finite life n 2 at the original endurance limit stress will be shorter than that predicted by Palmgren-Minor and the endurance limit of the damaged material, sigma e super 1, will be smaller than that calculated on the previous slide.

37. Prob. – 3 Solution – page 4 The reduction in n 2 and sigma e super 1 will now be recalculated incorporating the Mason modification. An equation for n 2m is again developed using the ratio of the sides of two similar triangles. Click on the graphic button to see these triangles on the Mason modified diagram. The green triangle is used for the left side of the equation and the red triangle for the right side. The only unknown in this equation is n 2m the finite life at the original endurance limit. Substituting the known parameters and carrying out the indicated numerical manipulations gives a result for the log of n 2m equal to 5.42. This converts to a finite life of 263 thousand cycles or a reduction of almost 60 % from that predicted by the Palmgren-Minor theory. This large reduction is a consequence of the effect of the log scale.




38. Prob. – 3 Solution – page 5 To determine the reduced endurance limit as affected by the Mason modification the ratio of the sides of similar triangles are again used from the damaged S-N diagram. Click on the graphic button to bring up a diagram showing these triangles. The green triangle is used for the left side of the equation and the red triangle is used for the right side. The only unknown in this equation is the reduced endurance limit, sigma e super 1. Substituting the appropriate parameter values into the equation and solving for the log of sigma e super 1 gives a value of 4.53. Hence the value of the reduced endurance limit is 33,880 psi. This represents an 11 percent reduction in the value as predicted by the Palmgren-Minor method. The use of the Mason modification is considered to be a more accurate method of analyzing cumulative damage and is generally recommended and used in fatigue design considerations where appropriate.


39. Review Exercise In this review exercise the items in the list on the left are to be matched with the mathematical relationships on the right. Place the cursor over an item on the left and hold down the left button. A pencil will appear that can be dragged to one of the green dots on the right. If the right choice is made the arrow will remain. If the selection is incorrect the arrow will disappear. After the exercise is completed proceed to the next page.



40. Off Line Exercise A vertical power screw used to raise and lower a load is supported by a single bearing as shown in the linked drawing. Below the bearing the shaft has a reduced diameter with a fillet radius at the transition. The applied axial load varies between 12,000 and 8,000 lbs in phase with a driving torque that fluctuates between 1000 in lbs and –500 in lbs. The material has a tensile strength of 150 kpsi. At the fillet the stress concentration for the axial load is 1.23 and for the torsional load is 1.35. Estimate the factor of safety for this design. When you have finished with this problem statement click on the exit button or main menu button to leave the chapter.

(Linked Drawing on Page 220) (Solution in Appendix)


_____________________________________________________________________________________Fluctuating Load Analysis - 213 - C.F. Zorowski 2002

Chapter 9 Fluctuating Load Analysis

Problem Solutions

Screen Titles

Problem 1 Solution Problem 2 Solution Problem 2 Solution (cont.) Problem 3 Solution Problem 3 Solution (cont.) Problem 3 Solution (cont.) Problem 3 Solution (cont.)



1. Problem 1 - Solution The equation of a straight line can be expressed as x over a constant "a" plus y over a constant "b" is equal to one. In this case the variable x is sigma m and y is sigma a. To determine the constants "a" and "b" use the two conditions that x or sigma m is equal to sigma u, the tensile strength, when y or sigma a is zero and y or sigma a is equal to sigma e, the endurance limit, when x or sigma m is zero. This gives "a" equal to sigma u and "b" equal to sigma e. The final equation becomes the ratio of sigma m to sigma u plus the ratio of sigma a to sigma e equal to one.

2. Problem 2 - Solution Begin with the Goodman equation for failure and the Gerber equation including a factor of safety. Since n as a function of the ratio of sigma m, the mean stress, to sigma u, the tensile strength, is desired it is necessary to eliminate sigma a over sigma e between the two equations. This is done by solving the Goodman equation for sigma a over sigma e and substituting into the Gerber equation which contains the factor of safety.



3. Problem 2- Solution (cont.) The substitution discussed on the previous page results in a quadratic equation in n with coefficients that are dependent on the ratio of sigma m to sigma u. The table gives values of n for values of sigma m over sigma u from zero to one at increments of 0.2. It is observed that using the Gerber equation provides at most a factor of safety of 20 % over the failure prediction by the Goodman diagram.

4. Problem 3 – Solution At the transition point the maximum axial force is 1300 lbs while the minimum axial force is -300 lbs due to the two axial components. Thus the mean value of fluctuating force is the average of these two values or 800 lbs. and the alternating component is 500 lbs. The area of the cross section acted on by the forces is calculated using the larger diameter of 3/16 inch. This gives an area of .026 square inches. Then the mean and alternating normal stress are calculated to be 30,769 psi and 19,230 psi respectively.



5. Problem 3 – Solution (cont.) The correction factors for the part endurance limit are now determined. With a ground finish ka is calculated to be .902. The size factor Kb for a 3/16-diameter rod is given by 1.05 and is taken to be 1 to be conservative. The load factor kc for an axial load is 0.923 since sigma u is less than 220 kpsi. The temperature factor kd is unity and the ke factor due to the stress concentration factor of 1.15 is .87. Multiplying these factors together times the uncorrected endurance limit gives a part endurance limit of 47.1 kpsi.

6. Problem 3 – Solution (cont.) Shown on this page is the Goodman diagram for this problem with the relevant parameters already calculated. To determine the largest possible alternating component of stress, sigma a prime, with sigma m held constant an equation can be developed from the geometry of the diagram making use of the vertical and horizontal sides of similar triangles. Thus the left side of the equation involving sigma a prime is obtained from the small green triangle. The left side of the equation comes from the large red triangle. This equation can then be solved for sigma a prime since all the other parameters are known.



7. Problem 3 – Solution (cont.) By substituting the known values from the diagram into the equation for sigma a prime gives a result for this unknown quantity of 33.33 kpsi. This value of sigma a prime, which is the maximum value the alternating stress can achieve, is 1.73 times the actual value of the alternating stress. Then F a prime becomes 865 lbs. resulting in an F max of 1665 lbs and an F min of – 65 lbs. Finally, with F1 held constant at 500 lbs. F 2 prime can vary from +1165 lbs. to – 565 lbs. before failure will occur.



Figure for Screen 13





Graphic for Screen 33


_____________________________________________________________________________________Limit Loading - - C.F. Zorowski 2002 223

Chapter 10 Limit Loading

Screen Titles

Basis of Limit Load Analysis Geometry of Bending Strain and Stress Distributions Equilibrium Conditions Stress-Strain Relations Axial Equilibrium Moment Equilibrium Plastic Moment Alternate Solution Plastic Hinge Behavior Ultimate Load Example Sample Problem - 1 Problem –1 Solution Sample Problem - 2 Problem – 2 Solution Sample Problem – 3 Problem – 3 Solution Review Exercise Off Line Exercise



1. Title page Chapter 10 introduces the concept of limit loading and applies the theory to a number of practical problems. The subject of limit loads and their prediction addresses the fact that much of the static load carrying capacity of mechanical elements produced from a reasonably ductile material resides in its properties and behavior beyond the elastic limit. Based on an idealized behavior for a ductile material this chapter develops a model process of solution that permits the determination of limit loads on mechanical beam elements subjected to transverse bending. A number of simple problems are used to demonstrate the application of the solution process. This includes statically indeterminate problems and beams of curved geometry. Several exercises are also included to provide the reader with practice in the application of the solution process.




3. Basis of Limit Load Analysis Virtually all mechanical element stress analysis models assume a linear stress strain behavior for the material. Once the internal stresses created by the external loads are determined the various theories of static strength and fatigue are applied to determine the acceptability of the design. In static strength determination this approach neglects the great potential for a material to carry an external load that exists beyond the small range of its elastic behavior. For example consider how small the area representing stored energy is under the linear portion of the stress strain diagram as compared to the material’s capacity to absorb work and energy as depicted under the plastic behavior portion of the stress strain diagram. Once the yield strain is exceeded most reasonably ductile materials can continue to deform and experience strains of several hundred times the yield strain before fracture takes place. Estimating the limit loads associated with this extended behavior is the subject of this chapter.

4. Geometry of Bending The developments to follow will emphasize determining the limit loads for beam type elements subject to bending by transverse loading. From experimentation it is well established that plane cross sections in a beam remain plane after bending. Thus, two parallel cross sections before bending will be inclined with respect to each other when the beam is subjected to curvature. Thus, longitudinal element fibers at the top of the beam will under go compression while element fibers at the bottom surface are extended. Hence, there exists some location through the thickness of the beam where there is no strain applied to a longitudinal element. This location is referred to as the neutral axis of the beam. The extensional strain of a fiber at some distance y below the neutral axis can be determined using the geometry of the bent beam on this page. The strain epsilon is equal to the arc ac minus the arc ab divided by the fiber’s original length ab. But the arc ac is just the radius of curvature rho plus y times the angle between the two original parallel planes, delta theta. Similarly the arc ab is simply the radius of curvature times the angle change. Substituting these two analytical expressions for the arc lengths and simplifying the relationship results in the strain, epsilon, equal to the distance from the neutral axis to the fiber of interest divided by the radius of curvature of the beam. Note that this results in compressive strains above the neutral axis and positive tensile strains in the plus y direction.



5. Strain and Stress Distributions The linear strain distribution model from the previous page results in some form of normal stress distribution across the depth of the beam. This distribution may be linear as for an elastic material or in the most general case it may be non linear as also depicted on this page. Irrespective of the form of the distribution the resultant force and moment of this stress distribution must satisfy the effect of the external loading at that cross section. If the beam is subjected to only transverse loading the axial force must be zero and the moment of the stress distribution about the neutral axis must be equal to the bending moment at that location.

6. Equilibrium Conditions The resultant axial force equilibrium condition is expresses as the integral of the stress over the area of the cross section. For simplicity the cross section will be taken to be rectangular of width b and height h. Thus, the integral of sigma da over the total area can be written as b times the integral of sigma dy over the height. Making use of the expression that the strain is equal to y over rho the integral can then be written as sigma d epsilon from minus epsilon two to epsilon one. The moment equilibrium condition is initially expressed as the integral of sigma times y da over the cross section. This can then be rewritten for the rectangular section as b times the integral of sigma y dy from minus h2 to h 1. Again substituting the relationship between epsilon and y gives finally rho squared times b times the integral of sigma epsilon d epsilon from minus epsilon two to epsilon one all equal to the bending moment M.



7. Stress Strain Relation To evaluate the equilibrium integrals on the previous page it is now necessary to assume some representation for the stress strain relationship for the material. A simple, yet reasonable model, depicting the behavior of an idealized ductile material, is now assumed. As shown in the figure this model assumes that the yield stress in tension and compression is equal in magnitude and that the tensile modulus and compressive modulus in the elastic region are equal. Once the yield stress is reached it is further assumed that the stress remains constant independent of strain and that the strain is unlimited. Although idealized this model has proven to be quite useful in limit load analysis.

8. Axial Equilibrium The model from the previous page is substituted into the axial force equilibrium equation. This gives rise to three separate integrals that must be evaluated. Note that the stress in the second integral is expressed in terms of the linear relationship that it has with the strain in terms of the modulus in the elastic region. Evaluating the integrals and combining the results gives a final equation that indicates epsilon one must be equal to epsilon two for the rectangular cross section. This in turn implies that h1 must be equal to h2, which further implies that the neutral axis must pass through the centroid of the cross section just as in simple elastic beam theory.



9. Moment Equation Substituting the idealized ductile stress strain relationship into the moment equilibrium equation again results in three distinct integrals. This set of integrals lead to quadratic and cubic terms in strain. Recognizing that epsilon one is equal to epsilon two from the previous page and expressing the modulus as the yield stress, sigma y, divided by the yield strain, epsilon y, a final relationship for the moment divided by b rho squared is obtained as sigma y times the quantity epsilon one squared minus one third epsilon y squared. As indicated earlier and expressed in terms of the idealized stress strain relationship epsilon y is significantly smaller than epsilon one so that its contribution can be neglected. Recall that it was earlier pointed out that the total strain that can be sustained by a ductile material can be several hundred times the yield strain. Thus the final equation for the bending moment is taken to be sigma y times b times the product of epsilon rho quantity squared. The task that now remains is to relate this result to linear beam theory.

10. Plastic moment The moment expression from the previous page is designated the plastic moment indicated by the sub p. The epsilon rho term is eliminated by substituting h over two rho for epsilon one. The plastic moment then becomes one-fourth sigma y times b h squared. This will now be compared to the moment in an elastic beam when the maximum stress is equal to sigma y. From the familiar formula that sigma is equal to M c over I the moment at the inception of yielding can be written as sigma y times I over c. For a rectangular cross section I is simply b h cubed over twelve and c is h over two. Substituting I and c into the Me expression gives one-sixth sigma y times b h squared. It is seen finally that the ratio of the plastic moment to the elastic moment is three halves or 1.5. This indicates that that moment that initiates yielding can be increased by 50 % before the cross section has become completely plastic as defined by the assumed material model.



11. Alternate Solution The final analytic result for the magnitude of the plastic moment could have been obtained more easily with the model shown on this page. Assuming that the neutral axis does pass through the centroid of the cross section and the stress distribution is constant over the cross section at sigma y then the resultant force above and below the neutral axis is just sigma y times half the area of the section. These two resultant forces are equivalent to a couple of magnitude F1 times h over two, which in fact is the bending moment on the section. Hence Mp becomes sigma y times b h squared over four as determined earlier. This method can be used to very quickly determine the magnitude of the plastic moment for any cross sectional shape provided it is assumed that the stress above and below the neutral axis is equal to a constant value, sigma y.

12. Exercise Problem - 1 In this exercise you are asked to determine an expression for the plastic moment in a solid circular cross section using the technique demonstrated on the previous page. Compare this result with the elastic moment for the same cross section. That is, the moment that first initiates yielding in the section. You will find that the ratio of the plastic moment to the elastic moment is greater than for a rectangular cross section. Can you explain why this is so? When you have completed your solution click on the solution button to check your result before proceeding further in the chapter.




13. Plastic Hinge Behavior The top figure on this page shows a beam fixed at one end, simply supported at the other and loaded with a concentrated force along its length. This is a statically indeterminant problem as the reactions F1, F2 and the moment, M, cannot be determined from the equations of equilibrium alone. However, it is possible to possible to sketch what the bending moment diagram will look like shown directly below the drawing of the beam. The curvature of the beam is also shown as would be anticipated providing that elastic behavior is taking place. Keep in mind that the scale of the indicated displacements are very exaggerated compared with the length of the beam. Now consider what happens as the load F is increased. The magnitude of the bending moment will increase at both the location of the load and at the fixed end. Assuming the idealized stress strain behavior assumed earlier for a ductile material the maximum bending moment that can be achieved at these two points will be the plastic moment for the cross section of the beam. When this occurs the material in the beam at these locations can undergo extensive yielding so the beam appears to be hinged at these points with the remainder of it appearing straight since the elastic deformations away from these points will be very small. These two locations are referred to as plastic hinges and can be used to determine the magnitude of the ultimate load F that caused them to occur.

14. Ultimate Load – Sample Problem The concept of the occurrence of plastic hinges due to the limit loading of a beam will now be applied to the previous example problem. All forces acting on a free body diagram of the entire beam must first satisfy equilibrium. This gives rise to equation 1 that states that the sum of the support reactions, F1 and F2, must be equal to the applied load F. Equation 2 comes from summing all moments about the left end of the beam. Note that the moment reaction at the left end is set equal to the plastic moment, Mp, since a plastic hinge is required at this location if a plastic hinge exists at the location of the load F. These two equations contain three unknowns, the reactions F1 and F2 together with the unknown limit load F. Hence a third equation is required. This is obtained by recognizing the plastic moment at the location, a, of the load, F, can be written as F1 times the distance a. This then establishes the value of this reaction permitting F and F2 to be determined from equations 1 and 2.



15. Problem –1 Solution Equations 2 and 3 from the previous page can be combined and solved directly for F in terms of the plastic moment Mp and the geometry of the beam. This expression gives the value of the limit load F that will produce plastic hinges at its position and the fixed end resulting in the shape of the deformation curve on the previous page. This model will not predict the magnitude of the deflection at point a. The assumption of unlimited material strain following yielding in the idealized material stress strain behavior results in the deflection being indeterminant. With F, the limit load, determined the problem is essentially solved. However, the remaining reaction F2 can also now be determined as

16. Exercise Problem - 2 In this exercise you are asked to use the reaction results of the previous problem solution to determine the value of the bending moment at the simply supported end of the beam. If the solution is correct the bending moment should be zero. After you have completed your analysis click on the solution button for verification before proceeding on with the rest of the chapter.




17. Problem – 1 Solution The answer for the limit load in the previous problem is expressed in terms of the length of the beam and the location of the load F. This will now be examined in terms of the ratio of the distance a to the length of the beam c. This is first made possible by rearranging the equation for F into the dimensionless representation involving the ratio of F times c over Mp and the ratio of a to c. Now numerical values can be calculated for the dimensionless limit load. The table presents numerical values of the limit load parameter for values of dimensionless load location for increments of a over c of 0.1. It should be noted that if a over c is either 0 or 1 the load parameter is infinite. This corresponds to the load being over a support, which could result in a very large load if the support were strong enough. As a over c increases the load parameter first decreases and then increases. As it gets closer to the fixed end its values are higher than at the simply supported end as would be expected. Also its smallest value occurs just short of the center of the beam from the right end.

18. Exercise Problem - 3 In this exercise you are asked to use the load limit analysis process just presented to determine the limit load that can be carried by a simply supported beam as a function of the position of the force from one end of the beam. How do the results differ from the solution to the previous problem? After completing your analysis click on the solution button to check your result before proceeding on to the remained of the chapter.




19.Sample Problem -2 The fixed and simply supported beam will now be re analyzed assuming the beam is subjected to a uniformly distributed loading system. Again the limit load, w, is desired. The solution is begun by applying the equations of equilibrium to a free body diagram of the entire beam. Equation 1 satisfies vertical equilibrium while equation 2 comes from summing moments about the right end. A third equation is obtained by assuming that a plastic hinge occurs at some yet to be determined distance a from the left end and writing an equation summing the moments on a free body diagram of the section of length a. These three equations involve four unknowns, F1, F2, w and a. To obtain a deterministic solution it is necessary to solve for F1, F2 and w in terms of a and then to vary the value of a to obtain the minimum value of w. This will be the limit load that produces the plastic hinge behavior.

20. Problem – 2 Solution The second equilibrium equation is first solved for F2 in terms of Mp and w. This result is then substituted into equation 3, which is the plastic hinge equation, and introduces the parameter a, the location of the plastic hinge. This equation must then be solved for w in terms of Mp and a.



21. Problem –2 Solution The last equation on the previous page is solved for w and put into a non dimensional form involving the load parameter w c squared over Mp as a function of the ratio of a over c. Numerical values of the load parameter are now determined along the length of the beam at increments of a over c of 0.1. The table presents the results of these calculations. Note that a over c values of 0 and 1 are not included since a plastic hinge cannot occur over a simple support. Again the load parameter is seen to decrease and then increase as the position parameter increases. The minimum value of the load parameter from this table is 11.67 and occurs at a over c of 0.6. Thus the correct value of the limit load parameter is about 11.5 and the previously unknown hinge position occurs just a little past the center from the fixed end. This position appears quite reasonable. To obtain a more exact solution the position of the zero slope of the load parameter curve as a function of the hinge position parameter could be determined analytically.

22. Exercise - 4 In this exercise you are asked to determine the limit load for a uniformly distributed load on a beam simply supported at each end. Then repeat the solution for a beam that is fixed at both ends. Compare the results. Do they appear reasonable? Check your answers by clicking on the solution button before going on to the rest of the chapter.




23. Sample Problem – 3 The problem of determining the concentrated load required to collapse a circular ring will now be analyzed. In this solution it is assumed that the thickness of the ring is small compared to its radius so that simple bending theory can be applied to any cross section. It is then assumed that collapse requires the occurrence of four plastic hinges, two on the horizontal diameter and two on the vertical diameter. To determine the limit load F only one quarter of the ring need be analyzed due to the horizontal and vertical symmetry of the system. Vertical equilibrium of the quarter of the ring in the second quadrant requires that the half of the external vertical load at the top must be balance by an equal compressive axial force at the cut cross section on the horizontal diameter. At the same time plastic moments, Mp, must also act at the horizontal cut cross section and the vertical cross section at the top.

24. Problem 3 - Solution A hinge moment equation for the plastic bending moment can now be written for either the top or horizontal diameter cross sections. Both of these equations are exactly the same and give a final result of the limit load being equal to 4 times the plastic moment Mp divided by the ring radius R. How would the problem of the limit load for a ring collapsed by both horizontal and vertical concentrated forces of the same magnitude be solved? Would it require more than four plastic hinges?



25. Review Exercise In the review exercise simply click on the appropriate answer. You will be provided with an immediate feedback. If you select the wrong answer try again. The hot words in the questions will link back to the page in the chapter that deals with the issue raised in that question. When you have completed the quiz go on to the next page.

26. Off Line Exercises Use the limit load analysis procedure presented in chapter 10 to solve these two problems. When you are finished with these problem statements click on the main menu or exit buttons to leave the chapter.


_____________________________________________________________________________________Limit Loading - 239 - C.F. Zorowski 2002

Chapter 10 Limit Loading

Problem Solutions

Screen Titles

Problem 1 Solution Problem 1 Solution (cont..) Problem 2 Solution Problem 3 Solution Problem 3 Solution (cont..) Problem 3 Solution (cont..) Problem 4 Solution Problem 4 Solution (cont.)



1. Problem 1 - Solution The centroid of the top half of the circular cross section is located at 2 R over Pi above the horizontal diameter. This is the point through which the resultant axial force for the top half of the cross section acts. Therefore, the plastic moment is equal to the couple of the F resultants which is F times 2 times 2R over Pi. With F equal to sigma y times Pi R squared over 2 the final expression for the plastic moment, Mp, is given by 2 sigma y times R cubed.

2. Problem 1 - Solution (cont.) This plastic moment is now to be compared to the elastic moment at the onset of yielding. Begin with the expression for stress as M c over I. For a circular cross section I is equal to Pi R fourth over 4 and c is equal to R. Solving for M and substituting sigma y for sigma together with I and c in terms of R results in Me equal to sigma y R cubed times the quantity Pi over 4. Thus the ratio of Mp to Me becomes 8 over Pi or 2.55. This is significantly greater than the value of this ratio for a rectangular cross section. The reason for this is that more of the material of the cross section is closer to the horizontal axis where the constant yield stress has more of an effect compared to the linear distribution. When you are finished with this solution click on the return button to go back to chapter 10.



3. Problem 2 – Solution Summing moments about the right end of the beam consists of three contributions. The first is the plastic moment, which is considered positive. The second is a negative contribution of F2 times the length of the beam and the third is a positive contribution of F times its moment arm a. Substituting the determined values of F2 and F into the equation permits Mp to be divided out as a common factor. Combining the remaining terms over a common denominator results in a numerator in which all the terms cancel each other. Thus the left side of the equation is zero which satisfies the condition that the right end is simply supported.

4. Problem 3- Solution A simple supported beam acted on by a concentrated load will only posses one plastic hinge under the location of the vertical load. This is where the bending moment diagram will have its largest value. This plastic moment will be produced by the two end reactions on the beam times the moment arm from their location to where the vertical force F acts. This is expressed analytically by the first two equations at the top of the page. With F1 and F2 known the applied limit load F can be determined as their sum. This result in an equation for F in terms of the plastic moment Mp and the geometry of the beam and loading represented by c and a. For a numerical comparison with the solution for the fixed – simply supported beam already solved this final expression is put into a similar dimensionless form.



5. Problem 3 – Solution (cont.) Substituting numerical values of the dimensionless load location, a over c, at increments of 0.1 from the end to the center of the beam gives the listed table of values for the dimensionless load parameter, F times c over Mp. At a over c equal to zero the load is over the end support, which is represented mathematically by infinity and is not listed in the table. Note that the load parameter decreases from near the end support to the center by almost a factor of three. For the second half of the beam the values would be repeated since this load function in terms of its location parameter will be symmetric about the center of the beam.

6. Problem 3 – Solution (cont.) This page compares the results of the solution of the concentrated load on a simply supported beam at both ends to a concentrated load on a beam that is fixed at one end and simply supported at the other. Note that at all locations the fixed beam will support a greater load than the one that is simply supported at both ends, as one would expect. Also note that even at the center of both beams the SS-Fixed beam will carry 50 % more load than the beam with simple supports at both ends. All of these effects are due to the clamped end support, which provides for greater strength. When you have finished with this page click on the return button to go on to the next page in chapter 10.



7. Problem 4– Solution A simply supported beam carrying a uniform distributed load, w, will result in the occurrence of a plastic hinge at its center from simple considerations of symmetry. In this instance the two support reactions are equal and given by one half the total load represented by w times c. Either of the support reactions can now be used to calculate the value of the plastic moment at the center of the beam. The contribution of the support reaction given by F times c over 2 minus the contribution of the distributed load which in terms of its resultant is w times c over 2 times the moment arm c over 4 must be equal to Mp. Solving for w gives 8Mp divided by c squared.

8. Problem 4 – Solution (cont.) Now consider the beam from the previous page to be fixed at both ends under the same distributed load, w. In this case there must be a plastic hinge at both ends as well as in the center. From vertical equilibrium and symmetry the two support reactions are again equal and have a magnitude of wc over 2. The plastic hinge equation is similar to that developed on the previous page with the exception that a negative moment Mp must be included on the left side representing the plastic moment reaction at the fixed end of the beam. Solving this revised hinge moment equation gives a final value for w of 16 Mp over c squared or twice the limit load for the simply supported beam. When finished with this solution click on the return button to go back to the next page in chapter 10.

Design for Strength and Endurance - Appendix

Appendix - 245 - C. F. Zorowski 2002

Appendix

Chapter Quizzes Chapter Quiz Solutions Final Course Project Off Line Exercise Solutions Topical Index


Appendix - 246 - C. F. Zorowski 2002


_____________________________________________________________________________________Course Project - - C.F. Zorowski 2002 267

Course Project Instructions

Purpose The purpose of this end of course project is to conduct an evaluation of the design of a significantly loaded and critical existing mechanical part by applying the knowledge contents of “Design for Strength and Endurance” to perform a complete and detailed strength and fatigue life analysis of the component. Project Subject The component to be investigated is a pedal crank arm from a relatively inexpensive model of adult bicycle. Since the product appears quite rugged and appropriately sized it may be significantly over designed for its intended purpose and life. You are asked to evaluate its current design strength and endurance with the thought of a possible design to reduce its dimensions and subsequently its weight and cost. On the following page is a photograph of the part showing both the plan and side views with a number of significant dimensions. The crank appears to be a drop forged steel element that has been finished with some degree of surface grinding. The holes on the crank have been machined and the hole at the pedal end ids threaded to receive the foot pedal. The hole on the side view is used to key the crank to the shaft that dives the chain sprocket. It also appears that the crank has been lightly plated. Repeated hardness testing of the material resulted in Rockwell B multiple values of 72 on the side and values of 76, 74 and 75 on the top surface. One of the more difficult tasks to be undertaken in this evaluation is a rational estimate of the appropriate loading that this element will be subjected to together with its expected product life. Some research into

this subject will need to be undertaken before any quantitative analysis can be undertaken. Hint: (The areas of man machine interfacing and ergonomics might prove to be helpful and the internet is a good source for all kinds of information.) Proposed Tasks Following is a list of suggested tasks to be undertaken and completed in arriving at a justifiable conclusion as to the overall performance of the part.

1. Establish a simple physical model for the crank element.

2. Assume an appropriate loading system in terms of direction and loading.

3. Determine what type of internal stresses will result from the loading.

4. Calculate the maximum stresses due to the applied loading.

5. Estimate the static and dynamic strength properties of the material.

6. Apply an appropriate theory of failure to determine static strength.

7. Conduct a fatigue analysis to determine expected life.

8. Draw conclusions as to the adequacy of the design and the potential for improvement.

Summary Since this is a real problem requiring a number of assumptions that will affect the results obtained by different analysts no one specific solution is offered for comparison. However, it is recommended that you evaluate your conclusion in terms of what appears reasonable from your own personal experience with this rather common element that is normally encountered indirectly in some fashion in most every ones life.


Course Project C.F.Zorowski 2002 268

Bicycle Crank Dimensions

(units of inches)

Top View

Side View

6 ½

7 ½

5/8 in. ½ in.dia.

½ ½ 19/32 23/32 in.

29/64 ½

3/8

29/32 in.A = 1 3/32 B = 61/64

AB

Design for Strength and Endurance – Appendix

Index C. F. Zorowski 2002

Index

A Actual Load Distribution 152 Audio (CD) 10 Annealing 106

B Beams

Bending Geometry 226 Bending Moment 18,19 Bending Moment Distribution 28 Plastic Moment 229 Shear Force 18,19 Shear Force Distribution 27 Shear and Moment Diagrams 28 Support Reactions 27

Brittle Failure 133

C Case Hardening 109 Casting Processes

Die 102 Investment 101 Sand 100 Shell Molding 101

CD Audio 10 Installation 9 Navigation 10 Operation 9 Special Features 10

Cold Working Processes Rolling 105 Drawing 105 Heading 105 Stamping 105 Roll Threading 105

Combined Fatigue Loading 203 Content and Format 5 Coulomb Mohr Theory 131,132 Course Objectives 3 Cumulative Fatigue Damage

Palmgren-Minor Theory 206 Mason Modification 210

D

Design Issues 124 Designer’s Dilemma 125

Die Casting 102 Ductile Failure 132 Ductility 83

E Elastic Behavior 81 Elastic Modulus 17,64,66,81

Common Values 81 Endurance Limit (Fatigue)

Behavior 175 Equations 176 Modifying Factors 177

Engineering Stress Strain Diagram 83 Exercise Problems

Chapter 1 18 Chapter 2 34,37,39,42 Chapter 3 58,61,63,65,66 Chapter 4 822,84,89 Chapter 5 _ Chapter 6 128,130,134 Chapter 7 150,151,159,160 Chapter 8 174,176 Chapter 9 195,197,202 Chapter 10 230,232,233,235

Extrusion 103

F Factor of Safety (nominal)

Definition 148 Generic Values 151 Items Effecting 148 Modified Failure Theories 150 Variation Scenarios 149

Factor of Safety (statistical) Actual Load Distribution 152 Factor of Safety Calculation 155,158 Load Capability Distribution 152 No Failure Scenario 159 Percentage Failure Interpretation 152 Percentage Failure Calculation 155,156

Failure Applicable Theories 133,134 Brittle 133 Ductile 132

Fatigue Endurance Limit Equations 176



Endurance Limit Behavior 175 Failure Process 172 Failure Testing 173 S-N Diagram 173

Fatigue Failure Theories Gerber Theory 197 Goodman Theory 196 Soderberg Theory 197 Strength Equation 174,175

Fluctuating Stresses 192 Generic Behavior 192 Stress-Time Relations 193 Mean/Fluctuating Stress Diagram 195

Forging 103 Fracture 83 Fracture Stress 83

G Gaussian Distribution 154 Geometry of Bending 226 Gerber Fatigue Theory 196 Goodman Fatigue Theory 196

H Hardness Test

Generic 80,87 Brinell 88 Rockwell 88 Scale Comparisons 89

Heat Treatment Annealing 106 Case Hardening 109 Generic 105 Normalizing 106 Tempering 108 Quenching 107

Hooke's Law 17,64 Hot Working Processes

Rolling 102 Extrusion 103 Forging 103

I

Installation and Operation (CD) 9 Investment Casting 101

J

K

L Load Capability Distribution 152 Load Factor (Fatigue) 179 Limit Loading

Analysis Basis 226 Geometry of Bending 226 Non-Linear Stress Distribution 227,228 Plastic Hinge 231 Plastic Moment 229 Ultimate Load 231

M

Mason Modification 210 Material Processing 100 Maximum Normal Strain 62 Maximum Normal Stress

Two Dimensions 37 Three Dimensions 41

Maximum Normal Stress Theory 126 Maximum Shear Strain 62 Maximum Shear Stress

Two Dimensions 36 Three Dimensions 41

Maximum Shear Stress Theory 127 Mean/Fluctuating Stress Diagram 195 Mechanical Tests

Compression 80 Tensile 80 Hardness 80,87

Modified Goodman Diagram 194 Modifying Factors (Fatigue) 177

Load 179 Size 178 Surface 177 Temperature 180 Miscellaneous 180

Mohr’s Circle for Strain Graphical Construction 62 Sample Problems 63 Mohr’s Circle for Stress

Analytical development 35,36 Graphical construction 38 Sample problems 37,39, 42 Three Dimensions 41



Mohr’s Theory of Failure 131

N Navigation (CD) 10 No Failure Scenario 159 Nominal Stress Strain Diagram 83

O Off line exercises

Chapter 1 23 Chapter 2 43 Chapter 3 68 Chapter 4 90 Chapter 5 110 Chapter 6 138 Chapter 7 161 Chapter 8 184 Chapter 9 212 Chapter 10 237

Operation (CD) 9

P Palmgren-Minor Theory 206 Percentage Failure Calculation 155,156 Percentage Failure Interpretation 152 Percent Elongation 86 Permanent Set 82 Plain Carbon Steel

Effects of Carbon Content Yield Stress 87 Tensile Strength 87 Elongation 87 Hardness Values 89

Plastic Behavior 82 Plastic Hinge 231 Poisson’s Ratio 65

Common Values 81 Principle Strains

Two Dimensions 62 Principle Strain Axes Orientation 62 Principle Stress Axes Orientation

Two dimensions 38 Three dimensions 40

Principle Stresses Two dimensions 37 Three dimensions 41

Q

Quenching 107

R Relation between E and G 67 Ring Collapse 236 Review Exercises

Chapter 1 23 Chapter 2 43 Chapter 3 67 Chapter 4 90 Chapter 5 109 Chapter 6 138 Chapter 7 160 Chapter 8 183 Chapter 9 211 Chapter 10 237

Rolling Cold 105 Hot 102 Material Property Comparison 104

S

Sample Problems Chapter 1 18 Chapter 2 34,37,39,42 Chapter 3 63,66 Chapter 4 82,84,89 Chapter 5 - Chapter 6 128,129,130,135 Chapter 7 150,151,157,159,160 Chapter 8 174,176,181 Chapter 9 198,204,208 Chapter 10 230,231,233,234,236

Sand Casting 101 Shafts

Shear Stress 21,22 Angular Twist 22

Shear Modulus 17,64,67 Common Values 81

Shell Molding Casting 101 Soderberg Fatigue Theory 196 Size Factor (Fatigue) 178 Strain

Definitions Rectilinear Components 56 Simple tension, compression 15



Simple Shear 16 True Strain 84

Strain Rosette (45 degrees) 64 Strain Systems (2 dimensions) About Rotated Axes Normal Strain Components 56,57 Shear Strain Components 59,60 Total Rotated Normal Strain 58 Total Rotated Shear Strain 60 Total Rotated Strain Equations 61 Graphical Interpretation 61 Stress Concentration 181 Stresses Definitions Bending Stress 19 Cross shear 32 Normal stress 32 Rectilinear components 32 Shear stress 32 Shear Stress in Bending 20,21 Simple Tension, Compression 15 Simple Shear 16 True Stress 85 Stress Strain Diagram (Tensile Test) Nominal 83 True 85 Stress-Strain Relations 65 Stress Systems (2 dimensional) About rotated axes 33 X’ direction 33 Y’ direction 34 Double angle formulation 35 Study Guide 7 Surface Factor (Fatigue) 177

T Temperature Factor (Fatigue) Tempering

Effects on Properties 108 Generic 108

Tensile Properties 86 Tensile Strength 85 Tensile Test 80,81 Theories of Failure 125

Coulomb Mohr 131,132 Distortion 128,129,130 Maximum Normal Stress 126 Maximum Shear Stress 127

Mohr 131

Torsional Fatigue 202 True Strain 84 True Stress 85 True Stress Strain Diagram 85

U Ultimate Load 231 Ultimate Stress 83,86

V

W Work Hardening 82

X

Y Yield Stress 82,86 Yielding in Pure Shear 128,130

Z

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