The a992 steel shaft has a diameter

PowerPoint Slides Of Chapter 3

1-Introduction

2- definition

3- Application in industry

4- example with picture

5- solve a problem from the book chapter 3 with an explanation

6- conclusion

Fundamental Equations of Mechanics of Materials Axial Load

Normal Stress

s = P A

Displacement

d = L L

0 P(x)dx A (x)E

d = ! PL AE

dT = a "TL

Torsion

Shear stress in circular shaft

t = Tr

J

where

J =

p

2 c4 solid cross section

J =

p

2 (co

4 - ci 4) tubular cross section

Power

P = Tv = 2pf T

Angle of twist

f = L L

0 T(x)dx J(x)G

f = !

TL JG

Average shear stress in a thin-walled tube

tavg =

T 2tA m

Shear Flow

q = tavg t =

T 2Am

Bending

Normal stress

s =

My

I Unsymmetric bending

s = -

Mz y

Iz +

Myz

Iy , tan a =

Iz Iy

tan u

Shear

Average direct shear stress

tavg = V A

Transverse shear stress

t =

VQ It

Shear flow

q = tt =

VQ I

Stress in Thin-Walled Pressure Vessel

Cylinder

s1 =

pr

t s2 =

pr

2t

Sphere

s1 = s2 =

pr

2t

Stress Transformation Equations

sx# = sx + sy

2 +

sx - sy 2

cos 2u + txy sin 2u

tx#y# = - sx - sy

2 sin 2u + txy cos 2u

Principal Stress

tan 2up = txy

(sx - sy)>2 s1,2 =

sx + sy 2

{ A asx - sy2 b2 + txy2 Maximum in-plane shear stress

tan 2us = - (sx - sy)>2

txy

tmax = A asx - sy2 b2 + t2xy savg =

sx + sy 2

Absolute maximum shear stress

tabs max

= smax

2 for smax, smin same sign

tabs max

= smax - smin

2 for smax, smin opposite signs

Geometric Properties of Area Elements Material Property Relations

Poisson’s ratio

n = - Plat

Plong

Generalized Hooke’s Law

ex =

1 E

3sx - n(sy + sz)4

ey = 1 E

3sy - n(sx + sz)4

ez = 1 E

3sz - n(sx + sy)4

gxy = 1 G

txy, gyz = 1 G

tyz, gzx = 1 G

tzx

where

G =

E 2(1 + n)

Relations Between w, V, M

dV dx

= w(x), dM dx

= V

Elastic Curve

1 r

= M EI

EI

d4v

dx4 = w(x)

EI

d3v

dx3 = V (x)

EI

d2v

dx2 = M(x)

Buckling Critical axial load

Pcr =

p2EI

(KL)2

Critical stress

scr =

p2E

(KL >r)2 , r = 2I>A Secant formula

smax =

P A

c 1 + ec r2

sec a L 2r

A PEA b d Energy Methods Conservation of energy

Ue = Ui

Strain energy Ui =

N2L 2AE

constant axial load

Ui = L

L

0

M2dx 2EI

bending moment

Ui = L

L

0

fsV 2dx

2GA transverse shear

Ui = L

L

0

T2dx 2GJ

torsional moment

xh

y A = bh

b

C

Rectangular area

Ix = bh 31

12

Iy = hb 31

12

Ix = bh 3x

h

A = bh

b

C

Triangular area

1 36h13

1 2

xh

A = h(a + b)

b

a

C

Trapezoidal area

h13 2a + b a + b

1 2

Ix = pr 4

x

y

C

Semicircular area

1 8

A = pr 2

2

4r 3p

Iy = pr 41

8

r

Ix = pr 4

x

y

C

Circular area

1 4

A = pr2

Iy = pr 41

4

r

A = ab

C

Semiparabolic area

2 3

a25

b38 a zero slope

b

Exparabolic area

a34 a

b310 zero slope

b C

A = ab3

Average Mechanical Properties of Typical Engineering Materialsa

(SI Units)

Materials Density R (Mg/m 3 )

Moduls of Elasticity E

(GPa)

Modulus of Rigidity G

(GPa)

Yield Strength (MPa) SY

Tens. Comp.b Shear

Ultimate Strength (MPa) Su

Tens. Comp. b Shear

%Elongation in 50 mm specimen

Poisson’s Ratio N

Coef. of Therm. Expansion A

(10–6)/°C

Metallic

Aluminum 2014-T6 Wrought Alloys 6061-T6

2.79 73.1 27 414 414 172 469 469 290 10 0.35 23

2.71 68.9 26 255 255 131 290 290 186 12 0.35 24

Cast Iron Gray ASTM 20 Alloys Malleable ASTM A-197

7.19 67.0 27 – – – 179 669 – 0.6 0.28 12

7.28 172 68 – – – 276 572 – 5 0.28 12

Copper Red Brass C83400 Alloys Bronze C86100

8.74 101 37 70.0 70.0 – 241 241 – 35 0.35 18

8.83 103 38 345 345 – 655 655 – 20 0.34 17

Magnesium [Am 1004-T61]

Alloy 1.83 44.7 18 152 152 – 276 276 152 1 0.30 26

Structural A-36

Steel Structural A992

Alloys Stainless 304

Tool L2

7.85 200 75 250 250 – 400 400 – 30 0.32 12

7.85 200 75 345 345 – 450 450 – 30 0.32 12

7.86 193 75 207 207 – 517 517 – 40 0.27 17

8.16 200 75 703 703 – 800 800 – 22 0.32 12

Titanium [Ti-6Al-4V]

Alloy 4.43 120 44 924 924 – 1,000 1,000 – 16 0.36 9.4

Nonmetallic

Concrete Low Strength

High Strength

2.38 22.1 – – – 12 – – – – 0.15 11

2.37 29.0 – – – 38 – – – – 0.15 11

Plastic Kevlar 49

Reinforced 30% Glass

1.45 131 – – – – 717 483 20.3 2.8 0.34 –

1.45 72.4 – – – – 90 131 – – 0.34 –

Wood Douglas Fir Select Structural White Spruce Grade

0.47 13.1 – – – – 2.1c 26d 6.2 d – 0.29e –

3.60 9.65 – – – – 2.5 c 36 d 6.7 d – 0.31 e –

a Specific values may vary for a particular material due to alloy or mineral composition,mechanical working of the specimen,or heat treatment. For a more exact value reference books for the material should be consulted.

b The yield and ultimate strengths for ductile materials can be assumed equal for both tension and compression. c Measured perpendicular to the grain. d Measured parallel to the grain. e Deformation measured perpendicular to the grain when the load is applied along the grain.

Average Mechanical Properties of Typical Engineering Materialsa

(U.S. Customary Units)

Materials Specific Weight (lb/in 3 )

Moduls of Elasticity E

(103) ksi

Modulus of Rigidity G (103) ksi

Yield Strength (ksi) SY

Tens. Comp.b Shear

Ultimate Strength (ksi) Su

Tens. Comp. b Shear

%Elongation in 2 in. specimen

Poisson’s Ratio N

Coef. of Therm. Expansion A

(10–6)/°F

Metallic

Aluminum 2014-T6 Wrought Alloys 6061-T6

0.101 10.6 3.9 60 60 25 68 68 42 10 0.35 12.8

0.098 10.0 3.7 37 37 19 42 42 27 12 0.35 13.1

Cast Iron Gray ASTM 20 Alloys Malleable ASTM A-197

0.260 10.0 3.9 – – – 26 96 – 0.6 0.28 6.70

0.263 25.0 9.8 – – – 40 83 – 5 0.28 6.60

Copper Red Brass C83400

Alloys Bronze C86100

0.316 14.6 5.4 11.4 11.4 – 35 35 – 35 0.35 9.80

0.319 15.0 5.6 50 50 – 35 35 – 20 0.34 9.60

Magnesium [Am 1004-T61]

Alloy 0.066 6.48 2.5 22 22 – 40 40 22 1 0.30 14.3

Structural A-36

Steel Structural A992

Alloys Stainless 304

Tool L2

0.284 29.0 11.0 36 36 – 58 58 – 30 0.32 6.60

0.284 29.0 11.0 50 50 – 65 65 – 30 0.32 6.60

0.284 28.0 11.0 30 30 – 75 75 – 40 0.27 9.60

0.295 29.0 11.0 102 102 – 116 116 – 22 0.32 6.50

Titanium [Ti-6Al-4V]

Alloy 0.160 17.4 6.4 134 134 – 145 145 – 16 0.36 5.20

Nonmetallic

Concrete Low Strength

High Strength

0.086 3.20 – – – 1.8 – – – – 0.15 6.0

0.086 4.20 – – – 5.5 – – – – 0.15 6.0

Plastic Kevlar 49

Reinforced 30% Glass

0.0524 19.0 – – – – 104 70 10.2 2.8 0.34 –

0.0524 10.5 – – – – 13 19 – – 0.34 –

Wood Douglas Fir Select Structural White Spruce Grade

0.017 1.90 – – – – 0.30c 3.78d 0.90 d – 0.29e –

0.130 1.40 – – – – 0.36 c 5.18 d 0.97 d – 0.31 e –

a Specific values may vary for a particular material due to alloy or mineral composition,mechanical working of the specimen,or heat treatment. For a more exact value reference books for the material should be consulted.

b The yield and ultimate strengths for ductile materials can be assumed equal for both tension and compression. c Measured perpendicular to the grain. d Measured parallel to the grain. e Deformation measured perpendicular to the grain when the load is applied along the grain.

MECHANICS OF MATERIALS

MECHANICS OF MATERIALS NINTH EDITION

R. C. HIBBELER

Prentice Hall Boston Columbus Indianapolis New York San Francisco Upper Saddle River Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montréal Toronto Delhi Mexico City São Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo

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Credits and acknowledgments borrowed from other sources and reproduced, with permission, in this textbook appear on appropriate page within text (or on page xix).

Copyright © 2014, 2011, 2008, 2005, 2003, 2001 by R. C. Hibbeler. Published by Pearson Prentice Hall. All rights reserved. Manufactured in the United States of America. This publication is protected by Copyright, and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. To obtain permission(s) to use material from this work, please submit a written request to Pearson Education, Inc., Permissions Department, 1 Lake Street, Upper Saddle River, NJ 07458.

Many of the designations by manufacturers and seller to distinguish their products are claimed as trademarks. Where those designations appear in this book, and the publisher was aware of a trademark claim, the designations have been printed in initial caps or all caps.

10 9 8 7 6 5 4 3

Library of Congress Cataloging-in-Publication Data on File.

ISBN 10: 0-13-325442-9 ISBN 13: 978-0-13-325442-6

Prentice Hall is an imprint of

www.pearsonhighered.com

To the Student With the hope that this work will stimulate

an interest in Engineering Mechanics and provide an acceptable guide to its understanding.

PREFACE

It is intended that this book provide the student with a clear and thorough presentation of the theory and application of the principles of mechanics of materials. To achieve this objective, over the years this work has been shaped by the comments and suggestions of hundreds of reviewers in the teaching profession, as well as many of the author’s students. The eighth edition has been significantly enhanced from the previous edition, and it is hoped that both the instructor and student will benefit greatly from these improvements.

New to This Edition

• Preliminary Problems. This feature can be found throughout the text, and is given just before the Fundamental Problems. The intent here is to test the student’s conceptual understanding of the theory. Normally the solutions require little or no calculation, and as such, these problems provide a basic understanding of the concepts before they are applied numerically. All the solutions are given in the back of the text.

• Updated Examples. Some portions of the text have been rewritten in order to enhance clarity and be more succinct. In this regard, some new examples have been added and others have been modified to provide more emphasis on the application of important concepts. Included is application of the LRFD method of design, and use of A992 steel for structural applications. Also, the artwork has been improved throughout the book to support these changes.

• New Photos. The relevance of knowing the subject matter is reflected by the real-world applications depicted in over 30 new or updated photos placed throughout the book. These photos generally are used to explain how the relevant principles apply to real-world situations and how materials behave under load.

• Additional Fundamental Problems. These problem sets are located just after each group of example problems. In this edition they have been expanded. They offer students simple applications of the concepts covered in each section and, therefore, provide them with the chance to develop their problem-solving skills before attempting to solve any of the standard problems that follow. The fundamental problems may be considered as extended examples, since the key equations and answers are all listed in the back of the book. Additionally, when assigned, these problems offer students an excellent means of preparing for exams, and they can be used at a later time as a review when studying for the Fundamentals of Engineering Exam.

VI I I PREFACE

• Additional Conceptual Problems. Throughout the text, usually at the end of each chapter, there is a set of problems that involve conceptual situations related to the application of the principles contained in the chapter. These analysis and design problems are intended to engage the students in thinking through a real-life situation as depicted in a photo. They can be assigned after the students have developed some expertise in the subject matter and they work well either for individual or team projects.

• New Problems. There are approximately 31%, or about 460, new problems added to this edition, which involve applications to many different fields of engineering.

Contents The subject matter is organized into 14 chapters. Chapter 1 begins with a review of the important concepts of statics, followed by a formal definition of both normal and shear stress, and a discussion of normal stress in axially loaded members and average shear stress caused by direct shear.

In Chapter 2 normal and shear strain are defined, and in Chapter 3 a discussion of some of the important mechanical properties of materials is given. Separate treatments of axial load, torsion, and bending are presented in Chapters 4 , 5 , and 6 , respectively. In each of these chapters, both linear- elastic and plastic behavior of the material are considered. Also, topics related to stress concentrations and residual stress are included. Transverse shear is discussed in Chapter 7 , along with a discussion of thin-walled tubes, shear flow, and the shear center. Chapter 8 includes a discussion of thin-walled pressure vessels and provides a partial review of the material covered in the previous chapters, where the state of stress results from combined loadings. In Chapter 9 the concepts for transforming multiaxial states of stress are presented. In a similar manner, Chapter 10 discusses the methods for strain transformation, including the application of various theories of failure. Chapter 11 provides a means for a further summary and review of previous material by covering design applications of beams and shafts. In Chapter 12 various methods for computing deflections of beams and shafts are covered. Also included is a discussion for finding the reactions on these members if they are statically indeterminate. Chapter 13 provides a discussion of column buckling, and lastly, in Chapter 14 the problem of impact and the application of various energy methods for computing deflections are considered.

Sections of the book that contain more advanced material are indicated by a star (*). Time permitting, some of these topics may be included in the course. Furthermore, this material provides a suitable reference for basic principles when it is covered in other courses, and it can be used as a basis for assigning special projects.

PREFACE IX

Alternative Method of Coverage. Some instructors prefer to cover stress and strain transformations first, before discussing specific applications of axial load, torsion, bending, and shear. One possible method for doing this would be first to cover stress and its transformation, Chapter 1 and Chapter 9, followed by strain and its transformation, Chapter 2 and the first part of Chapter 10. The discussion and example problems in these later chapters have been styled so that this is possible. Also, the problem sets have been subdivided so that this material can be covered without prior knowledge of the intervening chapters. Chapters 3 through 8 can then be covered with no loss in continuity.

Hallmark Elements Organization and Approach. The contents of each chapter are organized into well-defined sections that contain an explanation of specific topics, illustrative example problems, and a set of homework problems. The topics within each section are placed into subgroups defined by titles. The purpose of this is to present a structured method for introducing each new definition or concept and to make the book convenient for later reference and review.

Chapter Contents. Each chapter begins with a full-page illustration that indicates a broad-range application of the material within the chapter. The “Chapter Objectives” are then provided to give a general overview of the material that will be covered.

Procedures for Analysis. Found after many of the sections of the book, this unique feature provides the student with a logical and orderly method to follow when applying the theory. The example problems are solved using this outlined method in order to clarify its numerical application. It is to be understood, however, that once the relevant principles have been mastered and enough confidence and judgment have been obtained, the student can then develop his or her own procedures for solving problems.

Photographs. Many photographs are used throughout the book to enhance conceptual understanding and explain how the principles of mechanics of materials apply to real-world situations.

Important Points. This feature provides a review or summary of the most important concepts in a section and highlights the most significant points that should be realized when applying the theory to solve problems.

Example Problems. All the example problems are presented in a concise manner and in a style that is easy to understand.

Homework Problems. Apart from the preliminary, fundamental, and conceptual problems, there are numerous standard problems in the

X PREFACE

book that depict realistic situations encountered in engineering practice. It is hoped that this realism will both stimulate the student’s interest in the subject and provide a means for developing the skill to reduce any such problem from its physical description to a model or a symbolic representation to which principles may be applied. Throughout the book there is an approximate balance of problems using either SI or FPS units. Furthermore, in any set, an attempt has been made to arrange the problems in order of increasing difficulty. The answers to all but every fourth problem are listed in the back of the book. To alert the user to a problem without a reported answer, an asterisk (*) is placed before the problem number. Answers are reported to three significant figures, even though the data for material properties may be known with less accuracy. Although this might appear to be a poor practice, it is done simply to be consistent, and to allow the student a better chance to validate his or her solution. A solid square (!) is used to identify problems that require a numerical analysis or a computer application.

Appendices. The appendices of the book provide a source for review and a listing of tabular data. Appendix A provides information on the centroid and the moment of inertia of an area. Appendices B and C list tabular data for structural shapes, and the deflection and slopes of various types of beams and shafts.

Accuracy Checking. The Ninth Edition has undergone a rigorous Triple Accuracy Checking review. In addition to the author’s review of all art pieces and pages, the text was checked by the following individuals:

• Scott Hendricks, Virginia Polytechnic University • Karim Nohra, University of South Florida • Kurt Norlin, LaurelTech Integrated Publishing Services • Kai Beng Yap, Engineering Consultant

Acknowledgments Over the years, this text has been shaped by the suggestions and comments of many of my colleagues in the teaching profession. Their encouragement and willingness to provide constructive criticism are very much appreciated and it is hoped that they will accept this anonymous recognition. A note of thanks is given to the reviewers.

S. Suliman, Penn State C. Valle, Georgia Institute of Tech C. Sulzbach, Colorado School of Mines K. Cook-Chennault, Rutgers University J. Ramirez, Purdue University J. Oyler, University of Pittsburg

PREFACE XI

P. Mokashi, Ohio State Y. Liao, Arizona State University P. Ziehl, University of South Carolina

There are a few people that I feel deserve particular recognition. A long- time friend and associate, Kai Beng Yap, was of great help to me in checking the entire manuscript and helping to prepare the problem solutions. A special note of thanks also goes to Kurt Norlin in this regard. During the production process I am thankful for the assistance of Rose Kernan, my production editor for many years, and to my wife, Conny, and daughter, Mary Ann, for their help in proofreading and typing, that was needed to prepare the manuscript for publication.

I would also like to thank all my students who have used the previous edition and have made comments to improve its contents; including those in the teaching profession who have taken the time to e-mail me their comments, notably S. Alghamdi, A. Atai, S. Larwood, D. Kuemmerle, and J. Love.

I would greatly appreciate hearing from you if at any time you have any comments or suggestions regarding the contents of this edition.

Russell Charles Hibbeler hibbeler@bellsouth.net Mastering Ad to come

your work...

your answer specific feedback

®

XIV PREFACE

Resources for Instructors • MasteringEngineering. This online Tutorial Homework program allows you to integrate dynamic homework with automatic grading and adaptive tutoring. MasteringEngineering allows you to easily track the performance of your entire class on an assignment-by- assignment basis, or the detailed work of an individual student. • Instructor’s Solutions Manual. An instructor’s solutions manual was prepared by the author. The manual includes homework assignment lists and was also checked as part of the accuracy checking program. The Instructor Solutions Manual is available at www.pearsonhighered.com . • Presentation Resources. All art from the text is available in PowerPoint slide and JPEG format. These files are available for download from the Instructor Resource Center at www.pearsonhighered.com . If you are in need of a login and password for this site, please contact your local Pearson representative. • Video Solutions. Developed by Professor Edward Berger, University of Virginia, video solutions located on the Companion Website offer step-by-step solution walkthroughs of representative homework problems from each section of the text. Make efficient use of class time and office hours by showing students the complete and concise problem solving approaches that they can access anytime and view at their own pace. The videos are designed to be a flexible resource to be used however each instructor and student prefers. A valuable tutorial resource, the videos are also helpful for student self-evaluation as students can pause the videos to check their understanding and work alongside the video. Access the videos at www.pearsonhighered.com/ hibbeler and follow the links for the Mechanics of Materials text.

Resources for Students • Mastering Engineering. Tutorial homework problems emulate the instrutor’s office-hour environment. • Companion Website —The Companion Website, located at www.pearsonhighered.com/hibbeler includes opportunities for practice and review including: • Video Solutions —Complete, step-by-step solution walkthroughs of representative homework problems from each section. Videos offer: students need it with over 20 hours helpful review.

An access code for the Mechanics of Materials, Ninth Edition companion website was included with this text. To redeem the code and gain access to the site, go to www.pearsonhighered.com/hibbeler and follow the directions on the access code card. Access can also be purchased directly from the site.

CONTENTS

1 Stress 3

Chapter Objectives 3 1.1 Introduction 3 1.2 Equilibrium of a Deformable Body 4 1.3 Stress 22 1.4 Average Normal Stress in an Axially

Loaded Bar 24 1.5 Average Shear Stress 32 1.6 Allowable Stress Design 46 1.7 Limit State Design 48

2 Strain 67

Chapter Objectives 67 2.1 Deformation 67 2.2 Strain 68

3 Mechanical Properties of Materials 83

Chapter Objectives 83 3.1 The Tension and Compression Test 83 3.2 The Stress–Strain Diagram 85 3.3 Stress–Strain Behavior of Ductile and

Brittle Materials 89 3.4 Hooke’s Law 92 3.5 Strain Energy 94 3.6 Poisson’s Ratio 104 3.7 The Shear Stress–Strain Diagram 106 *3.8 Failure of Materials Due to Creep

and Fatigue 109

4 Axial Load 121

Chapter Objectives 121 4.1 Saint-Venant’s Principle 121 4.2 Elastic Deformation of an Axially Loaded

Member 124 4.3 Principle of Superposition 138 4.4 Statically Indeterminate Axially Loaded

Member 139 4.5 The Force Method of Analysis for Axially

Loaded Members 145 4.6 Thermal Stress 153 4.7 Stress Concentrations 160 *4.8 Inelastic Axial Deformation 164 *4.9 Residual Stress 166

5 Torsion 181

Chapter Objectives 181 5.1 Torsional Deformation of a Circular

Shaft 181 5.2 The Torsion Formula 184 5.3 Power Transmission 192 5.4 Angle of Twist 204 5.5 Statically Indeterminate Torque-Loaded

Members 218 *5.6 Solid Noncircular Shafts 225 *5.7 Thin-Walled Tubes Having Closed Cross

Sections 228 5.8 Stress Concentration 238 *5.9 Inelastic Torsion 241 *5.10 Residual Stress 243

XVI CONTENTS

6 Bending 259

Chapter Objectives 259 6.1 Shear and Moment Diagrams 259 6.2 Graphical Method for Constructing Shear

and Moment Diagrams 266 6.3 Bending Deformation of a Straight

Member 285 6.4 The Flexure Formula 289 6.5 Unsymmetric Bending 306 *6.6 Composite Beams 316 *6.7 Reinforced Concrete Beams 319 *6.8 Curved Beams 323 6.9 Stress Concentrations 330 *6.10 Inelastic Bending 339

7 Transverse Shear 363

Chapter Objectives 363 7.1 Shear in Straight Members 363 7.2 The Shear Formula 365 7.3 Shear Flow in Built-Up Members 382 7.4 Shear Flow in Thin-Walled Members 391 *7.5 Shear Center for Open Thin-Walled

Members 396

8 Combined Loadings 409

Chapter Objectives 409 8.1 Thin-Walled Pressure Vessels 409 8.2 State of Stress Caused by Combined

Loadings 416

9 Stress Transformation 441

Chapter Objectives 441 9.1 Plane-Stress Transformation 441 9.2 General Equations of Plane-Stress

Transformation 446 9.3 Principal Stresses and Maximum In-Plane

Shear Stress 449 9.4 Mohr’s Circle—Plane Stress 465 9.5 Absolute Maximum Shear Stress 477

10 Strain Transformation 489

Chapter Objectives 489 10.1 Plane Strain 489 10.2 General Equations of Plane-Strain

Transformation 490 *10.3 Mohr’s Circle—Plane Strain 498 *10.4 Absolute Maximum Shear Strain 506 10.5 Strain Rosettes 508 10.6 Material-Property Relationships 512 *10.7 Theories of Failure 524

11 Design of Beams and Shafts 541

Chapter Objectives 541 11.1 Basis for Beam Design 541 11.2 Prismatic Beam Design 544 *11.3 Fully Stressed Beams 558 *11.4 Shaft Design 562

CONTENTS XVI I

12 Deflection of Beams and Shafts 573

Chapter Objectives 573 12.1 The Elastic Curve 573 12.2 Slope and Displacement by

Integration 577 *12.3 Discontinuity Functions 597 *12.4 Slope and Displacement by the Moment-

Area Method 608 12.5 Method of Superposition 623 12.6 Statically Indeterminate Beams

and Shafts 631 12.7 Statically Indeterminate Beams and

Shafts—Method of Integration 632 *12.8 Statically Indeterminate Beams and

Shafts—Moment-Area Method 637 12.9 Statically Indeterminate Beams and

Shafts—Method of Superposition 643

13 Buckling of Columns 661

Chapter Objectives 661 13.1 Critical Load 661 13.2 Ideal Column with Pin Supports 664 13.3 Columns Having Various Types

of Supports 670 *13.4 The Secant Formula 682 *13.5 Inelastic Buckling 688 *13.6 Design of Columns for Concentric

Loading 696 *13.7 Design of Columns for Eccentric

Loading 707

14 Energy Methods 719

Chapter Objectives 719 14.1 External Work and Strain Energy 719 14.2 Elastic Strain Energy for Various Types

of Loading 724 14.3 Conservation of Energy 737 14.4 Impact Loading 744 *14.5 Principle of Virtual Work 755 *14.6 Method of Virtual Forces Applied

to Trusses 759 *14.7 Method of Virtual Forces Applied

to Beams 766 *14.8 Castigliano’s Theorem 775 *14.9 Castigliano’s Theorem Applied

to Trusses 777 *14.10 Castigliano’s Theorem Applied

to Beams 780

Appendix A. Geometric Properties of an Area 788 B. Geometric Properties of Structural

Shapes 804 C. Slopes and Deflections of Beams 812

Solutions and Answers for Preliminary Problems 814

Fundamental Problems Partial Solutions and Answers 825

Selected Answers 844

Index 866

Cover. Construction cranes. Mushakesa / Shutterstock. Chapter 1 Opener. Structural steel framework on new industrial unit. Copyright by Joe Gough/Shutterstock. Chapter 2 Opener. A copper rod after necking and fracture. sciencephotos / Alamy. Chapter 3 Opener. Earthquake damage on Olfusarbru bridge, South Coast, Iceland. ARCTIC IMAGES / Alamy. Page 63. Photo for Prob. C1-3 is from the Deutsches Museum. Used with permission. Chapter 4 Opener. Oil Rig. George Hammerstein /Corbis. Chapter 5 Opener. Piling machine drilling foundation on building site. HAWKEYE / Alamy. Chapter 6 Opener. Fragment view of the road under reconstruction. Lev Kropotov/Shutterstock. Chapter 7 Opener. Edmonton, Alberta, Canada. Design Pics / LJM Photo/LJM Photo/Newscom. Chapter 8 Opener. Tram on mountain, Snowbird Ski Resort, Wasatch Mountains, Utah, United States. Tetra Images / Alamy. Chapter 9 Opener. Industrial texture of power station generator turbine. Yurchyks/Shutterstock. Chapter 10 Opener. Support for Frederick Douglass-Susan B. Anthony Memorial Bridge, Rochester, NY USA. Russell C. Hibbeler. Chapter 11 Opener. Zinc steel structure of light fl oor. Jarous/Shutterstock. Chapter 12 Opener. Pole vaulter in midair. Corbis Flirt / Alamy. Chapter 13 Opener. Water tower against blue sky. VanHart/Shutterstock Chapter 14 Opener. An Olympic worker cleans a diving board as an Olympic athlete dives in practice at the Aquatics Center ahead of competition in the London Olympics in London, Britain, 24 July 2012. BARBARA WALTON/EPA/Newscom.

All other photos provided by the author.

CREDITS

MECHANICS OF MATERIALS

The bolts used for the connections of this steel framework are subjected to stress. In this chapter we will discuss how engineers design these

connections and their fasteners.

Chapter 1

3

1.1 Introduction Mechanics of materials is a branch of mechanics that studies the internal effects of stress and strain in a solid body that is subjected to an external loading. Stress is associated with the strength of the material from which the body is made, while strain is a measure of the deformation of the body. In addition to this, mechanics of materials includes the study of the body’s stability when a body such as a column is subjected to compressive loading. A thorough understanding of the fundamentals of this subject is of vital importance because many of the formulas and rules of design cited in engineering codes are based upon the principles of this subject.

CHAPTER OBJECTIVES

■ In this chapter we will review some of the important principles of statics and show how they are used to determine the internal resultant loadings in a body. Afterwards the concepts of normal and shear stress will be introduced, and specific applications of the analysis and design of members subjected to an axial load or direct shear will be discussed.

Stress

4 CHAPTER 1 STRESS

1 Historical Development. The origin of mechanics of materials dates back to the beginning of the seventeenth century, when Galileo performed experiments to study the effects of loads on rods and beams made of various materials. However, at the beginning of the eighteenth century, experimental methods for testing materials were vastly improved, and at that time many experimental and theoretical studies in this subject were undertaken primarily in France, by such notables as Saint-Venant, Poisson, Lamé and Navier.

Over the years, after many of the fundamental problems of mechanics of materials had been solved, it became necessary to use advanced mathematical and computer techniques to solve more complex problems. As a result, this subject expanded into other areas of mechanics, such as the theory of elasticity and the theory of plasticity . Research in these fields is ongoing, in order to meet the demands for solving more advanced problems in engineering.

1.2 Equilibrium of a Deformable Body Since statics has an important role in both the development and application of mechanics of materials, it is very important to have a good grasp of its fundamentals. For this reason we will review some of the main principles of statics that will be used throughout the text.

External Loads. A body is subjected to only two types of external loads; namely, surface forces and body forces, Fig. 1–1 .

Surface Forces. Surface forces are caused by the direct contact of one body with the surface of another. In all cases these forces are distributed over the area of contact between the bodies. If this area is small in comparison with the total surface area of the body, then the surface force can be idealized as a single concentrated force , which is applied to a point on the body. For example, the force of the ground on the wheels of a bicycle can be considered as a concentrated force. If the surface loading is applied along a narrow strip of area, the loading can be idealized as a linear distributed load , w ( s ). Here the loading is measured as having an intensity of force/length along the strip and is represented graphically by a series of arrows along the line s . The resultant force FR of w(s) is equivalent to the area under the distributed loading curve, and this resultant acts through the centroid C or geometric center of this area. The loading along the length of a beam is a typical example of where this idealization is often applied.

w(s)

Concentrated force idealization

Linear distributed load

Surface force

Body force

s

C

G

FR W

Fig. 1–1

1.2 EQUILIBRIUM OF A DEFORMABLE BODY 5

1

Many machine elements are pin connected in order to enable free rotation at their connections. These supports exert a force on a member, but no moment.

Body Forces. A body force is developed when one body exerts a force on another body without direct physical contact between the bodies. Examples include the effects caused by the earth’s gravitation or its electromagnetic field. Although body forces affect each of the particles composing the body, these forces are normally represented by a single concentrated force acting on the body. In the case of gravitation, this force is called the weight of the body and acts through the body’s center of gravity.

Support Reactions. The surface forces that develop at the supports or points of contact between bodies are called reactions . For two-dimensional problems, i.e., bodies subjected to coplanar force systems, the supports most commonly encountered are shown in Table 1–1 . Note carefully the symbol used to represent each support and the type of reactions it exerts on its contacting member. As a general rule, if the support prevents translation in a given direction, then a force must be developed on the member in that direction. Likewise, if rotation is prevented, a couple moment must be exerted on the member . For example, the roller support only prevents translation perpendicular or normal to the surface. Hence, the roller exerts a normal force F on the member at its point of contact. Since the member can freely rotate about the roller, a couple moment cannot be developed on the member.

F

F

Type of connection Reaction

Cable

Roller

One unknown: F

One unknown: F

F Smooth support One unknown: F

External pin

Internal pin

Fx

Fy

Fx

Fy

Two unknowns: Fx, Fy

Fx

FyM

Fixed support Three unknowns: Fx, Fy, M

Two unknowns: Fx, Fy

Type of connection Reaction

u u

u

TABLE 1–1

6 CHAPTER 1 STRESS

1 Equations of Equilibrium. Equilibrium of a body requires both a balance of forces , to prevent the body from translating or having accelerated motion along a straight or curved path, and a balance of moments , to prevent the body from rotating. These conditions can be expressed mathematically by two vector equations

!F = 0 !MO = 0

(1–1)

Here, ! F represents the sum of all the forces acting on the body, and ! MO is the sum of the moments of all the forces about any point O either on or off the body. If an x, y, z coordinate system is established with the origin at point O , the force and moment vectors can be resolved into components along each coordinate axis and the above two equations can be written in scalar form as six equations, namely,

!Fx = 0 !Fy = 0 !Fz = 0

!Mx = 0 !My = 0 !Mz = 0 (1–2)

Often in engineering practice the loading on a body can be represented as a system of coplanar forces . If this is the case, and the forces lie in the x – y plane, then the conditions for equilibrium of the body can be specified with only three scalar equilibrium equations; that is,

!Fx = 0 !Fy = 0

!MO = 0 (1–3)

Here all the moments are summed about point O and so they will be directed along the z axis.

Successful application of the equations of equilibrium requires complete specification of all the known and unknown forces that act on the body, and so the best way to account for all these forces is to draw the body’s free-body diagram .

In order to design the horizontal members of this building frame, it is first necessary to find the internal loadings at various points along their length.

1.2 EQUILIBRIUM OF A DEFORMABLE BODY 7

1

Internal Resultant Loadings. In mechanics of materials, statics is primarily used to determine the resultant loadings that act within a body. For example, consider the body shown in Fig. 1–2 a , which is held in equilibrium by the four external forces. * In order to obtain the internal loadings acting on a specific region within the body, it is necessary to pass an imaginary section or “cut” through the region where the internal loadings are to be determined. The two parts of the body are then separated, and a free-body diagram of one of the parts is drawn, Fig. 1–2 b . Notice that there is actually a distribution of internal force acting on the “exposed” area of the section. These forces represent the effects of the material of the top part of the body acting on the adjacent material of the bottom part.

Although the exact distribution of this internal loading may be unknown , we can use the equations of equilibrium to relate the external forces on the bottom part of the body to the distribution’s resultant force and moment , FR and MRO, at any specific point O on the sectioned area, Fig. 1–2 c . It will be shown in later portions of the text that point O is most often chosen at the centroid of the sectioned area, and so we will always choose this location for O , unless otherwise stated. Also, if a member is long and slender, as in the case of a rod or beam, the section to be considered is generally taken perpendicular to the longitudinal axis of the member. This section is referred to as the cross section .

* The body’s weight is not shown, since it is assumed to be quite small, and therefore negligible compared with the other loads.

section

F4

F2 (a)

F1

F3

F1 F2

(b)

FR

F1 F2

O

MRO

(c)

Fig. 1–2

8 CHAPTER 1 STRESS

1

Three Dimensions. Later in this text we will show how to relate the resultant loadings, FR and MRO, to the distribution of force on the sectioned area, and thereby develop equations that can be used for analysis and design. To do this, however, the components of FR and MRO acting both normal and perpendicular to the sectioned area must be considered, Fig. 1–2 d . Four different types of resultant loadings can then be defined as follows:

Normal force, N. This force acts perpendicular to the area. It is developed whenever the external loads tend to push or pull on the two segments of the body.

Shear force, V. The shear force lies in the plane of the area, and it is developed when the external loads tend to cause the two segments of the body to slide over one another.

Torsional moment or torque, T. This effect is developed when the external loads tend to twist one segment of the body with respect to the other about an axis perpendicular to the area.

Bending moment, M. The bending moment is caused by the external loads that tend to bend the body about an axis lying within the plane of the area.

In this text, note that graphical representation of a moment or torque is shown in three dimensions as a vector with an associated curl. By the right-hand rule , the thumb gives the arrowhead sense of this vector and the fingers or curl indicate the tendency for rotation (twisting or bending).

O

(c)

MRO

F1 F2

FR

(d)

O

F1 F2

N

T

M V

Torsional Moment

Bending Moment

Shear Force

MRO

FR

Normal Force

Fig. 1–2 (cont.)

The weight of this sign and the wind loadings acting on it will cause normal and shear forces and bending and torsional moments in the supporting column.

1.2 EQUILIBRIUM OF A DEFORMABLE BODY 9

1

Coplanar Loadings. If the body is subjected to a coplanar system of forces , Fig. 1–3 a , then only normal-force, shear-force, and bending-moment components will exist at the section, Fig. 1–3 b . If we use the x, y, z coordinate axes, as shown on the left segment, then N can be obtained by applying !Fx = 0, and V can be obtained from !Fy = 0. Finally, the bending moment MO can be determined by summing moments about point O (the z axis), !MO = 0, in order to eliminate the moments caused by the unknowns N and V .

section

F4

F3F2

F1

(a)

O

V MO

N x

y

Bending Moment

Shear Force

Normal Force

(b)

F2

F1

Fig. 1–3

Important Points

• Mechanics of materials is a study of the relationship between the external loads applied to a body and the stress and strain caused by the internal loads within the body.

• External forces can be applied to a body as distributed or concentrated surface loadings , or as body forces that act throughout the volume of the body.

• Linear distributed loadings produce a resultant force having a magnitude equal to the area under the load diagram, and having a location that passes through the centroid of this area.

• A support produces a force in a particular direction on its attached member if it prevents translation of the member in that direction, and it produces a couple moment on the member if it prevents rotation .

• The equations of equilibrium !F = 0 and !M = 0 must be satisfied in order to prevent a body from translating with accelerated motion and from rotating.

• When applying the equations of equilibrium, it is important to first draw the free-body diagram for the body in order to account for all the terms in the equations.

• The method of sections is used to determine the internal resultant loadings acting on the surface of the sectioned body. In general, these resultants consist of a normal force, shear force, torsional moment, and bending moment.

10 CHAPTER 1 STRESS

1 Procedure for Analysis

The resultant internal loadings at a point located on the section of a body can be obtained using the method of sections. This requires the following steps.

Support Reactions. • First decide which segment of the body is to be considered. If the

segment has a support or connection to another body, then before the body is sectioned, it will be necessary to determine the reactions acting on the chosen segment. To do this draw the free-body diagram of the entire body and then apply the necessary equations of equilibrium to obtain these reactions.

Free-Body Diagram. • Keep all external distributed loadings, couple moments, torques,

and forces in their exact locations , before passing an imaginary section through the body at the point where the resultant internal loadings are to be determined.

• Draw a free-body diagram of one of the “cut” segments and indicate the unknown resultants N, V, M , and T at the section. These resultants are normally placed at the point representing the geometric center or centroid of the sectioned area.

• If the member is subjected to a coplanar system of forces, only N, V , and M act at the centroid.

• Establish the x, y, z coordinate axes with origin at the centroid and show the resultant internal loadings acting along the axes.

Equations of Equilibrium. • Moments should be summed at the section, about each of the

coordinate axes where the resultants act. Doing this eliminates the unknown forces N and V and allows a direct solution for M (and T ).

• If the solution of the equilibrium equations yields a negative value for a resultant, the directional sense of the resultant is opposite to that shown on the free-body diagram.

The following examples illustrate this procedure numerically and also provide a review of some of the important principles of statics.

1.2 EQUILIBRIUM OF A DEFORMABLE BODY 11

1 EXAMPLE 1.1

Determine the resultant internal loadings acting on the cross section at C of the cantilevered beam shown in Fig. 1–4 a .

(a)

A B

C 3 m 6 m

270 N/m

Fig. 1–4

SOLUTION Support Reactions. The support reactions at A do not have to be determined if segment CB is considered.

Free-Body Diagram. The free-body diagram of segment CB is shown in Fig. 1–4 b . It is important to keep the distributed loading on the segment until after the section is made. Only then should this loading be replaced by a single resultant force. Notice that the intensity of the distributed loading at C is found by proportion, i.e., from Fig. 1–4 a , w >6 m = (270 N>m)>9 m, w = 180 N>m. The magnitude of the resultant of the distributed load is equal to the area under the loading curve (triangle) and acts through the centroid of this area. Thus, F = 12(180 N>m)(6 m) = 540 N, which acts 13(6 m) = 2 m from C as shown in Fig. 1–4 b .

Equations of Equilibrium. Applying the equations of equilibrium we have

S + !Fx = 0; -NC = 0 NC = 0 Ans. + c!Fy = 0; VC - 540 N = 0 VC = 540 N Ans. a+ !MC = 0; -MC - 540 N(2 m) = 0

MC = -1080 N # m Ans. NOTE: The negative sign indicates that MC acts in the opposite direction to that shown on the free-body diagram. Try solving this problem using segment AC , by first obtaining the support reactions at A , which are given in Fig. 1–4 c .

180 N/m

540 N

2 m 4 mVC

MC

NC

(b)

BC

1.5 m 0.5 m

1 m

180 N/m90 N/m

540 N 135 N

VC

MC

NC

(c)

1215 N

3645 N"m CA

12 CHAPTER 1 STRESS

1 EXAMPLE 1.2

The 500-kg engine is suspended from the crane boom in Fig. 1–5 a . Determine the resultant internal loadings acting on the cross section of the boom at point E .

SOLUTION Support Reactions. We will consider segment AE of the boom, so we must first determine the pin reactions at A . Notice that member CD is a two-force member. The free-body diagram of the boom is shown in Fig. 1–5 b . Applying the equations of equilibrium,

a+ !MA = 0; FCD1352(2 m) - [500(9.81) N](3 m) = 0 FCD = 12 262.5 N

S + !Fx = 0; A x - (12 262.5 N)1452 = 0 A x = 9810 N

+ c!Fy = 0; -A y + (12 262.5 N)1352 - 500(9.81) N = 0 A y = 2452.5 N

Free-Body Diagram. The free-body diagram of segment AE is shown in Fig. 1–5 c .

Equations of Equilibrium.

S + !Fx = 0; NE + 9810 N = 0

NE = -9810 N = -9.81 kN Ans.

+ c !Fy = 0; -VE - 2452.5 N = 0

VE = -2452.5 N = -2.45 kN Ans.

a+ !ME = 0; ME + (2452.5 N)( 1 m) = 0

ME = -2452.5 N # m = -2.45 kN # m Ans.

A 1 m1 m1 m

1.5 m

E

C

B

D

(a)

A

1 m2 m

500(9.81) N

Ay

Ax

FCD

(b)

3 4

5

9810 N

2452.5 N

VE

ME

NE

(c)

EA

1 m

Fig. 1–5

1.2 EQUILIBRIUM OF A DEFORMABLE BODY 13

1 EXAMPLE 1.3

Determine the resultant internal loadings acting on the cross section at G of the beam shown in Fig. 1–6 a . Each joint is pin connected.

(a)

300 lb/ft

2 ft 2 ft 6 ft

1500 lb

A

B

G D

C

3 ft E

3 ft

6 ft (6 ft) ! 4 ft

(6 ft)(300 lb/ft) ! 900 lb

1500 lb

Ey ! 2400 lb

Ex ! 6200 lb

FBC ! 6200 lb

(b)

2 3

1 2

6200 lb

3 4

5

(c)

B

FBA ! 7750 lb FBD ! 4650 lb

(d)

NG

MGVG2 ft

3 4

5

7750 lb1500 lb

A G

SOLUTION Support Reactions. Here we will consider segment AG . The free-body diagram of the entire structure is shown in Fig. 1–6 b . Verify the calculated reactions at E and C . In particular, note that BC is a two-force member since only two forces act on it. For this reason the force at C must act along BC, which is horizontal as shown.

Since BA and BD are also two-force members, the free-body diagram of joint B is shown in Fig. 1–6 c . Again, verify the magnitudes of forces FBA and FBD .

Free-Body Diagram. Using the result for FBA , the free-body diagram of segment AG is shown in Fig. 1–6 d .

Equations of Equilibrium.

S + !Fx = 0; 7750 lb1452 + NG = 0 NG = -6200 lb Ans. + c !Fy = 0; -1500 lb + 7750 lb1352 - VG = 0 VG = 3150 lb Ans.

a+ !MG = 0; MG - (7750 lb)1352(2 ft) + 1500 lb(2 ft) = 0 MG = 6300 lb # ft Ans.

Fig. 1–6

14 CHAPTER 1 STRESS

1 EXAMPLE 1.4

Determine the resultant internal loadings acting on the cross section at B of the pipe shown in Fig. 1–7 a . End A is subjected to a vertical force of 50 N, a horizontal force of 30 N, and a couple moment of 70 N # m . Neglect the pipe’s mass.

0.75 m

50 N

1.25 m

B

A

0.5 m

C

D

70 N!m

(a)

30 N

1.25 m

70 N·m 30 N

(b)

y

A

50 N

0.5 m

x

z

B

(FB)z (MB)z

(MB)x (FB)x

(MB)y

(FB)y

Fig. 1–7

SOLUTION The problem can be solved by considering segment AB , so we do not need to calculate the support reactions at C .

Free-Body Diagram. The x, y, z axes are established at B and the free-body diagram of segment AB is shown in Fig. 1–7 b . The resultant force and moment components at the section are assumed to act in the positive coordinate directions and to pass through the centroid of the cross-sectional area at B .

Equations of Equilibrium. Applying the six scalar equations of equilibrium, we have *

"Fx = 0; (FB)x = 0 Ans.

"Fy = 0; (FB)y + 30 N = 0 (FB)y = -30 N Ans.

"Fz = 0; (FB)z - 50 N = 0 (FB)z = 50 N Ans.

"(MB)x = 0; (MB)x + 70 N # m - 50 N (0.5 m) = 0 (MB)x = -45 N # m Ans. "(MB)y = 0; (MB)y + 50 N (1.25 m) = 0

(MB)y = -62.5 N # m Ans. "(MB)z = 0; (MB)z + (30 N)(1.25) = 0 Ans.

(MB)z = -37.5 N # m NOTE: What do the negative signs for (FB)y, (MB)x , (MB)y and (MB)z indicate? The normal force NB = ! (FB)y ! = 30 N, whereas the shear force is VB = 2(0)2 + (50)2 = 50 N. Also, the torsional moment is TB = ! (MB)y ! = 62.5 N # m and the bending moment is MB = 2(45)2 + (37.5)2 = 58.6 N # m.

* The magnitude of each moment about an axis is equal to the magnitude of each force times the perpendicular distance from the axis to the line of action of the force. The direction of each moment is determined using the right-hand rule, with positive moments (thumb) directed along the positive coordinate axes.

1.2 EQUILIBRIUM OF A DEFORMABLE BODY 15

1

P1–1. In each case, explain how to find the resultant internal loading acting on the cross section at point A . Draw all necessary free-body diagrams, and indicate the relevant equations of equilibrium. Do not calculate values. The lettered dimensions, angles, and loads are assumed to be known.

P

B Au

2a aa

(a)

C

D

B

a a

w

A C

a

P

(b)

B CA

P

u

a

(c)

a/2 a/2

M

O P

B

r

A

(d)

f

u

(e)

a

a

2a

A

C

B

P

u

a

a a

a 3a

a

P (f)

B

A

D

C

PRELIMINARY PROBLEMS

It is suggested that you test yourself on the solutions to these examples, by covering them over and then trying to think about which equilibrium equations must be used and how they are applied in order to determine the unknowns. Then before solving any of the Problems, build your skills by first trying to solve the Preliminary Problems, which actually require little or no calculations, and then do some of the Fundamental Problems given on the following pages. The solutions and answers to all these problems are given in the back of the book . Doing this throughout the book will help immensely in understanding how to apply the theory, and thereby develop your problem-solving skills.

16 CHAPTER 1 STRESS

1

F1–1. Determine the internal normal force, shear force, and bending moment at point C in the beam.

A B

C

2 m 2 m1 m

60 kN!m

1 m

10 kN

F1–1

F1–2. Determine the internal normal force, shear force, and bending moment at point C in the beam.

A B C

1.5 m 1.5 m

100 N/m 200 N/m

F1–2

F1–3. Determine the internal normal force, shear force, and bending moment at point C in the beam.

A

B

2 m 2 m 2 m

C

20 kN/m

F1–3

F1–4. Determine the internal normal force, shear force, and bending moment at point C in the beam.

A C

B

3 m3 m

10 kN/m

F1–4

F1–5. Determine the internal normal force, shear force, and bending moment at point C in the beam.

300 lb/ft

3 ft 3 ft 3 ft

A BC

F1–5

F1–6. Determine the internal normal force, shear force, and bending moment at point C in the beam.

3 m

2 m 2 m 2 m

A

D

C B

5 kN/m

F1–6

FUNDAMENTAL PROBLEMS

1.2 EQUILIBRIUM OF A DEFORMABLE BODY 17

1 PROBLEMS

1–1. The shaft is supported by a smooth thrust bearing at B and a journal bearing at C . Determine the resultant internal loadings acting on the cross section at E .

A E DB C

4 ft

400 lb 800 lb

4 ft 4 ft 4 ft

Prob. 1–1

1–2. Determine the resultant internal normal and shear force in the member at (a) section a – a and (b) section b – b , each of which passes through point A . The 500-lb load is applied along the centroidal axis of the member.

30!

A

ba

b a

500 lb500 lb

Prob. 1–2

1–3. The beam AB is fixed to the wall and has a uniform weight of 80 lb>ft. If the trolley supports a load of 1500 lb, determine the resultant internal loadings acting on the cross sections through points C and D.

D

5 ft 20 ft

3 ft 10 ft

C

BA

1500 lb

Prob. 1–3

*1–4. The shaft is supported by a smooth thrust bearing at A and a smooth journal bearing at B . Determine the resultant internal loadings acting on the cross section at C .

A DB

C

900 N

1.5 m

600 N/m

1.5 m1 m1 m1 m

Prob. 1–4

1–5. Determine the resultant internal loadings in the beam at cross sections through points D and E . Point E is just to the right of the 3-kip load.

6 ft 4 ft

A

4 ft

B CD E

6 ft

3 kip

1.5 kip/ft

Prob. 1–5

1–6. Determine the normal force, shear force, and moment at a section through point C . Take P = 8 kN.

1–7. The cable will fail when subjected to a tension of 2 kN. Determine the largest vertical load P the frame will support and calculate the internal normal force, shear force, and moment at the cross section through point C for this loading.

0.75 m

C

P

A

B

0.5 m 0.1 m

0.75 m 0.75 m

Probs. 1–6/7

18 CHAPTER 1 STRESS

1 1–11. The forearm and biceps support the 2-kg load at A . If C can be assumed as a pin support, determine the resultant internal loadings acting on the cross section of the bone of the forearm at E. The biceps pulls on the bone along BD.

75!

230 mm 35 mm35 mm

C E B

D

A

Prob. 1–11

*1–12. The serving tray T used on an airplane is supported on each side by an arm. The tray is pin connected to the arm at A , and at B there is a smooth pin. (The pin can move within the slot in the arms to permit folding the tray against the front passenger seat when not in use.) Determine the resultant internal loadings acting on the cross section of the arm through point C when the tray arm supports the loads shown.

9 N

500 mm

12 N

15 mm 150 mm

60!

AB

C

T

VC MC

NC

100 mm

Prob. 1–12

*1–8. Determine the resultant internal loadings on the cross section through point C . Assume the reactions at the supports A and B are vertical.

1–9. Determine the resultant internal loadings on the cross section through point D . Assume the reactions at the supports A and B are vertical.

0.5 m 0.5 m 1.5 m1.5 m

C A B

3 kN/m 6 kN

D

Probs. 1–8/9

1–10. The boom DF of the jib crane and the column DE have a uniform weight of 50 lb>ft. If the hoist and load weigh 300 lb, determine the resultant internal loadings in the crane on cross sections through points A , B , and C .

5 ft

7 ft

C

D F

E

B A

300 lb

2 ft 8 ft 3 ft

Prob. 1–10

1.2 EQUILIBRIUM OF A DEFORMABLE BODY 19

1 1–17. Determine resultant internal loadings acting on section a – a and section b – b . Each section passes through the centerline at point C .

45!

1.5 m

1.5 m

3 m

45!

A

C

B

b a

a b

5 kN

Prob. 1–17

1–18. The bolt shank is subjected to a tension of 80 lb. Determine the resultant internal loadings acting on the cross section at point C .

A B

C

90! 6 in.

Prob. 1–18

1–19. Determine the resultant internal loadings acting on the cross section through point C . Assume the reactions at the supports A and B are vertical.

*1–20. Determine the resultant internal loadings acting on the cross section through point D . Assume the reactions at the supports A and B are vertical.

3 ft 3 ft

DC A B

6 ft

6 kip/ft6 kip/ft

Probs. 1–19/20

1–13. The blade of the hacksaw is subjected to a pretension force of F = 100 N. Determine the resultant internal loadings acting on section a – a that passes through point D .

1–14. The blade of the hacksaw is subjected to a pretension force of F = 100 N. Determine the resultant internal loadings acting on section b – b that passes through point D .

A B

C

D

F F

a

b

b a

30!

225 mm

150 mm

Probs. 1–13/14

1–15. A 150-lb bucket is suspended from a cable on the wooden frame. Determine the resultant internal loadings on the cross section at D.

*1–16. A 150-lb bucket is suspended from a cable on the wooden frame. Determine the resultant internal loadings acting on the cross section at E .

2 ft

2 ft

3 ft

1 ft1 ft

1 ft E

D C

B

AI

30! G

H

Probs. 1–15/16

20 CHAPTER 1 STRESS

1 *1–24. Determine the resultant internal loadings acting on the cross section of the semicircular arch at C .

C

A B

w0

u

r

Prob. 1–24

1–25. Determine the resultant internal loadings acting on the cross section through point B of the signpost. The post is fixed to the ground and a uniform pressure of 7 lb> ft2 acts perpendicular to the face of the sign.

4 ft

z

y

6 ft

x

B

A

3 ft

2 ft

3 ft

7 lb/ft2

Prob. 1–25

1–21. The forged steel clamp exerts a force of F = 900 N on the wooden block. Determine the resultant internal loadings acting on section a – a passing through point A .

200 mm

a

a F ! 900 N

F ! 900 N

30" A

Prob. 1–21

1–22. The metal stud punch is subjected to a force of 120 N on the handle. Determine the magnitude of the reactive force at the pin A and in the short link BC . Also, determine the internal resultant loadings acting on the cross section passing through the handle arm at D .

1–23. Solve Prob. 1–22 for the resultant internal loadings acting on the cross section passing through the handle arm at E and at a cross section of the short link BC .

60! 50 mm

100 mm

200 mm

300 mm B

C

D

120 N

50 mm 100 mm

E

30! A

Probs. 1–22/23

1.2 EQUILIBRIUM OF A DEFORMABLE BODY 21

1 *1–28. The brace and drill bit is used to drill a hole at O . If the drill bit jams when the brace is subjected to the forces shown, determine the resultant internal loadings acting on the cross section of the drill bit at A .

z

x y

AO

9 in. 6 in.

6 in. 6 in.

9 in.3 in.

Fx ! 30 lb

Fy ! 50 lb

Fz ! 10 lb

Prob. 1–28 1–29. The curved rod AD of radius r has a weight per length of w . If it lies in the vertical plane, determine the resultant internal loadings acting on the cross section through point B . Hint : The distance from the centroid C of segment AB to point O is OC = [2 r sin (u>2)]>u.

O

r

C

B

D

A

u

u 2

Prob. 1–29 1–30. A differential element taken from a curved bar is shown in the figure. Show that dN >du = V , dV >du = -N, dM >du = -T, and dT >du = M.

M V

N du

M ! dM T ! dT

N ! dN V ! dV

T

Prob. 1–30

1–26. The shaft is supported at its ends by two bearings A and B and is subjected to the forces applied to the pulleys fixed to the shaft. Determine the resultant internal loadings acting on the cross section located at point C. The 300-N forces act in the -z direction and the 500-N forces act in the +x direction. The journal bearings at A and B exert only x and z components of force on the shaft.

y

B

C

400 mm

150 mm

200 mm

250 mm

A

x

z

300 N 300 N

500 N

500 N

Prob. 1–26

1–27. The pipe assembly is subjected to a force of 600 N at B . Determine the resultant internal loadings acting on the cross section at C .

A C

B

yx

z

400 mm

150 mm

500 mm

600 N

150 mm

30"

60"

Prob. 1–27

22 CHAPTER 1 STRESS

1 1.3 Stress It was stated in Section 1.2 that the force and moment acting at a specified point O on the sectioned area of the body, Fig. 1–8 , represents the resultant effects of the actual distribution of loading acting over the sectioned area, Fig. 1–9 a . Obtaining this distribution is of primary importance in mechanics of materials. To solve this problem it is necessary to establish the concept of stress.

We begin by considering the sectioned area to be subdivided into small areas, such as !A shown in Fig. 1–9 a . As we reduce !A to a smaller and smaller size, we must make two assumptions regarding the properties of the material. We will consider the material to be continuous , that is, to consist of a continuum or uniform distribution of matter having no voids. Also, the material must be cohesive , meaning that all portions of it are connected together, without having breaks, cracks, or separations. A typical finite yet very small force !F, acting on !A , is shown in Fig. 1–9 a . This force, like all the others, will have a unique direction, but for further discussion we will replace it by its three components , namely, !Fx, !Fy, and !Fz, which are taken tangent, tangent, and normal to the area, respectively. As !A approaches zero, so do !F and its components; however, the quotient of the force and area will, in general, approach a finite limit. This quotient is called stress , and as noted, it describes the intensity of the internal force acting on a specific plane (area) passing through a point.

F1 F2

O

MRO FR

Fig. 1–8

F1 F2 F1

!F

!A

!F !Fz

z

yx !Fx !Fy z

(c)x y

(b)

zz

x y(a)x y

tyz

sy tyx

txz

sx txy

Fig. 1–9

1.3 STRESS 23

1 Normal Stress. The intensity of the force acting normal to !A is defined as the normal stress , s (sigma). Since !Fz is normal to the area then

sz = lim!A S0 !Fz !A (1–4)

If the normal force or stress “pulls” on !A as shown in Fig. 1–9 a , it is referred to as tensile stress , whereas if it “pushes” on !A it is called compressive stress .

Shear Stress. The intensity of force acting tangent to !A is called the shear stress , t (tau). Here we have shear stress components,

tzx = lim!A S0 !Fx !A

(1–5)

tzy = lim!A S0 !Fy !A

Note that in this subscript notation z specifies the orientation of the area !A , Fig. 1–10 , and x and y indicate the axes along which each shear stress acts.

General State of Stress. If the body is further sectioned by planes parallel to the x – z plane, Fig. 1–9 b , and the y – z plane, Fig. 1–9 c , we can then “cut out” a cubic volume element of material that represents the state of stress acting around the chosen point in the body. This state of stress is then characterized by three components acting on each face of the element, Fig. 1–11 .

Units. Since stress represents a force per unit area, in the International Standard or SI system, the magnitudes of both normal and shear stress are specified in the basic units of newtons per square meter (N>m2). This unit, called a p ascal (1 Pa = 1 N>m2) is rather small, and in engineering work prefixes such as kilo- (103), symbolized by k, mega- (106), symbolized by M, or giga- (109), symbolized by G, are used to represent larger, more realistic values of stress. * Likewise, in the Foot-Pound-Second system of units, engineers usually express stress in pounds per square inch (psi) or kilopounds per square inch (ksi), where 1 kilopound (kip) = 1000 lb.

x y

z

Tzx Tzy

sz

Fig. 1–10

x y

z

z

zx zy

yz

yx

xz

x xy

y

s

s s

t

tt

t

t t

Fig. 1–11

* Sometimes stress is expressed in units of N>mm2, where 1 mm = 10-3 m. However, in the SI system, prefixes are not allowed in the denominator of a fraction and therefore it is better to use the equivalent 1 N>mm2 = 1 MN>m2 = 1 MPa.

24 CHAPTER 1 STRESS

1 1.4 Average Normal Stress in an Axially Loaded Bar

In this section we will determine the average stress distribution acting on the cross-sectional area of an axially loaded bar such as the one shown in Fig. 1–12 a . This bar is prismatic since all cross sections are the same throughout its length. When the load P is applied to the bar through the centroid of its cross-sectional area, then the bar will deform uniformly throughout the central region of its length, as shown in Fig. 1–12 b , provided the material of the bar is both homogeneous and isotropic.

Homogeneous material has the same physical and mechanical properties throughout its volume, and isotropic material has these same properties in all directions. Many engineering materials may be approximated as being both homogeneous and isotropic as assumed here. Steel, for example, contains thousands of randomly oriented crystals in each cubic millimeter of its volume, and since most problems involving this material have a physical size that is very much larger than a single crystal, the above assumption regarding its material composition is quite realistic.

Note that anisotropic materials such as wood have different properties in different directions, and although this is the case, if the anisotropy is oriented along the bar’s axis (as for instance in a typical wood rod), then the bar will also deform uniformly when subjected to the axial load P .

Average Normal Stress Distribution. If we pass a section through the bar, and separate it into two parts, then equilibrium requires the resultant normal force at the section to be P , Fig. 1–12 c . Due to the uniform deformation of the material, it is necessary that the cross section be subjected to a constant normal stress distribution , Fig. 1–12 d .

P

P (a) (b)

P

P

Region of uniform deformation of bar

P

P

External force

Cross-sectional area

Internal force

(c) (d)

P

!F " s!A

P

y

x

x

z

y

A!

s

Fig. 1–12

1.4 AVERAGE NORMAL STRESS IN AN AXIALLY LOADED BAR 25

1 As a result, each small area !A on the cross section is subjected to a

force !F = s !A , and the sum of these forces acting over the entire cross-sectional area must be equivalent to the internal resultant force P at the section. If we let !A S dA and therefore !F S dF , then, recognizing s is constant , we have

+ c FRz = "Fz; LdF = LAs dA P = s A

s = P A (1–6)

Here

s = average normal stress at any point on the cross-sectional area

P = internal resultant normal force , which acts through the centroid of the cross-sectional area. P is determined using the method of sections and the equations of equilibrium

A = cross-sectional area of the bar where s is determined

Since the internal load P passes through the centroid of the cross-section, the uniform stress distribution will produce zero moments about the x and y axes passing through this point, Fig. 1–12 d . To show this, we require the moment of P about each axis to be equal to the moment of the stress distribution about the axes, namely,

(MR)x = " Mx; 0 = LA y dF = LA ys dA = sLA y dA

(MR)y = " My; 0 = - LA x dF = - LA xs dA = -sLA x dA

These equations are indeed satisfied, since by definition of the centroid, 1y dA = 0 and 1x dA = 0. (See Appendix A .)

Equilibrium. It should be apparent that only a normal stress exists on any small volume element of material located at each point on the cross section of an axially loaded bar. If we consider vertical equilibrium of the element, Fig. 1–13 , then apply the equation of force equilibrium,

"Fz = 0; s(!A ) - s#(!A ) = 0

s = s#

!A

s

s¿

Fig. 1–13

26 CHAPTER 1 STRESS

1

In other words, the two normal stress components on the element must be equal in magnitude but opposite in direction. This is referred to as uniaxial stress .

The previous analysis applies to members subjected to either tension or compression, as shown in Fig. 1–14 . As a graphical interpretation, the magnitude of the internal resultant force P is equivalent to the volume under the stress diagram; that is, P = s A (volume = height * base). Furthermore, as a consequence of the balance of moments, this resultant passes through the centroid of this volume .

Although we have developed this analysis for prismatic bars, this assumption can be relaxed somewhat to include bars that have a slight taper . For example, it can be shown, using the more exact analysis of the theory of elasticity, that for a tapered bar of rectangular cross section, for which the angle between two adjacent sides is 15°, the average normal stress, as calculated by s = P>A , is only 2.2% less than its value found from the theory of elasticity.

Maximum Average Normal Stress. In our analysis both the internal force P and the cross-sectional area A were constant along the longitudinal axis of the bar, and as a result the normal stress s = P>A is also constant throughout the bar’s length. Occasionally, however, the bar may be subjected to several external loads along its axis, or a change in its cross-sectional area may occur. As a result, the normal stress within the bar could be different from one section to the next, and, if the maximum average normal stress is to be determined, then it becomes important to find the location where the ratio P>A is a maximum . To do this it is necessary to determine the internal force P at various sections along the bar. Here it may be helpful to show this variation by drawing an axial or normal force diagram . Specifically, this diagram is a plot of the normal force P versus its position x along the bar’s length. As a sign convention, P will be positive if it causes tension in the member, and negative if it causes compression. Once the internal loading throughout the bar is known, the maximum ratio of P > A can then be identified. This steel tie rod is used as a hanger to suspend a portion of a staircase, and as a result it is subjected to tensile stress.

!

P

PP

P

Tension Compression

s

s

s

P— A!s

P— A

Fig. 1–14

1.4 AVERAGE NORMAL STRESS IN AN AXIALLY LOADED BAR 27

1 Important Points

• When a body subjected to external loads is sectioned, there is a distribution of force acting over the sectioned area which holds each segment of the body in equilibrium. The intensity of this internal force at a point in the body is referred to as stress .

• Stress is the limiting value of force per unit area, as the area approaches zero. For this definition, the material is considered to be continuous and cohesive.

• The magnitude of the stress components at a point depends upon the type of loading acting on the body, and the orientation of the element at the point.

• When a prismatic bar is made from homogeneous and isotropic material, and is subjected to an axial force acting through the centroid of the cross-sectional area, then the center region of the bar will deform uniformly. As a result, the material will be subjected only to normal stress . This stress is uniform or averaged over the cross-sectional area.

Procedure for Analysis

The equation s = P>A gives the average normal stress on the cross-sectional area of a member when the section is subjected to an internal resultant normal force P . For axially loaded members, application of this equation requires the following steps.

Internal Loading. • Section the member perpendicular to its longitudinal axis at the

point where the normal stress is to be determined and use the necessary free-body diagram and force equation of equilibrium to obtain the internal axial force P at the section.

Average Normal Stress. • Determine the member’s cross-sectional area at the section and

calculate the average normal stress s = P>A . • It is suggested that s be shown acting on a small volume element

of the material located at a point on the section where stress is calculated. To do this, first draw s on the face of the element coincident with the sectioned area A. Here s acts in the same direction as the internal force P since all the normal stresses on the cross section develop this resultant. The normal stress s on the other face of the element acts in the opposite direction.

28 CHAPTER 1 STRESS

1

The bar in Fig. 1–15 a has a constant width of 35 mm and a thickness of 10 mm. Determine the maximum average normal stress in the bar when it is subjected to the loading shown.

(b)

9 kN

9 kN

12 kN

12 kN

PAB ! 12 kN

PBC ! 30 kN

PCD ! 22 kN 22 kN

P (kN)

x 12 22 30

(c)

12 kN 22 kN 9 kN

9 kN

4 kN

4 kN 35 mm

A DB C

(a)

EXAMPLE 1.5

(d)

30 kN

85.7 MPa35 mm

10 mm

Fig. 1–15

SOLUTION Internal Loading. By inspection, the internal axial forces in regions AB, BC , and CD are all constant yet have different magnitudes. Using the method of sections, these loadings are determined in Fig. 1–15 b ; and the normal force diagram, which represents these results graphically, is shown in Fig. 1–15 c . The largest loading is in region BC , where PBC = 30 kN. Since the cross-sectional area of the bar is constant , the largest average normal stress also occurs within this region of the bar.

Average Normal Stress. Applying Eq. 1–6 , we have

sBC = PBC A

= 30(103) N

(0.035 m)(0.010 m) = 85.7 MPa Ans.

NOTE: The stress distribution acting on an arbitrary cross section of the bar within region BC is shown in Fig. 1–15 d . Graphically the volume (or “block”) represented by this distribution of stress is equivalent to the load of 30 kN; that is, 30 kN = (85.7 MPa)(35 mm)(10 mm).

1.4 AVERAGE NORMAL STRESS IN AN AXIALLY LOADED BAR 29

1 EXAMPLE 1.6

The 80-kg lamp is supported by two rods AB and BC as shown in Fig. 1–16 a . If AB has a diameter of 10 mm and BC has a diameter of 8 mm, determine the average normal stress in each rod.

A

60! B

C

3 4

5

(a) (b)

60!

FBA FBC

y

x

80(9.81) " 784.8 N

B 3

4

5

Fig. 1–16

SOLUTION Internal Loading. We must first determine the axial force in each rod. A free-body diagram of the lamp is shown in Fig. 1–16 b . Applying the equations of force equilibrium,

S+ !Fx = 0; FBC1452 - FBA cos 60" = 0 + c !Fy = 0; FBC1352 + FBA sin 60" - 784.8 N = 0 FBC = 395.2 N, FBA = 632.4 N

By Newton’s third law of action, equal but opposite reaction, these forces subject the rods to tension throughout their length.

Average Normal Stress. Applying Eq. 1–6 ,

sBC = FBC A BC

= 395.2 N

p(0.004 m)2 = 7.86 MPa Ans.

sBA = FBA A BA

= 632.4 N

p(0.005 m)2 = 8.05 MPa Ans.

NOTE: The average normal stress distribution acting over a cross section of rod AB is shown in Fig. 1–16 c , and at a point on this cross section, an element of material is stressed as shown in Fig. 1–16 d .

632.4 N

8.05 MPa

8.05 MPa

(c)(d)

30 CHAPTER 1 STRESS

1 EXAMPLE 1.7

The casting shown in Fig. 1–17 a is made of steel having a specific weight of gst = 490 lb>ft3. Determine the average compressive stress acting at points A and B .

0.75 ft

0.75 ft

2.75 ft

y

z

x (a)

A

B0.75 ft 0.4 ft

2.75 ft

(b)

A

P

(c)

9.36 psi

B

Wst

Fig. 1–17

SOLUTION Internal Loading. A free-body diagram of the top segment of the casting where the section passes through points A and B is shown in Fig. 1–17 b . The weight of this segment is determined from Wst = gst Vst . Thus the internal axial force P at the section is

+ c !Fz = 0; P - Wst = 0 P - (490 lb>ft3)(2.75 ft)3p(0.75 ft)24 = 0

P = 2381 lb Average Compressive Stress. The cross-sectional area at the section is A = p(0.75 ft)2, and so the average compressive stress becomes

s = P A

= 2381 lb

p(0.75 ft)2 = 1347.5 lb>ft2

s = 1347.5 lb>ft2 (1 ft2>144 in2) = 9.36 psi Ans. NOTE: The stress shown on the volume element of material in Fig. 1–17 c is representative of the conditions at either point A or B . Notice that this stress acts upward on the bottom or shaded face of the element since this face forms part of the bottom surface area of the section, and on this surface, the resultant internal force P is pushing upward.

1.4 AVERAGE NORMAL STRESS IN AN AXIALLY LOADED BAR 31

1 EXAMPLE 1.8

Member AC shown in Fig. 1–18 a is subjected to a vertical force of 3 kN. Determine the position x of this force so that the average compressive stress at the smooth support C is equal to the average tensile stress in the tie rod AB . The rod has a cross-sectional area of 400 mm2 and the contact area at C is 650 mm2.

x

A

B

C

200 mm

(a)

3 kN

(b)

x

3 kN

A

200 mm

FAB

FC

Fig. 1–18 SOLUTION Internal Loading. The forces at A and C can be related by considering the free-body diagram for member AC , Fig. 1–18 b . There are three unknowns, namely, FAB , FC , and x . To solve this problem we will work in units of newtons and millimeters. + c !Fy = 0; FAB + FC - 3000 N = 0 (1) a+ !MA = 0; -3000 N(x) + FC(200 mm) = 0 (2) Average Normal Stress. A necessary third equation can be written that requires the tensile stress in the bar AB and the compressive stress at C to be equivalent, i.e.,

s = FAB

400 mm2 =

FC 650 mm2

FC = 1.625FAB

Substituting this into Eq. 1 , solving for FAB , then solving for FC , we obtain FAB = 1143 N FC = 1857 N The position of the applied load is determined from Eq. 2 ,

x = 124 mm Ans. NOTE: 0 6 x 6 200 mm, as required.

32 CHAPTER 1 STRESS

1 1.5 Average Shear Stress Shear stress has been defined in Section 1.3 as the stress component that acts in the plane of the sectioned area. To show how this stress can develop, consider the effect of applying a force F to the bar in Fig. 1–19 a . If the supports are considered rigid, and F is large enough, it will cause the material of the bar to deform and fail along the planes identified by AB and CD . A free-body diagram of the unsupported center segment of the bar, Fig. 1–19 b , indicates that the shear force V = F>2 must be applied at each section to hold the segment in equilibrium. The average shear stress distributed over each sectioned area that develops this shear force is defined by

tavg = V A

(1–7)

Here

tavg = average shear stress at the section, which is assumed to be the same at each point located on the section

V = internal resultant shear force on the section determined from the equations of equilibrium

A = area at the section

The distribution of average shear stress acting over the sections is shown in Fig. 1–19 c . Notice that tavg is in the same direction as V , since the shear stress must create associated forces all of which contribute to the internal resultant force V at the section.

The loading case discussed here is an example of simple or direct shear , since the shear is caused by the direct action of the applied load F . This type of shear often occurs in various types of simple connections that use bolts, pins, welding material, etc. In all these cases, however, application of Eq. 1–7 is only approximate . A more precise investigation of the shear-stress distribution over the section often reveals that much larger shear stresses occur in the material than those predicted by this equation. Although this may be the case, application of Eq. 1–7 is generally acceptable for many problems in engineering design and analysis. For example, engineering codes allow its use when considering design sizes for fasteners such as bolts and for obtaining the bonding strength of glued joints subjected to shear loadings.

(b)

(c)

F

F

VV

tavg

F

(a)

B D

A C

Fig. 1–19

B C A

The pin A used to connect the linkage of this tractor is subjected to double shear because shearing stresses occur on the surface of the pin at B and C. See Fig 1–21c.

1.5 AVERAGE SHEAR STRESS 33

1

Shear Stress Equilibrium. Fig. 1–20 a shows a volume element of material taken at a point located on the surface of a sectioned area which is subjected to a shear stress tzy . Force and moment equilibrium requires the shear stress acting on this face of the element to be accompanied by shear stress acting on three other faces. To show this we will first consider force equilibrium in the y direction. Then

force ƒiiƒ

stress area ƒiƒ ƒiiiƒ

!Fy = 0; tzy1"x "y2 - t=zy "x "y = 0 tzy = t=zy

In a similar manner, force equilibrium in the z direction yields tyz = t=yz . Finally, taking moments about the x axis,

moment ƒiiiƒ

force arm ƒiiƒ ƒ

stress area ƒ ƒiiƒ ƒiiiƒ ƒiƒ

!Mx = 0; -tzy1"x "y2 "z + tyz1"x "z2 "y = 0 tzy = tyz so that

tzy = t=zy = tyz = t=yz = t

In other words, all four shear stresses must have equal magnitude and be directed either toward or away from each other at opposite edges of the element , Fig. 1–20 b . This is referred to as the complementary property of shear , and under the conditions shown in Fig. 1–20 , the material is subjected to pure shear .

Pure shear

(a) (b)

!

Section plane

x

y

z

!y

!z

!xt¿zy

tzy

t¿yz

tyz

t

t

t

t

Fig. 1–20

34 CHAPTER 1 STRESS

1 Important Points

• If two parts are thin or small when joined together, the applied loads may cause shearing of the material with negligible bending. If this is the case, it is generally assumed that an average shear stress acts over the cross-sectional area.

• When shear stress t acts on a plane, then equilibrium of a volume element of material at a point on the plane requires associated shear stress of the same magnitude act on three adjacent sides of the element.

Procedure for Analysis

The equation tavg = V>A is used to determine the average shear stress in the material. Application requires the following steps.

Internal Shear. • Section the member at the point where the average shear stress is

to be determined.

• Draw the necessary free-body diagram, and calculate the internal shear force V acting at the section that is necessary to hold the part in equilibrium.

Average Shear Stress. • Determine the sectioned area A , and determine the average

shear stress tavg = V>A. • It is suggested that tavg be shown on a small volume element of

material located at a point on the section where it is determined. To do this, first draw tavg on the face of the element, coincident with the sectioned area A . This stress acts in the same direction as V . The shear stresses acting on the three adjacent planes can then be drawn in their appropriate directions following the scheme shown in Fig. 1–20 .

1.5 AVERAGE SHEAR STRESS 35

1 EXAMPLE 1.9

Determine the average shear stress in the 20-mm-diameter pin at A and the 30-mm-diameter pin at B that support the beam in Fig. 1–21 a .

SOLUTION

Internal Loadings. The forces on the pins can be obtained by considering the equilibrium of the beam, Fig. 1–21 b .

a+ !MA = 0;

FBa 45 b 16 m2 - 30 kN12 m2 = 0 FB = 12.5 kN S+ !Fx = 0; 112.5 kN2 a 35 b - Ax = 0 Ax = 7.50 kN + c !Fy = 0; Ay + 112.5 kN2 a 45 b - 30 kN = 0 Ay = 20 kN Thus, the resultant force acting on pin A is

FA = 2A 2x + A 2y = 2(7.50 kN)2 + (20 kN)2 = 21.36 kN The pin at A is supported by two fixed “leaves” and so the free-body diagram of the center segment of the pin shown in Fig. 1–21 c has two shearing surfaces between the beam and each leaf. The force of the beam (21.36 kN) acting on the pin is therefore supported by shear force on each of these surfaces. This case is called double shear . Thus,

VA = FA 2

= 21.36 kN

2 = 10.68 kN

In Fig. 1–21 a , note that pin B is subjected to single shear, which occurs on the section between the cable and beam, Fig. 1–21 d . For this pin segment,

VB = FB = 12.5 kN

Average Shear Stress.

1tA2avg = V AA A = 10.6811032 Np 4 10.02 m22 = 34.0 MPa Ans.

1tB2avg = V BA B = 12.511032 Np 4 10.03 m22 = 17.7 MPa Ans.

4 m

(a)

2 m

30 kN

A B

C

3 45

4 m

(b)

2 m

30 kN

A

3 45

FB

Ax

Ay

(c)

VA VA

FA ! 21.36 kN

(d)

VB

FB ! 12.5 kN

Fig. 1–21

36 CHAPTER 1 STRESS

1 EXAMPLE 1.10

If the wood joint in Fig. 1–22 a has a width of 150 mm, determine the average shear stress developed along shear planes a – a and b – b . For each plane, represent the state of stress on an element of the material.

0.1 m 0.125 m

(a)

6 kN 6 kN

b b

a a

(b)

6 kN

F

F

Fig. 1–22

SOLUTION

Internal Loadings. Referring to the free-body diagram of the member, Fig. 1–22 b ,

S+ !Fx = 0; 6 kN - F - F = 0 F = 3 kN

Now consider the equilibrium of segments cut across shear planes a – a and b – b , shown in Figs. 1–22 c and 1–22 d .

S+ !Fx = 0; Va - 3 kN = 0 Va = 3 kN

S+ !Fx = 0; 3 kN - Vb = 0 Vb = 3 kN

Average Shear Stress.

1ta2avg = VaAa = 311032 N10.1 m2 10.15 m2 = 200 kPa Ans. 1tb2avg = VbAb = 311032 N10.125 m2 10.15 m2 = 160 kPa Ans. The state of stress on elements located on sections a – a and b – b is shown in Figs. 1–22 c and 1–22 d , respectively.

3 kN

(c)

Vata ! 200 kPa

3 kN

(d)

Vbtb = 160 kPa

1.5 AVERAGE SHEAR STRESS 37

1 EXAMPLE 1.11

The inclined member in Fig. 1–23 a is subjected to a compressive force of 600 lb. Determine the average compressive stress along the smooth areas of contact defined by AB and BC , and the average shear stress along the horizontal plane defined by DB .

1 in.

3

45

600 lb

1.5 in. 3 in. 2 in.

A C

B D

(a) Fig. 1–23 SOLUTION Internal Loadings. The free-body diagram of the inclined member is shown in Fig. 1–23 b . The compressive forces acting on the areas of contact are

S+ !Fx = 0; FAB - 600 lb1352 = 0 FAB = 360 lb + c !Fy = 0; FBC - 600 lb1452 = 0 FBC = 480 lb Also, from the free-body diagram of the top segment ABD of the bottom member, Fig. 1–23 c , the shear force acting on the sectioned horizontal plane DB is

S+ !Fx = 0; V - 360 lb = 0 V = 360 lb

Average Stress. The average compressive stresses along the horizontal and vertical planes of the inclined member are

sAB = FAB AAB

= 360 lb11 in.2 11.5 in.2 = 240 psi Ans.

sBC = FBC ABC

= 480 lb12 in.2 11.5 in.2 = 160 psi Ans.

These stress distributions are shown in Fig. 1–23 d . The average shear stress acting on the horizontal plane defined

by DB is

tavg = 360 lb13 in.2 11.5 in.2 = 80 psi Ans.

This stress is shown uniformly distributed over the sectioned area in Fig. 1–23 e .

(b)

3

45

600 lb

FAB

FBC

(c) V

360 lb

(d)

3

45

600 lb

160 psi

240 psi

(e)

360 lb

80 psi

38 CHAPTER 1 STRESS

1 PRELIMINARY PROBLEMS

P1–2. In each case, determine the largest internal shear force resisted by the bolt. Include all necessary free-body diagrams.

6 kN 2 kN

(a)

5 kNA

B

10 kN 4 kN

20 kN

(b)

6 kNA

B

C 8 kN

P1–2

P1–3. Determine the largest internal normal force in the bar.

10 kN 6 kN2 kN5 kN

D

F

C B A

P1–3

P1–4. Determine the internal normal force at section A if the rod is subjected to the external uniformally distributed loading along its length of 8 kN>m.

8 kN/mA

2 m 3 m

P1–4

P1–5. The lever is held to the fixed shaft using the pin AB . If the couple is applied to the lever, determine the shear force in the pin between the pin and the lever.

20 N

A

B

20 N

0.2 m 0.2 m

0.1 m

P1–5

P1–6. The single-V butt joint transmits the force of 5 kN from one plate to the other. Determine the resultant normal and shear force components that this force creates on the face of the weld, section AB .

30! 30!

5 kN

5 kN

20 mm

100 mm

A B

P1–6

1.5 AVERAGE SHEAR STRESS 39

1 FUNDAMENTAL PROBLEMS

F1–7. The uniform beam is supported by two rods AB and CD that have cross-sectional areas of 10 mm2 and 15 mm2 , respectively. Determine the intensity w of the distributed load so that the average normal stress in each rod does not exceed 300 kPa.

w

A C

B D

6 m

F1–7

F1–8. Determine the average normal stress developed on the cross section. Sketch the normal stress distribution over the cross section.

300 kN

100 mm

80 mm

F1–8

F1–9. Determine the average normal stress developed on the cross section. Sketch the normal stress distribution over the cross section.

4 in. 1 in.

1 in. 4 in. 1 in.

15 kip

F1–9

F1–10. If the 600-kN force acts through the centroid of the cross section, determine the location y of the centroid and the average normal stress developed on the cross section. Also, sketch the normal stress distribution over the cross section.

80 mm

300 mm

60 mm

–y 80 mm

600 kN

x y60 mm

F1–10

F1–11. Determine the average normal stress developed at points A, B, and C . The diameter of each segment is indicated in the figure.

2 kip3 kip 8 kip9 kip

1 in. 0.5 in. 0.5 in.

A B

C

F1–11

F1–12. Determine the average normal stress developed in rod AB if the load has a mass of 50 kg. The diameter of rod AB is 8 mm.

8 mm

A

D

B

C

5 4

3

F1–12

40 CHAPTER 1 STRESS

FUNDAMENTAL PROBLEMS1 PROBLEMS

1–31. The supporting wheel on a scaffold is held in place on the leg using a 4-mm-diameter pin as shown. If the wheel is subjected to a normal force of 3 kN, determine the average shear stress developed in the pin. Neglect friction between the inner scaffold puller leg and the tube used on the wheel.

3 kN

Prob. 1–31

*1–32. The lever is held to the fixed shaft using a tapered pin AB, which has a mean diameter of 6 mm. If a couple is applied to the lever, determine the average shear stress in the pin between the pin and lever.

20 N 20 N

250 mm 250 mm

12 mm

A

B

Prob. 1–32

1–33. The bar has a cross-sectional area A and is subjected to the axial load P . Determine the average normal and average shear stresses acting over the shaded section, which is oriented at u from the horizontal. Plot the variation of these stresses as a function of u 10 … u … 90!2.

P

u

P

A

Prob. 1–33

1–34. The built-up shaft consists of a pipe AB and solid rod BC . The pipe has an inner diameter of 20 mm and outer diameter of 28 mm. The rod has a diameter of 12 mm. Determine the average normal stress at points D and E and represent the stress on a volume element located at each of these points.

C

ED

A 4 kN

8 kN B 6 kN

6 kN

Prob. 1–34

1–35. If the turnbuckle is subjected to an axial force of P = 900 lb, determine the average normal stress developed in section a – a and in each of the bolt shanks at B and C. Each bolt shank has a diameter of 0.5 in.

*1–36. The average normal stresses developed in section a – a of the turnbuckle, and the bolts shanks at B and C, are not allowed to exceed 15 ksi and 45 ksi, respectively. Determine the maximum axial force P that can be applied to the turnbuckle. Each bolt shank has a diameter of 0.5 in.

a

A

B P PC

a

1 in.

0.25 in.

Probs. 1–35/36

1.5 AVERAGE SHEAR STRESS 41

1 *1–40. Determine the average normal stress in each of the 20-mm diameter bars of the truss. Set P = 40 kN.

1–41. If the average normal stress in each of the 20-mm diameter bars is not allowed to exceed 150 MPa, determine the maximum force P that can be applied to joint C .

1–42. Determine the average shear stress developed in pin A of the truss. A horizontal force of P = 40 kN is applied to joint C . Each pin has a diameter of 25 mm and is subjected to double shear.

P

1.5 m

2 m

C

A B

Probs. 1–40/41/42

1–43. The 150-kg bucket is suspended from end E of the frame. Determine the average normal stress in the 6-mm diameter wire CF and the 15-mm diameter short strut BD .

*1–44. The 150-kg bucket is suspended from end E of the frame. If the diameters of the pins at A and D are 6 mm and 10 mm, respectively, determine the average shear stress developed in these pins. Each pin is subjected to double shear.

0.6 m0.6 m

1.2 m

D E

F

C

B

A

0.6 m

30!

Probs. 1–43/44

1–37. The plate has a width of 0.5 m. If the stress distri bution at the support varies as shown, determine the force P applied to the plate and the distance d to where it is applied.

4 m

30 MPa

P d

" (15x1/2) MPas

x

Prob. 1–37

1–38. The two members used in the construction of an aircraft fuselage are joined together using a 30° fish-mouth weld. Determine the average normal and average shear stress on the plane of each weld. Assume each inclined plane supports a horizontal force of 400 lb.

800 lb 800 lb

30!

1 in. 1 in.

1.5 in. 30!

Prob. 1–38

1–39. If the block is subjected to the centrally applied force of 600 kN, determine the averege normal stress in the material. Show the stress acting on a differential volume element of the material.

50 mm

150 mm

150 mm 50 mm

100 mm 100 mm

600 kN150 mm

150 mm

Prob. 1–39

42 CHAPTER 1 STRESS

1 *1–48. The beam is supported by a pin at A and a short link BC. If P = 15 kN, determine the average shear stress developed in the pins at A, B, and C. All pins are in double shear as shown, and each has a diameter of 18 mm.

C

B A

0.5 m 1 m 1.5 m 1.5 m

0.5 m P 4P 4P 2P

30!

Prob. 1–48

1–49. The joint is subjected to the axial member force of 6 kip. Determine the average normal stress acting on sections AB and BC . Assume the member is smooth and is 1.5-in. thick.

60!

20! 4.5 in.

1.5 in.

A

B

C

6 kip

Prob. 1–49

1–50. The driver of the sports car applies his rear brakes and causes the tires to slip. If the normal force on each rear tire is 400 lb and the coefficient of kinetic friction between the tires and the pavement is µ k = 0.5, determine the average shear stress developed by the friction force on the tires. Assume the rubber of the tires is flexible and each tire is filled with an air pressure of 32 psi.

400 lb

Prob. 1–50

1–45. The pedestal has a triangular cross section as shown. If it is subjected to a compressive force of 500 lb, specify the x and y coordinates for the location of point P ( x , y ), where the load must be applied on the cross section, so that the average normal stress is uniform. Compute the stress and sketch its distribution acting on the cross section at a location removed from the point of load application.

3 in. 6 in.x

y

500 lb

P(x,y)12 in.

Prob. 1–45

1–46. The 20-kg chandelier is suspended from the wall and ceiling using rods AB and BC , which have diameters of 3 mm and 4 mm, respectively. Determine the angle u so that the average normal stress in both rods is the same.

1–47. The chandelier is suspended from the wall and ceiling using rods AB and BC, which have diameters of 3 mm and 4 mm, respectively. If the average normal stress in both rods is not allowed to exceed 150 MPa, determine the lar gest mass of the chandelier that can be supported if u = 45!.

B

A

C

30!

u

Probs. 1–46/47

1.5 AVERAGE SHEAR STRESS 43

1 1–54. When the hand is holding the 5-lb stone, the humerus H , assumed to be smooth, exerts normal forces F C and F A on the radius C and ulna A , respectively, as shown. If the smallest cross-sectional area of the ligament at B is 0.30 in 2 , determine the greatest average tensile stress to which it is subjected.

14 in. 2 in.

0.8 in.

B

A

C GFB

FC

FA

75!

H

Prob. 1–54

1–55. The 2-Mg concrete pipe has a center of mass at point G . If it is suspended from cables AB and AC , determine the average normal stress developed in the cables. The diameters of AB and AC are 12 mm and 10 mm, respectively.

*1–56. The 2-Mg concrete pipe has a center of mass at point G . If it is suspended from cables AB and AC , determine the diameter of cable AB so that the average normal stress developed in this cable is the same as in the 10-mm diameter cable AC .

A

C

GB

30! 45!

Probs. 1–55/56

1–51. During the tension test, the wooden specimen is subjected to an average normal stress of 2 ksi. Determine the axial force P applied to the specimen. Also, find the average shear stress developed along section a – a of the specimen.

P

P

1 in. 2 in.

4 in.

4 in.

a

a

Prob. 1–51

*1–52. If the joint is subjected to an axial force of P = 9 kN, determine the averege shear stress developed in each of the 6-mm diameter bolts between the plates and the members and along each of the four shaded shear planes.

1–53. The average shear stress in each of the 6-mm diameter bolts and along each of the four shaded shear planes is not allowed to exceed 80 MPa and 500 kPa, respectively. Determine the maximum axial force P that can be applied to the joint.

P

P

100 mm

100 mm

Probs. 1–52/53

44 CHAPTER 1 STRESS

1 1–59. The jib crane is pinned at A and supports a chain hoist that can travel along the bottom flange of the beam, 1 ft … x … 12 ft. If the hoist is rated to support a maximum of 1500 lb, determine the maximum average normal stress in the 34 -in. diameter tie rod BC and the maximum average shear stress in the 58 -in. -diamater pin at B.

C

10 ft

x

A

B

30!

D

1500 lb

Prob. 1–59

*1–60. If the shaft is subjected to an axial force of 5 kN, determine the bearing stress acting on the collar A .

1–61. If the 60-mm diameter shaft is subjected to an axial force of 5 kN, determine the average shear stress developed in the shear plane where the collar A and shaft are connected.

A 2.5 mm

2.5 mm

100 mm60 mm 5 kN

15 mm

Probs. 1–60/61

1–57. If the concrete pedestal has a specific weight of g , determine the average normal stress developed in the pedestal as a function of z .

r0

2r0 h

z

Prob. 1–57

1–58. The anchor bolt was pulled out of the concrete wall and the failure surface formed part of a frustum and cylinder. This indicates a shear failure occurred along the cylinder BC and tension failure along the frustum AB . If the shear and normal stresses along these surfaces have the magnitudes shown, determine the force P that must have been applied to the bolt.

30 mm4.5 MPa

3 MPa 3 MPa

P

50 mm

A

25 mm 25 mm

B

C

45!45!

Prob. 1–58

1.5 AVERAGE SHEAR STRESS 45

1 1–66. Determine the largest load P that can be applied to the frame without causing either the average normal stress or the average shear stress at section a – a to exceed s = 150 MPa and t = 60 MPa , respectively. Member CB has a square cross section of 25 mm on each side.

2 m

B

A C

1.5 m

a

a

P

Prob. 1–66

1–67. The pedestal in the shape of a frustum of a cone is made of concrete having a specific weight of 150 lb>ft3 . Determine the average normal stress acting in the pedestal at its base. Hint: The volume of a cone of radius r and height h is V = 13pr2h .

*1–68. The pedestal in the shape of a frustum of a cone is made of concrete having a specific weight of 150 lb>ft3. Determine the average normal stress acting in the pedestal at its midheight, z = 4 ft . Hint: The volume of a cone of radius r and height h is V = 13pr2h.

z

y

x

8 ft

z ! 4 ft

1 ft

1.5 ft

Probs. 1–67/68

1–62. The crimping tool is used to crimp the end of the wire E . If a force of 20 lb is applied to the handles, determine the average shear stress in the pin at A . The pin is subjected to double shear and has a diameter of 0.2 in. Only a vertical force is exerted on the wire.

1–63. Solve Prob. 1–62 for pin B . The pin is subjected to double shear and has a diameter of 0.2 in.

A

20 lb

20 lb

5 in. 1.5 in. 2 in. 1 in.

E C

B D

Probs. 1–62/63

*1–64. A vertical force of P = 1500 N is applied to the bell crank. Determine the average normal stress developed in the 10-mm diamater rod CD , and the average shear stress developed in the 6-mm diameter pin B that is subjected to double shear.

1–65. Determine the maximum vertical force P that can be applied to the bell crank so that the average normal stress developed in the 10-mm diameter rod CD, and the average shear strees developed in the 6-mm diameter double sheared pin B not exceed 175 MPa and 75 MPa respectively.

450 mm

300 mm 45!

P

A

B

CD

Probs. 1–64/65

46 CHAPTER 1 STRESS

1 1.6 Allowable Stress Design To ensure the safety of a structural member or mechanical element, it is necessary to restrict the applied load to one that is less than the load the member (or element) can fully support. There are many reasons for doing this. For example, the intended measurements of a structure or machine may not be exact, due to errors in fabrication or in the assembly of its component parts. Unknown vibrations, impact, or accidental loadings can occur that may not be accounted for in the design. Atmospheric corrosion, decay, or weathering tend to cause materials to deteriorate during service. And finally, some materials, such as wood, concrete, or fiber-reinforced composites, can show high variability in mechanical properties.

One method of specifying the allowable load for a member is to use a number called the factor of safety. The factor of safety (F.S.) is a ratio of the failure load Ffail to the allowable load Fallow. Here Ffail is found from experimental testing of the material, and the factor of safety is selected based on experience so that all the above mentioned uncertainties are accounted for when the member is used under similar conditions of loading and geometry. Stated mathematically,

F.S. = Ffail

Fallow (1–8)

If the load applied to the member is linearly related to the stress developed within the member, as in the case of using s = P>A and tavg = V>A, then we can also express the factor of safety as a ratio of the failure stress sfail (or tfail ) to the allowable stress sallow (or tallow ). * Here the area A will cancel and so,

F.S. = sfail sallow

(1–9)

or

F.S. = tfail tallow

(1–10)

* In some cases, such as columns, the applied load is not linearly related to stress and therefore only Eq. 1–8 can be used to determine the factor of safety. See Chapter 13.

Cranes are often supported using bearing pads to give them stability. Care must be taken not to crush the supporting surface, due to the large bearing stress developed between the pad and the surface.

1.6 ALLOWABLE STRESS DESIGN 47

1 In any of these equations, the factor of safety must be greater than 1 in

order to avoid the potential for failure. Specific values depend on the types of materials to be used and the intended purpose of the structure or machine. For example, the F.S. used in the design of aircraft or space- vehicle components may be close to 1 in order to reduce the weight of the vehicle. Or, in the case of a nuclear power plant, the factor of safety for some of its components may be as high as 3 due to uncertainties in loading or material behavior. In many cases, the factor of safety or the allowable stress for a specific case can be found in design codes and engineering handbooks. Once it is obtained, a design that is based on an allowable stress limit is called allowable stress design (ASD). This method is intended to ensure a balance between both public and environmental safety on the one hand and economic considerations on the other.

Simple Connections. By making simplifying assumptions regarding the behavior of the material, the equations s = P>A and tavg = V>A can often be used to analyze or design a simple connection or mechanical element. In particular, if a member is subjected to normal force at a section, its required area at the section is determined from

A = P

sallow (1–11)

On the other hand, if the section is subjected to an average shear force , then the required area at the section is

A = V

tallow (1–12)

As discussed in Sec. 1.6 , the allowable stress used in each of these equations is determined either by applying a factor of safety to the material’s normal or shear failure stress or by finding these stresses directly from an appropriate design code.

Three examples of where the above equations apply are shown in Fig. 1–24 .

P

V ! P

Assumed uniform shear stress

P

A !

tallow

P tallow

B

("b)allow

P

Assumed uniform normal stress distribution

A ! P

("b)allow

The area of the column base plate B is determined from the allowable bearing stress for the concrete.

P

d

Assumed uniform shear stress

P

tallow

l ! —————P tallowpd

The embedded length l of this rod in concrete can be determined using the allowable shear

stress of the bonding glue.

P

P

The area of the bolt for this lap joint is determined from the shear stress which is largest between the plates. Fig. 1–24

48 CHAPTER 1 STRESS

1 1.7 Limit State Design We have stated in the previous section that a properly designed member must account for uncertainties resulting from the variability of both the material’s properties and the loading that must be supported. Since uncertainty can be considered using statistics and probability theory, in structural engineering there has been an increasing trend to separate load uncertainty from material uncertainty.* This method of design is called limit state design (LSD), or more specifically, in the United States it is called load and resistance factor design (LRFD). We will now discuss how this method is applied. Load Factors. Various types of loads R can act on a structure or structural member, and each is multiplied by a load factor g (gamma) that accounts for its variability. The loads include dead load , which is the fixed weight of the structure, and live loads , which involve people or vehicles that move about. Other types of live loads include wind, earthquake, and snow loads. The dead load D is multiplied by a small factor such as g

D = 1.4 ,

since it can be determined with greater certainty than, for example, the live load L caused by people. It can have a load factor of gL = 1.6 .

Building codes often require a structure to be designed to support various combinations of the loads, and when applied in combination, each type of load will have a unique load factor. For example, one load combination of dead (D), live (L), and snow (S) loads is

R = 1.2D + 1.6L + 0.5S The load factors for this combined loading reflect the probability that the total loading R will occur for all the events stated. Here, however, the load factor gS = 0.5 is small because of the low probability that a maximum snow load will occur simultaneously with the maximum dead and live loads. Resistance Factors. Resistance factors f (phi) are determined from the probability of material failure as it relates to the material’s quality and the consistency of its strength. These factors will differ for different types of materials. For example, concrete has smaller factors than steel because engineers have more confidence about the behavior of steel under load than they do about concrete. For example, a typical resistance factor f = 0.9 is used for a steel member in tension. Design Criteria. Once the load and resistance factors g and f have been specified using a code, then proper design of a structural member requires that its strength be greater than the load it supports. Thus, the LRFD criterion can be stated as fPn Ú !gi Ri (1–13) Here Pn is the nominal strength of the member, meaning the load, when applied to the member, causes it either to fail (ultimate load), or deform to a state where it is no longer serviceable or becomes unsuitable for its intended purpose. In summary then, the resistance factor f reduces the nominal strength of the member and requires it to be equal to or greater than the applied load or combination of loads calculated using the load factors g. * ASD combines these uncertainties by using the factor of safety or defining the allowable stress.

1.7 LIMIT STATE DESIGN 49

1 Important Point

• Design of a member for strength is based on selecting either an allowable stress or a factor of safety that will enable it to safely support its intended load (ASD), or using load and resistance factors to modify the strength of the material and the load, respectively (LRFD).

Procedure for Analysis

When solving problems using the average normal and shear stress equations, a careful consideration should first be made as to choosing the section over which the critical stress is acting. Once this section is determined, the member must then be designed to have a sufficient area at the section to resist the stress that acts on it. This area is determined using the following steps.

Internal Loading. • Section the member through the area and draw a free-body

diagram of a segment of the member. The internal resultant force at the section is then determined using the equations of equilibrium.

Required Area. • Provided either the allowable stress or the load and resistance

factors are known or can be determined, then the required area needed to sustain the calculated load or factored load at the section is determined from A = P>s or A = V >t.

Appropriate factors of safety must be considered when d esigning cranes and cables used to transfer heavy loads.

50 CHAPTER 1 STRESS

1 EXAMPLE 1.12

The control arm is subjected to the loading shown in Fig. 1–25 a . Determine to the nearest 14 in. the required diameters of the steel pins at A and C if the allowable shear stress for the steel is tallow = 8 ksi.

(a)

C

3 5

42 in.3 in.

8 in.

Double shear

A

C

3 kip 5 kip

B

Single shear

A

3 5

42 in.3 in.

8 in.

Cx

3 kip 5 kip

FAB

Cy (b)

C

Fig. 1–25

SOLUTION

Pin Forces. A free-body diagram of the arm is shown in Fig. 1–25 b . For equilibrium we have

a+ !MC = 0; FAB18 in.2 - 3 kip 13 in.2 - 5 kip 1352 15 in.2 = 0 FAB = 3 kip

S+ !Fx = 0; -3 kip - Cx + 5 kip 1452 = 0 Cx = 1 kip + c !Fy = 0; Cy - 3 kip - 5 kip1352 = 0 Cy = 6 kip The pin at C resists the resultant force at C, which is

FC = 211 kip22 + 16 kip22 = 6.083 kip Pin A. This pin is subjected to single shear, Fig. 1–25c, so that

A = V

tallow ; pa dA

2 b2 = 3 kip

8 kip >in2; dA = 0.691 in. Use dA = 34 in. Ans.