Mechanics of Materials 8th Edition R.C. Hibbeler

R. C. HIBBELER

MECHANICS
OF MATERIALS

MECHANICS
OF MATERIALS

EIGHTH EDITION

R. C. HIBBELER

Prentice Hall
Isar imprint of

PEARSON
Bm en pcarsoohghere.com

ISBN 10:04

ISBN 13:978

To the Student

With the hope that this work will st

st in Engineering Mechanic

and provide an acceptable guide to its understandin:

|_| PREFACE

ded that this book provide the student with a clear and

chanics of materials. To achieve this

jestive, over the years this

students. The eighth edition has be
ditishoped that x

New to This Edition

+ Updated Content. Some portions of the text have been rewriten

in onder fo enbance clarity and be more succinct. I this regard, som

de more emphasis on the application of imps

À to support

Also, the artwork has been improved throughout the bo

these chang

+ New Photos. ‘The relevance of knowing the subject matter is

Feflected by the real-world applications depicted in over 44 new or
os placed throughout the book. These photos generally

ate used to explain how the relevant principles apply to real-workd

Situations and how materials behave under load
+ Fundamental Problems. These problem sets are located jus

provide them with the chance to develop their problem-solving skill

fore attempting to solve any ofthe standard problems that follow
The fundamental problems may be considered as extended examples.
since the Key equations and answers are al listed in the back of th
book. Additionally, when assigned, hese problems offer students an

ing for exams, and they can be used a alte
time asa review when studying for the Fundamentals of Engineering

+ Conceptual Problems. Throughout the tex, usually atthe end
din the
‘chapter. These analysis and design problems are intended to engag
the students in thinking through a reife situation as depicted in a
oto, They can be assigned after the students have developed som
tise in the subject matter and they work well either or individual

Situations related to the application of the principles contain

+ New Problems. ‘There are approximately

problems added to this edition, which involve applications to many
ferent fields of engineering. Also, this new edition now has

+ Problems with Hints. With ı

additional homework problems i

his new edition, every problem indicated with a bullet (+) before the
problem number includes a suggestion, Key equation, or addtional
numerical result hat & given along withthe answer in the back of the
book. These problems further encourage students o solve problems on

Contents

The subject matter i organized into 14 chapters. Chapter 1 begins wi

finition of both normal and hear stress, and a discussion of norm:
sin axially loaded members and average shear Stress caused by
In Chapter 2 normal and shear srain are defined, and in Chapie
¿Sacussion of some of the important mechanical properties of materials
given. Separate treatments of axial load, torsion, and bending are

cated in Chapters 4,5. and 6, respectively. In each of these chapters,
both linear-clstic and plastic behavior

Also, topics relat

included: Transse

the material are considered.
I acer. dincasied in Chapter 7, along, wich

sion of thin-walled tube, shear flow, and the shear center. Chapter $
lesa discussion of thin-walled pressure vessels and provides

view ofthe material covered inthe previous chapters, sich th

application of varios theres of file. Chapter 11 provides a mean for
à further summary and review uf previous material by covering des
applications of beams and shafts. In Chapter 12 various methods for
Omputing deflections of beams and shafts ae covered. Also included is

ndicated by a star (*), Time permitting. some of these topics may be
pcloded la the coutée, Fertbermere; this material provides a suitable

Alternative Method of Coverage. Some instructor prefer to
stress and strain transformations frst, before discussing specie

method for doing this woukl be fist to cover stress

transformation, Chapter 1 and Chapter 9,

styled so that this is possible. Also, the problem sets have been
subdivided so that this material can be covered without prior knowl
the intervening chapters. Chapte

Hallmark Elements
Organization and Approach. The contents of each chapter ar

ems. The topics within cach section are plac
defined by tiles, The purpose of this is
nt far later reference and review

into subgroups

Chapter Contents. Fach chapter begins with a fulls
¡lustration that indicates à broad-range application of the material
within the chapter The "Chapter Objectives” are then provided to give à

Procedures for Analysis. Found after many ofthe sections of th
book, this unique feature provides the student with a logical and order
method to follow when appivin the theory: The example problems ar

using this oullined method in order to clariy ls numerical

principles have been mastered and enough confidence and judgment
Photographs. Many photographs are usc th the book

Important Points. Ths feat

ifianı points that should be realized when apply
Example Problems. Alithe example problems ace presented in à

Homework Problems. Numc

provide a means for developing the skill 10 reduce any such problem
mits physical description to a model ora symbolic representation &
which principles may x Throughout the book there is an

approximate balance of problems using either SI or FPS unit

biens in order of increasing dificult. The answers to all

+ Problems with Hints. With ı

additional homework problems i

his new edition, every problem indicated with a bullet (+) before the
problem number includes a suggestion, Key equation, or addtional
numerical result hat & given along withthe answer in the back of the
book. These problems further encourage students o solve problems on

Contents

The subject matter i organized into 14 chapters. Chapter 1 begins wi

finition of both normal and hear stress, and a discussion of norm:
sin axially loaded members and average shear Stress caused by
In Chapter 2 normal and shear srain are defined, and in Chapie
¿Sacussion of some of the important mechanical properties of materials
given. Separate treatments of axial load, torsion, and bending are

cated in Chapters 4,5. and 6, respectively. In each of these chapters,
both linear-clstic and plastic behavior

Also, topics relat

included: Transse

the material are considered.
I acer. dincasied in Chapter 7, along, wich

sion of thin-walled tube, shear flow, and the shear center. Chapter $
lesa discussion of thin-walled pressure vessels and provides

view ofthe material covered inthe previous chapters, sich th

application of varios theres of file. Chapter 11 provides a mean for
à further summary and review uf previous material by covering des
applications of beams and shafts. In Chapter 12 various methods for
Omputing deflections of beams and shafts ae covered. Also included is

ndicated by a star (*), Time permitting. some of these topics may be
pcloded la the coutée, Fertbermere; this material provides a suitable

Alternative Method of Coverage. Some instructor prefer to
stress and strain transformations frst, before discussing specie

method for doing this woukl be fist to cover stress

transformation, Chapter 1 and Chapter 9,

styled so that this is possible. Also, the problem sets have been
subdivided so that this material can be covered without prior knowl
the intervening chapters. Chapte

Hallmark Elements
Organization and Approach. The contents of each chapter ar

ems. The topics within cach section are plac
defined by tiles, The purpose of this is
nt far later reference and review

into subgroups

Chapter Contents. Fach chapter begins with a fulls
¡lustration that indicates à broad-range application of the material
within the chapter The "Chapter Objectives” are then provided to give à

Procedures for Analysis. Found after many ofthe sections of th
book, this unique feature provides the student with a logical and order
method to follow when appivin the theory: The example problems ar

using this oullined method in order to clariy ls numerical

principles have been mastered and enough confidence and judgment
Photographs. Many photographs are usc th the book

Important Points. Ths feat

ifianı points that should be realized when apply
Example Problems. Alithe example problems ace presented in à

Homework Problems. Numc

provide a means for developing the skill 10 reduce any such problem
mits physical description to a model ora symbolic representation &
which principles may x Throughout the book there is an

approximate balance of problems using either SI or FPS unit

biens in order of increasing dificult. The answers to all

ported to three significant figures,
ven though the data for material properties may be known with les
sccuracy. Although this might appear to be a pour practice, iis

problems that require a numerical analysis or a computer applicat
Appendices. The appendices of the book provide a source f

view and a tabular data, Appendix A prov

m the cen moment of inertia of an aren, Appendices B and
(Cis tabular data for structural shapes, and the deflection and slopes
various types of beams and shafts

Accuracy Checking. The Eighth Edition has undergone ou

ns Triple Accuracy Checking review. In addition to the author
individ

® Scott Hendricks, Virginia Polytechnic University
+ Karim Nohra, University of South Florida

® Kurt Norhin, Laurel Tech Integrated Publishing Services
+ Kai Beng Yap, Engineering Consultan

Acknowledgments
een

much appreciated and it hoped that they wil accept this anonymous
cognition. A note of thanks is given ta the reviewers.
Akthen Al Manasee
Yabin Liao, Arizon
Clif Lissenden, Pe

Gregory M. Ode Technological University
John Oyler. Univers ah
Row Xu, Vanderbilt Universi
Paul Zichl, University of South Carolin

There are à few people that I feel deserve particular recognit

ime friend and asociate, Kai Beng Yap, was of great help 4
checking the entire manuseript and helping to prepare the problem
solutions. A special note of thanks also goes to Kurt Norlin of Law

h Integrated Publishing Services in this regard. During the
duction process | am thankful forthe asstance of Rose Kerrun, my

Mary Ann, for their help in proofreading and typing, that was needed
prepare the manuscript for publicat
Tid also ik o thank all my students who have used the previous

Resources for Instructors

+ Instructor's Solutions Manual. An instrictor’ solutions mania
was prepared by the author. The manual includes homework assignment

sand was also checked as part of the accuracy checking program
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|_| CONTENTS

4

Stress 3

Chapter Objectives 3
1.1. introduction 3

1.2 Equilbrium of

Deformable Body 4

Stress in an Axialy

1.5 Average Shear Stress 32
1.6 Allowable Stress 46
1.7 Design of Simple Connections 47

Strain 65

Chapter Objectives 65
2.1 Delormation 65
22 Strain 66

3

Mechanical Properties
of Materials 21

Chapter Objectives 81
3.1. The Tension and Compression Test
32 The Stess-Strain Disgram_ 83

33 Stress-Strain Behavior of Ductile and

Brittle Materials 87
3.4 Hooke’ Law 90

35 Strain Energy 92

3.6 Poisson's Ratio 102

37 The Shear Stress-Strain Diagram 10

of Materials Da

and Fatigue

to Creep

4

Axial Load 119

45
46

47
“48

5)

Torsion 179

se
59

Chapter Objective
Saint-Venant’s Principle 119
Elastic Deformation of an Axial
Loaded Member 122

Principle of Superposition 136
cal indeterminate Axilly
Loaded Member 137
The Force Method of Analysis for
Axially Loaded Members 143
Thermal Stress. 151
Deformstion 162
Residual Stress 164

Chapter Obj
rear Shaft 179
The Torsion Formula 182
Power Transmission 190
Angle of Twist 200
Statically Indetermi
Members 214
Solid Noncicular
‘Shafts 221
"Thin Welle Tubes Having Closed
Cross Sections 224
Stress Concentration 234
Inelastic Torsion 237

Torque-Loaded

+510 Residual Stress 239

6

Bending 255

Chapter Objectives 255

61 Shear and Moment Diagrams 255

6.2. Graphical Method for Constructing Shear
and Moment Diagrams. 262

63 Banding Deformation of à Straight
Member 281

64 The Flexure Formula 285

65. Unsymmetric Bending 302

+66 Composite Beams 312

+67 Reinforced Concrete

+68 Curved Beams 319

6.9 Stress Concentrations 326

+610 Inelastic Bending 335

|

7

Transverse Shear 359

Chapter Objectives 359
7.1. Shear in Straight Members 359

7.2 The Shear Formula 361

7.3. Shear Flow in Buit-Up Members 378

7.4. Shear Flow in Thin-Walle
Members 387
7.5 Shear Center For Open Thin Walled

Members 392

8
Combined Losdings «es (Me

Chapter Objectives. 40

8.1. ThinWalled Pressure Vessels 405

8.2. State of Stress Caused by Combined
Loadings 412

LA

Stress Transformation 437
Chapter Objectives 43

9.4 Plane-Siress Transformation

9.2 General Equations of Plane-Stress.
Transformation. 442

9.3 Principal Stresses and Maximum In-Plane

9. 461

95

x
10 1
Strain Transformation asc Moe

Chapter Objectives 485
10.1 Plane Strain 485

10.2 Genera Equations of Plane-Strin
Transformation 486

“10.3 Mohrs Circle—Plane St

“10.4 Absolute Maximum Sheat
Strain 502

105

106

“107

11
Design of Beams
and Shafts 537

5

Chapter Objectives. 537
11.4. Basis for Beam Design 5

11.2 Prismatic Beam Design 540
“113 Fully Suessed Beams 554

“114 Shaft Design

12
Deflection of Beams
and Shafts 569

Chapter Objectives
12.1 The Elastic Curve 569
12.2 Slope and Displacement
by Integration 573
123 Discontinuity Fun
124

593
Slope and Displacement by the
Moment-Ares Method 604
12.5 Method of Superposition 619
12.6. Staticall indeterminate Beams
fis 627
Inceterminste Beams and
Shafts Method of integration 628
"12.8 Staticall Indeterminate Beams
and Shafts —Moment-Area
Method 633
12.9. Statcall Indeterminate Beams and
Shaftz--Method of Superposition 639

127

13 |
Buckling of Columns 657

Chapter Objectives 657

12.1 Crfical Load 657

13.2 Ideal Column with Pin
Supports. 660

13.3. Columns Having Various Types
of Supports 666

"13.4 The Secant Formula 678

“1315 Inelastic Buckling 684

“13.6 Design of Columns for Concentric
Leading 692

+13.7 Design of Columns for Eccentrie
Loading 703

14

Energy Methods 715

tna

Chapter Objectives 715

14: Estemal Work and San Energy, 715
14.2. Elastic Strain Energy or Various Types
oftonding 720
143. Conservation of Energy 733
164 Impr Loading 740
1455 Principle of Vital Work 751
148 Method of Vital Forces Appied
to Tunes 758
“147 Method of Vital Forces Apple
to Beams. 762
148. Catigiano's Theorem 771
149. Casgiano'sTheotem Applied
to Tunes 773
14.10 Castglano's Theorem Appled
to Bears 776
Appendices
Geomet Poperies ofan Ares 784
Ar Convo of vee
AZ, Moment of Ine for an Ares 787
AB Product of tne for area 791
AR. Mamani varía fr on Ave
about inined Ares 794
AS. Mohrs Cicle for Moments oflnentin 797
B. _ Geometrie Properties of Structural
Shapes 900
Slopes and Deletions of Beams 808

Fundamental Problems Partial Solutions
and Answers 810
Answers to Selected Problems 828

index 854

|_| CREDITS

Chapter 1, Close up of ron girders Jack Sullivan\Alamy Images
Chapter 2, Photoelasic phenamena: tension in a serew mount. Alfred

PasickatAlamy Images

Chapter 3, À woman stands near a collapsed bridge in one of the wors
of Yingxiu town in Wenchuan county, in Chinas

southwestern province of Sichuan on June 2,2008. UN Secretary of St

carthquake-hit are

Condoleezza Rice on June 29 met children made homeless by th

devastating earthquake that hit sn
Ihe country’s response to the disaster. LIU JIN/StringeriGetty Images

Invest China last month and praised

Inc AFP
(Chapter 3 text, Cup and cone steel Alamy Images
Chapter 4, Rotary bit on portable oil drilinz rig. © Lowell Georg

CORBIS. All Rights Reserved
Chapter § Steam rising from soils and blurred spinning hollow stem
auger. Alam Images

Chapter 6 Stel framework ato
Chapter 7, Train wheels on track. il Stephenson) Alam Images

struction site. Corbis RE

Chapter 7 text, Highway flyover: Gari Wyn Wiliams\Alanıy Images
Chapter 8, Ski ft with snow coves
Shutterstock

Chapter 9, Turbine blades. Chris PeatsallAlamy Images

Chapter 10, Complex stresses developed within an aiplıne wing.

Courtesy of Measurements Group, Inc. Raleigh, North Carolina, 2761
Chapter I, Metal frame and yellow eran. Stephen Fina\Alamy Im:
Chapter 12, Man pole vaulting in desert. © Patrick Giardino'CORBIS

All Rights Reserved
Chapter 13, Water storage tower, J

Chapter 14, Shot of jack-up-pile-driver and floating crane. John
MacCooeytAlamy Images

MECHANICS
OF MATERIALS

Stress

CHAPTER OBJECTIVES
In this chopter we wil review some of the important principles

dings in a body, A

stress mil be introduced, a

design of mombe

1.1 Introduction

Mechanics of materials isa branch
effects of stress and strain in a slid

od that subjected to an external

ding. Stres is associated withthe strength of the

the body is made, while strain is a measure

body. In addition to this mechanics of materials incl
lity when a b

compressive loading. A thorough understandin

his subject is ol vital importance because many

umn is subjected o
Y he fundamentals of
mulas and ules

the body dy such as a

of design cited in engineering e

Historical Development. The origin of mechanics of

mathematical and computer techniques to solve more complex problems.
A result this subject expanded int other arcas of me

1.2 Equilibrium of a Deformable Body

of mechanics of materials its very important to have a good grasp of its
External Loads. A bod ly two types of extern
Surface Forces. Surface forces are caused contact of one

body withthe surface of another. In al ease
over the area of contact between the bodies

he Body then the surface force
an be idealized a à single concentrated force, which applied 0 ap

bieyle can be considered as a concentrated force. the surface loading i
linear distributed load, w(s) Here the losdi

Jas having a
presented graphically
series of arrows along the line 3, The resultant force En of ws) is
equivalent 10 the arca under the distributed loading curve, and this
resultan acts through the centroid C or geomelri center ofthis area. The
ing along the length of a beam is atypical example of where this

en applied

à Drome Bo 5

Body Forces. A body foreisdeveloped when one body exerts a force
ntact between the bodies Exam

another body without direct physical
include the effects caused by the earths gra
el, Although body frees affect each othe particles composing the body.

hese forces are normal represented by a single concentrated

ation or ts clectromagne

1 the had: the case of gravitation, lis force is called the weight of th

Support Reactions. The surface forces that develop atthe supports
oF points of contact between bodies are called reactions. For two:
dies subjected to coplanar force systems,

y encountered are shown in Table 1-1. Not
represent cach support and the type of

ns it exerts on is contacting member. As a general rule, if the
support prevents translation In a given direction, thea a force must be
developed on the member in that direcion. Likewise, Y rotation i
prevented, a couple moment must be exerted on the member For example
he roller support only prevents translation perpendicular or normal to
Fon the member a

the surface. Hence. the roller exerts a normal
i point of contact, Since the member can

"ple moment cannot be developed un the member

E Equations of Equilibrium. Equilibrium of a body requires both
x balance of forces. wo prevent the body from translating or having
scceleraed motion along a sgh or carved path and à bslence of

‘moments. to prevent the body from rotating. These conditions can be

expressed mathematically by two vector equations

My =0

acting om the body, and

Here, SF represents the sum of all the {

YM, is the sum of the moments of all the forces about any

Often in engin cie the loading na body can be represented

26, of equilibrium requires
plete specification of al the known and unknown forces that act on
he body and so the Bestway to account forall these forces i 10 draw
the body fre-body diagram

ae & fre

Three Dimensions. Later in ti
nes Fe and M,
ed area, and thereby develop equ

can by
ot Fy and M,

both normal and perpendicular 10 the sec
red, Fig 1-24. Four diferent types of

Normal force, N. This force acts perpendicular to the area. I is
developed whenever the external loads tend to push ar pull on the two

Shear force, V. The sheat force lies in the plane of the area and is
developed when the external luads tend to cause the two segments of
the body to side over une anothe

Torsional moment or torque, T. This effect is

Bending moment, M. The bending moment is caused by the

In this text note that graphical representation of a moment or torque is

hand ral, the thumb gives the arrowhead sense of this vector and the

Coplanar Loadings. I the body is subjected 0 a coplanar syste

imponents will exist at the section, Fig. 1-35. If we use the 2, 3

vorinate ates ar shawn on the left segment, then Neaa be obtained by

bending moment Mp can be determined by summing moments about
point O (the 2 axis), SM = 0, in order to eliminate the moments

fased by the unknowns N and V

|_| important Points

+ Mechanics of materials su study ofthe relationship between the

ternal load applied 0 a body and the Sres and stain caused

| Bine internal ads within the body
ody 35 distributed

forces tha act

External forces can be applied to a bi

concentrated surface loadings
| throughout the volume of the bod}

+ Linear distributed loadings produce a resultant force having a
inde equal to the area under the oa diagram, and having a

aon tht passes through the centroid of ts ate.

force in a particular direction on its
ple moment on the member if it

librium SE = 0 and IM = 0 must be
satisfied in order to prevent a body from translating with

+ ‘The equations of

oven ap pira (a's
ody diagram

fist draw the fe the body in order
for all the

2 The method of sections is used to determine the internal
sultant loadings ating tioned body. I
general, these resullants consist of a normal force, shear force,
torsional moment, and bending moment

nthe surface ofthe

The resultant interna loadings at a point located on the section of à
an be obtained using the method of sections This requires

lowing steps

Support Reactions

+ First decide which segment ofthe bo
Segment hasa support or connection to another body, then I
the body is sectioned, it will be necessary to determine the

m the chosen segment. To do this draw the rec
body diagram of the entre body and then apply the necessary

equations of equilibrium to obtain these reactions.

listo be considered. Wthe

Free Body Diagram

+ Keep all external distributed londings, souple moments, torques,
and forces in their exact locations, before passing an imaginary
1 trough the body atthe point where the resultant internal

loadings are to be determine

+ Draw a tree-body diagram of
indicate the unknown resultant N, V, M,

sd Ta the tection:
These resultante are normally placed at the point representing

sealer or conoid ofthe sectioned

+ Ifthe memberis subjected toa copla
and M act at the centroid

+ Establish the x.y atthe
and show the resultant internal loadings acting along the avec

= coordinate axes with origin roid

Equations of Equilibrium

+ Moments should be summed at the section, about each of the

coordinate axes where the resultante act. Doing this eliminates
the unknown forces N and V and allows a diret solution for M
(aad)

2 IC the solution of the equilibrium equations yields a m
value for a resultant, the

sultant i opposite to that shown on the

assumed dinectional sense of the

e-body diagram,

The following example illus ssumericlly and al

tse proce

Bo "

a y

Determine the es

at internal loadings acting on the eros section
at Cof the cantilevered beam shown in Fig. Ida.

| 1 I [Wetec
——

Support Reactions. The support reactions
determined i segment CB is considered
Free-Body Diagram. “Thefrec:body diagram of segment CBisshown
in Fig. 1-4. Its important to keep the distributed loading om the
segment until afer the section is made, Only then should this loan,
be replaced by a single resultant force. Notice that the intensity ofthe
distributed loading at C is found by proportion, Le. from Fig. 1-4
N/m)/9m. w = 180N/m. The magnitude of the
résultant of the distributed load is equal to the area under the

lo not have to be

loading curve (angle) and act through the centroid of ths
Thus. F = 180 N/m)(6m) = SION, which acts $(6m) = 2m fro
Cas shown in Fig. 1-4

Equations of Equilibrium. Applying the equations of equilibrium

we have
SF, = 0 Ne=0

Ne=0 An
tur, = 0 Ve-SWN=0

son Ans
Me = 0, Mc ~ SONG m) = 0

NOTE: The negative si
direction to that shown 0
problem using segment AC
A. which ae given in Fig

1080 Nm An:

indicates that Mc acts in the opp
the free-body diagram, Try solving this
by first obtaining the support reactions at
se

12

AA

Determine the resultant intemal
C of the machine shaft shown in Fig. 1-Sa. The shaft is supported by
Jouimal bearings at A and B, which only exert vertical

soLuTIO
We wil
Support Reactions. The free-body diagram of the entire shatt is
shown in Fig. 1-5), Since segment AC is to be

reaction at À hae to be determined. Why

je this problem us

nsidered, only the

+ 3 Mp = 0:—A(0:400 m) + 120N(0.125 m)~ 225 N(0.100 m) = 0

A, = —1875N
won ae The negative sign indicates that A, acts inthe opposite sense 0 that

| | shown on the frec-body diagram
N° FreeBody Diagram. The free mn of segment AC is

Ebo showa ia Fg
¡ us 1 N Equations of Equilibrium.
Ss = SF, = 0 Ne=0 An
o o 1875N —40N — Ve = 0

Ve ==588N Ans

Me = Mc + 40N(0025m) + 1875N(0250m) = 0

Me = -S49N-m Ans

NOTE: The negative signs for Ve and Me indicate they act inthe
posite directions un the freebody diagram. As an exercise,
calculate the reaction at B and try 10

segment CBD of the shat,

The S00 kg engine is suspended from the crane boom in Fi

Determine the resaltan internal loadings ating on Ihe cross seston
the boom at point E

SOLUTION >
Support Reactions. We will consider segment AE of the boom s

we must fist determine the pin reaction

CD is a eos

member The free-body diagram of the b
shown in Fig. 1-6. Appising the equations of equlibeiu
SM = @ — Feo(3)(2m) — (800981) NIGm)
Fey = 122425 N
0 (122025 N)fE) = 0
A, = 9810 N
TSF,=0 4,
Free-Body Diagram. The frec-binly diagram of segment AE
Equations of Equilibrium.
Nr + 9810N=0
Ny = -9810.N = ~981 kN
TSF,=0; Vi U2SN=0
Van “24525 N= -245 KN
SM 0: Mp + (24525 Nm) = 0
My = ~2452.5 Nom = 245 KN=m

1 A. Notice that enema

14

Determine the resultant internal loadings acting on the cross section

at G of the beam shown in Fig. 17a. Each joints pin connected

sown

Support Reactions. Here we will consider segment AG. The
free-bovly diagram ofthe entire structure is shown in Fig 1-75. Verify
the calculated reactions at E and C. In particular note that BC is

re member Sine only two forces at on i. For this reason the

force at C must act along BC, which i horizontal as shown
Since BA and BD are also two-orse members the Iree-body

© iagram of joint Bis shown in Fig. 1-7c. Again, verify the magnitudes
at furces Fp and Fan
A Free-Body Diagram. Using the result for Eng, the free-body
| ‘ok diagram of segment AG is shown in Fig 1-7
Equations of Equilibrium.
A ©. OM) + Na=0 Na=-620b Ans
Han LE ib
Va= 31501 Ans
LHEMG= 0; Ma ~ (775010)(3)(2 4) + 15001 o

Determine the resultant internal loadings acting on the cross section
at B ofthe pipe shown i Fig. 1-Ba.The pipe has a mass of 2 kg/m and
is subjected to both a vertical force of SON and a couple moment af
TON matits end A. Is fixed tothe wall at C

seed to calculate the sup

Free-Body Diagram. The x, y, z ates are established at and th

Im ofsegment AB is shown in Fig.1-8b The evant

are assumed to act in
passthrough the centroid of
the cross ectional area at B. The weight of each segment of pipe is

Wan = (2k

Wan = (2 kg/em)(1.25 m)( 9S N/Kg) = 24525 N
These forces act through the center of gravity of each se

Equations of Equilibrium. Applying the six scalar equations of
equilibrium, we hare»

SF, = 0 (Fn). = 0 An
SF, = 0 (En, = 0 An
EF. = 0 ISIN — 24505N — SON -0
Fp) = AN A
S(Mp)=0% (Mp) + 70N+m — SON Sm)
24525 N (05m) ~ 981N(0.25m) = 0
(Ma), = 303 N-m
E(Ma}y = 0; (Ma), + 24.525 N (0625m) + SON (125m) = 0
(My), = -T78N-m An
SEM), = 0. (My).=0 An
NOTE: What do he negative signs for (Mp), and (Ma), indicate
Note thatthe normal forse Na = (Fa), = 0; whereas the shear fore

is Va B43. Also, the torsional m

Ta= (Ms), =778N-m and the bending moment is Mn

TOF = 303N-m.

Figs

pil = Tt
ir = ni

124. Determine the resaltan internal normal ore acting 1-3. Determine the reutan internal or

we] E — “ab

N

a Wan rn

Normal Stress. The

acting normal to AA is

defined asthe normal stress, (sigma). Since AR, is normal to the area
he

If the normal force or stress “pulls” on AA as shown in Fig. 1-104, i
referred to as tensile stress, whereas if "pushes” on AA itis called

Shear Stress. The intensity of

% acting tangent to AA is called
he shear stress, (au), Here we have shear stress components
roe gm SF
Ruy
ar d
aA
Note that in this subscript notation z species the orientation of th
area AA, Fig 1-1, and x and indicate the axes along which each she
General State of Stress. If the body is further sectioned hh
planes parallel to the 3—2 plane, Fig. 1-10), andthe 3-2 lane, Fig I-10

he state of stress actin

Units. Since st
International Standard or Sl system, the magnitudes of

point inthe body This st

represents a force per unit ares, in the

N/m). This unit, called a pascal (1 Pa = 1 N/m) is rather small

k

32 (10%), symbolized by M, or giga- (10), ssmbolized by Gare used

psi) or Klopounds per square inch (ks), wher

1.4 Average Normal Stress in an
Axially Loaded Bar

Homogeneous material base
throughout its volume, and kotrop
n all directions Many engineerin

tniformly when subjected to the axa load P

Average Normal Stress Distribution. If we pass a section
through and separate i into two pa

Asa result, each small area A A on the
wee AF =a AA, and the sum of these forces acti
nd therefore A

2 e
Here
7 = average normal stress at any point on the cross setional ae
P = internal resultant normal force, which acts through the centroid of

Since the internal load P passes th

on the uniform stress distribution wil produce 2
axes passing through this point, Fi. 1-13d. To show

moment of P about each axis to be equal to the moment of

soda = x | yaa

une zn; 0e fxar ef mine

Pl dre fetta Ge AA
Equilibrium. etc be apparent at nl a soma sete loc:
hu is sl ul an ae Ka bs If re cet vr
ae at 1 au soy en
ecules

|
s an (89)

menu

In other words the to normal stress components un the clement must
be equal in magnitude but opposite in direction. This is refered to as.
"The previous analysis applies to members subjected 10 either tension
‘or compression, as shown in Fig 1-15. Asa graphical interpretation, the
‘magnitude of the intemal resultan force Pis equivalent o the volume
under the stress diagram: that is, P = «A (volume = height base)
balance of moments this resultant

nd this analysis for prismatic bars, this

Furthermore, as a consequence
‘passes through the centoid ofthis
Although we have devel

taper. For example, it can be shown, using the more exact analysis of the
ent sides is 15° the ave

stress, as calculated by or = P/ A, i only
from the theory of elasticity

Maximum Average Normal Stress. In our analysis both the

intemal force Pand the cross-sectional area A were constan along the
Tongitudinal axis ofthe bar, nd asa resul the normal stress = P/A is
alo comstont throughout the bar length, Occasionally, however, the ba

en mal area may occur. As a result, the normal stress within the

ba

ould be diferent from ane section to the next and, if the mazimum

rage normal stress is to be determined, then it becomes important

to find the

ation where the ratio P/A is a maximum. To do this it is

necessary to determine the intemal force Pat Var

bar. Here it may be helpful to show this variation by drawing an axial or

normal force diagram, Speciic sal
P versus its position x along the bar' length. Asa sign convention,

P will be pos

known the maximum ratio of P/A can then be identified

Important Points

= When a body subjected to external loads is sectioned, there is a

distribution of force acting over the sectioned arca which holds
each segment of the body in equilibrium. The intensity of this
intemal force at a point in the body is re

| + stress is the limiting value of force per unit area, as the area
approaches zero. For this definition, the material is considered to

= ‘The magnitude ofthe stress components at a point depends upon
the type of
clement at the point

nly and the orientation of the

ang acting on the b

When a prsmatc bar ls made trom. homogene
material and is subjected to an axial force acting through the

of the cro sectional arca. hea
ar will ly Aca result the material will be
subjected only to normal stress. This stress is uniform or averaged
‘oer the cross sectional area

The equation = P/A gives the average normal stress on the cross.
sectional area of a member when the section is subjected 10 an

internal resultant normal force P. For axially k
application of this equation requires the following step,

lcd members

+ 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 farce Pat the section.

Average Norm

© Determine the member's crosesecional atest the section and
caleulate the average normal stress = P/A

n 2 small volume element

here stress is

eral located at a point om the se
do this first draw o on the face of the
ident with the sectioned area A. Here a acs in the sume

direction as the internal force P since all the n
the cross section develop this resultant. The normal
the other face ofthe element acs inthe opposite direcion.

20 Garni 5

EA

The bar in Fig. 1-16 has a constant width of 35 mm and a thickness
of 10 mm, Determine the maxim

n average normal stress in the ba

When itis subjected to theo

SOLUTION
Internal Loading. By inspection the internal axial forces in regions

B. BC, and CD are all constant yet have different magnitudes Usin
the method of sections, these loadings are determined in Fig 1-16),

and the normal force diagram which represents these results graphically
fs shown in Fig. 1-16 The la

loading is in region BC. where

Pac = SON, Since the erose-sctional area of the bar is constant, the

Average Normal Stress. Applying Ey.1-6,we have

3010) N
Pre = 857MPa Ant

NOTE: The tres distribution acting on an arbitrary cross section of
the bar within segon B
Fig 16 the load of 30 KN; that is 30 EN = (857 MPa)(3S mm) {10 mm)

bi is shown in Fig. 1-164. Graphically the volume

The S0-kg lamp is supported by two rods AB and BC as shown in
Fig 1-170. IF AB has a diameter of 10 mm and BC has a diameter of
$ mn, determine the average normal stress in each rod

Internal Loading. We must first determine the axial force in each
roa. free-boay diagram of the lamp is shown in Fig 1-170. Appiin
the equations of force equilibrium

SrR=0 FF Epa cos 0? = 0
TEE, = 0; Fye(3) + Fay sin 60° — 7848N = 0

En

their length

By Newton's third law of action, equal but opposi
forces subject the rods to tension throu

Average Normal Stress. Applying Eu. 1-6,

on 0 = 7.86 MPa Ans

NOTE: The average normal stress distribu
section of sod AB is shown in Fig 1-17e, and at a point on this ross aux

section, an element of materials sessed as shown in Fig. 1-174. w

30 Garni 5

ET

The casting shown in Fi. L-IRu is made of steel having a specific
weight of ya = 490 b/ft, Determine the average compressive Stress
ting a points A and B

SOLUTION

Internal Loading. A frec-body diagram of the top segment of the
casting where the hh points A and B is shown in
Fig. 1-18). The weight of Uns segment determined from Wa = ya Y

Thus he internal axial force P at the section is
0 P-W.=0
P = (40010/40)(275 ff 2(07S 13] = 0
2381 th
Average Compressive Stress. The crusesectional area atthe sec
tion is. A = (075 ft), amd s0 the average compressive stress becomes
ge Le arsine
a= 1375 1/1 (110/144ia") = 9.3 p dine

NOTE: The stress shown on the volume element of material in
Fig. 1-18c is representative ofthe condit point À or B
Notice that this stress act upward on the bottom or shaded face ofthe

the resultant neral force Ps pshing
pad

Member AC shown in Fig 1-19a is subjected to a vet

EN. Determine the position x of this force so that the avera

th support Cs equal u
tensile stress in the tie rod AB. The rod has a crosssecti
400mm? and the contact area at Ci 650 mn

the avera

Internal Loading. The forces at A and Ccan be related by
the free-body diagram for member AC, Fig 1-19, Th
unknowns namely; Fan, Fe, and x. To solve this proble
work in units of ne

ns and millimeters
TEE, = 0; Fan + Fe = SON = 0
EMA = 0 —S000N(x) + Fe(200.mm) = 0
Average Normal Stress. À necessary third equation a
that requires the tensile stress in the bar AB and the €
tres at Cto be equivalent, ie
F
Tom’ © EU mm
Fo = LA2SP

Substituting this into Ea, 1, solving

Y Fag, then soli

Fay = 1
Fe = ISIN

he postion of the applied load is determined from Ed. 2

x = 124mm
NOTE: < x < 200 mm, a required

32

E 1.5 Average Shear Stress
F a

develop comer the effect af

m | If the supports are considered rigid, and F is Large eno wil cause

og, inst Sarto arm anda sg te ples denied

VE EI? must d

en defined in Section 13 a the stress component that
sing a force Fo the Darin Fig 1-2.

shear sires distributed over each sectioned area that develops this shea

=] in

the shea must reste associated forces all of which contribute t
the internal resultat force Y atthe seston,

The loading case discussed here is an example of simple or direct

shear, since the shears caused by the direct action ofthe applied load F

hat use bolts, pins Welding material, ete In all these cases however
application of Eq. 1-7 i only approximate. À more precise investigator

citation. Although this may be the cate; application of Eq. 1-7 a
‘design sizes for fasteners such as bots and for obtaining the b

Srength of glued joints subjected to shear loadings.

En

Shear Stress Equilibrium. Figure -21asho
Y material taken at

SF, » 6 axa axay= a
Ina similar manner, force equilibrium in the = direction yields
Fill taking moment about the à ex

LE T
rq Ft

x“ rele A) ar Hrsfaxdody=0

“other words, all four shear stresses must hare equal magnitude and
be directed either toward or away from each other at opposite edges of
the element Fig 1-21h This referred to asthe complementary propern

shear. and under the conditions shown in Fi, 1-21, the material is

‘subjected to pure shear

34 arent 5

E Important Points

+ IC two parts are thin or small when joined together the applied
loads may cause shearing of the material with nepligible bending
this isthe case, iis generally assumed that an average shear

When shear stress 7 acts om a plane, then equilibrium ofa 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

The equation Tay = V/A is used to determine the average shear

stress the material Application requires the following steps

internal Shear,

+ Section the member atthe point where the average shear stress is

tobe determined.

* Daw the necessary free-body diagram, and calculate the internal
shear force V acting atthe section that is necessary to hold the
part in eguilbriun,

Au

age Shea Stross

= Determine the sectioned
shear stress Tay = V/A.

wea A, and determine the average

+ tis suggested that 7.4 be shown on a small volume element of
material located at a point on the section where ts determined
To do this ist draw 7 on the face ofthe element, coincident
With the sectioned area A, This tres acts in the same direction
as V. The shear st
thea be drawn in their appropriate directions following the
scheme shown in Fig 1-21

sses acting on the three adjacent planes can

Determin
pin at A and ıh
beam in Fig

the 20-mm -diametes
B that supp

the average shear stress in

mm diamete

soLuric

Internal Loadings. The force
comsidering the equilibrium of

EM, =0; Fa(2)(6m)-30kN(2m)=0 Fy=125 kN

Sr, 0, 0 A-7008

SF, =o, WEN = 0
ET

Thus the resultat force ating om pin A is

Fy = Val + a} = VG@SOEN) > QOEN) = 21.362N

ied “leaves” and so the
the cente the pin shown in Fig. |
Shearing surfaces between the beam and each leaf The force of th
beam (21.36 KN) acting on the pin is therefore supported by shea
force om each ofthese surfaces This cae is called double shear. Thus

A is supported by two

E

Va 10681

In Fig. 1-224, note that pin Bs subjested to single she, which ocsurs

on the section between the cable and beam, Fig I-22 For this pin
Vp = Fy = I25KN
Average Sheer Stress.
Y, _ 1068(10)N
aly = oe = © 34.0 MPa An
Va SIÓN ze ,

36 Garni 5)

E EEE)

1a bas a width of 150 mm, determine the
and hb For
the material

joint in Fig.

shear stress deve shear planes

each pl

SOLUTION
Internal Losdings. Referring to the free-boly diagram of the

member, Fig

LEFL=0; 6kN-F-F=0 Feen

and by shown in ga 1-28 and 1332
= S9F,=0; V,-3RN=0 V,=3N
SRN-Vy=0 Vis
Average Shear Stress.
© on
ña) N = mr Ans
I LG 100 kPa An

‘om elements located om sections a-a and Bb is
and 1-23 respectively

The state
shove in Figs 1-2:

7

EA y

The inclined member in Fig. 1-2
of 60 Ib: Determ

ted to compressive force
the average compressive stress along the smooth
reas of contact defined by AB and BC, and the average shear sires
along the horizontal plane defined by DB.

Internal Loadings. The free-body diagram ofthe inclined membe
is shown in Fig. 1-240, The compressive forces acting on the areas of

RSF, =0; Fan ~ 600th £

“ob
TEFL=0; Feb) =0 Fe = 48000

Also, from the frec-body diagram of the 1

‘segment ABD of th

bottom member, Fig. 1-24c, the shear force acting un the sectioned
horizontal plane DB is
SF, 0 360 Ib
Average Stress. The average compressive stresses along th
horizontal and vertical planes ofthe inclined member are
Fan >
ea E “ps An
En 48010
ee 60 ps Ans

Ave” eines
These stress distributions are shown in Fig. 1-244,
The average shear stres acting on the horizontal plane defined by

Duis

in

This stress is shown uniformly distributed over the sectioned area in

4 ;

E

ZN

46

1.6 Allowable Stress

To ensure this safety itis therefore necessary to choose an allowable

member can fully support. There are many reasons for doing ths. F
example, the load for which the member is designed may be different
rom actual loadings placed on The intended measurements of

Tosdiags ean our that may not be accounted for in the

n. Atmospheric corrosión, decay, or weathering tend to cause

vaterals o deteriorate during service. And lastly. some materia, su
wood, cone iberreinforeed composites, can show
variability in mechanical propertic

unber called he factor of Se: The factor of safety (ES) i ato ol
the fare oad Fit the allowable Wad Paes Here Fouad rom
mental testing ofthe materia, and the Tatoo safety

If the load applied 10 the member is linearly related to the stress
developed within the member, as in the case of using ı = P/A and
E A, hen we can also expres the factor of safety as a ratio of the

fare stress (or Fu) tothe allowable tree a (0 Fae th i
FS (4410)

values are intended to form a balan
mental safety and providing à reas

1.7 Design of Simple Connections

One es and tn bc an erg sew for SS

as Guarren 1 Srmess

E Important Point

+ Design of a member for strength is bad on selecting an
allowable stress that will enable i to safely Support is intended
load. Since there are many unknown factors that
the actual sizes in a member, then depending upon the intended
tse of the member, a factor of safety i applied to obtain the
allowable loa the member can sopport

When solving pr

a influence

blems using the average normal and shear stress

m should first be made as to choose
the section over which the critical stress i ating. Once thie section
is determined, the member mist then be designed to have à
sufficient area atthe section 0 resist the stress that acts on it. This
area is determined using the following steps

internat Losding,

+ Section the member through the area and draw a free-body
diagram of segment of the member. The internal resultat force at
| the section is them determined using the equations of equilibrium

Required Area

| + Provited the allowable stress is known or can be determined,
the required area needed to sustain the load atthe sections then

determine from A = Pam or A
| =

EE y

The control arm is subjected to the loading shown in Fig. 1-260

Determine to the nearest + in. the required diameter of the steel pin

at Cif the allowable shear stress forthe steel i rage = SS

Internal Shear Force. À tree-bodly’digram of the arm is shown in
Fig 1-260, For equilibrium we hav
HEMe= 0, Fax(B in.) ~ 3kip (Gin

Fa = 3kip
AEF, =0 Skip - Ci + Skip({)=0 C= 1hip
1EF,=0; C-3kip-Skpf)-0 C,=6kip

Fe = Vile Ep
Since the pin is subjected to double shear, a shear force of 3.41 kip

6.052 kip

is cros-sectional area between the arm and each support
leaf forthe pin, Fig 1-26
Required Area. We have

Use a pin having a diameter of

d = Yin, = 0750in An

UE A

The sus d-connected circular

der rod is supported at is end by

Ithe rod passes through a M-mm diameter
hole, determine the minimum required diameter of the rod an
the mess of the disk needed to support the 20-KN load
The allowable normal stress forthe Tod is gue, = 60 MPa, and the
allowable shear stress forthe disk e 35 MPa

SOLUTION
Diameter of Red. By inspection, the axial force inthe rod is 20 kN
Thus the required

2, 210)N
on N
so that

À = 0.0206m = 206 mm Ans

Thickness of Disk. As

mwa on the tree-body diagram in
atthe sectioned area of the disk must resist
fe I this
shear stress is astamed to be uniformly distributed over the sestioned

area. then, ince V = 20 UN, we have

EA y

The shaft shown in ais supported by the collar at €, which is
tached tothe shaft and located on the right side of the bearing a
Determine the largest value of P forthe axial forces at E and F so
thatthe bearing stress on the collar does not exceed an allowabl
stress of (a) = 75 MPa and the average normal stress in the shaft
bes not exceed an allowable stress of (0; unm = 35 MPa.

Ta soe te problem we wil determine P foreach posible al
odio. Then we wil choose the smallest value. Wi?
Normal Stress. Usage method of tector; the ana ood wäh
region FE ofthe shafts 2 whereas the largest ail force AP. occur
within region EC. Fig. 1-28, The variation of the internal loading i
Steal shown on the norma force diagram, Fig 1-2e Since the cos.
the maximum average normal stress Appying Eq I-11,wc have
r= 2(00 my? =
= ET

P= SLRKN am
eering Stress. As shown onthe fee body diogram in Fig. 1-284 "Er
area ot As = [AO — #0 m} = 219910") m Tb a
a. E 219910) m? = IE _

Fe) Na
P=SS04N

mparison. the largest foal that can be applied o the shaft is
SLR KN, since any load larger than this will cause the allowabl
NOTE: Here we have not considered a possible shear failure of th
collar as in Example 114

52 En

E EE

The rigid bar AB shown in Fig, 1-204 i supported by a steel rod AC
having a diameter of 20 mm and an aluminum block having a cross
sectional area of 1800 mm. The IS-mm-diameter pins at A and Care
subjected to sngl sear Ifthe failure stress forthe steel and aluminum
8 (ws = 6SOMPa and (oy) = 70MPa, respectively, and the
he bar Apply factor ofsafety ofS, = 2

failure shear sires for each pin is,
a Pthatcan be applied

SOLUTION
Using Eqs 1-9 and 1-10, the allowable stresses are

(reat SSOMPA | 09 se,
(ham = MPa
(ala _TOMP! _ 56 spy

am = LE MPa 35M

ua _ WOMPA _ gay,

~~ ze - 20 à MPa

The fre-body diagram of the bar is shown in Fig. 1-29, There are
three unknowns Here

il apply the moment equations

E

equilibrium
in terms the applied load P We have

m) ~ Fac m) = 0 «

in ordertoexpress Fac an
GSM = 0: Pa

WEM, = 0: F (2m) — P(0.7S m) = 0 e

‘We will now determine each value of P that creates the allowable

stress in the rod, block, and
Rod AC. This requi

Fac = (ra) ac) = 340(10°) N/a? [m
Using Eq.

1 0)"] = 1068kN

(1068 N) (29
er

Block B. In his case,
An = 38(10°) N/m [1800 mm (104) Jm = 650KN

(SONNE m)

168 KN

Pin A or €. Due to single shea

Fre V = tae

35010") N/a f(00K9 m] = 1145 EN

From Eq
AS EN (2m
ee issn

By comparison, as Preaches iss ue (168KN),theallowable

P= 168kN An

TT À
HR <
Aa

¡A a"

a

62 Guarren 1 Sim

E | [REVIEW PROBLEMS

66 Guarren 2

2.2 Strain

Normal Strain. I we define the

Fig 2-16 The change in
Ay — As. If we define the average normal stain

doser and lose Act length ofthe ln
shorter, such that As — 0, Alo, this causes $10
Ay 0. Comequentÿ. in the limit the normal

10m. In the Foot-Pound-S
a units of inches per inch (in fin.) So

imental work

22 sra 67
1 = 01%. As an example

Shear Strain. Deformations

se

Notice thatthe normal strains cause a chan

Cartesian Strain Components. Using the definitions of norma
deformation ofthe body in Fig 2.30: To de

show how they enn be used to describe th
imagine the b

Subuivided into small elements such as th
This element is rectangular, has undeformed dimensions Ax, Ay. and
ate in the neighborhood of a point in he body. Fig 2-3

les of each side. For example, Ax

des e,AX, so lts mew length is Ax + eAx. Therefore, the

‚pproximate lengths ofthe three ses of the parallelepiped are
Lee) ax a (+e)4

shear strains cause a change in ls shape, Of course, both of

fete occur simultaneously during the deformation

In summary then, the sae of strain at à point in a body requires
specifying three normal strains e,.€,, €,, and three shear strains

These strains completely describe the deformation of à

volume element of material located at the point and

Small Strain Analysis. Most engineering design involves

applications for which only small deformations are allowed. In this ext
Therefore, we wil assume thatthe
body are almost infinitesimal fn particular the normal stuns occurrin

within the material are very smal! compared Lo I, x0 that e <= 1. This
ferred 10 as 3 small spain analysis I can be used, for example, to

Important Points

sy wil uadergo déplacement or changes poston

a sal ine segment in the body wherens she
isa measure ofthe change in angle that occurs between two small
line sepmens that are orginally perpendicular o one another

+ The state

strain at a point is characterized by six strain

components: three normal strains £,. #,, €, and three she

su nts depend upon the orginal
tation ofthe line segments and their location in the body

i Yay. Yo» Yas These Compe

Strain is the geometrical yeasured
experimental techniques. Once obtained the stress in th
can then be determined from material property relations, as
discussed inthe next chapter

Mos engineering materials undergo very small deformations and so
the normalstrain e <= 1. Thisassumption of “small rain analysis
allows the calculations for normal stain to be simplified, since fist
heer approximations canbe made about thet

that take place within a

69

70 Gwarren 2 Sina

The slender rod shown in Fig 2-4 i subjected tw an increase
temperature along its axis which creates a normal stain in the

le = 40(10 3)2°%. where à is measured in meters Determine (a) the
displacement of he end B of the rod dueto the temperature increase

nnd (9) the average normal stain inthe rod

SOLUTION

art (a). Since the normal strain i reported

à each point along the
od. a diferer
‘deformed length that can be determined from Ey 2-1; that i

segment de. located at position z Fig. 2-4, has

The sum of these segments along the avs yield the deformed
ofthe rod.

020

The displacement of the end of the rod is therefore

Part (bj. The averag
Eu. 2-1. which assumes
Length 0200 mm and a change in length

is determined from

9 mm. Hence

20119 m/m Ans
5 20mm

ABC in Fig 2-Sa,th
wise about pin A through an angle of O15

Determine the normal strain developed in wire BD.

se P is applied tothe rig lever arm

some
em nf) ron

then

Lo Vin + L AL an Lande

Normal Strain,

Lo = Lap „ 403991 mm — 300 mm _ ggg,
ay = HD (0.00116 m/m

the elongation of wire BD as AL yp shown in Fig.2-Sb. Here,

tained by approx

okay = [ES

at,
in

te

7

XAMPLE

Due he plate isd
in Fig. 2-60, Determine (a) the average normal strain along the side

AB, and (b) the average shear strain in the plate at À relative tothe
‘rand y ates

ormed into the dashed shape shown

SOLUTION

Part (al. Line AB, coincident with the y axis becomes line AB" after
formation, as shown in Fig 2-6h, The length of AB" is

The average normal strain for AB is therefore

AB = AB _ 248018 mm = 250mm
AB

a indicate (be strain causes a contention of AB,

The negative sg

Part (b). As noted in Fig, 2-66 the once 00
sides ofthe plate at A changes ta # due tothe displacement of 10
Br. Since yay = 7/2 — 0, then y,, is the angle shown in the figure
Thus,

BAC between the

The plate shown in Fig. 2-Tu is fixed connected along AB and held in

the horizontal guides at its top and bottom, AD and BC. If its right
side CD is given a uniform horizontal displacement of 2 mn
determine (a) the average normal strain along the diagonal AC. and

(b) the shear strain at E relative to he x,y axes

Part (al. When the plate is deformed, the diagonal AC bec
AC’, Fig. 2-Th The length of diagonals AC and AC? can be found
from the Pythagorean theorem We hav

Ac’ = VOI + (OSE

Part (b). ‘To find the shear strain at E relative 10 the x and y axes it
is ist mation, Fig. 2-7. W

= 902759 = (= ) = 1.58404 rad

Applying Eu 2-3, the shear strain at Eis therefore

indicates that Ihe angle 9 is greater han 9

NOTE: Ifthe x and y axes were

horizontal and vertical at point E
0" ansle between these axes
deformation, and so y,, = Dat point E

7 Chers sn

| [FUNDAMENTAL PROBLEMS

Mechanical Properties
of Materials

CHAPTER OBJECTIVES

form a tension or compression test specimen of the material is

3.2 The Stress-Strain Diagram

pare atest specimen to match the sí
each structural member. Rather, the test results must be e
hey apply to a

mber of any size To achieve this, the load and
à are used to calculate various Values of

he stress and corresponding strain in the specimen. A plot ofthe results
he sres-strain dingram There are two ways in

corresponding deformation

Conventional Stress-Strain Diagram. We can determine th

nominal ur engineering stress by dividing the applied load P by th

wer the ross section and throughout the gauge

Likewise, the nominal or engineering strain is found directly from th

length. 5, by the specimen's original gauge length La. Here the strain
Thus

If the corresponding values of er and e are plotted so that the vertical
axis isthe stress andthe horizontal axis isthe traia, the resulting curve

called a conventional stress-strain diagram. Realize however, that 1

mperfections the way i is manufactured, the rate of loading, and the

imen is shown in F

< depending on the amount
strain induced i the material

84

ben

Elastic Behavior. Elastic behavior of the material occurs when

Fig. 3-4, Here the curve is actually a staight line throughout most of

this regian.so that the stress is proportional to the strain. The material
inthis region is said to be Linear elastic The upper tres limit to this
linear relationship is called the proportional limit. 0. IC the stress
slightly exce

fl

etl, Upon reaching this
bil stil return back to is original shape. Normally for steel, howeve
the elastic mined

proportional limit and therefore rather difficult o detect

I the proportional limit, the curve tends to bend and
< the elastie

Yielding. À slight increase in stres above the elastic limit will result
This behavior is called ylelding, and i is indicated by
the yield stress or yield point, ay, and the deformation that occurs

à plastic deformation. Although not shown in Fig. 3-4 or low
n steels or those that are hot rolled, the yield point is often

distinguished by two values. ‘The upper yield point
lowed by a sudden decrease in load-carrying capacity to a lower
yield point Notice that once the yield point is reach

la Pla: 4 dba speci

will continue to elongate (train) without an
increase in load. When the material isin this tate, it oflen referred (a

as being perfectly plastic

Strain Hardening. Wen ye

but becomes flatter un it reaches a maximum stress
ultimate sires, The is in the curve inthis manner is called strain

hardening, and itis identified in Fi.3-4 as the region in light green
Necking. Up to the ultimate stress as the specimen elongates its

s-sectional are se This decrease i faily uniform over th

imen's entire pause fever just after, at the ultimat
stress the erost-seetional area will begin to decrease in a localized

n of the specimen, As a result. a constriction or “neck” ten

m in this region as the specimen elongates further, Fig. Sa. This

{gion the curve due 10 necking i indicated in dark green
Here sede util the sp

true Stan’ Era Diagram. sacaba lo th ri Zu

values iscalled res-sirain diagram
ted it has a form shown by the light-bl

When this d

is a large divergence within the necking region. Here i can b

supports à decreasing load, since Ay is constant when calcula

86

when the load is removed, The true strain up to the elastic limit wil
of rand i very small (about 0.1%) compared with ther tae value
This is one of the primary reasons for using conventional tress-stra

E The above concepts can he summarized with reference to Fig. 3-6
which shows an actual conventional stress-strain diagram for amid st
Specimen, In order to enhance the detail the elastic region ofthe curve

reached at try = 35 ksi (241 MPa), where ey = .0012in,/in. This à
int of (ayy 38 (262 MPa), u

si 248 M

|
\ », 0%

3.3 Stress-Strain Behavior of Ductile
and Brittle Materials

Materials can be classified as either being ductile o brite, depending on

Ductile Materials. Any material that can be subjected o lan
ins before it fractures i called a ductile material. Mild ste
discussed prev sample Engineers often choose dtl

materials fo
shock or energ and if they become overloaded, they will usually extibi

‘One way to specify the ductility of a mate report its percent
clongation or pereent reduction in arca at the time of fracture. Th

percent longetion is Uhe specinen's [meluie rita expressed 48 à

percent. Thus if the specimen orginal gauge length is La and its length

Ass
steel specim
The percent redaction in area is another way ti

since « his value would be 38% fora mild

defined within the region of nesking a follows
Percent reduction of area = 22 (100% +4)

Here Anis the specimen’s original crosssecional area and Ay isthe area

of he neck at fracture. Mild stel has a typical value of

Besides steel, other metals such as brass, molybdenum, and zinc may

strain hardening, and finally necking ual fracture. In most meta
however, constant yielding will nor occur beyond the elastic range. Oni

à well-defined sil point and consequently ts standard
fine a veld strength usine a graphical procedure called th
Offset method. Normally a 02% strain (0.002 in Ji.) is chosen, and from
his point on

of the stres-strain diagram is drawn, The point where (is lin

al = = Ñ y

Brittle Materials.

Materials hat exhibit lite o no yielding beto
5s britile materials. Gray cast ron is an example

having a stress-strain diagram in tension as showa by portion AB of th
curve in Fig. 3-9, Here fracture at a = 22 ksi (152 MPa) took pli
initially a an imperfection or microscopie crack and then spread rapid

across the specimen, causing complete fracture. Since the appe
ack in a specimen is quite random, brite materials d

Im à set of observed tests is generally reported. A typical failed
‘Compared with their behavior in tension, brtle material, such as
evidenced by portion AC ofthe curve in Fig 3-9, For this case any cracks
or imperfections in the specimen tend to close up. and as the load
increases the material will generally bulge or become barrel shaped as
he strains become large, Fig. +10

Like

also has a low strength capacity in tension. The characteristics of is
sani, gravel, and cement) and the time and temperature of curing. A
pical example of a “complete” stress-strain diagram for concrete &
ven in Fi By inspection, its maximum compress

eed with steel bars or rous whenever is designed

ductile This

fect is shown in F

3.4 Hooke's Law

strain within the elastic region. Consequently. an increase in stress causes
à proportionate increase in stain This fact was discovered by Robert
Louke in 1676 using springs and is known as Hooke’ law. lt may be
expressed mathematically as

[ee +5)
¡pez detras a

modulus of elasticity or Young's modulus. named after Th
sho published an account of tin 1807
Equation 3-5 actually represents the equation of the initial straight

À portion of the stress-strain diagram up to the proportional limit

Furthermore, the modulus of elisticty represents the slope of (his line
Since strain is dimensionless, from Ey, 3-5, E wil have the Same units as
xs. such as pst, KS or pascals AS an example of it calculation,
ider the stress-strain diagram for steel shown in Fig. 3-6. Here
DIE a
As shown in Fig. 3-13, the proportional Hm type
steel alloy depends on its carbon content however, most grades of see,

be Eu = 29(10) ki or
200 GPa Values of E for other commonly used engineering matenals are

n tabulated in engineering codes and reference books Representative

dues are also listed on the inside back cover of this book. It should
be noted that the modulus of clatcity is a mechanical propenty that
indicates the stiffness of a material. Materials that are very sf, such

have large values of E [Eu = 20(10") ksi or 200 GPa]. whereas
spongy materials such as vulcanized rubber may have low values
E, = 0.10 ks or 0.70 MPa]

The modulus of elasticity is one of the most important mechanical

properties used in the development of equations presented inthis tex. It
must always be remembered, though, that E can be used only f a

Strain Hardening. 114 specimen of ductile material. such as see

he plastic region and then unluaded, elastic strain à
ine turns o is equilibrium state. The pl
strain remains, however, and as a result the material subjected to a

à loaded int

permanent set. For example, a wire when bent (plastically) wll spring
Back a litle (elastically) when the lond is removed; however, it wil n
lly return to its original position. T

he tres atrain diagram shown in Fig. 3-1da. Here the specimen i fr

oaded beyond its yield point A to point A’. Since interatomic forces

same forces pull the atoms back together when the load is removed
Fig 3-Hde. Consequently, the modulus of elasticity E, is the same, and
erefure the slope of line 0° isthe same as line OA

I the loads reapplied the atomsin the material will pain be displaced
until yielding occurs a or near the stress A’, and the stress-srain diagram
tines along the same path as before, Fig. 3-145. It should be noted
however, that this new stress-strain diagram, defined by O°A’B, now has

92

3.5 Strain Energy

strains in the material, i is referred to as stain energy. To obtain this
Strain energy consider a volume element of material from a tenso
est specimen. Its subjected to uniaval stress as shown in Fig 3-15,
This sress develops a force AF = 7 AA = @(Ax Ay) on the top and
bottom faces ofthe element after the element of length Az undergoes

final magnitude AF

when the displacement e Az is attained, the work done on the element
by the force & equal o the average farce magnitude (AF/2) times the
displacement e Az. This “external work” on the element i equivalen 0
the “internal work” or strain energy stored in the element -assuming
hat no energy is fost in the Form of heal. Consequentiy. ib strain energy
AU is AU = (RAF) € A = (La Ax Ay) e Az. Since Ihe Volume uf the

density and it can be expressed as
su _1 a
Ab tue 6)

a = Ee, and therefore we can exp

Modulus of Resilience. In particular, wh

the proportional limit. the strain-energy density. as 6
3-7, referred to as the modulus of resilience.

hat 1, is equivalent to the shaded triangular area under the diagram,
Physically a materials resilience represents the ability ofthe material t
absorb energy without any permanent damage to the mate

Modulus of Toughness. À:

material is the modulus of roughness, This quantity
tire area under the stress-strain dlagram, Fig. 3-1

|_| important Points

on dd gage [ei te Specie
+ À ductile m 4 steel. has four distinct behaviors as

itis loaded They are elastic behavior yielding, rain hardening. and »
+ A material is car ela he stress is proportional tothe strain

within he elastic region This behavior i described by Hooke’ lam

a = Fe, where the modulus of els

fe isthe slope of the ne

+: Important points on the stress-strain diagram are the proportional
limit yield ses ultimate sires, and fracture a
: a material can be specified by the specimen's

nt or the percent reduction in area

a | LE

hy can tache modes

material This is done by staining the material beyond the elastic

limit then releasing the load. The modulus of elasticity remains
the same; however, the material ductility decreases,

Sirain energy is cacegystoredin a material due tots deformati

This energy per unit volume is called strain-energy density. I a
it is measured up to the proportional limit, iis referred to as Bent
the modulus of resilience. and if itis measured up to the p
fracture, itis called. the modulus of toughness. It can be specie el
determined from the area under the oe diagram tc ae acted by >

9

EXAMPLE

A tension test for a steel alloy results in the stress-strain diagram
shown in Fig. 3-18 Calculate the modulus of elasticity and the
Yield strength based on a 02% offet. Identity un the graph the

SOLUTION
‘Modulus of Elasticity, We must alculato the slope of the initia
straight-line portion of the graph. L magnified curve and

point A, which has coordinates of approximately (0.0016in,in

50 ki) Therefore

1.2(10°) ksi Ans

Yield Strength. For a 02% offset, we begin at a strain of 02
‘or 00020 in in. and graphically extend a (dashed) line par
OA until it intersects the ae curve at A". The yield strength is
approximately

ays = 68 ksi An
Ultimate Stress. This is defined by the peak of the ae graph
point Bin Fig 318.

= 108 ksi An

Fracture Stress. When the specimen is stained to is maximum of
ej = 023 in in, it fractures at point C. Thus,

7 = Wks An

95

EX

The strest-strain diagram for an aluminum alloy that is used for
making, aircraft p m in Fig. 3-19, Ifa specimen of this
material i stressed to 600 MPa, determine the permanent stain hat
remains in the specimen when the load is relessed. Als, find the

modulus of resilience both before and afer the load application

’ermanent Strain. When the specimen is subjected to the load
ie strain-hardens until point B e reach cam. Ti
strain a this p Jy 0.023 mm/mm. When the lo

released, the material behaves by following the siaight line BC
Which is parallel to line OA. Since both Ines have Ihe same slope, th
strain at point C can be determined analytically The slope of ine OA

isthe modulus of elasticity Le oars
50 MPa
Tam 500
From triangle CRD, we requir
ab e004 uh) Pa
au 501101) Pa = SDP
© Cia cD

CD = 1.008 mmm
This strain represents the amount of recovered elastic strain. The
permanent strain, soc. thus

oc = 0423 mm/mm — 0.008 mam/mm

00150 mm/mm A

Note: If gauge marks un the specimen were originally 50 mm apart
then after the load is released these marks will be SO mm

Modulus of Resilience. Applying Eq.3-8:w have

tr asa = ze, = (450 MPa)(0.06 rum /mm
135 Mm) a
1 1

a = Lo en = (600 MPa)(0.008 mm/mm
240 NU Ans

NOTE: By comparison, the effect of strain-handening he material has
caused am increase in the modas of reience: however, nate that th
moulu of toughness or the material has deceased since the area unde
the orginal curve, OABF larger than the area under curve CBF

9 En

EN

En An aluminum rod shown in Fig. 3-20 has a circular ero
Subjected to an axial load of 10 KN, I portion of the stress-strain
¿lagram is shown in Fig. 3-20, determine the app
the rod wh

section and is

the load is applied. Take Eu = 70 GPa,

LUTION
For the analysis we will neglect the localized deforma

at the point
‘of load application and where the od's eross-sectional area suddenly
change (These effects will be discusced in Sections 4.1 and 47)
Throughout the midsection gres and

Tn order to find the elongation of the rod, we must fist obtain the

diagram. The normal stress within each sepment is

Po 10010)
Cr. spa

100) N

5659 MPa

ry = 40 MPa. Using Hooke's law

sale 0004547 mam om

The material within segment BC is strained plastically: since
ane > oy = A0 MPa. From the graph, for dc = 56.59 MPa, €
(0045 mm/mm, The approximate elongation ofthe rod is the

à = Sel. = 010004847(600 mm) + 0.0450(400 mm
18

+31. A cnet cinder having diameter o 6 3-4. A tension ct was performed cm a apccimcn hain
Mrs ropoind a the ab a ad venis 2 SO mm Te data are kt in the table. aa
ely the modula o classy N Dam = 50 MP. 105 m/e
Toad hip) [Contraction (in q A tie tt e nl leg
is Toad UN — [lomas tm)
en int abi The cure ilinarbetwecn the origin and Fr e
heist pot, Pot he digram, and determine the modas 3 tins
in the table. The cave ls near betwesn the oral Prot 3-4
ad the first poi. Pot the diagram, and determi
aaa Go in. Determine th el
oa | ot by the member determine required cosaciona ate.

3.6 Poisson's Ratio

A bar made 0
ial force of P

init length and the change in the d

eel has the dimens
80 KN is applied to the bar, d

SOLUTION
p__ RON

nt 2 160
A NA

10) Pa

the table on the inside back cover far A-3ósteel Eu = 200 GPa
the strain in the = direction is
7. _ 169(10') Pa
e E EE - 01) mn
The axial elomgation of the bari then
Le = [80101 Sm) = 120 um An

Using Eq. 3-9, where »

vats = 0328010)
eb, —RSS(10 NM)
ab, = -256(10) (0.05 m)

128)m An

104 cum m

=

3.7 The Shear Stress-Strain Diagram

In See. 15 it was shown that when a small element of material is

‘must be developed on four faces of the element. These stresses 7,, must

element uniformly. Fig 3-23b. As mentioned in See. 2.2 the shear sra
y measures the angular distortion ofthe clement relative to the sides
The bel

ading. If measurements are made of the a

orque and the resulting angle of twist, then by the methods to be

example of such a diagram lor a ductile material is shown in Fig 3-2
Like the tension fst this material when subjected to shear wil exh
near. savior and it will have a defined proportional lt

4 ly the material will begin to lose is shear strengt
intl reaches a point where i fractures 7

Fur most engineering materials ke the one just described, the elastic

is called the shear modulus of elasticity ur the modulus of
is value represents the slope of the line on the 7-y diagram,

Typicil Values for common engineering materials are

ted on the inside Dack cover, Notice th

For examp E A236 steel. En = 29107) ks and
Ga = 11.0010) ksi.so that from Ba. 3-11.44 = 032

EX E

À specimen of titanium alloy tested in torsion and the shear stress- 7
strain diagram is shown in Fig, 3-2Sa, Determine the shear modulus

tional limit, and the ultimate shear stress Also

d thatthe top of a block of this

material, shown in Fg. 3-25p, could be displace horizontal if the

by a sheas force Y

material behaves elastically when acted up
What ic the magnitade of V necessary o causo this displacement?

Shear Modulus. This value represents the slope of the straight-line
portion OA of the ry diagram, The coordinates of point A ate
(0.008 rad, 52 ksi). Thus

ski

The equation of line OA is therefore 7 = Gy = 6800y, which is
Hooke's law for shea

Proportional Limit. By inspection the graph ceases tobe linear at
point A. Thus

Ultimate Stress. This value represents the maximum shear stress
point From the graph,

Maximum Elastic Displacement and Shear Force. Since the

maximum elastic shear strain s 0.08 rad, a very small angle, the top
Of the block in Fig. 3-256 will be displaced horizontally

à
d = 0016%n An
The corresponding average shear stress in the block i ry = $2 ks

Thus the shear force V needed to cause the displacement i

106 cua

EA

CON

Since o < uv = 40 MPa, the material behaves elastically. The
modulus of elasticity is therefore

| 361(10') Pa
E A = 710 GPA Ans
Re Contraction of Diameter. Fist we will determine Poisson's sat

Since gg = 00480 mm/mm, then by Ea. 3-9

The contraction of

*3.8 Failure of Materials Due to Creep
and Fatigue

Since they are given special treatment in design

Creep. When a material has to support a load for avery lng pe
of ime, it may continue to deform untit a sudden fracture occurs o

known as veep. Normally creep is considered mhen metals and cami

to high temperatures For some materials however such as polymers
compasite materials including wood or concrete temperature i
application. As a typical example, consider the fact that a rubbe

band will not return to it original shape after being released from a
stretshod position in which fl was held for a ve

In the general sense, therefore both stress and/or temperamre play a

For practical purposes, when ereep becomes important, a metab
asuall de y strain Tor a given pe

ime. À ty that fused in this re
called the cn represents the highest stress th
material can withstand peciicd time without excceding an

lowable gih will vary with temperature
and allowabl

creep st

in must all be specied. Fo
«4 for tel in bolts and piping

ubjected 10 a different axial stress. By measuring the length of tim

each specimen, a curve of stress versus time can be established
Normally these tests are run to a maximum of 1000 hours An exampl
the results for stainless steel at a temperature of 1200 and
prescribed creep strain of 1% is shown in Fig 3-27. AS noted, this
material has à yield strength of 40 ksi (276 MPa) a zoom temperatu
(02% offset) and the crecp strength at 1000 his found to b

107

108

In general, the creep strength will decrease or higher temperatures o
for higher applied sreses For anger periods
the curves must be made. To de

Fatigue. When a metal is subjected to repeated cles of ess
stain e causes its structure to break down, ultimately leading ı
acre, This behavior called Jargue, and iti usualy responsible

fora larg percentage of fallre in connecting ro and erankshats o

in al
The nature of hi fire ap

where the localized stress becomes much greater than the average stress

acting over the cross section. As this higher stress seed it ead tothe

mation of minute cracks. Ocuurrence of thee cracks causes a furthe

increase of sress at their tips or boundaries, which in turn causes a

further extension of the cracks int the material as the stress continues

Eventually the cross-sectional area of the member is
reduced tothe point where the

jure occur The material, even though known 10 be

tie, behaves as if it Were brite
In order to specify a safe strength for a metallic material unde

ling. i is necessary 10 determine a limit below which no
evidence of failure can be detected after applying a load fora specified

peated

Y this purpose a series of specimens are

ota a specified stress and eyele to failure. The results are
lotied asa graph representing the stress S (or 27) om he vertical
and the number of eyeles-to failure Non the horizontal axic This rapt
is called an S-N disgram or stess-cyele diagram, and most often the
values of N are plotted on logarithmic scale since they are generally
quit a

Examples of S-N diagrams for two common engineering metals are
shown in Fig. 3-28. The endurance limit is usually identified as the
stress for which the S-V graph becomes horizontal or asymptotic. AS

27 ksi (156 MPa

ted, it has a well-defined value of (Sa) steel

is normally specified as the areas having a mit of 509 milion cycle,
Sau = 19ksi (131 MPa), Once a particular value is ob

infinite. and therefore the number of cycles to failure i

important Point:

+ Poisson’ ratio, visa ratio of he L
its longitudinal strain. Generally these
Strains are of opposite signs that iif one isan elongation, the

The shear stress-strain diagram is a plot of the shear stress
versus the shear strain, Ifthe mi

isotropic, and is also linear elastic, the slope of the str

shear modulos. G

+ There is mathematical relationship between G, E, and »

stress and/or temperature play an important role, Members

during a specified time without exceeding a specified
+ Fasigue en the stress or strain ie cycled. This
henomenon causes britl fracture of the material Members
are designed to resist fatigue by ensu

member does not exceed its endurance or fatigue limit This
value is determined from an S-N diagram as the maximum

ss the material can resist when subjected to a specified
‘number alles of loading

TE est D:
ae ee Hel ‘4
pe et tas] Bey

Her Hose IB ji:
Bass À u | SEE 3
freee HAE iget

— |

® ma

|_| CHAPTER REVIEW

116

| [REVIEW PROBLEMS

Axial Load

CHAPTER OBJECTIVES
In Chapter 1, we developed the method for finding the normal stress

sialy loaded members. ln this chopter we will discuss how

mine the deformation ofthese members, and we willalso develop

method for finding the suppor reactions when these reactions cannot
ects of thermal stress, stress concentrations, inlast
sons, and residual stress wil also be discuss

4.1 Saint-Venant's Principle

the force distribution within a
he type of material
material behaves in linear elastic manner, then Hooke’ law applies
And there isa proportional relationship between tres and strain

om which the body is made. In par

ne

ê tl
e
E if Bi

Es

dL

(lus yy eo

4.2 Elastic Deformation of an Axially
Loaded Member

Changes, From Saint: Venant% principle, thete effects occur within

de and cross-sectional area A(x) i isolated from the bar at he arbitrar
position x. The rec-body diagram of this element is shown in Fig 4-26

«riu load bar This

Pix) di

Far the entro length ofthe bar, we mist islote this exprestion t
ind 8. This yells

L = original length of ba

Pa) = internal axial force atthe section, located a distance from
one end
E = modulus of elasticity forthe material

Constant Load and Cross-Sectional Area. In many cas

he bar will have a constant cross-sectional area At and the material will
be homogeneous so E is constan, Furthermore, fa constant external
force is applied at each end, Fig 4-3, then the intemal force P
Iiroughout the len bar isalko constant Asa result, Eq. 4-1 can
7

an fk a
the baris subjected to several different ax gts length,
e f the bar 10 the next, the above equation can be applied
la each segment of the bar where these quantities remain constant. Th

displacement uf one end ofthe bar with respect to the other is then found
om the algebraic addition

124

a z

Sign Convention. In order to apply Eq. +
tion forthe internal axial force and th

For example he bar shown in Fig 4-5a, The N
rces "P are determined by the method of sections f nt
Fig 4-5b, They are Pan = +SKN, Pac = ~SKN. F AN. This

variation in asia load i shown om the axial
the bar, Fi. +

throughout the
is determined t

2e

IF the other data are substituted and a positive answer is caculated, it
means that end A will move away D (the bar elongates),
Whereas a negative result would indicate that end A moves toward
end D (the bat dt
indicate this relative displacement (Sp); however, if the displacement
ple, if D i

na), The double subseript notation is u

neo Mes 125

MOT

«Saint Venant prine

ple states that both the localized deformation and stress which occur within the

regions of load application or at the support tend to “even out” at a distance sulfiienty removed from
these regions

+ The displacement of one end of an axially loaded member relative to the other end is determined by
relating the applied femal load to the sess using r = P/ A and relating the displacement to the strain using
= db/dx. Finally ese two equations are combined using Hooke lan, = Ee, which yields Eq. 4-1

+ Since Hooke law has been used in the development of the displacement equation, ts important that no

The relative displacement between any two points À and B on an axially loaded member can be determined
by applying E 4-1 (or Eq. 4-2), Application requires the

Use the method of sections to determine the internal axial force P within the membe

‘If his force varies along the member’ length due to an external distributed loading, a section should be
ade atthe arbitrary location x (ram one end ofthe member and the forse represented as funcio

between any two external forces must be determined.
+ For any segment, an intemal tensile force is postive and an internal

pressive force is negative. For
te internal loading can be shown graphically by constructing the normal

+ When the members cross-sectional area sure al
its position x, Le. Als).

ng its length, the area must be expressed as a function of

+ Mabe cross-sectional area, the modulus of elasticity. the internal loading suddenly changes, then Ea. 4-2
should be applied to each segment for which these qui

# Wen substituting the data into Eqs. 4-1 through 4-3, be sure to account forthe proper sign for the

eral force P- Tensile loadings ae postive and compressive loadings are negative. Also, use a consistent
set of units For any segment, if

results positive numerical quantity. it indicate elongation: ts
negative, it indicates a contruction.

126 cm Axını Low

À bar shown in Fig. 4-60 is made fi
areas of Aus = Lim

ment of end A and the displacement

SOLUTION

Internal Foree. Dueto the application ofthe external the
miemal axial forces in regions AB, BC. and CD will al be different
These forces are obtained by applying the method of sections and the
equatio Fig. 4-60. This

“TMI Hp aw

(TO

sb
ip. #
AcE ROT

Here B moves away from C. since the segment elongates

127

The assembly shown in Fig. 4-7a consists of an aluminum tube AB
having. cross sectional area of 40 mm. A steel rod having a diamete
of 10mm is attached to a rigid collar and passes through the tube. Ifa
tensile load of 8 KN is applied tothe rod, determine the displacement
of the end € of the rok. Take Ey = 200 GPa, Ey = 70 GPa

SOLUTION
De

body diagram of he tube

Internal Force. The fr
ed toa tension of $0 KN and

in Fig 4-7b, shows that the rod i sub
à compression ol 80 KN

tube is subject

Displacement. We will first determine the displacement of end C

with respect to end 2 Working in unite of newtone and meters. we have

vo = Pi. os NOS) Fr
Fem = AE TOMS my TOIT) NT

The pobiive ips indicates thal end C mot
end B, since the bar elongates

The displacement of end B with respect 10 the ed end A is
m [-80(10') N04 m)
ETS]

001143 m = (001143 m—=

ce sign indices that the tube shortens, and so B

Here the nega
moves tothe right relative to À
Since both diplacements are 10 the right, the displacement of C
relative to the fixed end A i therefore

128 cum Axını Low

EX Rig beam AB rests on the two short post shown in Fe 4-8 AC à
SS made of steel and has a diameter of 20 mm. and BD is made
4 Aluminum and has a diameter of 40 mn. Determine the displacement
: ‘ot point Fon AB ita vertical load of 3 kN is applied over this poin
Take Ey = 200GPa, Ey = 10 GPa
ë ol soumon

Internal Force. The compressive forces acting at the top of euch
post are determined from the equilbsium of member AB, Fig. 48h
nal forces in each post, Fig. 4-8

These forces are equal to the
E Displacement. The displacement of the top uf exch put is

Post Ac:

+ + 0.286 mm
Puma 7 a FOO m 70010) NF le
0.102 mm |

the beam is shown in Fig. 4-84, By pr
triangle, the displacement of point Fis thereto

portion ol the blue shad.

EXAMPLE

129

A member is made from a material that has a specific weight y and
modulus of elasticity Ef tis in the fy
dimensions shown in Fig 4-94, dete

nine how ar
avisy when itis suspended inthe vertical p

me having th
end is displaced

N

Internal Force. The inte

member
since itis dependent on the weight W(s) of a segment of the membe
below any section, Fiz 4

4-9), Hence to calculate the displacement
we must Use Eq 4-1, At the section

ocated a distance y from its free
end. the radius x of the cone as a function of y is determined by
proportion; ke

a
"The volume of a cone having a base of

Since W becomes
SF, » 0: PO) =
placement. The area of the cruss section is alko a function of
Fig 4-9, We hav

Applying Eu 4-1 between the limits y
“Py

er.

[vw

yl
6E

Gand y = L yield

NOTE:

is result, notice how the units o th

Asa partial check
Ms, when canceled, give the displacement in units of length a8
expected,

le

"we If |
E

130 Cuarren 4 Axis LOA

| [FUNDAMENTAL PROBLEMS

136 Arten a Axis LOM
4.3 Principle of Superposition
de principle of superposition Snes that the restant tress
displacement atthe pant an be determined y algebraically suming
The following twa conditions mus be satisfied if the principle
super pain io be apple
1. The loading mus be linanly related tothe stress or displacement
thar i tobe determined. For example the equatons = P/A an
PLSAE involve a inar relatonshipbstwecn Panda 0
2. The loading mast not significantly change he orginal geometry or
direction and location of the applied foes and their moment
S wi change Fr example consider the ender rod show
1 which eset tothe lad PI Fe 4-10) Pis replaced
br o ot componente P= Py + Ps I cases thera to defect
sarge amount as shown, the moment tthe Toad aba is suppor
Tr, sot equal the sum ofthe maments of emp
PA à Poly + Pads beca dy rd
principle wil be used thoughout this text whenee
Slaw apis and ale. the uds ha are cod
Change in orton and ection of the lading wil be insignias! and
~ Tg

4.4 Statically indeterminate Axially
Loaded Member
Conse the bar shown in Fig 4-114 which is (ke supported at both ,
From the fee buy dlgram Fig 4-11). quilt require 5
0 Fat Fa-P=0 i
This type of problem is called statically indeterminate, since Un ‘ r
a compaibli or Einemarle condom. Inthe cate à

ae pr

which depends

<5 = PL/AE can
tha the internal force in segment AC is +F a, and in
nal force is —F Fig 4-11e, the above equation can

be used. Realia

Assuming that AE is constant, then Fy = Fa(Len/L ac}, so that sin

shawn correctly onthe free-body diagram,

The principle of superposition is sometimes used to simpli
stress and displacement problems having complicated loadings
This i done by subdividing the loading into Components. then
Superposition requires thatthe loading belincary elated to the

are not suffiien to determine al the reactions on a meme

occur atthe supports or ether po

|
ii The support reactions for sal indeterminate problems are

+ Draw a free-body diagram of the member in order to identify all

y the forces that acto

7 + Witte

ons of equilibrium for the membe

ci amd cs ic toc mascar | ® Consider drawing a displacement diagram inorder to investigate
rin spring te api od, the | the Way the member will elongate or contract When subjected to
the external loads

Express the compatibility conditions in terms of the

displacements caused by the lading

® Use a fond displacement relation, such as 8 = PL/AE, to relate
the unknown displacements o the reactions.

Solve the equilibrium and ompatibilty equations for the
reaction. I any of the results has a negative numerical value, t
indicates that this force act i the opposite sense of direction to
that indicated om the f

#4. Srareau oetemanare Ana LOADED Mes 139

The steel rod shown in Fig. 4-12a hs a diameter of 10 mm. is fixed
tothe wallat A and before its loaded. there isa gap of 2 mm between
the wal at Band the rod. Determine the reactions at A and. ith

P = 20KN asshown, Neplect the

SOLUTION

Equilibrium. As shown on the frec-body diagram, Fig, 4-12, we

will assume that force P is large enough to cause the rod' end B to

contact the wall at 8 The problem is statically indeterminate since chemin

there are wo unknowns and only one equation of equilibrium. y ¥
BSF, =0 Fa = En + 00) N = 0 w 0

Compatibility. The force P causes point B to move to with

further displacement. Therefore the compatibility condition for the

rodis

This displacement can be expressed in terms of the unknown

reactions using the load displacement relationship. Eq. 4-2, applied

to segments AC and CB, Fig 4-12. Working in unis of newtons and
= Fa.
(04m
DE = sm) 200107) Nav]

ROOKIE) Ne]

Fa04m) - Ex(08m

Sem o

Solving Egs 1 and 2 yields
AS An

Since the answer for Fis positive, indeed end B contacts the wall at

Bas originally assumed.

NOTE: If Fa were a negative quan
statically determinate,so that Fy ~ Oand F = 20 KN

140 cua

EXAMPLE

The aluminum post shown in Fig. 4-134 is reinforced with a brass
nd of P = 9kip,

core Ifthis assembly supports an axial compressv
applied to the rigid cap. determine the average m
the aluminum and the brass. Take Es = 10(10")ksi and
Ey = 15(10") ki.

SOLUTION
Equilibrium. The free-body diagram of the post is shown in
Fi 4-13), Here the resultant axial force atthe base is represented by

the unknown components carried by the aluminum, Fa, and bras,
Fa. The problem is statically indeterminate Why
"Vertical force equilibrium requires

13F,=0% Pip + Fu+ Fy a

Compatibility. The
aluminum and bras to diplace the same amount, Therefore

cap atthe top of the post catacs both the

Using the load- displacement relationships

Solving Eqs 1 and 2

Skip Fue = Skip

‘Since the results are positive, indeed the stress will be
The average normal stress in the aluminum and brass is therefore

Skip 637 ksi An

NOTE: Using these results the stress distributions are shown in
Fed

The three A-36 steel bars shown in Fig. 4-14a are pin connected to à
rigid member. I the applied load on the member is 15 KN, determin
the force developed in each bar, Bars AB and EF cach have a cross
sectional arca of SO mm", and bar CD has a cros-<ectional arca

Equilibrium. The free body diagram of the rigid member is shown

in Fig 4-140, This problem i statically indeterminate since there are
three unknowns and only two available equilibrium equations

TIE, <6; Fat For Fr—1SkN=0 w

Me = 0, —E,(04m) + ISEN(O2m) + F(04m) =O (2)

Compatibility. The applied load will cause the horizontal line ACE

shown in Fig. 4-14e to move to the Inclined line A'C'E”. The
its À, Can E can be related by similar tran

‘equation that relates these displacements i

ofp

Fel if Fat i Fel

2. Summ’jE 30mm}

mn)

Fe=0364+0 o
Solving Es 1-3 simultaneousIy yields
Fy = 952kN An
Fe = 3464N Ar
Fy = 202kN An

142 cua axis

The bot shown in Fig. 4-15 is made of 2014-76 aluminum
tightened sit compresses a cylindrical tube made of Am 1004-761
‘magnesium alloy The tube hasan

that both the inner radius ofthe tube and
y and x

1d amd have a negligible thickness Initially the nut is hand tightened

ter radis of in. ,andit is assumed
radius of the bat

i of the tube are considered to be

the bolt has 20 threads per inch, determine the stress in the bolt

SOLUTION

Equilibrium. The free-hods diagram of section ofthe bolt and he

nsidered in 0 late the foree in the bat

tube, Fig 4-15), e
Fi to tha in the tube, F. Equilibrium requires

1EF, = 0 E =E=0 w

Compatibility. When the nu nthe bolt, the tube
will shorten d,, and the bolt wil elongote ba, Fig. 4-1Se, Since the
ful undergoes one-half tuen, it advances a distance of (E in)
(0025 in. along the boll Thus the compatibilay of these displacements

is tightened

requires

Taking the mo

doli of elasiiy from the table on the inside back
caver, and applying Eq 4-2, yields

Fin
15m) = 02510) [65107]
m Fin)
Sr Tr
O78S05F, = 28 — LAUF, o

Solving Eqs.1 and 2 simultaneously, we get
Fam Fm 1122Kip

The stresses in the b

tan tube are therefore

aa u; a
B i
i
A aS - (03071
These rete ae es an reported ed ste für ach materia

4.5 The Force Method of Analysis
for Axially Loaded Members

To show how it is app
he support at B as "redundant

bar, then the bar will be
ns in Fig. 4-16b, By using the principle uf

position, we

back the unknown redundant load Ej, as «hown in Fig 4-166

lead P causes B to be displaced downward by an amount Sp, the

action Es place end B of the bar apıwurd by an amount
such that mo displacement occurs at 3 when the two loadings ar
This equation represents the compatibility equation for displacements at
point B, for which we have assumed that displacements are postiv

Appiving the load-displacemen relation

bp = PLa

AE and ón = Fal

we have applied

1 of compatibility 1
um condition to obtain the

=:

144 cum Axını Low

Procedure for Analysi:

The f

ce method of analysis requires the fo

ne one of the supports as redundant and write the equation of compatible. To do this the known
displacement at the redundant support, which is usually zero, i equated 10 the displacement at the
support caused only by the extemal loads acting on them

ber plus (vectoriall) the displacement at this
Support caused ony by the redundant reaction acting on the membe

© Express the extemal load and redundant displacements in terms of the loadings by using a load.
displacement relationship, sich as 8 = PL/AE

Equlibrium

then be solved for the magnitude ofthe redund

Draw frce-body diagram and write the
calculated result for the redundant, Solve these equations for any other reaction

propriate equations of equilibrium for the member using the

EA

The A-36 steel rod shown in Fig 4-17a has a diameter of 10 mm.Icis
fixed to the wall at A, and before iis loaded there sa gap

the wall at” and the rod of 02 mm, Determine the reactions
amd, Neglect the size of the collar at C Take Eu = 200 GPa

Compatibility. Here we will consider the support at 8 as

redundant. Using the principle of superposition, Fig 4-17, we have

) 002m = dp — By 0)

The deflections 3 and Sy are determined

Eq,

PL {20(10") N](0. mi ere
AE (0005 m} 1200107) N/ar] rh
nan m in

AE SR a TAO Fa

Substituting into Eg. 1,we get

Oe sg 00002 m = 0.509310) m ~ 783944(10-") Fy
Sr e Fy = 408/10) N = 405 kN Ans
> Equilibrium. From the free-body diagram, Fig. 4-17¢

ea ASF = 0; E, + 20KN —40SKN=0 Fa =160KN An

4.6 Thermal Stress

A change in temperature can cause a body to change its dimensions
Generally. ifthe temperature increases, the body will expand, whereas if
he temperature decreases it will contract, Ordinaril this expansion
contraction is linearls related to the temperature increase or decreas
hat oseurs. If this 8 the case, and the material is homogeneous a

isotropic, thas been found from e
th L can be cale

a a property ol the material referred 10 as th linear coefficient
of thermal expansion, The units measure strain per degre
temperature, They are 1/°F (Fahrenheit) in the FPS.
and 1/°C (Celsius) or 1/K (Kelvin) in the SI system Ty

AT = the algebraic change in temperature of the membe
L = the original length of the membe or R

The change ia length determinate membe
calelated using Eq, 4-4,since the member is free

in tate member, these thermal displacements will be constrained Long exiemons of ducs an
by the supports, thereby producing dhermal stressed that must à

ved in design. Determining these thermal tresses is possi
he methods cutlined in

“The A-36 steel bar shown in Fig, 4-186 is constrained to just fit

between two fixed supports when 7, = 60°F. Ifthe tempe
aised to Ta = 120

determine the average normal thermal stress

‘developed inthe bar

SOLUTION

Equilibrium. ‘The free-body diagram of the bar is shown in
Fig 4-18h Since there sno external load, the force at À is equal but
opposite tothe force at B that i

EF, =0 Fa

F

ate since this force cannot be
determined from equilibrium

Compatibility. Since 5,» = 0. the thermal displacement dr at A
that occur Fig. 4-18c, is counteracted by the force Fthat is required
push the bar 6, back to its orginal npatility

Applying the thermal and load: displacement

FL
0= art - À
AE

Thus from the data om the inside back cove

F = aSTAE

2871 i

20°F = 60°F) 0. in,)[29(10") kip/io"]

Since F also represents the intemal axial force within the har, the
average normal compressive stress is thu

F _ 287 kip
ER hi sk An

NOTE: From the magnitude of Fit should be apparent that changes

in temperature can cause large reaction forces in statically
indeterminate members

EXAMPLE

‘The rigid beam shown in Fig. 4-194 is fixed to the top ofthe three |

posts steel and 2014-Té aluminum. he posts each have TTT TULL IT
mm when no load is applied to the beam, and th

temperature is T; = FC. Determine the force supported by each

post ifthe bari subjected to a uniform distributed load of 150 kN/m

and the temperature is raised 10T = 80°C

Equilibrium. “The fre-body diagram of the beam isshown in Fig 4-19)

foment equilibrium about the ean’ center require the force in the
ted posts to be equal. Summing forces on the free-body diagram
we nave

ter, =0 LF + Fy ion = 0 w

Compatibility. Due to load, geometry. and material sommer. the
top of each post is displaced by an equal amount. Hence

i

is equal tots displacement
caused By the temperature increase, plus its displacement caused by 0
the internal axial compressive force Fig. 4-19 Thus, for the steel and

aluminum post, we have

y Ba = (Eur + (Bade ‘
Applying Eu 2 gives
Using Eqs 4-2 and 4-4 and the material properties on the inside back

Fai0250m) Es
DOS) Om FRIO) NAT
Fat.
Dan BAC) NA
Fa = L216F,, — 1659(10 o

[2(10-*y/rcKs0re

(23410 *y/PC(a0"C — 20°C)(0250

ical data has been expressed in terms
Ege 1 and

recs Celsius. Solin

Fg =16ARN Fy = 1234N An

‘The negative value for Fy indie

that shown in Fig. 4190. In other words, the steel posts are in tens

Nm
i

SOLUTION Set ata

154 cum Axını Low

A 2014-T6 aluminum tube having a erus-sectional area of 600 mm? is
das a deeve for an A-36 steel bolt having a cross-sectional area of
00 mar, Fig 4-20. When the temperature is T; = 15"C, the aut
the assembly in a smug p
le. the temperature increases to Ta = 80°C
im the bolt and sleeve

ion such thatthe aval force in the

SOLUTION
Equilibrium. The fx
assembly shown in Fig. 4-208, The forces F and F, are produce

Since the sleeve has ab thermal expansion than the
bolt and therefore the

ve wil expand more when the temperature
is increased. ts require hat

body diagram of a top segment of the

BF, <0, FF Mm
‘ Compatibility. The temperature increase cautekthe sleeve and bolt
Y Expand (6,) and (3), Fig. 4-20c- However the redundant forces

and F elongate the bolt and shorten the sleeve. Consequently the
end of the assembly reaches a final position, which isnot the same as

(400 rm") (10) w/a?) 20010") Na?)

NOTE: Since linear elastic material behavior was assum.
kd be

4.7 Stress Concentrations

mn

itn requires the magnitade of th
at P.in

P must act through
ncering practice, he actual
have to be determined. Instead, only the m

=
fal
E

o MS

160 cum ax L

encrally reported in handbooks related to
Hand 4-25. Note that K

Specific values of K a
is independent of the

frmined that a sharp comer, Fig 4-23a, produces à stress

cross section, However, this can be reduced to. s4 LS by
strodueing a filet, Fig. 4.230. À fu
i. In all

The stres-concentration fact
iermined on the basis of static loading, with the assumption that
le stress in the material does not exceed the proportional limit Hf the
material is very brite, the proportional limit may be at the fr
stress, and so for this material, failure will begin at the point of stress
oncentration, Essentially a crack begins 10 form at this point, and a

result in a crack, Instead
ve strength due to yielding and stain
hhardening In the next section we wil

he material i due or brit. Here the material local
the crack remain in a brite state, and o the crack continues ti
re fracture. As a result

om 161

in x
T +
1 4 i
TEEPE
RTE
payer) F.2s

Important Points

| # Stress concentrations oscur at sections where the cross-sectional
area suddenly changes The more severe the change, the larger
the stress concentration.
|
+ For desi

area, This

it is only necessary u
0 the smalles ross ecto
iress concentration factor, K, that

is done w = been
de

‘Normally the stress concentration in a ductile specimen that is
subjected to à satis loading will mot have to be considered in
design; however, i the material is bridle, or subjected to fatigue
Toadings then stress concentrations become important

162

Inelastic Axial Deformation

behave ©
150 that the loading causes the
manently deform. Su

as annealed low

dicated
increased to P,
hen again P

ly defined. Instead, a he ins
er. Fi
ate) such thatthe

cd to as being elastie perfect plastic
‘whichis subjected tothe axial load P. 1 the la
be developed in th

al exhibits perfectly plastic
m ill coatinue indefinitely with
dn reality. howeve

a au

| |
|

164

*4.9 Residual Stress

are removed The reason for this has todo with the elastic recovery of the

at occurs during unlcading To show this, consider a prismatic

removed, the material wil respond elastically and follow the line CD

‘deformed so that Ihe permanent set or strain inthe member is €
Te the member is sa

recovery CD. Since these forces wil constrain the member from
«covery, they will induce residual stresses in the member. To sol

problem of this kind, the complete cycle of loading and then unloadin
nf the member can be considered as the uperpositon of a postive load
(oading) on a negative load (unloading). The loading, Oto C. results in
plastic stress distribution whereas the unloading, along CD. results only

The bar in Fig 4-240 is made of steel that is assumed to be elastic
perfectly plastic. with ay = 250 MPa, Determine (a) the maximum

the applied load P that can be applied without causing th
steel to yield and (b) the maximum

of Pthat the bar can Support

Skeich the stress distribution tthe critical section for each case

Part (a). When the mat ally, we must use à
vom Fig. 4-24 that is unique

for the bars geometry Here

E sas
i” Tam ETS
w 10 mm =
From the figure K = 1.75, The maximum load, without caus

yielding, occur when To y. The average normal stress is
an = PLA Using Eg. 4-60 ba

er eke)

À
254410") Pa = 1.75

Py = VEN An

This load has been calculated using the smallest cross section, The
resulting stress distributo Fig. 4-29, For equilibeium
the “volume” contained within this distribution must equal 9.14 KN.

art (b). The maximum load sustained by the bar wil cause al th

at the smallest cross section to yield. Therefore, as P is

increased to the plastic load Py. it gradually changes the stress
distribution from the elastic stat shown in Fig. 4-29) 10 the plas

State shown in Fig. 4-296 We require

250 1h) Pa =

Py = 160KN An

Here P, equals the “volume” contained within the stress distribution
which in this case is P, = y À

166

The rod shown in Fig 4-30 has a radius of $ mm and is made of
an clastic perfectly plastic material for which oy = 420 MPa, E

à GPa, Fig. 4-H Ia force of P= 6 KN is applied tothe ro
then removed, determine the residual stress in the rod

SOLUTION
The free-body diagram ofthe r

shown in Fig 4-305 Application
of the load P will cause one of three possiblities. samely

ah
Segments AC and CB remain elastic, AC 8 plastic while CB is clastic
or both AC and CB are plastic

An elas analysis similar to that discussed in Sec. 4.4, will produce
Fa 5 EN and Fy = ISKN atthe supports However. this results i

oe ENT u 573 MPa (compresion) > oy = 420 MPa

bec
Far this ease, Ihe maximum possible

‚mes plastic while CB remains elastic:

ee developed in AC is
(Faly = oy = 320010) N/m (0005 m)] = 33 0kN

and from the equilibrium o

Fy = GOK — 330KN

MPa (compress

ZN

EFT 544 MPa (tension) < 40 MPa (OK)

ain the residual stress itis also
nding Sins

Residual Stress. In ander to o
necessary to know the train in each segment due 10 te
CH responds elastically

Fnten 27.0 N) (1300 m rat

ey =e. OUR _ open
tga B= MM m
z 10m

Here the yield stain is

er:

ZUNE

Therefore, when P is applied, the _stree-traio behavior for the
material in segment CB moves from Oto A”. Fig 430k

strain behavior forthe material in segment AC moves from O to Bt
the load P is applied inthe reverse direction, in other words, the load

and the stress

{= 45EN and Fy = ISKN must be applied to each segment.

AS: caleuted coc focos now pi
$73 MPa (tension) and acy = 191 MPa (compression) and as

o, 420 MPa + 573 MPa = 153 MPa Ans

ae = 344 MPa — 191 MPa = 153 MPa An

This residual stress is the same for both segments, which is to be
expected. Also note that the stress-strain behavior for segment AC
moves from 8 to D'in Fig 43e, while the stress-strain behavior fo

the material in segment CB moves from a’ ta C° when the load

168 cum Axını Low

Two steel wires are used to lift the

weight of Skip. Fig. 41e Wire AB has

«4 length of 2000 fr and

Wire AC has an unstretched length of
03 ft

sectional area of 0.05 in, and the te]

can be considered elastic perfectly
Plastic a6 shown by

Fis 431, determine
ire and its elongation

he free in

4h TT Ole i red th

depends on the corresponding strain

is There are three possibilities, namely the strains in both wires are
elasic, wire AB i plastically strained while wire AC is elastically
au ‘trained, or both wires ae plastically strained. We will asume that AC

Investigation of the free-body
Fie 4 le indicates that the pr

am of the suspended weigh,
‚m is statically indeterminate, The

ter, 0 Tao + Tac 3Mp=0 0)

Since AB becomes plastically strained then it must support its
maximum load

Tas = 07 Aan = Uk (OS in) = 2.50kip Am
Therefore, from Eu 1
Tac = 0.500 kip Ans

Note that wire AC remains elasticas assumed since the stress in the
wireisa ae = 0.500 kip/(105 in? = 10 ksi < SO ksi The corresponding
elastic strain à determined by proportion. Fig. 4-310: Le

0017
Inka ~ SO ksi
la ac = 0000340
mes The clongation of ACs thus
Bc = (0.000340)(20.8 1) = 0.00681 f Ans

And from Fig 4-3, the elongation of AB is then

|_| CHAPTER REVIEW

Mechanics of Materials 8th Edition R.C. Hibbeler

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Mechanics of Materials 8th Edition R.C. Hibbeler

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