126233678-ls-srinath

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Contents

Preface vList ofSvmbols XUISI' Ul1US {Systeme International d'Unit'es) xvr_"pical Physical COI1S(Cll1ts (As all Aid to Solving Problems) .XVI1. Analysis of Stress 1

'1.1 Introduction 11.2 Body Force, Surface Force and Stress Vector 213 The State of Stress at a Point 4'1.4 'N'ormal aDd Shear Stress Component's 41.5 Rectangular Stress (.;oolponent.s 41.6 Stress Co,mponeots on an Arbitrary Plane 61.7 Digression on [deal Fluid 1.11.8 Eqa]ity of Cross Shears 1 ,1.9 A 1-1Qre(Jeneral Theorenl 131.10 Principal Stresses 141 11 Siress Invariants /61.12 Pri.ncipal Planes are ()rthogonal 171.13 (;ubic .Equation has -rhree Real Roots J 71.'14 Particular Cases 191.15 Recapitulation 191.16 The State ofSlress Referred to Principal Axes 241.17 Mohr's Circles for the Three-Dimensjonal State of Stress 251.18 Mohr's Stress Plane 261.'19 Planes of Maximum Shear 281.20 (lctahedral Stresses 291.21 The State of Pure Shear 311.22 Decomposition into Hydrostatic and .Pure Shear States 3.11.23 Caucby's Stress Quadric 341.24 LaIne' s Ellipsoid 361.25 The Plane State of Stress 38.1.26 'Differential Equations of ,Equilibrium 401.27 Equil.ibrium Equations for Plane Stress State 421.28 Boundary Conditions 451.29 Equations ofEq-uilihr.i:um in Cylindrical Coordinate-s 45

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'vl Contents

1.30 Axisymmetric Case and Plane Stress Case 48Problems 49Append.i.x 1 Mohr's Circles 54Appendix 2Appendix 3

Tbe State of Pure Shear 56Stress Quadric of Cauchy 60

2. A.nalysis of Strain2.1 Introduction 6322 Deformations 642.3 Deformation in the Neigbbourhood of a Point 652.4 Change in Length of a Linear Element 672.5 Cbange in Length of a Linear Element-Linear Components 692.6 Rectangular Strain CODJpOnents 702,7 'fhe State of Strain at a Point 702,8

63

79~,Interpretation of yxy, Yyz, Yxz as Shear Strain ComponentsC:hange in Direction of a Linear Element 73

71

710 Cubical Dilatation 742,11 Change in the Angle between 'Two Line Elements 772.12 Principal Axes of Strain and Prine.ipai Strains 78') 13 Plant~Siale 1)1" Strain 8 ~2.14 The Pri.ncipal .t\xes of Strain R.elllain Orthogonal after Strain 84

2.16 Coolpatibility C;onditions 862.17 Strain Deviator and its Invariants 90

Prob/ell/.I 91Appendix on Compatibility Conditions 94

3. Stress-Strain Relations for Linearly Elastic Solids 973, I Introduction 973,2 Generalised Staiemeni:of flQoke's Law 973,3 Stress-Strain Relations for Isotropic Materials 983,4 Modulus of ,Rigidity .993,5 Bulk M'odulus 1013,6 Young's M.odulus and Poisson's Rati.o 1023.7 Relations between. the Elastic Coostant's 1023,8 DisplacCIllcnt Equations ofEguilibri.lllll .I 04

Problems 10Z

4. '("heories of .-ailure or Yield Criteria and Introduction toldeaUv Plastic Solid 1094,I '1ntroductjoo , 094,2 Theories of Failure 1104,3 Si.guifi.cance of the l~he()'ri.esof Failure III4,4 Use of Factor of Safety in Design 1214.5 A Note on the usc of Factor of Safety 1244.6 Mohr's Theory of Failure 1294.7 Ideally Plastic Solid 1324.8 Stress Space and Strain Space J 34


Contents vii

49 General Nature of the Yield Locus /354.10 Yield Surfaces of Tresca and Von Mises 1364.11 Stress-Strain Relations (Plastic Flow) 1374.12 Prandtl-Reuss Equations J394.13 Saint Venant-Von Mises Equations 140

Problems 140

5. Energy Methods5.1 Introduction 14352 Hooke's Law and the Principle of Superposition 1435.3 Corresponding Force and Displacement or Work-Absorbing

Component of Displacement 1455.4 Work Done by Forces and Elastic Strain Energy Stored 1465.5 Reciprocal Relation 1475.6 Maxwell-Betti-Rayleigh Reciprocal Theorem 1485.7 Generalised Forces and Displacements 1495.8 Begg's Deformeter 1525.9 First Theorem of Castiglia no /535.10 Expressions for Strain Energy 1555.11 Fictitious Load Method /615.12 Superposition of Elastic Energies 1635.13 Statically Indeterminate Structures 1645:14 TheQrem of Virtual Work 16(i5.15 Kirchhoff ts Theorem 1695.16 Second Theorem of Castigliano or Menabrea's Theorem 1705.17 Generalisation of Castiglia no's Theorem or Engesser's

Theorem 1735.18 Maxwell-Mohr Integrals 176

Problems 181

6. Bending of Beams 1896.1 Introduction 18962 Straight Beams and Asymmetrical Bending 1906.3 Regarding Euler-Bernoulli Hypothesis 1986.4 Shear Centre or Centre of Flexure 20 J6.5 Shear Stresses in Thin-Walled Open Sections: Shear Centre 2026.6 Shear Centres for a Few Other Sections 2086.7 Bending of Curved Beams (Winkler-Bach Formula) 2096.8 Deflections of Thick Curved Bars 216

Problems 223

143

7. Torsion7.1 Introduction 23072 Torsion of General Prislnatic Bars-Sol id Sections 2327.3 .Alternative Approach 2367.4 Torsion of Circular and Elliptical Bars 2407.5 Torsion of Equilateral Triangular Bar 2437.6 Torsion of Rectangular Bars 245

230 \

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'iill'. . Contents7.7 Membrane Analogy 2487.8 Torsion of Thin-Walled Tubes 2497.9 Torsion of Thin-Walled Multiple-Cell Closed Sections 2517.10 Torsion of Bars with Thin Rectangular Sections 2557.11 Torsion of Rolled Sections 2567.12 MUltiply Connected Sections 2597.13 Centre of Twist and Flexural Centre 264

Problems 265

8. Axisymmetric Problems 2698.1 Introduction 2698.2 Thick-Walled Cylinder Subjected to Internal and External

Pressures-Lame's Problem 27 J8.3 Stresses in Composite Tubes-Shrink Fits 2808.4 Sphere with Purely Radial Displacements 2878.5 Stresses Due to Gravitation 2928.6 Rotating Disks of Uniform Thickness 2948.7 Disks of Variable Thickness 2988.8 Rotating Sbafts and Cylinders 3008.9 Summary of Results for use in Problems 303

Problems 305

9. Thermal Stresses 3109.1 Introduction 3 I09.2 Thermoelastic Stress-Strain Relations 3119.3 Equations of Equilibrium 3119.4 Strain-Displacement Relations 3129.5 Some General Results 3129.6 Thin Circular Disk: Temperature Symmetrical about Centre 3149.7 Long Circular Cylinder 3/69.8 The Problem of a Sphere 3209.9 Normal Stresses in Straight Beams due to Thermal Loading 3239.10 Stresses in Curved Beams due to Thermal Loading 325

Problems 328

10. Elastic Stability10.1 Euler'sBuckling Load 331

331

I. Beam Columns 33510.2 Beam Column 33510.3 Beam Column Equations 33510.4 Beam Column with a Concentrated Load 33610.5 Beam Column with Several Concentrated Loads 33910.6 Continuous Lateral Load 34010.7 Beam-Column with End Couple 342IL General Treatment of Column Stability Problems (As an

Eigenvalue Problem) 34410.8 General Differential Equation and Specific Examples 344


Contents "ix

109 Buckling Problem as a Characteristic Value (Eigenvalue)Problem 350

10.10 The Orthogonality Relations 352m Energy Methods for Buckling Problems 355

10.11 Theorem of Stationary Potential Energy 35510.12 Comparison with the Principle of Conservation of Energy 35710.13 Energy and Stability Considerations 35810.14 Application to Buckling Problems 35910.15 The Rayleigh-Ritz Method 36010.16 Timoshenko's Concept of Solving Buckling Problems 36410.17 Columns with Variable Cross-Sections 36610.18 Use of Trigonometric Series 368

Problems 371

11. Introduction to Composite Materials11.1 Introduction 37411.2 Stress-Strain Relal'jons 375

. 11.3 Basic Cases of Elastic Symmetry 37711.4 Laminates 38111.5 Ply Stress and Ply Strain 40411.6 Failure Criteria of Composite Materials 40611.7 Micromechanics of Composites 41111.8 Pressure Vessels 42111.9 Transverse Stresses 422

Problems 424

12. Introduction to Stress Concentration and Fracture Mechanics 428L

12.112.212.312.412.512.6D.

12.712.812.912.1012.1112.1212.1312.1412.1512.16

374

Stress Concentration 428Introduction 428Members under Tension 429Members under Torsion' 439Members under Bending 443Notch Sensitivity 445Contact Stresses 446Fracture Mechanics 457Brittle Fracture 457.Stress Intensity Factor 458Fracture Toughness 460Fracture Conditions 462Fracture Modes 464Plane Stress and Plane Strain 468Plastic Collapse at a Notch 47 JExperimental Determination ofKIc 475Strain-Energy Release Rate 476,Meaning of Energy Criterion 479


-:;iI.( . '. Contents._---- ------------

12.17 Design Consideration 48212.18 Elasto-Plastic Fracture Mechanics (EPFM) 48212.l9 Plane Body 48512.20 Green's Thorem 4861221 The l-Jntegral 48612.22 Path Independence of the l-lntegral 4871223 l-Integral as a Fracture Criterion 48912.24 ASTM-standard Test for llc 49012.25 Relationships of Kc, o.. and J 491

Problems 49 J

Appendix 494

5() J

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Preface

The present edition of the book is a completely revised version of the earlier two'editions. The second edition provided an opportunity to correct severaltypographical errors and wrong answers to some problems. Also, in addition,based on many suggestions received, a chapter on composite materials was alsoadded and this addition was well received. Since this is a second-level courseaddressed to senior level students, many suggestions were being received toadd several specialized topics. While it was difficult to accommodate allsuggestions in a book of this type, still, a few topics due to their importanceneeded to be included and a new edition became necessary. As in the earliereditions, the first five chapters deal with the genera] analysis of mechanics ofdeformable solids. The contents of these chapters provide a firm foundation tothe mechanics of deformable solids which will enable the student to analyse andsolve a variety of strength-related design problems encountered in practice. Thesecond reason is to bring into focus the assumptions made in obtaining severalbasic equations. Instances are many where equations presented in handbooksare used to solve practical problems without examining whether the conditionsunder which those equations were obtained are satisfied or not.

The treatment starts with Analysis of stress, Analysis of strain, and Stress-Strain relations for isotropic solids. These chapters are quite exhaustive andinclude materials not usually found in standard books. Chapter 4 dealing withTheories of Failure or Yield Criteria is a general departure from older texts. Thistreatment is brought earlier because, in applying any design equation in strengthrelated problems, an understanding of the possible factors for failure, dependingon the material properties, is highly desirable. Mohr's theory of failure has beenconsiderably enlarged because of its practical application. Chapter 5 deals withenergy methods, which is one of the important topics and hence, is discussed ingreat detail. The discussions in this chapter are important because of theirapplicability to a wide variety of problems. The coverage is exhaustive anddiscusses the theorems of Virtual Work, Castigliano, Kirchhoff, Menabria,Engesser, and Maxwell-Mohr integrals. Several worked examples illustrate theapplications of these theorems.


xii Preface

Bending of beams, Centre of flexure, Curved Beams, etc., are covered in Chapter 6.This chapter also discusses the validity of Euler-Bernoulli hypothesis in thederivations of beam equations. Torsion is covered in great detail in Chapter 7.Torsion of circular, elLiptical, equilateral triangular bars, thin-walled multiple cellsections, etc., are discussed. Another notable inclusion in this chapter is thetorsion of bars with multiply connected sections which, in spite of its importance,is not found in standard texts. Analysis of axisymmetric problems like compositetubes under internal and external pressures, rotating disks, shafts and cylinderscan be foundin Chapter 8.

Stresses and deformations caused in bodies due to thermal gradients needspecial attention because of their frequent occurrences. Usually, these problemsare treated in books on Thermoelasticity. Tbe analysis of thermal stress problemsare not any more complicated than the traditional problems discussed in bookson Advanced Mechanics of Solids. Chapter 9 in this book covers thermal stressproblems.

Elastic instability problems are covered in Chapter] O. In addition to topics onBeam Columns, this chapter exposes the student to the instability problem as aneigenvalue problem. This is an important concept that a student has to appreciate.Energy methods as those of Rayleigh-Ritz, Timoshenko, use of trigonometricseries, etc., to solve buckling problems find their place in this chapter.

Introduction to the mechanics of composites is found in Chapter 11. Modern-day engineering practices and manufacturing industries make use of a variety ofcomposites. This chapter provides a good foundation to this topic. The subjectmaterial is a natural extension from isotropic solids to anisotropic soLids.Orthotropic materials, off-axis loading, angle-ply and cross-ply laminates, failurecriteria for composites, effects of Poisson's ratio, etc., are covered with adequatenumber of worked examples.

Stress concentration and fracture are important considerations in engineeringdesign. Using the theory-of-elasticity approach, problems i.n these aspects arediscussed in books solely devoted to these. However, a good introduction tothese important topics can be provided in a book of the present type. Chapter 12provides a fairly good coverage with a sufficient number of worked examples.Several practical problems can be solved with confidence based. on the treatmentprovided.

While SI units are used in most of numerical examples and problems, a few canbe found with kgf, meter and second units. This is done deliberately to make thestudent conversant with the use of both sets of units since in daily life, kgf isused for force and weight measurements. In those problems where kgf units areused, their equivalents in SI units are also given.

The web supplements can be accessed at http://www.mhhe.com/srinathiams3eand it contains the following material:

For Instructors Solution Manual PowerPoint Lecture Slides


'~iii Preface

For Students MCQ's (interactive) Model Question Papers

I am thankful to all the reviewers who took out time to review this book andgave me their suggestions. Their names are given below.

K S R K Murthy Department of Mechanical Engineering,Indian Institute of Technology, Guwahati, Assam

P K Sarkar Department of Mechanical Engineering,Indian School of Mines, Dhanbad, Bihar

S R Pandey Department of Applied Mechanics,,VIT Jamshedpur, Jharkhand

Dr Amit Kumar Department of Mechanical Engineering,National Institute of Technology, Patna, Bihar

M K Singha Department of Applied Mechanics,Indian Institute of Technology, New Delhi

P Venkitanarayanan Department of Applied Mechanics,Indian Institute of Technology,Kanpur. Uttar Pradesh

G Rajesh Kumar Department of Mechanical Engineering,Rizvi College of Engineering,Mumbai, Maharashtra

C A Akhadkar Department of Mechanical Engineering,SSVPS's B SDeora College of Engineering,Dhule, Maharashtra

D Prasanna Venkatesh Department of Mechanical Engineering,SRM Institute of Science and Technology.Chennai, Tamil Nadu

EV M Sargunar Sree Sastha Institute ofEngineering andTechnology,Chennai, Tamil Nadu

In addition to this, I am also thankful to the staff at McGraw-Hill Education India,especially Ms Vibha Mabajan, Ms Shu.k:tiMukherjee, Ms Surabhi Shukla, Ms SobiniMukherjee and Mr P L Pandita, for their cooperation during the different stages of thisproject.

Lastly, I wish to thank lOY family members for their patience, support and love givento me during the preparation of tbis manuscript. .

Feedback and suggestions are always welcome at the website of the book.

L S SRlNATB


(J

F/I

TII

CF x.y,

l'X)I, yz, zx

nx' NY' n,CF" CF2, CF31,,12, I')(Joel

'rOCICFr, CFo, CF.

t.8, If>"eyo fyz, "ea.

ux' UY' lizEXX' EY>' s;

List of Symbels(In the order they appear

In ,the text)

normal stressforceforce vector on a plane with normal n

components of force vector in x, y, z directionsarea of sectionnormal to the sectionshear stressnormal stress on x-plane, y-plane, z-planeshear stress on x-plane in j-direction, shear stresson y-plane in a-direction, shear stress on z-plane inx-directiondirection cosines of n in x, y, z directionsprincipal stresses at a pointfirst, second, third invariants of stressnormal stress on octahedral planeshear stress on octahedral planenormal stresses in radial, circumferential, axial (polar)direction .spherical coordinatesshear stresses in polar coordinatesdisplacements in x, y, z directionslinear strains in x-direction, j-directiou, z-direction (withnon-linear terms)linear strains (with linear terms only)shear strain components (with non-linear terms)shear strain components (with linear terms only)rigid body rotations about x, y, z axescubical dilatationprincipal strains at a pointfirst, second, third invariants of strain

;u- tyy. e..EX). s.; s;YX)' Yyz, YzxW". WY' Wzll. ""Ea + l').y+ Er:" ~, ~J" J2, J3


Ust of Symbols

Ey. E6, e,A,J.lG=J.LJ.l.EKp1)

eryUU*

strains in radial, circumferential. axial directionsLame's constantsrigidity modulusengineering Poisson's ratiomodulus of elasticitybulk modulus; stress intensity factor \

\

(Jul

ere.aublj~. MY'.~8IX' ~Y'I:Ip/xy ~'"TIf'aQpv

pressurePoisson's ratioyield point stresselastic energydistortion energy; complementary energyultimate stress in uniaxial tensionultimate stress in uniaxial compressioninfluence coefficient; material constantcompliance componentmoments about x, y, z axesLinear deflection; generalized deflectionmoments of inertia about x, y, z axespolar moment of inertiaproducts of inertia about xy and yz coordinatestorque; temperaturewarping functioncoefficient of thermal expansionlateral loadaxial loadelastic potentialPoisson's ratio in r-direcrion due to stress in j-dire.ctionwidththicknesstheoretical stress concentration factornormal forcestream functionf ller radiusradiinotch sensitivityfracture toughness in mode /offset yield stressangular velocityfracture resistancefracture stressboundaryf-integral

Vijb, Wt

x,Nf/>PD,dqK", ;SyillRG;rrJ


51 Units,(Systeme International

,d/Unl~/es)(a) Base Units

Q,umtfty ':: .'..: PO Unit (Syf1iboQFI~~~~,,-~~.~=~---=----~~-----4

length meter(m)kilogram (kg)second (s)newton (N)pascal (Pa)

.mass.time'force ..pressure

force is a derived unit: kgm/S2pressure is force per unit area: N/m2: kg/ms?kilo-watt is work done per second: kNm/s

(b) Multiplesgiga(G)mega(M)kilo(k)milli(m)micro(Jt)nano (n)

10000000001000000

] 0000.001

0.0000010.000 000 00 1

(c) Conversion Factors. ' , -, i :.: . . .,.... . to :-:-To CotnWt ..'_ . M,!ltip/y by ..'

kgf newton 9.8066kgf/cm2 Pa 9.8066 x lif,kgf/cm2 kPa ' 98:066newton kg( 0.10197Pa N/m2 1kPa kgf/cm2 0.010197HP leW 0.746IW kNmls 0.746kW kNmls 1


Typical PhysicalConstants

(As an Aid to Solving Problems)

Material Ultimate Strength Yield Strength Elastic Poisson's CoefJ.I (NlPa) (MPa) Modulus Ratio Therl11 J (GPa) iE:lpalls. Tells. Comp Shear Tens or Shear Tells Shear per C

CO/liP x 10 ~Aluminium alloy 414 414 221 300 170 73 2& 0.334 23.2Cast iron, gray 210 825 - - - SXJ 41 0.211 10.4Carbon steel ~ 690 552 415 250 200 83 0.292 11.7Stainless steel 568 568 - 276 - 2IJ7 SXJ 0.291 17.0For more accurate values refer to hand-books on material properties

,

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Analysis of Stress. .,

,'INTRODUCTION'1oeJI>A.IIn this book we shall deal with the mechanics of deformable solids. The starting. point for discussion can be either the analysis of stress or the analysis of strain. Inbooks on the theory of elasticity, one usually starts with the analysis of strain,which deals with the geometry of deformation without considering the forces thatcause the deformation. However, one is more familiar with forces, though themeasurement of force is usually done through the measurement of deformationscaused by the force. Books on the strength of materials, begin' with the analysis ofstress. The concept of stress has already been introduced in the elementry strengthof materials. When a bar of uniform cross-section, say a circular rod of diameterd, is subjected to a tensile force F along the axis of the bar, the-average stressinduced across any transverse section perpendicular to the axis of the bar andaway from the region of loading is given by

(J' _ F __ 4;,,:.F7_ Area - 1!d2

It is assumed that the reader is familiar with the elementary flexural stressand torsional stress concepts. In general, a structural member or a machineelement will not possess uniform geometry of shape or size, and the loadsacting on it will also be complex. For example, an automobile crankshaft or apiston inside an engine cylinder or an aircraft wing are subject to loadings thatare both complex 'as well as dynamic in nature. In such cases, one will have tointroduce the concept of the state of stress at a point and its analysis, which willbe the subject of discussion in this chapter. However, we shall not deal withforces that vary with time. .

It will be assumed that the matter of the body that is being considered iscontinuously distributed over its volume, so that if we consider a small volumeelement of the matter surrounding a point and shrink this volume, in the limit weshall not come across a void. In reality, however, all materials are composed of--many discrete particles, which are often microscopic, and when an arbitrarilyselected volume element is shrunk, in the limit one may end up in a void. But in

our analysis, we assume that the matter is continuously distributed. Such a body,


2 Advanced Mechanics of Solids

is called a continuous medium and the mechanics of such a body or bodies iscalled continuum mechanics.

aonv FORCE, SURFACE FOR

Analysis of Stress 3

further, the moment of the forces acting on area L1A about any point within thearea vanishes in the limit. The limiting vector is written as

I Ilim L1T = dT = T (1.1)

AA .....O 6A dASimilarly, at point P', the action of part D on C as L1A' tends to zero, can berepresented by a vector

1 1lim f:.T' = dT' =.;,

A_4'--+() L1A' dA' (1.2)

I IVectors T and T' are called the stress vectors and they represent forces per

unit area acting respectively at P and P' on planes with outward drawn normalsJ J Inand n .

JWe further assume that stress vector T representing the action of Con D

Jat P is equal in magnitude and opposite in direction to stress vector T'representing the action of D on C at corresponding point P'. This assump-tion is similar to Newton's third law, which is applicable to particles. \Vethus. have

J 1T=-T' (1.3)

2t.T

If the body in Fig. 1.1 is cut by a different2

plane 2-2 with outward drawn normals nand2n' passing through the same point P, thenthe stress vector representing the action

2of C2 on D2 will be represented by T(Fig. (1.3, i.e.

22 L1T'T= lim -

LlA-40 L1A

IIn general, stress vector T acting at point

1P on a plane with outward drawn normal n

Fig. 1.3 Body cut by another plane 2will be different from stress vector T acting

2at the same point P, but on a plane with outward drawn normal 11. Hence thestress at a point depends not only on the location of the point (identified by coordi-nates x, y. z) but also on tbe plane passing through the point (identified by direc-tion cosines 11... lip liz of the outward drawn normal).


.4 .Advanced Mechanics of Solids

Since an infinite number of planes can be drawn through a point, we get aninfinite number of stress vectors acting at a given point, each stress vectorcharacterised by the corresponding plane on which it is acting. The totality ofall stress vectors acting on every possible plane passing through the point isdefined to be the state of stress at the point. It is the knowledge of this state ofstress that is of importance to a designer in determining the critical planes andthe respective critical stresses. It will be shown in Sec. 1.6 that if the stressvectors acting on three mutually perpendicular planes passing through the pointare known, we can determine the stress vector acting on any other arbitraryplane at that point.

. ; .NORMAL AND SHEAR STRESS C0MP0NENTS,. . .. ". .n

Let T be the resultant stress vector at point P acting on a plane whose outwarddrawn normal is n (Fig. 1.4). This can be resolved into two components, one along

the normal n and the other perpendicular to n. Thecomponent paraJlel to n is called the normal stress

IIand is generally denoted by (J'. The component per-pendicular to n is known as the tangential stress orshear stress component and is denoted by ~. Wehave, therefore, the relation:

n(J'

nT

n

(1.4)

where Ii- is the magnitude of the resultant stress.II

Stress vector T can also be resolved into threecomponents parallel to the x, y, z axes. If these

Fig. 1.4 Resultant stressvector, normaland shear stresscomponents

n n ncomponents are denoted by T x , T y, T., we have

1,,2 II II "

2 2 2T =Tx +Ty +Tz (1.5)

1.5, .RECTANGULAR STRESS COMPONENTSLet the body B. shown in Fig. 1.1, be cut by a plane parallel to the yz plane. Thenormal to this plane is parallel to the x axis and bence, the plane is called the x

xplane. The resultant stress vector at P acting on this will be T. This vector canbe resolved into three components parallel to the x, y, z axes. The componentparallel to the x axis, being normal to the plane, will be denoted by (J',. (instead of byx0-). The components parallel to the y and z axes are shear stress components andare denoted by 'fXY and Tn respectively (Fig. 1.5).


In the above designation, the firstsubscript x indicates the plane onwhich the stresses are acting and thesecond subscript (y or z) indicates the

l4::::::;;;:::::-_,,_ n direction of the component. For ex-ax ample, '(~yis the stress component on

the x plane in y direction. Similarly,'fxz is the stress component on thex plane in z direction. To maintainconsistency, one should have denotedthe normal stress component as 'fcr' Thiswould be the stress component on thex plane in the x direction. However, todistinguish between a normal stress and

a shear stress, the normal stress is denoted by CT and the shear stress by r,At any point P, one can draw three mutually perpendicular planes, the x plane,

the y plane and the z plane. Following the notation mentioned above, the normaland shear stress components on these planes are

y


-rxyxT

o x

zFig. 1.S Stress components Oil x plane

CT.r' f,:., 'f,"! on x plane~" ~'x' -r:v: on y planeCT" 7":.

"6 Advanced Mechanics of Solids

in the negative x, y and z axes. The positive stress components on these faceswill, therefore, be directed along the negative axes. For example, the bottom facehas its outward drawn normal along the negative y axis. Hence, the positive stresscomponents on this face, i.e., 0:,,, 'f)'x and 't)'z are directed respectively along thenegative y, x and z axes.

="'" STRESS COMPONENTS ON AN ARBITRARY PLANEIt was stated in Section 1.3 that a knowledge of stress components acting on threemutually perpendicular planes passing through a point will enable one to deter-mine the stress components acting on any plane passing through that point. Letthe three mutually perpendicular planes be the x, y and z planes and let thearbitrary plane be identified by its outward drawn normal n whose direction

cosines are nx' II)' and n.Consider a small tetrahedronat P with three of its facesnormal to the coordinateaxes, and the inclined facehaving its normal parallel ton. Let h be the perpendicu-lar distance from P to theinclined face. If the tetrahe-dron is isolated from thebody and a free-body dia-gram is drawn, then it willbe in equilibrium under theaction of the surface forcesand the body forces. Thefree-body diagram is shownin Fig. 1.7,

y

8 n

n

T"

Ax

cz

Fig. 1.7 Tetrahedron at point P

Since the size of the tetrahedron considered is very small and in the limit as weare going to make h tend to zero, we shall speak in terms of the average stresses

nover the faces. Let T be the resultant stress vector on face ABC. This can be

II II II

resolved into components T, T', T z, parallel to the three axes x, y and z, On thethree faces, the rectangular stress components are

Analysis of Stress ' 7

tetrahedron, the sum of the forces in x, y and z directions must individuallyvanish. Thus, for equilibrium in x direction

n 1TxA - CT.t An" - 1'),., An)' - 1'zxAn. + "3 Ahrx = 0

Cancelling A,n ITx = O:t nx + 1')'" II)' + Tzx nz - 3 hrx

.Similarly, for equilibrium in y and z directions,

(1.6)

and

n ITy = 'l'.,llx + O:yn)' + 1'.,1'= - 3 hr"

n IT== 1'dlx + 'l',,J!)' + 0:=,,= - 3 hr.

(1.7)

(1.8)

In the limit as h tends to zero, the oblique plane ABC will pass throughpoint P, and the average stress components acting on the faces will tend to theirrespective values at point P acting on their corresponding planes. Consequently,one gets from equations (1.6)-( I.8)

n

r, = nx O:x+ n)' ~vx+ II, 'l'zxn

Ty = II;: t:ty + II)' 0:)'+ 11z 1'z), (1.9)n

T, = nx t:tt + n)' 1'yz+ n, 0:.Equation (1.9) is known as Cauchy's stress formula. This equation shows that

the nine rectangular stress components at P will enable one to determine the stresscomponents on any arbitrary plane passing through point P. It will be shown inSec. 1.8 that among these nine rectangular stress components only six arc indepen-dent. This is because 't'.." = 1'yX' 1'ry = 1"7 and 1'u = 1".a' This is known as the equaLityof cross shears. In anticipation of this result, one can write Eq. (1.9) as

n11 = n.t 'fix + lIy 1'iy + II, 1';, = ~nj 'l'ij

J

where i and j can stand for x or y or z, and a, = 1'u' 0:), = 1'yy and 0:, = ru'n .If T is the resultant stress vector on plane ABC, we have

(1.10)

In2 ,,2 n2 n2T = T." + T y + T zIf a; and 'fn are the normal and shear stress components, we have

(LIla)

(1.11b)n

Since the normal stress is equal to the projection of T along the normal, it isn n It

also equal to the sum of the projections of its components Tx, 1';, and T, alongn, Hence,

II 12 nO:n = nxTx + II)'Ty + IIzT. (1.12a)


'8 .Advanced Mechanics of SolidsIt n u

Substituting for r., Ty and T:: from Eq. (1.9)CI"= n; o,+ n; o, + n; CIt+ 2n,lly rxy + 2nl'z 'fyz+ 2nllx T:u (l.12b)

Equation (1.J I) can then be used to obtain the value of r,.1= ,_ .. $$ pe __ '_ .. __ ."" __ _...__ ....." ", , ~ _Example 1.1 A rectangular steel bar having a cross-section 2 em x 3 em issubjected to a tensile force of 6000 N (612.2 kg[). If the, axes are chosen asshown in Fig. 1.8, determine the normal and shear stresses on a plane whosenormal has the following direction cosines:

(i) " x =n)'=Jz.1Iz= 0(ii)

(iii) 1f1=II=n=-.Y Y x fj

y FnT n

nT

/1 /,,I,,,II."--r-, ,, .

i..---j;-' /,

x (a) (b) (c)/z F

Fig. 1.8 Example 1.1,

Solution Area of section = 2 x 3 = 6 cnr', The average stress on this plane is6000/6 -' J 000 Nfcm2. This is the normal stress Clv' The other stress componentsare zero.

(i) Using Eqs (1.9), (1.11b) and (1.12a)

nTx = 0, Tn _ 1000

Y - fi 'nT; =0

a = 1000 = 500 Nfcm2n 2

2 1 " 12 2 2 4Tn = T - a; = 250,000 N fcrn

Tn = 500 N/cln2 (5] kgffcn12)


"9Analysis of Stress: "n

(ii) T x = 0, n 1000Ty=.fi'

U" = 500 N'Icm2, and T" = 500 N/cm2 (51 kgf/cm2)

nT, =0

n(iii) r, = 0, n 1000Ty=.fj'

nTz =0

Un = 10~0 N/cm2

Tn = 817 N/cm2 (83.4 kgflcmz)

i.III1'.....4~~_)...... ad IiI'R!'1I"';M!I!_ I a. 5ll:, $ 'S! $$" lIIzS'li&O'oIl". 9"'1 '7r~~

Example 1.2 At a point P in a body, O";r. = 10,000 Nlcrn2 (1020 kgflcm2), 0" =-5,000 Nlem2 (-510 kg/len!2), O"z = -5,000 Nlcm2, 'r:r.y = fyz = 1'n = 10,000 Nle"m2.Determine the normal and shearing stresses on aplane that is equally inclinedto all the three axes.

Solution A plane that is equally inclined to all the three axes will have

I. 2 2 2 1nx= ny;: nz;: .fj smce nx + ny + nz =

From Eq. (1.12)

. 0"" = i [10000 - 5000 - 5000 + 20000 + 20000 + 20000]= 20000 N/cm2

From Eqs (1.6}-{1.8)

Tx = ~ (10000 + 10000 + 10000) = 10000 .fj N/cm2II I t:Ty = .fj (10000 - 5000 + 10000) = -5000 ....,3N/cm2

11 J r:;T, = .fj (10000 - 10000 - 5000) = -5000 v3 N/cm2

..2n

T = 3 [(108) + (25 X 106) + (25 X 106)] N2/cm4

= 450 x 106 N2/cm4

r 2 = 450 X 106 - 400 X 106 = 50 X 106 N2/cm4n

or r, =7000 N/cm2 (approximately)~_iiEiIl~ n ,__ a '$ 4 ua '" tlG.W!!!. ". ==.i1"....,. di'S 'lL" ....~Example 1.3 Figure 1.9 shows a cantilever beam in the form of a trapezium ofuniform thickness loaded by a force P at the end. If it is assumed that the bendingstress on any vertical section of the beam is distributed according to the elementary


'10 Advanced Mechanics of Solidsflex lire formula, show that the normal stress (J on a section perpendicular tothe top edge of the beam at point A is CT~ , where (JI is the flexural stress

cos f)~c , as ShOlVII in Fig. 1.9(b).

A 0'1 ~E-ax ~

:::~~

p

y

r-~Ar--~=jF A 0'1C_jC

0'1 B

(b) , .B(a)

-~1

(c)

Fig. 1.9 Example 1.3

(d)

Solution At point A, let axes x and y be chosen along and perpendicular to theedge. On the x plane, i.e. the plane perpendicular to edge EF, the resultant stressis along tbe normal (i.e., x axis). There is no shear stress on this plane since the "top edge is a free surface (see Sec. 1.9). But on plane AB at point A there canexist a shear stress. These are shown in Fig. 1.9(c) and (d). The normal to planeAD makes an angle 9 with the x axis. Let the normal and shearing stresses on thisplane be (JI and 1'1'

We have

The direction cosines of the normal to plane AD are

IIx = cos f), n =0z

The components of the stress vector acting on plane AD areII

T; = CTI= IIx a, + lIy 'tyX + liz 1'ty = (J cos9n

1'." = /I" 1'X),+ 11)' CT.+ IIz 't"zy = 0nT, = II" 't"xz + Ily 1'y" + flz (J z = 0

n n "Therefore, the normal stress on plane AD = CTn = II x Tx + 11)' T)' + liz T z =(J cos? 9.


' Analysis of Stress '~"1'

Cfl MeCf= -

cos2 (J I cos2 (J

Further, the resultant stress on plane AB is

1,2 n n., IIT =T;+T;+T;=~COS2(J

T2 = 0'2 COgle _ cr 2"

= 0'2 cos2e - (12 cos4e. Hence

I .T= - O'sln 26

2_______ ..... .,_. _._._ ...... ... _'_.'_-.":. .... "#0 .,. ... _ mr '_',n __or

:' ;i.~~). 9IGRESSIO,N '.ON, IDEAL FLum , .By definition, an ideal fluid cannot sustain any shearing forces and the normalforce on any surface is compressive in nature. This can be represented by

nT=-pn, p>On n

The rectangular components of T are obtained by taking the projections of Talong the x, y and z axes. If nx, lIyand liz are the direction cosinesof n, then

nTy = -Plly'

nTz = -P"t ( 1.13)

Since all shear stress components are zero, one has from Eqs. (1.9),

(1.14)Comparing Eqs (1.13) and (1.14)

o,= O'y = a,= -pSince plane n was chosen arbitrarily, one concludes that the resultant stress

vector on any plane is normal and is equal to -p. This is t.he type of stress that asmall sphere would experience when immersed in a liquid. Hence, the state ofstress at a point where the resultant stress vector on any plane is normal to theplane and has the same magnitude is known as a bydrostatic or an isotropic stateof stress. The word isotropy means 'independent of orientation' or 'same in alldirections'. This aspect will be discussed again in Sec. 1.14.

u;;,o;,;;..., EQUALITY OF CROSS SHEARSWe shall now show that of the nine rectangular stress components o.; Txy' Txz' O'y,Tyx, T)''Z1 0'" 1':.< and Tzy' only six are independent This is because 'l'xy =-ry,t, 1',,,, = Tzyand 1'z.< = 1'xz. These are known as cross-shears. Consider an infinitesimal rectan-gular parallelpiped surrounding point P. Let the dimensions of the sides be I1x, l1yand I1z (Fig. 1.10).


.l~Advanced Mechanics of Solids

z

,

Fig. 1.10 Stress components on a rectangular element

Since the element considered is small, we shall speak in terms of average stressesover the faces. The stress vectors acting on the faces are shown in the figure. Onthe left x plane, the stress vectors are 'fu. t:xyand t:.... On the right face, the stressesare 'rn' + A'fXl" 'fxy + A'fxy and Tn + A1",a. These changes are because the right face isat a distance Ax from the left face. To the first order of approximation we have.

. OrXl' IJrxy iJcAr....= t7X Ax, AT~y = Ox Ax, Arn = 17; Ax

Similarly, the stress vectors on the top face are 'rY.l'+ A'fY.l" "yx + A'r)!> and'ryz+ A-r".".where

IJrAT = yz Ayyz oy

On the rear and front faces, the components of stress vectors are respectively

.'f,.. 'rn' 'rzy'f" + A 'fzz , 'fzx + A'rzx 'fzy + A 'fzy

where"'- IJrV'ZX A ... = zy A_A" r:r =~...O!.. AZ, Ll .zy ;:L u.e;OZ V~

For equilibrium, the moments of the forces about the x, y and z axes mustvanish individually. Taking moments about the z axis, one gets

'f.u Ay Az ~ - ('rn + A'f.o) Ay Az ~ +

('rxy+ A 'f;Cy)Ay Az Ax - 'fY.l'Ax A.z Ax +2

('fY.l'+ A'fY.l') Ax Az ~'( - , + A'ryx) Ax Az Ay +


Analysis of Stress :':l:t.1.x ..1.y ( .1.x

Try Ax .1.y T -T",. ax .1.y T - Tzy + .1.1'.),) Ax .1.y T +

(Tzr + .1.1'".y) .1.x.1.y .y;::: 0Substituting for .1... .0-, .1.~l)' etc., and dividing by Llx: .1.y &

0. .1.u 0. xv 0.>" .1.lJ~.t,,-X_2_ + r + .. Ax + / _'.I' _ox 2 xy ox oy 2

Or),X A. _ t7cZy Llx: + or;;x oy = 0yx - rl... ,-,y ~ 2 ., ?-: (/Z oz_

In the limit as .1.x, .1.y and & tend to zero, ~eabove equation gives 1'xy ;:::Tyx'Similarly, taking moments about the other two axes, we get 1'yz;::: Tzy and Tz.

'14 Advanced Mechanics of Solids

Using the result t'xy = 1'yx. 1',)'%= 'Tty and t'zx = 1'",

Similarly, ..n' .T ' , , (' ') n = a, 11x 11 x + (Iy lIy 11y + (Iz liz 11z + 1'x" "x 11" + II" 11 x +

t'y; ("" 11'. + liz lI'y) + 1'z:c (II, n'x + 11"11';)Comparing the above two expressions, we observe

\

n nT' n' = T' 11 (1.15)

Note: An important fact is that cross shears are equal. This can be used to provethat a shear cannot cross a free boundary. For example, consider a beam ofrectangular cross-section as shown in Fig. 1.12 .

..y

1/ La'-1-.-.-

L-- ____yll x(a) (b)

Fig. 1.12 (a) Element with free surface; (b) Crossshears being zero

If the top surface is a free boundary; then at point A, the vertical shearstres component 1'xy = 0 because if 'exy were not zero, it would call for acomplementary shear 1'yx on the top surface. But as the top surface is aounloaded or a free surface, t'yx is zero and hence, t'x)' is also zero (referExample 1.3).

PRINCIPAL STRESSES . .We have seen tbat the normal and shear stress components can be determined onany plane with normal n, using Caucby's formula given by Eqs (1.9). From thestrength or failure considerations of materials, answers to the following questionsare important:

(i) Are there any planes passing through the given point on which the result-ant stresses are wholly normal (in other words, the resultant stress vectoris along the normal)?

(ii) What is the plane on which the normal stress is a maximum and what is itsmagnitude?

(iii) What is the plane on which the tangential or shear stress is a maximumand what it is its magnitude?

Answers to these questions are very important in the analysis of stress, and thenext few sections will deal with these. Let us assume that there is a plane n with


,Anal ' f S ,"'4~1:"ySlS 0 tress. 'iI'.,

direction cosines nx' ny and nz on which the stress is wholly normal. Let (J be themagnitude of this stress vector, Then we have

"T =(InThe components of this along the x, y and z axes are

(1.16)

nr, = unz (1.11)

Also, from Cauchy's formula, i.e. Eqs (1.9),It

T, = o, 11x + 't"xy ny + 1'xz liz"T y = 't'xy n" + (Jy lIy + 't')'% ".It

T z = fyz nx + 't')'% lIy + (J. ".Subtracting Eq. (1.17) from the above set of equations we get

(o; -:-Cf) "x + 't"xy lIy + f,u liz = 0't"xy "x + (Cfy - Cf) ",v + 't"J'ZII, = 0Tn n,+ TYE "+ (o, - (J) liz = 0

We can view the above set of equations as three simultaneous equations involv-ing the unknowns "x' lIyand liz, These direction cosines define the plane on whichthe resultant stress is wholly normal. Equation (1.18)' is a set of homogeneousequations: ,The trivial solution is "x = lIy = liz = O.For the existence of a non-trivialsolution, the determinant of the coefficients of 11", lIyand liz must be equal to zero, i.e,

(l.18)

(ax-a) 1'xy Txz1"xy (O'y -0') 1")'% =0 (1.19)Txz 1')'% (O'z -0')

Expanding the above determinant, one gets a cubic equation in (J as

(J" - (Jx + (Jy + (Jz)(J2 + (o; O'y + O'y (Jz +o; (Jx - T;y - 1";,. - .; x ) Cf-

(120)

The three roots of the cubic equation can be designated as Cfl, Cf2 and Cf3 Itwill be shown subsequently that all these three roots are real. We shall later givea method (Example 4) to solve the above cubic equation. Substituting anyone ofthese three solutions in Eqs (1.18), we can solve for the corresponding nr lIy andliz. In order to avoid the trivial solution, the condition.

(1.21)

is used along with any two equations from the set of Eqs (1.18). Hence, with each_,(J there ",;11 be an associated plane. These planes on each of which the stressvector is wholly normal are called the principal planes, and the corresponding


:1~'Advanced Mechanics of Solids

stresses, the principal stresses. Since the resultant stress is along the normal, thetangential stress component on a principal plane is zero, and consequently, theprincipal plane is also known as the shearless plane. The normal to a principalplane is called the principal stress axis.

. .. . .~.~The coefficients of 0'2, o and the last term in the cubic Eq. (1.20) can be written asfollows:

(1.22)

(1.23)cr

(1.24)

Equation (1.20) can then be written as

d' _/10'2 + 120'-/3 = 0The quantities II' 12 and 13 are known as the first, second and third invari-

ants of stress respectively. An invariant is one whose value does not changewhen the frame of reference is changed. In other words if x', y', z', isanother frame of reference at the same point and with respect to this frameof reference, the rectangular stress competence are 0"... ,0",.., O"z" 1'x ' y" '(y'"and 'z'x" then the values of II' 12 and 13, calculated as in Eqs (1.22) - (1.24),will show that

.i.e,

, , ,O'x + O'y + O'z = O'x + O'y + O'z

II = Iiand similarly,

The reason for this can be explained as follows. The principal stresses at a pointdepend only on the state of stress at that point and not on the frame ofreference describing the rectangular stress components. Hence, if xyz andx'y'z' are two orthogonal frames of reference at the point .then the followingcubic equations

and

(13 -/10'2 + 120'-13 =00'3 -1'10'2 + 1'20'-1'3 =0

Copyrlqhted matenal

ANSHULHighlight

Analysis of Stress :':;1'7', .must give the same solutions for

plots j{ 0') for different values of(1 as shown in Fig. 1.13, the curvemust cut the (1 axis at least once asshown by the dotted curve and for

(1 this value of (1,/(1) will be equal tozero. Therefore, there is at least onereal root.

Let (13 be this root and n the as-sociated plane. Since the state ofstress at the point can . be

characterised by the six rectangular components referred to any orthogonal frameof reference, let us choose a particular one, x'y'z', where the z' axis is along nandthe other two axes, x' and y', are arbitrary. With reference to this system, the stressmatrix has the form,

:,_: Advanced Mechanics of Solids

f (a)

..,

B

Fig. 1.13 Plot all((1) versus (1

U:r' Tx'y' 0

'fX'),' Uy' 0

0 0 u3(1.27)

o

y'x:Fig. 1.14 Rectangular element

with faces normal, , ,tox,y,z axes

Figure 1.14 shows these stress vectorson a rectangular element, The shear stresscomponent's Tt,z' and Ty'z' are zero since thez' plane is chosen to be the principal plane.With reference to this system, Eq. (1.19)becomes

(1x' - u) Tx'y' 0~xy' (UY' - U) 0 =0 (1.28)0 0 (13 - U)

Expanding

This is a cubic in (1. One of the solutions is (1 = 0'3' The two other solutions areobtained by solving the quadratic inside the brakets. The two solutions are

I-2(1.29)

The quantity under the square root (power~) is never negative and hence,

(11and (12 are also real. This means that the curve for j( 0') in Fig. 1.13 will cut the(1 axis at three points A, Band C in general. In the next section we shall study afew particular cases.


Analysis of Stress :~tst. " . ~ ( ", ,. ._.~. 4': I ~'

(i) If 0'1' 0'2 and 0:1 are distinct, i.e. 0'1' 0'2 and 0'3 have different values, thenthe three associated principal axes n1, n2 and n3 are unique and mutuallyperpendicular. This follows from Eq. (1.25) of Sec. 1.12. Since 0'" 0'2 and0'3 are distinct, we get three distinct axes nJ, n2 and n3 from Eqs (1.18), andbeing mutually perpendicular they are unique.

(it) If 0'1 = 0'2 and 0'3 is distinct, theaxis of n3 is unique and everydirection perpendicular to "3 isa principal direction associatedwith 0'1 = 0'2' This is shown inFig. 1.15.

To prove this, let us choose aframe of reference Ox'y'z' suchthat the z' axis is along ") andthe x' and y' axes are arbitrary.From Eq. (1.29), if 0'1 = 0'2' then the quantity under the radical mustbe zero. Since this is the sum of two squared quantities, this can happenonly if

,

Fig. 1.15 Case with 0'1 = 0'2and 0'3 distinct

O'x = 0'1 and 'CxI = 0

But we have chosen x' and y' axes arbitrarily, and consequently theabove condition must be true for any frame of reference with the z' axisalong "3' Hence, the x' and y' planes are shearless planes, i.e. principalplanes. Therefore, every direction perpendicular to "3 is a principal direc-tion associated with 0'1 = 0'2'

(iii) If 0'1 = 0'2 = 0'3' then every direction is a principal direction. This isthe hydrostatic or the isotropic state of stress and was discussed inSec. 1.7. For proof, we can repeat the argument given in (ii). Choose acoordinate system Ox'y'z' with the z' axis along "3 corresponding to 0').Since 0'1 = 0'2 every direction perpendicular to n) is a principal direction.Next, choose the z' axis parallel to "2 corresponding to (12' Then everydirection perpendicular to n2 is a principal direction since 0') = 0'3'Similarly, if we choose the z' axis parallel to nl corresponding to 0'1'every direction perpendicular to ", is also a principal direction. Conse-quently, every direction is a principal direction.

Another proof could be in the manner described in Sec. 1.7. Choosing Oxyzcoinciding with" l' n2 and "3' the stress vector on any arbitrary plane n has value0', the direction of 0' coinciding with n. Hence, every plane is a principal plane.Such a state of stress is equivalent to a hydrostatic state of stress or an isotropicstate of stress.

. ~ .

.The material discussed in the last few sections is very important and it is worth-while to put it in the form of definitions and theorems.


'in' ,:Advanced Mechanics of SolidsDefinition

nFor a given state of stress at point P, if the resultant stress vector T on any planen is along n having a magnitude (1, then (1 is a principal stress at P, n is theprincipal direction associated with (1, the axis of (1 is a principal axis, and theplane is a principal plane at P.

TheoremIn every state of stress there exist at least three mutually perpendicular principalaxes and at most three distinct principal' stresses. The principal stresses (11' (12and u3 are the roots of the cubic equation

(13 -/1(12 + l-z(1-/3 = 0where II' 12 and 13 are the first, second and third invariants of stress. The principaldirections associated with 0'1' (12 and (13 are obtained by substituting (1;(i = 1,2,3)in the following equations and solving for n.ny and nz:

(a, - (1J n:t + 1'x)' n)' + 1'xz nz= 01'..,. nx + (1y - (11) "+ 1'yz n: = 0

n; + n; + n; = IIf (11' (12 and (13 are distinct, then the axes of nl, n2 and n3 are unique and

mutually perpendicular. If, say (11 = (12 :;; cr" then the axis of n3 is unique andevery direction perpendicular to n3 is a principal direction associated with(11 = (12' If (11 = (12 = (13' then every direction is a principal direction.Standard Method of Solution

Consider the cubic equation y3 +pi + qy + r = 0, where p, q and r are constants.1Substitute y = x -- p3

This gives ~ + ax + b =0

where a= ~(3q_p2), b=-f;(2p3_9pq+27r)Put

Determine ~, and putting g = 2 ~ a!3, the solutions are

YI = g cos i :3 3Y2 = gcos(~ + 1200) - ~

Y3 = g cos (~ + 2400) - ~


Analysis of Stress :~~"__----------- .......... _-----------,---, ...."'-_._.Example 1.4 At a point P, the rectangular stress components are

{1'x = 1,(1'), = -2, {1': = 4, t:xy = 2, r)lt = - 3, and rXl = 1

all in units ofkl'a. Find the principal stresses and check for invariance.

Solution The given stress matrix is

1 2 1[ 'iij ] = 2 -2 -3

1 -3 4From Eqs (1.22}-{1.24),

11=1-2+4=3/2 = (-2 - 4) + (-8 - 9) + (4 - 1) = -2013 = 1(-8-9)-2(8+3)+ 1 (--ti+2)=-43

f(a) = a3 - 3a2 - 20a+ 43 = 0...For this cubic, following the standard method,

y=o, p=-3, q=-20, r=43

a = ! (-60 - 9) = -233

b= J.. (-54-540+ 1161)=2127cos = - (-)

(12i17)1f2=-11940

The solutions are .0') =Yl = 4.25 + 1= 5.25 kPa0'2 = Y2 = -5.2 + 1= -4.2 kPaa3 = Y3 = 0.95 + 1= 1.95 kPa

Renaming such that 0', > 0'2 > 0'3 we have,al = 5.25 kPa, 0'2 = 1.95 kPa, a3 = -4.2 kPa

The stress invariants are11 =5.25 + 1.95-4.2=3.012 = (5.25 x 1.95) - (1.95 x 4.2) - (4.2 x 5.25) = -2013 = -(5.25 x 1.95 x 4.2) = -43

These agree with their earlier values.

Example I.S With respect to the frame of reference Oxyz, the following state ofstress exists. Determine the principal stresses and their associated directions. A/so,check on the invariances of 11''2' '3'

Copyrlqhted matenal

- >

~'" Advanced Mechanics of Solids

1 2 1['t'ij] = 2 I I

1 I I

Solution For this state1,=1+1+1=312=(1-4)+(1-1)+(1-1)=-313 = 1(1-1) - 2(2- I) + 1(2- 1)=-1

f(u) = U3 -II U2 + 12u-13 = 0i.e., u3 - 3ul - 3u + 1 = 0or (0'3 + 1)-3u(u+ 1)=0i.e., (o + 1)(ul - a + I) - 3u(u + I) = 0or (u + 1) (q2 - 4u + I) = 0Hence, one solution is U = -1. The other two solutions are obtained from thesolution of the quadratic equation, which are U = 2 .fj .:. ul=-I, U2=2+.fj, u3=2-.fjCheck on the invariance:With the set of axes chosen along the principal axes, the stress matrix will havethe form

-1 0o 2+J3o 0

oo

2-J3II = -1 + 2 + .fj + 2 - J3 = 312= (-2 - .fj)+ (4 - 3) + (-2 + .fj)=-31)=-1 (4-3)=-1

Directions of principal axes:(i) For ul= -I, from Eqs (1.18) and (1.21)

Hence,

(1 + I)n..+ 2ny + nz = 02nx + (1 + 1)ny + nz = 0nx+ny+(1 + l)n:=O

together with

From the second and third equations above, n, = O.Using this in the thirdand fourth equations and solving, Ilx = (1/.J2), ny = (I/.J2).Hence, ul =-1 is in the direction (+I/.J2,-I/.J2,O).It should be noted that the plus and minus signs associated with nx, nyand nz represent the same line.

Copyr ighted malerial

Analysis of Stress ',~"

(Ii) For 0'2= 2 + J3

(-1 - J3 )n. + 211y + nz = 0211x + (-I - J3 )lIy + liz = 0

nx+ ny (-I - J3)lIz=O

together with

Solving, we get

( )

1/2

n =n = I+....!.....r y J3

I11 = -_.::.__~z (3 + J3f/2

(iii) For 0'3= 2 - J3We can solve for nx' ny and liz in a manner similar to the preceding one orget the solution from the condition that nl, n2 and n3 form a right-angledtriad, i.e. n3 = n I x n2.The solution is

. 1/2

nz =g (1 + )3)------------------_..-----.-.~--------------------~~~----Example 1.6 For the given state of stress, determine the principal stresses and theirdirections.

o 1 1[fij] = 1 0 1

1 1 0

Solution

.

II = 0, 12= -3,13 = 2f(G) = _G3 + 3G + 2 = 0

= (_0'3 - 1) + (30' + 3)=-(G+ 1)(G2-0'+ 1)+3(0'+ 1)= (0' + I) (0' - 2) (0' + 1) = 0

0'1= 0'2= -I and 0'3= 2

Since two of the three principal stresses are equal, and GJ is different, the axis of0'3 is unique and every direction perpendicular to O'J is a principal direction asso-ciated with 0'1= 0'2' For O:J = 2

-2nx + ny + nz = 0fl.. - 2ny + ": = 0


''2;a: Advanced Mechanics of Solids. .n" + ny - 2nz =0222nx + " + nz = 1

Th . Iese give Il" = ny = fl. = .J3......... 7_~ ..__,_i>._% .._u ,...._, _ ....t.b_~.4. ... _ ............$ ....... It ......Example 1.7 The stale of stress at a point is such that

0'"= O'y= a, = 't'lJ' = "t)" = 't'ZJt = PDetermine the principal stresses and their directions

Solution For the given state,

I. = 3p, 12 = 0, 13 = 0

Therefore the cubic is 0'3 - 3p0'2 = 0; the solutions are 0'. - 3p, 0'20'3= O. For 0'. = 3p

(p - 3p)n" + pny + pn, = 0pn" + (p- 3p) ny + pnz=Opnx + pny + (p- 3p) n. = 0

or

The above equations give

-2n" + lIy + !Ir = 0nx - 2ny + n. = 0nx + ny - 211. = 0

WI'th 2 2 2 I 11 '-3nx + ny + n. = ,one gets nr = ny = liz = ,r~.Thus, on a plane that is equaUy inclined to xyz axes, there is a tensile stress of

magnitue 3p. This is the case of a uniaxial tension, the axis of loading makingequal angles with the given xyz axes. If one denotes this loading axis by z', theother two axes, x' and y', can be chosen arbitrarily, and the planes normal tothese, i.e. x' plane and y' plane, are stress free.

,.TEJt::SlA.'FE:QF srasss REFEIUtED:. .TOPIuNClPAL~S

In expressing the state of stress at a point by the six rectangular stress compo-nents, we can choose the principal axes as the coordinate axes and refer therectangular stress components accordingly ..We then have for the stress matrix

0'. 0 0[TijJ= 0 0'2 0

o 0 0'3(1.30)


Analysis of Stress (Y2 ~ (Y3

Mark off (YI' (Y2 and (Y3 along the (Y axis and construct three circles withdiameters (YI - (Yz), (Y2 - (Y3) and (YI - (Y3) as shown in Fig. 1.16.

It will be shown in Sec. 1.18 that the point Q( (Y, -r) for all possible n will liewithin the shaded area. This region is called Mohr's stress plane 1!and the threecircles are known as Mohr's circles. From Fig. 1.16, the following points can beobserved:

(i) Points A, Band C represent the three principal stresses and the associatedshear stresses are zero.


"26' Advanced Mechanics of Solids

Fig. 1.16 Mohr's stress plane

(ii) The maximum shear stress is equal to ~ (0"] - 0"3) and the associated nor-

mal stress is ~ (0"] + 0"3)' This is indicated by point D on the outer circle.(lit) Just as there are three extremum values 0], 0"2 and 0"3 for the normal

stresses, there are three extremum values for the shear stresses, theseCT - CT 0'2 - (j. (j. - (j.being ] 2 3, 2 3 and J 2 2. The planes on which these shear

stresses act are called the principal shear planes. While the planes onwhich the principal normal stresses act are free of shear stresses, theprincipal shear planes are not free from normal stresses. The normal stresses

. d . h h .. I h . I 0'1+ 0'3 0'2 + 0'3associate Wit, t e pnncipa s ears are respective y 2 ' 2

and 0'1+ 0'2 . These are indicated by points D, E and F in Fig. 1.16. It will be2

shown in Sec. 1.19 that the principal shear planes are at 45 to the principalnormal planes. The principal shears are denoted by 1'1' 12 and 1':1 where

2r3 =(0'] -0'2), 21'2 =(0']-0'3)' 2r] =(0'2 -0'3) (1.36)(iv)

circle and the shear stress on any plane will not exceed

~ (0"] - 0"2) according as 0"] = 0"2 or 0"2 = 0"3'

(v) When 0] = 0'2 = (13' the three circles collapse to a single point on the (1axis and every plane is a shearless plane .

.MOHR'S STRESS PLANEIt was stated in the previous section that when points with coordinates (0", 1') forall possible planes passing through a point are marked on the (1- l' plane, as inFig. 1.16, the points are bounded by the three Mohr's circles. In this, section weshall prove this.


Analysis of Stress '-'27Choose the coordinate frame of reference Pxyz such that the axes are along the

"principal axes. On any plane with normal n, the resultant stress vector T and theDonna I stress (Jare such that from Eqs (1.32) and (1.33)

n 2T (1.37)

(1.38)

and also (1.39)

The above three equations can be used to solve for n.;.n; and n; yielding"2 (0' - 0'2) (0' - 0'3) + r2n = -'--;-----=:."--'-,-,--:..!---.,..-x (0'1 - 0'2)(0'1 - 0'3)

2 (0'- 0'3)(0'- 0'1) + ?ny = (0'2 - 0'3)( 0'2 - 0'1)

2 (0'- 0'1)(0'- 0'2)+ ?11 =

Z (0'3 - 0'1)(0'3 - 0'2)

(1.40)

(1.41)

(1.42)

Since n;, n; and n; are all positive, the right-hand side expressions in the aboveequations must all be positive. Recall that we have arranged the principal stressessuch that 0'1~ 0'2> 0'3' There are three cases one can consider.

Case (i) 0'1> 0'2> 0'3

Case (ii) 0'2 > 0'3For this case, the denominator in Eq. (1.40) is positive and hence, the numeratormust also be positive. In Eq. (1.41), the denominator being negative, the numera-tor must also be negative. Similarly, the numerator in Eq. (1.42) must be posi-tive. Therefore.

(0' - 0'2)( (J - (J3) + 1'2~ 0(0' - 0'3)(0'- 0'1) + 'f,2S 0(0'- (11)(


According to the first of the above equations, the point (0', r) must lie on or

outside a circle of radius ~ (0'2 - 0'3) with its centre at ~ (0'2 + 0'3) along the 0' axis

(Fig. 1.16). This is the circle with BC as diameter. The second equation indicates

that the point (0', .) must lie inside or on the circle ADC with radius .!.. (0'1 - 0'3)2

and centre at ~ (0'1 + 0'3) on the 0' axis. Similarly, the last equation indicates that

the point (0', .) must lie 00 or outside the circle AFB with radius equal to .!.. (0'1 - 0'0I 2

and centre at 2 (0'1 + O'~.

Hence, for this case, the point Q(0', r) should lie inside the shaded area of Fig. 1.l6.

Case (ii) 0'1 = 0'2 > 0'3Following arguments similar to the ones given above, one has for this case fromEqs (1.40)-(1.42)

From the rust two of these equations, since 0'1 = 0'2' point (0', 1") must lie on the

circle with radius .!.. (0'1 - 0'3) with its centre at .!.. (0'1 + 0'3)' The last equation2 2

indicates that the point must lie outside a circle of zero radius (since 0'1 = 0'2)'.Hence, in this case, the Mohr's circles will reduce to a circle BC and a point circleB. The point Q lies on the circle BEC.

Case (iii) 0'1 = 0'2 = 0'3This is a trivial case since this is the isotropic or the hydrostatic state of stress.Mohr's circles collapse to a single point on the 0' axis.

See Appendix I for the graphical determination of the normal and shear stresseson an arbitrary plane, using Mohr's circles.

From Sec. 1.17 and also from Fig. 1.16 for the case 0'. >

Analysis of Stress {~'

Substituting these values in Eqs.(1.37)-( 1.39) in Sec. 1.18, one gets nx = .Jli2 ,ny = 0 and n. = l .fi. This means that the planes (there are two of them)on which the shear stress takes on an extremum value, make angles of 45and 1350 with the (1) and

:\0> Advanced Mechanics of Solids

or

O'OCI = ~ (0'1 + 0'2 + 0'3) = iIIT~ = ~ [(0'1 - 0'2)2 + (0'2 - 0'3}2 + (0'3 - 0'1)2]

9T:CL = 2(0'1 + 0'2 + 0'3)2 - 6(0'10'2 + 0'20'3 + 0'30'1)

'f =.fi (12 _ 31)1/2oct 3 I "2

(1.43)

and (1.44a)

or (1.44b)

(1.44c)

It is important to remember that the octahedral planes are defined with respectto the principal axes and not with reference to an arbitrary frame of reference.Since O'OCI and "roct have been expressed in terms of the stress invariants, one canexpress these in terms of 0'..., O'Y' 0;, "rxy> 1)..: and 'fzx also. Using Eqs (1.22) and (1.23),

(1.45)

9T:C, = (o; - O'y)2 + (O'y - O'J2 + (0': - O'x)2 + 6(T.;y + T! + T; ) (1.46)The octahedral normal stress being equal to 1/3 11, it may be interpreted as the

mean normal stress at a given point in a body. If in a state of stress, the firstinvariant (0'1 + 0'2 + 0'3) is zero, then the normal stresses on the octahedral planeswill be zero and only the shear stresses will act. This is important from the pointof view of the strength and failure of some materials (see Chapter 4).

____ __ ... '=e . ~~. __ _,. .. _ ... _ ... ........ 1 __ ,_',,_p'p __ ._

Example 1.8 The state of stress at apoint is characterised by the components0:.. = 100 MPa, O'y= -40 MPa, o,= 80 MPa,'fxy = fyz = 'tzx = 0

Determine the extremum values of the shear stresses, their associated normalstresses, the octahedral shear stress and its associated normal stress.

Solution The given stress components are the principal stresses, since theshears are zero. Arranging the terms such that 0'1 > 0'2 > 0'3'

0'1 = 100MPa, 0'2 = 80 MPa, 0'3 = -40 MPaHence from Eq. (1.36),

'tl= u2 -u3 = 80+40 = 60 MPa

2 2

'f3= UI-U2 ;100-80=10MPa2 2

The associated normal stresses are

.,

: ~Gf>yrightedmaterialI

Analysis of Stress '-31', .

0'; = 0'3 + 0'1 = -40 + 100 = 30 MPa2 2

------------"""' .."" ""'''''_ -- ------------~J THE STATE OF PURE SHEARThe state of stress at a point can be characterised by the six rectangular stresscomponents referred to a coordinate frame of reference. The magnitudes of thesecomponents depend on the choice of the coordinate system, If, for at least oneparticular choice of the frame of reference, we find that


The given state can be resolved into two different states, as shown:

ax 1'xy 1'x: p 0 0 ax - p 1'xy 1'xz

't'xy ay 'tyz - 0 P 0 + 'txy ay - p tyz (1.48)'Txz 1'yz a. 0 0 P T;a Tyz 0'% - P

The first state on the right-hand side of the above equation is a hydrostaticstate. [Refer Sec. t. I4(iii).]

The second state is a state of pure shear since the first invariant for thisstate is

1'1 = (o, - p) + (ay --p) + (

Analysis of Stress . '33'For the given state, the octahedral normal and shear stresses are:

(Joel =~ II = 6

FromEq. (1.44)

_ .fi ( 2 2)1/2Toet - "3 II - 3/2

= ~ [182 - 3 (20 - 16+ 12- 64 + 60 - 36) J1I2

=f (396)"2 = 2..fi2For the hydrostatic state, (Joel = 6, since every plane is a principal plane with (J = 6

and consequently, Toel = O.For the pure shear state, (Joe! = 0 since the first invariant of stress for the pure

shear state is zero. The value of the second invariant of stress for the pure shearstate is

1'2 = (-16 - 16+ 0 - 64 + 0 - 36 ) = -132Hence, the value of TOCI for the pure shear state is

't' =::Ii (396)112 = 2 ..fi2ocr 3Hence, the value of (jOCI for the given state is equal to the value of (joel for thehydrostatic state, and 1:0


The principal stresses when arranged such that 0'( > 0'2 > 0'3 arepd pd

cr( =y,; 0'2 = 4t; 0'3 =-pThe maximum shear stress is therefore,

'max =~(O'( -cr3)=!P{~+1)Substituting the values

= 1400 (1.8 x 100 1)= 35 700 kPma.~ 2 2 x 1.8 + ,. a_ ... _". _ ....._,_ ............. _. - ... r;;c.o......_._._. __ 7_ ..._;"~ __ __ _.,._ ....., 4'1'. _

1.f&'t CAUCHY'S STRESS QUADRICWe shall now describe a geometrical description of the state of stress at a point P.Choose a frame of reference whose axes are aJong the principal axes. Let 0'1' 0'2and 0'3 be the principal stresses. Consider a plane with normal n. The normalstress on this plane is from Eq. (1.33),

Along the normal n to the plane, choose a point Q such that

PQ=R= 1/#As different planes n are chosen at P, we get different values for the normal

stress 0' and correspondingly different PQs. If such Qs are marked for every planepassing through P, then we get a surface S. This surface determines the normalcomponent of stress on every plane passing through P. This surface is known as

the stress surface of Cauchy. Thishas a very interesting property. LetQ be a point on the surface,Fig. 1.20(a). By the previous defmi-tion, the length PQ = R is such that,the normal stress on the plane whosenormal-is along PQ is given by

m

nTn

T

(b)

(a)Fig. 1.20 (a) Cauchy's stress quadric

(b) Resultant stress vector andnormal stress component

(1.50)

1cr=- (1.51)R2If m is a norma] to the tangent

plane to the surface S at point Q,then this normal m is parallel to the

nresultant stress vectorT at P.

Since the direction of the result-n

ant vector T is known, and its COffi-n

ponent 0' along the normal is known, the resultant stress vector T can be easilydetermined, as shown in Fig. ] .20(b).


Analysis of Stress '35n

We shall now show that the normal m to the surface S is parallel to T, theresultant stress vector. Let Pxyz be the principal axes at P (Fig. l.21). n is thenormal to a particular plane at P. The normal stress on tbis plane, as before, is

2 2 2a = alnx + a2ny + a3n.

Y

I'- Q (x. Y. z)-9

n

p

zFig. 1.21 Principal axes at P and

n to a plane

when (J is tensile

If the coordinates of the point Q are(x, Y, z) and the length PQ = R, then

n =.!. (1.52)Z R

Substituting these in the above equa-tion for (J

(JR2 = (Jlx2 + (J21 + (J3z2

From Eq. (1.51), we have (JR2 == I. Theplus sign is used when (J is tensile and theminus sign is used when (J is compressive.Hence, tbe surface S has the equations(a surface of second degree)

Substituting for n." fly and liz from Eq. (1.52)

x

(1.53a)

(1. 53b)

We know from calculus that for a surface with equation F(x, y, z) = 0, thenormal to the tangent plane at a point Q on the surface has direction cosines

proportional to :': and 0: .From Fig. (1.20), m is the normal perpendicularto the tangent plane to S at Q. Hence, if m.., my, and "I, are the direction cosines ofm, then

ofm=a-r oz

From Eq. (l.53a) or Eq. (1.53b)In,= 2a(J.x, my= 2a(J2Y'

where a is a constant of proportionality.n n nT is the resultant stress vector on plane n and its components T x, T r and

nT z according to Eq. (1.31), are

(1.54)

when (J is compressive

Copyr ighled malerial


or

Substituting these in Eq, (1.54)n

mx=2aRTx,n

nly = 2a RTy,n

m. =2a RTzn n n

i.e. m... my and nlz are proportional to T x , T y and T z n

Hence, m and T are parallel.The stress surface of Cauchy, therefore, has the following properties:

(i) If Q is a point on the stress surface, then PQ =l/~ where (J is thenormal stress on a plane whose normal is PQ.

n(it) The normal to the surface at Q is parallel to the resultant stress vector T

on the plane with normal PQ.Therefore, the stress surface of Cauchy completely defines the state of stress

at P. It would be of interest to know the shape of the stress surface for differentstates of stress. This aspect will be discussed in Appendix 3.

LAME'S ELLIPSOJ])Let Pxyz be a coordinate frame of reference at point P, parallel to the principalaxes at P. On a plane passing through P with normal n, the resultant stress vector

nis T and its components, according to Eq. (1.31), are

Let PQ be along the resultant stress vector and its length be equal to its/I

magnitude, i.e. PQ=ITI. The coordinates (x, y, z) of the point Q are thenn

x=Tx,II

y=Ty,II

Z = Tz

Since n2 + n2 + n2 = I, we get from the above two equations.x y z

y

On

~T

Fig. 1.22 Lame's ellipsoid

(1.55)

This is the equation of anellipsoid referred to the princi-pal axes. This ellipsoid is calledthe stress ellipsoid or Lame'sellipsoid. One of its three semi-axes is the longest, the otberthe shortest, and the third in-between (Fig. 1.22). These arethe extermum values.

If two of the principalstresses are equal, for instance

Copyrlqhted matenal


O"J = 0"2' Lame's ellipsoid is an ellipsoid of revolution and the state of stress at agiven point is symmetrical with respect to the third principal axis Pz. If all theprincipal stresses are equal, O"J = 0"2 = 0"3' Lame's ellipsoid becomes a sphere.

Each radius vector PQ of the stress ellipsoid represents to a certain scale, theresultant stress on one of the planes through the centre of the ellipsoid. It can beshown (Example 1.11) that the stress represented by a radius vector of the stressellipsoid acts on the plane parallel to tangent plane to the surface called thestress-director surface, defined by

(1.56)

The tangent plane to the stress-director surface is drawn at the point of intersec-tion of the surface with the radius vector. Consequently, Lame's ellipsoid and thestress-director surface together completely define the state of stress at point P.____ ,... ........ ~_ ... _'.:roz'"','.n.r .... $e .... ~ .." _ ........'_..._._ ....

Example 1.11 Show that Lame 's ellipsoid and the stress-director surface togethercompletely define the state of stress at a point.

Solution If O"J' 0"2 and 0"3 are the principal stresses at a point P, the equation ofthe ellipsoid referred to principal axes is given by

') 2 2x- Y z-+-+-=12 2 2CTJ CT2 CT3

The stress-director surface has the equation,22

X- +L+._= ICTI CTZ CT3

It is known from analytical geometry that for a surface defined byFtx; y, z) =0, the normal to the tangent plane at a point (xo,Yo,Zo ) has direction cosines

proportional to 0;,::,:, evaluated at (xo, Yo, zo). Hence, at a point(xo,Yo> zo) on the stress ellipsoid, if m is the normal to the tangent plane (Fig.1.23), then

Ellipsoid Surface

Stress-directorSurface

Fig. 1.23 Stress director sutface and ellipsoid surface



Consider a plane through P with normal parallel to m. On this plane, the resultant"stress vector will be T with components given by

Substituting for my my and InzmTy=ayo,

mT = azo

i.e. the components of stress on the plane with normal m are proportional to thecoordinates (xo, Yo, zo). Hence the stress-director surface has the following property.Let L(xo, Yo, zo) be a point on the stress-director surface. Let m be the normal tothe tangent plane at L. On a plane through P with normal m, the resultant stress

mvector is T with components proportional to xo, Yo and zoo This means that the

m nJ mcomponents of PL are proportional to T, Ty and T.

nPQ being an extension of PL and equal to T in magnitude, the plane having

this resultant stress will have m as its normal.

}l:u .THE PLANE STATE OF STRESSIf in a given state of stress, there exists a coordinate system Oxyz such that forthis system

o,= 0, 'fu = 0, 't: = yz (1.57)then the state is said to have a 'plane state of stress' parallel to the xy plane. Thisstate is also generally known as a two-dimensional state of stress. All the forego-ing discussions can be applied and the equations reduce to simpler forms as aresult of Eq. (1.57). The state of stress is shown in Fig. 1.24.

(a)

O'y

1':.y

0'1( C1

1''')'

C1

I(

y

(b)

Fig. 1.24 (a) Plane state a/stress (b) Conventional representation

Consider a plane with the normal lying in the xy plane. If nx , " and n, arethe direction cosines of the normal, we have nx = cos e, n>, = sin e and nz = 0(Fig. 1.25). From Eq. (1.9)


Analysis of Stress ''''3~'O'y

'f>y, n,,,'\~f!._Ox . - ,,,,,,,,

'f>y

0'

n0/

y

(a) (b)

Fig. 1.25 Normal and shear stress components on an oblique plane

nT.. = C1x cos 8 + 'f,y sin 8"Ty = C1y sin 8+ fxy cos 8 {1.58}nT =0

The normal and shear stress components on this plane are from Eqs (1.11a)and (l.llb)

C1 = C1x cos' 8 + C1y sin2 8 + 2f

110'- Advanced Mechanics of SolidsIn the xy plane, the maximum shear stress will be

Tmax =1(0'1 - 0"2 )and from Eq. (1.61)

(1.63b)

1f2

(1.64)

So far, attention bas been focussed on the state of stress at a point. In general. thestate of stress in a body varies from point to point. One of the fundamental prob-lems in a book of this kind is the determination of the state of stress at every pointor at any desired point in a body. One of the important sets of equations used in theanalyses of such problems deals with the conditions to be satisfied by the stress

components when they varyfrom point to point. These con-ditions will be established whenthe body (and, therefore. everypart of it) is in equillibrium, Weisolate a small element of thebody and derive the equationsof equilibrium from ito; free-body diagram (Fig. 1.26). A simi-lar procedure was adopted inSec. 1.8 for establishing theequality of cross shears.

Consider a small rectangu-lar element with sides ax, l1yand Az isolated from its parentbody. Since in the limit, we aregoing to make ax, y and Aztend to zero, we shall deal withaverage values of the stresscomponents on each face.These stress components areshown in Fig. 1.27.

The faces are marked as I.,2, 3etc. On the left hand face, i.e. faceNo.1, the average stress com-ponents are O"x, Txy and Txz. Onthe right hand face, i.e. faceNo.2, the average stress com-ponents are

.....--..1,.--~ I

I ", I..''''-''~ ..----- ..--.....' ....--,,1"

(b)

(a)

Fig. 1.26 Isolated cubical elementin equilibrium

y

5/,---- -0'Ir1/i"Ox_-1----"'1- ,, 2

x

z

Fig. 1.27Oy

Variation ofstresses

iJrT + .

Analvsis of Stress . .,1'o

This is because the right hand face is .1.x distance away from the left hand face.Following a similar procedure, the stress components on the six faces of theelement are as follows:

00-. ACT. + :) uZ, uZ

Let the body force components per unit volume in the x, y and z directionsbe rx' 1')" and r,. For equilibrium in x direction

Face 1 CTx' '"XY' '"xz

Face 2 00- OrX)'CTx + 0; ~x, '"xy + ox ~x,Face 3 CTy' '"} , yzFace 4

~ "=, oryx ~y,CTy + oy ~y, '"ft + oyFace 5 fYz' TZK, Tzy

Face 6

OrT + :cz ~x

.CZ ox

(Or.r )'rx + 0; ~Z ~x~y- 'zx ~x~y+ Yx ~x ~y~z= 0

Cancelling terms, dividing by .1.x, ~y,~ and going to the limit, we get

ro-x Oryx Or z.x 0.... + "" + :) + Yx =ox ay az

Similarly, equating forces in the y and z directions respectively to zero, we gettwo more equations. On the basis of the fact that the cross shears are equal,i.e . xy = 'fy., 'fyz = 'fzy, 'fxz := 'fzx' we obtain the three differential equations ofequilibrium as

i}o- Or Orx+ X)'+ n+y=O

ox oy oz xi}o- iJr iJr-:-y::...+ xy of yz + y = 0oy ox oz y

ro- iJr Or,"--::-~z + xz + ,. + y. = 0oz ox oy

Equations (1.65) must be satisfied at all points throughout the volume of thebody. It must be recalled that the moment equilibrium conditions established theequality of cross shears in Sec.I.S.

(1.65)

Copyr ighled malerial

ANSHULHighlight

:.a2' Advanced Mechanics of SolidsI ....

The plane stress has already been defined. If there exists a plane stress state inthe xy plane, then C1z = Tzx = or)':= r. = 0 and only C1x C1y, Txy, r. and Yy exist. Thediffemetial equations of equilibrium become

ro Urxy0: + oy +r, =0roy UrX}'oy + ox +ry=O (1.66)

Example 1.12 The cross-section of the wall ofa dam is shown in Fig. 1.28. Thepressure of water onface OB is also shown. With the axes Ox and Oy, as shownin Fig. 1.28, the stresses at any point (x, y) are given by tr= specific weightof water and p = specific weight of dam material)

C1x = -ry

C1. - (p 2r) x + (r pJ yJ' - tan f3 - tan3 f3 tan2 f3 -

o ---~x

A

"rty='f'yx=- ~ xtan f3

't'yz = 0, 1'zt = 0, (1. = 0

Fig. 1.28 Example 1.12Check if these stress components satisfy the differ-ential equations of equilibrium. Also, verify if theboundary conditions are satisfied on face DB.

and

Solution The equations of equilibrium are

ro t7r-=-x"- + .


Solution Substituting in the differential equations of equilibrium

03 tP _ oJ tP = 0

oy2 ox oy2 ox

Example 1.14 Consider the rectangular bean' shown in Fig. J.29. According to theelementary theory of bending, the 'fibre stress' in the elastic range due to bending isgiven by

My 12Myu =--=--7-x I bh3

y

i --I h. {2~--- ---6

X Z I----+---t h/2r-"-'-'-'-'-'-'-'-'-

Fig.l.29 Exmaple 1.14

where M is the bending moment which is a function of x. Assume that(1, == 1'%.1: = 'fzy= 0 and also that 'fxy = 0 at the top and bottom. and further, that (1y= 0at the bottom. Using the differential equations of equilibrium, determine fAJ' and (1y.Compare these with the values given in the elementary strength of materials,

Solution From Eq. (L65)

or

ro- Or Or_-=-,,-x xy xz 0+ + =ox oy OZSince 't'xz = 0 and M is a function of x

_ 12y oM + OrAJ' = 0bll3 ox d y

o.xy 12 oMoy = bhl ox Y

Integrating 6 oM 2 j'()-3 ~ Y +C1 x +c2bh C/X

Copyr ighted malerial

,14' Advanced Mechanics of Solids..' .; "where f(x) is a function of x alone and c), c2 are constants. It is given that

"" = 0 at y = !!.'.T.f 2

6 h2 0}.1~3 -4 ~ = -c1f(x) - c2bl: o x

c f(x) + C = _ 3 oMI 2 2bh ox

3 oM (4y2 )'f.t)' = 2bh ox h2 - I

From elementary strength of materials, we have

or

V (h12 , ciA'f.ty = lb Jy Y

where V= - oM is the shear force. Simplifying the above expressionox

or 1: = 3 oM (4y2 -1)x.v 2bh ox h3

we get

i.e, the same as the expression obtained above.From the next equilibrium equation, i.e. from

ib'y or xy or ~_-::-'-+ + .' =0oy ox OZ3 (4y2 ) 02M

= - 2bh h2 - I ox2

.. 302M (4y3 )O'y= - 2bl 2 2 - Y + C3 F(x) + c4'ox 3h .

where F(x) is a function of x alone. It is given that O'y = 0 at y = - ~ .

Hence,

Substituting


Analysis of Stress '':45', .'At Y = +h/2, the value of CJy is

I a2M wCJy = b t7X2 =b

where w is the intensity of loading. Since b is the width of the beam, the stresswill be wI b as obtained above.'SA ctS#ilW ,:tCd!SS "'~f" au 4 :Mdll:S' 'Sf' ?,_~ ..q$".'I'P d ..... $ ...... 111 pt,m

BOUNDARY 'CONDITIONSEquation (1.66) must be satisfied throughout the volume of the body. When thestresses vary over the plate (i.e. the body having the pJane stress state), thestress components CJx' CJy and 1'...y must be consistent with the externally appliedforces at a boundary point.

Consider the two-dimensional body shown in Fig.I.30. At a boundary point P,the outward drawn normal is n, Let F, and F; be the components of the surfaceforces per unit area at this point.

yF y

~

,/n

Fx'fxy

~Gy

(b)

1>-------"'(-_ Xo

(a)

Fig. 1.30 (a) Element near a boundary point (b)Free body diagram

F.. and Fy must be continuations of the stresses CJx, CJ)' and 't'xy at the boundary,Hence, using Cauchy's equations

nTx = F x = CJxnx + 1'xylly/Ir, = Fy = 0'), fly + 'f,), IIx

If the boundary of the plate happens to be parallel to y axis, as at point PI' theboundary conditions become

P; = o, and Fy =

'46 Advanced Mechanics of Solidsexist where the bodies under discussion possess radial symmetry; for example, athick cylinder subjected to internal or external pressure. For the analysis of suchproblems, it is more convenient to use polar or cylindrical coordinates. In thissection, we shall develop some equations in cylindrical coordinates .. Consider an axisymmetric body as shown in Fig. 1.31(a). The axis of thebody is usually taken as the z axis. The two other coordinates are r and 8,where 8 is measured counter-clockwise. The rectangular stress components ata point per, B, z) are

z

,,........ -_ .. -:-;--'--- ...6.'" p

(a)

Fig. 1.31

These are shown acting onthe faces of a radial elementat point P in Fig.l.31(b).C1" C18 and C1. are called theradial, circumferential andaxial stresses respectively. Ifthe stresses vary from pointto point, one can derive theappropriate differential equa-tions of equilibrium, as inSec. 1.26. For this purpose,consider a cylindrical ele-ment having a radial lengthlY with an included angle l1Band a height llz, isolated fromthe body. The free-body dia-gram of the element is shownin Fig.l.32(b). Since tbe ele-ment is very small, we workwith the average stressesacting on each face.

The area of the face aa' d' dis r l1Bllz and the area of facebb' c' c is (r + lY) 118 llz. Theareas of faces dec'd' and abb' e'are each equal to lY llz.

The faces abed and a'b' c' d' have each an area (r + ~) l1BlY. The averagestresses on these faces (which are assumed to be acting at the mid point of eaceface) are

On face add' d

(b)

(a) Cylindrical coordinates of a point(b) Stresses on an element

normal stress C1rtangential stresses 'f" and. 'frO

On face bb' c'c

normal stress C1T+ ib',lYor


Analysis of Stress ~

z

y

c'~--------------~yx

(a)x

tl)

Fig. 1.32 (a) Geometry of cylindrical element (b) Variation of stresses IU'7VSS faces

tangential stresses Trz ., .~ I1r and

The changes are because the face bb' c' c is Ar distance away from the face aa'd' d.On face dcc'd'

normal stress Getangential stresses rre and 'fez

On face abb' a

&-normal stress Ge+ c: l!:..(}tangential stresses 'f,e + ~ 110 and 'fez +

The changes in the above components are because the face abb' a is separated byan angle A6 from the face dcc'd',

On face a'b'c'd'normal stress o,tangential stresses 1'17 and foz

On face abed

normal stress o,+

.I t7rtangenna stresses r + 1%rz OZ Az and 'rOt + .

Let Yr, Yo and y:: be the body force components per unit volume. If the element isin equilibrium, the sum of forces in r, 6 and z directions must vanish individuaUy,. Equating the forces in r direction to zero,

( CTr + ~~ Ar) (r +Ar) A(}A:z + (!n + ~ AZ) (r + ~) ll(}llrCopyrigtlted material

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SO'" Advanced Mechanics of Solids passing through the point. Tbe equations are

nM" =u; n" +My" ny+u; n:

n

Mz =u; n.t +My: " + Mu n:n n n R

Mx, My, Mz are the x, y and z components of the vector M acting onplane n,

12 A rectangular beam is subjected to a pure bending moment M. The cross-section of the beam is shown in Fig. 1.34. Using the elementary flexureformula, determine the normal and shearing stresses at a point (x, y) on theplane AB shown.

y

M

A"-'" M

~.-.-.~~~.-.-.-.-.=.-..j-x-,"8

Fig. 1.34 Problem 1.2

OJI~ ).1

b

[ Ans.

13 Consider a sphere of radius R subjected to diametral compression(Fig. 1.35). Let o; ulJand u; be the normal stresses and 1'r8' 1'~and. 1'90' theshear stresses at a point. At point P(o, y, z) on the surface and lying in theyz plane, determine the rectangular normal stress components U'" uy and u.in terms of the spherical stress components.

[Ans. U" = Ue; Uy= U; cos2 4'; a, = U; sin2 4']

tFig.l.35 Problem 1.3


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_______ ~... tl ' _ ..._"__ .. _,,_.r_ .........,.-__ aiJ""':;_I. '_""__ if;l!li"'~"'~_._1._U _

Appendix 1

Mohr's Circles,

It was stated in Sec. 1.17 that when points with coordinates (a, or)for all possibleplanes passing through a point are marked on the a-or plane, as in Fig. 1.16, thepoints are bounded by the three Mohr's circles. The same equations can be usedto determine graphicaUy the normal and shearing stresses on any plane withnormal II. Equations (1.40}-{I .42) of Sec. 1.IS are

2 ( 0" - 0"2) (0" - 0"3 ) + 't"2n = ~-~-.:..,....,----"';_'..,.....-x (0"1 - 0"2)(O"J -0"3)

(Al.I)

2 (0" - 0"3) (0" - O"t) + 't"2n = -'-:--....;;..:-.:..,....,--__;_,:..._~y (0"2 - 0"3) (0"2 - O"t)

2' (0" - 0"1) (0" - 0"2 ) + 't"2n =~-....:...:....!.:--;--=.,:_~Z (0"3 - 0"1) (0"3 - a2

126233678-ls-srinath

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