cambridge introduction to continuum mechanics

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INTRODUCTION TO CONTINUUM MECHANICS

This textbook treats solids and fluids in a balanced manner, using ther-modynamic restrictions on the relation between applied forces andmaterial responses. This unified approach can be appreciated by en-gineers, physicists, and applied mathematicians with some backgroundin engineering mechanics. It has many examples and about 150 exer-cises for students to practice. The higher mathematics needed for acomplete understanding is provided in the early chapters. This subjectis essential for engineers involved in experimental or numerical mod-eling of material behavior.

Sudhakar Nair is the Associate Dean for Academic Affairs of theGraduate College, Professor of Mechanical Engineering and Aero-space Engineering, and Professor of Applied Mathematics at the Illi-nois Institute of Technology in Chicago. He is a Fellow of the ASME,an Associate Fellow of the AIAA, and a member of the AmericanAcademy of Mechanics as well as Tau Beta Pi and Sigma Xi.

Introduction to ContinuumMechanics

Sudhakar NairIllinois Institute of Technology

CAMBRIDGE UNIVERSITY PRESS

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© Sudhakar Nair 2009

2009

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Contents

Preface page xi

1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1 Concept of a Continuum 11.2 Sequence of Topics 2

2 Cartesian Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2.1 Index Notation and Summation Convention 42.2 Kronecker Delta and Permutation Symbol 6

2.2.1 Example: Skew Symmetry 72.2.2 Example: Products 8

2.3 Coordinate System 82.4 Coordinate Transformations 92.5 Vectors 122.6 Tensors 14

2.6.1 Examples of Tensors 142.6.2 Quotient Rule 172.6.3 Inner Products: Notation 18

2.7 Quadratic Forms and Eigenvalue Problems 182.7.1 Example: Eigenvalue Problem 202.7.2 Diagonalization and Polar Decomposition 212.7.3 Example: Polar Decomposition 23

3 General Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

3.1 Vectors and Tensors 313.2 Physical Components 333.3 Tensor Calculus 333.4 Curvature Tensors 353.5 Applications 36

3.5.1 Example: Incompressible Flow 363.5.2 Example: Equilibrium of Stresses 37

v

vi Contents

4 Integral Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

4.1 Gauss Theorem 424.2 Stokes Theorem 44

5 Deformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

5.1 Lagrangian and Eulerian Descriptions 495.2 Deformation Gradients 51

5.2.1 Deformation Gradient Vectors 525.2.2 Curvilinear Systems 54

5.3 Strain Tensors 555.3.1 Decomposition of Displacement Gradients 565.3.2 Stretch 575.3.3 Extension 585.3.4 Infinitesimal Strains and Rotations 585.3.5 Deformation Ellipsoids 605.3.6 Polar Decomposition of the Deformation Gradient 635.3.7 Stretch and Rotation 645.3.8 Example: Polar Decomposition 655.3.9 Example: Square Root of a Matrix 67

5.4 Logarithmic Strain 685.5 Change of Volume 685.6 Change of Area 695.7 Compatibility Equations 705.8 Spatial Rotation and Two-Point Tensors 715.9 Curvilinear Coordinates 72

6 Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

6.1 Material Derivative 766.1.1 Some Terminology 776.1.2 Example: Path Line, Stream Line, and Streak Line 79

6.2 Length, Volume, and Area Elements 806.2.1 Length 816.2.2 Volume 816.2.3 Area 82

6.3 Material Derivatives of Integrals 826.3.1 Line Integrals 826.3.2 Area Integrals 836.3.3 Volume Integrals 83

6.4 Deformation Rate, Spin, and Vorticity 836.5 Strain Rate 866.6 Rotation Rate of Principal Axis 87

Contents vii

7 Fundamental Laws of Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . 90

7.1 Mass 907.2 Conservation and Balance Laws 91

7.2.1 Conservation of Mass 917.2.2 Balance of Linear Momentum 917.2.3 Balance of Angular Momentum 927.2.4 Balance of Energy 927.2.5 Entropy Production 92

7.3 Axiom of Material Frame Indifference 927.4 Objective Measures of Rotation 947.5 Integrity Basis 95

8 Stress Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

8.1 External Forces and Moments 988.2 Internal Forces and Moments 988.3 Cauchy Stress and Couple Stress Tensors 100

8.3.1 Transformation of the Stress Tensor 1018.3.2 Principal Stresses 1018.3.3 Shear Stress 1028.3.4 Hydrostatic Pressure and Deviatoric Stresses 1038.3.5 Objective Stress Rates 104

8.4 Local Conservation and Balance Laws 1068.4.1 Conservation of Mass 1068.4.2 Balance of Linear Momentum 1078.4.3 Balance of Moment of Momentum (Angular Momentum) 107

8.5 Material Description of the Equations of Motion 1088.5.1 First Piola–Kirchhoff Stress Tensor 1098.5.2 Second Piola–Kirchhoff Stress Tensor 110

9 Energy and Entropy Constraints . . . . . . . . . . . . . . . . . . . . . . . . . 114

9.1 Classical Thermodynamics 1149.2 Balance of Energy 1159.3 Clausius–Duhem Inequality 116

9.3.1 Fourier’s Law of Heat Conduction 1189.3.2 Newton’s Law of Viscosity 1199.3.3 Onsager’s Principle 1199.3.4 Strain Energy Density 1199.3.5 Ideal Gas 120

9.4 Internal Energy 1219.4.1 Legendre or Contact Transformation 1219.4.2 Surface Energy 123

9.5 Method of Jacobians in Thermodynamics 123

viii Contents

10 Constitutive Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

10.1 Invariance Principles 13010.1.1 Principles of Exclusion 13010.1.2 Principle of Coordinate Invariance 13110.1.3 Principle of Spatial Invariance 13110.1.4 Principle of Material Invariance 13210.1.5 Principle of Dimensional Invariance 13210.1.6 Principle of Consistency 132

10.2 Simple Materials 13210.3 Elastic Materials 134

10.3.1 Elastic Materials of Cauchy 13410.3.2 Elastic Materials of Green 135

10.4 Stokes Fluids 13710.5 Invariant Surface Integrals 13810.6 Singularities 140

11 Hyperelastic Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142

11.1 Finite Elasticity 14211.1.1 Homogeneous Deformation 14311.1.2 Simple Extension 14411.1.3 Hydrostatic Pressure 14511.1.4 Simple Shear 14511.1.5 Torsion of a Circular Cylinder 146

11.2 Approximate Strain Energy Functions 14811.2.1 Hookean Materials 14911.2.2 Small-Strain Approximation 14911.2.3 Plane Stress and Plane Strain 150

11.3 Integrated Elasticity 15011.3.1 Example: Incremental Loading 151

11.4 A Variational Principle for Static Elasticity 15211.5 Isotropic Thermoelasticity 154

11.5.1 Specific Heats and Latent Heats 15511.5.2 Strain Cooling 15611.5.3 Adiabatic and Isothermal Elastic Modulus 15611.5.4 Example: Rubber Elasticity 157

11.6 Linear Anisotropic Materials 15911.7 Invariant Integrals 160

12 Fluid Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164

12.1 Basic Equations 16412.2 Approximate Constitutive Relations 16512.3 Newtonian Fluids 16512.4 Inviscid Fluids 167

Contents ix

12.5 Shearing Flow 16712.6 Pipe Flow 16912.7 Rotating Flow 17212.8 Navier–Stokes Equations 17512.9 Incompressible Flow 17512.10 Compressible Flow 17612.11 Inviscid Flow 176

12.11.1 Speed of Sound 17712.11.2 Method of Characteristics 178

12.12 Bernoulli Equation 17912.13 Invariant Integrals 180

13 Viscoelasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184

13.1 Kelvin–Voigt Solid 18413.2 Maxwell Fluid 18513.3 Standard Linear Solid 18713.4 Superposition Principle 18913.5 Constitutive Laws in the Operator Form 19013.6 Three-Dimensional Linear Constitutive Relations 19013.7 Anisotropy 19113.8 Biot’s Theory 192

13.8.1 Minimum Entropy Production Rate 19413.9 Creep in Metals 19413.10 Nonlinear Theories of Viscoelasticity 19513.11 K-BKZ Model for Viscoelastic Fluids 196

14 Plasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200

14.1 Idealized Theories 20014.1.1 Rigid Perfectly Plastic Material 20014.1.2 Elastic Perfectly Plastic Material 20114.1.3 Elastic Linearly Hardening Material 201

14.2 Three-Dimensional Theories 20314.3 Postyield Behavior 206

14.3.1 Levy–Mises Flow Rule 20614.3.2 Prandtl–Reuss Flow Rule 207

14.4 General Yield Condition and Plastic Work 20814.4.1 Plane Stress and Plane Strain 20914.4.2 Rigid Plasticity and Slip-Line Field 20914.4.3 Example: Symmetric External Cracks 212

14.5 Drucker’s Definition of Stability 21314.6 Ilıushin’s Postulate 21514.7 Work-Hardening Rules 216

14.7.1 Perfectly Plastic Material 21614.7.2 Isotropic Hardening 216

x Contents

14.7.3 Kinematic Hardening 21714.7.4 Hencky’s Deformation Theory 219

14.8 Endochronic Theory of Valanis 21914.9 Plasticity and Damage 22114.10 Minimum Dissipation Rate Principle 224

Author Index 229

Subject Index 231

Preface

This text is based on a one-semester course I have been teaching at the Illinois In-stitute of Technology for about 30 years. Graduate students from mechanical andaerospace engineering, civil engineering, chemical engineering, and applied math-ematics have been the main customers. Most of the students in my course havehad some exposure to Newtonian fluids and linear elasticity. These two topics arecovered here, neglecting the large number of boundary-value problems solved inundergraduate texts. On a number of topics, it becomes necessary to sacrifice depthin favor of breadth, as students specializing in a particular area will be able to delvedeeper into that area with the foundation laid out in this course. Space and timeconstraints prevented the inclusion of classical topics such as hypoelasticity and elec-tromagnetic effects in elastic and fluid materials and a more detailed treatment ofnonlinear viscoelastic fluids.

I have included a small selection of exercises at the end of each chapter, andstudents who attempt some of these exercises will benefit the most from this text.Instructors may add reading assignments from other sources.

Instead of placing all the references at the end of the book, I have given thepertinent books and articles relevant to each chapter at the end of that chapter.There are some duplications in this mode of presentation, but I hope it is moreconvenient. The Introduction is followed by a list of books on continuum mechanicsintended for students looking for deeper insight. Students are cautioned that thereare a variety of notations in the literature, and it is recommended that they read thedefinitions carefully before comparing the equations. In particular, my definitionof the deformation gradient happens to be the transverse of that found in manybooks. The “comma” notation for partial derivatives, which is commonly used inmechanics, usually confuses students when it comes to the order of the indices in atensor component. I have tried to avoid the “commas” in the beginning while thestudents are learning to manipulate tensor components.

I am grateful to my colleagues: Michael Gosz, who used this text for a courseon continuum mechanics he taught, and Hassan Nagib and Candace Wark for try-ing out the chapter on Cartesian tensors on their fluid dynamics students. I am alsoindebted to Professors Sia Nemat-Nasser and Gil Hegemier for instilling in me an

xi

xii Preface

appreciation for continuum mechanics. A number of students in my classes took itupon themselves to prepare detailed errata for the notes I had distributed. I appreci-ate their input and have tried to incorporate all of their corrections (while, probably,adding some new errors).

The editorial help provided by Peter Gordon, Vicki Danahy, Barbara Walthall,and Cambridge University Press is highly appreciated.

Throughout the preparation of the manuscript my wife, Celeste, has providedconstant encouragement, and I am thankful to her.

Introduction to ContinuumMechanics

1 Introduction

Mechanics is the study of the behavior of matter under the action of internal andexternal forces. In this introductory treatment of continuum mechanics, we acceptthe concepts of time, space, matter, energy, and force as the Newtonian ideals. Hereour objective is the formulation of engineering problems consistent with the fun-damental principles of mechanics. To paraphrase Professor Y. C. Fung—there aregenerally two ways of approaching mechanics: One is the ad hoc method, in whichspecific problems are considered and specific solution methods are devised that in-corporate simplifying assumptions, and the other is the general approach, in whichthe general features of a theory are explored and specific applications are consideredat a later stage. Engineering students are familiar with the former approach fromtheir experience with “Strength of Materials” in the undergraduate curriculum. Thelatter approach enables them to understand an entire field in a systematic way in ashort time. It has been traditional, at least in the United States, to have a course incontinuum mechanics at the senior or graduate level to unify the ad hoc conceptsstudents have learned in the undergraduate courses. Having had the knowledge ofthermodynamics, fluid dynamics, and strength of materials, at this stage, we look atthe entire field in a unified way.

1.1 Concept of a Continuum

Although mechanics is a branch of physics in which, according to current develop-ments, space and time may be discrete, in engineering the length and time scalesare orders (and orders) of magnitude larger than those in quantum physics and weuse space coordinates and time as continuous. The concept of a continuum refers tothe treatment of matter as continuous. The justification for this, again, rests on thelength scales involved. For example, consider a large volume V of air under con-stant pressure and temperature. Within this volume, visualize a small volume �Vcentered at a fixed point in space. Let us denote by �M the mass of material in-side �V. The ratio �M/�V is the average density ρ. However, if we shrink �V,we can imagine a state in which molecules of oxygen and nitrogen pass through it,and the concept of density loses its meaning. If the dimension of �V is kept large

1

2 Introduction

�M�V

�V

ρ

ε

Figure 1.1. Density variation with volume.

compared with the mean free path of the molecules, we could use the concept ofdensity. Mathematically, we say, “let �V → 0,” but, physically, it is still kept abovesome value ε � 0. We may also think of the situation as the discrete particles of mat-ter are approximated as a continuous “smeared” state. Figure 1.1 shows a sketch ofthe limiting process.

To distinguish moving matter from its fixed-background space we use the termmaterial particle, which is not to be confused with molecules or atoms.

1.2 Sequence of Topics

Prior to the consideration of mechanics topics such as stress and strain, the mathe-matical apparatus needed for our work is briefly reviewed. The fixed space in whichthe continuum moves is a three-dimensional (3D) Euclidean space. Cartesian ten-sors are essential for describing the deformation, motion, and the forces in mechan-ics within a Cartesian coordinate system. This topic is considered in Chapter 2. Al-though general tensors are not required for our studies, an understanding of thistopic is worthwhile to appreciate its connection to geometry and mechanics. A num-ber of advanced works in continuum mechanics use the general tensor formulationthat is invariant under coordinate transformations involving curvilinear coordinates.Chapter 3 introduces some of the basic properties of general tensors.

Integral theorems of calculus, namely, the theorems of Gauss, Green, andStokes, are extremely useful for our study of mechanics. These theorems are knownto students from their studies of calculus. In Chapter 4 we visit these theorems byusing index notation.

The next two chapters, Chapter 5 and Chapter 6, deal with the descriptions ofthe geometry of deforming bodies. Various strain measures and strain rate quanti-ties are introduced in these chapters.

A separate chapter, Chapter 7, is devoted to a discussion of the fundamentalaxioms of mechanics. For a proper introduction of the stress tensor in Chapter 8,the fundamental axiom dealing with the balance of momentum is essential.

Suggested Reading 3

Thermodynamics plays an important role not only in restricting the form ofstress–strain relations and stress–strain rate relations but also in explaining ther-momechanical coupling. Chapter 9 refreshes the readers’ knowledge in thermo-dynamics.

General forms of constitutive relations and their admissible forms are discussedin Chapter 10.

Chapter 11 considers elastic materials. It starts with nonlinear elastic materials,presents some of the classic inverse solutions, and then proceeds to linear elasticmaterials. A section on rubber elasticity is included because of the importance ofthis topic in engineering applications.

Chapter 12 deals with fluid dynamics. Again, classic inverse solutions of non-linear fluids are presented first. Newtonian fluids and Navier–Stokes equations arebriefly discussed.

Most students are unfamiliar with viscoelasticity and plasticity. These two topicsare dealt with in Chapters 13 and 14. The treatment is of an elementary nature asthis is assumed to be the first exposure of these two topics.

A digression into the available numerical solution techniques applicable to non-linear problems is not made. The finite-element method has become the methodof choice to deal with the solutions of solid mechanics problems. Finite-differencemethods are often used in fluid dynamics problems. Other methods, such as molec-ular dynamics and Monte Carlo methods (see Frenkel and Smit, 2002), are actuallyoutside the domain of continuum mechanics. However, these two methods play cru-cial roles in illuminating the foundations of irreversible thermodynamics (see Your-grau, van der Merwe, and Raw, 1966) and continuum mechanics. Students are en-couraged to supplement the topics covered here with courses in statistical mechanics(see Chandler, 1987) and related numerical methods.

SUGGESTED READING

Batchelor, G. K. (1967). An Introduction to Fluid Dynamics, Cambridge University Press.Batra, R. (2005). Elements of Continuum Mechanics, AIAA Publishers.Chandler, D. (1987). Introduction to Modern Statistical Mechanics, Oxford University Press.Eringen, A. C. (1962). Nonlinear Theory of Continuous Media, McGraw-Hill.Frenkel, D. and Smit, B. (2002). Understanding Molecular Simulation, Academic.Fung, Y. C. (1965). Foundations of Solid Mechanics, Prentice-Hall.Jaunzemis, W. (1967). Continuum Mechanics, Macmillan.Malvern, L. E. (1969). Introduction to the Mechanics of a Continuous Medium, Prentice-Hall.Reddy, J. N. (2008). An Introduction to Continuum Mechanics. Cambridge University Press.Truesdell, C. (1960). Principles of Continuum Mechanics, Field Research Laboratory, Socony

Mobil Oil Co., Dallas, TX.Truesdell, C. and Noll, W. (1965). The nonlinear field theories of mechanics, in Encyclopedia

of Physics (S. Flugge, ed.), Springer-Verlag, Vol. 3/3.Truesdell, C. and Toupin, R. A. (1960). The classical field theories, in Encyclopedia of Physics

(S. Flugge, ed.), Springer-Verlag, Vol. 3/1.Yourgrau, W., van der Merwe A., and Raw, G. (1966). Treatise on Irreversible and Statistical

Thermodynamics, Macmillan.

2 Cartesian Tensors

When the coordinates used to describe the geometry and deformation of a con-tinuum and the forces involved are Cartesian, that is, three mutually orthogonal,right-handed coordinates with the Euclidean formula for distances, the quantitiesentering the equations of motion are conveniently described by use of the Cartesiantensors. Before we familiarize ourselves with these, let us examine a few related top-ics. These topics are included here primarily to establish our notation and to refreshthe concepts the students might have seen in other contexts.

2.1 Index Notation and Summation Convention

Index notation uses coordinates x1, x2, and x3 to denote the classical x, y, and zcoordinates, respectively. The components of a vector v would be v1, v2, and v3

(in three dimensions), instead of the conventional u, v, and w. As far as matrixelements are concerned, index notation, such as A23 to identify the element in thesecond row and third column, has been in use for some time. The advantage of indexnotation, in conjunction with the summation convention, is that we can shorten longmathematical expressions.

Consider a system of M equations, in N unknowns:

A11x1 + A12x2 + · · · + A1NxN = c1,

A21x1 + A22x2 + · · · + A2NxN = c2,

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . = , . . . ,

AM1x1 + AM2x2 + · · · + AMNxN = cM.

(2.1)

This system of equations can also be written as

N∑j=1

Ai j xj = ci (i = 1, 2, . . . , M; j = 1, 2, . . . , N).

In accordance with the Einstein summation convention we can further simplifythe notation by writing

Ai j xj = ci (i = 1, 2, . . . , M; j = 1, 2, . . . , N), (2.2)

4

2.1 Index Notation and Summation Convention 5

where summation on the repeated index j is implied. Here, i is called a free indexand j is called a dummy index. In dealing with three-dimensional (3D) Euclideanspace, we have indices ranging from 1 to 3 (i.e., M = N = 3). Whenever an index isrepeated once (and only once), the terms have to be summed with respect to thatindex. For this convention to be effective, extreme care must be taken to avoid theoccurrence of any index more than twice. As examples, we have

Aii = A11 + A22 + A33,

Ai j Bi j = Bi j Ai j = Bji Aji .

The symmetry of an array can be expressed as

Ai j = Aji . (2.3)

If an array is skew symmetric (also called antisymmetric),

Bi j = −Bji . (2.4)

An arbitrary array C can be expressed as the sum of a symmetric and a skew-symmetric array:

Ci j = Ai j + Bi j , (2.5)

where

Ai j = C(i j) = 12

(Ci j + C ji ), Bi j = C[i j] = 12

(Ci j − C ji ). (2.6)

The subscripts inside the parentheses and square brackets help us to avoid introduc-ing new variables A and B.

There are rare occasions when we would like to suppress the summation con-vention. Suppose we want to refer to A11, A22, or A33; if we use Aii we get the sum ofthe three terms. We may underline the repeated index to suppress the summation:

Aii = A11, A22, or A33. (2.7)

When we need to substitute one formula into another, we have to make sure thatthe dummy indices are distinct. For example,

ai = Ci j bj , bj = ci Dji . (2.8)

A direct substitution for bj in the first equation shows

ai = Ci j ci Dji , (2.9)

where i appears three times on the right-hand side, violating the summation rule.To avoid this, first we write

bj = ckDjk, (2.10)

and then substitute in the first equation to get

ai = Ci j ckDjk = Ci j Djkck, (2.11)

6 Cartesian Tensors

where j and k are summation indices (or dummy indices) and the free index i ap-pears on both sides. The dummy indices are similar to dummy variables used inintegration of functions.

2.2 Kronecker Delta and Permutation Symbol

These two notations are extremely useful in connection with Cartesian coordinates.The Kronecker delta is defined as

δi j ={

1, if i = j0, if i �= j

. (2.12)

This definition assumes that i and j are explicit integers, such as i = 1 and j = 3,and it does not imply δi i = 1. Elements of the Kronecker delta correspond to theelements of the identity matrix

I = [δi j ] =

⎡⎢⎣ 1 0 0

0 1 00 0 1

⎤⎥⎦ . (2.13)

With this,

δi i = 3, δi jδ jk = δik, δi j Ajk = Aik. (2.14)

The permutation symbol or alternator is defined as

ei jk =

⎧⎪⎨⎪⎩

1, if i, j, k are even permutations of 1, 2, 3−1, if i, j, k are odd permutations of 1, 2, 3

0, otherwise.(2.15)

From the preceding definition we have

ei jk = e jki = eki j = −eikj = −e jik = −ekji , (2.16)

ei j j = e ji j = e j ji = 0. (2.17)

Explicitly we have

e123 = e231 = e312 = 1, e213 = e321 = e132 = −1. (2.18)

Figure 2.1 shows the order of the indices for even and odd permutations.

1 1

23 23

Figure 2.1. Even and odd permutations of the integers 1, 2,and 3.

2.2 Kronecker Delta and Permutation Symbol 7

Using the permutation symbol, we may express the determinant of a 3 × 3 – matrixA as

det A = |A| = A11 A22 A33 + A12 A23 A31 + A13 A21 A32

− A11 A23 A32 − A12 A21 A33 − A13 A22 A31

= ei jk A1i A2 j A3k

= ei jk Ai1 Aj2 Ak3

= 16 ei jke�mn Ai� Ajm Akn.

(2.19)

We can see the second and third equations as the (first) row expansion and the(first) column expansion of the determinant and the last equation as the sum of allrow expansions and all column expansions (which add up to 6), divided by 6.

We can also express the determinant in terms of the cofactors of the matrix. Fora 3 × 3 matrix the cofactor of an element Ai j is the 2 × 2 determinant we obtain byeliminating the ith row and jth column and multiplying it by (−1)i+ j . Let us denotethis cofactor by A∗

i j . Then

|A| = Ai j A∗i j (no sum on i). (2.20)

Observing that A∗i j does not contain Ai j itself (recall, we eliminated a row and a

column), we find

A∗i j = ∂|A|

∂Ai j. (2.21)

When the inverse of the matrix exists, we have

A−1i j = A∗

j i

|A| = 1|A|

∂|A|∂Aji

. (2.22)

A relation between the permutation symbols and the Kronecker deltas, knownas the “e–δ identity,” is useful in algebraic simplifications:

ei jkemnk = δimδ jn − δinδ jm. (2.23)

From this, when n = j , we get

ei jkemjk = 2δim, (2.24)

and, further, when m = i ,

ei jkei jk = 6. (2.25)

2.2.1 Example: Skew Symmetry

If Ai j is a skew-symmetric matrix, solve the system of equations

ei jk Ajk = Bi . (2.26)

8 Cartesian Tensors

We use the “e–δ” identity to get

eimnei jk Ajk = eimn Bi ,

(δmjδnk − δmkδnj )Ajk = eimn Bi ,

Amn − Anm = eimn Bi ,

Amn = 12

eimn Bi . (2.27)

where we use the skew-symmetry property Anm = −Amn.

2.2.2 Example: Products

If Ai j is symmetric and Bi j is skew-symmetric, show that Ai j Bi j = 0.Let

S = Ai j Bi j . (2.28)

If we interchange the dummy indices i → j and j → i , we have

S = Aji Bji . (2.29)

Using the symmetry of A and the skew symmetry of B, we can write this as

S = Ai j (−Bi j ) = −Ai j Bi j = −S, 2S = 0, S = 0. (2.30)

2.3 Coordinate System

As shown in Fig. 2.2, we use a proper (right-handed) Cartesian coordinate systemwith x1, x2, and x3 denoting the three axes. A directed line segment from the originto any point in this 3D space is called a position vector r , with components x1, x2,

x1

x2

x3

r

Figure 2.2. Cartesian coordinate system.

2.4 Coordinate Transformations 9

and x3 along the three axes. We also use the notation x instead of r at times. Inmatrix notation, using a column vector, we have

r = x =

⎧⎪⎨⎪⎩

x1

x2

x3

⎫⎪⎬⎪⎭ . (2.31)

Defining unit vectors (or base vectors) e1, e2, and e3 as

e1 =

⎧⎪⎨⎪⎩

100

⎫⎪⎬⎪⎭ , e2 =

⎧⎪⎨⎪⎩

010

⎫⎪⎬⎪⎭ , e3 =

⎧⎪⎨⎪⎩

001

⎫⎪⎬⎪⎭ , (2.32)

we can write

r = x1e1 + x2e2 + x3e3. (2.33)

The dot and cross products of the unit vectors can be expressed with theKronecker delta and the permutation symbol in the form

ei · e j = δi j , ei × e j = ei jkek. (2.34)

2.4 Coordinate Transformations

Two types of coordinate transformations are encountered frequently in our studies:coordinate translation and coordinate rotation. If we denote the new coordinates ofa point P by x′

i , in the case of translation, the two systems are related in the form

x′i = xi + hi , (2.35)

where hi are constants. This is shown in Fig. 2.3.

P

x1

x2

x′1

x′2

h1

h2

Figure 2.3. Coordinate translation.

10 Cartesian Tensors

x1

x2

x′1

x′2

θ

r

Figure 2.4. Coordinate rotation.

Let us consider first a rotation of coordinates in two dimensions. The originremains fixed and we obtain the x′

1 axis by rotating the x1 axis counterclockwise byan angle, θ . As shown in Fig. 2.4, the two systems are related in the form

x′1 = x1 cos θ + x2 sin θ,

x′2 = −x1 sin θ + x2 cos θ.

(2.36)

We obtain the inverse of this transformation by replacing θ with (−θ), as

x1 = x′1 cos θ − x′

2 sin θ,

x2 = x′1 sin θ + x′

2 cos θ.

Equations (2.36) can also be written in matrix form:{x′

1

x′2

}=

[Q11 Q12

Q21 Q22

] {x1

x2

}, (2.37)

where

Q11 = Q22 = cos θ, Q12 = −Q21 = sin θ. (2.38)

Using column vectors x′ and x and the square matrix Q, we have

x′ = Qx. (2.39)

A note of caution is in order at this point. We have absolute vectors in space,such as r , and then we have column representations of the components in a cho-sen coordinate system, such as x and x′. When there are no coordinate rotationsinvolved, we do not have to distinguish these two representations. When there arecoordinate rotations, we use r for the absolute vector.

2.4 Coordinate Transformations 11

Another way to relate x′ to x is by use of two systems of unit vectors, e′i and ei .

Let us do this for the 3D case. The position vector r has two representations:

r = x′1e′

1 + x′2e′

2 + x′3e′

3 = x1e1 + x2e2 + x3e3. (2.40)

Because e′1, e′

2, and e′3 are mutually orthogonal, taking the dot product of the pre-

ceding relation with e′1 gives

x′1 = (e′

1 · e1)x1 + (e′1 · e2)x2 + (e′

1 · e3)x3. (2.41)

The elements of the 3 × 3 Q matrix associated with the rotation are the dot productsof the unit vectors:

Q11 = e′1 · e1, Q12 = e′

1 · e2, Q13 = e′1 · e3. (2.42)

Next, let us do this using the index notation and the summation convention:

r = x′ke′

k = xj e j . (2.43)

Dotting with e′i , we obtain

x′ke′

i · e′k = xj e′

i · e j , x′kδik = Qi j xj ,

or

x′i = Qi j xj , x′ = Qx, (2.44)

where

Qi j = e′i · e j . (2.45)

We may start with Eq. (2.43) and dot it with ei . This would lead to

xi = ei · e′j x

′j . (2.46)

This is the inverse of Eqs. (2.44):

xi = Qji x′j = QT

i j x′j , x = QT x′. (2.47)

Using Eqs. (2.47) in Eqs. (2.44), we find

x′ = QQT x′, (2.48)

where the superscript T denotes transpose. This shows

QQT = I or Q−1 = QT . (2.49)

Matrices that satisfy this relation are called orthogonal matrices. Another way tosee that Q is orthogonal is as follows:

The length of the vector r can be found from

r2 = x′i x

′i = xj xj . (2.50)


Substituting for x′i , we obtain

QimQinxmxn = xi xi or xmQTmi Qinxn = xi xi . (2.51)

We can rewrite this in matrix notation as

xT QTQx = xT x or QTQ = I, (2.52)

QTikQkj = δi j or Qki Qkj = δi j , (2.53)

where we have used the fact that x is arbitrary. Thus

QT = Q−1. (2.54)

As Qi j �= Qji , we associate the first index i with the e′ system and j with the esystem.

Also, the unit vectors e′i are related to e j in the form

e′i = Qi j e j , (2.55)

which we can verify by taking the dot product with ek.

2.5 Vectors

The mathematical definition of vectors as directed line segments with their startingpoint fixed at the origin enables us to express any vector v in a chosen coordinatesystem as

v = vi ei , (2.56)

where vi are the components of the vector in that system.In another system with unit vectors e′, we will have

v = v′i e

′i . (2.57)

From our examination of the coordinate transformation, it is clear that the twosets of components are related in the form

v′i = Qi jv j . (2.58)

The dot product (inner product or scalar product) of two vectors u and v is definedas

u · v = ui ei · v j e j = uiv j ei · e j = uiv jδi j = uivi . (2.59)

The cross product (vector product) of these vectors is defined as

u × v = ui ei × v j e j = uiv j ei × e j = uiv j ei jkek. (2.60)

In physics and engineering there are many 3 × 1 arrays that transform from onecoordinate system to another according to the transformation rule of Eq. (2.58).We call them vectors. Examples include position vectors, velocities, forces (relativeto their point of application), and gradients of scalar functions. That brings us to

2.5 Vectors 13

scalars. These are quantities with no associated directions. Examples include tem-perature, hydrostatic pressure, and density of matter.

A scalar function ψ can be written as

ψ = ψ(x1, x2, x3). (2.61)

In a different coordinate system, using

xi = Qji x′j , (2.62)

we obtain

ψ(x1, x2, x3) = ψ(Qj1x′j , Qj2x′

j , Qj3x′j ) = ψ ′(x′

1, x′2, x′

3), (2.63)

which shows that the functional forms of ψ and ψ ′ are different. The numerical valueof the function at a given point obtained from ψ(x′

1, x′2, x′

3) or ψ(x1, x2, x3) will bethe same.

When we examine the transformation of the derivatives of a scalar function, weuse the chain rule of differentiation to get

∂ψ ′

∂x′i

= ∂ψ

∂xj

∂xj

∂x′i. (2.64)

From Eq. (2.62),

∂xj

∂x′i

= Qi j . (2.65)

Thus,

∂ψ ′

∂x′i

= ∂ψ

∂xjQi j . (2.66)

So the array formed by the three derivatives of ψ obeys the transformation rulefor a vector. We define the gradient vector as

∇ψ =

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

∂ψ

∂x1

∂ψ

∂x2

∂ψ

∂x3

⎫⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎭

= e1∂ψ

∂x1+ e2

∂ψ

∂x2+ e3

∂ψ

∂x3. (2.67)

We can also introduce a vector operator,

∇ = ei∂i , (2.68)

where

∂i = ∂

∂xi. (2.69)

In the x′ system, we could use

∂ ′i = ∂

∂x′i. (2.70)


To simplify our labor, we can express functions of a set of independent variablesthat belongs to an unambiguous family by using the condensed form

ψ(x) = ψ(x1, x2, x3). (2.71)

2.6 Tensors

In 3D space, the nine quantities Ti j are called the components of a tensor T ofrank 2 if they obey the transformation rule,

T ′i j = QimQjnTmn, (2.72)

when the two coordinate systems are related by the rotation

x′i = Qi j xj .

The number of indices shows the rank of a tensor. We call T a second-rank tensor.In three dimensions a second-rank tensor has 32 components, and in two dimensionsit has 22 components.

We can introduce a tensor of rank r in three dimensions with 3r components,

Ti1i2...ir . (2.73)

The transformation rule for these components will have r number of Q’s in the form

T ′i1i2...ir

= Qi1 j1 Qi2 j2 · · · Qir jr Tj1 j2... jr . (2.74)

From a heuristic definition of vectors as directed line segments in space, it isclear that these are independent of the coordinate system. However, to define thecomponents of a vector we need a frame of reference. If we know the representationof a vector in one coordinate system, we can use the transformation rule to findthem in any other system. Similarly, the tensor T is an absolute quantity, and itscomponent representation depends on the coordinate system. Again, when we knowits components in one system, we use the transformation rule to obtain them inanother system of coordinates.

In continuum mechanics, most of the time, we deal with 3D tensors of rank 2.From the relation between rank r and the number of Q factors in the transforma-tion, it is clear that vectors are tensors of rank 1 and scalars are those of rank 0. Inour studies we refer to these as vectors or scalars, and we reserve the term “tensor”for tensors of rank 2 and higher.

When we encounter an array of 3r elements, without examining how they trans-form from one coordinate system to another, we cannot conclude we have a tensorof rank r . We often encounter various arrays in spreadsheets (for example, home-work grades); these are, obviously, not tensors!

2.6.1 Examples of Tensors

In this subsection we examine three examples of tensors.

2.6 Tensors 15

We may construct second-rank tensors from vectors by using the idea of tensorproducts. We have seen the dot product and the cross product of two vectors. Thetensor product of two vectors u and v is written as

u ⊗ v = ui ei ⊗ v j e j = uiv j ei ⊗ e j . (2.75)

This operation produces a 3 × 3 array, with the coefficient of, say, e2 ⊗ e3 enteringthe second row and third column. Let

w = u ⊗ v, wi j = uiv j . (2.76)

In a new coordinate system,

w′i j = u′

iv′j = QimumQjnvn = QimQjnwmn. (2.77)

Thus the components of w obey the transformation law for second-rank tensors. Inour studies, we use the dyadic notation that implicitly assumes the tensor productsymbol, ⊗, between two vectors if a dot or cross product symbol is absent. That is,

uv = uiv j ei e j = uiv j ei ⊗ e j = u ⊗ v. (2.78)

The transpose of the tensor w can be written as

wT = wTi j ei e j = w j i ei e j = wi j e j ei . (2.79)

In matrix notation, if we interpret u and v as column vectors, the tensor productgives the matrix

w = uvT . (2.80)

For tensors of a rank higher than 2, we do not have this type of correspondencewith matrices. Most of the operations with second-rank tensors we encounter canbe carried out in matrix notation.

A second example of constructing tensors is given by use of the gradient opera-tor ∇. We have

∇∇ψ = ei∂i e j∂ jψ = ei e j∂2ψ

∂xi∂xj. (2.81)

For a vector-valued function φ, we have

φ = ei φi ,

∇ · φ = ∂iφi (divergence),∇ ×φ = ∂iφ j ei jkek (curl),

∇φ = ∂iφ j ei e j (gradient).

(2.82)

In fluid dynamics the velocity vector v, its divergence ∇ · v, its curl ∇ × v, and itsgradient ∇v all play significant roles.

As a third example, we consider the stress tensor σ . In French, the word tensionmeans stress and the word tensor originated from tension to denote a set of quan-tities that transforms as stress components under a coordinate transformation. Forsimplicity, let us consider the classical force balance of a triangle OAB, as shown in


x1

x2

x′1

x′2

σ11

σ12

σ21

σ22

σ ′11

σ ′12

θ

A

B

O

Figure 2.5. Components of stress in two di-mensions.

Fig. 2.5. With OA = AB sin θ and OB = AB cos θ , the stress components σ11, σ12,σ21, and σ22 in the x1, x2 coordinate system are related to the stresses σ ′

11, σ ′12, σ ′

21,and σ ′

22, by the relations

σ ′11 = σ11 cos2 θ + σ22 sin2 θ + (σ12 + σ21) sin θ cos θ,

σ ′12 = σ12 cos2 θ − σ21 sin2 θ − (σ11 − σ22) sin θ cos θ,

σ ′22 = σ11 sin2 θ + σ22 cos2 θ − (σ12 + σ21) sin θ cos θ,

σ ′21 = σ21 cos2 θ − σ12 sin2 θ − (σ11 − σ22) sin θ cos θ, (2.83)

where we have assumed a unit thickness for the triangular lamina and we obtainthe last two relations from the first two by setting θ → θ + π/2. This system can bewritten in terms of matrices as[

σ ′11 σ ′

12

σ ′21 σ ′

22

]=

[cos θ sin θ

− sin θ cos θ

] [σ11 σ12

σ21 σ22

] [cos θ − sin θ

sin θ cos θ

], (2.84)

or

σ ′ = Qσ QT . (2.85)

Using index notation, we have

σ ′i j = Qimσmn QT

nj = QimQjnσmn. (2.86)

Thus the stress components do transform in accordance with the transformation lawfor Cartesian tensors of rank 2. Using tensor notation, we have

σ ′ = σi j e′i e

′j , σ = σi j ei e j , Q = Qi j ei e j , (2.87)

σ ′ = Q · σ · QT . (2.88)

2.6 Tensors 17

When the tensors involved are of the second rank or lower, we use matrix notationand write the preceding relation as

σ ′ = Qσ QT . (2.89)

The following results are useful in dealing with tensor equations:

1. If all the components of a tensor vanish in one coordinate system, they vanishin all coordinate systems.

2. The sum and difference of two Cartesian tensors of rank r in N dimensions arealso Cartesian tensors of rank r in N dimensions.

On the basis of the preceding results, the importance of tensor analysis may be sum-marized as follows: The form of an equation can have general validity with respectto any frame of reference only if every term in the equation has the same tensorcharacteristics.

2.6.2 Quotient Rule

Consider an array of N3 quantities A(i, j, k), with each index ranging over1, 2, . . . , N. We do not know if it is a tensor. Suppose we know the product ofA(i, j, k) with an arbitrary tensor behaves as a tensor. Then we can use the quotientrule to establish the nature of A(i, j, k). For example, let vi be components of avector, and if the product A(i, j, k)vi is known to yield a tensor of rank 2, Bjk, then

A(i, j, k)vi = Bjk. (2.90)

In a different coordinate system

A′(i, j, k)v′i = B′

jk. (2.91)

Using the transformation rule for the known tensors, we obtain

A′(i, j, k)v′i = QjmQkn Bmn

= QjmQkn A(l, m, n)vl

= QjmQkn A(l, m, n)Qilv′i ,

(2.92)

[A′(i, j, k) − QjmQkn Qil A(l, m, n)]v′i = 0. (2.93)

Because v′i are arbitrary, it follows that the quantity inside the brackets must be zero

for all values of i . That is,

A′(i, j, k) = QjmQkn Qil A(l, m, n). (2.94)

Thus A is a third-rank tensor. This is known as the quotient rule.


2.6.3 Inner Products: Notation

Consider two second-rank tensors, A and B. The product Ai j Bjk can be expressedin tensor notation as

Ai j Bjkei ek = A· B. (2.95)

In matrix notation, we have

Ai j Bjkei ek = AB. (2.96)

The matrix multiplication rule implements the inner product automatically. Theproduct Ai j Bji that we obtain by letting k = i can be written in the matrix nota-tion as

Ai j Bji = Tr(AB), (2.97)

and we use the double-dot notation

Ai j Bji = A : B (2.98)

in the tensor form.

2.7 Quadratic Forms and Eigenvalue Problems

Associated with a second-rank tensor (or matrix), a homogeneous quadratic formcan be defined:

A = Ai j xi xj = xT Ax. (2.99)

If Ai j is unsymmetrical, when we construct the quadratic form the coefficient of xi xj

comes out to (Ai j + Aji )/2. Thus we need to consider only symmetric matrices inconnection with quadratic forms. In other words, if an unsymmetrical matrix Ai j isgiven, first we convert it to a symmetric matrix A(i j) when dealing with quadraticforms. The function A is called a homogeneous function of the second degree as ittakes the value µ2 A if all xi are replaced with µxi . We often encounter problemsin which we have to find the extremum or stationary value of A with respect toxi with the added normalization constraint |x| = 1. We can perform constrainedextremization by defining a modified function A∗ using a Lagrange multiplier λ inthe form

A∗ = A − λ(xi xi − 1) (2.100)

and setting its derivatives with respect to xi to zero. This gives

∂A∗

∂xi= 2 (Ai j xj − λxi ) = 0. (2.101)

This system of equations can be put in the form

Ax = λx or [A− λI]x = 0. (2.102)

2.7 Quadratic Forms and Eigenvalue Problems 19

This homogeneous system of simultaneous equations always has a trivial solutionx = 0 that does not satisfy our normalization constraint |x| = 1. Recall that, for anontrivial solution, we require

det[A− λI] = 0. (2.103)

For a given matrix A, this is possible for only special, discrete values of λ. Assumingwe have a 3 × 3 matrix, when the preceding determinant is expanded, we get a cubicequation known as the characteristic equation of the matrix A:

−λ3 + IA1λ2 − IA2λ + IA3 = 0, (2.104)

where the coefficients IA1, IA2, and IA3 can be written as

IA1 = Aii , IA2 = 12

(Aii Aj j − Ai j Ai j ), IA3 = det A. (2.105)

These coefficients are also known as the three invariants of the matrix A. If wedenote the three roots of cubic equation (2.104) by λ1, λ2, and λ3, we see

IA1 = λ1 + λ2 + λ3,

IA2 = λ1λ2 + λ2λ3 + λ3λ1,

IA3 = λ1λ2λ3.

(2.106)

The significance of the term invariant will be clear if we use a new coordinate systemx′

i that we obtain by rotating the xi system. Let

x′i = Qi j xj or x = QT x′. (2.107)

Using this in the quadratic form, Eq. (2.99), we get

A = x′T QAQT x′ = x′T A′x′, (2.108)

where

A′ = QAQT or A = QT A′ Q. (2.109)

Substituting for A in Eq. (2.103), we get

det[ QT A′ Q − λI] = 0,

det[ QT A′ Q − λ QT I Q] = 0,

det[ QT (A′ − λI) Q] = 0,

det QT det(A′ − λI)det Q = 0,

det(A′ − λI) = 0,

(2.110)

where we have used det Q = det QT �= 0. From this, we have the characteristicequation

−λ3 + IA′1λ2 − IA′2λ + IA′3 = 0. (2.111)


Comparing this equation with Eq. (2.104) and noting that the three roots of thecubic are the same, we clearly see that

IA1 = IA′1, IA2 = IA′2, IA3 = IA′3. (2.112)

Thus these three quantities remain invariant under coordinate rotation.If we substitute λ = λ1 in Eq. (2.102), the determinant of the system is zero,

which implies that only two of the three equations are independent. We can solvethe system to get a nonunique solution with one arbitrary parameter. The normal-ization condition |x| = 1 determines the value of the parameter. The vector foundthis way is called an eigenvector of A. It is known that, for a symmetric matrix,all eigenvalues are real and the eigenvectors corresponding to distinct eigenvaluesare orthogonal. Further, even when the eigenvalues are not distinct, three mutuallyorthogonal eigenvectors can be constructed if the matrix is symmetric. These eigen-vectors are called the principal directions of matrix A, and the eigenvalues are theprincipal values. Corresponding to the three eigenvalues λi , the three eigenvectorsare denoted by x(i). An interesting but simple relation can be found between the ex-tremum values of the quadratic form A and the Lagrange multipliers (eigenvalues)as follows: Multiply the equation

Ax(1) = λ1x(1) (2.113)

by x(1)T to get

A = x(1)T Ax(1) = λ1x(1)T x(1) = λ1. (2.114)

Thus the three eigenvalues are the extremum values of the quadratic form along theprincipal directions. If all the eigenvalues are positive, the quadratic form is calledpositive definite and the matrix is called a positive-definite matrix. In the 3D space,we can visualize an ellipsoid corresponding to the quadratic form, with its majoraxis, minor axis, and an “in-between” axis lying along the principal directions.

An interesting property of any square matrix is given by the Cayley–Hamiltontheorem, which states that matrix A satisfies its own characteristic equation. For a3 × 3 matrix, we have

−A3 + IA1 A2 − IA2 A+ IA3 I = O, (2.115)

where the quantity on the right-hand side is a square null matrix. As a consequence,A3 can be expressed in terms of lower powers of A, and when this step is usediteratively, any matrix polynomial can be reduced to a quadratic for the 3 × 3 case.

2.7.1 Example: Eigenvalue Problem

Obtain the eigenvalues and eigenvectors of the matrix

A = 12

⎡⎢⎣ 7 1 0

1 7 00 0 4

⎤⎥⎦ . (2.116)


The characteristic equation is obtained from

det(A− λI) = 0, (2.117)

as

[(7 − 2λ)2 − 1][4 − 2λ] = 0, or − λ3 + 9λ2 − 26λ + 24 = 0. (2.118)

The invariants of the matrix are

IA1 = 9, IA2 = 26, IA3 = 24. (2.119)

From the first of the equations in Eq. (2.118), we see the eigenvalues are

λ1 = 4, λ2 = 3, λ3 = 2, (2.120)

and the corresponding eigenvectors are

x(1) = 1√2

⎧⎪⎨⎪⎩

110

⎫⎪⎬⎪⎭ , x(2) = 1√

2

⎧⎪⎨⎪⎩

−110

⎫⎪⎬⎪⎭ , x(3) =

⎧⎪⎨⎪⎩

001

⎫⎪⎬⎪⎭ . (2.121)

2.7.2 Diagonalization and Polar Decomposition

Using the three eigenvectors as the three columns, we can construct the modal ma-trix of the given matrix A. Let us denote the modal matrix by M. Then,

M = [x(1), x(2), x(3)]. (2.122)

Then it is clear that

MT AM =

⎡⎢⎣λ1 0 0

0 λ2 00 0 λ3

⎤⎥⎦ . (2.123)

This procedure amounts to choosing a new Cartesian system aligned with the prin-cipal directions of the quadratic. That is, we choose a principal coordinate systemx′

i , using

xi = Mi j x′j or x′ = MT x. (2.124)

Multiplying Eq. (2.123) by M from the left and by MT from the right, we canexpress matrix A as

A =3∑

i=1

λi x(i)x(i)T . (2.125)

This is called the spectral representation of A.So far we have been concerned with a symmetrical matrix A. The polar decom-

position applies to any square matrix B. Polar decomposition refers to factoring Bin the form

B = RU or B = VR, (2.126)

where U and V are symmetric matrices and R is an orthogonal (rotation) matrix.


Using RRT = I, we get

U2 = BT B and V2 = BBT . (2.127)

To find U and V, we need the square roots of the matrices on the right-hand sidesof the preceding equations. After U or V is found, R can be expressed as

R = BU−1 = V−1 B. (2.128)

This brings us to finding the square root of a symmetric matrix, say C or, ingeneral, any function F[C] of a matrix. We begin by assuming the function F[C]has a converging infinite series expansion in C:

F[C] =∑

0

ai Ci . (2.129)

The functions we have in mind are C1/2, sin[C], exp[C], etc. The corresponding func-tions of a single variable, say x, are x1/2, sin x, exp x, etc. The generic matrix functionF has the corresponding function of a single variable F . Our symmetric matrix Chas three orthogonal eigenvectors x(i) with the corresponding eigenvalues λi . Mul-tiplying Eq. (2.129) from the right by x(i), we see that F (λi ) is the eigenvalue of Fcorresponding to the eigenvector, x(i). The eigenvalues of [C]1/2, sin[C], and exp[C]are λ

1/2i , sin λi , and exp λi , respectively. Next, we use the Cayley–Hamilton theorem

to reduce Eq. (2.129) to a quadratic in C (provided we are still dealing with 3 × 3matrices):

F[C] = c0 I + c1C + c2C2. (2.130)

Then the eigenvalues satisfy

F (λi ) = c0 + c1λi + c2λ2i , i = 1, 2, 3. (2.131)

For example, to find the square root of C, we use

λ1/2i = c0 + c1λi + c2λ

2i , i = 1, 2, 3, (2.132)

and solve for ci . We substitute these coefficients in Eq. (2.130) to get C1/2. We mayanticipate complications when the eigenvalues are not distinct. If λ2 = λ1, we followthe Frobenius limiting process. We begin by assuming λ2 = λ1 + ε. Then, one of thethree equations for ci becomes

F (λ1 + ε) = c0 + c1(λ1 + ε) + c2(λ1 + ε)2. (2.133)

If we subtract the equation corresponding to λ1 and divide by ε, we have

F (λ1 + ε) − F (λ1)ε

= c1 + c2[(λ1 + ε)2 − λ21]/ε. (2.134)

As ε → 0, we obtain the needed additional equation,

F ′(λ1) = c1 + 2c2λ1. (2.135)


All of our discussions in the last two subsections were limited to 3 × 3 matrices.Extension of the preceding results for square matrices of any order is straightfor-ward.

2.7.3 Example: Polar Decomposition

Suppose we want to factor the matrix

B = 15

⎡⎢⎣ 17 −11 0

19 23 00 0 15

⎤⎥⎦ . (2.136)

The factors U, V, and R are obtained from

U2 = C ≡ BT B =

⎡⎢⎣ 26 10 0

10 26 00 0 9

⎤⎥⎦ , R = BU−1, V = BRT . (2.137)

The eigenvalues of C are

λ1 = 9, λ2 = 16, λ3 = 36.

Using the expansion √λi = c0 + c1λi + c2λ

2i ,

we have the simultaneous equations

3 = c0 + 9c1 + 92c2, 4 = c0 + 16c1 + 162c2, 6 = c0 + 36c1 + 362c2.

The solutions of this system are

c0 = 52/35, c1 = 23/126, c2 = −1/630.

Writing

C1/2 = c0 I + c1C + c2C2,

we get

U = C1/2 =

⎡⎢⎣ 5 1 0

1 5 00 0 3

⎤⎥⎦ ,

R = 15

⎡⎢⎣ 4 3 0

−3 4 00 0 5

⎤⎥⎦ ,

V = 125

⎡⎢⎣ 101 7 0

7 149 00 0 75

⎤⎥⎦ .


SUGGESTED READING

Fung, Y. C. (1993). First Course in Continuum Mechanics 3rd ed., Prentice-Hall.Jeffreys, H. (1931). Cartesian Tensors, Cambridge University Press.Knowles, J. K. (1998). Linear Vector Spaces and Cartesian Tensors, Oxford University Press.Temple, G. (1960). Cartesian Tensors: An Introduction, Dover.

EXERCISES

Use index notation and the summation convention wherever necessary in the fol-lowing problems.

2.1. If v is a vector and n is a unit vector, show that

v = (n · v)n + (n × v) × n.

2.2. With the help of the e–δ identity, show that any three vectors a, b, and c satisfy

a × (b × c) = (a · c)b − (a · b)c.

Verify your result by using the method of determinants for the cross products.

2.3. Complete the following identities by obtaining the missing part X.

(a) ∇·(u × v) = v · ∇ × u + X(b) ∇(u · v) = (u · ∇v) + X(c) ∇ × (u × v) = u∇ · v + X

2.4. If the three functions u, v, and w of x1, x2, and x3 satisfy

F (u, v, w) = 0,

show that

∇u · ∇v × ∇w = 0.

2.5. Prove that all the components of a tensor vanish in all Cartesian coordinatesystems if they vanish in one coordinate system.

2.6. Prove that the sum (and difference) of two tensors of rank r in an N-dimensional space is also a tensor of rank r .

2.7. For any four vectors s, t, u, and v, show that

(s × t) · (u × v) = (s · u)(t · v) − (s · v)(t · u).

2.8. Show that

∇ · (u × v) = v · ∇ × u − u · ∇ × v.

2.9. For a twice-differentiable, vector-valued function v, show that

∇ × (∇ × v) = ∇(∇ · v) − (∇ · ∇)v.

2.10. If r = ei xi and r2 = xi xi , show that (Fung, 1993)

(a) ∇ · (rnr) = (n + 3)rn,

(b) ∇ × (rnr) = 0,

Exercises 2.1–2.16 25

(c) ∇ · ∇rn = n(n + 1)r (n−2).

(d) If F is any differentiable function, show that

∇ × [F (r)r] = 0.

2.11. Show that the differential equation

(∇2 + k2)ψ = 0,

in three dimensions, admits the solution

ψ = e±ikr

r.

2.12. If A and B are tensors of rank m and n, respectively, in N dimensions, showthat

(a) the rank of A · B is m + n − 2,(b) the rank of A× B is m + n − 1, and(c) the rank of A

⊗B is m + n.

2.13. If T is a two-dimensional, second-rank, unsymmetrical tensor and n is a two-dimensional vector (two-vector), compare n · T · n, (T · n) · n, and n · (n · T).

2.14. Consider two Cartesian coordinate systems in two dimensions, x′i and xi , with

the x′1 axis rotated by angle θ in the counterclockwise direction from the x1 axis. For

a position vector r , which makes an angle α with the x1 axis, obtain the componentsin the two systems and relate them as

x′ = Qx.

Consider a new position vector s obtained by rotating r by an angle, θ . Obtain thecomponents of s in the x-coordinate system:

s = Px.

Compare the two rotation matrices Q and P.

2.15. Compute the eigenvalues and the normalized eigenvectors of the matrix

C =

⎡⎢⎣ 4 2 0

2 4 00 0 1

⎤⎥⎦ .

Construct the modal matrix M and show that MT CM is a diagonal matrix D, withthe eigenvalues along the diagonal. Obtain the square root of this diagonal matrixD1/2. Show that MD1/2 MT is the square root of C.

2.16. Obtain the factors U, V, and R in the polar decomposition of the matrix

B =

⎡⎢⎣

√3 1 0

0 2 00 0 1

⎤⎥⎦


in the form

B = RU = VR.

2.17. For the matrix

B =[

0 1−1 0

],

show that

eBθ = I cos θ + B sin θ,

where θ is a real number.

2.18. For

C =[

5 40 5

],

compute C10 and sin C.

2.19. In the case of an unsymmetric matrix A, show that A and AT have the sameeigenvalues, λ(i), i = 1, 2, 3, and if x(i) are three independent eigenvectors of A andy i , i = 1, 2, 3 are the eigenvectors of AT (these are also called the left eigenvectorsof A), show that the two sets of eigenvectors can be selected to have

x(i)T y( j) = δi j .

This relation illustrates biorthogonality of two families of vectors.

2.20. Using the notation of the preceding problem, show that A has the spectralrepresentation

A =3∑

i=1

λ(i)x(i)y(i)T ,

when there are three independent eigenvectors x(i).

2.21. Obtain the eigenvalues and the left and right eigenvectors of the matrix

A =

⎡⎢⎣ 1 −1 1

−1 1 −1−1 1 −1

⎤⎥⎦ .

What is the spectral representation for this matrix?

3 General Tensors

There are a number of continuum mechanics books in the literature, written withthe general tensor formulation. An essential characteristic of general tensors is theuse of curvilinear coordinates. We may imagine a Cartesian grid placed inside acontinuum. As the continuum deforms, the grid lines deform with it into a meshof curves. Thus general tensor formulations are convenient to describe continuummechanics. Although this chapter is not required to follow our presentation of con-tinuum mechanics in terms of Cartesian tensors, it will be of value to extend the stu-dents’ knowledge through additional reading. In modern computational mechanics,curvilinear grids are often used and the governing equations in general tensor for-mulation are needed. As an added bonus, students will be able to follow the theoryof shells and the general theory of relativity!

In the previous chapter we saw systems of Cartesian coordinates and transfor-mations of tensors between two coordinates that are related linearly. When we haveto deal with curvilinear coordinates, the coordinate transformations are in generalnonlinear and the coordinates may not form an orthogonal system. Representationof tensors in such a system depends on the directions of local tangents and normalsto the coordinate surfaces. It is conventional to use superscripts to denote the coor-dinate curves. The reason for this will be clarified later. Let us begin with a Cartesiansystem with labels x1, x2, and x3, transforming into curvilinear system ξ 1, ξ 2, and ξ 3.A two-dimensional (2D) illustration of this is shown in Fig. 3.1. We assume that therelation between xi and ξ j ,

xi = xi (ξ 1, ξ 2, ξ 3), (3.1)

and its inverse,

ξ j = ξ j (x1, x2, x3), (3.2)

are locally one-to-one. What is meant by “locally one-to-one” is that in the neigh-borhood of a point, (ξ 1, ξ 2, ξ 3), the increments dxi of the coordinates xi are linearlyrelated to the increments dξ j of ξ j , and these relations can be inverted.

27

28 General Tensors

x1

x2

ξ 1

ξ 2

AB

C

O

Figure 3.1. Distance element in a curvilinear system.

We have

dxi = ∂xi

∂ξ jdξ j (3.3)

and the matrix

L = [Lji ] =[

∂xi

∂ξ j

], (3.4)

which is evaluated at (ξ 1, ξ 2, ξ 3), is nonsingular. We use the notation

L = det[Lji ]. (3.5)

Referring to Fig. 3.1, the distance vector−→OC can be expressed in terms of the

increments in the Cartesian coordinates dxi , as well as in terms of the increments inthe curvilinear coordinates dξ i .

With−−→OA = g1dξ 1,

−−→OB = g2dξ 2, (3.6)

we have the vector sum,−→OC = dxi ei = gi dξ i . (3.7)

From this, the base vector, gi can be related to the Cartesian base vectors ei as

gi = ∂x j

∂ξ ie j = Li j e j . (3.8)

General Tensors 29

A base vector gi is tangential to the curve ξ i . Here the “curve ξ 1” refers to thecurve along which ξ 1 changes while ξ 2 and ξ 3 remain constant. Unlike Cartesiancoordinates, curvilinear coordinates may not have the dimension of length. As anexample, we use angles as coordinates in polar and spherical systems. The termgi dξ i must have the dimension of length, and if dξ i is nondimensional, gi has thedimension of length. In general, then, all the gi may not have the same dimensionsand they are not unit vectors.

The distance between O and C can be expressed in the form

|OC|2 = dxkdxk = gi · g j dξ i dξ j = gi j dξ i dξ j , (3.9)

where we have defined

gi j = gi · g j = ∂xk

∂ξ i

∂xk

∂ξ j, [g] = LLT . (3.10)

The elements gi j are called the components of the Riemann metric tensor g associ-ated with the curvilinear system. In the distance formula, when a Cartesian systemis used we have the Euclidean distance measure dxkdxk, and when a curvilinearsystem is used we have the Riemann measure gi j dξ i dξ j .

As mentioned earlier, g1 is tangent to the coordinate curve ξ 1, and g1 and g2

lie in the tangent plane of the surface formed by ξ 1 and ξ 2 curves. We can constructa normal g3 to this surface with its length adjusted to have g3 · g3 = 1. Completingthis scheme, we obtain three new vectors gi with the properties

gi · g j = δij , (3.11)

which is the Kronecker delta in the general tensor setting. In expanded form,

g1 · g1 = 1, g1 · g2 = 0, g1 · g3 = 0,

g2 · g1 = 0, g2 · g2 = 1, g2 · g3 = 0,

g3 · g1 = 0, g3 · g2 = 0, g3 · g3 = 1. (3.12)

We call gi the covariant base vectors or simply base vectors and gi the con-travariant base vectors or reciprocal base vectors (see Fig. 3.2). Because gi formnormal vectors to the tangent planes, we have

g1 = kg2 × g3, (3.13)

where the constant of proportionality k is found from

g1 · g1 = 1 = kg1 · g2 × g3. (3.14)

As the scalar product or triple product of three vectors is invariant with respect tothe permutations of the vectors, we find that the constant k, given by

k = (g1 · g2 × g3)−1, (3.15)

is invariant if we permute the integers in Eq. (3.13).

30 General Tensors

ξ 1

ξ 2

g1

g2

g1

g2

Figure 3.2. Base vectors and reciprocal base vectors.

Using the Cartesian representation of gi in Eq. (3.8), we find that the scalarproduct is the determinant of L, i.e.,

k = 1/L. (3.16)

Because

gi = Li j e j , (3.17)

we have

g = det[gi j ] = det[gi · g j ] = |LikLjk| = L2. (3.18)

Then

k = 1L

= 1√g, (3.19)

and

g1 = 1√g

g2 × g3, g2 = 1√g

g3 × g1, g3 = 1√g

g1 × g2. (3.20)

Using the permutation symbol, we can write the preceding result as

1√g

gi × g j = ei jkgk. (3.21)

As gi is a vector in the three dimension (3D) space, we can express it as a linearcombination of the base vectors g j ,

gi = gi j g j , (3.22)

3.1 Vectors and Tensors 31

where we find the coefficient gi j by using

gi · g j = δij = gikgkj . (3.23)

Thus the matrix [gi j ] is the inverse of the matrix [gi j ]. That is

gi j g jk = δki . (3.24)

Inverting the system, Eq. (3.22), we get,

gi = gi j g j . (3.25)

The array gi j is called the contravariant or reciprocal metric tensor, and gi are alsocalled the contravariant or reciprocal base vectors. We later verify that what we calltensors indeed follow the transformation law for tensors.

From the Cartesian permutation symbol, we construct the permutation tensors:

εi jk = √gei jk, εi jk = 1√

gei jk, (3.26)

where we have written the permutation symbol as ei jk to match the indices withthose on the left-hand side.

With these we get

gi × g j = εi jkgk, gk = 12εi jkgi × g j . (3.27)

3.1 Vectors and Tensors

As we have two systems of base vectors, gi and gi , any vector v can be written as

v = vi gi = v j g j , (3.28)

where vi are called the covariant components and v j are called the contravariantcomponents. The raising and lowering of indices, that is, the conversion of covariantto contravariant or vice versa can be performed by scalar multiplication of Eq. (3.28)by gk or by gk;

vi gi · gk = v j g j · gk,

viδik = v j g jk,

vi = gi jvj , or its inverse, vi = gi jv j . (3.29)

For a second-rank tensor A, we can have multiple representations:

A = Ai j gi ⊗ g j = Aij gi ⊗ g j = Ai

j gi ⊗ g j = Ai j gi ⊗ g j . (3.30)

We may relate the covariant, contravariant, and mixed components by using themetric tensor and its reciprocal in the form

Ai j = gikg jl Akl,

Ai j = gikg jl Akl,

Aij = gikg jl Ak

l . (3.31)

32 General Tensors

It has to be noted that in mixed components the horizontal positions of the super-script and subscript are important. If Ai j is symmetric in i and j , we can see that

Aij = Aj

i = Aji . (3.32)

Once we have learned the basic operations using the metric tensor and its reciprocal,we are ready to define covariant, contravariant, and mixed arrays properly. For thiswe introduce two curvilinear systems, xi and xi . The system xi , being curvilinear, isa temporary notation; we will make it Cartesian again afterward. If an array v(i) inthe xi system transforms into v(i) according to the rule

v(i) = v( j)∂x j

∂ xi, (3.33)

we call it a covariant array and use a subscript to write

vi = v j∂x j

∂ xi. (3.34)

It is a contravariant array if

v(i) = v( j)∂ x j

∂xi, (3.35)

and we write it with a superscript as

vi = v j ∂ xi

∂x j. (3.36)

Note that the coordinate differentials themselves transform as

dxi = dx j ∂ xi

∂x j, (3.37)

which is the rule for contravariant arrays: This is the reason for using superscriptsfor coordinates.

If we consider the partial derivatives of a scalar function

φ(x1, x2, x3) = φ(x1, x2, x3), (3.38)

we find that they follow the covariant rule,

∂φ

∂ xi= ∂φ

∂x j

∂x j

∂ xi. (3.39)

As we discussed in the last chapter, geometry or physical considerations determinethe tensor nature of a given array.

We may extend the preceding rules to second- or higher-rank tensors. Forexample,

covariant: Ai j = Amn∂xm

∂ xi

∂xn

∂ x j,

contravariant: Ai j = Amn ∂ xi

∂xm

∂ x j

∂xn,

mixed: Aij = Am

n∂ xi

∂xm

∂xn

∂ x j. (3.40)

3.3 Tensor Calculus 33

In Cartesian coordinates, the base vectors ei can be used to construct reciprocalbase vectors e i . Because of their orthonormality, we end up with

ei = ei . (3.41)

Thus subscripted quantities and the corresponding superscripted quantities areidentical; or, there is no need to distinguish covariant components from contravari-ant components. We have already used this idea when we referred to δi

j and ei jk.Another convention consistently used in general tensor analysis is that the sum-

mation convention applies when the repeated pair of indices occurs in subscript–superscript form. That is, we should write the distance formula as

(ds)2 = dxkdxk = gi j dξ i dξ j . (3.42)

3.2 Physical Components

As mentioned earlier, in general tensor analysis the base vectors may not be unitvectors, nor may they have the same dimensions. However, we may convert them tounit vectors by dividing them by their magnitude. For example, the relation

v = vi gi (3.43)

can be written as

v = v1√g11g1√g11

+ · · · + . (3.44)

From this we can extract the physical component:

v(1) = v1√g11. (3.45)

Similarly, for the covariant components in

v = vi gi , (3.46)

the physical components are of the form

v(1) = v1

√g11. (3.47)

3.3 Tensor Calculus

Tensor calculus deals with differentiating and integrating tensors in curvilinear co-ordinates. In the Cartesian system the base vectors are constants, and

∂u∂xk

= ei∂kui . (3.48)

In the case of general tensors in a curvilinear system, ξ , the base vectors are notconstants, and

∂u∂ξk

= gi∂kui + ui∂kgi . (3.49)

34 General Tensors

The derivatives of the base vectors in the second term can be written as a linearcombination of the three base vectors, to have

∂kgi = �jki g j , (3.50)

where the coefficients �jki are called the Christoffel symbols of the second kind. With

this, the derivative of a vector can be written as

∂u∂ξk

= (∂kui + u j�ijk)gi . (3.51)

We define the covariant derivative of a contravariant component ui as

Dkui = ∂kui + �ijku j . (3.52)

Similarly, the covariant derivative of a covariant component becomes

Dkui = ∂kui − �jiku j . (3.53)

The symbol Dk has different meanings, depending on the tensor components onwhich it operates.

The Christoffel symbols of the first kind are defined as

�i j,k = �li j gkl . (3.54)

Using the relation between the base vectors gi and the Cartesian unit vectors ei ,

gi = em∂i xm, (3.55)

we have

∂kgi = em∂k∂i xm = g j∂mξ j∂k∂i xm. (3.56)

Comparing this with Eq. (3.50), we get

�jki = ∂k∂i xm∂mξ j = �

jik. (3.57)

Using gi j = ∂i xm∂ j xm and ∂i xm∂mξ j = δji , we have

�i j,k = ∂i∂ j xm∂kxm, (3.58)

where the symmetry with respect to i and j is obvious. Differentiating gi j and per-muting the indices, we also have

�i j,k = 12

(∂i g jk + ∂ j gik − ∂kgi j ). (3.59)

Another useful result involves the derivative of√

g in the form

∂i√

g = ∂i L = ∂L∂x j ,k

∂i∂kx j = L∂ jξk∂i∂kx j = L�k

ik (3.60)

or

∂i log√

g = 1√g∂i

√g = �k

ik. (3.61)

3.4 Curvature Tensors 35

The parametric form of a curve in space is given by

ξ i = ξ i (a), (3.62)

where a is a parameter. The derivative with respect to a of a vector function v,defined in terms of the coordinates ξ i , can be written as

∂v

∂a= g j Div

j ∂ξ i

∂a, (3.63)

which is known as the absolute derivative of the vector function.For higher-rank tensors, we have

Di Ajk = ∂i Ajk + �jim Amk + �k

im Ajm,

Di Ajk = ∂i Ajk − �mi j Amk − �m

ik Ajm,

Di Ajk = ∂i Aj

k − �mik Aj

m + �jim Am

k. (3.64)

It is common practice to use the following alternative notation:

( ),i = ∂i ( ),

( );i = ( )|i = Di ( ),{ ijk

}= �i

jk,

{i j, k} = �i j,k. (3.65)

3.4 Curvature Tensors

In calculus, for a smooth function φ(ξ 1, ξ 2, ξ 3), we have

∂1∂2φ = ∂2∂1φ. (3.66)

This commutative property of partial differentiation does not apply to the covariantdifferential operator Di . We could show that for any vector-valued function Ai ,

(Dj Di − Di Dj )Ak = Rmki j Am, (3.67)

where

Rmkij = �l

jk�mil − �l

ik�mjl + ∂i�

mjk − ∂ j�

mik. (3.68)

This fourth-rank tensor is known as the Riemann–Christoffel curvature tensor. Ifthis curvature is identically zero, we can interchange the order of covariant differ-entiation. The curvature tensor is obtained from the derivatives of the metric tensor.If R is zero, we have a flat space. By lowering the index m, we obtain the Riemanncurvature as

Ri jkl = gimRmjkl . (3.69)

36 General Tensors

In two dimensions there is only one nonvanishing component of Ri jkl ,

R1212, (3.70)

and in three dimensions there are six, namely,

R1212, R1313, R2323, R1213, R2123, R3132. (3.71)

These curvatures are important when we consider the compatibility of strain com-ponents in Chapter 5.

The Riemann curvature satisfies the Bianchi identity:

DmRi jkl + DkRi jlm + Dl Ri jmk = 0. (3.72)

Two other curvature tensors of importance are the Ricci curvature,

Ri j = Rmi jm, (3.73)

and the Einstein curvature,

Gıj = Ri

j − 12δi

j Rkk, (3.74)

which plays an important role in general relativity.

3.5 Applications

Before we leave this chapter, it is instructive to consider some examples of tensorformulations of familiar mechanics concepts.

3.5.1 Example: Incompressible Flow

In an incompressible flow the density of the fluid is constant. If we consider an in-finitesimal volume in space with sides formed by the incremental curvilinear coor-dinates and base vectors g1dξ 1, g2dξ 2, and g3dξ 3 (see Fig. 3.3), the quantity of fluidentering it must be equal to the quantity leaving it.

g1dξ 1

g2dξ 2

g3dξ 3

Figure 3.3. Volume element in a curvilin-ear system.

3.5 Applications 37

Let

v = vi gi = vi gi (3.75)

be the fluid velocity vector in the contravariant and covariant representations. Con-sider the vector area

dS3 = g1dξ 1 × g2dξ 2. (3.76)

The amount of fluid leaving this area is

−Q3 = −v · dS3 = −v3 g3 · g1dξ 1 × g2dξ 2 = −√gv3dξ 1dξ 2. (3.77)

The amount of fluid leaving the opposite surface is

Q3 + dQ3 = Q3 + ∂3 Q3dξ 3 = Q3 + ∂3(√

gv3)dξ 1dξ 2dξ 3. (3.78)

For all six surfaces, the total outflow must be zero. After removing the commonfactor dξ 1dξ 2dξ 3, we get

∂i (√

gvi ) = 0. (3.79)

With the gradient operator

∇ = gi∂i , (3.80)

using Eq. (3.61), we can write the continuity equation as

∇ · v = 0. (3.81)

3.5.2 Example: Equilibrium of Stresses

Given an infinitesimal area dS, with unit normal n, the force per unit area (traction)acting on this surface is denoted by T(n). As we have seen in the preceding example,we consider three special infinitesimal areas:

dS1 = g2dξ 2 × g3dξ 3, dS1 =√

gg11dξ 2dξ 3,

dS2 = g3dξ 3 × g1dξ 1, dS2 =√

gg22dξ 3dξ 1,

dS3 = g1dξ 1 × g2dξ 2, dS3 =√

gg33dξ 1dξ 2, (3.82)

dV = √gdξ 1dξ 2dξ 3. (3.83)

Usually there are body forces such as gravity acting on this volume. Let f be thebody force per unit mass. To convert this to force per unit volume, we multiply it bythe density ρ.

Balancing the forces on the six surfaces gives (after canceling dV),

∂1

(√gg11T(1)

)+ ∂2

(√gg22T(2)

)+ ∂3

(√gg33T(3)

)+ √

gρ f = 0. (3.84)

38 General Tensors

We introduce the stress tensor σ through the relations√g11T(1) = σ 1 j g j ,√g22T(2) = σ 2 j g j ,√g33T(3) = σ 3 j g j . (3.85)

The equilibrium of forces can now be written as

∂i (√

gσ i j g j ) + √gρ f = 0. (3.86)

Using Eq. (3.61), we may write this equation in the form

∇ · σ + ρ f = 0 or Diσi j + ρ f j = 0. (3.87)

It remains to be shown that the stress tensor transforms as it should.

SUGGESTED READING

Fung, Y. C. (1965). Foundations of Solid Mechanics, Prentice-Hall.Sokolnikoff, I. S. (1951). Tensor Analysis, Theory and Applications, Wiley.Synge, J. L. and Schild, A. (1949). Tensor Calculus, University of Toronto Press.

EXERCISES

3.1. If a curvilinear system is orthogonal, show that the metric tensor in its matrixform is diagonal.

3.2. For a curvilinear system, the gradient operator is defined as

∇ = gi∂i .

Using this, deduce that

(a)

∇ · A = Dk Ak = 1√g

∂

∂ξk(√

g Ak),

(b)

∇ · ∇φ = ∇2φ = gi j Di∂ jφ,

= 1√g

∂

∂ξ i

(√ggi j ∂φ

∂ξ j

).

3.3. Show that

(a)

Dmgkl = 0,

(b)

�i j,k = 12

(∂i g jk + ∂ j gik − ∂kgi j ),

(c)

∂i g j = −�jikgk.


3.4. By computing

[Dj Di − Di Dj ]Ak,

obtain an expression for the Riemann–Christoffel curvature Rmki j .

3.5. A surface of revolution is given in terms of the Gaussian coordinates r and θ as

x = r cos θ,

y = r sin θ,

z = f (r).

Obtain the base vectors gα (α = 1, 2) and the metric tensor gαβ . Invert the metrictensor to get gαβ . An infinitesimal distance on this surface is given by the first fun-damental form,

(ds)2 = dr · dr = gαβdξαdξβ, ξ 1 = r, ξ 2 = θ.

The second fundamental form is given by

dr · dn = −bαβdξαdξβ,

where n is the unit normal to the surface. Obtain b in terms of r , θ , and the derivativeof f .

Find the mean and Gaussian curvatures given, respectively, by

K = 12

bαα,

K = 1g

det b.

3.6. Apply the preceding results for a sphere of radius a and for the paraboloidf (r) = r2/2a, where a is a constant.

3.7. Any point on the surface of a right circular cone can be described by its distancefrom the apex s and the meridian angle θ . If the semiapex angle of the cone is φ,obtain the base vectors and the metric tensor for this coordinate system.

3.8. For a space curve with position vector R(s), where s is the curve length, the unittangent vector t, the curvature κ , the unit normal n, the binormal b, and the torsionτ are given by

t = dRds

, κn = dtds

, b = t × n, τn = −dbds

.

Show that these quantities satisfy the Frenet–Serret formula,dnds

= −(κt + τ b).

Illustrate this for the helix given by

x1 = a cos θ, x2 = a sin θ, x3 = bθ,

where a and b are constants.

3.9. In polar coordinates, the unit vectors er and eθ are functions of θ and the gradi-ent operator is given by

∇ = er∂

∂r+ eθ

∂

r∂θ.

40 General Tensors

In the 2D case the physical components of the stress are

σ =[

σ rr σ rθ

σ rθ σ θθ

].

Expand the equilibrium equation

∇ · σ + ρ f = 0.

3.10. For the polar coordinates,

r =√

x21 + x2

2 , θ = arctan (x2/x1).

Obtain the base vectors, metric tensor, and the gradient operator for this system.

3.11. The spherical coordinates r, θ, and φ are related to the Cartesian coordinates, as

x1 = r sin θ cos φ,

x2 = r sin θ sin φ,

x3 = r cos θ.

Compute the base vectors, metric tensor, and gradient operator for the sphericalcoordinate system.

3.12. The base and a side of a parallelogram make an acute angle α. If a 2D systemof coordinates is constructed in such a way that ξ 1 is parallel to the base and ξ 2 isparallel to the side, find a relation between this system and the Cartesian system.Obtain the base vectors, metric tensor, and Laplace operator for this system. (In theanalysis of plates that are parts of a swept-wing airplane, parallelogram shapes areencountered.)

3.13. The Cartesian coordinates (x, y) and the “pseudoelliptical” coordinates (ξ, η)are related by

x = aξ cos η, y = bξ sin η,

where a and b are constants.Compute

(a) the base vectors,(b) the metric tensor,(c) the reciprocal base vectors,(d) the reciprocal metric tensor, and(e) the Laplace operator.

3.14. In the context of the examples of this chapter, referring to Fig. 3.4, let−−→OA = r1 = g1dξ 1,

−−→OB = r2 = g2dξ 2,

−→OC = r3 = g3dξ 3.

Using the cross products of vectors, show that the vector areas satisfy−−−→ABC = −−−→

OAB + −−−→OBC + −−−→

OC A,

where−−−→ABC has its normal outward from the volume element and the areas on the

right-hand side have their normals inward. If these areas have magnitudes dS, dS3,dS1, and dS2, respectively, and unit normals n, n3, n1, and n2, show that

ndS = ni dSi .


−T(1)

−T(2)

−T(3)

T(n)

O

A

BC

Figure 3.4. Tetragonal volume element withtraction vectors.

If the traction acting on dS is T(n), etc., in the limit as the volume of the tetrahedrongoes to zero, show that

T(n)dS = T(i)dSi .

3.15. In the preceding exercise, if the unit vector n is resolved as n = ni gi and intro-ducing the stress tensor σ i j as T (i)

√gii = σ i j g j (no sum on i), show that

T (n) j = σ i j ni .

3.16. Obtain the characteristic equation for the principal stresses by extremizingT(n) · n.

3.17. In Exercise 3.14, if we introduce a new coordinate system ξ i in such a way thatthe same area element ABC caps the three vectors,

−−→OA = g1dξ 1, etc.,

where O is distinct from O, show that

T(i)dSi = T(i)dSi .

Using the preceding results, prove that the stress tensor σ i j obeys the transformationlaw for a second-rank contravariant tensor.

3.18. In dynamics the Lagrangian L is defined as

L = K − V,

where K is the kinetic energy and V is the potential of the forces. For a single parti-cle of mass m moving in a curvilinear coordinate system without any external forces,obtain the equation of motion from

ddt

∂L∂ xi

− ∂L∂xi

= 0,

where xi (t) are the curvilinear coordinates of the particle.

4 Integral Theorems

To deal with the conservation and balance laws of mechanics, we use the integraltheorems of Gauss and Stokes frequently. Although most undergraduate calculuscourses cover this material, it is of interest to reconsider these theorems by usingthe index notation and the summation convention.

4.1 Gauss Theorem

Consider a convex region V bounded by a smooth surface S in three dimensions. Fora nonconvex region, if it can be divided into a finite number of convex subregions,we could still use the following theorem. Let A(x1, x2, x3) be a differentiable (asmany times as we need) function defined in V. We begin by considering the integral

I =∫∫∫

V

∂ A∂x1

dx1dx2dx3. (4.1)

As shown in Fig. 4.1, a differential tube of cross-sectional area dx2dx3 intersectsthe surface S at two places, dividing the total surface into S∗ and S∗∗. Integratingwith respect to x1, we have

I =∫∫

S∗Adx2dx3 −

∫∫S∗∗

Adx2dx3. (4.2)

With

dx2dx3 = n∗1dS on S∗, dx2dx3 = −n∗∗

1 dS on S∗∗, (4.3)

we get

I =∫

SAn1dS. (4.4)

Next, we generalize this to get∫V

∂i AdV =∫

Sni AdS. (4.5)

42

4.1 Gauss Theorem 43

x2

x3

x1

n∗n∗∗

ds∗∗ds∗

S∗S∗∗

Figure 4.1. Integration in three dimensions.

If there are a finite number of internal surfaces in V where A is not differentiable,but left and right limits for A exist as these surfaces are crossed, the contributionsfrom these surfaces have to be added to the preceding result. Such discontinuousbehavior occurs in situations involving shock waves and fracture surfaces in solids.

If we replace A with a component of a tensor, Ajkl . . . , we get∫V

∂i Ajkl . . . dV =∫

Sni Ajkl...dS. (4.6)

As special cases, we have the following relations:

1. ∫V

∂i Ai dV =∫

Sni Ai dS or

∫V

∇ · AdV =∫

Sn · AdS, (4.7)

which is the well-known divergence theorem.2. ∫

V∂i AdV =

∫S

ni AdS or∫

V∇AdV =

∫S

nAdS, (4.8)

where the equation on the left has been multiplied by ei to get the vector formof the equation on the right.

3. ∫V

∂i Aj dV =∫

Sni Aj dS or

∫V

∇ × AdV =∫

Sn × AdS, (4.9)

where we have used multiplication by ei jkek.

44 Integral Theorems

The preceding results can be summarized as∫V

∇ ∗ AdV =∫

Sn ∗ AdS, (4.10)

where ∗ stands for ·, ×, or ⊗, producing the dot product, cross product, or the tensorproduct, and A is a tensor of any rank. The Gauss theorem has also been attributedto Green and Ostrogradsky.

We can also see that the preceding result applies to converting area integralsinto contour integrals, instead of volume integrals into surface integrals.

4.2 Stokes Theorem

Consider a two-dimensional (2D) convex region S bounded by the curve C in thex1, x2 plane (see Fig. 4.2). We assume A is differentiable inside S. The area integral

I =∫∫

S

∂ A∂x1

dx1dx2 (4.11)

can be evaluated by integrating with respect to x1 as

I =∫

C∗Adx2 −

∫C∗∗

Adx2 =∮

CAdx2. (4.12)

Here, we have shown the two points P1 and P2, which are located at the minimumand maximum values of x2, dividing the curve C into C∗ and C∗∗. Similarly, for afunction B,

J =∫∫

S

∂ B∂x2

=∫

C ′Bdx1 −

∫C ′′

Bdx1 = −∮

CBdx1, (4.13)

where we find C ′′ and C ′ by dividing the curve C, using the minimum and maximumvalues of x1. Combining I and J , we obtain∫∫

S

[∂ A∂x1

− ∂ B∂x2

]dx1dx2 =

∮C

[Adx2 + Bdx1]. (4.14)

x1 x1

x2x2

C∗

C∗∗

S

P1

P2

C ′

C ′′SQ1

Q2

Figure 4.2. Integration in two dimensions.

4.2 Stokes Theorem 45

If we equate A and B to vector components,

A = A2, B = A1, (4.15)

we find ∫∫S

[∂ A2

∂x1− ∂ A1

∂x2

]dx1dx2 =

∮C

[A1dx1 + A2dx2]. (4.16)

This result is known as the Stokes theorem. The left-hand side of this equation hasthe e3 component of the curl of the vector A. We may write this as∫∫

Se3 · ∇ × Adx1dx2 =

∮C

A · dx. (4.17)

This shows that the integral of the normal component of the curl of a vector field onthe surface S is equal to the integral of the tangential component of the same fieldaround the closed curve C.

If we have a curved surface S in three dimensions bounded by a curve C, canwe relate the normal component of the curl to the tangential component as we havedone previously? The answer is “yes,” provided there is a mapping that transformsthe curved surface in three dimensions to a flat surface in two dimensions.

Let us assume a 2D plane (y1, y2), with every point in a convex region S mappedonto a point (x1, x2, x3), with mapping functions

xi = xi (y1, y2). (4.18)

Then S would be mapped into a surface S, as shown in Fig. 4.3. The boundary curveC would be mapped into C. Assume that there is a vector-valued function

A = A(y1, y2) (4.19)

that satisfies the Stokes theorem,∫∫S

[∂ A2

∂y1− ∂ A1

∂y2

]dy1dy2 =

∮C

[A1dy1 + A2dy2]. (4.20)

x1

x2

x3

y1

y2

Figure 4.3. Mapping a 2D surface onto a 3D surface.


To distinguish the partial derivatives with respect to the y coordinates from thosewith respect to the x coordinates, we use the notation

∂α = ∂

∂yα

, α = 1, 2; ∂i = ∂

∂xi, i = 1, 2, 3. (4.21)

Next, consider a three-vector Ai (x1, x2, x3), defined on the surface S, which has unitnormal n. Let us compute

I =∫

Sn · ∇ × AdS =

∫S

ni ei · e jkl∂k Ale j dS =∫

Seiklni∂k AldS. (4.22)

Let us add a y3 coordinate and three unit vectors e1, e3, and e3 to our y1, y2 plane.We know that the surface area dS with unit normal n is the map of the area

dy1dy2 with unit normal e3. Let dx be the map of dy1 e1 and dx∗ be the map of dy2 e2.We have

dSe3 = dy1 e1 × dy2 e2, dS = dy1dy2. (4.23)

The map of this vector area can be written as

ndS = dx × dx∗ (4.24)

or

ndS = ∂1xi ei dy1 × ∂2xj e j dy2. (4.25)

After the cross product is performed,

ndS = ∂1xi∂2xj ei jkekdy1dy2. (4.26)

The i component of this vector is

ni dS = ∂1xj∂2xkei jkdy1dy2. (4.27)

We substitute this result into the expression for I to get

I =∫

Seikl∂k Al∂1xj∂2xmei jmdy1dy2. (4.28)

Using the e–δ identity, we can simplify this to get

I =∫

S[∂k Al − ∂l Ak]∂1xk∂2xldy1dy2. (4.29)

So far we have kept two sets of functions, Aα and Ai . If we relate these two sets as

Aα = (∂αxi )Ai , (4.30)

we have

∂1 A2 = (∂1∂2xi )Ai + ∂ j Ai∂2xi∂1xj , (4.31)

∂2 A1 = (∂2∂1xi )Ai + ∂ j Ai∂1xi∂2xj , (4.32)

I =∫

S[∂1 A2 − ∂2 A1]dy1dy2. (4.33)


Finally, ∫S

n · ∇ × AdS =∮

Cdx · A. (4.34)

This result is known as the Kelvin transformation. Note that, in writing A on theright of the differential in the line integral, the Stokes theorem with the Kelvin trans-formation has been made applicable to tensors of ranks higher than one. The con-cept of mapping previously introduced anticipates deformation of bodies presentedin later chapters.

SUGGESTED READING

Greenberg, M. D. (1988). Advanced Engineering Mathematics, Prentice-Hall.Kreyzig, E. (2006). Advanced Engineering Mathematics, 9th ed. Wiley.Sokolnikoff, I. S. and Redheffer, R. M. (1966). Mathematics of Physics and Modern Engineer-

ing, McGraw-Hill.Thomas, G. B. and Finney, R. L. (1979). Calculus and Analytic Geometry, Addison-Wesley.

EXERCISES

4.1. Prove the Green’s theorem:∫V

[u∇2v − v∇2u]dV =∫

S[u∇v − v∇u] · ndS.

4.2. If a is a constant vector, show that∮C

a · dx = 0,

∮C

a × dx = 0.

4.3. Transform the surface integrals

I =∫

S(n × ∇) · udS, I =

∫S(n × ∇) × udS

into volume integrals and evaluate them.

4.4. Obtain the differential equations for the vector function φ and the scalar ψ

inside an arbitrary volume if the surface integrals∫S

n × (rφ)da = 0,

∫S

n · (rψ)da = 0,

where r = (xi xi )1/2 and r = x.

4.5. In the torsion of shafts, the St. Venant warping function φ satisfies the Laplaceequation in a simply connected 2D domain D in the x, y plane, representing thecross-section of the shaft and the boundary conditions,

n · ∇φ = k · n × r,

where n is the unit normal to the boundary, r = xi + y j, and k is the unit vectorperpendicular to the plane.Show the integrals representing the shear forces in the x and y directions,∫

D

(∂φ

∂x− y

)dA = 0,

∫D

(∂φ

∂y+ x

)dA = 0.


4.6. If

v = e3 × r/r2, r = x1e1 + x2e2, r = |r|,compute

I =∮

v · dx

around a unit circle, r = 1, directly. Compare your result with the value obtainedwith the Stokes theorem. Explain the reason for the difference.

4.7. In an infinite 3D medium, the function u satisfies Poisson’s equation,

∇2u = q(x),

with u → 0 as r = |x| → ∞. To obtain a general solution, we use the function

v(x) = 14π |x − x′| .

Show that v satisfies the Laplace equation everywhere except at x = x′. Remove asphere of radius ε centered at x = x′ and apply the Green’s theorem to u and v withthe volume integral over the infinite domain without the small sphere. In the limitε → 0, show that

u(x) = − 14π

∫V

q(x′)dV|x′ − x| ,

where V is the infinite domain. Here, v is called the Green’s function for Poisson’sequation.

4.8. Consider an arbitrary simple closed curve C inside a domain D in the complexplane z = x + iy. For a complex valued function, w(z) = u(x, y) + iv(x, y), if∮

Cwdz = 0

for all such curves, show that∂u∂x

= ∂v

∂y,

∂u∂y

= −∂v

∂x.

These are known as the Cauchy–Riemann equations. This result is known asMorera’s theorem.

4.9. In the 2D case, derive the Stokes theorem using the Gauss theorem.

5 Deformation

Consider a material continuum occupying a region B0 with volume V0 and surfacearea S0 at time t = 0. At a later time t , we see this body in a region B with volumeV and surface area S. We refer to these two configurations as undeformed and de-formed configurations, respectively. We use a fixed Cartesian system to describe thegeometry of the two configurations. As shown in Fig. 5.1, a generic particle in theundeformed configuration has the position X, and the same particle in the deformedconfiguration has the position x. We simultaneously develop descriptions of the ge-ometry of this mapping from the state B0 to B, using X1, X2, and X3 as independentvariables and using x1, x2, and x3 as independent variables. To distinguish the par-tial derivative operator ∂ in the two systems, we use Latin indices, i, j, k, . . . , for thecomponents of x and Greek indices, α, β, γ, . . . , for the components of X. For theindices of the unit vectors to match the components of the vectors, we denote e1, e2,and e3 by ei in one system and by eα in the other. Thus

X = Xαeα, x = xi ei . (5.1)

We introduce the partial derivatives

∂α = ∂

∂ Xα

, ∂i = ∂

∂xi. (5.2)

5.1 Lagrangian and Eulerian Descriptions

If we use Xα as independent variables and xi as dependent variables, i.e.,

xi = xi (X1, X2, X3) = xi (X), (5.3)

we have the Lagrangian description of deformation. The variables Xα are called thematerial coordinates or the Lagrangian coordinates.

49

50 Deformation

x

X

B0

B

u

U

Figure 5.1. Deformation of a body.

If xi are the independent variables and Xα are the dependent variables, i.e.,

Xα = Xα(x1, x2, x3) = Xα(x), (5.4)

we have the Eulerian description of deformation. The coordinates xi are called thespatial coordinates or the Eulerian coordinates.

Depending on the complexity of the differential equations, the boundary condi-tions, and the usefulness of the solution variable, one of these formulations may bepreferable over the other in a particular application. For example, in the mechanicsof solids, when we load a rectangular, uniformly thick plate, it deforms into a com-plex curved shape. From the simplicity of the initial geometry, we may prefer theLagrangian formulation for this case. For a fluid flow problem, we can observe theflow characteristics through a wind-tunnel window. We may not care where the var-ious fluid particles were at time t = 0. An Eulerian approach is often used in fluiddynamics.

In the Lagrangian formulation, we label a material particle and follow it as itmoves with time. In Eulerian formulation, we stand at a spatial point and observethe passing particles. As X and x are locations of the same material particle, thefunctional relations, Eqs. (5.1), can be inverted in the neighborhood of a point Xto obtain Eqs. (5.2). This one-to-one correspondence is known as the axiom of con-tinuity. This axiom implies (a) the indestructible nature of matter and (b) the im-penetrability of matter. According to (a), matter occupying a finite volume cannotbe deformed into zero volume, and, according to (b), the motion of a body car-ries every material line into another line and every surface into another surface. Asmooth curve cannot be deformed into an intersecting curve.

5.2 Deformation Gradients 51

The necessary and sufficient condition for the existence of the one-to-one cor-respondence of the functions Xα and xi in a small neighborhood is

det[∂αxi ] �= 0. (5.5)

The displacement vector is defined as the vectorial distance from the initial positionX to the final position x of a particle. When this vector is considered to be a functionof the material coordinates we denote it by U, and when it is a function of spatialcoordinates we use u:

U(X1, X2, X3) = u(x1, x2, x3) = x − X. (5.6)

In component form,

Uα = xiδiα − Xα, ui = xi − Xαδαi , (5.7)

where, for consistency of notation, the Kronecker delta is introduced.

5.2 Deformation Gradients

From Eqs. (5.3) and (5.4) we get

dxi = dXα∂αxi , dXα = dxi∂i Xα. (5.8)

The quantities ∂αxi and their inverses ∂i Xα are called the deformation gradients.Figure 5.2 shows the mapping of an element, dX1. Differentiating Eq. (5.3) by xj

and Eq. (5.4) by Xβ , we see

∂αxi∂ j Xα = δi j , ∂i Xα∂β xi = δαβ. (5.9)

Let us introduce the Green’s deformation gradient tensor F and the Cauchydeformation gradient tensor f , in matrix forms:

F = ∇x = [∂αxi ], f = ∇X = [∂i Xα]. (5.10)

These matrices are inverses of each other:

F f = I = f F. (5.11)

It should be noted that in the literature the transposes of these matrices are oftenused for the deformation gradients.

The axiom of continuity requires that these matrices be nonsingular. The de-terminants of these matrices are the Jacobian determinants of transformations (5.3)and (5.4):

J = |∂αxi |, j = |∂i Xα|, J j = 1. (5.12)

There will not be any problem in distinguishing the Jacobian “ j” from the integerindices.

52 Deformation

X1

X2

dX1

dx

dx1 = ∂1x1dX1

dx2 = ∂1x2dX1

X

x

BB0

Figure 5.2. Deformation of an element.

These determinants have the expansions

J = 16

ei jkeαβγ ∂αxi∂β xj∂γ xk, j = 16

eαβγ ei jk∂i Xα∂ j Xβ∂kXγ . (5.13)

Differentiating J by Fαi = ∂αxi and j by fiα = ∂i Xα , we get

∂ J∂ Fαi

= 12

ei jkeαβγ ∂β xj∂γ xk = cofactor of ∂αxi , (5.14)

∂ j∂ fiα

= 12

eαβγ ei jk∂ j Xβ∂kXγ = cofactor of ∂i Xα. (5.15)

Remembering the relation between cofactors and inverses, we can write ∂i Xα interms of ∂αxi , and vice versa, as

∂i Xα = 12J

ei jkeαβγ ∂β xj∂γ xk, (5.16)

∂αxi = 12 j

eαβγ ei jk∂ j Xβ∂kXγ . (5.17)

5.2.1 Deformation Gradient Vectors

We know that an element dX gets mapped into dx during deformation. In particu-lar, the element e1dX1 goes to

dx = ∂1xi ei dX1 = G1dX1, G1 = ∂1xi ei . (5.18)

5.2 Deformation Gradients 53

X1

X2

G1

G2

e1

e2

Figure 5.3. Deformation gradient vectors G1 and G2.

If we had marked e1 in the undeformed configuration, we would see its imageG1 in the deformed configuration. The three unit vectors have

Gα = ∂αxi ei = ∂αx (5.19)

as their images. These are known as the Green’s deformation gradient vectors (seeFig. 5.3).

Similarly, the unit vectors ei in the current configuration had their images

gi = ∂i Xαeα = ∂i X (5.20)

in the undeformed configuration. These are called the Cauchy deformation gradientvectors.

These two systems of vectors are intimately related to the base vectors we hadseen in Chapter 3. To see this, consider an arbitrary element dX getting mappedinto dx. We have

dx = ∂αxdXα = GαdXα, (5.21)

dX = ∂i Xdxi = gi dxi . (5.22)

The lengths of these elements can be expressed as

(dS)2 = dXαdXα, (ds)2 = dxi dxi . (5.23)

If we write the current length in terms of material coordinates, we have

(ds)2 = ∂αxdXα · ∂β xdXβ = Gα · GβdXαdXb = GαβdXαdXβ, (5.24)

where Gαβ is our old metric tensor, which is the dot product of the base vectors. Incontinuum mechanics we refer to G = [Gαβ] as the Green’s deformation tensor.

A parallel consideration in terms of spatial coordinates gives

(dS)2 = gi j dxi dxj , gi j = gi · g j . (5.25)

54 Deformation

The tensor g is called the Cauchy deformation tensor. We may express the defor-mation tensors, G and g, as

G = F FT , g = f f T . (5.26)

The tensors G and g are symmetric by virtue of their definition in terms of dotproducts. Because the squares of distances are positive, these tensors (matrices) arepositive definite.

5.2.2 Curvilinear Systems

The base vectors Gα or gi remind us of curvilinear coordinate systems. If we hadetched a rectilinear grid on the undeformed body, after deformation, we would seethe grid deformed into a curvilinear net. This is the curvilinear system associatedwith base vectors Gα and metric tensor G. In the context of general tensors, it isappropriate to use Xα and xi as the coordinates. We can imagine a similar picturefor a rectangular grid in the current configuration.

We may introduce reciprocal base vectors, Gα and gi , to go with the curvilinearcoordinates. These are defined as

Gα = ei∂i Xα, gi = eα∂αxi . (5.27)

Using the reciprocity of F and f , we see that

Gα · Gβ = δαβ , gi · g j = δi

j . (5.28)

Associated with the reciprocal base vectors are the reciprocal deformation tensorsGαβ and gi j , which are inverses of Gαβ and gi j , respectively. In matrix form,

G−1 = [Gαβ] = [∂i Xα∂i Xβ] = f T f ,

G = [Gαβ] = [∂αxi∂β xi ] = F FT ,

g−1 = [gi j ] = [∂αxi∂αxj ] = FT F,

g = [gi j ] = [∂i Xα∂ j Xα] = f f T . (5.29)

The reciprocal deformation tensors Gαβ and gi j are attributed to Piola and Finger,respectively. The characteristics of a deformed body in the neighborhood of a pointx can be described as

xi (X1 + dX1, . . . ,) = xi (X1, . . . ,) + dXα∂αxi (X1, . . . ,)

+ 12

dXαdXβ∂α∂β xi (X1, . . . ,) + · · · + .

In classical continuum mechanics it is postulated that the neighborhood can beas small as possible. Then, the second-order quantities dXαdXβ render the thirdterm negligible. As a result, the deformation gradients ∂αxi completely define thecharacteristics of the deformation. Materials satisfying this postulate are called sim-ple materials. To take into account the coarse microstructure of certain materials,we are not permitted to assume that the third term is negligible, and we may haveto include the second derivatives in the theories.

5.3 Strain Tensors 55

5.3 Strain Tensors

The lengths of elements before and after deformation are given by

(dS)2 = dXαdXα = gi j dxi dxj , (ds)2 = dxi dxi = GαβdXαdXβ. (5.30)

The strain tensors Eαβ and ei j are measures of the change in length:

(ds)2 − (dS)2 = 2EαβdXαdXβ = [Gαβ − δαβ]dXαdXβ

= 2ei j dxi dxj = [δi j − gi j ]dxi dxj . (5.31)

The strain tensor E is called the Green’s strain, and e is called the Almansi strain.The names Lagrange strain and Cauchy strain are also used, respectively, for thesetwo quantities. It is easy to see that both of these strains are symmetric tensors.

A relation between these two strain measures can be established as

ei j = Eαβ∂i Xα∂ j Xβ, Eαβ = ei j∂αxi∂β xj , (5.32)

which have the matrix forms

e = f E f T , E = FeFT . (5.33)

By use of

xi = [Xβ + Uβ]δiβ, Xα = [xj + u j ]δα j , (5.34)

the deformation gradients become

∂αxi = δαi + ∂αUβδiβ, ∂i Xα = δiα − ∂i u jδα j . (5.35)

The deformation tensors become

Gαβ = δαβ + ∂αUβ + ∂βUα + ∂αUγ ∂βUγ ,

gi j = δi j − ∂i u j − ∂ j ui + ∂i uk∂ j uk. (5.36)

Now the strain tensors can be written as

Eαβ = 12

[∂αUβ + ∂βUα + ∂αUγ ∂βUγ ],

ei j = 12

[∂i u j + ∂ j ui − ∂i uk∂ j uk]. (5.37)

56 Deformation

In engineering notation, using U, V, and W to represent Uα and X, Y, and Z torepresent Xα , we have

EXX = ∂U∂ X

+ 12

[(∂U∂ X

)2

+(

∂V∂ X

)2

+(

∂W∂ X

)2]

,

EYY = ∂V∂Y

+ 12

[(∂U∂Y

)2

+(

∂V∂Y

)2

+(

∂W∂Y

)2]

,

EZZ = ∂W∂ Z

+ 12

[(∂U∂ Z

)2

+(

∂V∂ Z

)2

+(

∂W∂ Z

)2]

,

EXY = 12

[∂U∂Y

+ ∂V∂ X

+ ∂U∂ X

∂U∂Y

+ ∂V∂ X

∂V∂Y

+ ∂W∂ X

∂W∂Y

],

EYZ = 12

[∂V∂ Z

+ ∂W∂Y

+ ∂U∂Y

∂U∂ Z

+ ∂V∂Y

∂V∂ Z

+ ∂W∂Y

∂W∂ Z

],

EZX = 12

[∂W∂ X

+ ∂U∂ Z

+ ∂U∂ Z

∂U∂ X

+ ∂V∂ Z

∂V∂ X

+ ∂W∂ Z

∂W∂ X

]. (5.38)

Similarly, the Almansi strain components can be written as

exx = ∂u∂x

− 12

[(∂u∂x

)2

+(

∂v

∂x

)2

+(

∂w

∂x

)2]

,

eyy = ∂v

∂y− 1

2

[(∂u∂y

)2

+(

∂v

∂y

)2

+(

∂w

∂y

)2]

,

ezz = ∂w

∂z− 1

2

[(∂u∂z

)2

+(

∂v

∂z

)2

+(

∂w

∂z

)2]

,

exy = 12

[∂u∂y

+ ∂v

∂x− ∂u

∂x∂u∂y

− ∂v

∂x∂v

∂y− ∂w

∂x∂w

∂y

],

eyz = 12

[∂v

∂z+ ∂w

∂y− ∂u

∂y∂u∂z

− ∂v

∂y∂v

∂z− ∂w

∂y∂w

∂z

],

ezx = 12

[∂w

∂x+ ∂u

∂z− ∂u

∂z∂u∂x

− ∂v

∂z∂v

∂x− ∂w

∂z∂w

∂x

]. (5.39)

5.3.1 Decomposition of Displacement Gradients

The displacement gradients ∂i u j and ∂αUβ can be resolved into symmetric and skew-symmetric tensors.

Let

∂i u j = ei j + ri j , ∂αUβ = Eαβ + Rαβ, (5.40)


where ei j and Eαβ are symmetric tensors called infinitesimal strains and ri j and Rαβ

are skew-symmetric tensors called infinitesimal rotations. These can be written as

ei j = 12

(∂i u j + ∂ j ui ),

Eαβ = 12

(∂αUβ + ∂βUα),

ri j = 12

(∂i u j − ∂ j ui ),

Rαβ = 12

(∂αUβ − ∂βUα). (5.41)

The Almansi strain components are related to the infinitesimal strains in theform

ei j = ei j − 12

[eik + rik][e jk + r jk]. (5.42)

When |ei j | and |ri j | are small, the Almansi strain is approximately equal to the in-finitesimal strain. Further, as

Eαβ = ei j∂αxi∂β xj , (5.43)

neglecting higher-order terms,

E11 = e11, . . . . (5.44)

In other words, when the displacement gradients and rotations are small, the mate-rial description and spatial description are identical.

5.3.2 Stretch

Consider an element dX deforming into dx. Let us attach unit vectors N to dX andn to dx. Then,

N = dXdS

, n = dxds

. (5.45)

In the component form,

Nα = dXα

dS, ni = dxi

ds. (5.46)

Using

(ds)2 = GαβdXαdXβ, (dS)2 = gi j dxi dxj , (5.47)

we get (dsdS

)2

= Gαβ

dXα

dSdXβ

dS= Gαβ Nα Nβ,

(dSds

)2

= gi jdxi

dsdxj

ds= gi j ni n j . (5.48)

58 Deformation

The ratio (ds/dS) is called the stretch. This ratio depends on the orientation of theelement. We have

�N = dsdS

= √Gαβ Nα Nβ, λn = ds

dS= 1√

gi j ni n j(5.49)

for the stretches in the material and spatial descriptions.As special cases, if N = e1, we get

N1 = 1, N2 = 0, N3 = 0, �1 =√

G11, (5.50)

and if n = e1,

n1 = 1, n2 = 0, n3 = 0, λ1 = 1√g11

. (5.51)

(Note that �1 �= λ1 as N = e1 does not deform into n = e1.) A similar considerationof unit normals in the other two directions yields the geometrical meaning of thediagonal components of the deformation tensors:

G11 = �21, G22 = �2

2, G33 = �23. (5.52)

To see the meaning of the off-diagonal elements, recall that eα in the undeformedbody maps onto Gα in the deformed body. If we take two orthogonal unit vectors e1

and e2, their images are G1 and G2. If θ12 denotes the angle between G1 and G2, wehave

cos θ12 = G1 · G2

|G1||G2| = G12√G11G22

. (5.53)

Thus G12 indicates the distortion of the original angle of 90◦.

5.3.3 Extension

The extensions EN and en are defined as

EN = �N − 1, en = λn − 1. (5.54)

Extension, stretch, deformation, and strain are related in the form

E1 = �1 − 1 =√

G11 − 1 =√

2E11 + 1 − 1,

e1 = λ1 − 1 = 1√g11

− 1 = 1√1 − 2e11

− 1. (5.55)

5.3.4 Infinitesimal Strains and Rotations

When the difference between ds and dS is small, for an element initially in the 1-direction, we find (

dsdS

)2

− 1 = ds − dSdS

ds + dSdS

= 2E11. (5.56)


dX1

dX2

A

BC

A′

B′ C ′

O

Figure 5.4. Infinitesimal deformation of a rectangle.

Using

ds + dSdS

≈ 2, (5.57)

we get

E11 = ds − dSdS

, (5.58)

which is the elementary definition of strain as the ratio of change in length to theoriginal length.

Consider a rectangular material element OACB with O at (X1, X2) and OA =dX1 and OB = dX2, as shown in Fig. 5.4. During deformation, O moves by (U1, U2)to O′ and A, B, and C end up at A′, B′, and C ′. Using a rigid body translation, we canmove the element back to make O′ coincide with O. Then the relative coordinatesof the other points are

A′ : (dX1 + ∂1U1dX1, ∂1U2dX1),B′ : (∂2U1dX2, dX2 + ∂2U2dX2),C ′ : (dX1 + ∂1U1dX1 + ∂2U1dX2, dX2 + ∂1U2dX1 + ∂2U2dX2).

Assuming small displacement derivatives, we find

E11 = ∂1U1, E22 = ∂2U2. (5.59)

The counterclockwise rotations θ12 of OA and θ21 of OB are obtained as

θ12 = ∂1U2, θ21 = −∂2U1. (5.60)

60 Deformation

The engineering shear strain �12 and the mathematical shear strain E12 are measuresof the change in angle from the original 90◦:

�12 = 2E12 = ∂2U1 + ∂1U2. (5.61)

The infinitesimal rotation R12 is defined as the average of the two rotations,

R12 = 12

[∂1U2 − ∂2U1]. (5.62)

It turns out that, if we consider an element with initial arbitrary orientation θ andfinal orientation θ + φ, the infinitesimal rotation R12 is the average of tan φ, withaveraging done over all values of θ ’s.

5.3.5 Deformation Ellipsoids

Consider a small sphere in the undeformed configuration dXαdXα = K2, with thecenter at X. In the deformed configuration, the center has moved to x. The map ofthe sphere can be found from

dXαdXα = ∂i Xα∂ j Xαdxi dxj = gi j dxi dxj = K2. (5.63)

As gi j is positive definite, the preceding quadratic form represents an ellipsoid,known as the material ellipsoid of Cauchy (see Fig. 5.5).

Similarly, a sphere dxi dxi = k2 in the deformed configuration is the map of anellipsoid,

dxi dxi = ∂αxi∂β xi dXαdXβ = GαβdXαdXβ = k2, (5.64)

in the undeformed configuration. This is known as the spatial ellipsoid of Cauchy(Fig. 5.5).

Let us concentrate on the spatial ellipsoid for the time being. The directions ofthe three axes of the ellipsoid correspond to the direction along which the stretch

X1

X

X2

x

x1

x2

dXαdXα = K2

dxi dxi = k2

GαβdXαdXβ = k2

gi j dxi dxj = K2

Figure 5.5. Spatial and material ellipsoids of Cauchy.


� = ds/dS is stationary. If N is a unit vector pointing in the direction of dX insidethe ellipsoid in the undeformed configuration, we have

Nα = dXα

dS. (5.65)

By dividing Eq. (5.64) by (dS)2, we find

Gαβ Nα Nβ = �2N. (5.66)

To find the directions N along which �2N is stationary, we have to extremize it sub-

ject to the condition that N is a unit vector. For such constrained extremization, asdiscussed in Chapter 2, we use the method of Lagrange multipliers and define a newfunction,

F (N1, N2, N3) = Gαβ Nα Nβ − G(Nα Nα − 1), (5.67)

where G is the Lagrange multiplier. Setting

∂ F∂ Nα

= 0, (5.68)

we obtain a system of simultaneous equations,

Gαβ Nβ − GNα = 0, (5.69)

where we have used the fact that Gαβ is symmetric.In expanded form, we have⎡

⎢⎣ G11 − G G12 G13

G12 G22 − G G23

G13 G23 G33 − G

⎤⎥⎦

⎧⎪⎨⎪⎩

N1

N2

N3

⎫⎪⎬⎪⎭ =

⎧⎪⎨⎪⎩

000

⎫⎪⎬⎪⎭ . (5.70)

This is a typical matrix eigenvalue problem. For a nontrivial solution N, we musthave

det(G − GI) = 0. (5.71)

We may expand the determinant to obtain the characteristic equation of the matrixG in the form

−G3 + IG1G2 − IG2G + IG3 = 0, (5.72)

where the coefficients of the cubic are the three invariants of the matrix G, namely,

IG1 = G11 + G22 + G33,

IG2 = G22G33 + G33G11 + G11G22 − G223 − G2

31 − G212

=∣∣∣∣∣ G22 G23

G32 G33

∣∣∣∣∣ +∣∣∣∣∣ G11 G31

G13 G33

∣∣∣∣∣ +∣∣∣∣∣ G11 G12

G21 G22

∣∣∣∣∣ ,

IG3 =

∣∣∣∣∣∣∣G11 G12 G13

G21 G22 G23

G31 G32 G33

∣∣∣∣∣∣∣ . (5.73)

62 Deformation

Let G(α) denote the three roots of cubic (5.72). They satisfy

IG1 = G(1) + G(2) + G(3),

IG2 = G(1)G(2) + G(2)G(3) + G(3)G(1),

IG3 = G(1)G(2)G(3). (5.74)

Substituting G = G(γ ) in Eq. (5.70) for each value of γ , we find a vector N(γ ) (afternormalizing it to have unit length). This gives us three directions along which �N

is stationary. These are called the principal axes of the ellipsoid. For a symmetricmatrix, it is known that these three eigenvectors are mutually orthogonal (or whenthe cubic has repeated roots, the directions can be selected to be orthogonal) andthe eigenvalues G(γ ) are real.

N(α) · N(β) = δαβ. (5.75)

Multiplying Gαβ N(γ )β = GN(γ )

α by N(γ )α (summation on α only), we obtain

G(γ ) = Gαβ N(γ )α N(γ )

β = �2(γ ). (5.76)

The eigenvalues happen to be the squares of the principal stretches. The deforma-tion tensor G has the spectral representation

G =3∑

γ=1

�2(γ ) N(γ ) N(γ )T . (5.77)

The analysis of the material ellipsoid gi j dxi dxj = K2 can be carried out in asimilar manner. The eigenvalues of g will be denoted by 1/λ2

(i) and the eigenvectors

by n(i). It has the spectral representation

g =3∑

i=1

1

λ2(i)

n(i)n(i)T . (5.78)

The extremum values of the strain tensors E and e and their directions (princi-pal strains and principal directions) are related to the principal stretches and theirdirections in a simple fashion.

In the eigenvalue problem,

Gαβ N(γ )β − G(γ ) N(γ )

α = 0, (5.79)

if we substitute

Gαβ = 2Eαβ + δαβ, (5.80)

we get

Eαβ N(γ )β − 1

2(G(γ ) − 1)N(γ )

α = 0. (5.81)


The principal directions of the strain tensor E are the same as N(γ ), and the principalstrains are

E(γ ) = 12

(G(γ ) − 1). (5.82)

Similarly,

e(i) = 12

(1 − g(i)). (5.83)

Thus the solution of the eigenvalue problem for G or g also gives the principalstrains and their directions. We note that the strain matrix is not positive definiteand the term “strain ellipsoid” will be incorrect.

5.3.6 Polar Decomposition of the Deformation Gradient

The deformation gradient tensor

Fαi = ∂αxi (5.84)

has the property

F FT = G, (5.85)

where, as we have seen, G is symmetric and positive definite, but F is neither.From the theory of matrices, we know that we can factor any real, square matrix

into an orthogonal matrix and a positive-definite symmetric matrix. The deforma-tion gradient matrix F can be expressed as

F = RU = VR, (5.86)

where R is an orthogonal matrix (rotation matrix) and U is a positive-definite sym-metric matrix known as the right-stretch matrix. The matrix V is called the left-stretch matrix. This factorization is called the polar decomposition. As discussed inChapter 2, the orthogonal matrix R satisfies

RT R = I, RRT = I, RT = R−1. (5.87)

Using these properties, from Eq. (5.86), we get

FT F = U2 = [gi j ] = [∂αxi∂αx j

],

F FT = V2 = [Gαβ] = [∂αxi∂β xi

].

We note that the eigenvectors N(γ ) of G are the same as those of V. The spectralrepresentation of V is

V =3∑

γ=1

�(γ ) N(γ ) N(γ )T . (5.88)

In a similar way, we may resolve the Cauchy deformation gradient f as

f = ru = vr, (5.89)

64 Deformation

where

rT r = r rT = I, (5.90)

and

f T f = u2 = [Gαβ] = [∂i Xα∂i Xβ

],

f f T = v2 = [gi j ] = [∂i Xα∂ j Xα] .

From F f = I,

Vu = I, Uv = I, Rr = I. (5.91)

In terms of the eigenvalues and eigenvectors of g, 1/λ(i), and n(i),

v =3∑

i=1

1λ(i)

n(i)n(i)T . (5.92)

5.3.7 Stretch and Rotation

In the relations

dx = FT dX, dX = f T dx, (5.93)

using the polar decomposition, we get

dx = URT dX = RT VdX, dX = urT dx = rTvdx (5.94)

The operations on dX show that, in the first case, the material element is rotated byRT and then stretched by U, and in the second case, it is first stretched by V andthen rotated by RT .

For an arbitrary unit vector N = dX/dS, we obtain

dx = RT VNdS.

Writing n = dx/ds, we have

n = dSds

RT VN. (5.95)

If N = N(γ ), a principal direction of G, VN(γ ) = �(γ ) N(γ ) and

n = RT N(γ ). (5.96)

Thus the unit vector along a principal direction undergoes a rigid body rotation. Atriad of mutually orthogonal unit vectors representing the three principal directionswill get transformed to a new triad of orthogonal unit vectors.

Using f = RT V−1, we obtain

g = RT V−2 R, (5.97)

gn = RT V−2 RRT N(γ ) = 1

�2(γ )

RT N(γ ) = n�2

(γ )

. (5.98)


Thus n is an eigenvector of g, and the new triad just mentioned points along theprincipal directions of the material ellipsoid of Cauchy. Also, the largest semima-jor axis of the spatial ellipsoid gets mapped into the smallest semimajor axis of thematerial ellipsoid.

To find the factors U and V, we need the square root of the positive-definitematrix G. We can use the Cayley–Hamilton theorem, which states that any squarematrix satisfies its own characteristic equation, to obtain the square root of a matrix.The characteristic equation for G is Eq. (5.72), and

P(G) ≡ −G3 + IG1G2 − IG2G + IG3 = 0. (5.99)

According to the Cayley–Hamilton theorem, the matrix G satisfies the matrix poly-nomial equation

P(G) ≡ −G3 + IG1G2 − IG2G + IG3 I = O, (5.100)

where O is a 3 × 3 matrix with all elements zero. We have also used the notation Pfor the matrix function obtained from the regular function P. This result implies thatany power of G higher than 2 can be expressed in terms of G2, G, and I. In fact, anymatrix function φ of G can be expressed the same way. Moreover, the eigenvectorsof G are also the eigenvectors of φ, and the eigenvalues of φ(G) are φ(G).

Thus

G1/2 = aG2 + bG + cI, (5.101)

where the coefficients a, b, and c have to be found.For the three eigenvalues G(γ ) of G, the eigenvalues of G1/2 are (G(γ ))1/2. Using

these, we get three equations,

(G( γ ))1/2 = a(G( γ ))2 + bG( γ ) + c, γ = 1, 2, 3, (5.102)

for a, b, and c.When there are repeated roots G(γ ), we may use the Frobenius method, in which

we form a new equation for a, b, c by differentiating Eq. (5.101) with respect to G.

5.3.8 Example: Polar Decomposition

Consider the deformation of a square region in the X1, X2 plane. First, let us stretchthis region in the X1 direction, doubling the length. Then, we rotate it by θ = 30◦

(see Fig. 5.6). The spatial coordinates of an initial point X1, X2 become

x1 = 2X1 cos θ − X2 sin θ,

x2 = 2X1 sin θ + X2 cos θ. (5.103)

The Green’s deformation gradient matrix is obtained as

F =[ √

3 1−1/2

√3/2

]. (5.104)

66 Deformation

X1

X2

x1

x2

θ Figure 5.6. Stretching and rotation.

The Green’s deformation matrix is obtained as

G = F FT =[

4 00 1

]. (5.105)

The characteristic equation for G is found as∣∣∣∣∣ 4 − G 00 1 − G

∣∣∣∣∣ = 0. (5.106)

The eigenvalues are

G(1) = 4, G(2) = 1. (5.107)

The corresponding eigenvectors are

N(1) ={

10

}, N(2) =

{01

}. (5.108)

The spatial ellipse for this plane deformation is given by(dX1

0.5

)2

+(

dX2

1

)2

= 1, (5.109)

where we have assumed the radius of the spatial sphere is k = 1. The square root ofthe diagonal matrix G is

V =[

2 00 1

], (5.110)

and, using this, we obtain

R =[ √

32

12

− 12

√3

2

]. (5.111)

This corresponds to the 30◦ rotation we imparted to our region.By inverting the matrix F, we get

f =[√

3/4 −1/21/4

√3/2

], (5.112)

g = f f T = 116

[7 −3

√3

−3√

3 13

]. (5.113)


The eigenvalues of g are

1

λ2(1)

= 14,

1

λ2(2)

= 1, (5.114)

and the corresponding principal directions are

n(1) ={ √

3212

}, n(2) =

{− 12√3

2

}. (5.115)

A circle in the undeformed configuration,

(dX1)2 + (dX2) = 1, (5.116)

gets mapped into the ellipse,

716

dx12 − 6

√3

16dx1dx2 + 13

16(dx2)2 = 1, (5.117)

which has a semimajor axis of 2 and semiminor axis of 1.

5.3.9 Example: Square Root of a Matrix

Consider the deformation gradient matrix

F =

⎡⎢⎣

√3 1 0

0 2 00 0 1

⎤⎥⎦ . (5.118)

The Green’s deformation matrix is obtained as

G = F FT =

⎡⎢⎣ 4 2 0

2 4 00 0 1

⎤⎥⎦ . (5.119)

The eigenvalues are obtained from

[(4 − G)2 − 4](1 − G) = 0, (5.120)

which has the roots

G(1) = 1, G(2) = 2, G(3) = 6. (5.121)

Writing

V = G1/2 = aG2 + bG + cI, (5.122)

we find the equations for a, b, and c as

12a + 1b + c = √1,

22a + 2b + c = √2,

62a + 6b + c = √6.

(5.123)

68 Deformation

Solution of these equations gives

a = 4 − 5√

2 + √6

20, b = −32 + 35

√2 − 3

√6

20, c = 48 − 30

√2 + 2

√6

20. (5.124)

Using these, we find

V = G1/2 = 1√2

⎡⎢⎣

√3 + 1

√3 − 1 0√

3 − 1√

3 + 1 00 0

√2

⎤⎥⎦ . (5.125)

The rotation matrix R and the left-stretch matrix U are found as

R = V−1 F = 1

2√

6

⎡⎢⎣

√3 + 1

√3 − 1 0

1 − √3

√3 + 1 0

0 0 2√

6

⎤⎥⎦ , (5.126)

U = RTF = 1

2√

2

⎡⎢⎣ 3 + √

3 3 − √3 0

3 − √3 3

√3 + 1 0

0 0 2√

2

⎤⎥⎦ . (5.127)

Of course, it is easy to see in the preceding calculation that we dealt with only a2 × 2 submatrix!

5.4 Logarithmic Strain

Apart from the Green’s and Almansi strain tensors, a third strain measure fre-quently used is the logarithmic strain. If we consider a bar of current length L,stretched by �L, the incremental strain is

�EL = �LL

, (5.128)

which has the infinitesimal form

dEL = dLL

. (5.129)

Integrating this, we find, when the length has changed from L0 to L,

EL = ln(L/L0). (5.130)

Using the left-stretch tensor V, we have the tensor version of this relation:

EL = lnV. (5.131)

5.5 Change of Volume

Let us consider a small parallelepiped formed by three material elements, dX, dX∗,and dX∗∗. The volume enclosed is the scalar product of these elements:

dV = dX · dX∗ × dX∗∗ = eαβγ dXαdX∗βdX∗∗

γ . (5.132)

5.6 Change of Area 69

Under deformation, these elements map onto dx, dx∗, and dx∗∗, respectively. Thenew volume is

dv = dx · dx∗ × dx∗∗ = ei jkdxi dx∗j dx∗∗

k

= ei jk∂αxi∂β xj∂γ xkdXαdX∗βdX∗∗

γ

= eαβγ dXαdX∗βdX∗∗

γ det[F ]

= JdV, (5.133)

where we have used

ei jk∂αxi∂β xj∂γ xk = eαβγ J. (5.134)

Using j = 1/J , we obtain

dV = jdv. (5.135)

If we multiply this by ∂mXα and use the relation

∂mXα∂αxi = δim, (5.136)

we get an expression for the cofactors in a determinant:

ei jk∂β xj∂γ xk = eαβγ J∂i Xα. (5.137)

5.6 Change of Area

Consider an element of area dA in the undeformed configuration, with a unitnormal N. Vectorial representation of this area element is

d A = NdA. (5.138)

In this representation, the direction of N is not unique. However, if A is viewed asa vector product of two material elements dX and dX∗, in the form

d A = dX × dX∗ = eαβγ dXβdX∗γ eα,

dAα = eαβγ dXβdX∗γ , (5.139)

the direction of N is unique.Deformation of the body maps dX onto dx and dX∗ onto dx∗, and the

deformed elements form the area

da = nda = dx × dx∗,

dai = dxj dx∗kei jk

= ∂β xj∂γ xkei dXβdX∗γ . (5.140)

Using the cofactor expansion, Eq. (5.137), we get

dai = J∂i XαdAα. (5.141)

70 Deformation

This relation has as its inverse

dAα = j∂αxi dai . (5.142)

5.7 Compatibility Equations

In three dimensions the strain tensor Eαβ has six components that are related to thethree displacement components Uα in the form

2Eαβ = Gαβ − δαβ = ∂αUβ + ∂βUα + ∂αUγ ∂βUγ . (5.143)

If the displacements are known, it is straightforward to get the strain componentsthrough differentiation. If we reverse this situation and assume that six functions aregiven for the strain components, we have to integrate the preceding six equations tofind three displacements. This poses a problem of an overdetermined system. Thegiven six strain functions have to be constrained in some way to guarantee single-valued continuous displacements. These required restrictions are called the compat-ibility equations. A crude incompatible strain distribution is shown in Fig. 5.7. Here,the undeformed body has been cut into small pieces. Prescribing a strain distribu-tion is the same as stretching and distorting each piece as we wish. Next, we try toassemble them into the deformed body. The arbitrary stretches and distortions re-sult in a body with holes and a shape that depends on the order of assembling thepieces.

In the case of infinitesimal strains, we have

2Eαβ = Gαβ − δαβ = ∂αUβ + ∂βUα. (5.144)

From these we can eliminate Uα to obtain six relations among Eαβ . A compact wayto write these six equations is

∇ × (∇ × E)T = O. (5.145)

In two dimensions, there is a single relation:

E11,22 + E22,11 = 2E12,12. (5.146)

Figure 5.7. Incompatibility of arbitrarystrains.

5.8 Spatial Rotation and Two-Point Tensors 71

For the case of finite strains, the elimination method is very cumbersome, if notimpossible. An alternative approach is to use the fact that the deformed body existsin the flat three-dimensional (3D) space; that is, the Riemann–Christoffel curvaturehas to be zero. With Gα as base vectors and Gαβ as the metric tensor, we can set

Rαβγ δ = 0. (5.147)

Recall the expression for R from Chapter 3. It turns out, if we linearize the curva-ture expression, we get back the six compatibility equations previously mentioned.

5.8 Spatial Rotation and Two-Point Tensors

So far we have used a single Cartesian coordinate system, and if we rotate this sys-tem, both Xα and xi change according to

X′α = Qαβ Xβ, x′

i = Qi j xj , (5.148)

where Q is a rotation matrix. Quantities such as Gαβ , Fαi , and gi j transform as

G′ = QGQT , F′ = QF QT , g′ = Qg QT . (5.149)

There are times when we would like to keep the material coordinate system fixed asXα and use a rotated Cartesian system for the spatial description. Then

x′i = Qi j xj . (5.150)

With this,

G′αβ = ∂x′

i

∂ Xα

∂x′i

∂ Xβ

= Qi j Qik∂x′

j

∂ Xα

∂x′k

∂ Xβ

= Gαβ, (5.151)

where we use Qi j Qik = δ jk,

F ′αi = ∂x′

i

∂ Xα

= Qi j Fα j , F′ = F QT , (5.152)

g′i j = ∂ Xα

∂xi

∂ Xα

∂xj= Qimgmn Qjn, g′ = Qg QT . (5.153)

We call G a Lagrangian tensor as it is invariant during the rotation of the spatialcoordinate system, and we call g an Eulerian tensor as it transforms according tothe rules for second-rank tensors during the rotation of the spatial frame. The de-formation gradient tensor F has a mixed behavior, as though it has one foot in theLagrangian system and the other in the Eulerian system. Such tensors are calledtwo-point tensors.

Transformation rules can be derived in a similar way if we keep the spatial framefixed and rotate the material frame.

72 Deformation

X2

X3x1

x2

x3

O

uO′

X

x

X1

Figure 5.8. Curvilinear coordinates.

5.9 Curvilinear Coordinates

This chapter concludes with a brief introduction to the description of deformationby use of general curvilinear coordinates. As shown in Fig. 5.8, we use two curvilin-ear coordinates, X α and xi , to describe the initial configuration B0 and the currentconfiguration B. To avoid any confusion with the deformation gradient vectors wesaw in the Cartesian description, let Cα and ci denote the base vectors for the twocurvilinear systems. The deformation gradient tensors are

Fαi = ∂xi

∂ Xα, fi

α = ∂ Xα

∂xi. (5.154)

With

Cαβ = Cα.Cβ, ci j = ci .c j , (5.155)

the initial and current distances are given by

dS2 = CαβdXαdXβ = Cαβ fiα f j

βdxi dx j , (5.156)

ds2 = ci j dxi dx j = ci j Fαi Fβ

j dXαdXβ. (5.157)

The Green’s strain tensor Eαβ and the Almansi strain tensor ei j are given by

ds2 − dS2 = 2EαβdXαdXβ = 2ei j dxi dx j . (5.158)

The displacement vector can be written as

U = u = CαUα = ci ui . (5.159)

With the preceding notation, it is easy to recast our earlier Cartesian description intothe general curvilinear form. Green and Zerna (1954), Green and Adkins (1960),and Eringen (1962) are excellent references for further study of the general formu-lations of deformation.


SUGGESTED READING

Eringen, A. C. (1962). Nonlinear Theory of Continuous Media, McGraw-Hill.Green, A. E. and Adkins, J. E. (1960). Large Elastic Deformations and Nonlinear Continuum

Mechanics, Oxford University Press.Green, A. E. and Zerna, W. (1954). Theoretical Elasticity, Oxford University Press.

EXERCISES

5.1. The deformation of a body is given by the mapping

x1 = X1 + εX2,

x2 = −εX1 + X2,

x3 = X3.

Obtain the following:

(a) the deformation gradient tensor Fαi

(b) the deformation tensor Gαβ

(c) the Jacobian J(d) the strain tensor, Eαβ

(e) the maximum and minimum stretches and their directions(f) the right-stretch tensor U, the left-stretch tensor V, and the rotation tensor

R(g) approximate expressions for the preceding quantities by neglecting

quadratic terms in ε

(h) the shape of a unit circle, X21 + X2

2 = 1, after the deformation using exactand approximate expressions

5.2. Show that an element dX, oriented in a principal direction N(γ ), undergoes apure rotation and a pure stretch during deformation. Use the polar decomposition tohelp with this deduction. Also show that any other element would undergo rotationduring the stretching operation in dX · V · R.

5.3. A cross-section of a thin rubber tube has the flat initial position

Y = 0, X = cos θ, 0 ≤ θ < 2π,

where we assume that the thickness of the tube is negligible. If it is deformed intoan ellipse,

x = a cos θ, y = b sin θ,

where a and b are constants. Obtain an expression for the tangential stretch of thetube wall. Find the locations of the maximum and minimum stretches.

5.4. In an experiment, a specimen undergoing finite plane strain showed stretches� = 0.8 along the X1 direction and � = 0.6 along the X2 direction. To deduce theshear strain, a third measurement of the stretch along the direction 45◦ between thetwo coordinate directions is made. This has a value of 0.5. Obtain all three compo-nents of strain for this case.

74 Deformation

5.5. Expand the compatibility equation

∇ × (∇ × E)T = O.

Specialize its components to the plane strain case.

5.6. A rubber tube has an annular cross-section with inner radius a and outer radiusb. The tube is radially sliced and rejoined in such a way that the old outer surfaceis now the inner surface. Assume that the midsurface radius and the tube thicknessremain constant. Using polar coordinates R, φ for the material description and r, θfor the spatial description, obtain deformation relations

r = r(R, φ), θ = θ(R, φ),

assuming the cut is made along φ = 0, for this eversion. Also, compute the Green’sdeformation and strain tensors using the R, φ coordinates.

5.7. A plane deformation is given by

x1 = X1 + X2, x2 = X2 − X1.

Obtain the deformation gradient F, deformation tensor G, and the factors of F,U, R, and V. A unit circle in the undeformed body with four points at θ = 0,

π/2, π, 3π/2, marked as A, B, C, D, deforms into a curve under this deformation.Obtain the equation for this curve and identify the locations of the four markedpoints.

x

y

R Figure 5.9. Circular ring.

5.8. A circular ring (see Fig. 5.9) has a “neutral” radius R, and its cross-section isspanned by the coordinates x, y. Suppose we apply a uniform, distributed momentaround the ring and rotate it by an angle φ. Compute the strain of a circular fiber inthe ring. Note that, when φ = π , the outermost fiber becomes the innermost duringthe deformation, and the fiber located at the neutral radius does not undergo anystrain. Linearize the strains assuming that the angle φ is small.

5.9. An infinite thin circular membrane has a hole of radius unity. The hole is closedby deforming it according to

x1 =(

R + 1R

)cos θ, x2 =

(R − 1

R

)sin θ,

where R and θ are the polar coordinates of a material particle.Obtain an expression for the stretch of an element tangential to the circle of radiusR as a function of θ . Locate the points in the undeformed membrane where thesestretches are a maximum and a minimum.


5.10. In a cylindrical coordinate system with coordinates r , θ , and z, and correspond-ing unit vectors, er , eθ , and k, the displacement components are u, v, and w. Obtainthe strain components in this system and reduce them to the small-strain versions:

err = ∂u∂r

, eθθ = 1r

∂v

∂θ+ u

r, ezz = ∂w

∂z,

2erθ = 1r

∂u∂θ

+ ∂v

∂r− v

r,

2eθz = ∂v

∂z+ 1

r∂w

∂θ,

2ezr = ∂w

∂r+ ∂u

∂z.

Further, if the deformation is axisymmetric, what are the expressions for the smallstrains?

5.11. In the general curvilinear coordinates, show that the Green’s strain and theAlmansi strain are given by

2Eαβ = DαUβ + DβUα + DαUγ DβUγ ,

2ei j = Di u j + Dj ui − Di ukDj uk,

where Dα and Di are the covariant differential operators in the X α and xi coordi-nate systems, respectively.

5.12. Formulate the eigenvalue problem in the general curvilinear setting for theprincipal strains when the components of the Green’s strain Eαβ or the Almansistrain ei j are given. Note that a unit normal vector N satisfies

Cαβ Nα Nβ = Nα Nα = 1.

6 Motion

In the last chapter we considered the initial position X and the final position x ofa material particle. The motion of the particle from X to x is a continuous process.This continuity is expressed, with the parameter t representing time, as

x = x(X1, X2, X3, t), X = X(x1, x2, x3, t). (6.1)

6.1 Material Derivative

In the motion of continuous media the time rates of changes of field quantities asso-ciated with the material particles play an important role. Let A, the component ofa tensor field, be a function of the material coordinates Xα and time t . The materialderivative or substantial derivative of A is defined as

DADt

= A = lim�t→0

1�t

{A X,t+�t − A X,t

}

= ∂ A∂t

(X, t), (6.2)

where X stands for X1, X2, X3.If A is a function of the spatial coordinates x and time t , we have

DADt

= A = lim�t→0

1�t

{A[xi (X, t), t]

X,t+�t− A[xi (X, t), t]

X,t

}

= lim�t→0

1�t

{A[xi (X, t) + ∂xi

∂t X,t�t + · · · , t + �t] − A[xi (X, t), t]

}

= lim�t→0

1�t

{A(xi , t + �t) + ∂ A

∂xi

∂xi

∂t�t − A(xi , t)

}

= ∂ A∂t xi

+ ∂ A∂xi

∂xi

∂t X,t.

76

6.1 Material Derivative 77

The time derivative of x, with the material coordinates fixed, is called the velocityof the particle:

v = ∂x∂t

. (6.3)

With this, for any tensor-valued function A, we have

DADt

= ∂ A∂t

+ v · ∇A. (6.4)

In the preceding equation, the first term represents the change in A that is due tothe change in time at a fixed spatial position. The second term represents the changethat is due to the change in x with time. The first term is called the local rate ofchange, and the second term the convective rate of change. Sums and products oftwo functions A and B obey

DDt

(A + B) = DADt

+ DBDt

,DDt

(AB) = DADt

B + ADBDt

. (6.5)

The acceleration a of a particle is defined as

a = v = ∂v

∂t+ v · ∇v, v = v(x, t),

= ∂v

∂t, v = v(X, t). (6.6)

If we know the Mth material derivative, we can find the (M + 1)th by the relation(DDt

)M+1

A = ∂

∂t

(DDt

)M

A + v · ∇(

DDt

)M

A. (6.7)

6.1.1 Some Terminology

At a spatial point x, if v = 0, that point is called a stagnation point.If at a point x the velocity v does not explicitly depend on time, that is,

v = v(x), (6.8)

the motion is called steady. However, if we express x in terms of X and t , we find

v = v[x(X, t)], (6.9)

with t appearing explicitly.A particle that was at X at time t = 0 is called particle X.Path Lines: A path line is a curve traversed by a particle X as time t varies. It

has the parametric equation

x = x(X, t); Xα : fixed, −∞ < t < ∞. (6.10)

When a velocity field is given as vi = vi (x, t), the path lines can be obtained bysolving the differential system

dxi

dt= vi , xi |t=0 = Xαδαi . (6.11)

78 Motion

X, t = 0

x, t

Time = tx∗

A

B

Figure 6.1. Path line, stream line, and streak line.

In Fig. 6.1, the curve on the left shows the path of a particle as time varies.Stream Lines: At a given time t , the curves tangential to the velocity vectors are

called stream lines. That is,

dx = kv (6.12)

or

dx1

v1= dx2

v2= dx3

v3. (6.13)

Obviously, at a given time, there is no velocity normal to the stream lines. InFig. 6.1, the middle diagram shows a stream line with tangential velocity vectors.

Streak Lines: Streak lines are curves representing the locus of material particlesat time t that would pass through or would have passed through a given point x∗.The particle that would be at x∗ at some time τ is identified as

X∗ = X(x∗, τ ). (6.14)

As the parameter τ traverses from the past to the future, we collect all the particlesthat would ever visit x∗. Their location at a particular time t is obtained as

x = x(X∗, t) = x[X(x∗, τ ), t]. (6.15)

In Fig. 6.1, the curve AB represents the location of the particles at time t thatwould pass through x∗ at some time, along with their paths.

Material curve: We may express parametrically a contiguous curve of materialparticles as

Xα = Xα(s), (6.16)

where s is a parameter. This set is called a material curve.Material surface: With two parameters, s1 and s2, we may define a material

surface:

Xα = Xα(s1, s2). (6.17)

These material curves and surfaces flow with the motion and occupy spatial curvesand surfaces given by

xi = xi [X(s), t)],

xi = xi [X(s1, s2), t]. (6.18)

6.1 Material Derivative 79

Transient spatial curve: An equation of the form

f (x, t) = 0 (6.19)

describes a spatial curve that changes with time. If time changes by �t , the coordi-nates have to increase by �xi to meet

�t∂ f∂t

+ �xi∂i f = 0. (6.20)

Dividing by �t and taking the limit, we find

∂ f∂t

+ ∂xi

∂t∂i f = 0. (6.21)

At any instant this gives the rate of change of the coordinates of a point on thiscurve. If it is a material curve, this rate of change will be the velocity of the materialparticle, that is,

∂xi

∂t= vi . (6.22)

From this we get the Lagrange criterion for a spatial curve to be a material curve:

DfDt

= ∂ f∂t

+ v · ∇ f = 0. (6.23)

6.1.2 Example: Path Line, Stream Line, and Streak Line

Consider the plane motion given by the equations

x1 = X1 cosh t + X2 sinh t, x2 = X1 sinh t + X2 cosh t. (6.24)

1. Obtain the path line of a particle that was at the point (0,1) at time t = 0.Using t = 0 and x1 = 0, x2 = 1 in Eqs. (6.24), we find X1 = 0, X2 = 1. The equa-tion for the path line (see Fig. 6.2) has the parametric form

x1 = sinh t, x2 = cosh t,

which represents the hyperbola

x22 − x2

1 = 1. (6.25)

x1

x2

Figure 6.2. Path line.

80 Motion

2. Obtain the stream line through (0,1) at time t = 0.From Eqs. (6.24), the velocity components are

v1 = X1 sinh t + X2 cosh t, v2 = X1 cosh t + X2 sinh t,

Eliminating X1 and X2, by using Eqs. (6.24) we obtain

v1 = x2, v2 = x1.

As stream lines are tangential to the velocity vector,

dx1

v1= dx2

v2or x1dx1 − x2dx2 = 0.

Integrating this, we find x22 − x2

1 = C. The condition x1 = 0, x2 = 1 yields thehyperbola

x22 − x2

1 = 1. (6.26)

3. Obtain the streak line through (0,1) when t = 1.To identify all the particles that would visit the spatial location (0,1) in time τ ,we invert Eqs. (6.24) and substitute x1 = 0, x2 = 1 to get

X1 = − sinh τ, X2 = cosh τ.

Their location at time t is given by

x1 = − cosh t sinh τ + sinh t cosh τ = sinh(t − τ ),

x2 = cosh t cosh τ − sinh t sinh τ = cosh(t − τ ).

Using t = 1 and eliminating (1 − τ ), we have, again,

x22 − x2

1 = 1. (6.27)

Of course, for a steady flow given by Eqs. (6.24), all three lines are identical. Thisexample illustrates the procedure involved in finding these lines from the givenmotion.

6.2 Length, Volume, and Area Elements

In this section we are interested in finding the material derivatives of elements ofarc length, area, and volume.

With xi = xi (X), we begin by computing

DDt

(∂αxi ) = ∂αvi = ∂ jvi∂αxj . (6.28)

By removing ∂α , we find

DDt

(dxi ) = ∂ jvi dxj . (6.29)

From the identity

∂i Xα∂β xi = δαβ,

6.2 Length, Volume, and Area Elements 81

we get

DDt

(∂i Xα∂β xi ) = 0,

∂β xiDDt

(∂i Xα) = −∂i Xα

DDt

(∂β xi ),

∂β xiDDt

(∂i Xα) = −∂ j Xα∂iv j∂β xi ,

DDt

(∂i Xα) = −∂iv j∂ j Xα. (6.30)

6.2.1 Length

We have

DDt

(ds)2 = DDt

(dxi dxi ) = 2∂iv j dxi dxj . (6.31)

The velocity gradient

li j = ∂iv j (6.32)

can be resolved into a symmetric and a skew-symmetric component as

∂iv j = li j = di j + wi j , (6.33)

where

di j = 12

(li j + l ji ), wi j = 12

(li j − l ji ). (6.34)

The tensors d and w are known as the deformation rate and the spin tensors, respec-tively. As dxi dxj is symmetric, we get

DDt

(ds)2 = 2di j dxi dxj . (6.35)

6.2.2 Volume

From

dv = J dV, (6.36)

we get

DDt

(dv) = DJDt

dV, (6.37)

where

DJDt

= ∂ J∂(∂αxi )

DDt

(∂αxj )

= ∂ J∂(∂αxi )

∂αxj∂ jvi . (6.38)

82 Motion

Observing that the partial derivative of J is the cofactor in the determinant expan-sion for J , we obtain

∂ J∂(∂αxi )

= J∂i Xα. (6.39)

Then

DJDt

= J∂ivi ,DDt

(dv) = ∂ivi dv. (6.40)

6.2.3 Area

The material derivative of an area element, dai , can be obtained as

DDt

(dai ) = DDt

(J dAα∂i Xα)

=[

DJDt

∂i Xα + JDDt

(∂i Xα)]

dAα

= ∂ jv j dai − ∂iv j da j . (6.41)

6.3 Material Derivatives of Integrals

We come across different types of integral quantities in continuum mechanics, andit is important to know how to calculate the rate of change of these integrals.

6.3.1 Line Integrals

Let C0 be a material curve at t = 0 and C be its deformed shape at time t . Thematerial derivative of the line integral of φ, which is a component of a tensor-valuedfunction, can be written as

DDt

∫C

φ(x)dxi = DDt

∫C0

φ(x)∂αxi dXα

=∫

C0

DDt

(φ∂αxi )dXα

=∫

CO

[∂αxi

Dφ

Dt+ φ∂ jvi∂αxj

]dXα

=∫

C

[Dφ

Dtdxi + φ∂ jvi dxj

]. (6.42)

If C is a fixed curve in space,

ddt

∫C

φdxi =∫

C

∂φ

∂tdxi . (6.43)

6.4 Deformation Rate, Spin, and Vorticity 83

6.3.2 Area Integrals

For a material surface S, using the transformation to the undeformed system, we getthe material derivative of the area integral of φ:

DDt

∫Sφ(x)dai =

∫S

[Dφ

Dtdai + φ(∂ jv j dai − ∂iv j da j )

]. (6.44)

If the surface S is fixed in space,

ddt

∫Sφdai =

∫S

∂φ

∂tdai . (6.45)

6.3.3 Volume Integrals

If V is a material volume, the material derivative of the volume integral is obtainedas

DDt

∫V

φ(x)dv =∫

V

[Dφ

Dt+ φ∂ivi

]dv

=∫

V

[∂φ

∂t+ ∂i (φvi )

]dv. (6.46)

For a spatial volume,

ddt

∫V

φdv =∫

V

∂φ

∂tdv. (6.47)

6.4 Deformation Rate, Spin, and Vorticity

We have introduced the deformation rate tensor d and the spin tensor w as

d = 12

[∇v + (∇v)T ], w = 12

[∇v − (∇v)T ]. (6.48)

The skew-symmetric tensor w has three nonzero components:

w12 = −w21, w23 = −w32, w31 = −w13. (6.49)

We can associate a vector ω, called a dual vector, with the skew-symmetric tensor,with its components defined as

ωi = ei jkw jk = ei jk∂ jvk (6.50)

or

ω = ∇ × v. (6.51)

The vector ω is called the vorticity vector. In particular,

ω1 = 2w23, ω2 = 2w31, ω3 = 2w12. (6.52)

84 Motion

For a physical interpretation of d, we begin by defining the stretching of an element,dx, oriented in the direction n in the current configuration as

dn = 1ds

DDt

(ds) = 12

1(ds)2

DDt

(ds)2,

= di jdxi

dsdxj

ds= di j ni n j . (6.53)

This shows that the diagonal elements d11, d22, and d33 are the rates at which ele-ments pointing in the 1-, 2-, and 3-directions stretch.

To see the significance of the off-diagonal elements of d, consider two materialelements dx and dx∗ oriented in the directions n and n∗, respectively (see Fig. 6.3).If θ is the angle between them,

cos θ = ni n∗i = dxi

dsdx∗

i

ds∗ , (6.54)

DDt

(cos θ) = DDt

(dxi

dsdx∗

i

ds

)

= dxi

dsds∗DDt

(dx∗i ) + dx∗

i

dsds∗DDt

(dxi )

− dxi dx∗i

(ds)2ds∗DDt

(ds) − dxi dx∗i

ds(ds∗)2

DDt

(ds∗)

= 2di jdxi

ds

dx∗j

ds∗ − [dn∗ + dn]dxi

dsdx∗

i

ds∗ ,

− sin θDθ

Dt= 2di j ni n∗

j − [dn + dn∗ ] cos θ. (6.55)

n

n∗

θ12

Figure 6.3. Shearing deformation of the angle between two material elements.

6.4 Deformation Rate, Spin, and Vorticity 85

In the special case, n = e1 and n∗=e2, we have θ = π/2 and

− Dθ

Dt= 2d12. (6.56)

Thus 2d12 represents the rate of closing of the 90◦ angle between e1 and e2. Werefer to d12 as the mathematical shearing strain. To interpret the spin components,we consider a fixed unit vector ν in space and a unit vector n formed by materialparticles. Let φ denote the angle between them. Then,

cos φ = νi ni = νidxi

ds(6.57)

DDt

cos φ = νiDDt

(dxi

ds

),

− sin φDφ

Dt= νi

[∂ jvi

dxj

ds− dxi

(ds)2

DDt

(ds)]

= νi [nj∂ jvi − ni dn]. (6.58)

There are two special cases:

1. Assuming

ν = e1, n = e2, (6.59)

we get

Dφ21

Dt= −∂2v1. (6.60)

2. Assuming

ν = e2, n = e1, (6.61)

we get

Dφ12

Dt= −∂1v2. (6.62)

From the expression for φ21 we observe that, if v1 increases with x2, the verticalmaterial element gets rotated clockwise. The velocity component v2 is responsiblefor rotating a horizontal element, and if it increases with x1 an anticlockwise rotationis imparted. Here, we use the notation φ21 to indicate the angle from fixed e1 torotating e2 and φ12 when the two unit vectors are interchanged (see Fig. 6.4).

With these, we can express the deformation rate and spin components in theform

2d12 = −(φ12 + φ21), 2w12 = φ21 − φ12. (6.63)

Figure 6.5 shows the velocity distribution of a flow:

v1 = −12�x2, v2 = 1

2�x1. (6.64)

86 Motion

n = e1

n = e2

ν = e1

ν = e2

φ12

φ21

Figure 6.4. Rotation of material elements relative to fixed directions.

x1

x2−v1

v2

Figure 6.5. Instantaneous rotation of a flow.

From this,

∂1v2 = 12�, ∂2v1 = −1

2�, (6.65)

d12 = 0, w12 = 12�. (6.66)

The corresponding dual vector (vorticity) is obtained as

� = �e3. (6.67)

6.5 Strain Rate

Let us begin by noting that

DDt

(ds)2 = 2di j dxi dxj , (6.68)

DDt

[(ds)2 − (dS)2] = 2EαβdXαdXβ,DDt

(dS)2 = 0. (6.69)

From these,

Eαβ = ∂αxi∂β xj di j . (6.70)


The material derivative of the Almansi strain follows from

DDt

[(ds)2 − (dS)2] = 2DDt

(ei j dxi dxj ), (6.71)

as

ei j = di j − ∂ivkekj − eik∂ jvk. (6.72)

We also have

Gαβ = 2Eαβ, gi j = −2ei j . (6.73)

6.6 Rotation Rate of Principal Axis

In Chapter 5 we saw that a principal direction of the spatial ellipsoid N gets mappedinto a principal direction of the material ellipsoid n. Using the polar decomposition,we obtain

n = RTN. (6.74)

The rotation rate of this unit vector is given by

n = RTN. (6.75)

Eliminating N, we have

n = RTRn = WT n, (6.76)

where WT is called the twirl tensor.

SUGGESTED READING

Batchelor, G. K. (1967). An Introduction to Fluid Dynamics, Cambridge University Press.Bird, R. B., Stewart, W. E., and Lightfoot, E. N. (1960). Transport Phenomena, Wiley.

EXERCISES

6.1. A steady flow is defined by the relation v = v(x). Show that the path lines,stream lines, and streak lines through a specified point coincide for this case.

6.2. Circulation of a fluid around a closed material curve C is defined as

� =∮

Cv · dx.

(a) Transform this integral into an area integral using Stokes theorem.(b) A motion is circulation preserving if � = 0. Show that, in that case,

∂ω

∂t+ ∇ × (ω × v) = 0.

(c) Deduce from this

∇ × v = 0.

88 Motion

(d) Show that the expression for acceleration,

v = ∇φ,

satisfies the preceding equation.

6.3. A flow is irrotational if w = 0. Show that this condition is satisfied if v is writtenas

v = ∇φ,

where φ is called the velocity potential.

6.4. A flow is incompressible if DJ/Dt = 0. Show that this condition is satisfied if v

is written as

v = ∇ × ψ .

6.5. For a plane (v3 = 0), incompressible flow, we may set ψ = ψe3, where ψ iscalled the stream function. Show that, for incompressible, irrotational flow,

∂1φ = ∂2ψ, ∂2φ = −∂1ψ,

∇2φ = 0, ∇2ψ = 0.

6.6. For a two-dimensional (2D) incompressible flow (v3 = 0) around a unit circlewith center at r = 0, the velocity components are

v1 = U(

1 − cos 2θ

r2

), v2 = −U

sin 2θ

r2,

where U is a constant representing the uniform velocity of the fluid in the far fieldand r and θ are polar coordinates. Show that the component of velocity normal tothe circular boundary is zero. Compute the rate of volume change and vorticity.Obtain the location of the maximum rate of change of the kinetic energy density,DK/Dt , for this flow. The kinetic energy density is defined as K = ρv2/2.

6.7. In a plane flow, fluid flows out of a source radially, with the radial velocity

vr = q2πr

,

where q is the source strength. An expanding space curve imposed on this flow hasthe equation

r − f (t) = 0.

Obtain f (t) so that this space curve is a material curve.

6.8. The equation for a material curve in a flow is given by

x − ty − t2 = 0,

where t represents time. The velocity component vy has a constant value, 2. Obtainthe component vx, the deformation rate tensor d, and the spin tensor w.

6.9. The motion of a fluid is described by

v1 = x2(V0 − �t), v2 = x1(−V0 + �t), v3 = 0.

Compute the deformation rate, spin, and vorticity for this flow. What is the equationfor the stream line through (1,1,0) at t = 1? Is this a steady flow?


6.10. In a plane flow the velocity components are

v1 = Ux2

a, v2 = −U

x1

a,

where U and a are constants. Obtain the stream line through the point (a, a). Attime t = 0, a material line in this flow has the equation x1 − x2 = 0. Obtain the equa-tion for this line at t = 1.

7 Fundamental Laws of Mechanics

The fundamental laws of mechanics dealing with mass, momentum, energy, and en-tropy are stated and briefly discussed in this chapter. These concepts have long his-tories, and it is not necessary or appropriate in our study of continuum mechanics toelaborate on these topics. Our discussion is based on the elementary ideas we havegathered from our earlier studies of statics, dynamics, and thermodynamics.

7.1 Mass

With each body we associate a measure called mass, which is nonnegative andadditive. It is assumed to be invariant under the class of motions encountered inengineering. With a lower limit on the length scales mentioned in Chapter 1, wetreat mass as continuous and define the mass density ρ as

ρ = lim�V→0

�m�V

. (7.1)

With this, the mass of a body occupying a volume V can be expressed as

m =∫

Vρdv, (7.2)

and integrals ∫m

f dm =∫

Vfρdv. (7.3)

The following mass integrals play important roles in mechanics:

1. Total mass:

m =∫

dm. (7.4)

2. Linear momentum:

L =∫

mvdm. (7.5)

90

7.2 Conservation and Balance Laws 91

3. Moment of momentum or angular momentum:

H =∫

mx × vdm. (7.6)

Although this relation is based on the assumption that the moment is takenabout the origin, we are free to select any convenient point.

4. Kinetic energy:

K = 12

∫m

v · vdm. (7.7)

5. Internal energy:

E =∫

mεdm, (7.8)

where ε is the internal energy density.6. Entropy:

S =∫

msdm, (7.9)

where s is the entropy density.

We will modify some of these integrals to include microstructural details with addi-tional degrees of freedom such as microrotations, on top of the three displacementcomponents.

7.2 Conservation and Balance Laws

7.2.1 Conservation of Mass

The law of conservation of mass states that total mass of a body remains invariantunder any motion:

m =∫

V0

ρ0dV0 =∫

VρdV. (7.10)

7.2.2 Balance of Linear Momentum

The time rate of change of linear momentum is equal to the applied force F actingon the body:

DLDt

= F ⇔ DDt

∫V

vρdV = F. (7.11)


7.2.3 Balance of Angular Momentum

The time rate of change of angular momentum about a fixed point is equal to theapplied moment about the same fixed point M:

DHDt

= M ⇔ DDt

∫V

[x × v + J · µ]ρdV = M. (7.12)

Here, we modify the classical moment of momentum, ρx × v, by adding a mi-crostructural spin angular momentum, ρ J · µ, with ρ J representing inertia andµ the angular velocity. We will discuss the significance of this complication inChapter 9. The two balance laws previously stated (without the complication) inthe global form are called the Euler equations of motion.

7.2.4 Balance of Energy

The time rate of change of the sum of the kinetic and the internal energies is equalto the rate of flow of energies into the body. Of all possible energies flowing into thebody, we concentrate mainly on the mechanical power input and the rate of heatenergy entering the body:

DDt

(K + E) = P + H. (7.13)

where P is the mechanical power input and H is the rate of flow of heat energy. Thisstatement is also known as the first principle of thermostatics.

7.2.5 Entropy Production

In any irreversible process taking place in a closed system, the rate of entropy pro-duced has to be positive. The total entropy change can be written as the sum of theentropy change that is due to reversible processes SE and irreversible processes SI :

S = SE + SI . (7.14)

The entropy production inequality or the second law of thermostatics states that

DSI

Dt≥ 0, (7.15)

with the equality applicable if there is no irreversible process involved. We will dis-cuss this principle in more detail in Chapter 9.

7.3 Axiom of Material Frame Indifference

We discussed the motion of a continuum with respect to a coordinate system fixedin space. In classical mechanics we consider the Newtonian ideal of absolute spaceand absolute time and an observer with a measuring tape and a clock. For everydayengineering applications (including trips to the Moon and beyond) this approachhas worked out satisfactorily. When the physical properties of materials are to be

7.3 Axiom of Material Frame Indifference 93

determined, we should know how the motion of the observer will affect the mea-surements. Although the values of the measured quantities may depend on the mo-tion of the observer, their interrelations must be independent of the observer. Forexample, the velocity and acceleration of a particle would appear differently to dif-ferent observers moving with their own velocities and accelerations. However, thedistance ds the particle covers in time dt must be independent of the observers.Quantities that are invariant under observer transformations are called frame indif-ferent. Other terms for this concept include objectivity and observer independence.An observer transformation is defined as

x′i = Qi j (t)xj + bi (t), (7.16)

where x and t represent an absolute Cartesian system of coordinates and time, as inclassical mechanics. The observer’s coordinate system x′ is translating and rotatingas a function of time. The rotation matrix Q(t) is orthogonal.

The distance relation is

(ds)2 = dx′i dx′

i = QimQindxmdxn = δmndxmdxn = dxmdxm. (7.17)

Thus the absolute distance and the distance measured by the observer are identical.It is useful to note that

QTQ = I, (7.18)

and, by differentiating, we obtain

QTQ = − QTQ. (7.19)

Including a constant shift in time t ′ = t − a, we define two motions, x′i (X, t ′) and

xj (X, t), as objectively equivalent if

x′i (X, t ′) = Qi j (t)xj (X, t) + bi (t). (7.20)

The relation between velocities,

DDt ′ x′

i (X, t ′) = Qi jDDt

xj (X, t) + Qi j x j (X, t) + bi (t),

v′i (X, t ′) = Qi jv j (X, t) + Qi j x j (X, t) + bi , (7.21)

shows that velocity components are not frame indifferent.Differentiating again, we have the acceleration

a′i = Qi j a j + 2Qi jv j + Qi j x j + bi , (7.22)

which is, again, not frame indifferent.An objective tensor would follow the transformation law for tensors under ob-

jective coordinate transformations. That is, Ai j is objective if

A′i j (X, t ′) = Qim(t)Qjn(t)Amn(X, t). (7.23)

Using the relations

dx′i = Qi j dxj , dxj = Qi j dx′

i , (7.24)


we can write the velocity gradient, which is not objective, as

∂ ′kv

′i = Qi j∂lv j

∂xl

∂x′k

+ Qi j∂xj

∂x′k

= Qi j Qkl∂lv j + Qi j Qkj . (7.25)

But the deformation rate becomes

d′ik = 1

2(∂ ′

kv′i + ∂ ′

i v′k) = Qi j Qkldjl + 1

2(Qi j Qkj + Qi j Qkj ), (7.26)

where the second term is zero. Thus d is objective as

d′i j = QimQjndmn. (7.27)

If we assemble the spin tensor, we find

w′i j = QimQjnwmn + 1

2(QikQjk − QikQjk), (7.28)

which is not objective.In matrix form

w′ = QwQT + 12

( QQT − QQT )

= QwQT + QQT. (7.29)

From this we obtain an expression for Q in terms of the spins in the two systems:

Q = Qw − w′Q. (7.30)

7.4 Objective Measures of Rotation

In applications involving viscoelastic polymers, objective measures of rotation havebeen found to be useful. One such measure is the objective part of the spin tensor(VanArsdale, 2003; Zhou and Tamma, 2003). Consider the velocity gradient in theform

∇v = ∂v j

∂xiei e j

= ∂ Xα

∂xi

DDt

∂xj

∂ Xα

ei e j

= f F (7.31)

= RTV−1[VR+ VR]

= RTR+ RT V−1VR. (7.32)

Using the transpose of this expression, we find

d = 12

RT [V−1V + VV−1]R, (7.33)

w = wr + wd, wr ≡ RTR, wd ≡ 12

RT [V−1V − VV−1]R, (7.34)

7.5 Integrity Basis 95

where wr represents the rigid body spin and wd the deformational contribution tothe overall spin; the latter part is a frame-independent quantity that is suitable foruse in describing material response to stresses.

Wedgewood and Geurts (1995) and Wedgewood (1999) introduced a rigid bodyrotation rate in a rotating–translating observer frame in such a way that the averagecomponents of the quantity v × v is set to zero. They assume the velocity distribu-tion in a small neighborhood in the form

v = r · ∇v. (7.35)

This approach leads to multiple values for the rotation rate in certain cases, and theboundary conditions have to be used to select the physically feasible solution.

Later, when we discuss the constitutive relations, we restrict the class of vari-ables used to be objective tensors, and these relations must satisfy the fundamentallaws of mechanics.

7.5 Integrity Basis

In continuum mechanics we encounter scalar-, vector-, and tensor-valued functionswith scalars, vectors, or tensors as arguments. The symmetry properties of the func-tions under coordinate rotations dictate the allowable forms for the arguments. Fora scalar function with scalar variables, we have the function and its variables in-variant under a rotation of coordinates. When a scalar function has vectors as ar-guments, the function can depend on only the invariants obtained from the vectorsunder arbitrary rotations. These invariants are the norms of the vectors, their innerproducts taken two at a time and their triple products taken three at a time. Thisinvariant group is called the irreducible integrity basis of the scalar function. Thisresult is attributed to Cauchy.

When a scalar function ψ depends on a second-rank tensor A, the argumentstake the form of the invariants of A, namely, IA1, IA2, and IA3. Because the firstinvariant of A2 or Tr A2 can be expressed in terms IA1 and IA2, and Tr A3 can beexpressed in terms of IA1, IA2, and IA3, any polynomial in A that is invariant underarbitrary rotation has the functional form

ψ = ψ(Tr A, Tr A2, Tr A3). (7.36)

These three traces form an irreducible basis for isotropic functions.With isotropic functions depending on two tensors, A and B, the irreducible

basis consists of

Tr A, Tr A2, Tr A3,

Tr B, Tr B2, Tr B3,

Tr (A · B), Tr (A · B2), Tr(A2 · B), Tr(A2 · B2). (7.37)


Results for tensor-valued functions that exhibit various symmetries such asorthotropy, transverse isotropy, etc., have been worked out by Smith and Rivlin(1964), Adkins (1960), Spencer and Rivlin (1959), and Rivlin and Smith (1969).

SUGGESTED READING

Adkins, J. E. (1960). Symmetry relations for orthotropic and transversely isotropic materials,Arch. Ration. Mech. Anal., 4, 193–213.

Dahler, J. S. and Scriven, L. E. (1963). Theory of structured continua. I. General considera-tion of angular momentum and polarization, Proc. R. Soc. London Ser. A, 275, 504–527.

Jaunzemis, W. (1967). Continuum Mechanics, Macmillan.Rivlin, R. S. and Smith, G. F. (1969). Orthogonal integrity bases for N symmetric tensors, in

Contributions to Mechanics, (S. Abir, ed.), Pergamon.Smith, G. F. and Rivlin, R. S. (1964). Integrity bases for vectors. The crystal classes, Arch.

Ration. Mech. Anal., 15, 169–221.Spencer, A. J. M. and Rivlin, R. S. (1959). Finite integrity bases for five or fewer 3 × 3 matri-

ces, Arch. Ration. Mech. Anal., 2, 435–446.VanArsdale, W. E. (2003). Objective spin and the Rivlin–Ericksen model, Acta Mech., 162,

111–124.Wedgewood, L. E. (1999). An objective rotation tensor applied to non-Newtonian fluid me-

chanics, Rheol. Acta, 38, 91–99.Wedgewood, L. E. and Geurts, K. R. (1995). A non-affine network model for polymer melts,

Rheol. Acta, 34, 196–208.Zhou, X. and Tamma, K. K. (2003). On the applicability and stress update formulations for

corotational stress rate hypoelasticity constitutive models, Finite Elem. Anal. Design, 39,783–816.

EXERCISES

7.1. For a fixed volume in space (control volume), we have material flowing in andout of its surface. Obtain the conservation of mass in this Eulerian description.

7.2. For the preceding problem, obtain the balance of linear and angular momen-tum.

7.3. Extend this approach to the balance of energy and the entropy production in-equality.

7.4. Evaluate the suitability of the following rates for use as objective quantities:

(a)

Ai j − Aik∂kv j − Ajk∂kvi ,

(b)

Ai j − wik Akj + Aikwkj ,

(c)

Ai j − Akj∂ivk − Aik∂kv j ,

(d)

Ai j + Akj∂ivk + Aik∂kv j ,


where Ai j is an objective tensor, ∂iv j is the velocity gradient, and wi j is the spintensor.

7.5. Consider a system of particles Pi (i = 1, 2, . . . , N). The particle Pi with mass mi

is subjected to an external force Fi . There are also interparticle forces fi j on Pi thatare due to Pj . Writing the second law of Newton for the motion of each particle,deduce the balance of linear and angular momentum for the system of particles.

8 Stress Tensor

We begin this chapter by recalling external forces and moments acting on a materialcontinuum. Generally, these forces and moments may be grouped into body forcesand body moments and contact forces and contact moments. The body forces andbody moments are those exerted by external agencies such as gravity and magneticfields on the body. They depend on the density of the material. The contact forcesand contact moments are applied through the surface area of the body by otherbodies. In elementary strength of materials, body moments are usually neglected.

8.1 External Forces and Moments

With the subscripts B for body and C for contact, we have

F = FB + FC, M = MB + MC . (8.1)

Introducing body force density f and moment density �, we have

FB =∫

Vf ρdv, MB =

∫V

�ρdv, (8.2)

where ρ is the mass density and V is the current volume.Similarly, the contact force and moment can be expressed as

FC =∫

Sσ (n)da, MC =

∫S

m(n)da, (8.3)

where σ (n) is the surface traction, m(n) is the surface moment on a unit area withnormal n. Again, in elementary courses, the surface moments are neglected. Thetotal external moment has to be calculated including the moment that is due to theexternal forces.

8.2 Internal Forces and Moments

We introduce an imaginary surface cutting the body into two parts: BI and BI I (seeFig. 8.1). At any point on this surface the outward normal of BI is the negative of

98

8.2 Internal Forces and Moments 99

n

−n

σ (n)

σ (n)

BI

BII

da

Figure 8.1. Traction vector on a plane.

the outward normal of BI I :

nI = −nI I . (8.4)

Concentrating on BI, the traction and surface moment acting on the newly createdsurface at a point x are σ (n) and m(n). These are called internal tractions and surfacemoments, respectively. These have dimensions of force per unit area and momentper unit area, respectively.

The traction acting on the same area of BII with normal −n is

σ (−n) = −σ (n). (8.5)

Of all the possible surfaces passing through x, we distinguish three coordinateplanes with normals ei and denote the tractions on these planes by σ (i) and surfacemoments by m(i). To relate the traction vectors on the coordinate planes passingthrough x and the traction vector σ (n) acting on an arbitrary plane with normal n,we construct an infinitesimal tetrahedron, as shown in Fig. 8.2, with the inclinedplane with normal n having an area �a. The areas �ai that are the projections ofthe area �a along the axes xi are related to �a in the form

n�a = ei�ai , �ai = ni�a. (8.6)

Denoting the volume of the tetrahedron by �v, we may balance the linear momen-tum of the tetrahedron as

σ (n)�a − σ (i)�ai + ρ�v f = DDt

(vρ�v), (8.7)

where the velocity v, body force intensity, and all the tractions have to be interpretedas mean values within the volume and the respective areas. When we divide thisequation by �a and let �a go to zero, the terms containing �v, being of a higher

100 Stress Tensor

σ (n)

n

−σ (1)−σ (2)

−σ (3)

x1

x2

x3

Figure 8.2. Traction vectors on the surfaces of a tetrahedron.

order in �a, go to zero. The mean values become the corresponding values at thepoint x, and we get

σ (n) = σ (i)ni . (8.8)

This result establishes the importance of the tractions on the three coordinateplanes: Given the normal n of any plane, the traction σ (n) can be found in termsof the tractions σ (i).

Through the construction of a similar tetrahedron, using the balance of angularmomentum, we can establish the surface moment relation:

m(n) = m(i)ni . (8.9)

In this process, we had to assume the angular momentum of the tetrahedron, netbody moment, and the moments that are due to the body forces and surface tractionsare of a higher order in �a.

8.3 Cauchy Stress and Couple Stress Tensors

The traction vectors on the coordinate planes can be represented in terms of theircomponents as

σ (i) = σi j e j , (8.10)

where σi j are the components of a second-rank tensor, called the Cauchy stress ten-sor. The first index, i , represents the coordinate plane (in terms of normals), and the

8.3 Cauchy Stress and Couple Stress Tensors 101

second index, j , represents the direction of the component. In matrix form we have

[σi j ] =

⎡⎢⎣σ11 σ12 σ13

σ21 σ22 σ23

σ31 σ32 σ33

⎤⎥⎦ . (8.11)

The components σ11, σ22, and σ33 are called the normal stresses, and the remainingsix off-diagonal components are called the shear stresses.

8.3.1 Transformation of the Stress Tensor

In Eq. (8.8), we may keep the plane fixed, maintaining the traction on it while a newcoordinate system, x′, is introduced. Denoting the stresses and the components ofthe normal n by σ ′

i j and n′i , respectively, in the new system, we get

σ (n) = σ ′i j n

′i e

′j = σlknl ek. (8.12)

Using nl = Qiln′i and ek = Qjke ′

j , we can establish the transformation law for thestress components as that for a second-rank tensor, namely,

σ ′i j = Qil Qjkσlk. (8.13)

A similar representation holds for the surface moments:

m(i) = mi j e j , (8.14)

where mi j are the components of the couple stress tensor. The couple stress tensoris also known as the Cosserat stress tensor in honor of the pioneering work on mo-ment stresses carried out by the Cosserat brothers in the early 20th century. Thetransformation law for mi j can be obtained to be that for a second-rank tensor, aswe have done previously.

8.3.2 Principal Stresses

On an arbitrary plane with normal n, the traction σ (n) can be obtained as

σ (n) = σi j ni e j . (8.15)

We may resolve this with one component perpendicular to the plane and the re-mainder tangential to the plane. Denoting the perpendicular component by N, wehave

N = n · σ (n) = σi j ni n j . (8.16)

For a given stress tensor σi j , as the direction of the plane changes, the value of Nchanges according to the preceding relation. We call N the normal stress on theplane. It is important to know the maximum value of the normal stress and thecorresponding plane when we design and analyze structures.

We should find the extremum value of N with respect to ni with the constraintni ni = 1. The procedure is identical to the one we used for extremum stretch, and

102 Stress Tensor

it results in an eigenvalue problem. However, as we will see, the stress tensor isnot symmetric when the body moments � are present. This fact appears to makethis eigenvalue problem different from the case of the maximum stretch. But aninspection of the quadratic form in ni shows that, because of the symmetry of ni n j ,its coefficient is σi j + σ j i . Thus we are free to use the symmetric form of the matrixσi j in computing the eigenvalues. Assuming the new matrix is the symmetric versionσ(i j), the eigenvalue problem results in the system of equations

σ(i j)nj = σni , (8.17)

where σ is the eigenvalue and n is the eigenvector. For nontrivial solutions of thishomogeneous system, we have to have

|σ(i j) − σδi j | = 0. (8.18)

Let σ (k) and n(k) be the eigenvalues and eigenvectors of this problem. MultiplyingEq. (8.17) by ni , we see that the eigenvalues are the stationary values of N, i.e.,

N = σ(i j)n(k)j = σ (k)n(k)

i n(k)i = σ (k). (8.19)

Again the three eigenvectors are mutually orthogonal, and these directions arecalled the principal directions and the corresponding normal tractions σ (k) are theprincipal stresses. The three invariants of the matrix σ(i j) are

Iσ1 = σ (1) + σ (2) + σ (3) = σ(i i),

Iσ2 = σ (1)σ (2) + σ (2)σ (3) + σ (3)σ (1) = 12

[σ(i i)σ( j j) − σ(i j)σ(i j)],

Iσ3 = σ (1)σ (2)σ (3) = |σ(i j)|. (8.20)

8.3.3 Shear Stress

The magnitude of the tangential component S of the traction vector on any planemay be computed as

S2 = σ (n) · σ (n) − (n · σ (n))2. (8.21)

On the principal planes n(k), we have σ (n) = n(k)σ (k) and

S2 = 0. (8.22)

Thus there is no shear stress on the principal planes.To obtain the planes on which the shear stress is stationary, it is convenient to

use a new coordinate system with its axes oriented along the principal directions. Inthis system the stress matrix can be written as

[σi j ] =

⎡⎢⎣σ1 0 0

0 σ2 00 0 σ3

⎤⎥⎦ , (8.23)


where we have used the notation σi = σ (i). On any arbitrary plane with normal n,the traction vector has the simple expression

σ (n) = n1σ1e1 + n2σ2e2 + n3σ3e3. (8.24)

The shear stress on this plane becomes

S2 = n2i σ

2i − (niσi )2. (8.25)

Adding the constraint ni ni = 1 and using a Lagrange multiplier �, we set up afunction:

F (n1, n2, n3) = n2i σ

2i − (niσi )2 − �(ni ni − 1). (8.26)

This function becomes stationary when ∂ F/∂ni = 0. This way, we obtain the threeequations

σ 21 n1 − 2Hσ1n1 − �n1 = 0,

σ 22 n2 − 2Hσ2n2 − �n2 = 0,

σ 23 n3 − 2Hσ3n3 − �n3 = 0, (8.27)

where H is defined as

H = niσi .

We may verify that the three solutions of these equations are

n1 = 0, n22 = n2

3 = 12,

n2 = 0, n23 = n2

1 = 12,

n3 = 0, n21 = n2

2 = 12, (8.28)

and the values of the stationary shear stresses for these three cases, respectively, are

S = |σ2 − σ3|2

,|σ3 − σ1|

2,|σ1 − σ2|

2. (8.29)

8.3.4 Hydrostatic Pressure and Deviatoric Stresses

It is known that many materials respond differently to hydrostatic pressure than toshear stresses. When the normal stresses on the faces of an infinitesimal cube differfrom each other, the hydrostatic pressure is defined as

−p = 13σi i = 1

3

(σ (1) + σ (2) + σ (3)

). (8.30)

If we remove the hydrostatic part from the stress tensor, what remains is called thedeviatoric stress s. Thus,

si j = σi j + pδi j = σi j − 13σkkδi j . (8.31)

104 Stress Tensor

8.3.5 Objective Stress Rates

There are materials that respond not only to the stress but also to the rate of changeof stress in time. For a stress rate to be an admissible variable in a constitutive law, ithas to be objective (frame indifferent). The obvious candidates such as ∂σi j/∂t andDσi j/Dt can be ruled out if we use the appropriate rate of the relation

σ ′i j = QimQjnσmn (8.32)

when the rotation matrix Q is time dependent.The apparent rate of change in stress is best illustrated if we consider a bar

under a constant stress σ0. If we use a rotating frame with the x1 axis along the axisof the bar at time t = 0, the frame shows σ11 = σ0. As the frame rotates, we see σ11

diminishing, and, eventually, after a 90◦ rotation, σ22 has the value σ0. During all thistime, the bar itself did not experience any change in stress.

As it happens, an objective stress rate is not unique; a number of candidateshave been proposed by various authors. A stress rate introduced by Jaumann in1911, which is called the Jaumann stress rate or corotational stress rate, appearsoften in constitutive relations. The physical meaning of objective stress rates maybe explored in the following way. At an arbitrary point P in a continuum, let usestablish the origin of two coordinate systems: one, the Eulerian system with unitvectors ei and coordinates xi , and the second, a rotating system with unit vectorse ′

i and coordinates x′i . At time t , the two systems overlap. We assume the second

system rotates with an angular velocity �. We will select � later. Our goal is tocompute the time rate of change of quantities from the point of view of an observersitting on the rotating system. In an incremental time �t , the base vectors of therotating system are related to those of the fixed system as

e′i = Pi j (t ′)ei , (8.33)

where P is an orthogonal matrix. When t ′ = t , the systems coincide and

Pi j (t) = δi j . (8.34)

The time rate of change of the base vectors e′i can be expressed as

e′i = −Wi j (t ′)e′

j , (8.35)

where W is a spin tensor related to � through

Wi j = ei jk�k. (8.36)

Taking the material derivative of Eq. (8.33), we find

e′i = Pik(t ′)ek = PikPjke′

j . (8.37)

From this we observe that the spin tensor is related to P in the form

W(t ′) = − P PT , W(t) = − P. (8.38)


When t ′ = t + �t ,

P(t ′) = I + P�t = I − W�t. (8.39)

Any second-rank tensor field Ai j that is time dependent has componentsA′

i j (t + �t) in the rotating system. We define its rate of change as observed in therotating system as

Ai j ≡ lim�t→0

1�t

[A′i j (t + �t) − A′

i j (t)]. (8.40)

As the two coordinate systems coincide at time t ,

A′i j (t) = Ai j . (8.41)

Now Ai j at t + �t is given by

Ai j (t + �t) = Ai j + Ai j�t. (8.42)

Using the transformation of coordinates and neglecting quadratic terms in �t ,

A′i j (t + �t) = PimPjn Amn(t + �t)

= (δim − Wim�t)(δ jn − Wjn�t)(Amn + Amn�t)

= Ai j + [Ai j − Wim Amj + AinWnj ]�t,

Ai j = Ai j − Wim Amj + AinWnj ,

A = A− W A+ AW. (8.43)

Having the expression for the rate of change of a tensor in a rotating coordinatesystem, we would like to see how this quantity transforms under an arbitrary time-dependent rotation Q such that x′

i = Qi j (t)xj .Starting with A′ = QAQT ,

A′ = A′ − W′A′ + A′W′

= QAQT + QAQ

T + QAQT − W′QAQT + QAQT W′

= Q [A− QT W′QA+ AQT W′Q + QTQA+ AQTQ] QT

= Q [A− WA+ AW] QT, (8.44)

where the last equation requires

W = QT W′Q + QTQ. (8.45)

For various choices of W satisfying the preceding condition, we obtain the corre-sponding objective stress rates σ . In the Jaumann stress rate, W is selected as thematerial spin tensor,

W = w, � = 12∇ × v, (8.46)

106 Stress Tensor

and the corresponding objective Cauchy stress rate is the Jaumann rate

σ = σ − wσ + σw. (8.47)

If we choose the rotation of the frame identical to the rotation R in the polar de-composition F = VR, we obtain

W = RRT , (8.48)

and the corresponding stress rate is known as the Green–McInnis–Naghdi stressrate.

The Truesdell stress rate defined by

σ = σ − ∇v · σ − σ · (∇v)T + σ∇ · v, (8.49)

and the Oldroyd stress rate defined by

σ = σ − ∇v · σ − σ · (∇v)T , (8.50)

are also well known in the literature as objective stress rates.Zhou and Tamma (2003) have shown that the logarithmic spin rate W, defined

by the equation

DDt

ln U − W ln U + (ln U)W = d, (8.51)

where U is the right stretch tensor and d is the deformation rate, leads to an ob-jective stress rate that has certain numerical advantages over other objective stressrates.

8.4 Local Conservation and Balance Laws

The global conservation and balance laws presented in Chapter 7 can be trans-formed into their local (differential equation) forms by use of the stress tensors. Weuse the Gauss theorem to achieve this transformation. The balance of energy willhave to wait until we discuss some concepts from thermodynamics (what Truesdellcalls thermostatics).

8.4.1 Conservation of Mass

The global conservation equation

DDt

∫V

ρdv = 0 (8.52)

can be written as ∫V

[ρdv + ρ(dv)•] = 0. (8.53)

From this, using (dv)• = ∂ivi dv, we have the local form:

ρ + ρ∂ivi = 0. (8.54)

8.4 Local Conservation and Balance Laws 107

8.4.2 Balance of Linear Momentum

From the global balance relation

DDt

∫V

vρdv = F, (8.55)

using the conservation of mass and the expression for the force F, we get∫V

vρdv =∫

Sσ (n)da +

∫V

f ρdv

=∫

Sniσ

(i)da +∫

Vf ρdv

=∫

V∂iσ

(i)dv +∫

Vf ρdv

=∫

V

[∂iσ

(i) + f ρ]

dv. (8.56)

As this equation holds for arbitrary volumes, the integrands must be equal. Thus,

∂iσ(i) + ρ f = ρv. (8.57)

Using the Cauchy stress tensor σ , we can also write this as

∇ · σ + ρ f = ρa, (8.58)

where a is the acceleration. The expanded form of this tensor equation is

∂1σ11 + ∂2σ21 + ∂3σ31 + ρ f1 = ρa1,

∂1σ12 + ∂2σ22 + ∂3σ32 + ρ f2 = ρa2,

∂1σ13 + ∂2σ23 + ∂3σ33 + ρ f3 = ρa3. (8.59)

8.4.3 Balance of Moment of Momentum (Angular Momentum)

The global balance of moment of momentum can be written as

DDt

∫V

[ J · µ + x × v]ρdv = M, (8.60)

where we have included the intrinsic angular velocity µ, which is distinct from thelocal angular velocity �, with a moment of inertia tensor ρ J . In many cases of prac-tical interest, we may need to include anisotropic characteristics in this inertia (con-sider a spinning ellipsoid, as an example). The integral on the left-hand side can bereduced to ∫

V[α + v × v + x × v]ρdv, (8.61)

where

α = DDt

( J · µ). (8.62)

108 Stress Tensor

The moment on the right-hand side can be written as

M =∫

S[m(n) + x × σ (n)]da +

∫V

[� + x × f ]ρdv. (8.63)

Introducing the moment stress vectors and Cauchy stress vectors, we get

M =∫

S[ni m(i) + ni x × σ (i)]da +

∫V

[� + x × f ]ρdv. (8.64)

Using the Gauss theorem, we transform the surface integral to a volume integral:

M =∫

V[∂i m(i) + x × ∂iσ

(i) + ei × σ (i)]dv +∫

V[� + x × f ]ρdv. (8.65)

When we equate the terms on the left-hand side to those on the right-hand side ofthe equal sign, the terms involving x cancel because of the balance of linear momen-tum. What remains is the relation

∂i m(i) + ρ� + ei × σ (i) = ρα ⇔ ∇ · m + ρ� + ei × σ (i) = ρα. (8.66)

In terms of the components, this becomes

∂1m11 + ∂2m21 + ∂3m31 + ρ�1 + σ23 − σ32 = ρα1,

∂1m12 + ∂2m22 + ∂3m32 + ρ�2 + σ31 − σ13 = ρα2,

∂1m13 + ∂2m23 + ∂3m33 + ρ�3 + σ12 − σ21 = ρα3.

These equations show that the Cauchy stress σ may not be symmetric when the bodymoment �, the couple stress m, or the intrinsic spin µ, is present. If all of these newquantities are absent, we obtain the classical result

σi j = σ j i (8.67)

as a consequence of the balance of angular momentum. The balance of linear andangular momentum equations are also called the equations of motion.

8.5 Material Description of the Equations of Motion

The equations of motion developed in the previous section are in terms of the spatialcoordinates as independent variables. All the stress components in these equationshave dimensions of force or surface moment per unit area of the current configura-tion. We may say that the Cauchy stress is the “true” stress experienced by the bodyat a given time. The body forces and the body moments have the dimensions of forceor moment per unit mass and, as such, they are easy to describe in terms of the mate-rial coordinates. There are two ways of introducing the so-called pseudostresses in amaterial description: (a) keeping the elemental force vector constant, we define thetraction vector by using the undeformed area, or (b) we treat the elemental forcevector as a material element and scale the area as in (a). In the latter descriptionthe elemental force vector stretches and rotates as the material deforms. These twodefinitions result in stress tensors called the first and second Piola–Kirchhoff stresstensors.

8.5 Material Description of the Equations of Motion 109

σ (n)daT(N)dA

dA

da

Figure 8.3. Traction vectors in the definition of the first Piola–Kirchhoff stress tensor.

8.5.1 First Piola–Kirchhoff Stress Tensor

We introduce a traction vector T(N), which keeps the elemental force vector con-stant during deformation. As shown in Fig. 8.3,

T(N)dA = σ (n)da. (8.68)

Here, σ (n) is transported parallel to itself and scaled with the ratio of the areas toobtain T(N). Using the tractions on the coordinate planes, we obtain

T(α) NαdA = σ (i)ni da,

T(α)dAα = σ (i)dai ,

T(α)dAα = σ (i) J∂i XαdAα,

T(α) = σ (i) J∂i Xα. (8.69)

The inverse form of this relation is

σ (i) = T(α) j∂αxi . (8.70)

Using the component form of these vectors,

T(α) = Tαi ei , σ (i) = σi j e j , (8.71)

we find

Tαi = J∂kXασki , σi j = j∂αxi Tα j . (8.72)

The tensor Tαi is called the first Piola–Kirchhoff stress or the Lagrange stress tensor.To rewrite the equations of motion, let us consider the three terms in the Cauchy

equations:

ρai = jρ0∂2xi

∂t2, ρ fi = jρ0 Fi , (8.73)

110 Stress Tensor

σ (n)danS(N)dA

Nθ

θ ′

dA

da

Figure 8.4. Traction vectors in the definition of the second Piola–Kirchhoff stress tensor.

where Fi = fi ;

∂kσki = ∂k( j∂αxkTαi )

= (∂k j)∂αxkTαi + j(∂k∂αxk)Tαi + j∂αxk(∂kTαi )

= j∂αTαi , (8.74)

where we have used

j = j(∂mXβ), ∂α∂mXβ = ∂m∂α Xβ = 0. (8.75)

Combining the preceding three results and canceling the factor j , we find that theequations of motion in material coordinates come out as

∂αTαi + ρ0 Fi = ρ0∂2xi

∂t2. (8.76)

8.5.2 Second Piola–Kirchhoff Stress Tensor

The first Piola–Kirchhoff stress tensor. We just defined may not be symmetric whenthe Cauchy stress is symmetric. The second Piola–Kirchhoff stress, also known asthe Kirchhoff stress, which is symmetric when the Cauchy stress is symmetric, isdefined as

Sαβ = J∂i Xα∂ j Xβσi j = Tαi∂i Xβ. (8.77)

We may associate traction vectors on the coordinate planes

S(α) = Sαβeβ = J∂i Xα∂ j Xβσi j eβ = Tαi∂i Xβeβ, (8.78)

and those on any plane

S(N) = S(α) Nα = J∂i Xα∂ j Xβσi j Nαeβ = ∂i Xβ Tαi Nαeα = T(N) · f , (8.79)

where f is the deformation gradient (not to be confused with the body force). Inmatrix notation we have

S = Tf = J f Tσ f . (8.80)

The last expression clearly shows the symmetry of S when σ is symmetric.


A physical interpretation of the traction S(N) may be obtained as follows.Consider a material element D attached to the area dA in the undeformed con-figuration. In the current configuration,

D = Dαeα = dXαeα = dxi∂i Xαeα = d · f . (8.81)

Comparing this withS(N) = T(N) · f (8.82)

shows that the Kirchhoff traction vector deforms as a material element to end up inthe current configuration as the Lagrange traction vector. This is shown in Fig. 8.4.Because the force vector in the undeformed configuration is treated as a materialelement, the angles θ and θ ′ are not the same. Angles between material elementsmay get distorted due to shearing.

The equation of motion can be written in terms of S as

∂α(Sαβ∂β xi ) + ρ0 Fi = ρ0∂2xi

∂t2. (8.83)

Because of the deformation gradient multiplying the stress tensor, these equationsare nonlinear for finite deformations. Using x1 = X1 + U1, etc., we can rewrite theseas

∂α[Sαβ(δβγ + ∂βUγ )] + ρ0 Fγ = ρ0Uγ . (8.84)

The material description of the angular momentum balance in terms of couplestresses can be carried out in a similar fashion.

SUGGESTED READING

Zhou, X. and Tamma, K. K. (2003). On the applicability and stress update formulations forcorotational stress rate hypoelasticity constitutive models, Finite Elem. Anal. Design, 39,783–816.

EXERCISES

8.1. At a point in a continuum, the stress matrix is given by

σ =

⎡⎢⎣ 4 3 0

0 −4 00 0 2

⎤⎥⎦ .

(a) Obtain the traction vector on a plane with normal vector (2/3,−2/3, 1/3).(b) Obtain the normal and tangential components of the traction on this plane.

8.2. The stress matrix at a point is given as

σ =

⎡⎢⎣ 0 2 1

2 σ22 21 2 0

⎤⎥⎦ .

(a) Obtain the value of σ22 if there is a traction-free plane passing through thispoint.

(b) Determine the unit normal to this plane.

112 Stress Tensor

8.3. In a medium supporting couple stresses, the Cauchy stress tensor is given as

σ =

⎡⎢⎣ 2 1 0

−3 2 00 0 3

⎤⎥⎦ .

Obtain the principal stresses and their directions.

8.4. A body is in plane stress if σ3i = 0. In addition, if the body force f = 0 and theacceleration a = 0, show that

σ11 = φ,22 , σ22 = φ,11 , σ12 = σ21 = −φ,12

satisfy the equations of motion. The function φ is called the Airy stress function.Obtain the traction vector on a boundary s = s(x1, x2) in terms of φ.

8.5. Obtain the principal stresses and their planes for

σ =

⎡⎢⎣ 1 0 0

0 4 −30 −3 4

⎤⎥⎦ .

What are the three invariants of this matrix?

8.6. Show that the second invariant of the deviatoric stress, si j , can be written as

(a)

Is2 = s11s22 + s22s33 + s33s11 − (s212 + s2

23 + s231),

(b)

Is2 = 12

(sii s j j − si j si j ),

(c)

Is2 = −16

[(σ11 − σ22)2 + (σ22 − σ33)2 + (σ33 − σ11)2]

−(σ 212 + σ 2

23 + σ 231).

8.7. For the deformation

x1 = X1 + kX2, x2 = X2, x3 = X3,

the Cauchy stress at a point is given as

σ =

⎡⎢⎣ 0 1 2

1 0 22 2 0

⎤⎥⎦ .

(a) Obtain the components of the first and second Piola–Kirchhoff stresses.(b) Obtain the traction vector on a plane in the current configuration with nor-

mal n = (1, 1, 0)/√

2.(c) Show corresponding traction vectors of the two Piola–Kirchhoff stresses.

8.8. Show that the Oldroyd stress rate and the Truesdell stress rate are objective.

8.9. Consider a simple shear flow, in which

v1 = γ x2, v2 = 0, v3 = 0,


and γ is a constant. Obtain the material spin tensor w. Also obtain the componentsof the objective Jaumann stress rate.

8.10. For a hyperelastic material there is a strain energy density function defined by

ρu = 12σi j ei j

per unit volume, and the total strain energy is given by

U =∫

Vρudv.

In the material description the strain energy density is defined as

ρ0u0 = 12�αβ Eαβ

per unit volume, where �αβ is a symmetric stress measure. Comparing the totalstrain energy, obtain an expression for the stress measure �αβ and relate it to theKirchhoff stress.

9 Energy and Entropy Constraints

In this chapter we discuss the constraints imposed on the mathematical formula-tions of material behavior by the balance of energy and the irreversible part of theentropy production. In other words, we discuss the implications of the firstand second laws of thermodynamics on the constitutive relations of continuummechanics.

9.1 Classical Thermodynamics

In classical thermodynamics, we call a collection of particles under consideration asystem. Let us denote this system by B, and let S be the surface enclosing it. Weassume B is isolated, meaning there is no mass moving across the surface S. Bymoving this surface, we can do work on the system. By bringing this surface intocontact with another surface belonging to a different system, heat energy can bemade to flow into the system or out of the system. For basic properties of the system,we have the extensive variables: V, the volume; E, the internal energy; and S, theentropy. We also have intensive variables: p, the pressure; T , the temperature; andthe chemical potential, etc. In continuum mechanics, we need to extend the idea of apressure p to the stress tensor σ and the idea of a specific volume to an appropriatestrain tensor. If the surface S is assumed to prevent heat flow across it, the system iscalled adiabatic.

When all the information needed to characterize a system (to the extent werequire) is known, we say we know the state of the system. The properties of thesystem in a particular state are called state variables. For example, in the case of anideal gas, we would like to know its volume, pressure, temperature, energy content,specific volume, and entropy content. These form the state variables. We do notcare to know the velocities and locations of the zillions of individual molecules, andwe do not include in the energy content the rest mass in the relativistic sense, as thisdoes not change in the processes we have in mind. The processes we have in mindinvolve taking the system from one state to another, in a slow fashion. The system issaid to be in equilibrium if the state variables have time-independent values. In ourexample of an ideal gas, if the volume is suddenly changed by �V by pushing the

114

9.2 Balance of Energy 115

surface, we can expect certain wave motions inside that would eventually decay andthe system will reach a new state of equilibrium.

If certain state variables can be expressed as functions of certain other statevariables, such functions are called equations of states. If the state variables areindependent of the space coordinates, the system is called homogeneous. If twosystems are brought into contact and if the surfaces are allowed to conduct heat,one of the state variables, namely the temperature, assumes a common value aftersufficient time.

9.2 Balance of Energy

The material continuum is endowed with two forms of energy–the kinetic energy,K, and the internal energy, E. We allow the flow of energy into the body in theforms of mechanical power P and heat power H. We briefly discuss other forms ofenergy and power input at the end of this section.

The kinetic energy K and the internal energy E are defined as

K = 12

∫V

[v · v + µ · J · µ]ρdv, E =∫

Vερdv. (9.1)

In the first of Eqs. (9.1), the kinetic energy density consists of the classical ρv2/2and a “microstructural” contribution ρ Jµ2/2, where ρ J represents the microstruc-tural moment of inertia. Further, we have introduced a certain anisotropy for themicrostructure and written the rotational energy in the tensor form by using themoment of inertia tensor ρ J . The vector µ is an independent kinematic variablerepresenting the microstructural angular velocity, which is not related to the localspin of classical continua. It is useful to think of steam, which is composed of H2Omolecules, as an example. These molecules have a common structure with an asso-ciated inertia tensor, but they may have independent angular velocities (spins). Ofcourse, we are not going to count individual molecules and their properties. Otherexamples are liquid crystals and granular materials (see Cowin, 1978; Christoffersen,Mehrabadi, and Nemat-Nasser, 1981). The global conservation of energy reads

DDt

(K + E) = P + H. (9.2)

The mechanical power input into the body consists of the rates of work done on thebody by surface tractions, surface moments, body forces, and body moments:

P =∫S

[σ (n) · v + m(n) · µ]da +∫

V[ f · v + � · µ]ρdv, (9.3)

where µ is the angular velocity vector conjugate to the surface moment vector.The heat power is due to the rate of heat flow into the body, q, and the rate of

heat generation inside the body, h:

H =∫S

q · nda +∫

Vhρdv. (9.4)


The next step in our calculation is to obtain the local form of the conservation ofenergy. We begin with

DKDt

= 12

∫V

[DDt

(vivi + Ji jµiµ j )]

ρdv =∫

V[aivi + αiµi ] ρdv, (9.5)

where we have used the definition

αi = DDt

Ji jµ j

and the conservation of mass for simplification.The rate of change of internal energy is obtained as

DEDt

=∫

Vερdv. (9.6)

Of the two forms of power input, first we consider the simpler expression,

H =∫S

qi ni da +∫

Vhρdv =

∫V

[∂i qi + ρh]dv. (9.7)

In the expression for P,

σ (n) · v = niσi jv j ,

m(n) · µ = ni mi jµ j . (9.8)

Using the Gauss theorem, we get

P =∫

V[∂i (σi jv j + mi jµ j ) + ρ( f jv j + � jµ j )]dv

=∫

V[σi j∂iv j + mi j∂iµ j + v j (∂iσi j + ρ f j ) + µ j (∂i mi j + ρ� j )]dv. (9.9)

We introduce the relations

∂iσi j + ρ f j = ρa j ,

∂i mi j + ρ� j = ρα j − e jklσkl

in Eq. (9.9) to obtain the local form of the conservation of energy:

ρε = ρh + ∂i qi + σi j∂iv j + mi j∂iµ j − e jklσklµ j . (9.10)

In this energy conservation relation, the third and the fourth terms are called thestress power and the couple stress power, respectively. Further simplification of thisrelation is possible if (a) the couple stresses are absent and (b) part of the stresstensor can be derived from a potential.

9.3 Clausius–Duhem Inequality

In classical thermostatics, the incremental form of the internal energy is written as

�E = δW + δQ, (9.11)

9.3 Clausius–Duhem Inequality 117

where δW is the work done on the system and δQ is the heat added. We have re-served the symbol � to denote change in a state variable. The quantities W and Qare not state variables, and the symbol δ in front of them just denotes “small” incre-ments. For our system, the increment of the state function, entropy, which is due toheat flow across the surface, is given by

�S = δQT

, (9.12)

where T is the absolute temperature.Substituting this in the energy equation, we find

�E = δW + T�S. (9.13)

In irreversible thermodynamics, the entropy production �S consists of twoparts. One is due to the entropy flow into the body, �SE , which is associated withthe heat flow, and the other, �SI , is due to irreversible processes taking place insidethe body. The total entropy S is considered to be a state variable, and Eq. (9.13) istaken as a fundamental relation, although its original derivation was based on heatflow only.

The work term in Eq. (9.13) is replaced with stresses τi j and thermodynamicmoment stresses λi j , producing incremental work per unit volume because of theincrements in a corresponding array of state variables. Without being specific, letεi j and νi j be these state variables. The work done by these stresses and momentstresses have to be reversible.

Introducing internal energy density ε and specific entropy s, we have the localform of the equation

ρ�ε = ρT�s + τi j�εi j + λi j�νi j . (9.14)

This is a version of the Gibbs equation in thermodynamics. In the limit, theGibbs equation gives the defining relations

T = ∂ε

∂s, τi j = ρ

∂ε

∂εi j, λi j = ρ

∂ε

∂νi j. (9.15)

Further, dividing the Gibbs eqution by δt and taking the limit, we find

ρ s = 1T

(ρε − τi j εi j − λi j νi j ). (9.16)

Substituting energy equation (9.10) into Eq. (9.16), we find that Eq. (9.16) becomes

ρ s = 1T

(ρh + ∂i qi + σi j∂iv j + mi j∂iµ j − e jklσklµ j − τi j εi j − λi j νi j ). (9.17)

At first glance, this equation appears to have too many variables! To reduce it to asimpler form, first we have to choose εi j and νi j , with their rates relating to ∂iv j and∂iµ j . Second, parts of the stresses σi j and mi j may contribute to reversible work, andwe can relate them to τi j and λi j .


The rate of global entropy gain that is due to heat input is seen to be

SE =∫

S

ni qi

Tda +

∫V

ρhT

dv. (9.18)

From this the local form is obtained as

ρ sE = ∂i

(qi

T

)+ ρh

T

= 1T

(∂i qi + ρh) − qi∂i TT2

. (9.19)

Subtracting this from entropy equation (9.17), we get

ρ sI = 1T

(σi j∂iv j + mi j∂iµ j − e jklσklµ j − τi j εi j − λi j νi j + qi∂i T

T

). (9.20)

As already mentioned, further simplification is achieved if we assume that theCauchy stress σ and the couple stress m can be resolved into a hyperelastic partthat can be derived from a potential and an irreversible part that contributes to theinternal entropy generation. The potential associated with the hyperelastic part isknown as the strain energy density function, u. This potential is a function of thestate variables εi j and νi j . This resolution is expressed as

σ = σ 0 + σ I , m = m0 + mI . (9.21)

Then we choose τi j and λi j to satisfy

σ 0i j (∂iv j − eki jµk) + m0

i j∂iµ j = τi j εi j + λi j νi j = ρu. (9.22)

This leaves the internal entropy production rate:

ρ sI = 1T

[σ I

i j (∂iv j − eki jµk) + mIi j∂iµ j + qi∂i T

T

]. (9.23)

The Clausius–Duhem inequality, in its local form, states that

sI ≥ 0, (9.24)

with the equality satisfied if all processes are reversible. It has been shown byPrigogine, using statistical mechanics, that the entropy production rate we havederived is valid only for what are called “first-order” processes, in which the irre-versible stresses and heat flux may be expressed as linear functions of the variables∂iv j , ∂iµ j , and ∂i T . The coefficient matrix in such a system of relations must bepositive semidefinite to satisfy the Clausius–Duhem inequality.

Let us illustrate this with two simple cases.

9.3.1 Fourier’s Law of Heat Conduction

In the generalized Fourier’s law for anisotropic materials, the heat flux can be ex-pressed as

qi = ki j∂ j T, (9.25)

9.3 Clausius–Duhem Inequality 119

where ki j is the conductivity matrix. If we assume a process in which there are nostresses, using this in the entropy production, we find

ki j∂i T∂ j T ≥ 0 (9.26)

for an arbitrary temperature gradient. For this to be true, the matrix k has to bepositive semidefinite.

9.3.2 Newton’s Law of Viscosity

We consider a fluid flowing over a plate with velocity parallel to the plate. In itssimplest form, with the x coordinate parallel to the plate and the y coordinate per-pendicular, the only velocity gradient is ∂yvx. Newton’s law of viscosity states

σxy = µ∂yvx, (9.27)

where µ is the viscosity. The Clausius–Duhem inequality requires that µ ≥ 0 if allother terms are absent.

9.3.3 Onsager’s Principle

The rate of specific entropy production can be written in terms of what are calledgeneralized forces Xk and fluxes Jk in the form

ρ sI = Xk Jk. (9.28)

In Eq. (9.23), we can identify

Xk ⇒ (σ Ii j , mI

i j , ∂i T/T),

Jk ⇒ (∂iv j − eki jµk, ∂iµ j , qi )1T

. (9.29)

At thermodynamic equilibrium, sI = 0, and all the fluxes and forces are also zero.Onsager has shown that, close to the equilibrium state, a system of linear relationsexists that relates the fluxes and forces:

Jk = LkmXm. (9.30)

Fourier’s law and Newton’s law, previously discussed, are examples of such linearrelations. Onsager’s principle states that the matrix Lkm is symmetric. Further, theClausius–Duhem inequality requires Lkm to be positive semidefinite.

9.3.4 Strain Energy Density

To simplify the form of the strain energy density function u, let us assume that, inEqs. (9.21),

m = 0, � = 0, µ = 0, (9.31)


which implies that the only stress remaining is the Cauchy stress σ, which is nowsymmetric, satisfying the balance of angular momentum. Let us also assume thatσ I

i j = 0.Let

u = u(Eαβ), εi j ⇒ Eαβ, νi j = 0. (9.32)

Then

ρu = ρ∂u

∂ Eαβ

Eαβ

= ρ∂u

∂ Eαβ

di j∂αxi∂β xj . (9.33)

In Eq. (9.22), using ∂iv j = di j + wi j and σ 0i jwi j = 0, we find

σ 0i j∂i Xα∂ j Xβ = ρ

∂u∂ Eαβ

,

S0αβ = ρ0

∂u∂ Eαβ

. (9.34)

We conclude by noting that the second Piola–Kirchhoff stress Sαβ is the stressmeasure τi j if Eαβ is the state variable εi j . We say that the stress S0 is conjugate tothe strain E.

9.3.5 Ideal Gas

A measure of volume change is the volume ratio J (this is the Jacobian determinant,not to be confused with the Onsager fluxes). If we assume the strain energy has theform

u = u(J ), (9.35)

we find

ρdudJ

J = σi j (di j + wi j ) = σi j di j . (9.36)

Using J = J dii , we have

ρ0dudJ

di jδi j = ti j di j . (9.37)

Thus the only stress sustainable in this medium is the hydrostatic pressure,

ρ0dudJ

δi j = ti j . (9.38)

We see the well-known state variable from classical thermodynamics, namely, thepressure p, given by

p = −ρ0dudJ

, (9.39)

emerging as the only stress in this system.

9.4 Internal Energy 121

9.4 Internal Energy

For the sake of simplicity, let us consider a homogeneous system with the internalenergy

E = E(S, ε), (9.40)

where the group of variables ε represents the specific volume in classical thermo-dynamics and the six strain components for the hyperelastic case previously con-sidered. This group will contain more strain measures if microstructural spins areallowed. In applications involving chemical mixtures, various chemical concentra-tions have to be added to this group. For granular materials, the void fraction is animportant state variable, which also has to be included. We may assume this groupcan be written as a linear array εi (we will not use the term vector, as our array doesnot transform according to the rules). We denote their conjugate variables, calledthermodynamic tensions, by the array τi , and the inner product of these two arrays,τiεi , has the dimension of energy and is invariant under translations and rotations.Equation (9.40) is called a caloric equation of state. We invert this relation to have

S = S(E, ε). (9.41)

At thermodynamic equilibrium, S = 0, and for an adiabatic process there is no en-tropy flow into the body. Internal processes leading to equilibrium may generateentropy, and at equilibrium the entropy reaches a maximum. This is called the max-imum entropy principle.

For a given entropy with fixed state variables ε, of all the values of the variableE, a state of equilibrium corresponds to the minimum energy. This is known as theminimum energy principle.

The Gibbs equation can be written as

dE = TdS + τi dεi , i = 1, 2, . . . , N, (9.42)

with

T = ∂ E∂S

, τi = ∂ E∂εi

. (9.43)

We define a thermodynamic path in the state space by using the parameter λ as

S = S(λ), εi = εi (λ). (9.44)

If S is independent of λ, the process is called isentropic. For an isothermal process,T = ∂ E/∂S has to be independent of λ.

9.4.1 Legendre or Contact Transformation

There are situations for which it is convenient to control the temperature T insteadof the entropy S or tensions τi , i = 1, 2, . . . , M ≤ N, instead of εi . We obtain newfunctions with T and τi as independent variables by using Legendre transformationsor contact transformations.


1

0

E(S)

E∗(T)

T

S

Figure 9.1. Legendre transform of E(S) to E∗(T).

As shown in Fig. 9.1, a plot of E as a function of S has to have a positive slope T ,the absolute temperature. If the slope is given, we can move a line with that pre-scribed slope to contact the curve, E vs. S. If the intercept of this line with the Eaxis is denoted by E∗(T), we have

E(S) = E∗(T) + T S, E∗(T) = E(S) − T S. (9.45)

From this, we obtain

∂ E∗

∂T= −S,

∂ E∂S

= T. (9.46)

This process of introducing T as the independent variable is called a Legendre orcontact transformation of E(S). The term “contact” refers to the construction ofE∗(T) by contacting the E–S curve by a straight line of prescribed slope. In addition,if we want to use τi , i = 1, 2, . . . , M as new variables, we introduce

E∗(T, τ1, τ2, . . . , τM) = E(S, ε1, ε2, . . . , εN) − T S −i=M∑i=1

τiεi . (9.47)

The following list shows some of the well-known functions in thermodynamics ob-tained by the Legendre transform. Here the summation on i from 1 to N is implied.

Potential Relation VariablesInternal energy E S, εi

Helmholtz free energy U = E − T S T, εi

Enthalpy H = E − τiεi S, τi

Gibbs free energy G = E − T S − τiεi T, τi (9.48)


The following differential relations can be obtained from our list:

dE = TdS + τi dεi ,

dU = −SdT + τi dεi ,

dH = TdS − εi dτi ,

dG = −SdT − εi dτi . (9.49)

We also use the notation for specific (per unit mass) quantities, ε for internalenergy, u for Helmholtz free energy, h for enthalpy, and g for Gibbs free energy.

The Helmholtz free energy U is convenient to describe an isothermal processas the temperature T is an independent variable. Similarly, in solid mechanics forconstant-stress processes, H and G are useful functions.

9.4.2 Surface Energy

In certain areas of continuum mechanics, the basic assumption of continuity of fieldvariables has to be relaxed to include the creation of new surfaces of discontinuityinside the body. These surfaces are assumed to be formed at the expense of energy,and this energy is taken to be stored in the newly formed surface Sc. The conserva-tion of energy has to be written, including the surface energy, as

DDt

(K + E +∫Sc

γ da) = P + H, (9.50)

where γ is the surface energy density. There is also an entropy production associatedwith the creation of new surfaces. To our previous entropy production rates we adda surface entropy term to get

DSDt

= DDt

(SE + SI +∫Sc

ηda), (9.51)

where η is the surface entropy density.

9.5 Method of Jacobians in Thermodynamics

A method of manipulating thermodynamic functions by use of Jacobians was pre-sented by Shaw in 1935 (see Margenau and Murphy, 1943). In the derivation ofmany thermodynamic relations, independent variables are changed frequently fromone set to another. The Jacobian determinant of such transformations can be ef-fectively used as an operational tool to achieve the desired result in a more directfashion.

Consider a mapping from a two-dimensional (2D) space (x, y) to another space(u, v). Assuming a one-to-one mapping, we have

dudv = J dxdy, (9.52)


where J is the Jacobian determinant:

J =

∣∣∣∣∣∣∣∣∣∣

∂u∂x y

∂u∂y x

∂v

∂x y

∂v

∂y x

∣∣∣∣∣∣∣∣∣∣. (9.53)

We use a shorthand notation:

J = [u, v][x, y]

. (9.54)

Although the individual bracket group in the numerator has no meaning withoutthe denominator, symbolically we have

[u, v] = J [x, y], (9.55)

and, based on the determinant,

[u, v] = −[v, u], [u, u] = 0, (9.56)

and as a product of two determinants,

[u, v][x, y]

[x, y][p, q]

= [u, v][p, q]

, (9.57)

where we have used two transformations

(u, v) ⇒ (x, y) ⇒ (p, q). (9.58)

In this notation, the partial derivative

∂u∂x y

= [u, y][x, y]

. (9.59)

A cyclic property of this notation can be obtained as follows: Let w = w(u, v). Then

dw = ∂w

∂u vdu + ∂w

∂v udv

= [w, v][u, v]

du + [w, u][v, u]

dv. (9.60)

Further, assume that u = u(x, y) and v = v(x, y):

∂w

∂x y= [w, v]

[u, v]∂u∂x y

+ [w, u][v, u]

∂v

∂x y,

[w, y][x, y]

= [w, v][u, v]

[u, y][x, y]

+ [w, u][v, u]

[v, y][x, y]

. (9.61)

After canceling [x, y], and changing y to x, we have the cyclic relation

[u, v][w, x] + [v,w][u, x] + [w, u][v, x] = 0. (9.62)

Having familiarized ourselves with this notation, let us apply it to thermodynamicvariables.


To begin with, we have the Gibbs relation

dε = Tds − pdv, T = ∂ε

∂s v, −p = ∂ε

∂v s, (9.63)

where v is the specific volume. Using the equality of mixed derivatives, we obtainthe Maxwell relation

∂T∂v s

= ∂(−p)∂s v

,

[T, s][v, s]

= − [p, v][s, v]

= [p, v][v, s]

,

[T, s] = [p, v]. (9.64)

This shows that, if we map from the T, s plane to the p, v plane, the area elementsremain constant.

Now we may derive further Maxwell relations:

∂p∂T s

= [p, s][T, s]

= [p, s][p, v]

= ∂s∂v p

,

∂p∂s T

= [p, T][s, T]

= − [p, T][p, v]

= −∂T∂v p

,

∂v

∂T s= [v, s]

[T, s]= [v, s]

[p, v]= − ∂s

∂p v

,

∂v

∂s T= [v, T]

[s, T]= [v, T]

[v, p]= ∂T

∂p v

. (9.65)

For an arbitrary fixed variable x, we have

[ε, x] = T[s, x] − p[v, x],

[h, x] = T[s, x] + v[p, x]. (9.66)

The specific heats at constant pressure cp and at constant volume cv can beexpressed as

cp = ∂h∂T p

= [h, p][T, p]

= T[s, p][T, p]

= T∂s∂T p

,

cv = ∂ε

∂T v= [ε, v]

[T, v]= T

∂s∂T v

. (9.67)

From these,

cp − cv = T[

[s, p][T, p]

− [s, v][T, v]

]

= T[s, p][T, v] − [s, v][T, p]

[T, p][T, v]. (9.68)

Using our cyclic identity

[p, v][T, s] + [v, T][p, s] + [T, p][v, s] = 0, (9.69)


we find

cp − cv = −T[p, v][T, p]

[p, v][T, v]

= T∂v

∂T p

∂p∂T v

. (9.70)

Similarly,

cp

cv

= [s, p][s, v]

[T, v][T, p]

= ∂p∂v s

∂v

∂p T. (9.71)

For an ideal gas,

pv = RT, (9.72)

and we have

cp − cv = R. (9.73)

SUGGESTED READING

Cengel, Y. A. and Boles, M. (2002). Thermodynamics, 4th ed., McGraw-Hill.Christoffersen, J., Mehrabadi, M. M., and Nemat-Nasser, S. (1981). A micromechanical de-

scription of granular material behavior, J. Appl. Mech., 48, 339–344.Cowin, S. C. (1978). Microstructural continuum models for granular materials, in Proceed-

ings of the U.S. Japan Seminar on Continuum Mechanical and Statistical Approaches in theMechanics of Granular Materials, Gakujatsu Bunken Fukyukai, pp. 162–170.

Margenau, H. and Murphy, G. M. (1943). The Mathematics of Physics and Chemistry, VanNostrand.

Sonntag, R. E., Borgnakke, C., and Van Wylen, G. (2003). Fundamentals of Thermodynamics,6th ed., Wiley.

EXERCISES

9.1. For an ideal gas,

ε = CT, pv = RT.

Obtain the functions s(T, v), s(T, p), and ε(s, v) for this substance, explicitly.

9.2. The internal energy and specific volume are given as

ε = ε(T, v), v = v(T, p).

(a) Derive expressions for specific heat at constant pressure cp and constantvolume cv , and show that

cp − cv = ∂v

∂T p

[p + ∂ε

∂v T

].

(b) Evaluate cp − cv for the ideal gas.(c) In the case of an adiabatic process (δQ = 0), show that an ideal gas satisfies

(cv + R)pdv + cvvdp = 0.

(d) From the preceding part, obtain pvγ = constant, where γ = cp/cv .


9.3. In linear, isotropic thermoelasticity the Helmholtz free energy is given by

ρu = Gεi jεi j + λ

2εi iε j j − 2G

1 + ν

1 − 2να(T − T0)εi i ,

where ρ, G, λ, ν, and α are constants. Assuming that εi j , the infinitesimal strain com-ponents, are the thermodynamic state variables, obtain the corresponding stresses,σi j .

9.4. In a Newtonian fluid, the hyperelastic part of the stress σ 0i j and the irreversible

part σ Ii j are given by

σ 0i j = −pδi j , σ I

i j = λdkkδi j + 2µdi j ,

where λ and µ are constants. Obtain the specific entropy production rate for thisflow at constant temperature T .

9.5. If we assume the strain energy density for a material in the current configuration(in the absence of couple stresses) is given as

u = u(∂1x1, ∂2x1, . . . , ∂3x3),

show that the first Piola–Kirchhoff stress tensor is obtained as

T0αi = ρ0

∂u∂xi,α

,

where we have used xi,α = ∂xi/∂ Xα.

9.6. Consider two neighboring equilibrium states I and II of a system. In state I,it has the state variables E, τ 0

i , εi , S, and T , and in state II, E + �E, τ 0i + �τ 0

i ,εi + �εi , S + �S, and T + �T .

(a) It is further assumed that no work is done on the system and no heat isadded. If we add an assumption to the second law that what is permittedby the second law would actually occur in nature, show that the entropy atstable equilibrium state is a maximum.

(b) If we assume E is a monotonically increasing function of S, and betweenstates I and II there is no change in entropy and �E �= 0, show that thestable equilibrium requires that E be a minimum.

9.7. An ideal rubber is considered to be incompressible, and its specific entropychange from a reference value of s0 is given as

ρ(s − s0) = −κ(λ21 + λ2

2 + λ23 − 3),

where ρ is the constant density and λi is the stretch in the xi direction. It is alsoknown that at constant temperature, ∂ε/∂λi = 0, where ε is the internal energy den-sity. Obtain a relation between the second Piola–Kirchhoff stress S11 in terms of λ1

for a band made of this material. The stress and stretch λ1 are along the length ofthe band.

9.8. For a one-dimensional bar the Gibbs function is given as

G = G(T, σ ), s = −GT , ε = −Gσ ,

where the subscripts indicate partial differentiation, T is the absolute temperature,σ is the tensile stress, and ε is the infinitesimal strain along the length of the bar. If


the bar has an initial length L and cross-sectional area of unity, show that the heatadded δQ at constant temperature for a stress increase δσ , change in length δL, andtemperature rise δT at constant stress are related in the form

δQδσ T

= TL

δLδT σ

.

9.9. Show that

(a)

[s, ε] = −p[s, v], [h, s] = v[p, s].

(b)∂ε

∂p s= − p

γ

∂v

∂p T,

∂ε

∂T s= pcv

T∂T∂p v

,

where γ = cp/cv .(c)

∂h∂v s

= γ v∂p∂v T

,∂h∂T s

= vcp

T∂T∂v p

.

(d)∂T∂s ε

= Tcv

[1 − T

p∂p∂T v

],

∂T∂s h

= Tcp

[1 − T

v

∂v

∂T p

].

(e)

∂cp

∂p T= −T

∂2v

∂T2 p,

∂cv

∂v T= T

∂2 p∂T2

v.

10 Constitutive Relations

Equations describing the response of specific materials to applied loads are calledconstitutive relations. So far, we have studied the local descriptions of deformation,motion, forces, and energy and entropy constraints. Subject to the same externalforces, different materials will respond differently even if they have the same massand geometric properties. This complex material behavior cannot be expressed by asingle universal equation. Excluding chemical and electrical variables, we have thefollowing conservation and balance relations:

∂ρ

∂t+ ∂i (ρvi ) = 0,

∂iσi j + ρ( f j − a j ) = 0,

∂i mi j + e jklσkl + ρ(l j − α j ) = 0,

ρε − σi j∂iv j − mi j∂iµ j − ∂i qi − ρh = 0, (10.1)

which form eight equations. We have ρ, vi , µi , σi j , mi j , qi , ε, and, implicitly, thetemperature field T as unknowns. These add up to 30 unknowns. The accelerationsai and αi are derived from vi and µi through the kinematic relations discussed inChapter 8. Of the 22 additional relations required for completing the problem, anequation of state for ε is one; we need 9 relations for σi j and 9 for mi j , which leaves 3.These are the relations for the flux qi . These 22 relations are called the constitutiverelations.

One approach to obtain these needed relations is to use the known laws relatingforces and displacement of molecules in a lattice structure or in a multiply collidingfluid situation. Progress in this direction is being made by use of molecular dynamicsand the Monte Carlo simulations. The number of particles one could introduce inthese simulations is still moderate because of computational limitations.

Another approach is through experiments. However, the number of experi-ments required is still prohibitively large. The preferred method is known as phe-nomenology. In the phenomenological approach, mathematical expressions relatingstress and deformation variables are proposed subject to certain invariance princi-ples required by continuum mechanics. These relations are tested and refined by a

129


limited number of experiments. They are further generalized to include observeddeviations in new materials.

10.1 Invariance Principles

The invariance principles may be grouped into six categories:

1. principles of exclusion,2. principle of coordinate invariance,3. principle of spatial invariance,4. principle of material invariance,5. principle of dimensional invariance, and6. principle of consistency.

We examine briefly each of these items.

10.1.1 Principles of Exclusion

These principles provide, on the basis of experience, certain rules for excluding alarge number of variables from constitutive relations. We may consider the follow-ing classes of exclusion principles under this heading.

1. Principle of heredity: The behavior of a material particle X at time t is de-pendent on only the past experience of the body. For example, for the Cauchystress,

σ =τ=tF

τ=−∞ [x(X, τ )], (10.2)

where F represents a general functional that may involve integrals in time. Thebasic idea is to exclude the future experience of the particle. As an example,consider a one-dimensional bar with the stress–strain relation

σ (t) = Eε(t) + α

∫ t

−∞e−β(t−τ ) dε

dτdτ. (10.3)

Here we have a kernel inside the integral that shows the material has a fadingmemory of the past strain rate.

2. Principle of neighborhood: The behavior of a particle, X, occupying a point,x, in the current configuration at time, t , is only affected by the history of anarbitrarily small neighborhood of x.

σ =τ=tF

τ=−∞ [dx(X, τ )],

=τ=tF

τ=−∞ [∂αx(X, τ ), ∂α∂β x(X, τ ), . . . , ]. (10.4)

These two principles are called the principles of determinism. Modern theoriesthat incorporate the influence of particles at a distance (nonlocal theories) have

10.1 Invariance Principles 131

been found useful in certain applications. The additional principles needed forelectromagnetic phenomena are not included here.

10.1.2 Principle of Coordinate Invariance

This simply restricts the variables to the tensor format, which automatically satisfiescoordinate invariance. For example, if we use the system x, we have

σi j =τ=tFi j

τ=−∞[∂αxk(X, τ )]. (10.5)

In a new system x′,

σ ′i j =

τ=tF ′

i jτ=−∞

[∂αx′k(X, τ )]. (10.6)

10.1.3 Principle of Spatial Invariance

The constitutive relations must be invariant under rigid body translations and rota-tions of the spatial coordinates. In other words, the tensors entering these equationsmust be objective tensors. For example, consider the case of a simple material forwhich dx ⇒ ∂αxk. We may use coordinate system x′ or x to describe the constitutiverelations. As the material responds, irrespective of the spatial system we use, firstwe have

Fi j = F ′i j . (10.7)

As far as the independent variables are concerned, these two systems are related as

x′i = Qi j xj , xi = Qji x′

j . (10.8)

The relation

σi j =τ=tFi j

τ=−∞[∂αxk, τ ] (10.9)

transforms to

σ ′i j =

τ=tFi j

τ=−∞[∂αx′

k(X, τ )],

QimQjntmn =τ=tFi j

τ=−∞[∂α Qkl xl(X, τ )],

tmn = QimQjn

τ=tFi j

τ=−∞[∂α Qkl xl(X, τ )]. (10.10)

Comparing this with Eq. (10.9), we find a constraint on the allowable functionalforms of Fi j ,

τ=tFmn

τ=−∞[∂αxk(X, τ )] = QimQjn

τ=tFi j

τ=−∞[Qkl∂αxl(X, τ )]. (10.11)


In tensor notation, we have (omitting the hereditary integral)

F [∇x] = QT · F [∇x · QT ] · Q. (10.12)

All variables have to be objective.

10.1.4 Principle of Material Invariance

We may transform the material coordinates Xα by using proper orthogonal matricesQ. This may be generalized to matrices R (not to be confused with R in polar de-composition), called the full orthogonal group of transformations (see Ludwig andFalter, 1987) if they impart not only rotations but also reflections of coordinates:

X′α = Rαβ Xβ. (10.13)

If the material has certain symmetry, such as isotropy, orthotropy, etc., the constitu-tive relations must be invariant with respect to transformations that obey this sym-metry. For example, isotropic constitutive relations must be invariant with respectto the full orthogonal group. Transversely isotropic materials allow rotation about afixed axis. There are crystals with hexagonal packing; these should have constitutiverelations that are invariant if the material coordinates are rotated by 60◦.

10.1.5 Principle of Dimensional Invariance

In their functional forms, constitutive relations must obey dimensional constraints.

10.1.6 Principle of Consistency

The basic principles of conservation and balance laws, as well as the entropy pro-duction inequality, must not be violated by constitutive equations.

The importance of the preceding principles can be appreciated if we visualizean experimentalist recording the response of one variable, say the length of a bar, tothe applied load. We readily see a plot of the current length versus the applied load.If we want to generalize this relation to bars of different cross-sectional areas andlengths, the concepts of stress and strain enter the scheme. Next, we want to general-ize this to three dimensions by incorporating anisotropy and nonlinear stress–strainrelations, subject to the principles just discussed. The situation is more complicatedfor fluids and viscoelastic materials.

10.2 Simple Materials

This group of materials is not as simple as the name implies! In the principle ofneighborhood, we include only the deformation gradient; all other higher deriva-tives are neglected. A large class of materials for which the stresses depend on the

10.2 Simple Materials 133

history of the deformation gradient F is included here. We begin with

σ =τ=tF

τ=−∞ [∂αxk]. (10.14)

This relation, when subjected to the principle of spatial invariance, gives

Q · σ · QT =τ=tF

τ=−∞ [F · QT ]. (10.15)

Next, we use the polar decomposition,

F = V · R, (10.16)

to write this relation as

Q · σ · QT =τ=tF

τ=−∞ [V · R · QT ]. (10.17)

The orthogonal, time-dependent tensor R is arbitrary. We may choose

Q = R (10.18)

to simplify the preceding relation to obtain

R · σ · RT =τ=tF

τ=−∞ [V] (10.19)

or

σ = RT ·τ=tF

τ=−∞ [V] · R. (10.20)

This shows that the Cauchy stress depends on the history of the left-stretch tensorV only and not on the history of the rotation R.

As

V = G1/2, (10.21)

the preceding functional form can also be written as

σ = RT ·τ=tG

τ=−∞[G] · R. (10.22)

A further change in the functional form can be obtained with the Lagrange strainE, defined by

G = 2E + I. (10.23)

Instead of the general history-dependent functionals, as limiting cases, we have G assimple functions. For example,

1. elastic materials: G is a function of E only;2. viscoelastic materials: G is a function of E and E;3. Stokes fluids: G is a function of E.


10.3 Elastic Materials

There are two ways of considering elastic materials: One is by using Cauchy’s defini-tion and the other is by using the Green’s definition. The couple stresses are omittedfor this discussion.

10.3.1 Elastic Materials of Cauchy

Special case 1, in the preceding list, can be written as

σ = G(∂αxk). (10.24)

We stipulate that there exists a natural state of the material that is distinguished byuniform conditions inside the body. In the natural state the deformation gradientis zero and the body is unstressed. Recent theories consider natural states evolvingwith the deformation history. We consider a unique natural state. This implies thatat time t = 0, the Green’s deformation tensor is the identity tensor, i.e.,

G(0) = I. (10.25)

As the couple stress tensor is zero, σ is symmetric and so is G. From our study ofspatial invariance, we can write

σ = RT · F(G) · R, (10.26)

where G(G1/2) = F(G), and R is the rotation tensor in the polar decomposition ofthe deformation gradient tensor F.

For anisotropic materials, the orientation of the undeformed material has tobe distinguished, and the function F would depend on the initial orientation. Aswe have seen, the initial orientation of the unit vector eα is now Gα in the currentconfiguration. We write this in the form

F = F(G, Gα). (10.27)

A third generalization is to make the material inhomogeneous. Now properties de-pend on the particle:

F = F(G, Gα, X). (10.28)

The arbitrary matrix function F , with the 3 × 3 matrix G as its argument, can berepresented by the Cayley–Hamilton theorem as

F = c′0 I + c′

1G + c′2G2, (10.29)

where the coefficients c′i are functions of the eigenvalues of G.

We may multiply Eq. (10.26) by f = F−1 = V−1 RT to obtain the form

S =√

IG3 f T · σ · f = c0G−1 + c1 I + c2G, (10.30)

where ci = √IG3c′

i and S is the second Piola–Kirchhoff stress.

10.3 Elastic Materials 135

In finite-strain elasticity, an area of intense research activity has been rubberelasticity. Rubber is considered to be an incompressible material up to moderatelyhigh strains. When a material is incompressible, there is no volume change andIG3 = 1. But it can support an arbitrary amount of pressure, which is called a re-active stress as it cannot be found from the constitutive relations. Then the stresscan be written as

S = −pI + c0G−1 + c1 I + c2G, (10.31)

where, now, ci = ci (IG1, IG2).

10.3.2 Elastic Materials of Green

According to the Green’s definition of elasticity, elastic materials have an internalenergy function called the strain energy, and the stresses can be derived from it byusing appropriate strain measures as state variables. This group of elastic materialsare also known as hyperelastic. For an isothermal case in which the temperature Tis fixed, the appropriate internal energy function is the Helmholtz free energy U. Ifεi are the state variables representing strain, the conjugate thermodynamic tensionsare

τ 0i = ρ

∂u∂εi T

, (10.32)

where u is the Helmholtz free energy density. The scalar function u has to be in-variant with respect to the rotation of the spatial coordinates. If we take ∂αxi as ameasure of the strain εi and assume u = u(∂αxi ), we see that a rotation of the spatialcoordinates would alter the form of u. Because dxi dxi = (ds)2 is an invariant, weassume

u = u(∂αxi∂β xi ) = u(Gαβ). (10.33)

This functional form for the Helmholtz free energy is known as the Boussinesq form.Then, with

εi ⇒ ∂αxi , εi ⇒ ∂ jvi∂αxj , (10.34)

we have

τ 0αi = 2ρ

∂u∂Gαβ

∂β xi . (10.35)

From Chapter 9 we have

σ 0i j∂iv j = σ 0

i j di j = ρ∂u∂εi

εi . (10.36)

Substituting for εi and εi , we have

σ 0i j∂iv j = 2ρ

∂u∂Gαβ

∂β xi∂αxj∂ jvi . (10.37)


The expression on the right-hand side is symmetric in α and β because of the sym-metry of Gαβ . Equating the coefficients of the velocity gradients from both sides, weobtain an expression for σ 0 as

σ 0i j = 2ρ

∂u∂Gαβ

∂β xi∂αxj . (10.38)

In terms of the second Piola–Kirchhoff stress, we have

Sαβ = 2ρ0∂u

∂Gαβ

. (10.39)

Further, if we take u as a function of the Green’s strain E,

Sαβ = ρ0∂u

∂ Eαβ

. (10.40)

For isotropic, homogeneous materials, rotations of the initial configuration shouldnot affect the form of the Helmholtz free energy function. Then u should depend onE only through the three invariants of E or those of G. The three invariants of Gare

IG1 = Gαβδαβ,

IG2 = 12

[GααGββ − GαβGαβ],

IG3 = 16

[eαβγ eµνλGαµGβνGγ λ]. (10.41)

The derivatives of these invariants with respect to Gαβ are obtained as

∂ IG1

∂Gαβ

= δαβ,

∂ IG2

∂Gαβ

= IG1δαβ − Gαβ,

∂ IG3

∂Gαβ

= IG3G−1αβ . (10.42)

The second Piola–Kirchhoff stress becomes

Sαβ = 2ρ0[u,1 δαβ + u,2 (IG1δαβ − Gαβ) + u,3 IG3G−1αβ ], (10.43)

where we have used the notation

u,i = ∂u∂ IGi

(10.44)

Finally, we rearrange the terms to get S in the form

S = g0G−1 + g1 I + g2G, (10.45)

where the coefficients are defined as

g0 = 2ρ0 IG3u,3 , g1 = 2ρ0(u,1 +IG1u,2 ), g2 = −2ρ0u,2. (10.46)

10.4 Stokes Fluids 137

In the natural state, G = I, and

IG1 = 3, IG2 = 3, IG3 = 1. (10.47)

For a stress-free natural state, for a Green’s elastic material,

(g0 + g1 + g2)|IG1=3,IG2=3,IG3=1 = 0 or u,1 +2u,2 +u,3 = 0, (10.48)

and for a Cauchy elastic material,

(c0 + c1 + c2)|IG1=3,IG2=3,IG3=1 = 0. (10.49)

The coefficients gi for the Green’s elastic material are obtained as derivatives of asingle function u, the Helmholtz free energy, and the coefficients ci in the Cauchyelastic case are three independent functions. Using the mixed derivatives, in thesense of Maxwell relations, we may relate the derivatives of gi .

For the incompressible case, u,3 = 0, as IG3 = 1. We add a hydrostatic pressureand write the stress as

S = −pI + g0G−1 + g1 I + g2G. (10.50)

10.4 Stokes Fluids

In Stokes fluids, the stress tensor is split into a reversible part that can be derivedfrom a potential and an irreversible part:

σ = σ 0 + σ I . (10.51)

As in an ideal gas,

σ 0 = −pI. (10.52)

where p is called the thermodynamic pressure. The irreversible part is assumed tobe a function of the deformation rate tensor d:

σ I = F(d ). (10.53)

As the deformation rate tensor d is an objective tensor, this form satisfies the rele-vant invariance requirements. Next, we use the Cayley–Hamilton theorem to repre-sent the function F(d ) as a polynomial in d:

F(d) = a0 I + a1d + a2d 2, (10.54)

where ai = ai (Id1, Id2, Id3), with Idi being the invariants of d.The stress σ can be written as

σ = (a0 − p)I + a1d + a2d 2. (10.55)

When there is no motion, all the invariants of d are equal to zero and the stress isthe hydrostatic stress that is due to the thermodynamic pressure. That is,

ai (0, 0, 0) = 0. (10.56)


For an incompressible fluid, Id1 = 0, and the pressure is a reactive quantity and itremains as a basic unknown. For this case, the stress can be written as

σ = −pI + a1d + a2d 2, (10.57)

where p is called the mechanical pressure, and

ai = ai (Id2, Id3). (10.58)

To satisfy the Clausius–Duhem inequality about the entropy production,

ρTs = σ Ii j di j > 0. (10.59)

Using the constitutive relation, this inequality becomes

a1dii + a2di j dji + a3di j djkdki > 0, (10.60)

where the three terms contain the first invariants of d, d 2, and d 3, respectively. Wecan show that

Id21 = I 2d1 − 2Id2, Id31 = I 3

d1 − 3Id1 Id2 + 3Id3. (10.61)

The entropy production constraint can be written in terms of these invariants as

a1 Id1 + a2(I 2d1 − 2Id2) + a3(I 3

d1 − 3Id1 Id2 + 3Id3) > 0. (10.62)

The mechanical pressure p can be expressed as

p = −13

tii = p − a1 − 13

a2 Id1 − 13

a3(I 2d1 − 2Id2). (10.63)

The mechanical pressure p is the pressure one measures as the average of threenormal stresses, and the thermodynamic pressure, as we have seen, is related to theinternal energy ε. In the static case they are equal. The coefficients ai , which partic-ipate in the production of entropy, are called phenomenological coefficients. Theyare in general functions of the state variables: temperature and specific volume.

We continue the discussion of elastic materials and Stokes fluids in the next twochapters.

10.5 Invariant Surface Integrals

The balance and conservation laws discussed in this chapter lead to certain invariantsurface integrals that have many applications, in particular in fracture mechanicsand fluid dynamics. We make the following simplifications at the outset: The Cauchystress tensor is symmetric, and the moment stresses and the intrinsic rotations areabsent.

The local forms of balance and conservation are∂ρ

∂t+ ∂i (ρvi ) = 0,

∂ jσi j + ρ( fi − ai ) = 0,

ρε − σi j∂ jvi − ∂i qi − ρh = 0. (10.64)

10.5 Invariant Surface Integrals 139

We anticipate applications for which, with respect to a steadily moving coordinatesystem, the field variables appear time independent. As an example, in front ofan extending crack (far away from any boundaries) the stress field can be ap-proximated as independent of time. An airfoil moving in a fluid will experience atime-independent velocity field around it. Let us introduce the moving coordinatesthrough

x′i = xi − Vi t. (10.65)

Then

∂

∂t= −Vi

∂

∂x′i,

DDt

= (vi − Vi )∂

∂x′i. (10.66)

Using the notation ∂ ′i = ∂/∂x′

i , conservation of mass, balance of momentum, andbalance of energy can be written as

∂ ′i [ρ(vi − Vi )] = 0,

∂ ′jσi j − ρ(v j − Vj )∂ ′

j (vi − Vi ) + ρ fi = 0,

ρ(vi − Vi )∂ ′i ε − ρh − ∂ ′

i qi − σi j∂′j (vi − Vi ) = 0. (10.67)

Next, we integrate these three expressions over a fixed, arbitrary volume V′ enclosedby a surface S′ (in the moving coordinate system), with the assumption that noneof the field quantities is singular inside this volume. We then convert the volumeintegrals into surface integrals by using the Gauss theorem. Thus the conservationof mass gives ∫

S′ρ(vi − Vi )n′

i dS′ = 0. (10.68)

The momentum equation gives∫S′

{[σi j − ρ(v j − Vj )(vi − Vi )]n′

j − φn′i

}dS′ = 0, (10.69)

where the potential φ is related to the body force through

ρ fi = −∂ ′i φ. (10.70)

Multiplying the local form of the momentum equation by (vi − Vi ), we find

(vi − Vi )∂ ′i σi j − (vi − Vi )∂ ′

i φ − 12ρ(v j − Vj )∂ ′

j [(vi − Vi )(vi − Vi )] = 0. (10.71)

Using the local forms of conservation of mass and the two forms of the momentumequations, we can write the energy equation as∫

S′

{ρ(vi − Vi )[ε + 1

2(v2 − V2)] + ψi + φi + viφ − qi − σi jv j

}n′

i dS′ = 0, (10.72)

where we have introduced two more potentials, φ and ψ , through the relations

φ∂ ′i vi = −∂ ′

j φ j , ρh = −∂ ′i ψi . (10.73)


We may obtain a simpler version of the invariant integral involving the energyunder the following assumptions: the heat generation, the heat flow, and the bodyforces are zero and the deformation gradient is small compared with unity. Theseare suitable for small deformations of solids:

vi = ui = (v j − Vj )∂ ′j ui or |vi | � |Vi |. (10.74)

Neglecting terms of the order of vi compared with those of the order of Vi , we findthat the energy equation gives∫

S′[ρ(ε + 1

2V2)n′

i − σkj∂′i u j n′

k]dS′ = 0, (10.75)

where we have canceled a common factor, (−Vi ). Noting that the integral involvingV2 is zero, we find ∫

S[ρεni − σkj∂i u j nk]dS = 0. (10.76)

In applications involving fluid mechanics, if we use the original coordinates, weget ∫

Svi [ρ(ε + 1

2v2)ni − σi j n j ]dS = 0. (10.77)

10.6 Singularities

It was mentioned earlier that the surface integrals, Eqs. (10.76) and (10.77), vanish ifthere are no singularities inside the surface. When there is a point singularity, theseintegrals may not be zero. If we consider two surfaces S1 and S2 enclosing the singu-larity, the integrals over these surfaces will be the same as there is no singularity inthe region between them. Thus we may shrink the surface to an arbitrarily small sizeto enclose the singularity. The integrals represent certain intrinsic properties of thesingularity. One may compare this situation with Cauchy’s residue theorem. By ex-amining Eq. (10.77), we see the integral is the loss of energy from the small volumeper unit time. This energy must equal the work done by the fluid on the singularity;in other words, the energy dissipates at the singularity. We will revisit this idea laterwhen we consider elasticity and fluid dynamics. The preceding discussion is basedon the theory presented by Cherepanov (1979).

SUGGESTED READING

Cherepanov, G. P. (1979). Mechanics of Brittle Fracture, McGraw-Hill.Ludwig, W. and Falter, C. (1987). Symmetries in Physics, Springer-Verlag.Malvern, L. E. (1969). Introduction to the Mechanics of a Continuous Medium, Prentice-Hall.Truesdell, C. (1960). Principles of Continuum Mechanics, Field Research Laboratory, Socony

Mobil Oil Co., Dallas, TX.Truesdell, C. and Noll, W. (1965). The nonlinear field theories of mechanics, in Encyclopedia

of Physics (S. Flugge, ed.), Springer-Verlag, Vol. 3/3.Truesdell, C. and Toupin, R. A. (1960). The classical field theories, in Encyclopedia of

Physics, (S. Flugge, ed.), Springer-Verlag, Vol. 3/1.

Exercises 10.1–10.9 141

EXERCISES

10.1. Express the first invariants of the matrices A2 and A3 in terms of the threeinvariants of the matrix A.

10.2. Obtain the Maxwell relations among the first-order derivatives of the coeffi-cients gi for the Green’s elastic materials.

10.3. For hyperelastic materials, if we assume the Neumann–Kirchhoff form,

u = u(∂αxi ),

obtain the corresponding stress.

10.4. Another form for u is given by the Hamel form,

u = u(∂i Xα).

Obtain the stress for this spatial description.

10.5. Yet another form for u is the Murnaghan form,

u = u(ei j ).

What is the stress for this case?

10.6. Based on the constitutive relations of Cauchy elastic materials, hyperelasticmaterials, and the Stokes fluids, show that the principal axes of the second Piola–Kirchhoff stress S coincide with those of the Green’s deformation tensor G andthose of the Cauchy stress σ coincide with those of the deformation rate tensor d.

10.7. In an isotropic elastic material, λi are the three principal stretches and thecorresponding principal stresses are σi , respectively. If the stretches are ordered as

λ1 < λ2 < λ3,

obtain the conditions necessary for a similar ordering of the principal stresses.

10.8. From experiments, one concludes that the second Piola–Kirchhoff stress canbe written in terms of the deformation gradient components as

Sαβ = Aαβγ i∂γ xi + Bαβγ δ∂γ xi∂δxi .

Using the principles of invariance, evaluate the suitability of this relation.

10.9. In a shear flow the only nonzero deformation rate component is ∂2v1 = d. Theshear stress σ21 = τ at a point was measured for various values of d. Based on thisstudy, it was suggested that

τ = µ d + ν d3,

where µ and ν are constants. How would you generalize this for the three-dimensional (3D) case for further experiments to verify your extrapolation?

11 Hyperelastic Materials

In the previous chapter we briefly examined the constitutive relations of hyperelas-tic materials. There has been a tremendous amount of research in the area of elasticmaterials, beginning with the early studies by Hooke, Navier, Cauchy, Germain,Kirchhoff, St. Venant, Airy, Love, and others. Most of these studies dealt with lin-ear stress–strain relations with the infinitesimal-strain assumption. With the adventof the finite-element method and high-speed computers, the idealized geometricalconfigurations of the elastic bodies as well as the constraints of linear elasticity ofthe past can now be relaxed whenever needed.

We begin this chapter with a discussion of finite elasticity in a general setting.Some of the well-known inverse solutions are introduced. A small section is devotedto linear elasticity, as most students have had multiple courses in this area. Thechapter concludes with the topic of linear thermoelasticity.

11.1 Finite Elasticity

The basic equations of finite elasticity, in terms of the Cauchy stress, can be listed asfollows:

conservation of mass: ρ = ρ0/√

I3,

linear momentum: ∇ · σ = ρ(a − f ),angular momentum: σ = σ T ,

constitutive relation: σ = −pI + FT · (a0G−1 + a1 I + a2G) · F, (11.1)

where the coefficients ai are given in terms of the partial derivatives of the strainenergy density function (Helmholtz free energy density), u(I1, I2, I3), by

a0 = 2ρ0

√I3u,3 ,

a1 = 2ρ0√I3

(u,1 +I1u,2 ),

a2 = − 2ρ0√I3

u,2 . (11.2)

142

11.1 Finite Elasticity 143

We have written the constitutive relation in a form applicable to both compressibleand incompressible materials. For compressible materials, p = 0, and for incom-pressible materials, a0 = 0, and a1 and a2 are functions of I1 and I2 only, as I3 = 1.We omit the subscript G for the invariants, as all invariants in this chapter pertainto the Green’s deformation tensor. We also have the definitions

G = FFT , Fαi = ∂αxi , a = x. (11.3)

At this stage, we keep the Helmholtz free energy density u as an unspecified func-tion. For rubberlike materials, which are assumed to be incompressible, u is oftenassumed in the form

u = A1(I1 − 3) + A2(I2 − 3), (11.4)

with Ai being constants. Materials with this Helmholtz energy function are known asMooney–Rivlin materials. If A2 = 0, we have what are known as Mooney materials.

The field equations, Eqs. (11.1), have to be solved subject to boundary and ini-tial conditions.

At every point on the boundary surface S, we have

xi = xi or σ(n)i = σ

(n)i , i = 1, 2, 3, (11.5)

where the overbar (¯) represents a given function.At time t = 0, the location and velocity of the material particles are given as

xi = Xαδαi , xi = vi . (11.6)

A number of solutions have been obtained for problems in finite-strain elasticityin the static case by use of the inverse method. This method assumes a displacementdistribution and attempts to find the corresponding stress distribution, which is inequilibrium. These results may be used in an empirical approach—in which we as-sume a form for the Helmholtz free energy density function and attempt to deducethe empirical constants from experiments.

11.1.1 Homogeneous Deformation

When all the strains in a body are constants, we refer to the deformation as homo-geneous. Here, we assume a deformation of the form

xi = Aiα Xα, (11.7)

where Aiα is a constant, nonsingular matrix. By selecting the Xα coordinates alongthe principal directions of A ( in which A is diagonal), we can have

x1 = λ1 X1, x2 = λ2 X2, x3 = λ3 X3, (11.8)


where λi are the stretches. The Green’s deformation tensor is represented by thediagonal matrix

G =

⎡⎢⎣λ2

1 0 00 λ2

2 00 0 λ2

3

⎤⎥⎦ , (11.9)

which has, as invariants,

I1 = λ21 + λ2

2 + λ23, I2 = λ2

1λ22 + λ2

2λ23 + λ2

3λ21, I3 = λ2

1λ22λ

23. (11.10)

As G is diagonal, no rotation is necessary, i.e., R = I, and the stress can be expressedas

σ = a0 I + a1G + a2G2, (11.11)

where

a0 = 2ρ0λ1λ2λ3u,3 ,

a1 = 2ρ0

λ1λ2λ3[u,1 +(λ2

1 + λ22 + λ2

3)u,2 ],

a2 = − 2ρ0

λ1λ2λ3u,2 . (11.12)

Explicitly, the normal components of the stress tensor are

σ11 = a0 + a1λ21 + a2λ

41,

σ22 = a0 + a1λ22 + a2λ

42,

σ33 = a0 + a1λ23 + a2λ

43, (11.13)

and the shear components are σ12 = σ23 = σ31 = 0.The equations of equilibrium are satisfied if there are no body forces or accel-

erations.For an incompressible material, I3 = 1, and

λ3 = 1λ1λ2

, (11.14)

and in the expressions for the normal stresses, we replace a0 with −p, the reactivehydrostatic pressure.

11.1.2 Simple Extension

An easy experiment involves stretching a circular bar in the x1 direction. From thesymmetry of the isotropic specimen, we conclude that

λ2 = λ3, (11.15)

and the axial component of the stress σ = σ11 is related to the axial stretch λ = λ1 as

σ = a0 + a1λ2 + a2λ

4. (11.16)


To satisfy the boundary conditions on the cylindrical surface, we have to have σ22 =σ33 = 0. That is,

a0 + a1λ22 + a2λ

42 = 0, (11.17)

which may be viewed as an equation for λ2 = λ3 [although Eq. (11.16) for σ is cou-pled with this equation].

For the incompressible case, λ22 = 1/λ, a0 = 0, and we substitute −p for a0 in

the preceding equation. Instead of the unknown λ2, now we solve for p.

11.1.3 Hydrostatic Pressure

If we set λ1 = λ2 = λ3 = λ, the state of stress corresponding to this dilatation is ahydrostatic pressure p, given by

p = −(a0 + a1λ2 + a2λ

4). (11.18)

11.1.4 Simple Shear

We assume a deformation of the form

x1 = X1 + γ X2, x2 = X2, x3 = X3, (11.19)

where γ , the shear angle, is a constant.From this, the deformation gradient F and the deformation tensor G are ob-

tained as

F =

⎡⎢⎣ 1 0 0

γ 1 00 0 1

⎤⎥⎦ , G =

⎡⎢⎣ 1 γ 0

γ 1 + γ 2 00 0 1

⎤⎥⎦ . (11.20)

The invariants of G are

I1 = 3 + γ 2, I2 = 3 + γ 2, I3 = 1. (11.21)

The stress tensor can be written as

σ = FT [a0G−1 + a1 I + a2G]F. (11.22)

After some matrix multiplications, we find

σ11 = 2ρ0[(1 + γ 2)u,1 +(2 + γ 2)u,2 +u,3],

σ22 = 2ρ0[u,1 +2u,2 +u,3],

σ33 = 2ρ0[u,1 +(2 + γ 2)u,2 +u,3],

σ12 = 2ρ0γ [u,1 +u,2],

σ13 = σ23 = 0. (11.23)

In linear elasticity, a block of material can be sheared with an engineering shearstrain γ by an applied shear stress σ12 = Gγ , where G is the modulus of rigidity. The


preceding stress expressions show that, in general, in addition to the shear stress, weneed normal stresses σ11 and σ22 to maintain the simple shear. This effect is calledthe Poynting effect. The further need for an out-of-plane stress σ33 to maintain thevolume of the specimen is attributed to the Kelvin effect.

For an incompressible material, the preceding expressions are replaced with

σ11 = 2ρ0γ2u,1,

σ22 = −2ρ0γ2u,2,

σ33 = −p + 2ρ0[u,1 +(2 + γ 2)u,2] = 0,

σ12 = 2ρ0γ [u,1 +u,2],

σ13 = σ23 = 0, (11.24)

where we have eliminated p from σ11 and σ22 using the third equation. As the hy-drostatic pressure adjusts itself to satisfy σ33 = 0, we do not have the Kelvin effectin this class of materials. The Poynting effect is still present. Another observationwe can make is that

σ11 − σ22 = γ σ12 (11.25)

for compressible as well as for incompressible materials.

11.1.5 Torsion of a Circular Cylinder

The inverse solution assumes the classical assumption of the torsion problem,namely, plane cross-sections of the shaft normal to its axis remain plane. Points onthe cross-section undergo a rigid body rotation, the rotation angle linearly varyingalong the axis of the shaft. Using polar coordinates (R, θ, Z) for the initial positionand (r, φ, z) for the final position, we have

r = R, φ = θ + α, z = Z, (11.26)

where α = γ Z is the angle of rotation. The Cartesian coordinates are given by

x1 = R cos φ, x2 = R sin φ, x3 = Z,

X1 = R cos θ, X2 = R sin θ, X3 = Z. (11.27)

We can obtain the deformation gradient (after we compute the derivatives by usingthe chain rule) as

F =

⎡⎢⎣ 1 0 0

0 1 0−γ R sin φ γ R cos φ 1

⎤⎥⎦ . (11.28)

It is more convenient to look at this tensor in a rotated coordinate system, with x′1

along r and x′2 along φ. The undeformed configuration is still described by the Xα


coordinate system. The rotation matrix is given by

Q =

⎡⎢⎣ cos φ sin φ 0

− sin φ cos φ 00 0 1

⎤⎥⎦ . (11.29)

As x′i = Qi j xj and F ′

αi = ∂αx′i = ∂α Qi j xj , the deformation gradient and the defor-

mation tensors in the new coordinate system can be obtained as

F′ = F QT =

⎡⎢⎣ cos φ − sin φ 0

sin φ cos φ 00 β 1

⎤⎥⎦ ,

G′ = F′ F′T =

⎡⎢⎣ 1 0 −β sin φ

0 1 β cos φ

−β sin φ β cos φ 1 + β2

⎤⎥⎦ , (11.30)

where

β = γ R = γ r. (11.31)

The invariants are seen as

I1 = 3 + β2 = I2, I3 = 1. (11.32)

In this new system, the stress is obtained as

σ ′ = F′T [a0G′−1 + a1 I + a2G′]F′. (11.33)

The stress components are found as

σ ′11 = σrr = a0 + a1 + a2,

σ ′22 = σφφ = a0 + (1 + β2)a1 + (1 + 3β2 + β4)a2,

σ ′33 = σzz = a0 + a1 + (1 + β2)a2,

σ ′32 = σzφ = β[a1 + (2 + β2)a2],

σ ′12 = σrφ = 0,

σ ′31 = σzr = 0. (11.34)

The equation of equilibrium in the r direction is given by

∂

∂r(rσrr ) − σφφ = 0. (11.35)

As β = γ r is a function of r , this equation cannot be satisfied in general. However,for incompressible materials (note that the deformation we have prescribed is vol-ume preserving), a0 is replaced with the hydrostatic pressure −p, and we could havep varying along the radius to satisfy the preceding equation. That is,

∂

∂r(r p) = ∂

∂r[r(a1 + a2)] − σφφ. (11.36)


The torque required to create a twist of γ is obtained as

T = 2π

∫ a

0r2σzφdr, (11.37)

where a is the radius of the shaft. To achieve the deformation without any stretchingin the axial direction, we see that an axial stress σzz is needed. This is due to thePoynting effect.

11.2 Approximate Strain Energy Functions

Because the strain energy density (Helmholtz free energy density) u is a functionof IG1, IG2, and IG3, we may attempt to expand it as a polynomial in its variables.However, these variables are not small as IG1 = IG2 = 3 and IG3 = 1 in the naturalstate. But the invariants of the Green’s strain E can be small. For potential use withsmall strains, we expand the Helmholtz free energy in the form

ρ0u = αE IE1 + 12

(λE + 2µE)I 2E1 − 2µE IE2 + �E I 3

E1 + mE IE1 IE2 + nE IE3 + · · · +,

(11.38)

where we have used the notation of Eringen. Note that if IE1 = O(ε) (order of ε,which is a small quantity), IE2 = O(ε2) and IE3 = O(ε3). We have kept terms up toO(ε3).

We may also use the Almansi strain e to express u as a polynomial in the invari-ants of e as

ρ0u = αe Ie1 + 12

(λe + 2µe)I 2e1 − 2µe Ie2 + �e I 3

e1 + me Ie1 Ie2 + ne Ie3 + · · · + . (11.39)

For the existence of a natural state, we must have u,1 = 0. Thus αE = 0 and αe = 0.The invariants of E and e are related as the eigenvalues of the two matrices satisfy

E1 = e1

1 − 2e1, e1 = E1

1 + 2E1, etc. (11.40)

Using binomial expansion and retaining terms up to O(ε3), we obtain

E1 = e1 + 2e21 + 4e3

1, e1 = E1 − 2E21 + 4E3

1 . (11.41)

The invariants can be obtained to the same degree of approximation as

Ie1 = IE1 + 2(I 2E1 − 2IE2) + 4(I 3

E1 − 3IE1 IE2 + 3IE3) + · · · +,

Ie2 = IE2 − 2IE1 IE2 + 6IE3 + · · · +,

Ie3 = IE3 + · · · + . (11.42)

The elastic constants in the preceding two representations are then related in theform

λe = λE, µe = µE, �e = �E,

me = mE − 4λE − 12µE, ne = nE + 12µE. (11.43)

11.2 Approximate Strain Energy Functions 149

The Cauchy stress and the second Piola–Kirchhoff stress are obtained from this as

σ = [λIe1 + (3� + me − λ)I 2e1 + (9me + ne)Ie2]I

+ [2µ − (me + ne + 2λ + 2µ)Ie1]e + (ne − 4µ)e2 + · · ·+,

S = [λIE1 + 3(� + mE)I 2E1 + (mE + nE)IE2]I

+ [2µ − (mE + nE)IE1]E + nE E2 + · · · + . (11.44)

11.2.1 Hookean Materials

The stress–strain relations previously given are explicit up to quadratic terms in thestrain variables. If we set the coefficients of the quadratic and higher-order terms tozero, we get

σ = λIe1 I + 2µe,

S = λIE1 I + 2µE. (11.45)

These linear relations are called Hooke’s law for isotropic materials. Materials obey-ing this law are called Hookean materials. The constants λ and µ are known as Lameconstants.

11.2.2 Small-Strain Approximation

If the strains and rotations are small compared with unity, the differences betweenE and e and those between S and σ vanish. We may write

ei j = 12

(∂i u j + ∂ j ui ). (11.46)

The equation of linear momentum becomes

(λ + µ)∇(∇ · u) + µ∇2u + ρ f = ρa, (11.47)

which is called the Navier equation. We may rewrite this as

(λ + µ)u j, j i + µui, j j + ρ fi = ρui . (11.48)

The stress–strain relation

σi j = λekkδi j + 2µei j (11.49)

can be inverted to give

ei j = 1E

[(1 + ν)σi j − νσkkδi j ], (11.50)


where E is Young’s modulus and ν is Poisson’s ratio, which are related to the Lameconstants in the form

λ = Eν

(1 + ν)(1 − 2ν), µ = G = E

2(1 + ν), (11.51)

where G is the modulus of rigidity.Explicitly, we have

e11 = 1E

[σ11 − ν(σ22 + σ33)],

e22 = 1E

[σ22 − ν(σ33 + σ11)],

e33 = 1E

[σ33 − ν(σ11 + σ22)],

e12 = 12G

σ12,

e23 = 12G

σ23,

e31 = 12G

σ31. (11.52)

11.2.3 Plane Stress and Plane Strain

In problems involving in-plane loading of a thin plate, we assume σ33 = σ13 =σ23 = 0, where the x3 axis is assumed to be perpendicular to the plate. This is calleda plane stress approximation.

When the dimension of the structure in the x3 direction is large compared withthat of the other two directions, the two end sections are constrained from moving,and the applied loads in the x1, x2 plane do not vary with x3, we have the plane strainapproximation: e33 = e13 = e23 = 0.

In these two cases, in static elasticity of Hookean materials under small-strainapproximation, the problem reduces to a two-dimensional (2D) problem. A num-ber of exact solutions are available for simple geometric configurations. Solutiontechniques using complex variables developed by Kolosoff and Mushkelishvili havebeen extensively used on this topic.

11.3 Integrated Elasticity

In linear elasticity, we assume that (a) the stress–strain relations are linear (Hooke’slaw), (b) the strain–displacement relations are linear (small strains and rotations),and (c) the entire load is applied at once. In integrated elasticity or incrementalelasticity, the load is increased in infinitesimal steps and for each load increment thegeometry is updated. In the finite-element literature, this approach is also known as

11.3 Integrated Elasticity 151

A B

C

D

�

L

P

P

θ

uFigure 11.1. Truss with one horizontal elastic member andfour rigid members.

the updated Lagrangian method. The incremental-loading scenario is illustrated bya simple example in the following subsection.

11.3.1 Example: Incremental Loading

Consider the five-bar truss shown in Fig. 11.1. Except for the horizontal bar AB, allthe other four bars are rigid with length L, and all the joints are pin connected. Thebar AB is linear elastic with Young’s modulus E and constant cross-sectional areaA. We start applying the load P gradually, starting when the angle θ = 45◦. At anyangle θ , the compressive force in the member BC is given by

2F1 = Pcos θ

,

and the tensile force in the horizontal, elastic member AB is

F = P tan θ.

The change in this force is given by

�F = �(P tan θ).

The stress increment �σ = �F/A, and the strain increment �ε = �F/E A.The incremental strain is related to the change in the current length � as

�ε = ��

�= �F

E A. (11.53)

From the geometry of the truss,

sin θ = �

L, �� = Lcos θ�θ. (11.54)


0.02 0.04 0.06 0.08 0.1

0.02

0.04

0.06

0.08

0.1

0.12

0.14

p

µ

Figure 11.2. Force p = P/E A as a func-tion of µ = u/L. The lower curve shows theeffect of incremental change in geometry.

From Eqs. (11.53) and (11.54), we find

cos θ

sin θ�θ = �F

E A. (11.55)

Converting this into the differential form and integrating, we obtain

FE A

= log{√

2 sin θ}, (11.56)

where we have used the condition F = 0 when θ = 45◦. The displacement of the barBC is related to the angle θ in the form

uL

= sin θ − 1√2. (11.57)

The applied force P and the displacement u are related as

p =√

2 − (1 + µ)2

1 + µlog(1 + µ), (11.58)

where p = P/E A, and µ = √2u/L.

If we neglect the change in geometry, we get

p = µ. (11.59)

A plot of the preceding two relations is shown in Fig. 11.2. It can be seen thatthe stiffness of the structure decreases as the angle increases.

11.4 A Variational Principle for Static Elasticity

In an isothermal process, the energy stored in a body can be expressed in terms ofthe Helmholtz free energy as

U =∫

Vρu(IG1, IG2, IG3)dV, (11.60)

where, based on isotropy, the energy is assumed to depend on the invariants of theGreen’s deformation tensor G.

11.4 A Variational Principle for Static Elasticity 153

We consider a virtual work principle in which the virtual change in the internalenergy is equal to the virtual work done by the external forces. That is,∫

VρδudV =

∫Sσ

σ (n) · δxda +∫

Vf · δxdV, (11.61)

where Sσ is the part of the surface on which tractions are prescribed. This can bewritten as∫

VρdV

2∂u∂Gαβ

∂αxi∂βδxi −∫Sσ

σ(n)i δxi da −

∫V

ρdV fiδxi = 0, (11.62)

where we have used the symmetry of Gαβ . We have

∂βδxi = ∂ jδxi∂β xj . (11.63)

Using this, we rewrite the virtual work relation in the form∫V

ρdV[

2∂u

∂Gαβ

∂αxi∂β xj∂ jδxi − fiδxi

]−

∫Sσ

σ(n)i δxi da = 0. (11.64)

Integrating the first term inside the volume integral by parts, we get

−∫

VdV

[∂ j

{2

ρ∂u∂Gαβ

∂αxi∂β xj

}+ ρ fi

]δxi

+∫Sσ

{nj

[2

ρ∂u∂Gαβ

∂αxi∂β xj

]− σ

(n)i

}δxi da = 0, (11.65)

where we have set δxi = 0 on the part of the surface where the displacements areprescribed. To satisfy this equation, inside the domain we must have

∂ j

[2

ρ∂a∂Gαβ

∂αxi∂β xj

]+ ρ fi = 0, (11.66)

and on the boundary,

nj

[2

ρ∂a∂Gαβ

∂αxi∂β xj

]= σ

(n)i . (11.67)

The differential equation (Euler–Lagrange equation of the variational problem) canbe written as

∂ jσ j i + ρ fi = 0, (11.68)

where σi j is defined as

σi j = 2ρ∂u∂Gαβ

∂αxi∂β xj . (11.69)

The boundary condition simply says

σ(n)i = njσi j . (11.70)

The definition of the Cauchy stress and the relation between the traction and thestress tensor are exactly the ones at which we had arrived earlier.


If the applied traction and the body force are conservative, we have associatedpotentials � and ϕ such that

σ(n)i = −∂�

∂xi, fi = − ∂ϕ

∂xi, (11.71)

and the virtual work principle can be written as a variational principle, namely,

δ� = 0, (11.72)

where the potential energy � is defined as

� =∫

Vρ[u + ϕ]dV +

∫Sσ

�da. (11.73)

The variational and virtual work formulations allow us to approximate the solutionsof elasticity problems by assuming convenient sequences of displacement distribu-tions. For incompressible materials, the constraint IG3 = 1 may be introduced intothe variational theorem through a Lagrange multiplier function.

11.5 Isotropic Thermoelasticity

Let us assume that the coefficient of thermal expansion α is independent of strainand the total strain can be expressed as the sum of an expansion that is due to tem-perature change and one that is due to the state of stress. That is,

ei j = α(T − T0)δi j + 1E

[(1 + ν)σi j − νσkkδi j ], (11.74)

and its inverse

σi j = λekkδi j + 2µei j − (3λ + 2µ)α(T − T0)δi j , (11.75)

where T0 is a reference temperature. For a reversible, infinitesimal deformation, thedifferential form of the Helmholtz free energy density u can be written as

ρdu = −ρsdT + σi j dei j

= −ρsdT + λede + 2µei j dei j − (3λ + 2µ)α(T − T0)de, (11.76)

where we have used e = ekk.Integrating ρ∂u/∂ei j with respect to the strains, we have

ρu = λe2

2+ µei j ei j − (3λ + 2µ)α(T − T0)e + C(T), (11.77)

where the unknown function C(T) is related to the entropy density s through

ρs = −(

ρ∂u∂T

)ei j

= −dCdT

− e2

2dλ

dT+ e

ddT

[α(3λ + 2µ)(T − T0)] − ei j ei jdµ

dT. (11.78)

11.5 Isotropic Thermoelasticity 155

If we take s = s(T, ei j ), when ei j = 0 the volume is constant. Then the specific heatcv is obtained as

cv = δQρδT

= T(

∂s∂T

)ei j

= −Tρ

d2CdT2

. (11.79)

Assuming u and s are measured from the reference state, we have

C(T) = −∫ T

T0

dT∫ T

T0

ρcvdT/T. (11.80)

Using this, we can find the entropy density s and the internal energy:

ε = u + Ts. (11.81)

If we do a similar analysis with the Gibbs free energy g, with

ρdg = −ρsdT − ei j dσi j , (11.82)

we can express g and s in terms of the specific heat at constant stress, cp.

11.5.1 Specific Heats and Latent Heats

We defined two kinds of specific heats—cv for raising the temperature while thestrains are held constant and cp while the stresses are held constant. We definethe latent heat �i j as the amount of heat required to increase the strain ei j by unitywhile all other strains and the temperature are fixed. Similarly, the latent heat Li j isthe amount of heat required to increase the stress σi j by unity while all other stressesand the temperature are fixed:

δQ = ρTds = cvdT + �i j dei j

= cpdT + Li j dσi j . (11.83)

From the Helmholtz free energy we have

∂s∂ei j

= − 1ρ

∂σi j

∂T. (11.84)

Then,

�i j = −Tρ

∂σi j

∂T. (11.85)

Similarly, working with the Gibbs free energy, we have

∂s∂σi j

= − 1ρ

∂ei j

∂T, (11.86)

Li j = Tρ

∂ei j

∂T. (11.87)


For a constant-stress heat input, we may express the entropy change in two ways:

ds = cv

dTT

+ �i jdei j

dTdTT

= cpdTT

. (11.88)

From this we get the relation

cp − cv = ρ

TLi j�i j = −T

ρ

(∂σi j

∂T

)ei j

(∂ei j

∂T

)σi j

. (11.89)

For the linear thermoelasticity we have

�i j = −Tρ

(∂σi j

∂T

)ei j

= Tρ

α(3λ + 2µ)δi j ,

Li j = Tρ

(∂ei j

∂T

)σi j

= Tρ

αδi j , (11.90)

cp − cv = 3α2Tρ

(3λ + 2µ). (11.91)

11.5.2 Strain Cooling

For linear elastic materials, the effect of stretching under adiabatic conditions is adrop in temperature. A formula was derived by Kelvin to explain this cooling effectby using

δQ = cvdT + �i j dei j = 0 (11.92)

to have

dTdei j

= −�i j

cv

= −αT(3λ + 2µ)ρcv

δi j . (11.93)

All the parameters on the right-hand side are positive, and, because of the nega-tive sign in front, the temperature does decrease. This phenomenon is called straincooling.

11.5.3 Adiabatic and Isothermal Elastic Modulus

From our consideration of the Helmholtz free energy, the linear thermoelastic con-stitutive relation is given by

σi j = λekkδi j + 2µei j − (3λ + 2µ)α(T − T0)δi j . (11.94)

If the straining takes place rapidly, we may assume there is no heat flow into thespecimen and we have an adiabatic process. For this,

cvdT + �i j dei j = 0, �i j = T0

ρα(3λ + 2µ)δi j . (11.95)

Integrating this, we obtain

cv(T − T0) + �i j ei j = 0, (11.96)

11.5 Isotropic Thermoelasticity 157

where the constant of integration is taken as zero by use of the reference state. Fromthis we can solve for T − T0, as

T − T0 = −�i j ei j/cv = − T0

ρcv

α(3λ + 2µ)ekk. (11.97)

Substituting this in the stress–strain relation, we obtain

σi j = λ′ekkδi j + 2µei j , (11.98)

where

λ′ = λ + T0

ρcv

α2(3λ + 2µ)2. (11.99)

This shows that the adiabatic value of the Lame constant λ′ is greater than itsisothermal value λ.

11.5.4 Example: Rubber Elasticity

Natural rubber is known as polyisoprene. It belongs to a large group of chainmolecules known as elastomers. Chain molecules are made of a repeating unit calleda monomer. In addition to natural rubber, a variety of manmade elastomers areavailable from the chemical industry. The linear long-chain molecules can be cross-linked (typically, the linking takes place for every 100 or so monomer units) byadding sulfur. This is the vulcanization process invented by Goodyear. Now we havea tangled network in its natural state. As we apply tension, we expect the networkto get less tangled, as shown in Fig. 11.3(c).

Beginning with a rubber band of cross-sectional area A0 and length �0, we canexpress the total change in internal energy as

dE = TdS + F d�, (11.100)

CH3CH3CH3

CH2 CH2CH2 CH2CH2 CH2

HHH

C CC CC C

(a)

(b) (c)

Figure 11.3. (a) Three monomer units of a polyisoprene chain, (b) a tangled cross-linkednetwork before loading, (c) untangled network after loading.


where F is the force and d� is the change in length. It is assumed that the stretchingis sufficiently slow so that no irreversible entropy production occurs. The relationbetween the force F and the extension (� − �0) is generally nonlinear, as indicatedin Eq. (11.4). From the internal energy we get

∂ E∂S �

= T,∂ E∂� S

= F, (11.101)

and by considering the mixed second derivative, we get the Maxwell relation

∂T∂� S

= ∂ F∂S �

. (11.102)

The Helmholtz free energy is given by

U = E − T S, dU = −SdT + F d�. (11.103)

From this, we have

−S = ∂U∂T �

, F = ∂U∂� T

, (11.104)

and

−∂S∂� T

= ∂ F∂T �

. (11.105)

Substituting E − T S for U in the expression for F , we have

F = ∂ E∂� T

− T∂S∂� T

, (11.106)

which shows the effect of decreasing entropy on the force. In other words, we areapplying the force to keep the entropy down. The original, tangled network had ahigher entropy or more disorder than the straightened, stretched state.

We may rewrite the preceding equation in the form

∂ E∂� T

= F + T∂S∂� T

= F − T∂ F∂S �

. (11.107)

Experiments involving the quantities on the right-hand side have shown that thedependence of E on � is negligible. We define an ideal rubber with the property

∂ E∂� T

= 0. (11.108)

Then, the force is given by

F = −T∂S∂� T

, (11.109)

which shows that the entropy decreases with length at constant temperature. An-other assumption regarding ideal rubber is that the volume is preserved during elon-gation, that is,

V = V0. (11.110)

11.6 Linear Anisotropic Materials 159

If we pull the rubber band sufficiently fast, we can assume no heat input into thematerial. Then,

cvdT − F d� = 0, (11.111)

∂T∂� S

= Fcv

= − Tcv

∂S∂� T

. (11.112)

Because the partial derivative is negative, we find a temperature rise in the rubberband.

In our Mooney model, with λ = �/�0,

G11 = λ2, G22 = G33 = 1λ

, (11.113)

IG1 = λ2 + 2λ

, IG2 = 2λ + 1λ2

. (11.114)

For the ideal rubber, we assume the Mooney–Rivlin constant, A2 in Eq. (11.4), iszero, and, then,

U = ρ0V0 A1

(λ2 + 2

λ− 3

). (11.115)

The force becomes

F = ∂U∂�

= 2ρ0V0

�0A1

(λ − 1

λ2

). (11.116)

The second Piola–Kirchhoff stress S11 is given by

S11 = 2ρ0 A1

(λ − 1

λ2

). (11.117)

It has been shown by use of statistical mechanics that the constant A1 has the value

A1 = RT2zM

, (11.118)

where R is the universal gas constant, T is the absolute temperature, z is the num-ber of monomer units between cross-links, and M is the molecular weight of themonomer unit.

11.6 Linear Anisotropic Materials

When a material is anisotropic, the linear relation between the stress and strain canbe written as

σi j = Si jklekl , ei j = Ci jklσkl, (11.119)

where S is called the stiffness tensor and C is the compliance tensor. As the stressand strain tensors are symmetric, we note that

Si jkl = Sjikl = Si jlk, Ci jkl = C jikl = Ci jlk. (11.120)


The Helmholtz free energy density for this material has the form

ρu = 12

Si jklei j ekl , (11.121)

which shows a further symmetry:

Si jkl = Skli j . (11.122)

The same symmetry exists for the compliance tensor C.It is customary (and convenient) to use the Voigt notation, which associates a

linear array with a matrix, such as⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

σ11

σ22

σ33

σ23

σ13

σ12

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

σ1

σ2

σ3

σ4

σ5

σ6

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

, (11.123)

to write

σi = Si j e j , ei = Ci jσ j . (11.124)

Now, the stiffness and compliance matrices are symmetric, 6 × 6 matrices, with21 independent constants in each. Using material symmetry, such as orthotropy,transverse isotropy, and isotropy, we find that these constants reduce to 9, 5, and 2,respectively.

11.7 Invariant Integrals

In elasticity problems involving singularities, the invariant integral for small strainsand rotations (small deformation gradients) presented in Chapter 10 has found nu-merous applications. These singularities usually include crack tips with small plasti-cally deformed regions or inhomogeneities extending with a stationary field in themoving coordinate system. As mentioned earlier, when the closed surface surroundsa singularity, the integral may be nonzero, and we may write

�i =∫

S[ρεni − σkj∂i u j nk]dS. (11.125)

For isothermal cases, we use the Helmholtz free energy density u for the internalenergy ε to obtain

�i =∫

S[ρuni − σkj∂i u j nk]dS. (11.126)

Recalling that the singularity is moving through the medium with velocity compo-nents Vi , we see that material particles are crossing the surface S with velocity (−V).Then, �i Vi is the energy absorbed per unit time by the singularity, and �i have

Exercises 11.1–11.3 161

dimensions of a force. This force is known as a material force or Eshelby force.In the plane strain case, for a crack tip moving in the x1 direction, we have

J =∫

S[ρun1 − σkj∂1u j nk]dS, (11.127)

which is the well-known J integral introduced by Rice.Instead of singular points, we may also consider lines of singularities or surfaces

of singularities in the context of invariant integrals.

SUGGESTED READING

Antman, S. (2004). Nonlinear Problems in Elasticity, 2nd ed., Springer.Barber, J. R. (2002). Elasticity, 2nd ed., Kluwer Academic.Green, A. E. and Zerna, W. (1968). Theoretical Elasticity, 2nd ed., Clarendon.Landau, L. D. and Lifshitz, E. M. (1986). Theory of Elasticity, 3rd ed., Pergamon.Ogden, R. W. (1984). Nonlinear Elastic Deformation, Ellis Horwood.Timoshenko, S. P. and Goodier, J. N. (1970). Theory of Elasticity, 3rd ed., McGraw-Hill.

EXERCISES

11.1. A circular cylindrical shaft of radius a and length � is made of a Mooney ma-terial. One end of the shaft is rotated by an angle β with respect to the other end.Compute the torque T and the axial force N required for twisting the shaft underthe assumptions of rigid body rotation of each cross-section and no change in axiallength.

11.2. Obtain the balance of linear momentum for static elasticity for plane stressand plane strain:

∂1σ11 + ∂2σ12 + ρ f1 = 0,

∂1σ12 + ∂2σ22 + ρ f2 = 0.

Assume that the body forces are conservative and that they are obtained from apotential ϕ in the form

ρ f1 = −∂1ϕ, ρ f2 = −∂2ϕ.

Show that the Airy stress function φ, introduced in the form

σ11 = ∂22 φ + ϕ, σ22 = ∂2

1 φ + ϕ, σ12 = −∂1∂2φ,

satisfies the balance of linear momentum.Using these in the stress–strain relations for a Hookean material, obtain the gov-erning equations for φ for plane stress and plane strain. When fi = 0, show that, forboth plane stress and plane strain, φ satisfies the biharmonic equation

∇4φ = 0.

11.3. Transform the biharmonic equation in the preceding problem into the polarcoordinates r and θ . Obtain the functions Fn(r) in the separable solutions

φ = Fn(r)einθ ,

where i is the imaginary number and n is an integer.


11.4. By considering the dimensional changes of a parallelepiped, show that for in-finitesimal strains the volume change per unit volume is given by

�VV

= e = eii .

Using Hooke’s law, relate this to the hydrostatic pressure

p = −13σi i

in the form

e = − pK

,

where K is the bulk modulus. Express K in terms of E and ν and, also, in terms ofthe Lame constants λ and µ.

11.5. Consider linear, plane elasticity in the x1, x2 plane. Show that the J integraldefined by

J =∮

S[ρun1 − σkj∂1u j nk]dS,

where S is now a curve, is invariant with respect to any two choices, S = C1 and S =C2, as long as the stresses are nonsingular in the region between the two contours.

11.6. In the Navier equations, neglecting the body forces, assume that

u = sin k(x ± cLt), v = 0, w = 0,

where k is the wave number and c is the wave speed. Obtain an expression for cL interms of the Lame constants and the density. Waves with displacements oriented inthe direction of propagation are called longitudinal or P waves.

11.7. In the preceding exercise, if we assume that

u = 0, v = sin k(x ± cT t), w = 0,

we have shear (transverse) or S waves. Obtain the wave speed in this case.

11.8. In the Navier equations (without body forces), assume that

u = ∇φ + ∇ × ψ,

where φ and ψ are known as the scalar and vector potentials. Show that both ofthese functions satisfy the wave equation, with the wave speed cL for φ and cT forψ . Also show that the φ wave represents volume change and the ψ wave representsrotations.

11.9. In linear thermoelasticity, assuming that λ, µ, α, and the specific heats cv andcp are independent of temperature T , obtain the densities of Helmholtz free energyu, internal energy ε, entropy s, enthalpy h, and the Gibbs free energy, g.

11.10. Compute the ratio E′/E, where E′ and E are, respectively, the adiabatic andthe isothermal Young’s moduli.

Exercises 11.4–11.11 163

11.11. Assume a rubber band is under constant tension. Show that its length de-creases according to

∂�

∂T F= 1

k∂S∂� T

,

where

k = ∂ F∂� T

when it is uniformly heated.

12 Fluid Dynamics

Fluid dynamics is another active area of research because of its impact on aerody-namics, flows in turbomachinery, design of propellers, and applications in convectiveheat transfer. Our aim here is to discuss the basic concepts of fluid dynamics in thegeneral context of continuum mechanics. In Chapter 10 we briefly discussed Stokesfluids. Here we continue with some solutions by using the inverse method. The New-tonian fluid is considered next. We conclude with ideal fluids and thermodynamicimplications.

12.1 Basic Equations

The equations of conservation of mass and the balance of momenta and the consti-tutive relations are

∂ρ

∂t+ ∂i (ρvi ) = 0,

∂iσi j + ρ f j = ρv j ,

σi j = σ j i ,

σi j = (−p + a0)δi j + a1di j + a2dikdkj , compressible,

σi j = − pδi j + a1di j + a2dikdkj , incompressible, (12.1)

where the phenomenological coefficients ai are functions of the invariants of thedeformation rate tensor, i.e.,

ai = ai (I1, I2, I3), a0(0, 0, 0) = 0, (12.2)

p is the thermodynamic pressure, and p is the reactive hydrostatic pressure. All theinvariants pertain to the deformation rate tensor d. For the incompressible case wehave I1 = 0.

For nonnegative entropy production we must have

a0 I1 + a1(I 21 − 2I2) + a2(I 3

1 − 3I1 I2 + 3I3) > 0, (12.3)

164

12.3 Newtonian Fluids 165

which reduces to

−2a1 I2 + 3a2 I3 > 0 (12.4)

for the incompressible case.We also have the defining kinematic relations

vi = ∂vi

∂t+ v j∂ jvi . (12.5)

The thermodynamic pressure p is known from an equation of state,

p = p(ρ, T). (12.6)

These equations are supplemented with a system of boundary and initial conditions.At every point on the surface, three quantities have to be prescribed. These couldbe three components of the traction vector, three components of the velocity vector,or a mixture involving traction components and their complementary velocity com-ponents. At solid boundaries the velocity of the fluid should be equal to the velocityof the solid. This is known as the no-slip condition.

12.2 Approximate Constitutive Relations

The constitutive relations of Stokes fluids require specification of three functionsai , known as the phenomenological coefficients, which are functions of the threeinvariants of the deformation rate tensor d, as mentioned earlier. We may assumeI1 = O(ε), I2 = O(ε2), I3 = O(ε3), and expand the phenomenological coefficients.To include quadratic terms in ε in the stress expressions, let

a0 = λI1 + λ1 I 21 + λ2 I2,

a1 = 2µ + 2µ1 I1,

a2 = 4ν, (12.7)

where we have used the same constants as in Eringen. Of course, the possibility thatthere are real fluids with the preceding phenomenological constants is not precludedeven if I1 is not a small quantity. These fluids are called second-order fluids.

The stresses are obtained for compressible fluids as

σ = [(−p + λ)I1 + λ1 I 21 + λ2 I2]I + 2(µ + µ1 I1)d + 4νd 2, (12.8)

where the coefficient of I has to be replaced with − p for the incompressible case.Interestingly, the entropy production inequality shows that, for the incompressiblecase, ν = 0. Thus there is no second-order incompressible fluid.

12.3 Newtonian Fluids

We consider a class of fluids for which the stresses are linear functions of the defor-mation rate. These are called Newtonian fluids. From the second-order constitutive

166 Fluid Dynamics

relations we remove all second-order terms to have

a0 = λI1, a1 = 2µ, a2 = 0, (12.9)

and

σi j = (λI1 − p)δi j + 2µdi j , compressible,

σi j = − pδi j + 2µdi j , incompressible. (12.10)

The constants λ and µ are called the dilatational and shear viscosities. The entropyproduction rate is proportional to

σi j di j = λI 21 + 2µdi j di j = λI 2

1 + 2µ(I 21 − 2I2). (12.11)

To assess the constraints on the constants, let us express the invariants in terms ofthe eigenvalues of d, namely, the stretchings di :

I1 = d1 + d2 + d3, I2 = d1d2 + d2d3 + d3d1. (12.12)

Now,

σi j di j = (λ + 23µ)(d1 + d2 + d3)2 + 2µ[(d1 − d2)2 + (d2 − d3)2 + (d3 − d1)2].

(12.13)For nonnegative dissipation we must have

λ + 23µ ≥ 0, µ ≥ 0. (12.14)

The hydrostatic pressure for compressible Newtonian fluids can be found as

−13σi i = p − (λ + 2

3µ)I1, (12.15)

and for incompressible fluids,

−13σi i = p. (12.16)

If

λ + 23µ = 0, (12.17)

we see that the hydrostatic pressure and the thermodynamic pressure are equal forthe compressible Newtonian fluids. This condition is known as the Stokes condition.This approximation is often used for water and air.

By defining a deviatoric deformation rate,

d′i j = di j − 1

3dkkδi j , (12.18)

we can have

σi j = −pδi j + 2µd′i j (12.19)

for Newtonian fluids under the Stokes assumption.

12.5 Shearing Flow 167

12.4 Inviscid Fluids

From the first-order theory, previously described, by eliminating the linear terms indi j , we obtain a zeroth-order theory, where

σi j = −pδi j . (12.20)

These are called inviscid fluids. The thermodynamic pressure p is dependent on thedensity ρ and temperature T :

p = p(ρ, T). (12.21)

If p is independent of temperature, the fluid is called a barotropic fluid.For incompressible zeroth-order fluids,

σi j = − pδi j , (12.22)

and such a fluid is known as an ideal fluid. The pressure here is a reactive quantity.As mentioned earlier, let us consider a few classic flows by using the inverse

method. In the inverse method, we prescribe a velocity distribution and see if thebalance of momentum can be satisfied.

12.5 Shearing Flow

We assume the velocities are of the form

v1 = u(y), v2 = v3 = 0, (12.23)

where we use y = x2 and x = x1. For the time being, u is arbitrary. As shown in Fig12.1, the flow is confined between a solid static wall at y = 0 and a free surface or arigid plate at y = h. The deformation rate tensor is given by

d =

⎡⎢⎣ 0 u′/2 0

u′/2 0 00 0 0

⎤⎥⎦ , (12.24)

which has the invariants

I1 = 0, I2 = −(u′/2)2, I3 = 0, (12.25)

where u′ = du/dy. The stress is obtained as

σ = (a0 − p)

⎡⎢⎣ 1 0 0

0 1 00 0 1

⎤⎥⎦ + a1u′

2

⎡⎢⎣ 0 1 0

1 0 00 0 0

⎤⎥⎦ + a2(u′)2

4

⎡⎢⎣ 1 0 0

0 1 00 0 0

⎤⎥⎦ , (12.26)

where

ai = ai [0,−(u′/2)2, 0] = ai (y). (12.27)

168 Fluid Dynamics

x1

x2

u

Figure 12.1. Shearing flow over a fixed surface.

The nonzero stresses are

σ11 = σ22 = (a0 − p) + 14

a2(u′)2,

σ33 = a0 − p,

σ12 = 12

a1u′. (12.28)

Let us neglect the body force. For a steady flow, the acceleration

vi = ∂vi

∂t+ v j∂ jvi = 0. (12.29)

The equations of linear momentum give

−∂1 p + [12

a1u′]′ = 0,

[−p + a0 + 14

a2(u′)2]′ = 0, (12.30)

where a prime indicates differentiation with respect to x2 = y. Integrating the sec-ond equation gives

−p + a0 + 14

a2(u′)2 = g(x), (12.31)

where g(x) has to be found. Substituting this into the first of Eqs. (12.30), we obtain

dgdx

+ (12

a1u′)′ = 0. (12.32)

12.6 Pipe Flow 169

This equation is satisfied (by separation of variables) if

dgdx

= −C,

(12

a1u′)′

= C, (12.33)

where C is a constant. Integrating these, we obtain

g(x) = −Cx + D, a1u′ = 2Cy + E, (12.34)

where D and E are new constants. The stresses can be expressed as

σ11 = σ22 = −Cx + D, σ12 = Cy + E/2. (12.35)

We may consider two situations: (a) There is a free surface at y = h, and (b) thereis a rigid plate at y = h moving with a velocity U.

Case (a): The constants C = 0, D = 0 for σ22 = 0 for all x and E = 0 for σ12 = 0at y = h. The velocity distribution is obtained from the nonlinear differential equa-tion

a1u′ = 0, u(0) = 0, (12.36)

which allows u(y) = 0 as a trivial solution. Recall that the coefficient a1 is a functionof the unknown u′.

Case (b): The differential equation

a1u′ = 2Cy + E (12.37)

may have solutions.For a Newtonian fluid, a1 = 2µ, and

u′ = 12µ

(2Cy + E), u = 12µ

(Cy2 + Ey + G), (12.38)

where the constant G = 0 to have u(0) = 0. If we stipulate σ11 is independent of x,we get C = 0. This gives the simple linear distribution for the velocity:

u(y) = Uy/h . (12.39)

In addition, if there is a pressure drop in the x direction,

∂1σ11 = −C, u(y) = Uyh

+ C2µ

y(y − h). (12.40)

12.6 Pipe Flow

The axisymmetric flow in the direction of the axis of a uniform circular pipe is calleda Hagen–Poiseuille flow. This case is better suited for cylindrical coordinates, r , θ ,and z. Because of the axisymmetry, all quantities are independent of θ . We assumea velocity distribution of the form

vr = 0, vθ = 0, vz = w(r). (12.41)

170 Fluid Dynamics

The boundary condition on the pipe wall is

w(a) = 0, (12.42)

where a is the radius of the pipe.Using the gradient operator

∇ = er∂

∂r+ eθ

∂

r∂θ+ ez

∂

∂z, (12.43)

and

d = 12

[∇v + (∇v)T ], (12.44)

we find

d =

⎡⎢⎣ drr drθ drz

drθ dθθ dθz

drz dθz dzz

⎤⎥⎦ = 1

2w′

⎡⎢⎣ 0 0 1

0 0 01 0 0

⎤⎥⎦ . (12.45)

The invariants are

I1 = 0, I2 = −14

(w′)2, I3 = 0, (12.46)

and the functional dependence of the phenomenological coefficients can be seen as

ai = ai [0,−(w′)2/4, 0] = ai (r). (12.47)

The nonzero stresses are given by

σrr = σzz = −p + a0 + 14

a2(w′)2, σθθ = −p + a0, σrz = 12

a1w′. (12.48)

The equations of motion are

∂(rσrr )∂r

− σθθ = 0,

∂(rσrz)∂r

− r∂σzz

∂z= 0. (12.49)

Substituting the stress expressions, we find that the equations of motion give

−∂p∂r

+ a′0 + 1

4r[a2r(w′)2]′ = 0,

−∂p∂z

+ 12r

[a1rw′]′ = 0. (12.50)

Integrating the first equation, we have

−p + a0 +∫ r

0

14r

[a2r(w′)2]′ dr = f (z), (12.51)

where f is an unknown function. Substituting this in the second equation, we obtain

dfdz

+ 12r

[a1rw′]′ = 0. (12.52)

12.6 Pipe Flow 171

This equation is satisfied if

dfdz

= C,12r

[a1rw′]′ = −C, (12.53)

where the constant C has to be found. From these,

f = Cz + D, a1rw′ = −Cr2 + E, a1w′ = −Cr + E

r. (12.54)

For w′ to be finite at r = 0, we must have E = 0. Then the differential equation forw is given by

a1w′ = −Cr, (12.55)

which has to be integrated with the condition w(a) = 0. The stresses are obtained as

σrr = σzz = Cz + D −∫ r

0

14r

a2(w′)2dr,

σθθ = Cz + D − a2(w′)2

4−

∫ r

0

14r

a2(w′)2dr,

σrz = −12

Cr. (12.56)

For the Newtonian fluid,

w′ = − C2µ

r, w = C4µ

(a2 − r2), (12.57)

which is the well-known parabolic distribution of the velocity. The stresses are

σrr = σzz = σθθ = Cz + D,

σrz = −12

Cr. (12.58)

For a second-order fluid, a1 = 2µ + 2µ1 I1 = 2µ and a2 = 4ν. Then

w(r) = C4µ

(a2 − r2), (12.59)

and

σrr = σzz = Cz + D − C2νr2

8µ2,

σθθ = Cz + D − 3C2νr2

8µ2,

σrz = −12

Cr. (12.60)

The hydrostatic pressure can be found to be

p = −Cz − D + 5C2νr2

24µ2. (12.61)

172 Fluid Dynamics

The pressure difference between a point on the wall and another on the axis is seenas

p|a − p|0 = 5C2νa2

24µ2, (12.62)

which isolates the effect of the second-order constant ν. At the exit of a pipe, thispressure difference would cause the flow to “swell” as it encounters atmosphericpressure on the outer surface. This an indicator of a higher-order fluid. The constantC can be found from the pressure drop or from the volume flow rate,

Q =∫ a

02πrvzdr = πCa4

8µ. (12.63)

12.7 Rotating Flow

The flow in a circular cylindrical container with the velocity field

vr = 0, vθ = rω(r), vz = 0 (12.64)

is known as the Couette flow. Again, in polar coordinates, the deformation rate canbe written as

d =

⎡⎢⎣ 0 β 0

β 0 00 0 0

⎤⎥⎦ , (12.65)

where we use

β = rω′

2. (12.66)

The invariants are found as

I1 = 0, I2 = −β2, I3 = 0. (12.67)

The phenomenological coefficients are

ai = ai (0,−β2, 0) = ai (r). (12.68)

The stress tensor in polar coordinates is obtained as

σ = (a0 − p)

⎡⎢⎣ 1 0 0

0 1 00 0 1

⎤⎥⎦ + a1β

⎡⎢⎣ 0 1 0

1 0 00 0 0

⎤⎥⎦ + a2β

2

⎡⎢⎣ 1 0 0

0 1 00 0 0

⎤⎥⎦ . (12.69)

Explicitly, the nonzero stresses are

σrr = σθθ = a0 − p + a2β2,

σzz = a0 − p,

σrθ = a1β. (12.70)

12.7 Rotating Flow 173

Because of the centrifugal acceleration, we have the equations of motion,

∂rσrr = −ρrω2, (12.71a)

∂r (r2σrθ ) = 0, (12.71b)

∂zσzz = −ρg, (12.71c)

where we have assumed axisymmetry and a body force (weight of the fluid) in the zdirection which is assumed to be downward.

Integrating Eq. (12.71b), we get

σrθ = Cr2

, a1β = Cr2

, a1ω′ = 2C

r3, (12.72)

where C is a constant. The second relation verifies that C is indeed a constant andnot a function of z. The third relation is the nonlinear differential equation for ω(r).

The remaining two equations of motion give

−∂r p + (a0 + a2β2)′ = −ρrω2, (12.73a)

−∂z p = −ρg. (12.73b)

Integrating Eq (12.73b), we obtain,

p = ρgz + F (r), (12.74)

where we must find F (r) by substituting in Eq. (12.73a). This gives

F ′ = ρrω2 + (a0 + a2β2)′. (12.75)

Integrating this relation gives

F (r) =∫

ρrω2dr + a0 + a2β2 + D, (12.76)

p = ρgz +∫

ρrω2dr + a0 + a2β2 + D. (12.77)

The stresses are found as

σrr = σθθ = −ρgz −∫

ρrω2dr − D,

σzz = −ρgz −∫

ρrω2dr − a2β2 − D,

σrθ = Cr2

. (12.78)

The singularity in the shear stress at r = 0 can be avoided if C = 0 or if the flow isconfined between two concentric cylinders. Let us assume the inner cylinder has aradius of a and the outer cylinder has a radius of b. Further, we may take the outercylinder as fixed and the inner cylinder as rotating with an angular velocity of �.

174 Fluid Dynamics

The torque required for the rotation is given by

T =∫ 2π

0σrθr2dθ |r=a = 2πC. (12.79)

If there is a free surface, the normal stress on it must be zero. That is, σzz = 0. Thisgives

z = − 1ρg

[ρ

∫rω2dr + a2β

2 + D]

. (12.80)

Let us illustrate this with a second-order fluid. Because the invariant I1 = 0, we have

a1 = 2µ, a2 = 4ν. (12.81)

Then,

ω′ = Cµr3

, ω = C2µ

[1b2

− 1r2

], (12.82)

where we used ω(b) = 0. The condition ω(a) = � gives

C = 2µ�/

[1b2

− 1a2

]= T

2π. (12.83)

The integral ∫ r

bρrω2dr = ρC2

2b2

[r2

b2− b2

r2− 4 ln

rb

]. (12.84)

The stresses are obtained as

σrr = σθθ = −ρgz − D − ρC2

2b2

[r2

b2− b2

r2− 4 ln

rb

],

σzz = −ρgz − D − ρC2

2b2

[r2

b2− b2

r2− 4 ln

rb

]− C2ν

µ2r4,

σrθ = Cr2

. (12.85)

To maintain the velocity distribution assumed for the Couette flow, the normalstress, σzz, has to be applied. As we have seen, if there is a free surface, the equationfor the surface is given by

z = − 1g

{Dρ

+ C2

2b2

[r2

b2− b2

r2− 4 ln

rb

]+ C2ν

ρµ2r4

}, (12.86)

where the constant D can be selected to have z = 0 at any chosen radius. Remem-bering that the z axis points downward, the last term involving ν contributes to arise in the fluid level. This rise is higher at the inner wall than at the outer wall. Thiseffect is known as the Weissenberg effect. Usually this effect is demonstrated by in-serting a rotating rod into a second-order fluid in a cylindrical vessel to see the fluidrise around the rod.

12.9 Incompressible Flow 175

12.8 Navier–Stokes Equations

For Newtonian fluids we have

σi j = (−p + λI1)δi j + 2µdi j , compressible, (12.87)

σi j = − pδi j + 2µdi j , incompressible, (12.88)

where

I1 = ∂kvk, di j = 12

(∂iv j + ∂ jvi ). (12.89)

Substituting these in the equations of motion, we have

ρ(∂vi

∂t+ v j∂ jvi ) = ρ fi − ∂i p + ∂i (λ∂ jv j ) + ∂ j [µ(∂iv j + ∂ jvi )], (12.90)

ρ(∂vi

∂t+ v j∂ jvi ) = ρ fi − ∂i p + ∂ j [µ(∂iv j + ∂ jvi )], (12.91)

for compressible and incompressible fluids, respectively. These are the well-knownNavier–Stokes equations.

Next, let us consider the incompressible and compressible cases separately.

12.9 Incompressible Flow

The density is constant for incompressible flows, and the viscosity may dependon temperature. Assuming isothermal conditions, the Navier–Stokes equationsbecome

∂vi

∂t+ v j∂ jvi = fi − 1

ρ∂i p + ν∂ j∂ jvi , (12.92)

where we used ∂ jv j = 0, and ν is the kinematic viscosity defined as

ν = µ

ρ. (12.93)

Further, for steady flows, we get

v j∂ jvi = fi − 1ρ

∂i p + ν∂ j∂ jvi . (12.94)

We may scale this equation by using a velocity U and a length scale L, representingthe distance along which significant changes in quantities take place. Let us use thefollowing scaled replacements:

vi → vi/U, p → p/ρU2, fi → fi/U2, xi → xi/L (12.95)

The equations of motion can be written as

∂vi

∂t+ v j∂ jvi = fi − ∂i p + 1

R∂ j∂ jvi , (12.96)

where the nondimensional constant R is given by

R = ρUL/µ. (12.97)

176 Fluid Dynamics

This number is called the Reynolds number, which is a measure of the ratio of theinertia forces to viscous forces. The Reynolds number plays an important role in thetransition of laminar flows into turbulent flows.

12.10 Compressible Flow

Equation of motion (12.90) has to be supplemented with the equation of conserva-tion of mass,

∂ρ

∂t+ ∂i (ρvi ) = 0. (12.98)

The thermodynamic pressure is given by the equation of state,

p = p(ρ, T). (12.99)

With pressure and temperature variations, we need the energy equation

ρε = σi j di j + ∂i qi + ρh (12.100)

to supplement the balance and the conservation relations.Assuming an ideal gas with Fourier’s law of heat conduction, we find

ε = cv T, p = RρT, qi = k∂i T. (12.101)

Using these, we obtain

ρcv T − ∂i (k∂i T) − ρh = (−RρT + λdkk)dii + 2µdi j di j . (12.102)

The three equations of motions for vi have to be solved with the conservation ofmass for ρ and the energy equation for T , simultaneously.

12.11 Inviscid Flow

When the dilatational viscosity λ and the shear viscosity µ are negligible we call thefluid an inviscid fluid. The equations previously discussed for compressible fluidsbecome

∂vi

∂t+ v j∂ jvi = fi − 1

ρ∂i p,

∂ρ

∂t+ ∂i (ρvi ) = 0, (12.103)

ρcv T + RρT∂ivi − ∂i (k∂i T) − ρh = 0,

p = p(ρ, T). (12.104)

12.11 Inviscid Flow 177

The system of corresponding equations for incompressible fluids is

∂vi

∂t+ v j∂ jvi = fi − 1

ρ∂i p,

∂ivi = 0,

ρcT − ∂i (k∂i T) − ρh = 0, (12.105)

where c is the specific heat. In the incompressible case, the specific heats at constantvolume and at constant pressure are the same. The heat conduction equation can besolved separately in this case.

12.11.1 Speed of Sound

Equations (12.103) for inviscid, compressible fluids can be applied to small distur-bances in a still fluid of density ρ0 kept at a constant pressure p0 and temperatureT0. The small deviations from the quiescent state can be written as

vi = 0 + v′i , p = p0 + p′, ρ = ρ0 + ρ ′. (12.106)

Neglecting the body forces and the quadratic and higher-order terms, Eqs. (12.103)give

ρ0∂v′

i

∂t= −∂i p′ = −∂p

∂ρ ρ0

∂ρ ′

∂xi, (12.107)

∂ρ ′

∂t= −ρ0

∂v′i

∂xi. (12.108)

Eliminating the velocity components, we obtain the wave equation for ρ ′,

∂2ρ ′

∂t2= ∂p

∂ρ ρ0

∂2ρ ′

∂xi∂xi. (12.109)

A wave of density fluctuation moving in the direction n can be represented as

ρ ′ = Aeik(ni xi −ct), (12.110)

where k and c represent the wave number and the wave speed. From the wave equa-tion, we find

c2 = ∂p∂ρ ρ0

. (12.111)

Noting

p′ = c2ρ ′, (12.112)

we see that the velocity components v′i and the pressure p′ also satisfy the preceding

wave equation.

178 Fluid Dynamics

12.11.2 Method of Characteristics

In Eqs. (12.103) for inviscid compressible fluids, if the thermal gradients are absent,the flow is known as an isentropic flow. We briefly examine the one-dimensionalisentropic flow. The thermodynamic pressure can be assumed to be a function ofthe density,

p = p(ρ), (12.113)

and the only component of velocity is u in the x direction. Thus,

ρ∂u∂t

+ ρu∂u∂x

+ c2 ∂ρ

∂x= 0, (12.114)

∂ρ

∂t+ u

∂ρ

∂x+ ρ

∂u∂x

= 0, (12.115)

where

c2 = ∂p∂ρ ρ

. (12.116)

These are two nonlinear partial differential equations that come under the classifi-cation of quasi-linear equations because of the presence of the highest derivatives inthe linear form. One way to solve such systems is through the use of the method ofcharacteristics, provided we have a hyperbolic system of equations. Because the lin-ear wave equation with constant coefficients allows solutions in terms of character-istic variables x − ct and x + ct , we begin by assuming that there are characteristiccurves s(x, t), such that

ds = dx − cdt,∂

∂x= ∂

∂s,

∂

∂t= −c

∂

∂s. (12.117)

Along these curves we can replace ∂u/∂t with −c∂u/∂x. The isentropic flow equa-tions give

ρ(−c + u)ux + c2ρx = 0, (12.118)

ρux + (−c + u)ρx = 0, (12.119)

where ux = ∂u/∂x, etc. Setting the determinant of this homogeneous system to zero,we find

c = u + c, c = u − c, (12.120)

which show real slopes for the two sets of characteristic curves—a requirement forhyperbolic systems. Further, the eigenvectors of the system show that

ρux + cρx = 0, alongdxdt

= u − c, (12.121)

ρux − cρx = 0, alongdxdt

= u + c. (12.122)

12.12 Bernoulli Equation 179

These relations can be integrated to obtain the so-called Riemann invariants,

u +∫ ρ

ρ0

cdρ

ρ= C1, (12.123)

u −∫ ρ

ρ0

cdρ

ρ= C2, (12.124)

which are constants along the characteristic curves.

12.12 Bernoulli Equation

When the flow is steady, we may obtain a conservation principle known as theBernoulli equation under certain conditions. The convective acceleration term canbe written as

v j∂ jvi = v j (∂ jvi − ∂iv j ) + v j∂iv j = 2v jw j i + 12∂i (v2). (12.125)

We assume there is a potential ϕ for the (conservative) body force f such that

fi = − ∂ϕ

∂xi. (12.126)

We also assume the fluid is either incompressible or barotropic, that is, p = p(ρ).Then we have a function P such that

1ρ

∂i (p, p) = ∂ P∂xi

. (12.127)

With these, the equations of motion for an inviscid fluid can be expressed as

∇(ϕ + P + 12v2) = −2v · w. (12.128)

Now, if the flow is irrotational, w is zero, and

ϕ + P + 12v2 = constant. (12.129)

Even for a rotational flow, if we dot multiply Eq. (12.128) by v and use

v · (v · w) = 0, (12.130)

we get

vi∂i (ϕ + P + 12v2) = 0. (12.131)

Because the stream lines are parallel to v, this equation shows that the directionalderivative of the total potential along a stream line is zero. That is,

ϕ + P + 12v2 = constant along a stream line. (12.132)

180 Fluid Dynamics

12.13 Invariant Integrals

Following our discussion of the invariant surface integrals in Chapter 10, the mo-mentum flux exerts a force

F j =∫

S[ρviv j − σi j ]ni dS (12.133)

on a singularity in a fluid flow. As an example, consider a rotating cylinder (or avortex line) perpendicular to the plane of a two-dimensional (2D) flow with far-field velocity U in the x direction. Assuming the center of the cylinder is located atthe origin x = 0, y = 0, the velocity distribution is

vx = U − �y2πr2

, vy = �x2πr2

, (12.134)

where � is the circulation due to the rotating cylinder. For an incompressible fluid,

σi j = − pδi j = −[p0 − 12v2]δi j , (12.135)

where p0 is the stagnation pressure. When this is inserted into the integral takenfrom 0 to 2πr , the p0 term vanishes and

Fx = 12ρ

∫ 2πr

0[2(vxnx + vyny)vx − v2nx]ds, (12.136)

Fy = 12ρ

∫ 2πr

0[2(vxnx + vyny)vy − v2ny]ds. (12.137)

For the circle, nx = x/r and ny = y/r , and

vxnx + vyny = Unx. (12.138)

After integration, we get the famous result known as the Kutta–Joukowski theoremor the Magnus effect in the case of a rotating cylinder:

F1 = 0, F2 = ρ�U. (12.139)

The invariant integral involving the internal energy has applications in com-pressible fluids.

SUGGESTED READING

Batchelor, G. K. (1967). An Introduction to Fluid Dynamics, Cambridge University Press.Fredrickson, A. G. (1964). Principles and Applications of Rheology, Prentice-Hall.Gidaspow, D. (1994). Multiphase Flow and Fluidization, Academic.Levich, V. G. (1962). Physicochemical Hydrodynamics, Prentice-Hall.Panton, R. L. (2005). Incompressible Flow, Cambridge University Press.Schlischting, H. (1955). Boundary Layer Theory, McGraw-Hill.

Exercises 12.1–12.6 181

EXERCISES

12.1. A fluid has the phenomenological coefficients

a0 = 0, a1 = −2µI2, a2 = 0.

It is contained between two parallel plates separated by a distance h. Calculate thevelocity distribution and the flow rate if the rate of pressure drop is a constant.

12.2. Assuming the preceding fluid is flowing through a pipe of radius a, obtain thevelocity profile and the flow rate for a given rate of pressure drop.

12.3. For the rotating flow between two cylinders, assume the inner radius a = b/4,where b is the outer radius. Further, choose the arbitrary constant D such that thestress-free surface has zero height at r/b = 1. Introducing a nondimensional pa-rameter for z, obtain a plot of z versus r/b. Use an appropriate nondimensionalgroup to represent the coefficient ν. Use a coordinate system with the z axis pointingdownward.

12.4. An incompressible fluid obeys the constitutive law

d12 = 12µ

{σ12 if |σ12| > τ

0 if |σ12| ≤ τ,

where τ is a constant. Assuming it is flowing between two fixed parallel plates atx2 = ±a and the only nonzero velocity is u = v1 caused by a pressure gradient in thex1 direction, obtain the velocity profile, the domain in which d = 0, and the flow ratein terms of the pressure gradient. Such flows are known as “plug flows” in the liter-ature. This type of constitutive relation describes a special case of Bingham fluids,which are distinguished by a critical shear stress below which the rate of deformationis zero.

12.5. In a 2D flow in a plane normal to the x3 axis, if a Newtonian fluid is incom-pressible and irrotational, we have

∂1v1 + ∂2v2 = 0, ∂2v1 − ∂1v2 = 0.

If the velocity potential φ and the stream function ψ are defined as

v1 = ∂1φ = ∂2ψ, v2 = ∂2φ = −∂1ψ,

obtain the governing equations for φ and ψ .Show that the contour lines on which φ is a constant are perpendicular to the contourlines on which ψ is a constant. Also show that the velocity vector is tangential to theψ contours.

12.6. The equation of state for a van der Waals gas is

p = RTv − b

− av2

, v = 1ρ

.

Show that the energy equation can be written for this case as

ρcv T + ρRT1 − bρ

dii − ∂i (k∂i T) − ρh = λ(dii )2 + 2µdi j di j .

182 Fluid Dynamics

12.7. An incompressible, Newtonian fluid occupies the domain −∞ < x < ∞, 0 <

y < ∞. At time t = 0, the boundary y = 0 is given a velocity U. Assuming that itis a plane flow and there are no body forces or pressure drop, obtain the velocitydistribution as a function of time and location.

12.8. In the preceding exercise, instead of a suddenly applied velocity for the bound-ary y = 0, assume the boundary is oscillating with the time dependence, u(x, 0) =U cos ωt . Neglecting the transient effects, obtain the velocity distribution as a func-tion of t and y.

12.9. As a second modification to Exercise 12.7, assume there is a second, nonmov-ing boundary at y = h and the boundary y = 0 is oscillating as before. Again, obtainvelocity distribution.

12.10. A rod of radius a occupies the region −∞ < x < ∞, 0 < r < a. It is sur-rounded by an incompressible, Newtonian fluid. At t = 0, the rod is given a velocityU in the axial direction. To find the velocity of the fluid follow these instructions:

(a) Show that in cylindrical coordinates the axial velocity u satisfies

∂u∂t

= ν

[∂2u∂r2

+ 1r

∂u∂r

].

(b) After taking the Laplace transform

u(r, p) =∫ ∞

0u(r, t)e−pt dt,

obtain the solution u in terms of the modified Bessel function K0 as

u(r, p) = AK0(√

p/ν r),

and the other solution, I0, is not useful as it is unbounded as r → ∞.(c) Choose the constant A to satisfy the boundary condition at r = a.(d) In the Laplace inversion integral,

u(r, t) = 12π i

∫�

u(r, p)ept dp,

where � is a vertical line in the complex p plane, just right of the origin,deform the contour to go from −∞ to 0 just below the branch cut alongthe negative real axis and from 0 to −∞ just above the branch cut. Notethat the branch cut is necessary as K0 is not single valued without it.

(e) Using the substitutions p = νλ2eiπ for the integration above the branch cutand p = νλ2e−iπ for the integration below the branch cut, obtain

u(r, t) = 2Uπ

∫ ∞

0

J0(λr)Y0(λa) − Y0(λr)J0(λa)

J 20 (λa) + Y2

0 (λa)

e−νλ2t

λdλ.

(f) Obtain the viscous force exerted by the fluid on a unit length of the rod.

12.11. In the case of one-dimensional isentropic flow, if

p = Aργ ,

where A is a constant, obtain the characteristic slopes and the Riemann invariants.

Exercises 12.7–12.13 183

12.12. Equations (12.114) and (12.115) can be linearized by the so-called hodographtransformation in which we use u and ρ as independent variables and x and t asdependent variables. Obtain the linearized equations in this way.

12.13. In a plane flow with far-field velocity vx = U, there is a line source of strengthq at the origin (the line being normal to the plane of the flow). The velocity distri-bution that is due to the source is

vx = qx2πr2

, vy = qy2πr2

.

Find the force on the source using the invariant momentum flux.

13 Viscoelasticity

Materials in which the stress at a point depends on the entire history of the strainare called viscoelastic materials. Hyperelastic materials and Stokes fluids can beconsidered as special cases of viscoelastic materials. In general, viscoelastic mate-rials exhibit a combination of solidlike and fluidlike characteristics. Examples ofviscoelastic materials can be found in plastics, polymers, and metals at high tem-peratures. The nonlinear viscoelastic behavior of metals is commonly referred to ascreep. This chapter mainly deals with linear viscoelastic materials. Before we con-sider general constitutive relations, it is of interest from a historical perspective toexamine some one-dimensional classic models. As in the case of fluids, the behav-ior of viscoelastic materials under hydrostatic pressure differs from their behaviorunder shear stresses. This experimental evidence has to be incorporated when wegeneralize the one-dimensional behavior to three dimensions. We also refer to vis-coelastic solids and viscoelastic fluids on the basis of whether the displacement un-der a constantly applied load reaches a finite limit.

13.1 Kelvin–Voigt Solid

The one-dimensional force–displacement relation for the Kelvin–Voigt solid iswritten as

P = ku + µu, (13.1)

where k and µ are material constants, P is the uniaxial force, and u and u are theinfinitesimal displacement and displacement rate, respectively.

A mechanical model representing this constitutive relation has a spring withstiffness k and a dashpot with viscosity µ. This arrangement is shown in Fig. 13.1.Because of the linear nature of this force–displacement relation, we can obtain theresponse of the material to a given history of the load by superposing the response ofthe material to a load in the form of a unit step function. This superposition principleis applicable to all linear viscoelastic materials. We will demonstrate this using theLaplace transform later in this chapter. The response u of a viscoelastic material toa unit load input is called the creep compliance function J (t). For the Kelvin–Voigt

184

13.2 Maxwell Fluid 185

P

k

uµ

P

Figure 13.1. Mechanical model for a Kelvin–Voigt solid.

solid, we have to solve the differential equation

u + kµ

u = 1µ

H(t), (13.2)

where H is the Heaviside step function. Using the initial condition u(0) = 0, we findthat the solution of the differential equation is

J (t) = 1k

[1 − e−t/τ ], τ = µ

k, (13.3)

where the constant τ is called the relaxation time. For high values of the viscosity,the relaxation time can be seen to be large.

The load history needed to produce a constant displacement of unity is calledthe stress relaxation function G(t). Substituting

u(t) = H(t), (13.4)

into differential equation (13.1), we get

P(t) = kH(t) + µδ(t), (13.5)

where δ(t) is the Dirac delta function. Figure 13.2 shows sketches of the creep com-pliance and stress relaxation functions for a Kelvin–Voigt material.

13.2 Maxwell Fluid

The mechanical model for a Maxwell fluid consists of a spring and dashpot arrangedin a series, as shown in Fig. 13.3. The load displacement relation is given by

u = Pk

+ Pµ

, (13.6)

where, again, k and µ are constants. Because of the series arrangement, for aconstant applied load the material can flow to an unlimited extent, given enoughtime. The creep compliance and stress relaxation functions for the Maxwell fluid are

J (t) = [1k

+ tµ

]H(t), G(t) = ke−t/τ , τ = µ

k. (13.7)

Figure 13.4 shows sketches of these functions.

186 Viscoelasticity

P

P

1k

u

u

k

1 1

t

t

t

t

Figure 13.2. Creep compliance and stress relaxation functions for a Kelvin–Voigt solid.

P

k

u

µ

P

Figure 13.3. Mechanical model for a Maxwell fluid.

1 1P

P

1k

u

u

k

t t

t t

Figure 13.4. Creep compliance and stress relaxation functions for a Maxwell fluid.

13.3 Standard Linear Solid 187

13.3 Standard Linear Solid

Based on the mechanical model shown in Fig. 13.5, we have the constitutive relation

P + pP = k[u + qu], (13.8)

where

p = k2

µ, k = k1 + k2, q = k1k2

µ(k1 + k2). (13.9)

An alternative, three-parameter mechanical model that has the same constitutiverelation as in Eq. (13.8) is given in the Exercises.

With the initial conditions u(0−) = 0 or P(0−) = 0, we can solve the differentialequation to obtain the creep compliance or the stress relaxation function. Also, fromthe symmetry of the differential equation from the creep compliance function, bychanging the parameters we can obtain the stress relaxation function. For example,to obtain the creep compliance, we set P(t) = H(t) and the differential equationbecomes

u + qu = 1k

[δ(t) + pH(t)]. (13.10)

Integrating this equation from (0 − ε) to (0 + ε) and letting ε → 0, we obtain

u(0+) = 1k. (13.11)

For t > 0 we have

u + qu = pk

, (13.12)

which has the solution

u(t) = pqk

+ Ce−qt . (13.13)

Using the initial condition, Eq. (13.11), we find that

C = 1k

[1 − pq

], (13.14)

PP

k1

k2µu

Figure 13.5. Mechanical model for a standard linear solid.

188 Viscoelasticity

and J (t) = u(t) is given by

J (t) = 1k

[(1 − pq

)e−qt + pq

]. (13.15)

The stress relaxation function is obtained by the replacements

p → q, q → p, k → 1/k (13.16)

in J (t) as

G(t) = k[(1 − qp

)e−pt + qp

]. (13.17)

The creep compliance and stress relaxation functions show that the stress relaxationtime (1/p) and the strain relaxation time (1/q) are generally distinct. We define therelaxation times

τs = 1p, τe = 1

q(13.18)

for the stress and strain, respectively. The creep compliance and stress relaxationfunctions can be written as

J (t) = 1k

[(1 − τe

τs)e−t/τe + τe

τs

], (13.19)

G(t) = k[

(1 − τs

τe)e−t/τs + τs

τe

]. (13.20)

Sketches of these functions when τe > τs are shown in Fig. 13.6.

P

P

1 u

u

1

1k

k

t t

t t

kτs/τe

τe/kτs

Figure 13.6. Creep compliance and stress relaxation functions for a standard linear solid whenτe > τs .

13.4 Superposition Principle 189

13.4 Superposition Principle

Consider the case in which an arbitrary load history is given as shown in Fig. 13.7.Assume we have the creep compliance function for the material. As shown in thedisplacement diagram, if the load were kept constant at t = τ the displacement att would be u(t). Because of a step increase in the load at τ , the increment in u isgiven by

du(t) = J (t − τ )dP(τ ). (13.21)

Integrating this, we obtain

u(t) =∫ t

0J (t − τ )

dPdτ

dτ. (13.22)

Similarly, if the displacement history is known, the load history can be found withthe stress relaxation function G(t), as

P(t) =∫ t

0G(t − τ )

dudτ

dτ. (13.23)

These integral representations are known as the Boltzmann superposition inte-grals. Taking the Laplace transform of these convolution integrals, Eqs. (13.22) and(13.23), with transform variable s, we get

L[u] = L[J ]sL[P], L[P] = L[G]sL[u] (13.24)

or

s2L[J ]L[G] = 1. (13.25)

This relation shows the reciprocal property of the two functions, rate of creep com-pliance, and rate of stress relaxation in the Laplace domain.

P

dP

udu

ttτ τ

Figure 13.7. Superposition of strain by use of the creep compliance function.

190 Viscoelasticity

13.5 Constitutive Laws in the Operator Form

With the notation

D = ddt

, (13.26)

and

P(D) = pmDm + pm−1 Dm−1 + · · · + p0, Q(D) = qn Dn + qn−1 Dn−1 + · · · + q0,

(13.27)the load displacement relation can be written for a multiparameter model as

P(D)P(t) = Q(D)u(t). (13.28)

This representation is known as the operator form of the constitutive relation. Tak-ing the Laplace transform, with t → s and the assumption that the initial values ofP and all its time derivatives up to the order of (m − 1) and of u and its derivativesup to the order of (n − 1) are zero,

P(s)P = Q(s)u, (13.29)

where we have used the notation

P = L[P, t → s]. (13.30)

13.6 Three-Dimensional Linear Constitutive Relations

The relations previously discussed for viscoelastic materials, based on mechanicalmodels, can be generalized to stress–strain relations in three dimensions. This issimilar to generalizing a simple spring to the Hookean solid. First, we separate thehydrostatic parts of stress and strain in the form

p = −13σi i , ε = εi i , (13.31)

where, for infinitesimal strain εi j , the quantity ε, as defined here, has the significancethat it represents change in volume per unit volume.

For most materials, the assumption

p(t) = −Kε(t), (13.32)

where the bulk modulus K is taken as a constant, is satisfactory. If the response tohydrostatic pressure exhibits viscoelastic behavior, we may write

p(t) = −∫ t

0K(t − τ )

dε

dτdτ, (13.33)

where K(t) may be called a hydrostatic stress relaxation function.The deviatoric quantities

si j = σi j − 13σkkδi j , ei j = εi j − 1

3εkkδi j (13.34)

13.7 Anisotropy 191

may be related by use of the creep compliance and stress relaxation functions in theform

si j =∫ t

0G(t − τ )

dei j

dτdτ, ei j =

∫ t

0J (t − τ )

dsi j

dτdτ. (13.35)

The kernels K, G, and J are called hereditary kernels. These relations are applica-ble for isotropic, homogeneous, linear viscoelasticity. In terms of the Laplace trans-forms of quantities involved, we have

si j = sGei j , ei j = s J si j . (13.36)

If we had a relation between si j and ei j in the differential operator form,

P(s)si j = Q(s)ei j , (13.37)

then the stress relaxation and creep compliance functions G and J are given by

G = 1sQP

, J = 1sPQ

. (13.38)

We may factor P and Q in the form

sP = (s + 1/τs1)m1 (s + 1/τs2)m2 · · · (s + 1/τsp)mp, m1 + m2 + · · · + mp = m + 1,

(13.39)

sQ = (s + 1/τe1)n1 (s + 1/τe2)n2 · · · (s + 1/τeq)nq , n1 + n2 + · · · + nq = n + 1.

(13.40)

Using these factored forms in Eqs. (13.38), we express G and J as partial fractions.Noting

L[e−t/τ ] =∫ ∞

0e−(s+1/τ )t dt = 1

s + 1/τ, L[tne−t/τ ] = n!

(s + 1/τ )(n+1), (13.41)

we see multiple stress relaxation times τs j , j = 1, 2, . . . , p, and strain relaxationtimes τei , i = 1, 2, . . . , q, while inverting the partial fraction expressions for G and J .If we encounter factors in which 1/τ = 0, that is, 1/si is a factor, its inverse will give aterm like t (i−1). Such terms in J represent fluid-type motion without any decay withtime. We have seen this behavior in Maxwell fluids. We have tacitly assumed all re-laxation times obtained from our factoring P and Q are nonnegative. If any of themis negative (say in Q), a slight force can set off an exponentially growing displace-ment, violating material stability. We need the existence of nonnegative factors as aconstraint on the coefficients of the operational polynomials P and Q.

13.7 Anisotropy

In the anisotropic case, we use the Voigt notation introduced in Chapter 11 to relatethe deviatoric stress and strain in the form

si = sGi j e j , ei = s J i j s j , i, j = 1, 2, . . . , 6. (13.42)

192 Viscoelasticity

13.8 Biot’s Theory

We consider the interesting results obtained by Biot (1955) in irreversible thermody-namics with applications to linear viscoelasticity. Although rapidly evolving thermo-dynamic state variables are not uncommon in many chemical and physical systems(examples can be found in chemical reactions and turbulent flows), Biot’s theoryconcerns slowly evolving state variables close to their equilibrium values.

Consider a homogeneous thermodynamic system I, defined by n state variablesqi , which are measured from a reference state qi = 0. These variables may representdisplacements, shear angles, concentrations, temperature rises, etc. We associate aset of external forces Pi with qi as conjugate variables. Let SI represent the entropyof system I. As system I is not isolated, we create an isolated system by embeddingsystem I in a second system, II, which is a large reservoir at temperature T0; the com-bined system can be considered to be isolated. The total entropy of the combinedsystem is

S = SI + SII. (13.43)

When the two systems are brought into contact, let δQI and δQII be the heat addedto the two systems. From the balance of energy,

δQI = −δQII. (13.44)

For system I,

δQI = dEI − Pi dqi , (13.45)

δQII = −dEI + Pi dqi . (13.46)

The entropy change in the reservoir is then given by

dSII = 1T0

δQII = 1T0

[−dEI + Pi dqi ]. (13.47)

The total entropy change in the isolated system can be expressed as

dS = dSI + dS2 = dSI + 1T0

[−dEI + Pi dqi ]. (13.48)

According to the second law, dS > 0 when at least one dqi �= 0. As SI and EI arefunctions of the state variables, we have

Xi ≡ ∂SI

∂qi− 1

T0

[∂ EI

∂qi+ Pi

]> 0, (13.49)

where we have introduced the notation Xi associated with qi for the entropy-producing force separate from Pi . Note that

dS = Xi dqi , S = Xi qi , (13.50)

where qi is the flux associated with the generalized force Xi . We have seen this formof entropy production rate in Chapter 9.

13.8 Biot’s Theory 193

We define an equilibrium state in which all external forces are zero, i.e., Pi = 0,and the total entropy is

Seq = SI − EI

T0. (13.51)

At equilibrium the entropy of the system is a maximum, and for small values of qi

we may write

T0SI − EI = −12

ai j qi qj ≡ −U, (13.52)

where ai j is a symmetric, positive-definite matrix. We use the notation U for thequadratic form because of its similarity to the Helmholtz free energy. In terms of U,Eq. (13.48) can be expressed as

T0dS = −dU + Pi dqi . (13.53)

From the reference state qi = 0, if we slowly increase the values of qi in a quasi-static, reversible manner, at each equilibrium step we would have

T0 S = −U + Pi qi = 0 or T0∂S∂qi

= −∂U∂qi

+ Pi = 0. (13.54)

In the irreversible case, we use Onsager’s principle, and the entropy production rateis nonzero. That is,

T0∂S∂qi

= bi j qj , (13.55)

where bi j is a symmetric, positive-semidefinite (some state variables may not partici-pate in entropy production) matrix. Equation (13.54) can be written as the evolutionequations

∂U∂qi

+ bi j qj = Pi or bi j qj + ai j qj = Pi . (13.56)

Biot has introduced a dissipation function D as

D = 12

bi j qi qj , (13.57)

which can be introduced in the evolution equation, Eq. (13.56), to get

∂ D∂qi

+ ∂U∂qi

= Pi , (13.58)

which has a resemblance to the Lagrange equations of dynamics. Also,

T0∂S∂t

= T0∂S∂qi

qi = bi j qi qj = 2D. (13.59)

In terms of the dissipation function D, the entropy-producing forces Xi inEq. (13.50) can be written as

∂ D∂qi

= Xi , (13.60)

194 Viscoelasticity

and the power input as

Xi qi = 2D. (13.61)

13.8.1 Minimum Entropy Production Rate

For a given power input Xi qi , one may ask the question: Of all possible sets of valuesfor qi , which set makes the entropy production rate a minimum?

Using Eq. (13.59), we want to minimize D subject to the constraint Xi qi = k.With the help of a Lagrange multiplier λ, we define a modified function

D∗ = D − λ(Xi qi − k), (13.62)

and minimize it to get

∂ D∂qi

= λXi , (13.63)

as the equations for the general directions of qi . If we multiply these by qi (and sumon i), we find

2Dmin = λk, (13.64)

which simplifies to λ = 1 when Eq. (13.61) is used. Thus, our evolution equation,Eq. (13.60), corresponds to minimum entropy production rate for a given powerinput.

Biot (1955) and Fung (1965) are highly recommended for further reading onthis topic. These suggested readings also include a method for eliminating hiddenvariables qi (which do not have any associated generalized external forces Pi ).

13.9 Creep in Metals

In uniaxial experiments conducted at high temperatures, it is seen that the strain ratedepends on the stress in a nonlinear fashion. Generally, the strain versus time plotcan be divided into three domains. These are called the primary creep, secondarycreep, and tertiary creep domains. A sketch of the three domains is shown in Fig.13.8. In the case of primary creep, the following constitutive law is often used todescribe the creep behavior:

e = Asntm, (13.65)

where A, n, and m are constants. For a multiaxial state of stress, using deviatoricquantities, we can have

ei j = AI n−1s2 tmsi j , (13.66)

where Is2 is the second invariant of the deviatoric stress si j . The assumption that thematerial is incompressible is usually applied.

13.10 Nonlinear Theories of Viscoelasticity 195

e

t

Primary

Secondary

Tertiary

Figure 13.8. Three domains of creep behavior.

For the secondary creep, the preceding law with m = 0 is used. This domainis also known as the domain of stationary or steady-state creep. The tertiary creepbehavior is seen just before the specimen fails.

Another aspect of creep in materials involves differing strain rate responses totensile and compressive loads. That is, the creep constants, A, n, and m for compres-sion, are different from those for tension.

13.10 Nonlinear Theories of Viscoelasticity

The infinitesimal strains used in the preceding sections are not applicable when largedeformations are involved. Also, the material response itself may depend on stressesand strains in a nonlinear fashion. The elementary constitutive relations previouslydiscussed have been extended, in a number of ways, for nonlinear viscoelastic solidsand fluids. An extensive classification of viscoelastic materials can be found in thebook by Eringen (1962). A brief introduction to this topic is subsequently given.

We begin by introducing an intermediate time τ between the initial time τ = 0and the current time τ = t . We also have an intermediate location χ correspondingto the intermediate time. Then

X = χ(x, 0), χ = χ(x, τ ), x = χ(x, t). (13.67)

Extending the Greek indices for the material variables and Latin indices for thecurrent variables, we use a, b, c, etc., for the intermediate variables:

χ = χaea . (13.68)

196 Viscoelasticity

The Cauchy deformation gradient f and the Green’s deformation gradient F aredefined for the intermediate time τ with the components

fia = ∂χa

∂xi, Fai = ∂xi

∂χa, (13.69)

and the corresponding deformation tensors,

gi j = fia f ja, Gab = Fai Fbi . (13.70)

The Almansi and Lagrange strains are given by

ei j = 12

[δi j − gi j ], Eab = 12

[Gab − δab]. (13.71)

As can be seen from the preceding definitions, what we have is an Eulerian de-scription with the intermediate position in the role of the initial position X we hadused before. Next, we introduce higher-order intermediate strain rate and stressrate variables. These are generalized versions of the deformation rate and Jaumannstress rates. We define (

DDτ

)M

(dχadχa) = A(M)ab dχadχb, (13.72)

where A(M)ab is called the Rivlin–Ericksen tensor of the order of M. When M = 1 we

get the intermediate deformation rate tensor. Starting with

s(0)ab = σab, (13.73)

we define higher-order objective rates as

s(N)ab = s(N−1)

ab + s(N−1)ac ∂bvc + s(N−1)

bc ∂avc, (13.74)

where vc is the velocity component at time τ . It can be shown that these stress ratesare objective. With this, the general constitutive relations for viscoelastic materialscan be expressed as

τ=tFcdτ=0

[∂αχa, A(1)

ab (τ ), A(2)ab (τ ), . . . , A(M)

ab (τ ), s(0)ab (τ ), s(1)

ab (τ ), . . . , s(N)ab (τ )

]= 0. (13.75)

From this, we can obtain various subclasses of materials, including Stokes fluid andCauchy elastic materials, by specifying the values of M and N and the functionalforms.

13.11 K-BKZ Model for Viscoelastic Fluids

Among the theories of viscoelastic fluids, the K-BKZ theory has found widespreadapplications in modeling polymer fluids. This model is based on two independent

Suggested Reading 197

papers by Kaye (1962) and Bernstein, Kearsley, and Zapas (1963). Our subsequentdiscussion is based on a review of the theory by Tanner (1988).

We begin with the hereditary integral representation of the shear stress in aone-dimensional pure shear experiment, as

s(t) =∫ t

−∞G(t − τ )Edτ, (13.76)

where E is the shear strain. Integrating by parts and assuming G(0) = 0 andG(−∞) = 0, we have

s(t) =∫ t

−∞G(t − τ )Edτ. (13.77)

Nonlinear dependence on E is then introduced by replacing G with ∂U/∂ E, whereU = U[E(τ ), t − τ ]. The original K-BKZ model assumes incompressibility with anassociated hydrostatic pressure. The 3D version of the model is

si j (t) =∫ t

−∞Fai

∂U∂ Eab

Fbj dτ. (13.78)

The invariance requirements restrict, for isotropic, incompressible fluids, the formof the memory kernel U to

U = U(IE1, IE2). (13.79)

As we have done in the case of finite elasticity, if U is assumed to be a function ofthe invariants of the deformation gradients Gab,

si j (t) =∫ t

−∞Fai

∂U∂Gab

Fbj dτ. (13.80)

Tanner (1988) gives a number of special functional forms for U and their successesand failures when applied to various polymeric fluids.

The original isothermal theory was extended by Bernstein, Kearsley, and Zapas(1964) in a subsequent paper.

SUGGESTED READING

Bernstein, B., Kearsley, E. A., and Zapas, L. J. (1963). A study of stress relaxation with finitestrain, Trans. Soc. Rheol., 7, 391–410.

Bernstein, B., Kearsley, E. A., and Zapas, L. J. (1964). Thermodynamics of perfect elasticfluids, J. Res. Natl. Bur. Stand., 68B, No. 3, 103–113.

Biot, M. A. (1955). Variational principles in irreversible thermodynamics with applicationsto viscoelasticity, Phys. Rev., 97, 1463–1469.

Eringen, A. C. (1962). Nonlinear Theory of Continuous Media, McGraw-Hill.Fung, Y. C. (1965). Foundations of Solid Mechanics, Prentice-Hall.Kaye, A. (1962). Non-Newtonian flow in incompressible fluids, Part I: A general rheological

equation of state, Part II: Some problems in steady flow, Note No. 134, College of Aero-nautics, Cranfield, UK.

Rabotnov, Y. N. (1969). Creep Problems in Structural Members, North-Holland.Tanner, R. I. (1988). From A to (BK)Z in constitutive relations, J. Rheol., 32, 673–702.

198 Viscoelasticity

EXERCISES

13.1. Consider a bar of cross-sectional area A, length L, and density ρ. One end ofthe bar, x = L, is fixed to a wall and the other end, x = 0, is subjected to an oscillat-ing force, F (t) = F0eiωt . Assuming the material is a Kelvin–Voigt solid satisfying

σ = Eε + µε,

obtain the displacement under the load, neglecting changes in the cross-sectionalarea and the inertia forces. Reduce the displacement to its steady-state form. Definethe so-called complex modulus Ec, which is a complex number involving E, µ, andω, by comparing the solution with the elastic solution.

13.2. Assume the bar in the preceding exercise extends from x = 0 to ∞. If a stressσ0 is suddenly applied at x = 0, obtain the displacement u(x, t) as a real integral.Neglect the change in the cross-sectional area.

13.3. Calculate the response of a standard linear solid when

P = P0t H(t).

13.4. A mechanical model is shown in Fig. 13.9. Obtain the viscoelastic model forthis setup.

P P

k1

k

µ

Figure 13.9. A mechanical model for a viscoelasticsolid.

13.5. The time history of the applied load on a viscoelastic bar is shown in Fig. 13.10.Using the superposition integral, obtain the displacement history if the material is(a) a Maxwell fluid, and (b) a standard linear solid.

P/P0

1

1

2

3−1t

Figure 13.10. History of applied load.

13.6. For a 3D viscoelastic body, write the field equations and boundary and initialconditions under the assumption of infinitesimal strains. Using the Laplace trans-form, convert these equations into equations in the transformed variables. Comparewith the equations of linear elasticity and establish the correspondence principle.Use Eqs. (13.33) and (13.35) as constitutive laws. Comment on the boundary condi-tions for which the correspondence principle fails.

Exercises 13.1–13.8 199

13.7. A cube of side length a is made of a Maxwell material. It is placed on a fric-tionless surface at time t = 0. Because of the weight of the material, there is a com-pressive stress at any cross-section. Assuming a specific gravity of γ , compute thecompressive stress at a height x from the bottom of the cube. Obtain the height ofthe cube at any time t . Neglect changes in the cross-sectional area.

13.8. A Maxwell fluid with the constitutive relation∂u∂y

= 1G

τ + 1µ

τ

occupies the region 0 < y < h, −∞ < x, z < ∞. At the boundary y = 0, the veloc-ity u = 0, and at y = h, u = Ueiωt . Neglecting any body forces, obtain the velocitydistribution u(y, t).

14 Plasticity

The time-independent, permanent deformation in metals beyond the elastic limit isdescribed by the term plasticity. As shown in Fig. 14.1, the elastic part of the uni-axial stress–strain curve OA is reversible. When we unload from any point beyondpoint A, corresponding to the zero-stress state, there is a permanent deformation inthe specimen. Several theories describing the plastic behavior of metals have beenproposed, but none is totally satisfactory. Ideally, the stress σ , separating the elasticpart OA from the inelastic part, is called the yield stress σ0 of the material. How-ever, it is difficult to obtain this special point accurately from experiments, and itoften depends on the history of the loading the specimen has undergone.

When a specimen is unloaded from a point C to zero stress and reloaded, per-manent deformation usually begins at a stress level σC . In other words, the yieldstress is higher after the specimen has undergone a certain amount of permanentdeformation. This is known as work hardening or strain hardening.

Another feature of metal deformation is that, from a strained state, beyond theyield point, reversal of loading causes the compressive yield stress to be differentfrom the tensile yield stress. It is usually lower in magnitude than the tensile yieldstress. This effect is called the Bauschinger effect.

Before we consider modern theories of plasticity that attempt to describe themultiaxial stress–strain behavior, let us review some idealized classical theories inthe next few sections of this chapter.

14.1 Idealized Theories

14.1.1 Rigid Perfectly Plastic Material

In a uniaxial stress state the material is assumed to be rigid up to a value of stressσ = σ0 (see Fig. 14.2). At σ0, the material flows. Because of this limiting upperbound for the stress, it is to be kept in mind that all plasticity experiments (includingthought experiments) are done with displacement control. The term perfectly plas-tic refers to the absence of work hardening. In applications such as metal forming,in which the elastic strains are small, this idealization is still used to a great extent.

200

14.1 Idealized Theories 201

0σ

σ

ε

A

B C

O

Figure 14.1. Stress–strain relation for met-als in uniaxial tension.

14.1.2 Elastic Perfectly Plastic Material

When σ < σ0 the material obeys Hooke’s law, and when σ ≥ σ0 the material flows(see Fig. 14.3). In problems concerned with residual stresses the elastic effects aresignificant and this idealization is often used.

14.1.3 Elastic Linearly Hardening Material

The uniaxial stress–strain behavior previously described can be extended to work-hardening models through bilinear relations, as shown in Fig. 14.4.

σ

ε

Figure 14.2. Rigid perfectly plastic model.

202 Plasticity

σ

ε

Figure 14.3. Elastic perfectly plastic model.

σ

ε

Figure 14.4. Linear work-hardening models.

Plastic flow problems can be generally classified as contained plastic flow anduncontained plastic flow. These two cases are illustrated by Fig. 14.5, by consideringthe tip of a crack in a plate.

In uncontained plastic flow, the work hardening is important and the elasticeffects may be negligible. However, in contained plastic flow the elastic effects aresignificant.

Figure 14.5. Contained and uncontained plastic domains.

14.2 Three-Dimensional Theories 203

14.2 Three-Dimensional Theories

When we attempt to extend the uniaxial behavior to the three-dimensional (3D)stress state, an important factor to observe is that the material behavior under hy-drostatic stress is significantly different from that under pure shear. The onset ofpermanent deformation is governed by the state of shear stress in most polycrys-talline, metallic materials. Spitzig and Richmond (1984) provide Aluminum 1100as an example of pressure-sensitive plastic yield behavior. In this chapter we ex-clude pressure-dependent yielding. We classify the elastic behavior of materials ascompressible or incompressible on the basis of the stress–strain behavior under hy-drostatic pressure. The relation for the hydrostatic pressure can be written as

−p = Kε, compressible, (14.1)

p = p, incompressible, (14.2)

where

p = −13σi i , ε = εi i . (14.3)

We denote the deviatoric stress by

si j = σi j − 13σkkδi j , (14.4)

and the deviatoric strain by

ei j = εi j − 13εkkδi j . (14.5)

To distinguish the range of si j , in which the material behaves elastically, we needa yield condition. If, under uniaxial tension, the yield stress is σ0, the correspondingshear stress τ0 is given by

τ0 = σ0/2, (14.6)

and we would like to extend this as a condition on the stress state in the 3D case. Thisconstraint should involve only the deviatoric stresses. For most metals, experimentsindicate that the following two yield conditions may be used as approximations:

f (s) = 12

max|(σ1 − σ2), (σ2 − σ3), (σ3 − σ1)| − k1 = 0, (14.7)

f (s) = 12

si j si j − k2 = 0. (14.8)

These two conditions are known as the Tresca condition and the von Mises con-dition, respectively. The Tresca condition states that plastic flow commences whenthe maximum shear stress reaches a specified value. The von Mises condition isbased on the distortion energy density stored in the material. Distortion energy isthe difference between the total strain energy and the work done by the hydrostaticpressure.

204 Plasticity

To examine these conditions closely, it is advantageous to use the stress tensorreferred to as the principal axes:

σ =

⎡⎢⎣σ1 0 0

0 σ2 00 0 σ3

⎤⎥⎦ , (14.9)

where σi are the principal stresses and σ1 ≥ σ2 ≥ σ3.Here, the hydrostatic stress is given by

p = −13

(σ1 + σ2 + σ3). (14.10)

Then,

s = 13

⎡⎢⎣ 2σ1 − σ2 − σ3 0 0

0 2σ2 − σ3 − σ1 00 0 2σ3 − σ1 − σ2

⎤⎥⎦ , (14.11)

and the Tresca criterion reads

σ1 − σ3 − 2k1 = 0, (14.12)

and the von Mises criterion reads

Jσ2 = Is2 = 12

si j si j = 16

[(σ1 − σ2)2 + (σ2 − σ3)2 + (σ3 − σ1)2] − k2 = 0, (14.13)

where we have introduced the notation Jσ i = Isi relating the invariants of the devi-atoric stress and the full stress tensor. Noting that, when the material yields underuniaxial tension, the stress state is

σ1 = σ0, σ2 = 0, σ3 = 0, (14.14)

we find the constants

k1 = σ0/2, k2 = σ 20 /3. (14.15)

In the stress space, with the axes σ1, σ2, and σ3, we have

σ1 + σ2 + σ3 = −3p, (14.16)

representing planes with unit normal

n = 1√3

(1, 1, 1)T. (14.17)

These planes are called the π planes. Because the yield condition does not de-pend on any of these parallel planes, it represents a surface normal to these planes.In other words, the yield surface is an orthogonal cylinder with its axis inclined at anangle θ with respect to the three principal directions, where cos θ = 1/

√3, as shown

14.2 Three-Dimensional Theories 205

σ1

σ2

σ3

σ1 σ2

σ3

π plane

Yield surface

Figure 14.6. Yield surfaces in the stress space.

in Fig. 14.6. A stress state σ , which is a vector in the stress space with σ1, σ2, and σ3

as coordinates, can be resolved as

σ = −pn + τ , τ · n = 0. (14.18)

The three components of τ are

τ1 = 13

(2σ1 − σ2 − σ3), τ2 = 13

(2σ2 − σ3 − σ1), τ3 = −(τ1 + τ2). (14.19)

The yield surface is a closed curve in the two-dimensional (2D) τ1, τ2 space. Insteadof τ1 and τ2, we may use the more familiar quantities σ1 − σ3 and σ2 − σ3 for plottingthe yield curve.

The Tresca condition becomes an elongated hexagon in our plot and the vonMises condition an elongated ellipse. In the π plane, these figures will lose theirelongations and become a perfect hexagon and a circle.

Combined torsion and tension of thin-walled tubes is often used to generatea multiaxial stress state for determining whether the Tresca condition or the vonMises condition is a better approximation for the yield condition for a particularmaterial.

With tensile stress σ and the shear stress tensor τ , the stress tensor can be writ-ten as

σ =

⎡⎢⎣σ τ 0

τ 0 00 0 0

⎤⎥⎦ , (14.20)

with the principal stresses

σ1,3 = 12

[σ ±√

σ 2 + 4τ 2], σ2 = 0 (14.21)

and the maximum shear

τmax = 12

√σ 2 + 4τ 2. (14.22)

206 Plasticity

τ/σ0

σ/σ0

von Mises

Tresca

1/√

3

1/2

1

Figure 14.7. Yield curves in the tension–torsion of thin-walled tubes.

The value of J2 is obtained as

J2 = 13

[σ 2 + 3τ 2]. (14.23)

A plot of Tresca and von Mises Conditions for the tension-torsion experimentis shown in Fig. 14.7.

Once a yield condition has been established for a work-hardening material, σ0

will change with strain history. In materials such as soil and other granular materials,the yield condition is known to depend on the hydrostatic pressure p. The higher thevalue of p, the larger will be the yield stress.

In many applications, plane stress or plane strain approximations are satisfac-tory and it is worthwhile to examine the yield conditions for these cases. To do this,first we need to consider the relation between plastic strain and the stress state.

14.3 Postyield Behavior

Uniaxial experiments show that, under controlled strain rates, the stress increases tothe yield stress and a further increase in strain does not increase the stress substan-tially. The rate of loading has negligible effect on the stress at room temperature.Thus plastic flow is independent of time, unlike fluid flow, in which the shear stressis proportional to the strain rate. We have to keep in mind that the “loading” is ap-plied through controlled displacements on the boundary. Two distinct constitutiverelations are in use, depending on the rigid perfectly plastic or the elastic perfectlyplastic approximation. These are known as the flow rules.

14.3.1 Levy–Mises Flow Rule

Following the rigid perfectly plastic approximation, the elastic strains are neglected.Furthermore, the material is assumed to be incompressible. Thus,

εi j = ei j = epi j , (14.24)

14.3 Postyield Behavior 207

where ei j represents the inelastic strain. The constitutive relation is given by

dei j =√

3[12

dekldekl ]1/2 si j

σ0. (14.25)

Using

dλ =√

3J2(de), (14.26)

we have

dei j = dλsi j/σ0. (14.27)

From dissipation considerations, it can be shown that the von Mises criterion is thenatural criterion to use with the Levy–Mises flow rule.

For plane strain deformations, de33 = 0 and σ33 = (σ11 + σ22)/2. Using this andthe time derivatives

dei j = dei j

dtdt, (14.28)

we find, for both plane stress and plane strain,

e11

σ11 − σ22= e22

σ22 − σ11= e12

σ12. (14.29)

14.3.2 Prandtl–Reuss Flow Rule

We begin with a decomposition of the strain increment as

dεi j = dεei j + dε

pi j , (14.30)

where dεe indicates the elastic component of the strain. We assume plastic incom-pressibility, that is,

dεpii = 0 or dε

pi j = dep

i j . (14.31)

The flow rule states that

depi j = dλsi j/σ0, (14.32)

where dλ is a plastic strain increment-dependent scalar. In contrast with the Levy–Mises rule, in which the total strain increment is used to define dλ, the Prandtl–Reuss flow rule has

dλ =√

3J2(dep). (14.33)

The elastic strain increment is given by

dεei j = dsi j/2G, εe = −p/K, (14.34)

where K is the bulk modulus given by K = E/3(1 − 2ν).

208 Plasticity

14.4 General Yield Condition and Plastic Work

The yield condition can be represented by a hypersurface in the six-dimensionalstress space σi j . Because the hydrostatic pressure has no effect on yielding, this hy-persurface is not closed. If we use the five-dimensional space spanned by si j , we havea closed surface. The material behavior is elastic if the current stress state is insidethis surface or if unloading takes place from the yield surface.

We insert a parameter k, representing the variable yield stress in the case ofhardening plasticity, to express the yield condition as

f (s; k) = 0. (14.35)

The parameter k is considered to be an internal variable that depends on the historyof the plastic strains, and an experimenter has no direct means to alter it. We haveelastic constitutive relations if

f (s; k) < 0 or f (s; k) = 0, ∇ f · ds < 0, (14.36)

and plastic flow if

f (s; k) = 0 and ∇ f · ds ≥ 0. (14.37)

This stress space description of the onset of plastic flow has to be considered inthe context of strain-controlled experiments for clarity. If there is no hardening, thestress state is inside the yield surface or on it (see Fig. 14.8).

n = ∇ f|∇ f |

s

f (s, k) = 0

Figure 14.8. Yield surface in the deviatoric stress space.

14.4 General Yield Condition and Plastic Work 209

14.4.1 Plane Stress and Plane Strain

For plane stress, the principal stress σ3 = 0 and the von Mises condition becomes

σ 21 − σ1σ2 + σ 2

2 = σ 20 . (14.38)

In terms of a mean stress σ and a shear stress τ , defined by

σ ≡ σ1 + σ2

2, τ = σ1 − σ2

2, (14.39)

where we assume σ1 > σ2, we find

τ =√

(σ 20 − 2σ 2)/6. (14.40)

For plane strain, the plastic strain has to be confined to a plane normal to the x3 axisand we get

τ = σ0

2. (14.41)

The Tresca condition for plane strain gives

τ = σ0

2, (14.42)

the same as the von Mises condition. However, for plane stress, we have differentpossibilities depending on the edge of the hexagon where the state of stress at yieldhappens to be.

14.4.2 Rigid Plasticity and Slip-Line Field

In many applications involving metal processing, the plastic strains are orders ofmagnitude larger than elastic strains and, if residual stresses are not of concern, therigid plasticity approximation can be used. In the regions where metal flow is takingplace, the maximum shear τ = σ0/2 under the plane strain approximation. As shownin Mohr’s circle in Fig. 14.9, the stresses can be written as

σ11 = σ + τ sin 2θ,

σ22 = σ − τ sin 2θ,

σ12 = −τ cos 2θ, (14.43)

where

σ = 12

(σ1 + σ2), τ = 12|σ1 − σ2| = 1

2σ0. (14.44)

210 Plasticity

τ

θ

2θ

τ

τ

σ

σ

σ

σ11

σ12

F

α

β

σ1σ2 Figure 14.9. Mohr’s circle and slip lines.

Using these in the equilibrium equations,

σ11,1 + σ12,2 = 0, σ12,1 + σ22,2 = 0, (14.45)

we find

σ,1 + σ0[cos 2θ θ,1 + sin 2θ θ,2] = 0, (14.46)

σ,2 + σ0[sin 2θ θ,1 − cos 2θ θ,2] = 0. (14.47)

These are two nonlinear equations for the mean stress σ and the angle of themaximum shear plane(s) θ . Following the method of characteristics introduced inChapter 12, we define a characteristic variable α through

dα = dx1 − cdx2 or ∂2 = −c∂1. (14.48)

As we will see, c has two values, and the corresponding two characteristic curves aredenoted as α and β lines. The equilibrium equations give[

1 σ0(cos 2θ − c sin 2θ)−c σ0(sin 2θ + c cos 2θ)

] {σ,1

θ,1

}=

{00

}. (14.49)

The characteristic equation of this eigenvalue problem is

c2 − 2ccos 2θ

sin 2θ− 1 = 0, (14.50)

which has two real solutions,

c = − tan θ, c = cot θ. (14.51)

14.4 General Yield Condition and Plastic Work 211

These orthogonal directions are shown in Fig. 14.9. Using the eigenvectors corre-sponding to the two values of c, we obtain the Riemann invariants,

σ

σ0+ θ = Cβ, along the α line:

dx2

dx1= − tan θ, (14.52)

σ

σ0− θ = Cα, along the β line:

dx2

dx1= cot θ. (14.53)

The velocity gradients are obtained from the flow rule of Eq. (14.29) as

v1,1

σ0 sin 2θ= − v2,2

σ0 sin 2θ= −v1,2 + v2,1

2σ0 cos 2θ. (14.54)

These relations can also be written as

v1,1 + v2,2 = 0, v1,2 + v2,1 − 2v2,2 cot 2θ = 0. (14.55)

Introducing the characteristic curves, with

∂2 = −c∂1, (14.56)

we get [1 −c

−c 1 + 2c cot 2θ

] {v1,1

v2,1

}=

{00

}. (14.57)

The characteristic equation of this eigenvalue problem is the same as the one forstresses,

c2 − 2ccos 2θ

sin 2θ− 1 = 0, (14.58)

with the two real solutions, c = cot θ and c = − tan θ .For c = cot θ , we have

v1,1 = cot θv2,1, (14.59)

and for c = − tan θ ,

v1,1 = − tan θv2,1. (14.60)

These are the incompressibility conditions expressed along the characteristics.If α and β are the curvilinear coordinates along the two characteristics, the flow ruleshows that, along the maximum shear directions,

σαα = σββ = σ, (14.61)

εαα = εββ = 0. (14.62)

In a curvilinear system, using velocity components tangential to the characteristics,we get

Dαvα = 0, Dβvβ = 0, (14.63)

where D is the covariant differential operator.

212 Plasticity

These can also be expressed as

dvα = vβdθ, dvβ = −vαdθ, (14.64)

along the α line and the β line, respectively. This form of the incompressibility con-ditions is known as the Geiringer equations. Components of velocity normal to acharacteristic curve should be continuous across it; the tangential component maybe discontinuous. This possible sliding of the material along a characteristic in plas-tic flow is the reason why the characteristics are called slip lines.

14.4.3 Example: Symmetric External Cracks

Consider a bar in plane strain with two cracks at the onset of yielding between thetwo crack tips. A possible slip-line field is shown in Fig. 14.10.

This example illustrates some of the important features of the orthogonal net-work formed by the slip-line field. Outside the orthogonal network, the material hasmaximum shear stresses below the yield value. Between the two crack tips, the max-imum shear stress is at its yield value τ = σ0/2. Symmetry with respect to the planeof the two cracks establishes the direction of the slip lines at 45◦, as shown in thefigure. The crack surfaces are traction free with one principal stress, normal to thesurface, being zero and the other parallel to the surface. Again, along the line AOthe maximum shear planes are at 45◦ and σ = ±σ0/2. We choose the positive sign

P

P

A

B C

DO

Figure 14.10. Slip-line field for a bar withsymmetric cracks.

14.5 Drucker’s Definition of Stability 213

for the solid under tensile loading. Inside the triangle ABO, the slip lines are straightlines with the α lines at 45◦ and the β lines at 135◦. Inside this triangle the value ofσ is constant. There is another constant-stress triangle, OC D, with unknown meanstress, say, σ ∗. In between the two triangles there is the region BCO, with O be-ing the crack tip. The crack tip is a singular point as the mean stress jumps from itsvalue in ABO to its value in C DO. However, away from the tip, the mean stresscontinuously changes along the circular arcs in the area BCO. This area is called afan. Most slip-line fields involving straight boundaries are made of constant-stresstriangles and fans. To relate the constant mean stresses in the two triangles, we fol-low the characteristic line ABC D. From B to C the angle θ changes by −π/2, and,using this in Eq. (14.52), we see

σB

σ0= σC

σ0− π

2or σC = σD = σ ∗ = σ0

(π + 1

2

). (14.65)

Along OD, if the horizontal and vertical stresses are σ11 and σ22, we have

σ22 − σ11 = σ0, σ22 + σ11 = σ0(π + 1). (14.66)

From these equations,

σ22 = π + 22

σ0, σ11 = π

2σ0. (14.67)

Note that these results are at variance with the common approximations,

σ22 = σ0, σ11 = 0. (14.68)

Without going into detail, the velocity distributions are such that the blockC DO moves to the right as a rigid body with a velocity of, say, U, and the blockABO has a horizontal velocity of U and a vertical velocity of 2U. Inside the fan, thevelocity components are functions of the angle.

In many cases, slip-line field construction is not unique. There is an underly-ing assumption that outside the field the stresses satisfy the equilibrium equationswith the maximum shear stress below the yield value. There are virtual work prin-ciples distinguishing the best approximation among multiple approximations to thevelocity fields.

Books on plasticity cited at the end of this chapter have explicit solutions usingthe slip-line field for many metal-processing cases.

14.5 Drucker’s Definition of Stability

In a one-dimensional plot of s against the total deviatoric strain e, a work-hardeningconstitutive relation would appear as shown in Figure 14.11.

The incremental plastic work done on the body is given by

dWp = sdep. (14.69)

214 Plasticity

s s s

e ep ep

de

dep

dsdep/2 dsdep/2

Figure 14.11. Work hardening/softening and plastic work.

We have

d2Wp = dsdep/2. (14.70)

The material is said to be stable if d2Wp > 0. This is known as Drucker’s inequalityfor stable plastic materials. For perfectly plastic materials, we have neutral stabilitybecause d2Wp = 0. If we take a point s1 on the stress–strain curve and another points0 in the interior in a partially unloaded state, for stable materials,

(s1 − s0)dep > 0. (14.71)

Let us consider a cyclic loading situation (see Fig. 14.12), starting from an equi-librium, elastic state s0 at t = 0. At t = τ1, we are at s1 on the yield surface. At t = τ2,the stress state is s2 outside the original yield surface. In this process the yield sur-face has expanded because of strain hardening. Subsequent unloading leads to s0 attime τ3. With

dei j = deei j + dep

i j , (14.72)

s0

s1

s2

Figure 14.12. Cyclic loading with work hardening.

14.6 Ilıushin’s Postulate 215

the work done in this cyclic process is given by

�W =∫ τ1

0si j ee

i j dt +∫ τ2

τ1

si j (eei j + e p

i j )dt +∫ τ3

τ2

si j eei j dt

=∮

si j eei j dt +

∫ τ2

τ1

si j epi j dt

=∫ τ2

τ1

si j epi j dt. (14.73)

The work done by the initial loading, if held constant, is written as

�W0 =∫ τ2

τ1

s0i j epi j dt. (14.74)

The work done by an external agency is the difference between �W and �W0:

�Wex = �W − �W0 =∫ τ2

τ1

(si j − s0i j )e pi j dt. (14.75)

For this to be nonnegative, the integrand must be nonnegative. That is,

(si j − s0i j )e pi j > 0. (14.76)

In the limit when s0i j is on the yield surface,

si j epi j > 0. (14.77)

On the yield point s, if we draw the plastic strain rate ep, we must have

(s − s0) · ep > 0. (14.78)

The yield surface must be convex for this to be true for all s0. A second conclusionthat follows from this is that, if ep is not known, ep must be normal to the yieldsurface in order for Eq. (14.78) to be true for all s0. Now

dep = dλ∇ f. (14.79)

Thus the Prandtl–Reuss flow rule is the associated flow rule for the von Mises con-dition. When Eq. (14.79) is expressed as

depi j = dλ∂ f /∂si j , (14.80)

we have epi j derivable from a potential, which is also the yield function. Nonasso-

ciative flow rules in which plastic strain components are derivative of a potentialseparate from the yield function are also found in the literature (Lubliner, 1990).

14.6 Ilıushin’s Postulate

In view of the fact that, in perfect plasticity, stress cannot be changed arbitrarily,the use of a strain space yield condition has been recommended by many authors.Extending the single internal variable k to multiple internal variables representedby a vector q, the strain space yield condition reads

f (e, q) = 0. (14.81)

216 Plasticity

Ilıushin’s postulate states that, for any closed curve in strain space,∮σi j dεi j ≥ 0, (14.82)

where the equality holds if the process is entirely elastic.

14.7 Work-Hardening Rules

The manner in which the yield function evolves with plastic deformation can bedescribed by a number of rules.

14.7.1 Perfectly Plastic Material

For this class of materials, it is assumed that the yield function does not change withdeformation. From Eq. (14.79)

(dλ)2 = dep · dep/|∇ f |2. (14.83)

For the von Mises criterion

f (s) = 12

si j si j − 13σ 2

0 = 0,

|∇ f |2 = si j si j = 23σ 2

0 ,

dλ =√

3J2(dep)/σ0. (14.84)

Then Eq. (14.80) reads

depi j =

√3J2(dep)si j/σ0. (14.85)

14.7.2 Isotropic Hardening

If the yield surface is assumed to maintain its shape while its size changes with plas-tic deformation, we then have isotropic hardening. We note that, in the case ofcyclic loading, this assumption implies an increased yield strength on the compres-sive portion of the tension-compression loading. This is opposite to the observedBauschinger effect. A sketch of the yield surfaces in isotropic hardening is shown inFig. 14.13. To generalize the uniaxial tensile test results, it is customary to define aneffective stress σ as

σ =√

3J2(σ ) =√

3si j si j/2, (14.86)

which reduces to σ11 in the one-dimensional case. The effective plastic strain incre-ment is defined as

¯dep =√

4J2(dep)/3 =√

2depi j e

pi j/3, (14.87)

which reduces to dep11 in the one-dimensional case.

14.7 Work-Hardening Rules 217

Figure 14.13. Growth of the yield surfacein isotropic hardening.

The uniaxial test data can be written as

σ = H(∫

¯dep) or dσ = h(∫

¯dep) ¯dep. (14.88)

The flow rule reads

depi j = dλsi j , dλ = 3dσ /2σh. (14.89)

This version of the isotropic-hardening rule is also known as the strain-hardeningtheory. A second version assumes that the hardening depends on the plastic workdone. That is

σ = G(Wp), (14.90)

where

Wp =∫

si j depi j . (14.91)

14.7.3 Kinematic Hardening

A hardening rule in which the yield surface does not change in either shape or sizebut simply translates in the direction of the normal at the current stress state wasintroduced by Prager in 1955 (see Fig. 14.14). The initial yield surface

f (s) = 0 (14.92)

is changed, during plastic flow, into

f (s − α) = 0. (14.93)

218 Plasticity

Figure 14.14. Translation of the yield sur-face in kinematic hardening.

Although originally it was proposed that the incremental change in the rigid bodytranslation is

dα = c dep, (14.94)

this relation leads to some confusion in the six-dimensional stress spaces. In modernapplications, Ziegler’s modification in the form

dα = dµ(s − α) (14.95)

is always used.To determine the scalar quantity dµ, we note that, from Eq. (14.93),

∂ f∂si j

dsi j + ∂ f∂αi j

dαi j = 0, (14.96)

as the shape of f does not change. From Eq. (14.93) we also have

∂ f∂si j

= − ∂ f∂αi j

. (14.97)

From these,

∇ f · (ds − dα) = 0, (14.98)

∇ f · ds = dµ∇ f · (s − α), (14.99)

dµ = ∇ f · ds∇ f · (s − α)

. (14.100)

Two other recent theories of hardening are the Mroz theory and the sublayertheory. No attempt is made here to describe these theories.

14.8 Endochronic Theory of Valanis 219

14.7.4 Hencky’s Deformation Theory

The plasticity theories we have discussed so far deal with incremental plastic defor-mation. These are also called incremental or flow theories of plasticity. In the de-formation or total strain theory, the strain is expressed as a function of the stress asin nonlinear elasticity for loading situations. A linear relation is used for unloading.When the loading is proportional, that is, all components of the stress tensor areproportionally increased, the results obtained with this simpler theory agree withthose obtained with the incremental theory. Using the infinitesimal strain measures

εi j = εei j + ε

pi j , (14.101)

we have

eei j = si j

2G, εe

ii = 13K

σi i , (14.102)

epi j = φsi j , when f (s) = 0, (14.103)

where

φ2 = 3e p/2σ , e p =√

4J2(ep)/3, σ =√

3J2(si j ). (14.104)

For von Mises criterion, with Prandtl–Reuss equations for the flow rule, if theloading is proportional, it is possible to show that Hencky’s theory is identical to thePrandtl–Reuss theory.

14.8 Endochronic Theory of Valanis

In most metals it is difficult to determine the precise location of the yield point onthe stress–strain diagram. In monotonic loading a small error in the yield stress maynot affect the end result. However, when cyclic loadings are involved, the yield stressalong with the hardening rules could lead to large errors in the end result. Also, thetime-independent plasticity effects are hard to separate from viscoelastic effects. In1971, Valanis proposed a unified theory of viscoplasticity without a yield surface.The basis of this theory is an intrinsic “time” measure that governs the inelasticdeformation of a neighborhood. This intrinsic time depends on deformation, and itmust be an increasing function of strain. We define

dξ = √dεi j dεi j (14.105)

as an intrinsic measure of strain. When real-time dependence is involved, we definean endochronic time scale

dζ =√

α2dξ 2 + β2dt2, (14.106)

where α and β are material constants. A monotonically increasing material functionz is then introduced as

z = z(ζ ), z′ > 0. (14.107)

220 Plasticity

Whenever the body is subjected to strain, z will be increasing. The stress–strain re-lations are given by

si j =∫ z

z0

2µ(z − z′)∂ei j

∂z′ dz′, (14.108)

−p =∫ z

z0

[K(z − z′)∂εkk

∂z′ + D(z − z′)∂θ

∂z′ ]dz′, (14.109)

where θ is the coefficient of thermal expansion and

2µ = 2µ0 H(z) + 2µ1e−α1z + 2µ2e−α2z + · · · +, (14.110)

K = K0 H(z) + 2K1e−α1z + 2K2e−α2z + · · · +, (14.111)

D = D0 H(z) + 2D1e−α1z + 2D2e−α2z + · · ·+, (14.112)

and µ0, µ1, . . . , K0, K1, . . . , and D0, D1, . . . are constants.If the material is plastically incompressible we have

−p = Kεkk (14.113)

in place of Eq. (14.109). If µ contains only a single term of the form

2µ = 2µ1e−α1z, (14.114)

we have

si j = 2µ1

∫ z

z0

e−α1(z−z′)dei j (z′), (14.115)

dsi j

dz= 2µ1

dei j

dz− α1si j , (14.116)

dei j = 12µ1

dsi j + α1dz2µ1

si j . (14.117)

If we divide the strains into elastic and inelastic parts we have

dei j = deei j + dep

i j . (14.118)

Then

depi j = α1dz

2µ1si j , dee

i j = dsi j

2µ1. (14.119)

The first of these is the Prandtl–Reuss equation of plasticity. Also

α1dz2µ1

=√

J2(dep)J2(s)

, (14.120)

dz = 2µ1

α1

√J2(dep)

J2(s), (14.121)

depi j =

√J2(dep)J2(s)

. (14.122)

14.9 Plasticity and Damage 221

σ

ε

Figure 14.15. Closed loop in a viscoplasticstress–strain diagram.

The original endochronic theory just described has been subjected to a series ofcriticisms. One of these deals with the predictions of the theory concerning the be-havior under unloading–reloading. The theory violates Drucker’s postulate of ma-terial stability. Because Drucker’s postulate is not based on thermodynamics, thiscriticism is not considered relevant. A more serious criticism has to do with theopen loop predicted by the theory in small amounts of unloading and reloading.Experiments indicate a closed loop, as shown in Fig. 14.15.

To overcome this inadequacy, Valanis proposed a modified theory in 1980. Forsimplicity, we confine ourselves to a version of the modified theory applicable toplastically incompressible materials.

The intrinsic time measure is defined with the plastic strain increments as

dξ 2 = depi j e

pi j , (14.123)

instead of the total strain increments. The deviatoric stresses are given by the inte-gral representation

si j =∫ z

0ρ(z − z′)

∂epi j

∂z′ dz′, (14.124)

where, to have the material yield at the onset of loading, it is necessary to have

ρ(0) → ∞. (14.125)

The modified theory proposed by Valanis also incorporates plastic compress-ibility and coupling between dilatation and distortion.

14.9 Plasticity and Damage

The characterization of damage in materials has been an ongoing effort in the gen-eral area of continuum mechanics. In polycrystalline materials, one observes a highconcentration of dislocations and grain boundary sliding at the microscopic levels.

222 Plasticity

In brittle solids these singularities are replaced with microcracks. Applied stressesat the macroscopic levels are the driving forces responsible for the movement of dis-locations and the extension of microcracks—phenomena that are, at least partially,irreversible. The cumulative effect of these irreversibilities is the deterioration ofthe strength of the material and eventual failure under repeated loadings.

Rice (1971, 1975) introduced an internal variable approach to treat solids under-going irreversible thermodynamic processes. The internal variables are supposed todescribe the microcrack extensions, dislocation movements, etc. It is important tonote that internal variables are not directly observable, unlike macroscopic vari-ables: strain, stress, and temperature. Our subsequent discussion follows the workof Yang, Tham and Swoboda (2005), which specializes the results obtained by Riceby assuming homogeneous functional forms for certain quantities. Although thecontext of the original papers by Rice was irreversible deformations in plasticity,we adopt the formalism for a general treatment of damage in materials. A key as-sumption of the theory is that the increments of internal kinematic variables suchas microcrack lengths are driven by associated thermodynamic microforces that arefunctions of macroforces, namely, the applied stresses, and these increments do notdepend on the external stresses directly.

Consider a unit volume of a material sample in which the macrovariables, strain,stress, and temperature, are uniform and the collection of internal variables qi

(i = 1, 2, . . . , n) are denoted by H. These variables are not state variables in thethermodynamic sense, and the damage at any time is history dependent. Helmholtzfree energy U and Gibbs free energy G are expressed as

U = U(e, T, H), G = G(σ , T, H), (14.126)

where e and σ are strain and its conjugate stress, respectively, and T is the temper-ature. At a fixed value of H, we have the elastic relations

σ = ∂U∂e

, e = −∂G∂σ

. (14.127)

As mentioned earlier, a change in H corresponds to changes in qi , and associatedwith each increment dqi , there is a microforce Xi . The balance of energy states

Xi dqi = − ∂U∂ H

dH ≡ −dU P

= − ∂G∂ H

dH ≡ −dGP. (14.128)

The driving forces Xi are functions of the macrovariables and the extent of damage

Xi = Xi (e, T, H), (14.129)

where dependence on H is through the history of all the qi ’s. The irreversible en-tropy production rate

SI = 1T

Xi qi ≥ 0. (14.130)

14.9 Plasticity and Damage 223

We introduce a potential B(Xi , T, H) to define the direction of the damage flow:

qi = ∂ B∂ Xi

, B =∫ X

qi dXi , (14.131)

where we have used vector notation for the n-dimensional quantities Xi , and theintegration has to be carried out with fixed T and H. These are the evolution equa-tions for the kinetics of the damage variables. The function B may be highly nonlin-ear. The direction of the vector q is along the normal to the surface B.

The inelastic strain increment can be written as

deP = e(σ , T, H + dH) − e(σ , T, H)

= − ∂2G∂σ∂ H

dH

= −∂dGP

∂σ

= ∂ Xi

∂σdqi . (14.132)

The strain rate becomes

eP = ∂ Xi

∂σqi = ∂ B

∂ Xi

∂ Xi

∂σ= ∂ B

∂σ. (14.133)

The dissipation rate D(Xi , T, H) is defined as

D = Xi qi = Xi∂ B∂ Xi

= −∂U∂qi

qi = −∂U∂t T,e

. (14.134)

At this point we make two assumptions about the potential B:

1. B is a homogeneous function of degree r in its variables Xi .2. B is also a homogeneous function of degree s in its variables σi j .

Our first assumption is more general than the one used in Yang et al. (2005); theyassume each qi is a homogeneous function of all Xj . As a consequence of Euler’stheorem concerning homogeneous functions,

Xi qi = Xi∂ B∂ Xi

= r B = D. (14.135)

As D is positive semidefinite, so is B. From differentiating

D = Xi∂ B∂ Xi

, (14.136)

we find

rqi = qi + Xj∂2 B

∂ Xi∂ Xj,

qi = Ji j Xj , (14.137)

224 Plasticity

where Ji j are the Onsager coefficients relating fluxes and forces, obtained here as

Ji j = 1r − 1

∂2 B∂ Xi∂ Xj

. (14.138)

These coefficients come out to be symmetric because of the symmetry of the mixedpartial derivatives. Also, we need r ≥ 2.

As a consequence of the second assumption, Eq. (14.133) gives

σi j ePi j = s B = s

rD, (14.139)

which is positive semidefinite.We note that it has been recognized that all the internal variables are difficult

to account for and, in all practical cases, they have to be replaced with a certainaverage or equivalent set of a smaller number of variables.

14.10 Minimum Dissipation Rate Principle

Instead of prescribing a flow rule as in Eq. (14.131), we may use the the principle ofminimum dissipation rate to derive the flow rule.

For a given power input for creating damage,

Xi qi = k, (14.140)

we may ask: What values of Xi do we need to have the dissipation a local minimum?For this extremum problem, we set up a modified function using a Lagrange

multiplier λ as

D∗ = D − λ(Xi qi − k). (14.141)

Minimization of D∗ leads to

λqi = ∂ D∂ Xi

. (14.142)

These relations show that q is directed normal to the surface D. If we further imposethe condition that D is a homogeneous function of degree r , multiplying Eq. (14.142)by Xi , we get

λXi qi = Xi∂ D∂ Xi

= r Dmin. (14.143)

Then, λ = r . As D is a highly nonlinear function, unlike a quadratic, we may havemultiple local minima. Often a condition of convexity is imposed on D to avoidthis. This minimum principle uses the generalized forces as variables. If we use thecomplement of D with the conjugate variables qi , we get a maximum dissipationprinciple equivalent to the choice of strain rate direction with respect to a yieldsurface in plasticity.

The application of these ideas to damage is still at its infancy. The paper byKrajcinovic (2004) is recommended for further reading.

Exercises 14.1–14.3 225

SUGGESTED READING

Hill, R. (1950). The Mathematical Theory of Plasticity, Oxford University Press.Johnson, W. and Mellor, P. B. (1973). Engineering Plasticity, Van Nostrand Reinhold.Kachanov, L. M. (2004). Fundamentals of the Theory of Plasticity, Dover.Krajcinovic, D. (2004). Damage mechanics: Accomplishments, trends and needs, Int. J. Solids

Struct., 37, 267–277.Lubliner, J. (1990). Plasticity Theory, Macmillan.Nemat-Nasser, S. (2004). Plasticity: A Treatise on Finite Deformation of Heterogeneous In-

elastic Materials, Cambridge University Press.Prager, W. (1955). The Theory of Plasticity: A Survey of Recent Achievements, Proc. Instn.

Mech. Engrs., 41–57.Rajagopal, K. R. and Srinivasa, A. R. (2004). On thermo-mechanical restrictions of continua,

Proc. R. Soc. London Ser. A, 460, 631–651.Rice, J. R. (1971). Inelastic constitutive relations for solids: An integral variable theory and

its applications to metal plasticity, J. Mech. Phys. Solids, 19, 433–455.Rice, J. R. (1975). Continuum mechanics and thermodynamics of plasticity in relation to mi-

croscale deformation mechanisms, in Constitutive Equations in Plasticity (A. S. Argon, ed.),MIT Press, 23–79.

Spitzig, W. A. and Richmond, O. (1984). The effect of pressure on the flow stress of metals,Acta Metal., 32, 457–463.

Ulm, F.-J. and Coussy, O. (2003). Mechanics and Durability of Solids, Prentice-Hall, Vol. 1.Valanis, K. C. (1971). A theory of viscoplasticity without a yield surface, Part I. Theory, Arch.

Mech., 23, 517–533.Valanis, K. C. (1971). A theory of viscoplasticity without a yield surface, Part II. Application

to mechanical behavior of metals, Arch. Mech., 23, 535–551.Valanis, K. C. (1980). Fundamental consequence of a new intrinsic time measure plasticity as

a limit of the endochronic theory, Arch. Mech., 32, 171–191.Yang, Q., Tham, L. G., and Swoboda, G. (2005). Normality structures with homogeneous

kinetic rate laws, J. Appl. Mech., 72, 322–329.Ziegler, H. (1959). A Modification of Prager’s Hardening Rule, Quat. Appl. Math., 17, 55–65.

EXERCISES

14.1. A thin-walled tube with circular cross-section is subjected to combined tensionand torsion. If �, r , t , and θ denote, respectively, the current length, mean radius,thickness, and angle of twist of the tube, determine the infinitesimal-strain incre-ments in a further deformation defined by δ�, δr , δt , and δθ .

14.2. A rod is extended in simple tension by gradually increasing the tensile stressσ . If the yield stress is a linear function of the physical longitudinal strain ε, withslope h, determine the plastic work done in producing the strain ε beyond the yieldlimit in terms of h, Young’s modulus E, and the initial yield stress σ0.

14.3. A metal plate occupies the region −a < x < a, −b < y < b, and −t < z < t . Itis subjected to proportional loading along its two axes by increasing the parameterα, with σxx = 2α, and σyy = α. All other stress components are zero. If yielding isgoverned by von Mises condition, obtain the stresses at yield. If an additional plasticstrain of ε

pxx = 0.1 is imparted through the same proportional loading, obtain the

plastic strains in the other directions, assuming incompressibility of plastic strains.

226 Plasticity

14.4. A thin-walled circular cylindrical tube of elastic-plastic strain-hardening ma-terial is stressed in simple tension to the point of yielding and then, holding theaxial stress constant, it is twisted by a continuously increasing end torque. If thestrain-hardening law is √

3J2 = σ0

[1 + 3

4ε p

],

where σ0 is the initial yield stress and the material satisfies the von Mises yield con-dition and the associated flow rule, obtain the relative angle of twist.Let L, R, and G represent the initial length, radius of the tube, and the shear mod-ulus, respectively.

14.5. Consider a thin-walled tube of mean radius R, length L, and thickness t , madeof a rigid plastic hardening material. We obtain the hardening rule from uniaxialexperiments by fitting the curve

σ = σ0[0.99 +√

0.0001 + ε],

where σ0 is the initial yield stress and σ and ε are effective stress and strain, respec-tively. Assume von Mises yield condition and the associated flow rule.

The tube is first subjected to a shear stress of σ0/√

3 through an applied torque.Keeping the shear stress constant, a tensile load σ � σ0 is applied. Obtain the per-centage changes in the dimensions of the tube.

14.6. Assume the thin-walled tube in the previous exercise is made of an elastic,isotropic linear hardening material with the uniaxial stress–strain relation

σ = Eε, σ < σ0, σ = (1 − α

E)σ0 + αε, σ ≥ σ0.

Using the von Mises condition and associative flow rule for yielding and Poisson’sratio ν, convert the uniaxial law to a relation between effective stress and strain.

The tube is twisted to a point of initial yield and then, keeping the torque con-stant, it is stretched to a length of (1 + β)L. What are the axial stress and the angle oftwist at this point? Next, the torque is removed and then the axial tension is reducedto zero. What are the final dimensions of the tube?

14.7. A tensile test specimen with initial length L0 and cross-sectional area A0 ismade from an incompressible rigid hardening plastic material. The plastic strain isrelated to the true stress in the form

ε =∫ L

L0

dLL

= Kσ n,

where K and n are constants. Compute the force P required to have a displacementU. Sketch P as a function of U and find the location of the maximum load. What isthe physical significance of a maximum load?

14.8. The Stefani model for plasticity (see Fig. 14.16) involves a spring with stiffnessE in series with two friction and spring elements (from Ulm and Coussy, 2003). Ob-tain the load displacement diagram for this model during a complete displacementcycle: 0 → U → 0 → −U → 0. Obtain an expression for the energy dissipated in

Exercises 14.5–14.12 227

K E

k

k

h

h

Figure 14.16. The Stefani modelfor one-dimensional plasticity.

the process. What is the constraint on the constants in order to have positive dissi-pation? Assume K > 2k.

14.9. A cantilever beam of rectangular cross-section has height 2c and length L.Under an applied load P at the end x = L, the longitudinal strain is given by

ε = κy,

where κ is the curvature of the neutral axis and −c < y < c. Assuming Young’smodulus is E and the material is elastic perfectly plastic with a yield stress of σ0, ob-tain the load Pp at the onset of plastic flow. What is the load Pc when the cantilevercollapses? Obtain the equation y = f (x), describing the elastic-plastic boundarywhen Pp < P < Pc. If the load is removed after the initial yield, sketch the distri-bution of the residual stress.

14.10. For a rigid plastic material in plane strain, if the yield stress τ depends on themean stress σ = (σ1 + σ2)/2, show that the condition for the equilibrium equationsto be hyperbolic at yield is dτ/dσ < 1. Derive the equations for the characteristiccurves and see if they are orthogonal.

14.11. Consider a rigid plastic tensile specimen in plane strain with two symmetric,sharp notches with angle γ , instead of the sharp crack problem described in thischapter. Obtain the slip-line field and the stress distributions in the yielded regions,using the constant-stress triangles and fans.

14.12. A wedge with an apex angle of 2γ is in plane strain. On one side of the wedgea pressure p is applied to a distance of a from the tip. Construct slip-line fields byusing constant-stress triangles and fans to obtain the value of p when the tip beginsto yield. Consider γ < π/4 and γ > π/4 separately.

Author Index

Adkins, J. E., 73, 96Antman, S., 161

Barber, J. R., 161Batchelor, G. K., 3, 87, 180Batra, R., 3Bernstein, B., 197Biot, M. A., 197Bird, R. B., 87Boles, M., 126Borgnakke, C., 126

Cengel, Y. A., 126Chandler, D., 3Cherepanov, G. P., 140Christoffersen, J., 126Coussy, O., 225Cowin, S. C., 126

Dahler, J. S., 96

Eringen, A. C., 3, 73, 197

Falter, C., 140Finney, R. L., 47Fredrickson, A. G., 180Frenkel, D., 3Fung, Y. C., 3, 24, 38, 197

Geurts, K. R., 96Gidaspow, D., 180Goodier, J. N., 161Green, A. E., 73, 161Greenberg, M. D., 47

Hill, R., 225

Jaunzemis, W., 3, 96Jeffreys, H., 24Johnson, W., 225

Kachanov, L. M., 225

Kaye, A., 197Kearsley, E. A., 197Knowles, J. K., 24Krajcinovic, D., 225Kreyzig, E., 47

Landau, L. D., 161Levich, V. G., 180Lifshitz, E. M., 161Lightfoot, E. N., 87Lubliner, J., 225Ludwig, W., 140

Malvern, L. E., 3, 140Margenau, H., 126Mehrabadi, M. M., 126Mellor, P. B., 225Murphy, G. M., 126

Nemat-Nasser, S., 126, 225Noll, W., 3, 140

Ogden, R. W., 161

Panton, R. L., 180Prager, W., 225

Rabotnov, Y. N., 197Rajagopal, K. R., 225Raw, G., 3Reddy, J. N., 3Redheffer, R. M., 47Rice, J. R., 161, 225Richmond, O., 225Rivlin, R. S., 96

Schild, A., 38Schlischting, H., 180Scriven, L. E., 96Smit, B., 3Smith, G. F., 96

229

230 Author Index

Sokolnikoff, I. S., 38, 47Sonntag, R. E., 126Spencer, A. J. M., 96Spitzig, W. A., 225Srinivasa, A. R., 225Stewart, W. E., 87Swoboda, G., 225Synge, J. L., 38

Tamma, K. K., 96Tanner, R. I., 197Temple, G., 24Tham, L. G., 225Thomas, G. B., 47Timoshenko, S. P., 161Toupin, R. A., 3, 140Truesdell, C., 3, 140

Ulm, F.-J., 225

Valanis, K. C., 225van der Merwe, A., 3Van Wylen, G., 126VanArsdale, W. E.,

96

Wedgewood, L. E., 96

Yang, Q., 225Yourgrau, W., 3

Zapas, L. J., 197Zerna, W., 73, 161Zhou, X., 96Ziegler, H., 225

Subject Index

acceleration, 88, 93, 107, 112, 168centrifugal, 173convective, 179

adiabatic, 114, 121adiabatic process, 156Adkins, 72, 96aerodynamics, 164Airy, 142Almansi strain, 55alternator, 6angular momentum, 91, 92

balance of, 120angular velocity, 173anisotropic, 159anisotropy, 132antisymmetric, 5apex, 39array, 5, 121associated flow rule, 215atoms, 2axiom of continuity, 50axisymmetric, 75axisymmetric flow, 169

bar, 127circular, 144

barotropic, 179base vector

contravariant, 29covariant, 29reciprocal, 29

Bauschinger effect, 200, 216Bernoulli equation, 179Bernstein, 197Bianchi identity, 36biharmonic equation, 161bilinear relations, 201Bingham fluids, 181binormal, 39biorthogonality, 26Biot, 192, 194

Biot’s theory, 192Boltzmann superposition integrals, 189boundary, 47Boussinesq form, 135bracket group, 124bulk modulus, 162, 190

caloric equation of state, 121cantilever, 227Cauchy, 95, 142Cauchy–Riemann equations, 48Cayley–Hamilton theorem, 20, 22, 65, 134,

137chain molecules, 157chain rule, 13Chandler, 3characteristic equation, 21, 61, 65chemical potential, 114Cherepanov, 140Christoffel symbols, 34Christoffersen, 115Clausius–Duhem inequality, 118, 119, 138cofactor, 7, 52, 69, 82colliding, 129commutative property, 35compatibility, 36, 70compatibility equations, 71complex modulus, 198component

physical, 33tangential, 45

compressible, 143, 175concentrations, 192conductivity, 119configuration

deformed, 49undeformed, 49

conjugate, 135, 222consistency, 130constitutive relations, 129contact transformations, 121

231

232 Subject Index

contravariant component, 31convective heat transfer, 164convective rate, 77convex, 42convexity, 224coordinate

Cartesian, 2curvilinear, 54cylindrical, 75principal, 21

coordinate transformation, 27coordinates

curvilinear, 2material, 53polar, 39, 146, 172reflections of, 132spatial, 53spherical, 40

corotational, 104correspondence principle, 199Cosserat stress, 101Couette flow, 172counterclockwise, 59couple stress, 101Coussy, 226covariant, 211covariant components, 31covariant derivative, 34Cowin, 115crack tips, 212creep, 184

primary, 194secondary, 194tertiary, 194

creep compliance function, 184cross-linked, 157cross-products, 9cubic, 61curl, 15curvature, 39, 227

Einstein, 36Riemann–Christoffel, 71

cylindrical shaft, 161

damage in materials, 221dashpot, 184deformation, 2, 47, 129deformation gradient, 51, 111, 133, 141, 145

Cauchy, 196Green’s, 196

deformation rate, 81density, 1, 13derivative, 13

absolute, 35determinant, 7, 19deviatoric, 166differential tube, 42dilatation, 145dimension, 1

directional derivative, 179discrete, 2dissipation, 227dissipation function, 193distortion, 58, 70, 111divergence, 15dot, 9double-dot notation, 18Drucker’s inequality, 214dummy index, 5dyadic notation, 15

eigenvalue, 20, 22, 102eigenvalue problem, 61eigenvector, 20Einstein summation convention, 4elastic limit, 200elastic perfectly plastic, 206elasticity, 135

finite, 142incremental, 150integrated, 150

elastomers, 157ellipsoid, 20, 62, 107

material, 87spatial, 87

ellipsoid of Cauchymaterial, 60spatial, 60

empirical approach, 143endochronic, 219endochronic theory, 221energy, 90, 129

balance of, 96, 114, 222distortion, 203internal, 115, 155kinetic, 115potential, 154strain, 112, 120surface, 123

enthalpy, 122entropy, 90, 117, 129, 193

specific, 117surface, 123

equationcharacteristic, 19cubic, 19Gibbs, 117, 121

equation of state, 129equations of motion, 4equilibrium, 38, 114equilibrium equation, 40Eringen, 72, 148, 165, 195Eshelby force, 161Euclidean, 2, 4Euclidean space, 5Euler’s theorem, 223Euler–Lagrange equation, 153Eulerian, 50

Subject Index 233

eversion, 74evolution equations, 193, 223exclusion, 130expansion

column, 7row, 7

extension, 58extremum, 18

factor, 23fading memory, 130fan, 213fiber, 74finger, 54finite-difference, 3finite-element, 3finite-element method, 142finite-strain, 135fluid, 129

barotropic, 167compressible, 176ideal, 167incompressible, 138inviscid, 167, 176Newtonian, 3, 127, 164, 165second-order, 165Stokes, 165, 184zeroth-order, 167

fluxes, 119force

oscillating, 198forces, 2

contact, 98external, 98, 129generalized, 119inertia, 176viscous, 176

Fourier’s law, 118, 176fracture mechanics, 138fracture surfaces, 43frame indifferent, 93free energy

Gibbs, 122, 155, 222Helmholtz, 122, 135, 155, 160, 222

free index, 5Frenet–Serret formula, 39Frenkel, 3Frobenius method, 65functionals, 130, 133fundamental form

first, 39second, 39

fundamental laws, 90, 95Fung, 1, 24, 194

Gauss, 2, 42Gauss theorem, 44, 106, 108, 116,

139Gaussian curvature, 39

Geiringer equations, 212general relativity, 36geometry, 2Germain, 142Geurts, 95global, 106Goodyear, 157gradient, 12gradient operator, 39, 40Green, 2, 44, 72grid lines, 27

Hagen–Poiseuille flow, 169Hamel form, 141Heaviside step function, 185helix, 39Hencky’s theory, 219hereditary integral, 197hereditary kernels, 191heredity, 130hexagonal packing, 132history, 200hodograph transformation,

183homogeneous, 115Hooke, 142Hookean solid, 190hyperelastic, 118, 121, 127, 135hypersurface, 208

ideal gas, 126, 176ideal rubber, 127identity, e–δ, 7Ilıushin’s postulate, 216incompressible, 127, 135, 143,

175incompressible flow, 36, 88index notation, 2, 4, 11indices

Greek, 49, 195Latin, 49, 195

inelastic, 200infinite domain, 48inhomogeneous, 134integral

area, 83hereditary, 132line, 82surface, 108volume, 83

integrands, 107integrity basis, 95intercept, 122intrinsic time, 221invariance

coordinate, 130dimensional, 130material, 130spatial, 130

234 Subject Index

invariant, 2, 19, 61, 90, 121, 132second, 112

invariant group, 95invariant surface integrals, 138inverse method, 143inverse solutions, 3inverses, 51inversion integral, 182irreversible, 114, 127, 137, 158irreversible process, 92irrotational, 88, 179isentropic, 121isentropic flow, 178isolated system, 192isothermal, 121, 135, 162, 175isotropic hardening, 216isotropy, 132, 160

J integral, 161, 162Jacobian determinant, 51Jacobians, 123Jaumann stress rate, 104

K-BKZ theory, 196Kayes, 197Kearsley, 197Kelvin, 156Kelvin–Voigt solid, 184kernel, 130kinematic hardening, 218kinematic viscosity, 175kinetic energy, 41Kirchhoff, 142Kolosoff, 150Krajcinovic, 224Kronecker delta, 6, 29, 51Kutta–Joukowski theorem, 180

Lagrangian, 41Lame constants, 149, 162lamina, 16laminar flows, 176Laplace transform, 189Legendre transformations, 121Levy–Mises flow rule, 207linear elastic, 3linear momentum, 90local form, 116logarithmic spin rate, 106Love, 142

Magnus effect, 180major axis, 20Margenau, 123mass, 1, 90

conservation of, 91, 96, 164material

Cauchy elastic, 137Green’s elastic, 137

Hookean, 161hyperelastic, 112simple, 131

material force, 161material particle, 2materials

anisotropic, 134Cauchy elastic, 141elastic, 3, 133, 135granular, 121Hookean, 149hyperelastic, 141, 142, 184isotropic, 149Mooney, 143Mooney–Rivlin, 143rubberlike, 143viscoelastic, 133, 184

matrixmodal, 21orthogonal, 63positive-definite, 20rotation, 63skew-symmetric, 7square, 10, 20, 63strain, 63

matrix notation, 15, 18matter, 1, 2maximum, 44mean free path, 2mechanics, 2

computational, 27continuum, 1solid, 123statistical, 3, 118, 159

memory kernel, 197meridian angle, 39metal, 200method of characteristics, 178, 210metric tensor, 39microcracks, 222microstructure, 54, 115minimum, 44minimum dissipation rate, 224minor axis, 20mixed component, 31mixtures, 121modified Bessel function, 182modified function, 18modulus of rigidity, 150Mohr’s circle, 209molecular dynamics, 3molecules, 1moment of momentum, 91moments

contact, 98surface, 115

momentum, 90momentum flux, 183monomer, 157

Subject Index 235

Monte Carlo methods, 3Morera’s theorem, 48motion, 2

equations of, 108Mroz theory, 218multiaxial state of stress, 194Murnaghan form, 141Murphy, 123Mushkelishvili, 150

natural state, 137Navier, 142Navier equation, 149Navier–Stokes equations, 3, 175neighborhood, 130Neumann–Kirchhoff form, 141neutral axis, 227neutral stability, 214Newton’s law of viscosity, 119no-slip condition, 165nonassociative flow rules, 215nonlinear, 132, 223nonsingular, 28, 51normal, 27

outward, 98normalization, 18numerical solution, 3

objective, 104, 131objectivity, 93observer independence, 93observer transformation, 93off-diagonal, 84off-diagonal components, 101off-diagonal elements, 58Oldroyd, 106one-to-one, 27, 123Onsager coefficients, 224Onsager’s principle, 119operator

gradient, 15orthogonal, 4, 20, 27, 62, 102, 132, 133orthogonal group, 132orthotropy, 96, 132, 160Ostrogradsky, 44

parabolic distribution, 171paraboloid, 39parallelepiped, 68parallelogram, 40particle, 41path line, 77perfectly plastic, 200permanent deformation, 200permutation symbol, 6, 30, 31permutation tensors, 31permutations

even, 6odd, 6

phenomenological coefficients, 138, 164, 165,172

phenomenology, 129physical components, 40Piola, 54Piola–Kirchhoff stress

first, 109second, 110

pipe, 170plane

coordinate, 99inclined, 99

plastic flow, 206contained, 202uncontained, 202

plasticity, 3, 200Poisson’s equation, 48Poisson’s ratio, 150polar coordinates, 40polycrystalline, 203, 221polyisoprene, 157polymers, 94polynomial, 148positive definite, 54, 63positive semidefinite, 118potential, 41power

mechanical, 115stress, 116

Poynting effect, 146, 148Prager, 217Prandtl–Reuss flow rule, 207pressure, 1

hydrostatic, 13, 103, 120, 147, 190, 197,203

mechanical, 138thermodynamic, 137, 165, 166

Prigogine, 118principal, 102principal values, 20principle

invariance, 129maximum entropy, 121minimum energy, 121

productcross, 12, 15, 44dot, 11, 12, 15, 44inner, 12, 18, 95, 121scalar, 12, 29, 68tensor, 15, 44triple, 29vector, 12

propellers, 164proportional loading, 225pure shear, 203

quadratic form, 18quasi-linear equations, 178quotient rule, 17

236 Subject Index

rank, 14, 44Raw, 3reciprocal deformation tensors, 54reference state, 155relations

constitutive, 3, 129kinematic, 129, 165stress–strain, 142

relaxation time, 185reversible, 117Reynolds number, 176Ricci curvature, 36Rice, 222Richmond, 203Riemann curvature, 35Riemann invariants, 179, 182, 211Riemann metric tensor, 29Riemann–Christoffel curvature, 35, 39rigid body, 95Rivlin, 96Rivlin–Ericksen tensor, 196rotation, 57, 106, 132

rigid body, 146rubber, 157

scalar function, 32semimajor axis, 65semiminor axis, 67Shaw, 123shear, 146shear flow, 112shear strain

engineering, 60mathematical, 60

shock waves, 43shorthand notation, 124simple materials, 54singularities, 140skew symmetric, 5slip-line field, 213smeared, 2Smit, 3Smith, 96solid mechanics, 3space, 1specific heat, 177specimen, 200spectral representation, 21, 63spin, 85, 94spin tensor, 81Spitzig, 203spring, 184square root, 22St. Venant, 47, 142stagnation point, 77standard linear solid, 188stationary, 18, 103steady flow, 87Stefani model, 226stiffness, 160

Stokes, 2, 42Stokes fluids, 133Stokes theorem, 87strain, 2, 63

Almansi, 148finite, 71inelastic, 223Lagrange, 133logarithmic, 68plane, 150

strain cooling, 156strain hardening, 200, 214strain tensors, 55streak lines, 78stream function, 181stream line, 78strength of materials, 1stress, 2

Cauchy, 100, 106, 110, 130, 133couple, 116deviatoric, 103, 112, 203hydrostatic, 137, 190Kirchhoff, 110, 113Piola–Kirchhoff, 108, 120, 127, 134plane, 150shear, 141

stress rateGreen–McInnis–Naghdi, 106Jaumann, 105, 196objective, 104Oldroyd, 112Truesdell, 106, 112

stress relaxation function, 185stress–strain rate relations, 3stress–strain relations, 3, 132stretch, 58stretching, 84, 166sublayer theory, 218submatrix, 68substantial derivative, 76summation convention, 4, 11superposition principle, 184superscripts, 27surface, 42

cylindrical, 145swell, 172swept-wing airplane, 40Swoboda, 222system, 115

Cartesian, 27curvilinear, 27

Tamma, 94, 106tangent, 27tangent plane, 29Tanner, 197temperature, 1, 13, 129, 135, 154, 159, 167, 175, 222tensor, 14

Almansi strain, 72compliance, 159

Subject Index 237

Eulerian, 71general, 2, 27Lagrangian, 71metric, 40objective, 95, 137second-rank, 14spin, 83, 104, 105, 112strain, 55, 114stress, 2, 15, 38, 114stretch, 133two-point, 71

tensor calculus, 33tetrahedron, 41, 99Tham, 222theory of relativity, 27thermodynamic tensions, 135thermodynamics, 1, 106, 114

classical, 114irreversible, 3, 117

thermoelastic, 156thermoelasticity, 127, 142, 156, 162thermomechanical, 3thermostatics, 92thin-walled tubes, 205, 225torque, 148, 174torsion, 39, 47, 205traction, 37, 99tractions

surface, 115transpose, 11, 15transverse isotropy, 96, 160triad, 65triple, 95trivial solution, 19Truesdell, 106truss, 151turbomachinery, 164turbulent flows, 176twirl, 87twist, 148

Ulm, 226uniaxial tension, 203unit step function, 184universal gas constant, 159updated Lagrangian method,

151

Valanis, 219, 221van der Merwe, 3van der Waals gas, 181VanArsdale, 94variable

internal, 208variables

complex, 150conjugate, 121extensive, 114

independent, 131intensive, 114intermediate, 195internal, 222macroscopic, 222state, 114, 192thermodynamic, 124

variational principle, 154variational problem, 153vector, 12

absolute, 10base, 9Cauchy stress, 108column, 9, 10displacement, 51distance, 28force, 108gradient, 13moment stress, 108position, 8, 11, 12, 39traction, 99, 102, 112unit, 9velocity, 15, 37

vector area, 46vector function, 35, 47velocity, 77

angular, 115virtual work, 153viscoelasticity, 3viscoplasticity, 219viscosity, 175, 184

dilatational, 176shear, 176

void fraction, 121Voigt notation, 160, 191volume, 1volume integral, 48vorticity, 83, 86vulcanization process, 157

warping function, 47Wedgewood, 95Weissenberg effect, 174work hardening, 200, 202

Yang, 222yield condition

Tresca, 203von Mises, 203

yield function, 216yield stress, 200Young’s modulus, 150Yourgrau, 3

Zapas, 197Zerna, 72Zhou, 94, 106Ziegler, 218

cambridge introduction to continuum mechanics

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