35737861 Schaum Continuum Mechanics Mase

SCHAUM'S OUTLINE O F

THEORY AND PROBLEMS

CONTINUUM MECHANICS

GEORGE E. MASE, Ph.D. Professor o f Mechanics

Michigan State University

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ISBN 07-04M63-4

Preface

Because of its emphasis on basic concepts and fundamental principles, Continuum Mechanics has an important role in modern engineering and technology. Severa1 undergraduate courses which utilize the continuum concept and its dependent theories in the training of engineers and scientists are well established in today's curricula and their number continues to grow. Graduate programs in Mechanics and associated areas have long recognized the value of a substantial exposure to the subject. This book has been written in an attempt to assist both undergraduate and first year graduate students in understanding the fundamental principles of continuum theory. By including a number of solved problems in each chapter of the book, i t is further hoped that the student will be able to develop his skill in solving problems in both continuum theory and its related fields of application.

In the arrangement and development of the subject matter a sufficient degree of continuity is provided so that the book may be suitable as a text for an introductory course in Continuum Mechanics. Otherwise, the book should prove especially useful as a supplementary reference for a number of courses for which continuum methods provide the basic structure. Thus courses in the areas of Strength of Materials, Fluid Mechanics, Elasticity, Plasticity and Viscoelasticity relate closely to the substance of the book and may very well draw upon its contents.

Throughout most of the book the important equations and fundamental relationships are presented in both the indicial or "tensor" notation and the classical symbolic or "vector" notation. This affords the student the opportunity to compare equivalent expressions and to gain some familiarity with each notation. Only Cartesian tensors are employed in the text because i t is intended as an introductory volume and since the essence of much of the theory can be achieved in this context.

The work is essentially divided into two parts. The first five chapters deal with the basic continuum theory while the final four chapters cover certain portions of specific areas of application. Following an initial chapter on the mathematics relevant to the study, the theory portion contains additional chapters on the Analysis of Stress, Deforma- tion and Strain, Motion and Flow, and Fundamental Continuum Laws. Applications are treated in the final four chapters on Elasticity, Fluids, Plasticity and Viscoelasticity. At the end of each chapter a collection of solved problems together with severa1 exercises for the student serve to illustrate and reinforce the ideas presented in the text.

The author acknowledges his indebtedness to many persons and wishes to express his gratitude to a11 for their help. Special thanks are due the following: to my colleagues, Professors W. A. Bradley, L. E. Malvern, D. H. Y. Yen, J. F. Foss and G. LaPalm each of whom read various chapters of the text and made valuable suggestions for improvement; to Professor D. J. Montgomery for his support and assistance in a great many ways; to Dr. Richard Hartung of the Lockheed Research Laboratory, Palo Alto, California, who read the preliminary version of the manuscript and gave numerous helpful suggestions; to Professor M. C. Stippes, University of Illinois, for his invaluable comments and suggestions; to Mrs. Thelma Liszewski for the care and patience she displayed in typing the manuscript; to Mr. Daniel Schaum and Mr. Nicola Monti for their continuing interest and guidance throughout the work. The author also wishes to express thanks to his wife and children for their encouragement during the writing of the book.

Michigan State University June 1970

C O N T E N T S

Chapter 1 1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

1.10

1.11

1.12

1.13

Page MATHEMATICAL FOUNDATIONS

Tensors and Continuum Mechanics .................................... 1

General Tensors . Cartesian Tensors . Tensor Rank ...................... 1

Vectors and Scalars .................................................. 2

. . . . . . . . . . . . . . . . . . Vector Addition . Multiplication of a Vector by a Scalar 2

Dot and Cross Products of Vectors .................................... 3

Dyads and Dyadics .................................................... 4

.................. . . Coordinate Systems Base Vectors Unit Vector Triads G

Linear Vector Functions . Dyadics a s Linear Vector Operators .......... 8

Indicial Notation . Range and Summation Conventions .................. 8

Summation Convention Used with Symbolic Convention .................. 10

Coordinate Transformation . General Tensors .......................... 11

The Metric Tensor . Cartesian Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

Transformation Laws for Cartesian Tensors . The Kronecker Delta . Orthogonality Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

Addition of Cartesian Tensors . Multiplication by a Scalar ................ 15

Tensor Multiplication .................................................. 15

Vector Cross Product . Permutation Symbol . Dual Vectors . . . . . . . . . . . . . . 16

Matrices . Matrix Representation of Cartesian Tensors .................. 17

............................ Symmetry of Dyadics. Matrices and Tensors 19

Principal Values and Principal Directions of Symmetric Second-Order Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

............ Powers of Second-Order Tensors . Hamilton-Cayley Equation 21 .................................. Tensor Fields . Derivatives of Tensors 22

...................................... Line Integrals . Stokes' Theorem 23

The Divergence Theorem of Gauss ...................................... 23

Chapter 2 ANALYSIS OF STRESS

The Continuum Concept ................................................ 44

Homogeneity . Isotropy . Mass-Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

Body Forces . Surface Forces .......................................... 45 . . . . . . . . . . . . . . . . . . . . . . . . . . . . Cauchy's Stress Principle . The Stress Vector 45

................................ State of Stress a t a Point . Stress Tensor 46 . . . . . . . . . . . . . . . . . . . . . . . . . . The Stress Tensor-Stress Vector Relationship 47

. . . . . . . . . . . . . . Force and Moment . Equilibrium . Stress Tensor Symmetry 48 .......................................... Stress Transformation Laws 49

Stress Quadric of Cauchy .............................................. 50

. . . . . . . . . . . . . . . . . . Principal Stresses . Stress Invariants . Stress Ellipsoid 51 . . . . . . . . . . . . . . . . . . . . . . . . . . . . Maximum and Minimum Shear Stress Values 52

Mohr's Circles fo r Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

Plane Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Deviator and Spherical Stress Tensors 57

C O N T E N T S

Chapter 3 3.1

3.2

3.3

3.4

3.5

3.6

3.7

3.8

3.9

3.10

3.11

3.12

3.13

3.14

3.15

3.16

Page DEFORMATION AND STRAIN Particles and Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

Coritinuum Configuration . Deformation and Flow Concepts . . . . . . . . . . . . . . 77

Position Vector . Displacement Vector .................................. 77

Lagrangian and Eulerian Descriptions .................................. 79

Deformation Gradients . Displacement Gradients ........................ 8 0

Deformation Tensors . Finite Strain Tensors ............................ 81

Small Deformation Theory . Infinitesimal Strain Tensors ................ 82

Relative Displacements . Linear Rotation Tensor . Rotation Veztor . . . . . . . . 83

Interpretation of the Linear Strain Tensors ............................ 84

Stretch Ratio . Finite Strain Interpretation ............................ 86

Stretch Tensors . Rotation Tensor ...................................... 87

Transformation Properties of Strain Tensors ............................ 88

Principal Strains . Strain Invariants . Cubical Dilatation ................ 89

Spherical and Deviator Strain Tensors .................................. 91

Plane Strain . Mohr's Circles fo r Strain ................................ 91

.............................. Compatibility Equations for Linear Strains 92

Chapter 4 4.1 4.2

4.3

4.4

4.5

4.6

4.7

MOTION AND FLOW . . .................................... Motion Flow Material Derivative 110

.................... Velocity . Acceleration . Instantaneous Velocity Field 111

. . .............................. Path Lines Stream Lines Steady Motion 112

........................ . . Rate of Deformation Vorticity Natural Strain 112

. . . . . . Physical Interpretation of Rate of Deformation and Vorticity Tensors 113

.................. Material Derivatives of Volume. Area and Line Elements 114

.............. Material Derivatives of Volume, Surface and Line Integrals 116

Chapter 5 5.1

5.2

FUNDAMENTAL LAWS OF CONTINUUM MECHANICS Conservation of Mass . Continuity Equation ............................ 126

Linear Momentum Principle . Equations of Motion . Equilibrium Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

Moment of Momentum (Angular Momentum) Principle . . . . . . . . . . . . . . . . . . . . 128

Conservation of Energy . Fi rs t Law of Thermodynamics . Energy Equation .................................................. 128

Equations of State . Entropy . Second Law of Thermodynamics . . . . . . . . . . . . 130

The Clausius-Duhem Inequality . Dissipation Function . . . . . . . . . . . . . . . . . . . . 131

Constitutive Equations . Thermomechanical and Mechanical Continua . . . . . . 132

Chapter 6 LINEAR ELASTICITY . . . . . . . . . . . . . . . . . . . . . . . 6.1 Generalized Hooke's Law Strain Energy Function 140

................................ . . 6.2 Isotropy Anisotropy Elastic Symmetry 141

. .................................... 6.3 Isotropic Media Elastic Constants 142

. . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Elastostatic Problems Elastodynamic Problems 143

6.5 Theorem of Superposition . Uniqueness of Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . S t Venant's Principle 145

6.6 Two-Dimensional Elasticity . Plane Stress and Plane Strain . . . . . . . . . . . . . . 145

C O N T E N T S

Page

6.7 Airy's Stress Function ................................................ 147

6.8 Two-Dimensional Elastostatic Problems in Polar Coordinates . . . . . . . . . . . . . . 148

6.9 Hyperelasticity . Hypoelasticity ........................................ 149

6.10 Linear Thermoelasticity ................................................ 149

Chapter 7 FLUIDS 7.1 Fluid Pressure . Viscous Stress Tensor . Barotropic Flow ................ 160

7.2 Constitutive Equations . Stokesian Fluids . Newtonian Fluids ............ 161

7.3 Basic Equations fo r Newtonian Fluids . Navier-Stokes-Duhem Equations .... 162

7.4 Steady Flow . Hydrostatics . Irrotational Flow .......................... 163

7.5 Perfect Fluids . Bernoulli Equation . Circulation ........................ 164

7.6 Potential Flow . Plane Potential Flow .................................. 165

Chapter 8 8.1

8.2

8.3

8.4

8.5

8.6

8.7

8.8

$8.9

8.10

8.11

Chapter 9 9.1

9.2

9.3

9.4

9.5

9.6

9.7

9.8

PLASTICITY ........................................ Basic Concepts and Definitions 175

.............................................. Idealized Plastic Behavior 176

........................ Yield Conditions . Tresca and von Mises Criteria 177

............................ Stress Space . The 1I.Plane . Yield Surfaces 173

Post-Yield Behavior . Isotropic and Kinematic Hardening ................ 180

Plastic Stress-Strain Equations . Plastic Potential Theory ................ 181

.................. Equivalent Stress . Equivalent Plastic Strain Increment 182

............................ Plastic Work . Strain Hardening Hypotheses 182

Total Deformation Theory .............................................. 183

................................................ Elastoplastic Problems 183

.................... Elementary Slip Line Theory for Plane Plastic Strain 184

VISCOELASTICITY Linear Viscoelastic Behavior ............................................ 196

............................................ Siniple Viscoelastic Models 196 . . . . . . . . . . . . . . Generalized Models . Linear Differential Operator Equation 198

Creep and Relaxation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199 . . . . . . . . . . . . . . Creep Function . Relaxation Function . Hereditary lntegrals 200

...................................... Complex Moduli and Compliances 202

Three Dimensional Theory .............................................. 203 . . . . . . . . . . . . . . . . . . Viscoelastic Stress Analysis . Correspondence Principle 204

Chapter 1

Mathematical Foundations

1.1 TENSORS AND CONTINUUM MECHANICS

Continuum mechanics deals with physical quantities which are independent of any particular coordinate system that may be used to describe them. At the same time, these physicaI quantities a re very often specified most conveniently by referring to a n appropriate system of coordinates. Mathematically, such quantities are represented by tensors.

As a mathematical entity, a tensor has an existence independent of any coordinate system. Yet i t may be specified in a particular coordinate system by a certain set of quantities, known as its components. Specifying the components of a tensor in one coordinate system determines the components in any other system. Indeed, the law of transformation of the components of a tensor is used here as a means fo r definicg the tensor. Precise statements of the definitions of various kinds of tensors are given a t the point of their introduction in the material that follows.

The physical laws of continuum mechanics a re expressed by tensor equations. Because tensor transformations are linear and homogeneous, such tensor equations, if they are valid in one coordinate system, are valid in any other coordinate system. This invarz'unce of tensor equations under a coordinate transformation is one of the principal reasons for the usefulness of tensor methods in continuum mechanics.

1.2 GENERAL TENSORS. CARTESIAN TENSORS. TENSOR RANK.

In dealing with general coordinate transformations between arbitrary curvilinear coordinate systems, the tensors defined are known as general tensors. When attention is restricted to transformations from one homogeneous coordinate system to another, the tensors involved are referred to as Cartesian tensors. Since much of the theory of continuum mechanics may be developed in terms of Cartesian tensors, the word "tensor" in this book means "Cartesian tensor" unless specifically stated otherwise.

Tensors may be classified by rank , or order, according to the particular form of the transformation law they obey. This same classification is also reflected in the number of components a given tensor possesses in an n-dimensional space. Thus in a three-dimensional Euclidean space such as ordinary physical space, the number of components of a tensor is 3N, where N is the order of the tensor. Accordingly a tensor of order zero is specified in any coordinate system in three-dimensional space by one component. Tensors of order zero are called scalars. Physical quantities having magnitude only are represented by scalars. Tensors of order one have three coordinate components in physicaI space and are known as vectors. Quantities possessing both magnitude and direction are represented by vectors. Second-order tensors correspond to dyadics. Severa1 important quantities in continuum mechanics are represented by tensors of rank two. Higher order tensors such as triudics, or tensors of order three, and tetradics, or tensors of order four are also defined and appear often in the mathematics of continuum mechanics.

2 MATHEMATICAL FOUNDATIONS [CHAP. 1

1.3 VECTORS AND SCALARS Certain physical quantities, such as force and velocity, which possess both magnitude

and direction, may be represented in a three-dimensional space by directed line segments that obey the parallelogram law of addition. Such directed line segments are the geometrical representations of first-order tensors and are called vectors. Pictorially, a vector is simply an arrow pointing in the appropriate direction and having a length proportional to the magnitude of the vector. Equal vectors have the same direction and equal magnitudes. A unit vector is a vector of unit length. The nu11 or zero vector is one having zero length and an unspecified direction. The negative of a vector is that vector having the same magnitude but opposite direction.

Those physical quantities, such as mass and energy, which possess magnitude only are represented by tensors of order zero which are called scalars.

In the symbolic, or Gibbs notation, vectors are designated by bold-faced letters such as a, b, etc. Scalars are denoted by italic letters such as a, b, A, etc. Unit vectors are further distinguished by a caret placed over the bold-faced letter. In Fig. 1-1, arbitrary vectors a and b are shown along with the unit vector ê and the pair of equal vectors c and d.

Fig. 1-1

The magnitude of an arbitrary vector a is written simply as a, or for emphasis it may be denoted by the vector symbol between vertical bars as Ia].

1.4 VECTOR ADDITION. MULTIPLICATION OF A VECTOR BY A SCALAR Vector addition obeys the parallelogram law, which defines the vector sum of two vectors

as the diagonal of a parallelogram having the component vectors as adjacent sides. This law for vector addition is equivalent to the triangle rzde which defines the sum of two vectors as the vector extending from the tail of the first to the head of the second when the summed vectors are adjoined head to tail. The graphical construction for the addition of a and b by the parallelogram law is shown in Fig. 1-2(a). Algebraically, the addition process is expressed by the vector equation

a + b = b f a = c (1.1)

Vector subtraction is accomplished by addition of the negative vector as shown, for example, in Fig. 1-2(b) where the triangle rule is used. Thus

a - b = - b + a = d (1 -2) The operations of vector addition and subtraction are commutative and associative as illustrated in Fig. 1-2(c), for which the appropriate equations are

( a+b) + g = a + (b+g) = h (1.3)

Fig. 1-2

CHAP. 11 MATHEMATICAL FOUNDATIONS 3

Multiplication of a vector by a scalar produces in general a new vector having the same direction as the original but a different length. Exceptions are multiplication by zero to produce the nu11 vector, and multiplication by unity which does not change a vector. Multi- plication of the vector b by the scalar m results in one of the three possible cases shown ip Fig. 1-3, depending upon the numerical value of m.

O < m < l

Fig. 1-3

Multiplication of a vector by a scalar is a,ssociative and distributive. Thus

m(nb) = (mn)b = n(mb) (1.4)

In the important case of a vector multiplied by the reciproca1 of its magnitude, the result is a unit vector in the direction of the original vector. This relationship is expressed by the equation A

b = blb (1.7)

1.5 DOT AND CROSS PRODUCTS OF VECTORS

The dot or scalar product of two vectors a and b is the scalar

= a - b = boa = abcos8

in which 6' is the smaller angle between the two vectors as shown in Fig. 1-4(a). The dot product of a with a unit vector ê gives the projection of a in the direction of ê.

Fig. 1-4

The cross or vector product of a into b is the vector v given by

in which O is the angle less than 180" between the vectors a and b, and ê is a unit vector perpendicular to their plane such that a right-handed rotation about ê through the angle 0 carries a into b. The magnitude of v is equal to the area of the parallelogram having a and b as adjacent sides, shown shade'd in Fig. 1-4(b). The cross product is not commutative.


The scalar tr iple product is a dot product of two vectors, one of which is a cross product.

a . ( b x c ) = ( a x b ) * c = a . ( b x c ) = h (1.10)

As indicated by (1.10) the dot and cross operation may be interchanged in this product. Also, since the cross operation must be carried out first, the parentheses are unnecessary and may be deleted as shown. This product is sometimes written [abc] and called the box product. The magnitude h of the scalar triple product is equal to the volume of the parallelepiped having a, b, c as coterminous edges.

The vector triple product is a cross product of two vectors, one of which is itself a cross product. The following identity is frequently useful in expressing the product of a crossed into b x c.

a x (b x c) = (a c)b - (a b)c = w (1.11)

From (1.11), the product vector w is observed to lie in the plane of b and c.

1.6 DYADS AND DYADICS The indeterminate vector product of a and b, defined by writing the vectors in juxtaposi-

tion as ab is called a dyad. The indeterminate product is not in general commutative, i.e. ab # ba. The first vector in a dyad is known as the antecedent, the second is called the consequent. A dyadic D corresponds to a tensor of order two and may always be represented as a finite sum of dyads

D = albi + azbz + . . + a ~ b ~ (1.12)

which is, however, never unique. In symbolic notation, dyadics are denoted by bold-faced sans-serif letters as above.

If in each dyad of (1.12) the antecedents and consequents are interchanged, the resulting dyadic is called the conjugate dyadic of D and is written

Dc = blal + bzaz + + b ~ a ~ (1 .I 3)

If each dyad of D in (1.12) is replaced by the dot product of the two vectors, the result is a scalar known as the scalar o f t h e dyadic D and is written

Ds = ai-bl + az-bz + + a ~ * b ~ (1 . l4)

If each dyad of D in (1.12) is replaced by the cross product of the two vectors, the result is called the vector o f t h e dyadic D and is written

Dv = a 1 x b l + a z x b z + + a , v X b ~ (1.15)

I t can be shown that Dc, D, and Dv are independent of the representation (1.12).

The indeterminate vector product obeys the distributive laws

a(b + c) = ab + ac (1.16)

(a + b)c = ac + bc (1 .l7)

(a + b)(c + d) = ac + ad + bc + bd (1.18)

and if h and p are any scalars, (h + p)ab = Aab + pab (1.19)

(ha)b = a(hb) = hab (1 .20)


If v is any vector, the dot products v D and D v are the vectors defined respectively by

v ' D = (v.a1)bl+(v.az)bz+ + ( v * ~ N ) ~ N = i.i (1.21)

In (1.21) D is called the postfactor, and in (1.22) it is called the prefactor. Two dyadics D and E are equal if and only if for every vector v, either

v*D = v.E or D.v = E-v (1.23)

The unit dyadic, or idemfactor I, is the dyadic which can be represented as

I = êl& + i%, + êS3 (1 -24)

where 21, ê2, ê3 constitute any orthonormal basis for three-dimensional Euclidean space (see Section 1.7). The dyadic I is characterized by the property

1.v = v - I = v (1.25) for a11 vectors v.

The cross products v x D and D x v are the dyadics defined respectively by

v x D = (v X a1)bl + (vX az)bz + + (vX a ~ ) b ~ = F (1.2 6)

D X v = al(b1 X v) + az(bz X v) + + a ~ ( b ~ X v) = G (1.27)

The dot product of the dyads ab and cd is the dyad defined by

ab* cd = (b. c)ad (1.28)

From (1.28), the dot product of any two dyadics D and E is the dyadic

D E = (albl+ azbz + . - + aNb~) (cidi + czdz + . - + C N ~ N )

= (bi ci)a1dl + (bi cz)ald~ + . . ( b ~ C N ) ~ N ~ N = G (1 -29)

The dyadics D and E are said to be reciproca1 of each other if

E - D = D - E = I (1 30)

Fór reciprocal dyadics, the notation E = D-I and D = E-' is often used.

Double dot and cross products are also defined for the dyads ab and cd as follows,

ab : cd = (a c)(b d) = A, a scalar (1.31)

ab cd = (a x c)(b d) = h, a vector (1.32)

ab cd = (a. c)(b x d) = g, a vector (1.33)

ab c cd = (a x c)(b x d) = uw, a dyad (1.34)

From these definitions, double dot and cross products of dyadics may be readily developed. Also, some authors use the double dot product defined by

a b - . c d = (b*c)(a.d) = A, ascalar (1.35)

A dyadic D is said to be self-conjugate, or symmetric, if

D = D,

and anti-self-conjugate, or anti-symmetric, if

Every dyadic may be expressed uniquely as the sum of a symmetric and anti-symmetric dyadic. For the arbitrary dyadic D the decomposition is

D = ?j(D + Dc) + ?j(D - Dc) = G + H (1.38)


for which G, = &(D, + (D,),) = +(D, + D) = G (symmetric) (1.39)

and H, = *(Dc - (D,),) = +(D, - D) = -H (anti-symmetric) (1.40)

Uniqueness is established by assuming a second decomposition, D = G* + H*. Then

and the conjugate of this equation is G * - H * = G - H

Adding and subtracting (1.41) and (1.42) in turn yields respectively the desired equalities, G* = G and H* = H.

1.7 COORDINATE SYSTEMS. BASE VECTORS. UNIT VECTOR TRIADS

A vector may be defined with respect to a particular coordinate system by specifying the components of the vector in that system. The choice of coordinate system is arbitrary, but in certain situations a particular choice may be advantageous. The reference system of coordinate axes provides units for measuring vector magnitudes and assigns directions in space by which the orientation of vectors may be determined.

The well-known rectangular Cartesian coordinate system is often represented by the mutually perpendicular axes, Oxyz shown in Fig. 1-5. Any vector v in this system may be expressed as a linear combination of three arbitrary, nonzero, noncoplanar vectors of the system, which are called base vectors. For base vectors a, b, c and suitably chosen scalar coefficients h, p, V the vector v is given by

V = ha + pb + vc (1.43)

Base vectors are by hypothesis linearly independent, i.e. the equation

ha + pb + vc = O (1.44)

is satisfied only if h = p = V = O. A set of base vectors in a given coordinate system is said to constitute a bash for that system. Fig. 1-5

The most frequent choice of base vectors for the rectangular Cartesian system is the A A A

set of unit vectors i, j, k along the coordinate axes as shown in Fig. 1-5. These base vectors constitute a right-handed unit vector triad, for which

$ $ A A A C A A h

i x j = k, j x k = i, k x i = j (1 .45)

A A A A A A

and i . ; j - j = k . k = 1

Such a set of base vectors is often called an orthonormal bash. A A A

In terms of the unit triad i, j, k, the vector v shown in Fig. 1-6 below may be expressed by

in which the Cartesian components


are the projections of v onto the coordinate axes. The unit vector in the direction of v is given according to (1.7) by

ê" = VIU = (COS a)? + (COS P ) + (COS y ) k (1.48)

Since v is arbitrary, i t follows that any unit vector will have the direction cosines of that vector as its Cartesian components.

In Cartesian component form the dot product of a and b is given by

A A A

a * b = ( a , ? + a , ~ + a , k ) * ( b , i + b , j + b ~ )

= axbx + ayby + a,b, (1.49)

For the same two vectors, the cross product a x b is

Fig. 1-6

A A A

a x b = (ayb, - a,by) i + (a,b, - a,b,) j + (a,b, - aybx)k

This result is often presented in the determinant form

in which the elements are treated as ordinary numbers. The triple scalar product may also be represented in component form by the determinant

In Cartesian component form, the dyad ab is given by c ab = (a,i +a,; + a,k)(b,? + b,; + b,k)

A A A A 9" = a,b,i i + axbyi j + a,b,i k

+ ayb,S$ + a , b y ~ ~ + ayb,$k A A + a,b,kT + a,bykj + a,b,kk (1.53)

[abc] =

Because of the nine terms involved, (1.53) is known as the nonion form of the dyad ab. It is possible to put any dyadic into nonion form. The nonion form of the idemfactor in

A A A

terms of the unit triad i, j , k is given by A A A A A A

I = i i + j j + k k (1.54)

In addition to the rectangular Cartesian coordinate system already discussed, curvilinear coordinate systems such as the cylindrical (R, O, z) and spherical ( r , O, C$) systems shown in Fig. 1-7 below are also widely used. Unit triads ($R , ês, ê,) and (ê,, 60, ê+) of base vectors illustrated in the figure are associated with these systems. However, the base vectors here do not all have fixed directions and are therefore, in general, functions of position.

a, a, a,

bx b~ b~ C , C y c ,


(a) Cylindrical (b) Spherical Fig. 1-7

1.8 LINEAR VECTOR FUNCTIONS. DYADICS AS LINEAR VECTOR OPERATORS

A vector a is said to be a function of a second vector b if a is determined whenever b is given. This functional relationship is expressed by the equation

a = f(b) (1.55)

The function f is said to be linear when the conditions

f(b + c) = f(b) + f(c) (1.56)

are satisfied for a11 vectors b and c, and for any scalar A.

Writing b in Cartesian component form, equation (1.55) becomes

c a = f(b,i + b,; + b,k) (1.58)

which, if f is linear, may be written

a = b,f(?) + b,f(;) + b,f(k) (1.59) A .? A

I n (1.59) let f ( i ) =u, f ( j ) = v , f ( k ) = w, so that now

which is recognized as a dyadic-vector dot product and may be written

a = D - b (1.61) A A A

where D = u i + v j + wk. This demonstrates that any linear vector function f may be expressed as a dyadic-vector product. In (1.61) the dyadic D serves as a linear vector operator which operates on the argument vector b to produce the image vector a.

1.9 INDICIAL NOTATION. RANGE AND SUMMATION CONVENTIONS

The components of a tensor of any order, and indeed the tensor itself, may be represented clearly and concisely by the use of the indicial notation. In this notation, letter indices, either subscripts or superscripts, are appended to the generic or kernel letter representing the tensor quantity of interest. Typical examples illustrating use of indices are the tensor symbols

ai, bj, Tij, pij, ' i j k , R'>q


In the "mixed" form, where both subscripts and superscripts appear, the dot shows that j is the second index.

Under the rules of indicial notation, a letter index may occur either once or twice in a given term. When an index occurs unrepeated in a term, that index is understood to take on the values 1,2, . . . , N where N is a specified integer that determines the range of the index. Unrepeated indices are known as free indices. The tensorial rank of a given term is equal to the number of free indices appearing in that term. Also, correctly written tensor equations have the same letters as free indices in every term.

When an index appears twice in a term, that index is understood to take on a11 the values of its range, and the resulting terms summed. In this so-called summation convention, repeated indices are often referred to as dummy indices, since their replacement by any other letter not appearing as a free index does not change the meaning of the term in which they occur. In general, no index occurs more than twice in a properly written term. If it is absolutely necessary to use some index more than twice to satisfactorily express a certain quantity, the summation convention must be suspended.

The number and location of the free indices reveal directly the exact tensorial character of the quantity expressed in the indicial notation. Tensors of first order are denoted by kernel letters bearing one free index. Thus the arbitrary vector a is represented by a symbol having a single subscript or superscript, i.e. in one or the other of the two forms,

The following terms, having only one free index, are also recognized as first-order tensor quantities:

aijbj, Fikk, rijkUjZ)k

Second-order tensors are denoted by symbols having two free indices. Thus the arbitrary dyadic D will appear in one of the three possible f o m s

In the "mixed" form, the dot shows that j is the second index. Second-order tensor quantities may also appear in various f o m s as, for example,

Aijip, B?. jk , 8ifikvk

By a logical continuation of the above scheme, third-order tensors are expressed by symbols with three free indices. Also, a symbol such as X which has no indices attached, represents a scalar, or tensor of zero order.

In ordinary physical space a basis is composed of three, noncoplanar vectors, and so any vector in this space is completely specified by its three components. Therefore the range on the index of ai, which represents a vector in physical three-space, is 1,2,3. Accordingly the symbol ai is understood to represent the three components a1,az,a3. Also, ai is sometimes interpreted to represent the ith component of the vector or indeed to represent the vector itself. For a range of three on both indices, the symbol Aij represents nine components (of the second-order tensor (dyadic) A). The tensor Aij is often presented explicitly by giving the nine components in a square array enclosed by large parentheses as

Ai1 A12 A13

Aij = (1 .62)

1 0 MATHEMATICAL FOUNDATIONS [CHAP. 1

In the same way, the components of a first-order tensor (vector) in three-space may be displayed explicitly by a row or column arrangement of the form

ai = (ai, az, as) or ai = (1.63) a3

In general, for a range of N, an nth order tensor will have Nn components.

The usefulness of the indicial notation in presenting systems of equations in compact form is illustrated by the following two typical examples. For a range of three on both i and j the indicial equation

Xi = Ci j2 j (1.64)

represents in expanded form the three equations

For a range of two on i and j, the indicial equation . Aij = BipCjqDpq (1.66)

represents, in expanded form, the four equations

Ali = BiiCiiDii + BliC12D12 + B12CilD21 + B12Ci2D22

Ai2 = BiiC2iDii + BiiC22Di2 + BIZCZID~I + Bi2CmD22 (1.67)

A21 = BSICIIDI~ + B ~ I C I ~ D I ~ + B22CiiD21 + Bz2C12D22

For a range of three on both i and j, (1.66) would represent nine equations, each having nine terms on the right-hand side.

1.10 SUMMATION CONVENTION USED WITH SYMBOLIC NOTATION

The summation convention is very often employed in connection with the representation of vectors and tensors by indexed base vectors written in the symbolic notation. Thus if the rectangular Cartesian axes and unit base vectors of Fig. 1-5 are relabeled as shown by Fig. 1-8, the arbitrary vector v may be written

V = viêi + v2ê2 + v3ê3 (1.68)

in which vi, v2, v3 are the rectangular Cartesian components of v. Applying the summation convention to (1.68), the equation may be written in the abbreviated form

A V = Viei (1.69)

where i is a summed index. The notation here is essentially symbolic, but with the added feature of the summation convention. In such a "combination" style of notation, tensor character is not given by the free indices rule as it is in true indicial notation. Fig. 1-8

CHAP. I] MATHEMATICAL FOUNDATIONS 11

Second-order tensors may also be represented by summation on indexed base vectors. Accordingly the dyad ab given in nonion form by (1.53) may be written

ab = (aiêi)(b$j) = aibjêiêj (1.70)

I t is essential that the sequence of the base vectors be preserved in this expression. In similar fashion, the nonion form of the arbitrary dyadic D may be expressed in compact notation by

1.11 COORDINATE TRANSFORMATIONS. GENERAL TENSORS Let xi represent the arbitrary system of coordinates xl, x2, x3 in a three-dimensional

Euclidean space, and let Bi represent any other coordinate system B1, 02, O3 in the same space. Here the numerical superscripts are labels and not exponents. Powers of x may be expressed by use of parentheses as in ( x ) ~ or (x )~ . The letter superscripts are indices as already noted. The coordinate transformation equations

assign to any point (xl, x2, x3) in the xi system a new set of coordinates (O1, 02, B3) in the Oi system. The functions Bi relating the two sets of variables (coordinates) are assumed to be single-valued, continuous, differentiable functions. The determinant

or, in compact form,

is called the Jacobian of the transformation. If the Jacobian does not vanish, (1.72) possesses a unique inverse set of the form

The coordinate systems represented by xi and Bi in (1.72) and (1.75) are completely general and may be any curvilinear or Cartesian systems.

From (l.72), the differential vector dOi is given by

This equation is a prototype of the equation which defines the class of tensors known as contravariant vectors. In general, a set of quantities bi associated with a point P are said to be the components of a cont rava~ant tensor of order one if they transform, under a coordinate transformation, according to the equation

where the partia1 derivatives are evaluated at P. In (1.77), bj are the components of the tensor in the xj coordinate system, while bt i are the components in the Oi system. In general


tensor theory, contravariant tensors are recognized by the use of superscripts as indices. I t is for this reason that the coordinates are labeled xi here rather than Xi, but it must be noted that it is only the differentials dxi, and not the coordinates themselves, which have tensor character.

By a logical extension of the tensor concept expressed in (l.77), the definition of contravariant tensors of order two requires the tensor components to obey the transformation law

Contravariant tensors of third, fourth and higher orders are defined in a similar manner.

The word contravariant is used above to distinguish that class of tensors from the class known as covariant tensors. In general tensor theory, covariant tensors are recognized by the use of subscripts as indices. The prototype of the covariant vector is the partia1 derivative of a scalar function of the coordinates. Thus if + = +(xl, x2, x3) is such a function,

In general, a set of quantities bi are said to be the components of a covariant , temor of order one if they transform according to the equation

In (1.80), b: are the covariant components in the ei system, bi the components in the xi system. Second-order covariant tensors obey the transformation law

Covariant tensors of higher order and mixed tensors, such as

are defined in the obvious way.

1.12 THE METRIC TENSOR. CARTESIAN TENSORS

Let xi represent a system of rectangular Cartesian coordinates in a Euclidean three- space, and let 8' represent any system of rectangular or curvilinear coordinates (e.g. cylindrical or spherical coordinates) in the same space. The vector x having Cartesian components xi is called the position vector of the arbitrary point P(xl, x2, x3) referred to the rectangular Cartesian axes. The square of the differential element of distance between neighboring points P(x) and Q(x +dx) is given by

From the coordinate transformation xi = xi(ol, 02, e3)

relating the systems, the distance differential is


and therefore (1.83) becomes

where the second-order tensor g,, = (dxi/dOp)(dxi/d$q) is called the metric tensor, or fundamental tensor of the space. If Oi represents a rectangular Cartesian system, say the x" system, then

where S,, is the Kronecker delta (see Section 1.13) defined by S,, = O if p # q and S,, = 1 if p = q .

Any system of coordinates for which the squared differential element of distance takes the form of (1.83) is called a system of homogeneous coordinates. Coordinate transformations between systems of homogeneous coordinates are orthogonal transformations, and when attention is restricted to such transformations, the tensors so defined are called Cartesian tensors. In particular, this is the case for transformation laws between systems of rectangular Cartesian coordinates with a common origin. For Cartesian tensors there is no distinction between contravariant and covariant components apd therefore it is customary to use subscripts exclusively in expressions representing Cartesian tensors. As will be shown next, in the transformation laws defining Cartesian tensors, the partia1 derivatives appearing in general tensor definitions, such as (1.80) and (1.81), are replaced by constants.

1.13. TRANSFORMATION LAWS FOR CARTESIAN TENSORS. THE KRONECKER DELTA. ORTHOGONALITY CONDITIONS

Let the axes Oxixzxs and Ox;xtx$ represent two rectangular Cartesian coordinate systems with a common origin a t an arbitrary point O as shown in Fig. 1-9. The primed system may be imagined to be obtained from the unprimed by a rotation of the axes about the origin, or by a reflection of axes in one of the coordinate planes, or by a combination of these. If the symbol aij

denotes the cosine of the angle between the ith primed and jth unprimed coordinate axes, i.e. aij = cos (xl, xj), the relative orientation of the individual axes of each system with respect to the other is conveniently given by the table

or alternatively by the transformation tensor

Fig. 1-9


From this definition of aij, the unit vector along the x: axis is given according to (1.48) and the summation convention by

An obvious generalization of this equation gives the arbitrary unit base vector as

In component form, the arbitrary vector v shown in Fig. 1-9 may be expresse(; in the unprimed system by the equation

v = 2 ) . , ê, . (1.90)

and in the primed system by V = vi ei (1.91)

Replacing in (1.91) by its equivalent form (1.89) yields the result

Comparing (1.92) with (1.90) reveals that the vector components in the primed and unprimed systems are related by the equations

I v . - a, . 3 - %,Vi (1.93)

The expression (1.93) is the transformation law for first-order Cartesian tensors, and as such is seen to be a special case of the general form of first-order tensor transformations, expressed by (1.80) and (1.77). By interchanging the roles of the primed and unprimed base vectors in the above development, the inverse of (1.93) is found to be

I t is important to note that in (1.93) the free index on aij appears as the second index. In (1.94), however, the free index appears as the first index.

By an appropriate choice of dummy indices, (1.93) and (1.94) may be combined to produce the equation

v j = a.. tjazkVk . (1.95)

Since the vector v is arbitrary, (1.95) must reduce to the identity vj = vj . Therefore the coefficient aijaik, whose value depends upon the subscripts j and k , must equal 1 or O according to whether the numerical values of j and Ic are the same or different. The Kronecker delta. defined b'r

may be used to represent quantities such as aijaik. Thus with the help of the Kronecker delta the conditions on the coefficient in (1.95) may be written

In expanded form, (1.97) consists of nine equations which are known as the orthogonality or orthonormality conditions on the direction cosines aij. Finally, (1.93) and (1.94) may also be combined to produce vi = &ijakjvk from which the orthogonality conditions appear in the alternative form

aijakj = s i k (1.98)

A linear transformation such as (1.93) or (1.94), whose coefficients satisfy (1.97) or (1.98), is said to be an orthogonal transformation. Coordinate axes rotations and reflections of the axes in a coordinate plane both Iead to orthogonal transformations.


The Kronecker delta is sometimes called the substitution operator, since, for example,

8ijbj = 8iibi + 8i2b2 f 8i3b3 = bi and, likewise,

8 i j F i k = 8ijFik + 8 2 j F 2 k $. 83jF3k = F j k

It is clear from this property that the Kronecker delta is the indicial counterpart to the symbolic idemfactor I, which is given by (1.54).

According to the transformation law (1.94), the dyad UiVj has components in the primed coordinate system given by

U I V ~ = ( U i p ~ ) ( ~ j q ~ q ) = UipUjqUpVq (I .I 01)

In an obvious generalization of (1.101), any second-order Cartesian tensor Tij obeys the transf ormation Iaw TC = aipajq Tpq

With the help of the orthogonality conditions i t is a simple calculation to invert (1.102), thereby giving the transformation rule from primed components to unprimed components:

The transformation laws for first and second-order Cartesian tensors generalize for an Nth order Cartesian tensor to

T~&. . . = aipajqalcm . . . Tpqm.. . (I J0-4)

1.14 ADDITION OF CARTESIAN TENSORS. MULTIPLICATION BY A SCALAR Cartesian tensors of the same order may be added (or subtracted) component by com-

ponent in accordance with the rule

The sum is a tensor of the same order as those added. Note that like indices appear in the same sequence in each term.

Multiplication of every component of a tensor by a given scalar produces a new tensor of the same order. For the scalar multiplier A, typical examples written in both indicial and symbolic notation are

bi = xai or b = Xa (1.1 06)

1.15 TENSOR MULTIPLICATION The outer product of two tensors of arbitrary order is the tensor whose components

are formed by multiplying each component of one of the tensors by every component of the other. This process produces a tensor having an order which is the sum of the orders of the factor tensors. Typical examples of outer products are

(a) aibj = Tij (C) DijTkrn = Qiikrn

(b) viFjk = Nijk (d) €i jkVm = @ijkm

As indicated by the above examples, outer products are formed by simply setting down the factor tensors in juxtaposition. (Note that a dyad is formed from two vectors by this very procedure.)


Contraction of a tensor with respect to two free indices is the operation of assigning to both indices the same letter subscript, thereby changing these indices to dummy indices. Contraction produces a tensor having an order two less than the original. Typical examples of contraction are the following.

(a) Contractions of Tij and uivj Tii = Til + TZZ + T33

UiVi = U i V i + U2V2 + U3V3

(b) Contractions of Eijak E.. va3 . = bi

Eijai = cj

Eiiak = dk

(C) Contractions of EijFkm

EijFim = Gim EijFkk = Pij

EijFki = Hik EijFjm = Qim

EiiFkm = Kkm EijFkj = Rik

An Znner product of two tensors is the result of a contraction, involving one index from each tensor, performed on the outer product of the two tensors. Severa1 inner products important to continuum mechanics are listed here for reference, in both the indicial and symbolic notations.

Outer Product Inner Product Indicial Notation Symbolic Notation

1. aibj aibi a * b

2. aiEjk aiEik = f k a.E = f

aiEji = hj E-a = h

3. Eij Fkm EijFjm = Gim E . F = G

4. Eij Ekm Eij Ejm = Bim E . E = (E)2

Multiple contractions of fourth-order and higher tensors are sometimes useful. Two such examples are

1. Eij Fkm contracted to Eij Fij, or E : F

2. Eij Ekm E,, contracted to Eij Ejm Emq, or ( E ) 3

1.16 VECTOR CROSS PRODUCT. PERMUTATION SYMBOL. DUAL VECTORS

In order to express the cross product a x b in the indicial notation, the third-order tensor eij,, known as the permutation symbol or alternating tensor, must be introduced. This useful tensor is defined by

I 1 if the values of i , j, k are an even permutation of 1,2,3 (i.e. if they appear in sequence as in the arrangement 1 2 3 12) .

-1 if the values of i , j , k are an odd permutation of 1,2,3 (i.e. if ' i jk - they appear in sequence as in the arrangement 3 2 1 3 2).

I O if the values of i , j, k are not a permutation of 1 ,2,3 (i.e. if two or more of the indices have the same value).


From this definition, the cross product a x b = c is written in indicial notation by

Using this relationship, the box product a x b . c = h may be written

Since the same box product is given in the form of a determinant by (1.52), it is not surprising that the permutation symbol is frequently used to express the value of a 3 X 3 determinant.

I t is worthwhile to note that cijk obeys the tensor transformation law for third order Cartesian tensors as long as the transformation is a proper one (det aij = 1) such as arises from a rotation of axes. If the transformation is improper (det aij = -I), e.g. a reflection in one of the coordinate planes whereby a right-handed coordinate system is transformed into a left-handed one, a minus sign must be inserted into the transformation law for cijk.

Such tensors are called pseudo-tensors.

The dual vector of an arbitrary second-order Cartesian tensor Tij is defined by v. - c . . T.

2 - < ~ k i k (1.110)

which is observed to be the indicial equivalent of T,, the "vector of the dyadic Tu, as defined by (1.15).

1.17 MATRICES. MATRIX REPRESENTATION OF CARTESIAN TENSORS

A rectangular array of elements, enclosed by square brackets and subject to certain laws of combination, is called a matrix. An M x N matrix is one having M (horizontal) rows and N (vertical) columns of elements. In the symbol Aij, used to represent the typical element of a matrix, the first subscript denotes the row, the second subscript the column occupied by the element. The matrix itself is designated by enclosing the typical element symbol in square brackets, or alternatively, by the kernel letter of the matrix in script. For example, the M x N matrix 4, or [Aij] is the array given by

A matrix for which M = N, is called a square matrix. A 1 x N matrix, written [alk], is called a row matrix. An M x 1 matrix, written [akl], is called a column matrix. A matrix having only zeros as elements is called the zero matrix. A square matrix with zeros everywhere except on the main diagonal (from A11 to ANN) is called a diagonal matrix. If the nonzero elements of a diagonal matrix are a11 unity, the matrix is called the unit or identity mat.ix. The N X M matrix CAT, formed by interchanging rows and columns of the M x N matrix 4 , is called the transpose matrix of 4.

Matrices having the same number of rows and columns may be added (or subtracted) element by element. Multiplication of the matrix [Aij] by a scalar h results in the matrix [AAij]. The product of two matrices, cA%, is defined only if the matrices are conformable, i.e. if the prefactor matrix cA has the same number of columns as the postfactor matrix 23 has rows. The product of an M X P matrix multiplied into a P X N matrix is an M X N matrix. Matrix multiplication is usually denoted by simply setting down the matrix symbols in juxtaposition as in


Matrix multiplication is not, in general, commutative: cA% z LBd.

A square matrix cA whose determinant \Aij\ is zero is called a singular matriz. The cofactor of the element Aij of the square matrix 4 , denoted here by AT~, is defined by

* Aij = (-l)i+jMij (1.113)

in which Mij is the minor of Aij; i.e. the determinant of the square array remaining after the row and column of Aij are deleted. The adjoint matrix of O4 is obtained by replacing each element by its cofactor and then interchanging rows and columns. If a square matrix e4 = [Aij] is non-singular, i t possesses a unique inverse matrix 4-' which is defined as the adjoint matrix of O4 divided by the determinant of 04. Thus

From the inverse matrix definition (1.114) it may be shown that

where 4 is the identity matrix, having ones on the principal diagonal and zeros elsewhere, and so named because of the property

4cA = CAJ = (1.116)

I t is clear, of course, that 4 is the matrix representation of 6,,, the Kronecker delta, and of I, the unit dyadic. Any matrix cA for which the condition cAT = cA-I is satisfied is called an orthogonal matrix. Accordingly, if O4 is orthogonal,

As suggested by the fact that any dyadic may be expressed in the nonion form (1.53), and, equivalently, since the components of a second-order tensor may be displayed in the square array (1.62), it proves extremely useful to represent second-order tensors (dyadics) by square, 3 x 3 matrices. A first-order tensor (vector) may be represented by either a 1 x 3 row matrix, or by a 3 x 1 column matrix. Although every Cartesian tensor of order two or less (dyadics, vectors, scalars) may be represented by a matrix, not every matrix represents a tensor.

If both matrices in the product 4 % = C are 3 x 3 matrices representing second-order tensors, the multiplication is equivalent to the inner product expressed in indicial notation by

where the range is three. Expansion of (1 .I 18) duplicates the "row by column" multiplication of matrices wherein the elements of the ith row of the prefactor matrix are multiplied in turn by the elements of the kth column of the postfactor matrix, and these products summed to give the element in the .ith row and kth column of the product matrix. Severa1 such products occur repeatedly in continuum mechanics and are recorded here in the various notations for reference and comparison.

(a) Vector dot product

a b = b * a = A [aij][bjl] = [A]


( b ) Vector-dyadic dot product a * E = b a& = CB

(c) Dyadic-vector dot product E * a = c &a = c

E.. . - tja3 - Ci [Eij] [ajl] [cii]

1.18 SYMMETRY OF DYADICS, MATRICES AND TENSORS

According to (1.36) (or (1.37)), a dyadic D is said to be symmetric (anti-symmetric) if it is equal to (the negative of) its conjugate D,. Similarly the second-order tensor Dij is symmetm'c if D.. - D..

a3 - 3% (1 .I 22)

and is anti-symmetric, or skew-symmetric, if

Therefore the decomposition of Dij analogous to (1.38) is

or, in an equivalent abbreviated form often employed,

where parentheses around the indices denote the symmetric part of Dii, and square brackets on the indices denote the anti-symmetric part.

Since the interchange of indices of a second-order tensor is equivalent to the interchange of rows and columns in its matrix representation, a square matrix cA is symmetric if it is equal to its transpose CAT. Consequently a symmetric 3 x 3 matrix has only six independent components as illustrated by

Ali Ai2

(1.126)

A23 A33

An anti-symmetric matrix is one that equals the negative of its transpose. Consequently a 3 x 3 anti-symmetric matrix CB has zeros on the main diagonal, and therefore only three independent components as illustrated by


Symmetry properties may be extended to tensors of higher order than two. In general, an arbitrary tensor is said to be symmetric with respect to a pair of indices if the value of the typical component is unchanged by interchanging these two indices. A tensor is anti- symmetric in a pair of indices if an interchange of these indices leads to a change of sign without a change of absolute value in the component. Examples of symmetry properties in higher-order tensors are

(a) Rijkm = Rikjm (symmetric in k and j )

( b ) cijk = -c ki i (anti-symmetric in k and i)

(c) Gijkm = Gjimk (symmetric in i and j; k and m)

(d ) P i j k = Pikj = P k j i = pjik (syrnmetric in a11 indices)

1.19 PRINCIPAL VALUES AND PRINCIPAL DIRECTIONS OF SYMMETRIC SECOND-ORDER TENSORS

In the following analysis, only symmetric tensors with real components are considered. This simplifies the mathematics somewhat, and since the important tensors of continuum mechanics are usually symmetric there is little sacrifice in this restriction.

For every symmetric tensor Tij, defined a t some point in space, there is associated with each direction (specified by the unit normal ni) a t that point, a vector giveii by the inner p r oduct

Here Tij may be envisioned as a linear vector operator which produces the vector vi conjugate to the direction ni. If the direction is one for which vi is parallel to ni, the inner product may be expressed as a scalar multiple of ni. For this case,

and the direction ni is called a principal direction, or principal axis of Tij. With the help of the identity ni = Sijnj, (1.129) can be put in the form

which represents a system of three equations for the four unknowns, ni and h, associated with each principal direction. In expanded form, the system to be solved is

Note first that for every A, the trivial solution ni = O satisfies the equations. The purpose here, however, is to obtain non-trivial solutions. Also, from the homogeneity of the system (1.131) it follows that no loss of generality is incurred by restricting attention to solutions for which nini = 1, and this condition is imposed from now on.

For (1.130) or, equivalently, (1.131) to have a non-trivial solution, the determinant of coefficients must be zero, that is,

]Tij - ASijl = O (1.1 32)

Expansion of this determinant leads to a cubic polynomial in h, namely,

A3 - ITX' + IITA - IIIT = O (I .133)

which is known as the characteristic equation of Tij, and for which the scalar coefficients,

CHAP. I] MATHEMATICAL FOUNDATIONS 2 1

IT = Tii = t r Tij (trace of Tij) (1.1 34)

IIIT = ITijl = det Tij (1.1 36)

. are called the first, second and third invariants, respectively, of Tij. The three roots of the cubic (1.133), labeled X(I), h ( ~ , A(,), are called the principal values of Tij. For a symmetric tensor with real components, the principal values are real; and if these values are distinct, the three principal directions are mutually orthogonal. When referred to principal axes, both the tensor array and its matrix appear in diagonal form. Thus

If X ( I ) = h(z), the tensor has a diagonal form which is independent of the choice of X ( I ) and htz) axes, once the principal axis associated with XG) has been established. If a11 principal values are equal, any direction is a principal direction. If the principal values are ordered, it is customary to write thern as h( , ) , h(II), h(,,,) and to display the ordering as in A o , > A,,,, > ~ ( I I I , .

For principal axes labeled OxTx2x8, the transformation from Oxixzx3 axes is given by the elements of the table

in which nlj) are the direction cosines of the jth principal direction.

~1

2;

E;

1.20 POWERS OF SECOND-ORDER TENSORS. HAMILTON-CAYLEY EQUATION By direct matrix multiplication, the square of the tensor Tij is given as the inner

product TikTkj; the cube as TikTkrnTmj; etc. Therefore with Tii written in the diagonal form (1.137), the nth power of the tensor is given by

$1

all = n(l)

aZ1 = n(2)

a31 = ni3)

A comparison of (1.135) and (1.137) indicates that Tij and a11 its integer powers have the same principal axes.

Since each of the principal values satisfies (1.133), and because of the diagonal matrix form of Tn given by (1.138), the tensor itself will satisfy (1.133). Thus

x2

a12 = n(1) 2

aZ2 = n(2)

=

in which ,4 is the identity rnatrix. This equation is called the Hamilton-Cayley equation. Matrix multiplication of each term in (1.139) by I produces the equation,

x3

al, = ni1)

(L*, = n(32)

a,, = n 3 )


Combining (1.140) and (1.139) by direct substitution,

Continuation of this procedure yields the positive powers of 7 as linear combinations of T2, T and 8.

1.21 TENSOR FIELDS. DERIVATIVES OF TENSORS

A tensor field assigns a tensor T(x, t) to every pair (x, t) where the position vector x varies over a particular region of space and t varies over a particular interval of time. The tensor field is said to be continuous (or differentiable) if the components of T(x, t) are continuous (or differentiable) functions of x and t. If the components are functions of x only, the tensor field is said to be steady.

With respect to a rectangular Cartesian coordinate system, for which the position vector of an arbitrary point is A x = xiei (1 .1 42)

tensor fields of various orders are represented in indicial and symbolic notation as follows,

(a) scalar field: + = +(xi, t) or $J = +(x, t) (1.143)

(b) vector field: vi = vi(x, t) or V = V(X, t) (1 ,144)

(c) second-order tensor field: Tij = Tij (x, t) or T = T(x, t) (1.145)

Coordinate differentiation of tensor components with respect to xi is expressed by the differential operator aldxi, or briefly in indicial form by a;, indicating an operator of tensor rank one. In symbolic notation, the corresponding symbol is the well-known differential vector operator V, pronounced de1 and written explicitly

êi a = A V = eidi aXi

(1 .I 46)

Frequently, partia1 differentiation with respect to the variable xi is represented by the comma-subscript convention as illustrated by the following examples.

avi aTij (b) - = vi,i axi (e) ã2* = T i!, k

avi (c) - = vi,j

a2Tij - ax ( f ) a s r d z , - km

From these examples i t is seen that the operator ai produces a tensor of order one higher if i remains a free index ( (a) and (c) above), and a tensor of order one lower if i becomes a dummy index ((b) above) in the derivative.

Severa1 important differential operators appear often in continuum mechanics and are given here for reference.

divv = V - v or divt = visi (1.148)

curl v = V x v O r ~ijkajvk = 'ijkvk,i (1 .I 49)


1.22 LINE INTEGRALS. STOKESy THEOREM In a given region of space the vector function of position, F = F(x), is defined a t every

point of the piecewise smooth curve C shown in Fig. 1-10. If the differential tangent vector to the curve a t the arbitrary point P is dx, the integral

X F - ~ X = L ~ F - ~ X (1.151)

taken along the curve from A to B is known as the line .integral of F aIong C. In the indiciaI notation, (1.151) becomes ( z ~ ) s

S ,F idx i J (zi)* Fidxi (1 .I 52)

Fig. 1-10 Fig. 1-11

Stokes' theorem says that the line integral of F taken around a closed'reducible curve C, as pictured in Fig. 1-11, may be expressed in terms of an integral over any two-sided surface S which has C as its boundary. Explicitly,

in which 6 is the unit normal on the positive side of S , and dS is the differential element of surface as shown by the figure. In the indicial notation, (1.153) is written

1.23 THE DIVERGENCE THEOREM OF GAUSS The divergence theorem of Gauss relates a volume integral to a surface integral. In

its traditional form the theorem says that for the vector field v = v(x),

where 2 is the outward unit normal to the bounding surface S , of the volume V in which the vector field is defined. . In the indicial notation, (1.155) is written

i vi,i dV = L vini dS (1.156)

The divergence theorem of Gauss as expressed by (1.156) may be generalized to incor- porate a tensor field of any order. Thus for the arbitrary tensor field Tijk.. . the theorem is written

Tijk . . . p dV = Tijknp dS (1.157)


Solved Problems ALGEBRA OF VECTORS AND DYADICS (Sec. 1.1-1.8)

Determine in rectangular Cartesian form the unit vector which is (a) parallel to the vector v = 2? + 3 3 - 6g, (b ) along the line joining points P(1,0,3) and Q(O, 2 , l ) .

(a ) Ivl = v = 1/ (2)~ + (3)2 + (-6)2 = 7 A A A A v = VIV = (217) i + (317) j - (617)k

(b ) The vector extending from P to Q is A

u = ( O - I ) ? + ( 2 - O ) ; + ( 1 - 3 ) k A A A

= - i + 2 j - 2 k

u = d(-1)2 + (2)2 + (-2)2 = 3 Fig. 1-12 A A

Thus u = -(1/3) i + (213); - (213) k directed from P to Q

or 6 = (1/3)? - (213) 3 + ( 2 / 3 ) k directed from Q to P

1.2. Prove that the vector v = a? + b i + c k is normal to the plane whose equation is ax + by + cz = A. I z

Let P ( x l , yl, z l ) and Q(x2, y2, z2) be any two points in the plane. Then as l + byl + czl = X and ax2 + by2 + cz2 =

A and the vector joining these points is u = (x2 - x l ) i +

A A

(v2 - y l ) j + (2 , - 2,) k. The projection of v in the direction of u is u * v - 1 A A - - u u - [ ( x z - x I ) ~ + A ( v 2 - y 1 ) j

+ (2 , - 2,) k ] [a? + b 3 + c c ] 1 X - X Z = - (as2 + by2+ cz2- axl - byl - cz,) = - = O U U

Since u is any vector in the plane, v is I to the plane. Fig. 1-15

1.3. If r = x? + y3 + z c is the vector extending from the origin to the arbitrary point P(x, y, z) and d = a? + b j + c k is a constant vector, show that (r - d) r = O is the vector equation of a sphere.

Expanding the indicated dot product, A A A A A A

( r - d ) * r = [ ( x - a ) i + ( y - b ) j + ( z - c ) k ] * [ x i + y j + z k ]

= x 2 + y 2 + z 2 - a x - b y - c z = O

Adding d2/4 = (a2 + b2 + c2)/4 to each side of this equation gives the desired equation

( x - a/2)2 + ( y - b/2)2 + ( z - c/2)2 = (d/2)2

which is the equation of the sphere centered a t d/2 with radius d/2.

1.4. Prove that [a b X c]r = (a- r)b x c + (b r)c x a + (c r)a x b. Consider the product a X [ ( b X c ) X r] . By direct expansion of the cross product in brackets,

a X [ ( b X c ) X r] = a X [ ( b r)c - (c r)b] = - (b . r)c X a - (c r)a X b

Also, setting b X c = v ,

a x [ ( b x c ) x r ] = a x ( v x r ) = ( a W r ) b x c - ( a . b x c ) r


Thus - (b r)c x a - ( c r )a X b = ( a r ) b X c - (a b X c)r and so

( a * b x c)r = ( a * r ) b x c + ( b . r ) c X a + ( c * r ) a X b

This identity is useful in specifying the displacement of a rigid body in terms of three arbitrary points of the body.

1.5. Show that if the vectors a, b and c are linearly dependent, a b x c = O . Check the linear dependence or independence of the basis

A A A

u = 3 i + j - 2 k A A A

v = 4 i - j -k A A A

w = i - 2 j + k The vectors a, b and c are linearly dependent if there exist constants h, p and v, not a11 zero,

such that ha + pb + vc = O. The component scalar equations of this vector equation are

ha, + pbY + V C , = O

ha, + pb, + vc, = O

which is equivalent to a b X c = O. For the proposed basis u, v, w,

This set has a nonzero solution for h, p and v provided the determinant of coefficients vanishes,

Hence the vectors u, v, w are linearly dependent, and indeed v = u + w.

a, b, c,

a y by cY

a, b, c,

1.6. Show that any dyadic of N terms may be reduced to a dyadic of three terms in a form having the base vectors êl, ê2, ê3 as (a) antecedents, (b) consequents.

= O

Let D = albl + a2b2 + - - - + aNbN = aibi (i = 1,2, . . . , N ) . A A A A

(a ) In terms of base vectors, ai = al ie l + azie2 + a3ie3 = aj ie j and so D = a-2.b. 3 1 3 ~ = ê.(a . .b . ) 3 t t = $.c. 1 1 3

with j = 1,2,3.

(b ) Likewise setting bi = bj iê j i t follows that D = aibjiêj = (bjiai)êj = g j ê j where j = 1 ,2 ,3 .

1.7. For the arbitrary dyadic D and vector v, show that D - v = v . D,.

Let D = albl + a2b2 + . - - + aNbN. Then

D v = a l ( b , v) + a,(b2 v ) + . . . + aN(bN v )

= ( V . b , )a , + ( v . b2)a2 + . . + ( v . b N ) a N = v . D ,

1.8. Prove that ( D , D ) , = D, . D.

From (1.71), ü = Dijê iê j and D , = Djiê iê j . Therefore

D,.D = D..ê .ê . .D ê ê A A A A

j t 1 I p q p q = DjiDpq(ej 'ep)eieq A A A A A A A A A A

and (o, o) , = Dji Dpq ( e j e,) e,ei = Dpqeq ( e , e j ) e i D j i = D,,ê$, . ~ ~ ~ e ~ e ~ = D , . D


1.9. Show that (D x v), = -v X Dc.

D X v = al(bl X V) + a,(b2 X V) + ' ' ' + aN(bN x V)

(D X v), = (bl X v)al + (b, x v)a2 + - + (bN X v)aN

= -(V X b,)al - (v x b,)a, - ..- - (v X bN)aN = -v X D,

A A A A A A A A A

1.10. If D = a i i + b j j + c k k and r is the position vector r = x i + y j + xk, show that r D r = 1 represents the ellipsoid ax2 + by2 + cx2 = 1.

r - ~ . r = (s?+ y3+ zk) (a??+ b 3 3 + c k k ) (s?+ zk ) A A A A A A

= ( z i + y j + z k ) * ( a x i + b y j + c z k ) = as2+by2+cz2 = 1

A A A A A A A A A A

1.11. Forthedyadics D = 3 ? ? + 2 j j - j k + s k k and F = 4 i k - t - 6 3 3 - 3 k j + k k , compute and compare the double dot products D : F and D . . F.

From the definition ab : cd = (a c)(b d) i t is seen that D : F = 12 + 5 = 17. Also, from ab cd = (b e)(a d) i t follows that D F = 12 + 3 + 5 = 20.

1.12. Determine the dyadics G = D. F and H = F * D if D and F are the dyadics given in Problem 1.11.

From the definition ab cd = (b . c)ad,

O = (3?? + 2:; - 3 k + 5 k k ) . (4;: + 6:: - 3 k 3 + k k ) A A A A A A A A A A A A

= 1 2 i k + 1 2 j j + 3 j j - j k - 1 5 k j + 5 k k

A A A A A A A A A A A A A A

Similarly, H = ( 4 i k + 6 j j - 3 k j + k k ) . ( 3 i i + 2 j j - j k + 5 k k ) A A

= 2 0 i k + 1 2 3 ! - 6 3 k . - 6 k 3 + 8 k k

1.13. Show directly from the nonion form of the dyadic D that D = (Do?)? + (D-?)? + A A A A A A

( D 0 k ) k and also i w D . i = Dz,, i - D - j = D,,, etc. Writing D in nonion form and regrouping terms,

A A A A A A A A

II = (D,,i + D,,j + D,,k) i + (D,, i + D,,; + Dryk) j + (Dxzi + D,,? + D,$); A A A

= d l i + d 2 j +d ,k = ( o - ? ) ? + ( D * 3 ) 3 + ( ~ - 2 ) k A A e A A 6 A

Also now i - d l = i * ( ~ - i ) = i . ( D , , i + D Y , ~ +D,,k) = D,, A 6 A

j * d l = ; . ~ . i = D,,, j * d 2 = 3 . ~ . j = D,,, etc.

1.14. For an antisymmetric dyadic A and the arbitrary vector b, show that 2b. A = A, x b. From Problem 1.6(a), A = &cl + &c2 + $,c3; and because it is antisymmetric, 2A = (A - A,)

or A A A A A A

2A = ( elel + e2c2 + e3c3 - elel - c2e2 - c3e3) A A A A A = (êlcl - elel + e2c2 - c2e2 + esc3 - c3e3)

and so 2b A = [(b . $,)cl - (b . c,) ê,] + [(b . ê2)c2 - (b . C,) ê2] + [(b ê3)c3 - (b c3) $31

= [ ( ê , x c 1 ) x b + ( ê z X c 2 ) X b + ( ê 3 X ~ 3 ) X b ] = (AvXb)

A A A A A A A

1.15. If D = 6;; + 3 i j + 4 k k and u = 2 i + k, v = 5 j, show by direct calculation that D . ( u x v ) = ( D x u ) . ~ .

A A A A A

Since u X v = ( 2 i + k ) X 5 j = 1 0 k - 5 i , A A A A A A A A

D * ( U X V ) = ( 6 i i + 3 i j + 4 k k ) - ( - 5 i + l O k ) = -30?+40k A A A A A A A A A A

Next, D ~ U = ( 6 ? ? + 3 ? ; + 4 k k ) ~ ( 2 i + k ) = - 6 i k + 8 k j - 6 i j + 3 i i A A A A A A A A A A A

snd ( D X u ) * v = ( 3 i i - 6 i j - 6 i k + 8 k j ) . 5 j = - 3 0 i + 4 0 k

CHAP. 11 MATHEMATICAL F O U N D A T I O N S 27

Considering the dyadic A A A A A A

D = 31:-4:;+2ji+jj+kk

as a linear vector operator, determine vector r' produced when D operates

r = 4 ? + 2 3 + 5 k .

the on

Fig. 1-14

1.17. Determine the dyadic D which serves as a 'inea? vec^r operator for the vector function a = f(b) = b + b x r where r = x i + y j + xk and b is a constant vector.

In accordance with (1.59) and (1.60), construct the vectors A A A A A A

u = f ( i ) = i + i X r = i - z j + y k A A A A A A

v = f ( j ) = j + j X r = x i + j - x k A A A A A A

w = f ( k ) = k + k X r = - y i + x j + k A A A A A A A A A A A A A A A

Then ü = u i + v j + wk = ( i - z j + y k ) i + ( x i t j - x k ) j + ( - y i + x j + k ) k A A A

and a = D b = (b, + b,z - b,y) i + (-b,z + b, + b,x) j + (b,y - b,x + b,) k

AS a check the same result may be obtained by direct expansion of the vector function, A A A A A

a = b + b X r = b,i -i- b,3 + b,k + (b,z - b,y) i + (b,x - bxz) j + (bxy - b,x)k

1.18. Express the unit triad ê4, ê,, ê, in terms of A

A A A

i, j, k and confirm that the curvilinear triad A is right-handed by showing that ê, x ê, = e,.

By direct projection from Fig. 1-15, A A A A

e, = (COS g cos e) i + (cos g sin e) j - (sin g)k A A A e, = (- sin e) i + (cos e) j A A A A

e, = (sin g cos e) i + (sin g sin e) j + (cos @) k

and so Fig. 1-15

A A A = (sin @ cos e ) i + (sin g sin e) j + [(cosz e + sin2 e) cos g] k = 2,.

A e+ x ê, =

1.19. Resolve the dyadic D = 3;: + 4:: + 6 3 1 + 7;: + 10 ko;f 2 k3 into its symmetric and antisymmetric parts.

e A

1 k

cos g cos e cos g sin e - sin @

- sin e cos e O

Let D = E + F where E = E, and F = -F,. Then A A A A A A A A A A A A

E = (1/2)(D + D,) = (1/2)(6 i i + 4 i k + 4 k i + 6 j i + 6 i j + 14 j j

+ ioct + i o t k + 2 2 3 + 2;;) A A A A A A A A A A A A A A A A

= 3 i i + 3 i j + 7 i k + 3 j i + 7 j j + j k + 7 k i + k j = E,

A A A A A A A A A A A A A A A A

F = (1 /2) (~-o , ) = ( 1 / 2 ) ( 4 i k - 4 k i + 6 j i - 6 i j + 1 0 k i - 1 0 i k + 2 k j - 2 j k ) A A A A A A A A A A A A

= -3 i j - 3 i k + 3 j i - j k + 3 k i + k j = -F,


1.20. With respect to the set of base vectors ai, a2, a3 (not necessarily unit vectors), the set a', a2, a3 is said to be a reciprocal basis if ai- a3 = aij. Determine the necessary relationships for constructing the reciprocal base vectors and carry out the calculations for the basis

A A A

bl = 3 i + 4 3 , bz = - i + 2 ? + 2 k , bs = ? + $ + % By definition, al a1 = 1, a, a1 = O, a3 a1 = O. Hence a1 is perpendicular to both a, and

a,. Therefore i t is parallel to a, X a,, i.e. a1 = X(a2 X a3). Since al a1 = 1, al Xa2 X a, = 1 and h = l / (a l a, X a,) = l/[ala2a3]. Thus, in general,

For the basis bl , b2, b3, l / h = bi ba X b3 = 12 and so A A

bl = (b, x b3)/12 = ( j -k)/4 A A A

b2 = (b3 x bl)/12 = - i13 + j 14 + k/12

INDICIAL NOTATION - CARTESIAN TENSORS (Sec. 1.9-1.16)

1.21. For a range of three on the indices, give the meaning of the following Cartesian tensor symbols: Aii, Bijj, Rij, ai Tij, aibjSij.

Ai, repi.esents the single sum Aii = All + A,, + A,,.

Bijj represents three sums: (1) For i = 1, Bll l + B12, + B133.

(2) For i = 2, BZll + BZz2 + BZ33.

(3) For i = 3, B3,, + B322 + B333.

Rii represents the nine components Rll, R12, Ri3, R,,, Rz2, RZ3? R32> R33.

ai Tij represents three sums: (1) For j = 1, al Tll + a2 T2i + a3 T3i.

(2) For j = 2, a1 T12 + a2 TZ2 f a3 T3,.

(3) For j = 3, a1 T13 + a2 T23 + a3T33.

aibjSij represents a single sum of nine terms. Summing first on i, aibjSij = albjSlj + a2bjSzj + a3bjS3j. Now summing each of these three terms on j,

aibjSij = alblSll + alb2S12 + alb3S13 + a2b1S2, + a2b2Sz2

+ ~ 2 ~ 3 ~ 2 3 + ~ 3 ~ 1 ~ 3 1 + ~ 3 ~ 2 S 3 2 + ~ 3 ~ 3 ~ 3 3

122. Evaluate the following expressions involving the Kronecker delta aij for a range of three on the indices. (a) Sii = S l l + S z 2 + = 3

(b) Si jS i j = S l j S l j + 62jS2j + S3 jS3 j = 3

(C) S i j S i k S j k = S l j s l k 8 j k + 82j62k8jk + 83j83k8jk = 3

( d ) Gijsjk = s i l S l k + Si2S2k + 8i383k = Sik

(e) 6ifiik = S1#11, + 82jA2k + 83jA3k = Ajk

1.23. For the permutation symbol Cijk S ~ O W by direct expansion that (a) Cijk'kij = 6 , (b) ' i j k a j a k = 0.

(a) First sum on i, €.. i ~ k k i j .. = € l j k C k l j + '2 jkek2j + '3jkek3j


Next sum on j. The nonzero terms are

'ijk'kij = '12kEk12 + '13kCk13 + '21kek21 f '23kek23 + '31kCk31 + '32kEk32

Finally summing on k, the nonzero terms are - e.. ' .. - ~k k21 '123'312 + '132'213 + '213'321 + '231'123 + '312'231 + '321'132

= ( l ) ( l ) + (-I)(-1) + (-I)(-1) + ( l ) ( l ) + ( l ) ( l ) + (-I)(-1) = 6

( b ) Summing on j and k in turn,

Eijkajak = ' i l k a l a k + Ei2ka2ak + Ei3ka3ak

= ' i12ala2 + ' i 1 3 ~ 1 ~ 3 f ' i 2 1 ~ 2 ~ 1 + 'i23a2a3 f ' i 3 1 ~ 3 ~ 1 + Ei32a3a2

From this expression,

when i = 1, e l j k a j a k = a2a3 - a3a2 = O

when i = 2, € 2 j k a j a k = a i a 3 - a3a1 = 0

when i = 3, € 3 j k a j a k = a1U2 - a2al = 0

Note that eiikajak is the indicial form of the vector a crossed into itself, and so a X a = 0.

1.24. Determine the component f 2 for the vector expressions given below.

f i ' i j kT jk

f 2 = ' 2 jkTjk = € 2 1 3 ~ 1 3 + '231T31 = -T13 + T31

( b ) f i = C,, j b j - c j , i b j

f 2 = c2,1b1 + ~ 2 , 2 ~ 2 + ~ 2 , 3 ~ 3 - ~ 1 , 2 ~ 1 - ~ 2 . 2 ~ 2 - ~ 3 , 2 ~ 3

= fc2, 1 - ~ l , 2 ) ~ 1 + ( ~ 2 . 3 - ~ 3 , 2 ) ~ 3

( C ) f i = B i j f ;

f 2 = B21f: + B22f2 + B23f3*

1.25. Expand and simplify where possible the expression D i j x i x j for (a) D i j = D j i ,

(b) D i j = - D j i .

Expanding, Dijxixj = Dljx1xj + D2jx2xj + Dsjx3xj = D 1 1 x 1 x 1 + D 1 2 x 1 x 2 + 0 1 3 x 1 5 3 + D 2 1 ~ 2 ~ 1 + D 2 2 ~ 2 ~ 2

+ D23x2x3 + D31x3x1 + D32x3x2 + D33x3x3

(a) Dijxixj = D 1 1 ( x 1 ) 2 + D 2 2 ( x 2 ) 2 + D 3 3 ( x 3 ) 2 + 2D12x1x2 + 2D23x2x3 + 2D13x1x3

( b ) Dijxixj = O since Dll = -Dll, D12 = -D2,, etc.

1.26. Sh0w that c i jkckpq = - 8 i q s j p for (a) i = 1, j = q = 2, p = 3 and for (b) i = q = 1, j = p = 2. (It is shown in Problem 1.59 that this identity holds for every choice of indices.) (a) Introduce i = 1, j = 2, p = 3, q = 2 and note that since k is a summed index i t takes on

a11 values. Then - - 'ijk'kpq - '12kEk32 - E121E132 + '122'232 + '123'332 = O

and S i p S j q - aiqS, = 813S22 - S12S23 = O

(b) Introduce i = 1, j = 2, p = 2, q = 1. Then eijkekpq = ~123'321 = -1 and SipSjq - SiqSjp = S12S21 - S11S22 = -1.

1.27. Show that the tensor Bik = cijkaj is skew-symmetric. Since by definition of cijk an interchange o f two indices causes a sign change,

Bik = eijkaj = - (ek j ia j ) = - (Bk i ) = - Bki

30 M A T H E M A T I C A L F O U N D A T I O N S [CHAP. 1

1.28. If Bij is a skew-symmetric Cartesian tensor for which the vector bi = ( + ) E ~ ~ ~ B ~ ~ , show that Bpq = tpqi bi .

Multiply the given equation b y epqi and use the ident i ty given i n Problem 1.26.

~pqibi = & ~ ~ q i ~ i j k ~ j k = & ( a p j a q k - SpkSqj)Bjk = &(Bpq - B q p ) = &(Bpq + B p q ) = B p q

1.29. Determine directly the components of the metric tensor for spherical polar coordinates as shown in Fig. 1-7(b).

axi azi and label t h e co- W r i t e (1.87) as gpq =

; j ~ q ordinates as shown i n Fig. 1-16 (r = e,, @ = e,, e = e,). T h e n

x , = 8 , sin 0, cos 8,

x , = e , sin e, sin e ,

Fig. 1-16 Hence

ax1 - - - ax1 - sin O , cos 8 , - - 8x1 -

e , cos 0 , cos e3 - - ao1

-e, sin e, sin e, as2 863

8x2 - - - - 8x2 - sin e2 sin e, - - 0 , cos O , sin e , 8x2 - -

861 6 , sin e2 cos e,

302 as3

8x3 - - - -el sin e, 8x3 - = o as2 863

axi axi f r o m which g,, = -- - - sinz e, cos2 8 , + sin2 o, sin2 e, + cos2 e , = 1

as, asl

axi axi -- - g33 = ao3 ae, - eSsin2 e, sin2 e, + eSsin2 6 , cos2 e , = eSsin2 O ,

Also, gpq = O for p Z q. For example,

axi axi g12 = aBi dB., = (s in e, cos @,)(e, cos e , cos e,) . e + (s in O 2 sin e3)(e1 cos e, sin e3) - ( C O S eZ)(el sin e,)

= o

T h u s for spherical coordinates, (ds)2 = (de1) , + (el)2(de2)" (e1 sin de^)'.

1.30. Show that the length of the line element d s resulting from the curvilinear coordinate increment doi is given by ds = G d O i (no sum). Apply this result to the spherical coordinate system of Problem 1.29.

W r i t e (1.86) as (ds)2 = gpq de, de,. T h u s for the line element (de,, O , O ) , i t follows t h a t (ds) , = g l l (de l )2 and ds = \/g,, de,. Similarly for (0 , de,, O ) , ds = G d e 2 ;

and for (0 ,0 , de3), ds = de3. Therefore (Fig. 1-17),

CHAP. 11 MATHEMATICAL FOUNDATIONS 3 1

(1 ) For (del,O,O), ds = del = dr

(2) For ( O , de,, O ) , ds = e, de, = r d+

(3 ) For (O , O , de,), d s = e , sin 8 , de, = r sin + de

1.31. If the angle between the line elements represented by (doi, O, O) and (O , don,O) is

denoted by p,,, show that cosp,, = 8 1 2

66. Let ds, = 6 de, be the length of line element represented by (del , O , 0 ) and dsz = G d 6 2

axi be that of ( O , de,, O) . Write (1.85) as dxi = - dek, and since (A), = cos plz h, dsz,

a e k

ax , a%, as, ax, ax, ax3 - - - - de, de, + - - de, de2 i- - - de1 de2 = giz de1 dez ae , ae, ae, ae2 ae, as,

del de2 812 Hence using the result of Problem 1.30, tos plz = g,, - - =

ds1ds2 66

1.32. A primed set of Cartesian axes O X ~ X $ X ~ is obtained by a rotation through an angle e about the X , axis. Determine the transformation coefficients aij relating the axes, and give the primed components of the vector v = ~ $ 1 + vzêz + v&.

From the definition (see Section 1.13) ai; = cos (si, sj) and Fig. 1-18, the table of dire8ion cosiiies is

Thus the transformation tensor is

cos 6 sin 6 O

A = -sin e cos 6 O

X ;

xó

x3

Fig. 1-18

By the transformation law for vectors (1.94),

v ; = a I j v j = v , cose + v , sin 6

v i = aZjv j = -vl sin e + v , cos 6

v j = a3jvj = V ,

1.33. The table of direction cosines relating two sets of rectangular Cartesian axes is partially given as shown on the right. Determine the entries for the bottom row of the table so that 0 x i x ; x i is a right-handed system.

5 1

cos e

-sin e

o

2 2

sin e

cos 6

o

x3

O

O

1


The unit vector along the xí axis is given by the first row of the table as $I = (315) ê1 - (415) ê2 . A t Also from the table 2; = ê3. For a right-handed primed system 23 = Xêi, or e , =

[ (3 /5)ê1 - (4/5)$2] x 2, = (-3/5)& - (416)$~ and the third row is 1 x: 1 -415 1 -3/5 1 O I .

1.34. Let the angles between the primed and unprimed coordinate directions be given by the table shown on the right. Deter- mine the transformation coefficients aij

and show that the orthogonality conditions are satisfied.

The coefficients aij are direction cosines and may be calculated directly from the table. Thus

- 1 112 -112

aij -

-

o l l f i

1 112 -112

The orthogonality conditions aiiaik = S i , require:

1. For j = k = 1 that a l la l , + a2,a,, + a3,a3, = 1 which is seen to be the sum of squares of the elements in the first column.

2. For j = 2, k = 3 that a12a13+ aZ2a23 + a3,a3, = O which is seen to be the sum of products of corresponding elements of the second and third columns.

3. Any two columns "multiplied together element by element and summed" to be zero. The sum of squares of elements of any column to be unity.

For orthogonality conditions in the form ajiaki = Sjk, the rows are multiplied together instead of the columns. A11 of these conditions are satisfied by the above solution.

1.35. Show that the sum x A ~ ~ + pBij represents the components of a second-order tensor if Aij and Bij are known second-order tensors.

By (1.108) and the statement of the problem, Aij = a p i a q j ~ p q and Bii = apiaqjBiq. Hence

XAij + @Bii = ~ ( ~ ~ a ~ ~ ~ k ~ ) + y(apia,iBg,) = a p i a , i ( ~ ~ k q + yBP,)

which demonstrates that the sum transforms as a second-order Cartesian tensor.

1.36. S ~ O W that ( P i j k + P j k i + P j i k ) X i X j X k = 3Pijk xixjxk.

Since a11 indices are dummy indices and the order of the variables xi is unimportant, each tem of the sum is equivalent to the others. This may be readily shown by introducing new dummy variables. Thus replacing i, j, k in the second and third terms by p, q,r, the sum becomes

Pijkxixjxk + Pqrpxpxqxr + P q p r x p x q x ,

Now change dummy indices in these same terms again so that the form is

Pijk xixixk + Pijk + Pijk xjxixk = 3Pijk xixjxk

1.37. If Bij is skew-symmetric and Aij is symmetric, show that AijBij = 0.

Since A . = A , . and B. . = -B.. A. .B. . -A. .B. . or A..B.. + A..B.. = A..B.. + A B l t 31' 13 t3 I1 I1 1J 13 31 31 L3 i 1 P P PP = O.

Since a11 indices are dummy indices, ApPBpq = AijBij and so 2AijBij = O, or AijBij = 0.


1.38. Show that the quadratic form Dijxixj is unchanged if Dij is replaced by its symmetric part D(ij).

Resolving Dij into i ts symmetric and anti-symmetric parts,

Dij = D(;j) + DLij l = S(Di j + Dji) + &Dij - Dji)

Then D,i j lx ix j = #Dij + Dji)xixj = S(Dijx ix j + Dpqxqxp) = Dijxixj

1.39. Use indicial notation to prove the vector identities

(1) a x (b x c) = (a* c)b - (a . b)c, (2) a x b . a = O

(1) Let v = b X c. Then vi = eijkbjck; and if a X v = w, then

w p = ePqiaqeijkbjck

- ( S p j S q k - SpkSqj)aqbjck (see Problem 1.26)

= aqbpcq - aqbqcp

= (aqcq)b, - (aqbq)cp

Transcribing this expression into symbolic notation,

w = a X (b X c ) = ( a - c ) b - ( a . b )c

( 2 ) Let a X b = v. Thus vi = eij,ajbk; and if h = v a , then X = eiik(aiajbk). But eijk is skew- symmetric in i and i, while (aiajbk) is symmetric in i and j . Hence the product eijkaiajbk vanishes a s may also be shown by direct expansion.

1.40. Show that the determinant

may be expressed in the form ci jkA1iA2jA~k.

If now the substitutions ai = A l i , b, = AZi and ci = A s i are introduced,

From (1.52) and (1.109) the box product [abc] rnay be written

X = eijkaibjck = eijkAliAZjA3k

This result may also be obtained by direct expansion of the determinant. An equivalent expression for the determinant is ei jkAi lAj2Ak3.

x = a . b X c = [abc] = eijkaibjck =

1.41. If the vector vi is given in terms of base vectors a, b, c by vi = aiai + pbi + yCi,

a1 a2 a3

bl b2 b3

C 1 C 2 C 3


I v3 b3 c3 I eiikvibjck By Cramer's rule, a = and by (1.52) and (1 .I 09) , a = a1 bl c1 'pqrapbq~r '

a2 b2 c2

EijkaivjCk cijkaibjvk Likewise /3 =

epqrapbqcr = ~pqrapbqcr '

MATRICES AND MATRIX METHODS (Sec. 1.17-1.20)

1.42. For the vectors a = 3:+42, b = 2 3 - 6 2 and the dyadic D = 3 ? ? + 2 ? k - 43 5 - 5k ̂3, compute by matrix multiplication the products a D, D . b and a . D b.

Let a D = v; then [ v l , v2, v3] = [3,0,4] [ O - -4 ] = 9 , -20,6].

Let D . b = w; then

O -5 -6 -10

Let a D b = v . b = h; then [h] = [9, -20,6] ( 2 1 = [-761.

From (1.132), for principal values h,

1.43. Determine the principal directions and principal values of the second-order Cartesian tensor T whose matrix representation is

which results in the cubic equation h3 - 7h2 + 14h - 8 = ( h - l ) ( h - 2)(h - 4 ) = O whose principal values are h(1, = 1, = 2, h ( 3 ) = 4.

Next let n:l) be the components of the unit normal in the principal direction associated with

Ao , = 1. Then the first two equations of (1.231) give 2n:" - nil' = O and -n:" + 24 ' ' = 0 , from which n:" = nll) = 0; and from nini = 1, nB1) = *I.

For ht2) = 2, (1.131) yields ni2' - nl2' = 0 , -ny' + n f ' = O, and -ny' = O. Thus ny' = ni2' = %1/\/2, since nini = 1 and ni2' = 0.

For h,,, = 4, (1.131) yields -ni3' - n:' = 0, -ni3' - ns' = O , and 3 n F ) = O. Thus n33' = 0, n:3) = -nF' = TI/@.

[Tij] =

The principal axes x: may be referred to the original axes xi through the table of direction cosines

3 -1 o -1 3 O

O 0 1


from which the transformation matrix (tensor) may be written:

1.44. Show that the principal axes determined in Problem 1.43 form a right-handed set of orthogonal axes.

Orthogonality requires that the conditions aijaik = 6 j k be satisfied. Since the condition nini = 1 was used in determining the aii, orthogonality is automatically satisfied for j = k. Multiplying the corresponding elements of á n y row (column) by those of any other row (column) and adding the products demonstrates that the conditions for j + k are satisfied by the solution in Problem 1.43.

As indicated by the plus and minus values of aij in Problem 1.43, there are two sets of principal axes, x: and x:*. As shown by the sketch both sets a re along the principal directions with x: being a right-handed system, x;" a left-handed system.

Finally for the system to be right-handed, A n(2) ~ ( 3 ) = n(l). ~h~~

Fig. 1-19

A A A

e1 e2 e3

i i o

- 1 1 6 1 o

1.45. Show that the matrix of the tensor Tij of Problem 1.43 may be put into diagonal (principal) form by the transformation law T: = a;,ajqTPq, (or in matrix symbols T* = cATcAT).

A = (4 + 4) ê3 = e3


1.46. Prove that if the principal values h ( l ) , h (3) of a symmetric second-order tensor are a11 distinct, the principal directions are mutually orthogonal.

The proof is made for ht2) and h(3) . For each of these (1.129) is satisfied, so that T i j n j 2 ) = ~ ( , , n ~ ~ ' and ~ ~ ~ n j ~ ' = ~ ~ , n : ~ ) . Multiplying the first of these equations by ni3' and the second

Since Tij is symmetric, the dummy indices i and j may be interchanged on the left-hand side of the second of these equations and that equation subtracted from the first to yield

(A,,, - ~ ( ~ , ) n : ~ ' n : ~ ' = O

Since h(,, Z h(,,, their difference is not zero. Hence n:2'n:3' = 0 , the condition for the two directions to be perpendicular.

1.47. Compute the principal values of (T)2 of Problem 1.43 and verify that its principal axes coincide with those of T.

3 -1 3 -1 10 -6

[ T i . = [-i i !][-i i !] = [-: 1;

The characteristic equation for this matrix is

gn"' - Gn(l ' = 1 2

For h( , ) = 1, = n21' = 0 n ( l ) = 2 1 -6n:" $. 9n;" = O ' 3

10 - h -6 O

-6 10 - h O

o O l - h

6 n ( 2 ) - C n ( 2 ) = O 1 2

For A ( 2 , = 4, -Gn?' + 6nk2' = 0 = ni2' = k l f i , n F ) = 0

-Sn'2' - 3 - 0

= ( 1 - h)[(10 - h)2 - 361 = ( 1 - X)(h - 4)(h - 16) = O

For = 16,

-15ni3' = O

from which h(,) = 1, = 4, h (3) = 16. Substituting these into (1.131) and using the condition nini = 1,

which are the same as the principal directions of 1.

1.48. Use the fact that (T)2 has the same principal directions as the symmetrical tensor T to obtain fi when

First, the principal values and principal directions of T are determined. Following the procedure of Problem 1.43, the diagonal form of T is given by


with the transformation matrix being

l l f i lldz l l h

[aijl = [ 0

- 2 1 6 l l f i 11,h

o o Therefore fi = ;" ó 3 and using [aij; to relate this to the original axes by the

transformation fi = ~ ~ f i A, the matrix equation is

CARTESIAN TENSOR CALCULUS (Sec. 1.21-1.23)

1.49. For the function x = Aijxixj where Aij is constant, show that dhldxi = (Aij + Aji)xj and d2X/dxidxj = Aij + Aji. Simplify these derivatives for the case Aij = Aji.

a x axi a%, axi ax Consider - = A.. - X . + A,.%. - Since - z S i k , i t i~ seen that - = Akjz j + Aikzi =

axk %3axk 3 a> ' a x k ' Jxk J x k

( A + A jk)xj. Continuing the differentiation, - - 8% - a2A

- ( A k j + A . ) - - AkP+APk. If A . . = A . ax, axk lk axp 23 3 i 9

- = 2Akjxj and - - - 2APk. axk as, axk

1.50. Use indicial notation to prove the vector identities (a) V X V+ = O, (b ) V V X a = 0.

(a ) By (1.147), V + is written $,i and so v = V X V + has components vi = ~ ~ ~ ~ a ~ + , ~ = ~ i j k + , k j . But eijk is anti-symmetric in j and k, whereas +,kj is symmetric in j and k; hence the product E ~ ~ ~ + , ~ ~ vanishes. The same result may be found by computing individually the components of v. For example, by expansion vl = '123+,23 + ~ ~ 3 ~ + , ~ 2 = ($,23 - +, 32) = 0.

( b ) V V x e = h = ( ~ ~ ~ ~ a ~ , = eijkak,ji = O since ak,ij ak,ji and ~ i j k = -'jik.

1.51. Determine the derivative of the function X = ( X I ) ~ + 2 x 1 ~ ~ - ( ~ 3 ) ~ in the direction of the unit normal n = (2/7)& - (3/7)& - (6/7)ê3 or & = (261 - 3ê2 - 6ê3)/7.

a x A The required derivative is - = V h n = X,ini. Thus an


1.52. If Aij is a second-order Cartesian tensor, show that its derivative with respect to x k , namely Aij,k, is a third-order Cartesian tensor.

F o r the Cartesian coordinate systems xi and xl, xi = ajixi and âxildxi = ajt. Hence

which is the transformation law for a third-order Cartesian tensor.

1.53. If r2 = Xixi and f ( r ) is an arbitrary function of r, show that (a ) V ( f ( r ) ) = f'(r)xlr, and ( b ) V2( f ( r ) ) = f l r ( r ) + Zf'lr, where primes denote derivatives with respect to r.

a f ar ar (a) The components of V f a r e simply f,+ Thus f,i = - - ; and since = 2r -- = 2 ~ . . x . i t ar axi axj ax, 2 ~ a

ar 2 . a f ar follows t h a t - = 3. Thus f,i = - - = f'xi/r.

dxj r ar axi

Use the divergence theorem of Gauss to

S ~ O W that L X i n j d S = V6ij where nj d S

represents the surface element of S, the bounding surf ace of the volume V shown in Fig. 1-20. xi is the position vector of nj d S , and ni its outward normal.

= SijV Fig. 1-20

1.55. If the vector b = V x v, S ~ O W that i h b i s d S = i h,ibi d V where h = h(x i ) is a scalar function of the coordinates.

Since b = V X v, bi = eijkvk, and so

i hbini dS = i rijkhvk, jni d s

= & Li bi d~ since hrijkvk, ii = O


MISCELLANEOUS PROBLEMS

1.56. For the arbitrary vectors a and b, show that

A = (ax b) (a x b) + (a.b)2 = ( ~ b ) ~

Interchange the dot and cross in the first term. Then

h = a - b x ( a x b ) + (a .b ) (a0b)

= a [(b. b)a - (b.a)b] + (a . b)(a0b)

= (a .a ) (b -b ) - (b .a ) (aWb) + (a .b ) (a .b )

= ( a b ) 2

since the second and third terms cancel.

d 1.57. If ii = o x u and V = w x v, show that - (u X v) = ca X (u x v). d t

(a) In symbolic notation,

d z ( u x v ) = h X v + u x ; = ( m X u ) X v + u X ( m X v )

= ( V . m)u - (V U)OI + (U v)a - (U @)V

= (V ' ,)U - (U @)V = a X (U X V)

d ( b ) In indicial notation, let (u X V) = W. Then

d wi = ( E ~ ~ ~ u ~ v ~ ) = eijkGjwk + eijkuj;k

and since Aj = e jpqwpuq and ck = ekmnamVn,

W . i € . . y k E , 3pqapUqwk + eijk'kmnUjamvn ('ijkEkmn - ~ i n k ' k m j ) ~ ~ j a m w n

and using the result of Problem 1.59(a) below,

wi = ( S i m S j n - SinS jm - SimGjn + S i j S m n ) ~ j a m w n

= ( S i j S m n - 8 i n 8 j m ) u j ~ m ~ n E ~ i r n k ' ~ j ~ U j a r n w ~

which is the indicial form of m X (u X v).

. 1.58. Establish the identity cPqsrmnr =

Let the determinant of Aij be given by det A = or columns causes a sign change. Thus

and for an arbitrary number of row changes,

8 m p 8 m q S m s

S n p 8 n q 8 n s

8 r p 8 r q 8 r s

A 1 1 -412 A 1 3

AZl Az2 A 2 3

A 3 1 A 3 2 A 3 3

= -detA

. An interchange of rows

A12 A,, A 1 3

A,, A,, AZ3

A,, A 3 1 A 3 3

A,, A,, A 2 3

A,, A,, A,,

A 3 1 A,, A,,

A m l A m , A m 3

A n l A n a A n 3

A r 1 A, 4 3

=

- - em,, det A

4 0 MATHEMATICAL FOUNDATIONS [CHAP. 1

or column changes,

A,, A,, A I S

-42, - 4 2 , -4%

A,, - 4 3 , A,,

When Aij = S i j , det A = 1 and the identity is established.

- - epqS det A

Hence for a n arbi t rary row and column interchange sequence,

1.59. Use the results of Problem 1.58 to prove (a) cpqscsnr = S P n S ~ r - S p r a q n , ( b ) ~ P , S ~ S Q ~ = - 2 6 ~ ~ .

A,, A m , A m s

A n p A n q A n s

A, A r , A r s

Expanding the determinant of the identity in Problem 1 . 5 8 , -

'pqs'mnr - S m p ( S n q S r s - s n s a r q ) + O m q ( S n s S r p - O n p s r s ) + S m s ( S n p S r q - SnqSrp)

( a ) Identifying s with m yields,

'pqs'snr = S s p ( 8 n q s r s - S n s S r q ) f S s q ( 8 n s s r p - S n p S r s ) + S s s ( S n p ~ r q - S n q S r p ) - - s r p s n q - SpnSrq + S q n S r p - SnpSqr + 3 S n p S r q - 3 S n q s r p - - S n p a r q - S n q S r p

( b ) Identifying q with n in ( a ) , -

~pqs'sqr - S q p a r q - a q q s r p = S p r - 3 S p r = - 2 S p r

- - 'mnr'pqs det A

1.60. If the dyadic B is skew-symmetric B = -B,, show that B , x a = 2a B.

Writ ing B = blêl + b2ê2 + b3ê3 (see Problem 1 . 6 ) ,

B, = bl x ê1 + bz X êz + b3 X ê3

and B , X ~ = ( b l X ê l ) x a + ( b 2 x ê 2 ) x a + ( b 3 x ê 3 ) x a A A = ( a b l ) ê , - ( a . e l ) b l + ( a . b , ) ê z - ( a e z ) b 2 -t ( a b 3 ) ê 3 - ( a ê 3 ) b 3

= a ( b l ê l + bzêz + b 3 ê 3 ) - a . ( ê l b l + ê2b2 + ê 3 b 3 )

- - a . 0 - a . 0 , = 2 a . B

1.61. Use the Hamilton-Cayley equation to obtain ( B ) 4 for the tensor B = Check the result directly by squaring ( B ) 2 .

The characteristic equation for B is given by

By the Hamilton-Cayley theorem the tensor satisfies i ts own characteristic equation. Hence - 2 ( ~ ) ~ - 6 8 + 91 = 0, and multiplying this equation by B yields = 2 ( B ) 3 + 6 ( 8 ) 2 - 9 8 01

(B)4 = 1 0 ( B ) z + 38 - 181. Hence


Checking by direct matrix multiplication of (B)2,

(23)4 =

O 26

1.62. Prove that (a) Aii, (b) AijAij, ( c ) c i j k ~ k j p A t p are invariant under the coordinate transformation represented by (1.1 03), i.e. show that Aii = ~ : i , etc.

(a) By (1.103), Aij = aPiaqjAP,. Hence Aii = a,,aqiA;, = S,,A:, = A:, = Aii.

(b) AijAij = ap iaq jA~qamian jA~n = B ~ ~ s ~ ~ A P ~ A & ~ = ApqApq = A!.A!. 21 13

(C) cijk'kjpAip = €ijk'kjpamianp~hn = ( 8 i j s j p - ~ipsjj)amianpAkn - - (Smn - &mnsjj)Akn = (SmjSnj - ~ m n ~ j j ) ~ k n = ~mjk%jnALn

1.63. Show that the dual vector of the arbitrary tensor Tij depends only upon T~iji but that the product TijSij of Tij with the symmetric tensor Sij is independent of Triji.

By (1.110) the dual vector of Tij is vi = cijkTjk, or vi = eijk(TCjk) + T c j k l ) = ei jk T L j k l since cijkT(jk) = O (ci jk is anti-symmetric in j and k, T ( j k , symmetric in j and k).

For the product Ti jS i j = TCii>Sij + TLij1 Si? Here TLii, Sii = O and T,Sij = TCij,S,.

1.64 Show that D : E is equal to D 0 . E if E is a symmetric dyadic.

Write D = lIijêiêj and E = ~,,ê,ê,. By (1.31), D : E = D i j ~ , , ( ê i êp)(êj 2,). By (1.35), D..E = A A A

DijEpq ( g j - $,)(Si . 3,) = DijE,, (êj %)( ei e,) since E,, = E,,. Now interchanging the A A

role of dummy indices p and q in this last expression, D E = êq)( ei e,).

1.65. Use the indicial notation to prove the vector identity V x (a x b) = b Va - b(V a) + a(V*b) - a - V b .

Let V X (a x b ) = v; then v, = ~ ~ , ~ c ~ ~ ~ a ~ a ~ b ~ or

V p = 'pqi'ijk (ajbk),q = epqi'ijk(aj,qbk ajbk,q)

= ( S p j s q k - S p k s q j ) ( a j , q bk + ~ j ~ k , q ) = bq - aq,q bp + - aqbp,q

Hence v = b V a - b(V - a) + a(V b ) - a V b .

1.66. By means of the divergente theorem of Gauss show that & X (a X x) dS = 2aV

where V is the volume enclosed by the surface S having the outward normal n. The position vector to any point in V is x, and a is an arbitrary constant vector.

In the indicial notation the surface integral is cqPinpcijkajxkdS. By (1.157) this becomes s, the volume integral (rqPirijkajxk),, dV and since a is constant, the last expression becomes

q p j , p d = ( S q j s p k - s q k s p ~ a j x k , p dv = (aqxp,p - apxq,p) 'V

: L (aq S p p - a, a,,) dV = (3aq - a,) dV = Za,V


1.67. For the reflection of axes shown in Fig. 1-21 show that the transformation is orthogonal.

From the figure the transformation matrix is

The orthogonality conditions aijaik = S j k or ajiaki = S j k a r e clearly satisfied. In matrix form, by (1.117),

i - : :l[u-i 01 r: : 01 L 0 0 1 ] L 0 0 1 J L 0 0 1 J Fig. 1-21

1.68. Show that (I x v) D = v x D. A A A A A A A A A

I X v = ( i i + j j + k k ) X ( v , i + v , j + v , k ) A A A A A A A A

= i ( v , k - v , j ) + j ( - v , k + v , i ) + k ( v , j - v y i ) A A A A A A

= ( v X i ) i + ( v X j ) j + ( v X k ) k = v X I

Hence ( I X v ) - D = V X I * D = V X D .

Supplementary Problems A A A A A h

1.69. Show tha t u = i + j - k and v = i - j a r e perpendicular to one another. Determine w so

t h a t u, v, & forms a right-handed triad. Ans. w = ( - l / f i ) ( ? + 3 + 2 c )

1.70. Determine the transformation rnatrix between the u,v,W axes of Problem 1.69 and the coordinate directions.

1.71. Use indicial notation to prove (a ) V x = 3, (b) V X x = O, (c) a . Vx = a where x is the position vector and a is a constant vector.

1.72. Determine the principal values of the symmetric par t of the tensor Ti j =

Ans. h ( , , = -15, h ( Z ) = 5 , h,,) = 10 -3 -18


1.73. For the symmetric tensor Tij = determine the principal values and the directions of the principal axes.

Ans. hc l , = 2 , A ( z , = 7, A , , , = 12,

1.74. Given the arbi t rary vector v and any uni t vector ê, show t h a t v may be resolved into a component parallel and a component perpendicular to 2, i.e. v = (v *ê )ê + ê X (v X 3 ) .

1.75. If V . v = O, V x v = 6 and V X w = -c, show tha t V2v = y.

1.76. Check the result of Problem 1.48 by direct rnultiplication to show t h a t fifi = T.

1.77. Determine the square root of the tensor B =

* ( & + I ) # & - I ) 0

1.78. Using the result of Problem 1.40, det A = E ~ ~ ~ A ~ ~ A ~ ~ A ~ ~ , show t h a t det (AB) = det A det 0.

1.79. Verify t h a t (a) O3,,vp = v3, ( b ) 8siAji = Aj3, (C) Si j e i j k = 0 , ( d ) ai2aj3Aij = A23.

1.80. Let the axes 0 x ~ x ~ x ~ be related to Oxlx,x3 by the table

( a ) Show tha t ' the orthogonality conditions aijaik = S j k and apqaSq = a,, a r e satisfied.

( b ) What a r e the primed coordinates of the point having position vector x = 2 ê , - ê3?

( c ) What is the equation of the plane xl - x, + 3x3 = 1 in the primed system?

Ans. ( b ) ( 2 / 5 f i , 1115, - 2 1 5 f i ) ( c ) fi X ; - X ; - 2f i X ; = 1

h

1.81. Show t h a t the volume V enclosed by the surface S may be given a s V = Q

x ia the ponition vector and n the uni t normal to the surfaee. Hint: Write V = (116) 1 (xixi) , ini dS and use (1.157). ' S

Analysis of Stress

2.1 THE CONTINUUM CONCEPT

The molecular nature of the structure of matter is well established. In numero- --- --

investigations of material behavior, however, the individual molecule is of no conce- _and - -- ---- -only tKe35havior of &e material as a whole is-deemed . important. For these cases the observed macroscopic behavior is usually explained by disregardinfmolecular considerations and, instead, by assuming the material to be continuously distribuied throughout its volume and to completely - -- fi11 the space it occupies. This continuum concept of matter is the

-fundamental postulate of Continuum Mechanics. Within the limitations for which the continuum assumption is valid, this concept provides a framework for studying the behavior of solids, liquids and gases alike.

Adofion of t_hecontinuum viewpoint as the basis for the mathematical description of -- -- - - -- - - -- - - - -- - - . -. - material behavior means that - field _- quantities _ _ such as stress and displacement are expressed _ _ -- --_-_r _ -

functions of the space coordinates and time.

2.2 HOMOGENEITY. ISOTROPY. MASS-DENSITY A homogeneoousmaterial is-tne having-id~ntical propertiesat a11 poi&. With respect

to some property, a material is isotropic if that property is the same in a11 directions a t a , _- ____I_--- - --- -- -_ - - - -A - - - - point. A material is called anisotropic with gspect to those properties which are directional -- -- - - . - - - a t a pÒi$t. - - -

The concept of density is developed from the mass-volume ratio in the neighborhood of a point in the continuum. In Fig. 2-1 the mass in the small element of volume AV is denoted by AM. The average density of the material within AV is therefore

- anl P ( a v ) - AV (2.1)

The density a t some interior point P of the volume element AV is given mathematically in accordance with the continuum concept by the limit,

al)I = lim -- = -- d M

*"-O A V d!' (2.2)

Mass-density p is a scalar quantity. Fig. 2-1

CHAP. 21 ANALYSIS O F STRESS

2.3 BODY FORCES. SURFACE FORCES

Forces are vector quantities which are best described by intuitive concepts such as push or pull. Those forces which act on a11 elements of volume of a continuum are known as body forces. Examples are gravity and inertia forces. These forces are represented by the symbol bi (force per unit mass), or as pi (force per unit volume). They are related through the density by the equation

pbi = pi or pb = p (2.3)

Those forces which act on a surface element, whether it is a portion of the bounding surface of the continuum or perhaps an arbitrary interna1 surface, are known as surface forces. These are designated by f i (force per unit area). Contact forces between bodies are a type of surface forces. .

,'

2.4 CAUCHY'S STRESS PRINCIPLE. THE STRESS VECTOR

A material continuum occupying the region R of space, and subjected to surface forces f i and body forces bi, is shown in Fig. 2-2. .As a result of forces being transmitted from one portion of the continuum to another, the material within an arbitrary volume V enclosed by the surface S interacts with the material outside of this volume. Taking ni as the outward unit normal a t point P of a small element of surface AS of S, let Afi be the resultant force exerted across AS upon the material within V by the material outside of V. Clearly the force element Afi will depend upon the choice of AS and upon ni. I t should also be noted that the distribution of force on AS is not necessarily uniform. Indeed the force distribution is, in general, equipollent to a force and a moment a t P, as shown in Fig. 2-2 by the vectors Afi and AMi.

Fig. 2-2 Fig. 2-3

The average force per unit area on AS is given by Af,lAS. The Cauchy stress principie asserts that this ratio Afi/As tends to a definite limit df i /dS as AS approaches zero a t the point P, while a t the same time the moment of afi about the point P vanishes in-thgimiting - . - - __h

process. The resulting vector df i ldS (force per unit area) is called the stress vector ti"' -and i s shown in Fig. 2-3. If the moment a t P were not to vanish in the limiting process, a couple-stress vector, shown by the double-headed arrow in Fig. 2-3, would also be defined a t the point. One branch of the theory of elasticity considers such couple stresses but they are not considered in this text.

46 ANALYSIS O F STRESS [CHAP. 2

Mathematically the stress vector is defined by

dfi A O r f (n) = . Af

1im - = d f dS as-o AS

The notation tln' (or tt;)) is used to emphasize the fact that the stress vector a t a given point P in the continuum depends explicitly upon the particular surface element AS chosen there, a s represented by the unit normal ni (or ^n). For some differently oriented surface element, having a different unit normal, the associated stress vector a t P will also be different. The stress vector arising from the action across A S a t P of the material within

A

V upon the material outside is the vector -t:"'. Thus by Newton's law of action and reaction,

= - h

- t' - "' (2.5)

The stress vector is very often referred to as the t r a c t i o n vector .

2.5 STATE OF STRESS AT A POINT. STRESS TENSOR

At an arbitrary point P in a continuum, Cauchy's stress principie associates a stress A

vector t:"' with each unit normal vector ni, representing the orientation of an infinitesimal surface element having P as an interior point. This is illustrated in Fig. 2-3. The totality of a11 possible pairs of such vectors tiE;' and ni a t P defines the sta te of stress a t that point. Fortunately it is not necessary to specify every pair of stress and normal vectors to completely describe the state of stress a t a given point. This may be accomplished by giving the stress vector on each of three mutually perpendicular planes a t P. Coordinate transformation equations then serve to relate the stress vector on any other plane a t the point to the given three.

Adopting planes perpendicular to the coordinate axes for the purpose of specifying the state of stress a t a point, the appropriate stress and normal vectors are shown in Fig. 2-4.

Fig. 2-4

For convenience, the three separate diagrams in Fig. 2-4 are often combined into a single schematic representation as shown in Fig. 2-5 below.

Each of the three coordinate-plane stress vectors may be written according to (1.69) in terms of its Cartesian components as

CHAP. 21 ANALYSIS O F STRESS 47

Fig. 2-5 Fig. 2-6

The nine stress vector components, A t:ei) E u..

are the components of a second-order Cartesian tensor known as the stress tensor. The equivalent stress dyadic is designated by Z, so that explicit component and matrix representations of the stress tensor, respectively, take the forms

Pictorially, the stress tensor components may be displayed with reference to the coordinate planes as shown in Fig. 2-6. The components perpendicular to the planes (all, a2,, a,,) are called normal stresses. Those acting in (tangent to) the planes (al2, a,,, a,,, a,,, os?, o,,) are called shear stresses. A stress component is positive when it acts in the pgsitive direction of the coordinate axes, and on a plane whose outer normal points in one of the positive coordinate directions. The component uii acts in the direction of the jth coordinate axis and on the plane whose outward normal is parallel to the ith coordinate axis. The stress components shown in Fig. 2-6 are a11 positive.

2.6 THE STRESS TENSOR - STRESS VECTOR RELATIONSHIP

The relationship between the stress ten- h

sor uij a t a point P and the stress vector t,'"' on a plane of arbitrary orientation a t that point may be established through the force equilibrium or momentum balance of a small tetrahedron of the continuum, having its vertex a t P. The base of the tetrahedron is taken perpendicular to %, and the three faces are taken perpendicular to the coordinate planes as shown by Fig. 2-7. Des- ignating the area of the base ABC as dS, the areas of the faces are the projected areas, dSl = dS nl for face CPB, dS2 = dS n2 for face APC, dS3 = dS n3 for face BPA or Fig. 2-7

48 ANALYSIS OF STRESS [CHAP. 2

The average traction vectors -t:'&' on the faces and t:':' on the base, together with the average body forces (including inertia forces, if present), acting on the tetrahedron are shown in the figure. Equilibrium of forces on the tetrahedron requires that

tf(.) d S - tf'"' d s l - tf'"' d s 2 - tf'"' d s 3 + pbf d V = 0 (2.10)

If now the linear dimensions of the tetrahedron are reduced in a constant ratio to one another, the body forces, being an order higher in the small dimensions, tend to zero more rapidly than the surface forces. At the same time, the average stress vectors approach the specific values appropriate to the designated directions a t P. Therefore by this limiting process and the substitution (2.9), equation (2.10) reduces to

A

Cancelling the common factor d S and using the identity tje" =. uji, (2.11) becomes

Equation (2.12) is also often expressed in the matrix form

rt$i = Ln1k] [vki]

which is written explicitly 1 '=12 u13

A A A

The matrix form (2.14) is equivalent to the component equations

2.7 FORCE AND MOMENT EQUILIBRIUM. STRESS TENSOR SYMMETRY Equilibrium of an arbitrary volume V

of a continuum, subjected to a system of surface forces t:;' and body forces bi (including inertia forces, if present) as shown in Fig. 2-8, requires that the resultant force and moment acting on the volume be zero.

Summation of surface and body forces results in the integral relation,

1 ti" d~ + 1 pbi d V = O

or (2.1 6 ) L t ( ^ ' d ~ + L ~ b d ~ = O

A

Replacing tin' here by ujinj and converting the resulting surface integral to a volume integral by the divergence theorem o£ Gauss (1.157). equation (2.16) becomes Fig. 2-8

CHAP. 21 ANALYSIS OF STRESS 49

Since the volume V is arbitrary, the integrand in (2.17) must vanish, so that

ajijj + pbi = O or V I + pb = O (2.1 8)

which are called the equilibrium equations.

In the absence of distributed moments or couple-stresses, the equilibrium of moments about the origin requires that

in which xi is the position vector of the elements of surface and volume. Again, making h

the substitution tin' = ujini, applying the theorem of Gauss and using the result expressed in (2.18), the integrais of (2.19) are combined and reduced to

For the arbitrary volume V, (2.20) requires

- Equation (2.21) represents the equations u,, = a,,, a,, - a,,, a,, = V,,, or in a11

which shows that the stress tensor is szjmmetric. In view of (2.22), the equilibrium equations (2.18) are often written

o ~ ~ , ~ + pbi = O

which appear in expanded form as

2.8 STRESS TRANSFORMATION LAWS

At the point P let the rectangular Cartesian coordinate systems P X I X ~ X ~ and Px;x:x~ of Fig. 2-9 be related to one another by the table of direction cosines

Fig. 2-9


or by the equivalent alternatives, the transformation matrix [aijl, or the transformation dyadic

According to the transformation law for Cartesian tensors of order one (1.93), the A

components of the stress vector t ln) referred to the unprimed axes are related to the primed A

axes components tl'"' by the equation A A t ; ( P ) = a..t(n) or t '(n) -

J - A . (2.2 6 )

Likewise, by the transformation law (1.102) for second-order Cartesian tensors, the stress tensor components in the two systems are related by

O; = aiPajqap, or 1' = A . X A, (2.27)

In matrix form, the stress vector transformation is written A

[ti:")] = [aii] [t$] (2.28)

and the stress tensor transformation as

LaijI [a,] [apq] [aqjI (2.29)

Explicitly, the matrix multiplications in (2.28) and (2.29) are given respectively by

and

2.9 STRESS QUADRIC OF CAUCHY At the point P in a continuum, let the stress

tensor have the values uij when referred to directions parallel to the local Cartesian axes P61g253 shown in Fig. 2-10. The equation

uijgigj = f k2 (a constant) (2.32)

represents geometrically similar quadric surfaces having a common center a t P. The plus or minus choice assures the surfaces are real.

The position vector r of an arbitrary point lying on the quadric surface has components gi = rni, where ni is the unit normal in the direction of r. At the point P the normal component aNni of the

A

stress vector tin' has a magnitude

' J ~ = t i l ) n i = t ( D ) . n = ~ i j n i n j (2.33) Fig. 2-10

Accordingly if the constant k2 of (2.32) is set equal to uNr2, the resulting quadric

uij[icj = -iaNr2 (2.34)


is called the stress quadric of Cauchy. From this definition i t follows that the magnitude O, of the normal stress component on the surface element dS perpendicular to the position vector r of a point on Cauchy's stress quadric, is inversely proportional to r2, i.e. U , = i-k2/r2.

Furthermore i t may be. shown that the stress vector t$' acting on dS a t P is parallel to the normal of the tangent plane of the Cauchy quadric a t the point identified by r.

2.10 PRINCIPAL STRESSES. STRESS INVARIANTS. STRESS ELLIPSOID

At the point P for which the stress tensor com- A

ponents are aij, the equation (2.12), t:"' = uiini, as- A

sociates with each direction ni a stress vector t,'"'.

Those directions for which tiA and ni are collinear as shown in Fig. 2-11 are called principal stress directions. For a principal stress direction,

A h

t i n ) = &&i A

or t'"' - - a n (2.35)

in which a, the magnitude of the stress vector, is called a principal stress zjalue. Substituting (2.35) into (2.12) and making use of the identities ni = Siinj and uij = ajii results in the equations

A

(aij - Sija)ni = O o r (1 - lu) . n = 0 (2.36) Fig. 2-11

In the three equations (2.36), there are four unknowns, namely, the three direction cosines ni and the principal stress value O.

which upon expansion yields the cubic polynomial in a,

For solutions of (2.36) other than the trivial one nj = 0, the determinant of coefficients, luij - Sijal, must vanish. Explicitly,

where I, = aii = tr 1 (2.39)

lu.. - S . . I = 0 or 11 7 1 ~

11, = +(aiiujj - a. .c..) 21 li

111, = laijl = det

are known respectively as the first, second and third stress invariants.

a l l - "12 u13

a~~ u22 - u23

as 1 a32 O33 -

The three roots of (2.38), a,,,, a,,,, a(,) are the three principal stress values. Associated with each principal stress a,,,, there is a principal stress direction for which the direction cosines n,'k' are solutions of the equations

= 0 (2.,?7)


This resolution is shown in Fig. 2-13 where the axes are chosen in the principal stress directions and it is assumed the principal stresses are ordered according to U, > u,, > u,,,.

Hence from (2.12), the components of t$) are

ti.^' = u1n1 =

'-'II~z (2.48) h t(n) =

3 u111n3

and from (2.33), the normal component magnitude is

U, = u,n; + ~ , , n ; + ~,,,n,2 (2.49)

Substituting (2.48) and (2.49) into (2.47), the squared magnitude of the shear stress as a function of the direction cosines ni is given by Fig. 2-13

The maximum and minimum values of U, may be obtained from (2.50) by the method of Lagrangian rnultipliers. The procedure is to construct the function

F = U: - Anini (2.51)

in which the scalar A is called a Lagrangian multiplier. Equation (2.51) is clearly a function of the direction cosines ni, so that the conditions for stationary (maximum or minimum) values of F are given by JFldni = O. Setting these partials equal to zero yields the equations

n2{u TI - 2uI1(uIn; + ~ , , n ; + ~,,,n;) + A} = O (2.523)

n3{uSII - 2~, , , (~,n; + ~ , , n ; + ~,,,n;) + A} = O (2.52~)

which, together with the condition nini = 1, may be solved for A and the direction cosines nl, ne, na, conjugate to the extremum values of shear stress.

One set of solutions to (2.52), and the associated shear stresses from (2.50), are

n 1 = & 1 , n z = O , n 3 = 0 ; f o r w h i c h ~ , = O (2.53~)

nl = O, na = 11, n3 = 0; for which a, = O (2.533)

ni = O, nn = O, na = ?I; for which U, = O (2.53~)

The shear stress values in (2.53) are obviously minimum values. Furthermore, since (2.35) indicates that shear components vanish on principal planes, the directions given by (2.53) are recognized as principal stress directions.

A second set of solutions to (2.52) may be verified to be given by

nl = 0, n2 = - + I , n3 = +l/fi; - for which U, = (a,, - uIII)/2 (2.54a)

ni = *l/fi, n z = 0, n3 = *1/fi; for which a, = (oIII - uI)/2 (2.54b)

nl = 11/fi, nz = e l l f i , n3 = 0; for which U, = (aI - uII)/2 (2.54c)

Equation (2.54b) gives the maximum shear stress value, which is equal to half the difference of the largest and smallest principal stresses. Also from (2.543), the maximum shear stress component acts in the plane which bisects the right angle between the directions of the xhaximum and minimum principal stresses.


2.12 MOHR'S CIRCLES FOR STRESS

A convenient two-dimensional graphical representation of the three-dimensional state of stress a t a point is provided by the well- known Mohr's stress circles. In developing these, the coordinate axes are again chosen in the principal stress directions a t P as shown by Fig. 2-14. The principal stresses are assumed to be distinct and ordered according to

> OII > u~~~ (2.55) For this arrangement the stress vector t ln) has normal and shear components whose magnitudes satisfy the equations Fig. 2-14

U , = u1nS + uI1n; + uI1,n; (2.56)

Combining these two expressions with the identity nini = 1 and solving for the direction cosines ni. results in the eauations

These equations serve as the basis for Mohr's stress circles, shown in the "stress plane" of Fig. 2-15, for which the U , axis is the abscissa, and the U , axis is the ordinate.

In (2.58a), since U , - u,, > O and U , - uII1 > O from (2.55), and since (n~)' is non-negative, the numerator of the right-hand side satisfies the relationship

which represents stress points in the (V,, os) plane that are on or exterior to the circle

In Fig. 2-15, this circle is labeled CI.

Fig. 2-15


and, on the bounding circle arcs BC, CA and AB,

nl = cosx12 = O on BC, n2 = COSTIZ = O on CA, n3 = ~ 0 ~ x 1 2 = O on AB

According to the first of these and the equation (2.58a), stress vectors for Q located on BC will have components given by stress points on the circle CI in Fig. 2-15. Likewise, CA in Fig. 2-16 corresponds to the circle C2, and AB to the circle C3 in Fig. 2-15.

The stress vector components O, and O, for an arbitrary location of Q rnay be determined by the construction shown in Fig. 2-17. Thus point e may be located on C3 by drawing the radial line from the center of C3 a t the angle 2B. Note that angles in the physical space of Fig. 2-16 are doubled in the stress space of Fig. 2-17 (arc AB subtends 90" in Fig. 2-16 whereas the conjugate stress points O, and O,, are 180" apart on C3). In the same way, points g, h and f are located in Fig. 2-17 and the appropriate pairs joined by circle arcs having their centers on the O, axis. The intersection of circle arcs ge and hf

h

represents the components and O, of the stress vector tin) on the plane having the normal direction ni a t Q in Fig. 2-16.

Fig. 2-17

2.13 PLANE STRESS

In the case where one and only one of the principal stresses is zero a state of plane stress is said to exist. Such a situation occurs a t an unloaded point on the free surface bounding a body. If the principal stresses are ordered, the Mohr's stress circles will have one of the characterizations appearing in Fig. 2-18.

= O

Fig. 2-18


If the principal stresses are not ordered and the direction of the zero principal stress is taken as the X Q direction, the state of stress is termed plane stress parallel to the xlxz plane. For arbitrary choice of orientation of the orthogonal axes X I and xz in this case, the stress matrix has the form

ull @12 0 (2.65)

The stress quadric for this plane stress is a cylinder with its base lying in the X I X Z plane and having the equation

u,,xS + 2 u l z ~ , ~ , + u,,x~ = -Ck2 (2.66)

Frequently in elementary books on Strength of Materials a state of plane stress is represented by a single Mohr's circle. As seen from Fig. 2-18 this representation is necessarily incomplete since a11 three circles are required to show the complete stress picture. In particular, the maximum shear stress value a t a point will not be given if the single circle presented happens to be one of the inner circles of Fig. 2-18. A single circle Mohr's diagram is able, however, to display the stress points for a11 those planes a t the point P which include the zero principal stress axis. For such planes, if the coordinate axes are chosen in accordance with the stress representation given in (2.65), the single plane stress Mohr's circle has the equation

The essential features in the construction of this circle are illustrated in Fig. 2-19. The circle is drawn by locating the center C a t = (all + O,,)/:! and using the radius R = j/[(u1, - uzz) /2]2 + given in (2.67). Point A on the circle represents the stress state on the surface element whose normal is nl (the right-hand face of the rectangular parallelepiped shown in Fig. 2-19). Point B on the circle represents the stress state on the top surface of the parallelepiped with normal nz. Principal stress points U , and a,, are so labeled, and points E and D on the circle are points of maximum shear stress value.

Fig. 2-19

2.14 DEVIATOR AND SPHERICAL STRESS TENSORS

I t is very often useful to split the stress tensor uij into two component tensors, one of which (the spherical or hydrostatic stress tensor) has the form


where U , = - p = ukk/3 is the mean normal stress, and the second (the deviator s tress tensor) has the form

This decomposition is expressed by the equations

uij = sijUkk/3 + Sij or 1 = UMI + tD The principal directions of the deviator stress tensor Sij are the same as those of the

stress tensor uij . Thus principal deviator s tress values are -

S ( k ) - u ( k ) - a, (2.71)

The characteristic equation for the deviator stress tensor, comparable to (2 .38) for the stress tensor, is the cubic

I t is easily shown that the first invariant of the deviator stress tensor Ir, is identically zero, which accounts for its absence in (2 .72) .

Solved Problems

STATE OF STRESS AT A POINT. STRESS VECTOR. STRESS TENSOR (Sec. 2.1-2.6)

2.1. At the point P the stress vectors tin) A

and t:"*' act on the respective surface elements ni AS and n: AS*. Show that

the component of ti.^) in the direction of n: is equal to the component of t:^.*) in the direction of ni.

It is required to show that h

t:^.*'n, =

A

Frorn (2.12) tln*'ni = ojin;ni, and by (2.22) o. . = 0 . .

J i so that A

o..n*n. = (o..n.)n* = t:")n; 31 1 I 1J 1 1 Fig. 2-20


2.2. The stress tensor values a t a point P are given by the array

7 o -2

-2 o 4

Determine the traction (stress) vector on the plane a t P whose unit normal is íi = (213)êi - (213) $2 + (113) $3.

From (2.12), t';' = n. 1. The multiplication is best carried out in the matrix form of (2.13): 7 -

7 o -2 A A A

[ t p ) , t?', t!j")] = [2 /3 , -213, 1/31

h 1 0 A Thus tCn' = 4 e , - 3 e,.

2.3. For the traction vector of Problem 2.2, determine (a) the component perpendicular to the plane, (b) the magnitude of t i n ) , ( C ) the angle between tin) and G.

A - A ( a ) tin' a = (4 2, - , e,) (% 2, - ê, + Q ê 3 ) = 4419

A

( b ) jtin'( = d 1 6 + 10019 = 5.2

( c ) Since tia' 6 = t : ; ' [ cos O , cos O = (44/9) /5 .2 = 0.94 and e = 20'.

, , A

2.4. The stress vectors acting on the three coordinate planes are given by t le l ' , tlez' and t i ê3 ) . Show that the sum of the squares of the magnitudes of these vectors is independent of the orientation of the coordinate planes.

Let S be the sum in question. Then A A A A A A s = t;ei)t:ei) + t(.edt!ez) + t!edt(ed

t t a i

which from (2.7) becomes S uliuli + uziu2i + u3iu3i = ujiuji, an invariant.

2.5. The state of stress a t a point is given by the stress tensor

where a, b, c are constants and u is some stress value. Determine the constants a, b and c so that the stress vector on the octahedral plane (2 = (11fi)êl + ( 1 l f i ) ê ~ + (11fl )ê3) vanishes.

A

In matrix forrn, tin' = uijnj must be zero for the given stress tensor and normal vector.

Solving these equations, a = b = c = -1 /2 . Therefore the solution tensor is

-012 -012

-0/2 -012


2.6. The stress tensor a t point P is given by the 5 3

array

( 1 - i) t : = - 5 3 1

Determine the stress vector on the plane passing through P and parallel to the plane ABC shown in Fig. 2 - 2 1 .

B 5 2

The equation of the plane ABC is 3x1 + 6x2+ 22 , = 12, and the unit normal to the plane is therefore (see Prob- lem 1.2)

A n = 3ê1 + d ê + 2s 7 2 7 3 Fig. 2-21

From (2.14), the stress vector may be determined by matrix multiplication, -

Thus

2.7. The state of stress throughout a continuum is given with respect to the Cartesian axes 0 ~ 1 ~ 2 x 3 by the array

3 x i x 2 5 x ;

I: =

2 x 3

Determine the stress vector acting a t the point P ( 2 , 1 , fi) of the plane that is tangent to the cylindrical surface x ; + x3 = 4 a t P.

At P the stress components a re given by

The unit normal to the surface a t P is determined from grad + = V+ = ~ ( 2 ; + 23 - 4 ) . Thus V4 = 2x2ê2 + 223ê3 and so

ê ,h Therefore the unit normal a t P is = -ê,. 2 2 This may also be seen in Fig. 2-22. Finally the stress vector a t P on the plane I to R is given by

or t':' = 5 $,I2 + 3 ê2 + d ê 3 . Fig. 2-22

, CHAP. 21 ANALYSIS O F STRESS 61

EQUILIBRIUM EQUATIONS (Sec. 2.7)

2.8. For the distribution of the state of stress given in Problem 2.7, what form must the body force components have if the equilibrium equations (2.24) are to be satisfied everywhere.

Computing (2.24) directly from i given in Problem 2.7,

3x2 + 10x2 -k 0 + pbl = 0

O + O + 2 + p b 2 = O

O + O + O + p b 3 = O

These equations are satisfied when bl = -13x21p, b2 = -2Ip, b3 = 0.

2.9. Derive (2.20) from (2.19), page 49. Starting with equation (2.19),

substitute t:" = oiini in the surface integral and convert the result to a volume integral by (1.157):

( E ~ ~ ~ x ~ o ~ ~ ) ~ ~ d s = (eijkxjopk),p dV

Carrying out the indicated differentiation in this volume integral and combining with the first volume integral gives

Eiik(~j ,popk + ~ j ( ~ p k , p + pbk)} dV = O

- But from equilibrium equations, u ~ ~ , ~ + pbk = 0; and since xi,, - this volume integral

reduces to (2.20),

STRESS TRANSFORMATIONS (Sec. 2.8) \

2.10. The state of stress a t a point is given with respect to the Cartesian axes 0 ~ ~ x 2 ~ ~ by the array

I

Determine the stress tensor Z' for the rotated axes OxíxSxl related to the unprimed axes by the transformation tensor

l l f i l l f i

1 112

The stress transformation law is given by (2.27) as ali = aiPaiqopq O r 1' = A A,. The de- tailed calculation is best carried out by the matrix multiplication [ali] = [aip] [opq] [aqi] given in (2.29). Thus

l l f i - 1 l f i

[o$] =

1 112 -112

o


2.11. Show that the stress transformation law may be derived from (2.33), the equation uN = u i j n i n j expressing the normal stress value on an arbitrary plane having the unit normal vector ni.

Since U N is a zero-order tensor, i t is given with respect to an arbitrary set of primed or unprimed axes in the same form as

ohr = o!.n!nj = uijni n j a3 1

and since by (1.94) nE = aijnj,

U!.n! n! = o!.a. n a . n = E3 Z 3 21 IP P 54 q " N = "PqnPnq

where new dummy indices have been introduced in the last term. Therefore

("!.a. a . - 23 IP 3 9 O

and since the directions of the unprimed axes are arbitrary,

2.12. For the unprimed axes in Fig. 2-23, the stress tensor is given by

o o U i j = (i ; :)

Determine the stress tensor for the primed axes specified as shown in the figure.

I t is first necessary to determine completely the transformation matrix A. Since x; makes equal angles with the xi axes, the first row of the transformation table together with a,, is known. Thus Fig. 2-23

Using the orthogonality equations aijaik = Si,, the transformation matrix is determined by computing the missing entries in the table. I t is left as an exercise for the student to show that


l / , h l l f i - 1 l f i

1 1 6 1 1 6 l l f i

The result obtained here is not surprising when one considers the Mohr's circles for the state of stress having three equal principal stress values.

CAUCHY'S STRESS QUADRIC (Sec. 2.9)

2.13. Determine the Cauchy stress quadric a t P for the following states of stress:

(a) uniform tension o,, = o,, = o,, = a; o,, = u,, = a,, = O - - -- ( b ) uniaxial tension o,, = U ; o,, - V~~ - (r l2 - u13 = = O - ( C ) simple shear o,, = o,, = T ; a,, - a,, = = a,, = o,, = O

- - - - (d) plane stress with a,, = u = o,,; - o;, - T ; - 03, - = 0 . From (2.32), the quadric surface is given in symbolic notation by the equation { * I J = *k2.

Thus using the matrix form,

a ) [{I, [i E !] [!I = ~ 1 : + < + = *k2

2 = kk2lcr. Hence the quadric surface for uniform tension is the sphere 5: + 522 + c3

Hence the quadric surface for uniaxial tension i s a circular cylinder along the tension axis.

and so the quadric surface for simple shear is a hyperbolic cylinder parallel to the 5, axis.

(d) [Si? 52,531 [ [ ] = 05 ; + 2 ~ ( ~ { ; + 0 5 : = k 2

O 0 0

and so the quadric surface for plane stress is a general conic cylinder parallel to the zero principal axis.

2.14. Show that the Cauchy stress quadric for a state of stress represented by

= ia i is an ellipsoid (the stress ellipsoid) when a, b and c are a11 of the same sign.

The equation of the quadric is given by

53 - +k2 Therefore the quadric surface is the ellipsoid g+ g+ - - L bc a c a b a b c '


PRINCIPAL STRESSES (Sec. 2.10-2.11)

2.15. The stress tensor a t a point P is given with respect to the axes 0 x 1 ~ 2 ~ ~ by the values

Determine the principal stress values and the principal stress directions represented by the axes Oxtx,*x3.

From (9.37) the principal stress values o are given by

The roots are the principal stress values = -2, = 1, = 4. Let the x: axis be the direction of o ( l ) , and let n;" be the direction cosines of this axis. Then from (2.42),

(3 + 2)n i1) + nk l ) + n"' = 0 3

3 - 0 1 1

1 -a 2

1 2 -a

Hence n:" = O. n") = -ni" and since nini = 1, (ni1))2 = 112. Therefore nil) = 0, nil' = l l f i , ' 2

n i l ) = -i/dZ. Likewise, let x: be associated with a(2) . Then from (2.42),

Zn:2' + n ( 2 ) + n ( 2 ) = 0 2

= O or, upon expansion, (a + 2)(0 - 4) (o - 1 ) = O

Finally, let xB be associated with Then from (2.42),

-n(3) + n ( 3 ) + = 0 2

so that ni3' = - 2 / f i , ni3) = - 1 1 6 , n F ) = -1 / f i

2-16. show that the transformation tensor of direction cosines determined in Problem 2.15 transforms the original stress tensor into the diagonal principal axes stress tensor-

According to (2.29), [Gj] = [aip][aPq][aQj], which for the ~ rob lem a t hand becomeS

o -,h -2 o o

O 0 4


2.17. Determine the principal stress values and principal directions for the stress tensor

and therefore n:3' = n:' = ni3) = l l f i . For o(,) = = 0, (2.42) yield

From (2.37), 7 7 - a T

~ 7 ~ a L

which together with nini = 1 are insufficient to determine uniquely the first and second principal directions. Thus any pair of axes perpendicular to the n:3' direction and perpendicular to each other may serve a s principal axes. For example, consider the axes determined in Problem 2.12, for which the transformation matrix is

l l f i l l f i l l f i

[aiil

= O or (7 - a)[-27a + u2] + 27% = [37 - 0102 = 0.

According to the transformation law (2.29), the principal stress matrix [a*,] is given by

Hence = O , <rc2) = 0, o ( 3 ) = 37. For o(3) = 37, (2.42) yield

Show that the axes O x T x ~ x ~ , (where x,*, x3 and x: are in the same vertical plane, and xy, xi and x2 are in the same horizontal plane) ate also principal axes for the stress tensor of Problem 2.17.

The transformation matrix [aii] relating the two sets of axes clearly has the known elements

[aijl =

11,h 11,h l l f i

as is evident from Fig. 2-24. From the orthogonality conditions aijaik = Sjk, the remaining four elements are determined so that

h 1

Fig. 2-24


- 1 l f i 11d5

1 1 f i 1 6

1 1 6 1 1 a lld3

As before,

O I

2.19. Show that the principal stresses and the stress components oij for an arbitrary set of axes referred to the principal directions through the transformation coefficients

aij are related by oij = 5 ap,apjy . p = 1

* From the transformation law for stress (2.27), oij = apiaqjop,; but since o:, are principal stresses, there are only three terms on the right side of this equation, and in each p = q. There-

3

fore the right hand side may be written in form oij = 2 apiapjop. p=1

2.20. Prove that uijuikokj is an invariant of the stress tensor. By the transformation law (2.27),

f f / aiPikakj = ai~ajqupqairaksarsakmajnomn

= (aipair)(ajqajn)(aksakm)~pq"rs"mn

= aprsqnssmopqarsamn

= (sprupq)(aqnamn)(8smurs)

- - - arqoqmarm - aijoikokj

2.21. Evaluate directly the invariants IE, 111, 1111 for the stress tensor

Determine the principal stress values for this state of stress and show that the diagonal form of the stress tensor yields the same values for the stress invariants.

From (2.39), IE = oii = 6 + 6 f 8 = 20.

From (2.40), IIE = ( 1 / 2 ) ( ~ ~ ~ 0 ~ ~ - u ~ ~ u ~ ~ )

= 01ia22 + O22O33 + 033011 - 412a12 - 023423 - 031a31

= 36 + 48 + 48 - 9 = 123.

From (2.41), IIIz = (oij( = 6(48) + 3(-24) = 216.

The principal stress values of aij are o1 = 3, oI1 = 8, a111 = 9. In terms of principal values,

1, = a1 + o11 + 0111 = 3 + 8 + 9 = 20

IIE = o1011 + ~ 1 1 ~ 1 1 1 + ~ 1 1 1 ~ 1 = 24 + 72 + 27 = 123

1111 = O I O ~ I O I I I = (24)9 = 216


2.22. The octahedral plane is the plane which makes equal angles with the principal stress directions. Show that the shear stress on this plane, the so-called octahedral shear stress, is given by

UOCT = 3d(uI - ~ 1 1 ) ' -I- (011 - ~ 1 1 1 ) ' + ( ~ 1 1 1 - VI)'

With respect to the principal axes, the normal to the octahedral plane is given by

Hence from (2.12) the stress vector on the octahedral plane is

1 = - + 011ê2 + ~ ~ ~ $ 3 ) 6 and its normal component is

,,x = n. t ( n ) = h(o1 + o ~ I + o ~ ~ ~ )

Therefore the shear component is

Fig. 2-25

2.23. The stress tensor a t a point is given by uij = . Determine the maxi-

mum shear stress a t the point and show that it acts in the plane which bisects the maximum and minimum stress planes.

From ( 2 . 3 8 ) the reader should verify that the principal stresses are o, = 10, o,, = 5, = -15. From (2.51b) the maximum shear value is os = (oIII - 4 1 2 = -12.5. The principal

axes 0 x ~ x ~ x ~ are related to the axes of maximum shear 0 x ; x ~ x ~ by the transformation table below and are situated as shown in Fig. 2-26.

Fig. 2-26


The stress tensor referred to the primed axes is thus given by

1 o l l f i [a&] =

-12.5 O -2.5

The results here may be further clarified by showing the stresses on infinitesimal cubes a t the point whose sides are perpendicular to the coordinate axes (see Fig. 2-27).

Fig. 2-27

MOHR'S CIRCLES (Sec. 2.12-2.13)

2.24. Draw the Mohr's circles for the state of stress discussed in Problem 2.23. Label important points. Relate the axes 0 x 1 ~ ~ ~ 3 (conjugate to uij) to the principal axes OX? x2 x9 and locate on the Mohr's diagram the points giving the stress states on the coordinate planes of O X ~ X ~ X ~ .

The upper half of the symmetric Mohr's circles diagram is shown in Fig. 2-28 with the maximum shear point P and principal stresses labeled. The transformation table of direction cosines is

from which a diagram of the relative orientation of the axes is made as shown in Fig. 2-29. The x1 and 2; axes are coincident. x2 and x3 are in the plane of x: x; as shown. From the angles a = 36.8O and fl = 53.2O shown, the points A ( - 6 , 1 2 ) on the plane L to x2 and B ( 1 , 1 2 ) on the plane I to x , are located. Point C(5,O) represents the stress state on the plane L to x, .

Fig. 2-28 Fig. 2-29


2.25. The state of stress a t a point referred to axes 0 ~ 1 x 2 ~ s is given by

o

Determine analytically the stress vector components on the plane whose unit normal is íi = (2/3)ê1 + (1/3)ê2 + (2/3)&. Check the results by the Mohr's diagram for this problem.

From (2.13) and the symmetry property of the stress tensor, the stress vector on the plane of A n is given by the matr ix product

Thus t':) = -10ê1/3 - 10ê2 - 10ê3/3; and from @.33), UN = t ( : ) hn = -7019. From (2.47), a, = 70.719.

For oij the principal stress values a r e o, = 10, oI1 = -5, CIII = -15; and the principal axes a r e related to 0xlx2x3 by the table

O -315 415 213

Thus in the principal axes frame, n; = aijnj or [: i5 3uj Ili] = . Accordingly

tlie angles of Fig. 2-16 a r e given by 0 = ,8 = cos-I 213 = 48.2' and $ = cos-1113 = 70.5O, and the Mohr's diagram comparable to Fig. 2-17 is a s shown in Fig. 2-30.

Fig. 2-30


2.26. Sketch the Mohr's circles for the three cases of plane stress depicted by the stresses on the small cube oriented along the coordinate axes shown in Fig. 2-31. Determine the maximum shear stress in each case.

Fig. 2-31

The Mohr's circles are shown in Fig. 2-32.

( b )

Fig. 2-32

SPHERICAL AND DEVIATOR STRESS (Sec. 2.14)

12 4 o 2.27. Split the stress tensor aij = into i ts spherical and deviator parts and

show that the first invariant of the deviator is zero.

UM = akk/3 (12 + 9 + 3 ) / 3 = 8. Thus

uij = uMSij + ~ i j =

o -2 -5 and sii = 4 + 1 -.5 = 0.

2.28. Show that the deviator stress tensor is equivalent to the superposition of five simple shear states.

The decomposition is


where the last two tensors are seen to be equivalent to simple shear states by comparison of cases (a) and (b) in Problem 2.26. Also note that since sii = 0, -sll - s~~ = s ~ ~ .

2.29. Determine the principal deviator stress values for the stress tensor

10 -6 O

0 0 1

The deviator of oij is sij = (--i - g) and its principal values may be determined from the determinant O -6

Thus sI = 9, sII = -3, sIII = -6. The same result is obtained by first calculating the principal stress values of oij and then using (2.71). For qj, as the reader should show, a1 = 16, o11 = 4, oII1 = 1 and hence sI = 16 - 7 = 9, sI1 = 4 - 7 = -3, sII1 = 1 - 7 = -6.

2.30. Show that the second invariant of the stress deviator is given in terms of its principal stress values by 111, = (sIsI1 + SIISIII + SIIISI), or by the alternative form IIn, =

+ sS1 + sh).

Hence from (2.72), IInD = (sIsII + sI1sIII + sIIIsI). Since sI + sII + sIII = 0,

IInD = *(~SISII + 2~11~111 + 28111 81 - (SI + 811 + 8111 12) = -*(*I + SII + s :~~)

In terms of the principal deviator stresses the characteristic equation of the deviator stress tensor is given by the determinant

MISCELLANEOUS PROBLEMS 2.31. Prove that for any symmetric tensor such as the stress tensor uij, the transformed

tensor in any other coordinate system is also symmetric.

From (2.27), o; = aipajqo,,, = ajqaipaqp = aie

SI -s o o o SI,-s o O 0 SI11 - s

= (SI - s)(sII - s)(sIII - s) = o

= s3 + (81811 + SI I~ I I I + S I I I ~ I ) ~ - ~ 1 ~ 1 1 ~ 1 1 1


2.32. At the point P the principal stresses are such that 2 ~ , , = ul + uII I . Determine the unit normal ni for the plane upon which U , = u,, and U , = (u, - ~,,,)/4.

2 2 From (2.33), oN = nloI + ni(oI + a I I I ) / 2 + n,~,,, = (o, + a I I I ) / 2 ; and since n: + n: + ni = 1, these equations may be combined to yield n, = n3. Next from (2.47),

2 2 a i = n, a , + n i ( a , + a I I I ) 2 / 4 + n3ai1, - (oI + oI I I )2 /4 = (aI - a I I I )2/16

Substituting nl = n, and ni - 1 = -nf - n; = -2nS into this equation and solving for n,, the direc-

tion cosines are found to be nl = 1 / 2 f i , n2 = f i / 2 , n3 = 1 / 2 f i . The reader should apply these

results to the stress tensor oij =

2.33. Show that the stress tensor uij may be decomposed into a spherical and deviatoric part in one and only one way.

Assume two decompositions, <rij = Sijh -t sij = Si$* + sTi with sii = O and sTi = O. Then aii = 3X = 3h*, so X = h*; and from ASij + sij = hSij + s$ i t foilows that sij = s*,.

Prove that the principal stress values are real if 1 is real and symmetric. For real values of the stress components the stress invariants are real and hence the coefficients

in (2.38) are a11 real. Thus by the theory of equations one root (principal stress) is real. Let a ( , )

be this root and consider a set of primed axes a: of which xi is in the direction of ( r c3 ) . With respect to

aí1- . cri2 o such axes the characteristic equation i s given by the determinant o& a& - o O = O

o 0 a ( , ) - 0

12 or ( a c 3 ) - o) [(ai1 - o)(aS2 - a ) - a12] = O. Since the discriminant of the quadratic in square brackets

D = (ail + a i 2 ) 2 - 4[a;1a;12 - ( o i 2 ) 2 ] = (aíl - o i 2 ) 2 + 4aii > 0, the remaining roots must be real.

Use the method of Lagrangian multipliers to show that the extrema1 values (maximum and minimum) of the normal stress U , correspond to principal values.

From (2.33), U N = aijninj with qni = 1. Thus in analogy with (2.51) construct the function H = oN - A n i n i for which aHlani = O. Then

- - - aijni,pnj + aijninj,p - 2hni,pni 8 % ~

= aijSipnj + oijniS, - 2XSipni - - apjnj + a,ni - 2XSiPni = 2(api- XSip)ni = O

which is equivalent t o equation (2.36) defining principal stress directions.

2.36. Assume that the stress components uij are derivable from a symmetric tensor field +ij by the relationship uij = Q ~ ~ ~ c ~ ~ + ~ ~ , ~ ~ . Show that in the absence of body forces the equilibrium equations (2.23) are satisfied.

Using the results of Problem 1.58, the stress components are given by

o i j = S i j ( $ q q , p p - $qp ,qp ) $. $pi ,pj + +jp,pi - $pp, ji - @ji,pp

or explicitly 011 = $33.22 + #22,33 0 1 2 = a21 = -@33,21

- a 2 2 = $11,33+ @33,11 O23 = O32 - -@11,23

- O33 = $22.11 + $11.22 a31 = "13 - -@22,13


Substituting these values into q j , = 0,

"11.1 + "12.2 + "13.3 = $33,221 + $22,331 - $33,212 - g22.133 = O

"21 , l f "22,2 + "23,3 = -@33,211 + g11.332 + $33,112 - $11,233 = O

"31.1 + "32,2 f "33,3 = -@22,131 - g11.232 f g22.113 + +11,223 = O

2.37. Show that, as is asserted in Section 2.9, the normal to the Cauchy stress quadric a t the point whose position vector is r is parallel to the stress vector tln).

Let the quadric surface be given in the form $ = oijlirj 2 k2 = O. The normal a t any point is then V g or &$/ali = Hence $,, = u ~ ~ S ~ ~ ~ ~ + oijriSjp = 2uPiTi. Now since li = rni, this becomes

A

2uPimi or 2 9 - ( ~ ~ ~ ~ ) = 2 t ' n ) .

2.38. At a point P the stress tensor referred to axes 0 ~ 1 ~ 2 x 3 is given by

15 -10

c.. -

o 20 / / r If new axes O x l x 2 x 3 are chosen by a rotation about the origin for which the trans-

r315 O -4151

formation matrix is [aii] = I O 1 O I , determine the traction vectors on each

L _1

of the primed coordinate planes by projecting the traction vectors of the original axes onto the primed directions. In this way determine a;. Check the result by the transformation formula (2.27).

A

From (2.6) and the identity tte" = uij (2.7), the traction vectors on the unprimed coordinate

axes are t ' ; l ) = 15ê1 - 10ê,, t ( ê z ) = -10 ê1 + 5ê2, t' ;3) = 20 ê3 which correspond to the rows of the stress tensor. Projecting these vectors onto the primed axes by the vector form of (2.12),

A

t ( G ) = n l t ( e l ) + n 2 t ' e z ) + n 3 t ( e 3 ) , gives A A

which by the transformation of the unit vectors becomes

Likewise

and

so that


2.39. Slrow that the second invariant of the deviator stress tensor IIp, is related to the octahedral shear stress by the equation a,,, = \/-311r,.

From Problem 2.22, <rOCT = $ d ( o I - a I I ) 2 + (o11 - o I I I ) 2 + (oIII - a I ) 2 , and because o1 = U M + SI,

O I I = 6111 + sI1, etc.,

~ O C T = SI - ~ 1 1 ) ' + ( S I I - SIII) ' + ( S I I I - SI)^

Also s , + sII + s I I 1 = O and so (s l + s I I + s I I I ) 2 = O or

Hence

2.40. The state of stress throughout a body is given by the stress tensor

Cx3 o.. =

o -cx1

where C is an arbitrary constant. (a) Show that the equilibrium equations are satisfied if body forces are zero. (b) At the point P(4, -4,7) calculate the stress vector on the plane 2x1 + 2x2 - X Q = -7, and on the sphere ( ~ 1 ) ~ + ( ~ 2 ) ~ + ( ~ 3 ) ~ = (9)2. ( c ) Determine the principal stresses, maximum shear stresses and principal deviator stresses a t P . (d) Sketch the Mohr's circles for the state of stress a t P.

(a) Substituting directly into (2.24) from aij , the equilibrium equations are satisfied identically.

(b ) From Problem 1.2, the unit normal to the plane 22 , + 2x2 - x3 = -7 is fi = 22 3 1 + 2ê 3 2 - 1" 3e3.

Thus from (2.12) the stress vector on the plane a t P is

A A The normal to the sphere xixi = (9)2 a t P i s ni = $,i with $ = xixi - 81, or n = 4 e -. 9 1 $êz + $ê3. In the matrix form of (2.14) the stress vector a t P is

o1 = fi, oI1 = O , a I I 1 = -m. The maximum shear stress value is given by - (2.54b) as os = (oIII - aI ) /2 = *&. Since the mean normal stress a t P is = (oI 4 o,, + o I I 1 ) / 3 = 0, the principal deviator stresses are the same as the principal stresses.

( c ) From (2.37), for principal stresses o,

(d) The Mohr's circles are shown in Fig. 2-33. Fig. 2-33

-o 7 o 7 -a -4

O -4 -a

= o(a2 - 65) = 0; hence


Supplementary Problems .

2.41. At point P the stress tensor is aij = [ r 2 . Determine the stress vector on the

plane a t P parallel to plane ( a ) BGE, ( b ) BGFC of the small parallelepiped shown in Fig. 2-34.

Fig. 2-34

Ans. ( a ) t':' = 11 ê1 4- 12 ê2 + 9 ê 3 , ( b ) t '" = (21 2, f 14 ê2 + 21 ê , ) l f i

2.42. Determine the normal and shear stress components on the plane BGFC of Problem 2.41. Ans. U N = 6315, os = 37.715

2.43. The principal stresses a t point P are o, = 12, o,, = 3, o,,, = -6. Determine the stress vector and its normal component on the octahedral plane a t P.

Ans. t':' = ( 1 2 ê l + 3 ê 2 - 6 ê 3 ) / f i , oN = 3

2.44. Determine the principal stress values for

(a) oij = 1 0 1 (i 1) and (b) oij = 1 2 1

and show that both have the same principal directions.

(: 1:) Ans. ( a ) a1 = 2, o l ~ = olI1 = -1, ( b ) o, = 4 , 01, = o,,, = 1

3 -10

4 Decompose the stress tensor oij = O 3 : ) into its spherical and deviator par. and

30 -27 determine the principal deviator stresses. Ans. S I = 31, S I , = 8 , 8111 = -39

2.46. Show that the normal component of the stress vector on the octahedral plane is equal to one third the first invariant of the stress tensor.

o 1

2.47. The stress tensor a t a point is .ven as oij = 1 ozz :) with unspecified. Determine 0.2

2 1 o so that the stress vector on some plane a t the point will be zero. Give the unit normal for this traction-free plane.

Ans. oz2 = 1, f; = ( S I - 2 ê 2 + ê 3 ) / 6 .


2.48. Sketch the Mohr's circles and determine the maximum shear stress for each of the following stress states:

o ( a ) uij = i: : ) (b ) 0. = -7 :

0 0 0 -27

Ans. ( a ) U , = 7, (b ) os = 3712

2.49. Use the result given in Problem 1.58, page 39, together with the stress transformation law (2.27), page 50, to show that ~ ~ ~ ~ ~ ~ ~ ~ u ~ ~ u ~ ~ u ~ ~ is an invariant.

2.50. In a continuum, the stress field is given by the tensor

o

Determine (a) the body force distribution if the equilibrium equations are to be satisfied throughout the field, ( b ) the principal stress values a t the point P(a, 0 ,2&), (c) the maximum shear stress a t

P , (d) the principal deviator stresses a t P .

Ans. ( a ) b, = -4x,, (b ) a, -a, 8a, ( c ) 24.5~3, ( d ) -11~13, -5a/3,16a/3

Deformation and Strain

3.1 PARTICLES AND POINTS

In the kinematics of continua, the meaning of the word "point" must be clearly understood since it may be construed to refer either to a "point" in space, or to a "point" of a continuum. To avoid misunderstanding, the term "point" _will he ued exclusively t-o designate alõcation in fixed space. The word "particle" will denote a small vo1umetri.c 'element, or "material point", of a continuum. In brief, a point i s a place in space, a particle is a small part of a material continuum.

3.2 CONTINUUM CONFIGURATION. DEFORMATION AND FLOW CONCEPTS

At any instant of time t, a continuum having a volume V and bounding surface S will occupy a certain region R of physical space. The identification of the particles of the continuum with the points of the space it occupies a t time t by reference to a suitable set of coordinate axes is said to specify the configuration of the continuum a t that instant.

The term deformation refers to a change in the shape of the continuum between some ini'tial (undeformed~configuration ----- ãnd a subsequent -- (deformed) configuration. The emphasis in deformation studies is on the initial and final configurations. No attention is given to - -.------------I----- L-CI

\intermedia~configurations - - - - or to the particular sequence of configurations by which the -_ - I _ _ .__-_I___ _____- - -_ deformation o c c u r s X y - contrast, the word jlow -A----- is used to designate - the - - - - - continuigg-sme ---- of motion-of a continuum. Indeed, a configuration history is inherent in flow investigations -for which the specification of a time-dependent velocity field is given.

3.3 POSITION VECTOR. DISPLACEMENT VECTOR In Fig. 3-1 the undeformed configuratiofi of a material continuum a t time t = O is

shown together with the deformed configuration of the same continuum a t a later time t = t . For the present development i t is useful to refer the initial and final configurations to separate coordinate axes as in the figure.

Fig. 3-1

78 DEFORMATION AND STRAIN [CHAP. 3

Accordingly, in the initial configuration a representative particle of the continuum occupies a point PO in space and has the position vector

with respect to the rectangular Cartesian axes OXlX2X3. Upper-case letters are used as indices in (3.1) and will appear as such in severa1 equations that follow, but their use as sixmmation indices is restricted to this section. In the remainder of the book upper-case subscripts or superscripts serve as labels only. Their use here is to emphasize the connection of certain expressions with the coordinates (XiXaX3), which are called the material coordinates. In the deformed configuration the particle originally a t PO is located a t the point P and has the position vector

when referred to the rectangular Cartesian axes oxixnx3. Here lower-case letters are used as subscripts to identify with the coordinates (x1x2x3) which give the current position of the particle and are frequently called the spatial coordinates.

The relative orientation of the material axes OXlX2X3 and the spatial axes 0x1~2~3 is specified through direction cosines a,, and a,,, which are defined by the dot products of unit vectors as

A A A A - e, I, = I; e, = a,, - a,, (3.3)

No summ:,tion is implied by the indices in these expressions since k and K are distinct A h

indices. lnasmuch as Kronecker deltas are designated by the equations IK-IP = SKP and A A ekDep = 8kp, the orthoganality conditions between spatial and material axes take the form

Tn Fig. 3-1 the vector u joining the points Po and P (the initial and final' positions, respectively, of the particle), is known as the displacement vector. This vector may be expressed as

A u = ukek (3.5)

A

or alternatively as U = U,I,

in which the components UK and uk are interrelated through the direction cosines a,,. A

From (1.89) the unit vector êk is expressed in terms of the material base vectors IK as A

A

ek = ~ILKIK

Therefore substituting (3.7) into ( 3 4 ,

from which U~ = (3.9)

Since the direction cosines a,, are constants, the components of the displacement vector are observed from (3.9) to obey the law of transformation of first-order Cartesian tensors, as they should.

The vector b in Fig. 3-1 serves to locate the origin o with respect to O. From the geometry of the figure,

u = b + x - X (3.1 O)

CHAP. 31 DEFORMATION AND STRAIN 79

Very often in continuum mechanics it is possible to consider the coordinate systems O X I X Z X ~ and oxixexs superimposed, with b = O, so that (3.10) becomes

u = x - X (3.11)

In Cartesian component form this equation is given by the general expression

However, for superimposed axes the unit triads of base vectors for the two systems are identical, which results in the direction cosine symbols a,, becoming Kronecker deltas. Accordingly, (3.12) reduces to

U, = X , - X,$ (3.13)

in which only lower-case subscripts appear. In the remainder of this book, unless specifically stated otherwise, the material and spatial axes are assumed superimposed and hence only lower-case indices will be used.

3.4 LAGRANGIAN AND EULERIAN DESCRIPTIONS

When a continuum undergoes deformation (or flow), the particles of the continuum move along various paths in space. This motion may be expressed by equations of the f orm

which give the present location xi of the particle that occupied the point (XIX2X3) a t time t = O. Also, (3.14) may be interpreted as a mapping of the initial configuration into the current configuration. I t is assumed that such a mapping is one-to-one and continuous, with continuous partial derivatives to whatever order is required. The description of motion or deformation expressed by (3.14) is known as the Lagrangian formulation.

If, on the other hand, the motion or deformation is given through equations of the form

in which the independent variables are the coordinates xi and t , the description is known as the Eulerian formulation. This description may be viewed as one which provides a tracing to its original position of the particle that now occupies the location (xl , X Z , ~ 3 ) . If (3.15) is a continuous one-to-one mapping with continuous partial derivatives, as was also assumed for (3.14), the two mappings are the unique inverses of one another. A necessary and sufficient condition for the inverse functions to exist is that the Jacobian

should not vanish.

As a simple example, the Lagrangian description given by the equations

has the inverse Eulerian formulation, - X I + x2(et - 1)

Xl = 1 - et - e-t


3.5 DEFORMATION GRADIENTS. DISPLACEMENT GRADIENTS

Partial differentiation of (3.11) with respect to Xj produces the tensor dxildXj which is called the material de format ion gradient. In symbolic notation, dxildXj is represented by the dyadic

a in which the differential operator V, = --$i is applied from the right (as shown

dXi explicitly in the equation). The matrix form of F serves to further clarify this property of the operator V, when it appears as the consequent of a dyad. Thus

Partial differentiation of (3.15) with respect to x j produces the tensor dXi/dxj which is called the spatial de format ion gradient. This tensor is represented by the dyadic

having a matrix form

The material and spatial deformation tensors are interrelated through the well-known chain rule for partia1 differentiation,

Partial differentiation of the displacement vector ui with respect to the coordinates produces either the material displacement gradient a~Jax~, or the spatial displacement gradient duiIdxj. From (3.13), which expresses u, as a difference of coordinates, these tensors are given in terms of the deformation gradients as the material gradient

and the spatial gradient

In the usual manner, the matrix f o m s of J and K are respectively

and

CHAP. 31 DEFORMATION AND STRAIN 8 1

3.6 DEFORMATION TENSORS. FINITE STRAIN TENSORS

In Fig. 3-2 the initial (undeformed) and final (deformed) configurations of a continuum are referred to the superposed rectangular Cartesian coordinate axes OX1X2X3 and ox1x2x3. The neighboring particles which occupy points PO and &O before deformation, move to points P and Q respectively in the deformed configuration.

'*

Fig. 3-2

The square of the differential element of length between PO and QO is

(dX)2 .= d X * d X = dXi dXi = G i j dXi dXj (3.28)

From (3.15), the distance differential dXi is seen to be dXi

dXi = =dxj or d X = H - d x

so that the squared length ( d X ) 2 in (3.28) may be written

in which the second-order tensor

is known as Cauchy's deformation tensor. -

In the deformed configuration, the sqiare of the differential element of length between P and Q is

From (3.14) the distance differential here is

axi dx i = --dXj or d x = F-dX dXj

so that the squared length ( d ~ ) ~ in (3.32) may be written

d ~ k d ~ k ( d ~ ) ' = -- -- dXi dXj = Gij dXi d X j or (dx)' = d X . G . dX

dXi dX j (3.34)


is known as Green's deformation tensor


The difference (dx)' - (dX)' for two neighboring particles of a continuum is used as the measure o f deformation that occurs in the neighborhood of the particles between the initial and final configurations. If this difference is identically zero for a11 neighboring particles of a continuum, a rigid displacement is said to occur. Using (3.34) and (3.28), this difference may be expressed in the form

01- ( d ~ ) ' - (dx)' = dX . ( F , F - I) . dX = dX -2Lc dX (3.36)


L.. - - or LG = & ( F c F - I )

/ is called the Lagrangian (or Green's) finite strain tensor. *' /

Using (3.32) and (3.30), the same difference may be expressed in the form


is called the Eulerian (or Almansi's) finite strain tensor.

An especially useful form of the Lagrangian and Eulerian finite strain tensors is that in which these tensors appear as functions of the displacement gradients. 'rhus if dxi/dXj from (3.24) is substituted into (3.37), the result after some simple algebraic manipulations is the Lagrangian finite strain tensor in the form

In the same manner, if dXi/dxj from (3.25) is substituted into (3.39), the result is the Eulerian finite strain tensor in the form

The matrix representations of (3.40) and (3.41) may be written directly from (3.26) and (3.27) respectively.

3.7 SMALL DEFORMATION THEORY. INFINITESIMAL STRAIN TENSORS

The so-called small deformation theory of continuum mechanics has as its basic condition the requirement that the displacement gradients be small compared to unity. The fundamental measure of deformation is the difference (dx)' - (dX)', which may be expressed in terrns of the displacement gradients by inserting (3.40) and (3.41) into (3.36) and (3.38) respectively. If the displacement gradients are small, the finite strain tensors in (3.36) and (3.38) reduce to infinitesimal strain tensors, and the resulting equations represent small deformations.


In (3.40), if the displacement gradient components duildXj are each small compared to unity, the product terms are negligible and rnay be dropped. The resulting tensor is the Lagrangian infinitesimal strain tensor, which is denoted by

or i. = i(uvx + V, u) = S(J + Jc)

Likewise for duiIdxi < 1 in (3.41), the product terms rnay be dropped to yield the Eulerz'ccn infinitesimal strain t emor , which is denoted by

If both the displacement gradients and the displacements themselves are small, there is very little difference in the material and spatial coordinates of a continuum particle. Accordingly the material gradient components duildXj and spatial gradient components dui/dxj are very nearly equal, so that the Eulerian and Lagrangian infinitesimal strain tensors rnay be taken as equal. Thus

l i . = r i . or L = E (3.44)

if both the displacements and displacement gradients are sufficiently small.

3.8 RELATIVE DISPLACEMENTS. LINEAR ROTATION TENSOR. ROTATION VECTOR

In Fig. 3-3 the displacements of two neighboring Q particles are represented by the vectors u:"o' and uiQO' (see also Fig. 3-2). The vector

dui = u i < Q ~ ) - U ! P ~ ) or du = ~ ( Q o ) - u ( P o ) 1 dx

is called the relative displacement vector of the particle originally a t Qo with respect to the particle originally a t Po. Assuming suitable continuity conditions on the v P displacement field, a Taylor series expansion for uiPO) U ( P O )

rnay be developed in the neighborhood of Po. Neglect- po ing higher-order terms in this expansion, the relative displacement vector can be written as Fig. 3-3

Here the parentheses on the partia1 derivatives are to emphasize the requirement that the derivatives are to be evaluated a t point Po. These derivatives are actually the components of the material displacement gradient. Equation (3.46) is the Lagrangian form of the relative displacement vector.

It is also useful to define the uni t relative displacement vector d ~ l d X in which d X is the magnitude of the differential distance vector dXi. Accordingly if vi is a unit vector in the direction of d X i so that d X i = V , d X , then

Since the material displacement gradient duildXj rnay be decomposed uniquely into a symmetric and an antisymmetric part, the relative displacement vector dui rnay be written a s


The first term in the square brackets in (3.48) is recognized as the linear Lagrangian strain tensor lij. The second term is known as the linear Lagrangian rotation tensor and is denoted by

In a displacement for which the strain tensor lij is identically zero in the vicinity of point Po, the relative displacement a t that point will be an infinitesimal rigid body rotation. This infinitesimal rotation may be represented by the rotation vector

in terms of which the relative displacement is given by the expression

The development of the Lagrangian description of the relative displacement vector, the linear rotation tensor and the linear rotation vector is paralleled completely by an analogous development for the Eulerian counterparts of these quantities. Accordingly the Eulerian description of the relative displacement vector is given by

and the unit relative displacement vector by

Decomposition of the Eulerian displacement gradient auilaxj results in the expression

The first term in the square brackets of (3.54) is the Eulerian linear strain tensor e i j . The second term is the linear Eulerian rotation tensor and is denoted by

From (3.55), the linear Eulerian rotation vector is defined by

in terms of which the relative displacement is given by the expression

3.9 INTERPRETATION OF THE LINEAR STRAIN TENSORS

For small deformation theory, the finite Lagrangian strain tensor Lij in (3.36) may be replaced by the linear Lagrangian strain tensor lij, and that expression may now be written


(dx)' - (dX)' = ( d x - d X ) ( d x + d X ) = 2Lj d X i d X j

(dx)' - ( d X ) 2 = ( d ~ - d X ) ( d ~ + d X ) = d x 21. d x (3.58)

Since d x - d X for small deformations, this equation may be put into the form

The left-hand side of (3.59) is recognized as the change in length per unit original length of the differential element and is called the normal strain for the line element originally having direction cosines dXildX.

When (3.59) is applied to the differential line element PoQo, located with respect to the set of local axes a t PO as shown in Fig. 3-4, the result will be the normal strain fo r that element. Because POQO here lies along the X Z axis,

dXi ldX = dX3ldX = O, dXzldX = 1

and therefore (3.59) becomes

Fig. 3-4

Thus the normal strain for an element originally along the X z axis is seen to be the component 122. Likewise for elements originally situated along the X i and X 3 axes, (3.59) yields normal strain values l1l and 133 respectively. In general, therefore, the diagonal terms of the linear strain tensor represent normal strains in the coordinate directions.

Fig. 3-5

The physical interpretation of the off-diagonal terms of l i j may be obtained by a consideration of the line elements originally located along two of the coordinate axes. In Fig. 3-5 the line elements POQO and POMO originally along the X2 and Xs axes, respectively, become after deformation the line elements PQ and P M with respect to the parallel set of local axes with origin a t P. The original right angle between the line elements becomes the angle O . From (3.46) and the assumption of small deformation theory, a first order approximation gives the unit vector a t P in the direction of Q as


and, for the unit vector a t P in the direction of M, as

Therefore

or, neglecting the product temi which is of higher order,

duz dua COS o = - + - = 2JZ3

aX3 aX2

Furthermore, taking the change in the right angle between the elements as y2, = 712 - O, and remembering that for the linear theory r,, is very small, i t follows that

Y23 - sin yZ3 = sin ( ~ 1 2 - O ) = cos O = 24, (3.65)

Therefore the off-diagonal terms of the linear strain tensor represent one-half the angle change between two line elements originally a t right angles to one another. These strain components are called shearing strains, and because of the factor 2 in (3.65) these tensor components are equal to one-half the familiar "engineering" shearing strains.

A development, essentially paralleling the one just presented for the interpretation of the components of l i j , may also be made for the linear Eulerian strain tensor ei i . The essential difference in the derivations rests in the choice of line elements, which in the Eulerian description must be those that lie along the coordinate axes after deformation. The diagonal terms of cij are the normal strains, and the off-diagonal terms the shearing strains. For those deformations in which the assumption lij = aii is valid, no distinction is made between the Eulerian and Lagrangian interpretations.

3.10 STRETCH RATIO. FINITE STRAIN INTERPRETATION

An important measure of the extensional strain of a differential line element is the ratio dxldX, known as the stretch or stretch ratio. This quantity may be defined a t either the point PO in the undeformed configuration or a t the point P in the deformed configuration. Thus from (3.34) the squared stretch a t point PO for the line element along the unit vector i% = dXldX, is given by

Similarly, from (3.30) the reciproca1 of the squared stretch for the line element a t P along the unit vector íi = dxldx is given by

For an element originally along the local X2 axis shown in Fig. 3-4, = êz and therefore dXildX = dXJdX = O , dXzldX = 1 so that (3.66) yields for such an element

Similar results may be determined for A:;,, and A:;,).


For an element parallel to the xz axis after deformation, (3.67) yields the result

with similar expressions for the quantities 1 / ~ ~ ~ ~ , and 1 1 ~ : ~ ~ ) . In general, A,&, is not equal to A,;2, since the element originally along the X2 axis will not likely lie along the x2 axis after deformation.

The stretch ratio provides a basis for interpretation of the finite strain tensors. Thus the change of length per unit of original length is

and for the element POQO along the X2 axis (of Fig. 3-4), the unit extension is therefore

This result may also be derived directly from (3.36). For small deformation theory, (3.71) reduces to (3.60). Also, the unit extensions L(1) and L(,) are given by analogous equations in terms of Ltt and L33 respectively.

For the two differential line elements shown in Fig. 3-5, the change in angle y,, = ~ 1 2 - 6 is given in terms of A, ;~ , and A,;~, by

When deformations are small, (3.72) reduces to (3.65).

3.11 STRETCH TENSORS. ROTATION TENSOR

The so-called polar decomposition of an arbitrary, nonsingular, second-order tensor is given by the product of a positive symmetric second-order tensor with an orthogonal second- order tensor. When such a multiplicative decomposition is applied to the deformation gradient F, the result may be written

in which R is the orthogonal rotation tensor, and S and T are positive symmetric tensors known as the right stretch tensor and le f t stretch tensor respectively.

The interpretation of (3.73) is provided through the relationship dxi = (dxi /dXi)dXj given by (3.33). Inserting the inner products of (3.73) into (3.33) results in the equations

From these expressions the deformation of dXi into dxi a s illustrated in Fig. 3-2 may be given either of two physical interpretations. In the first form of the right hand side of (3.74), the deformation consists of a sequential stretching (by S ) and rotation to be followed by a rigid body displacement to the point P. In the second form, a rigid body translation to P is followed by a rotation and finally the stretching (by T). The translation, of course, does not alter the vector components relative to the axes Xi and X i .


3.12 TRANSFORMATION PROPERTIES OF STRAIN TENSORS

The various strain tensors Lij, Eij, lij and cij defined respectively by (3.379, (3.39), (3.42) and (3.43) are a11 second-order Cartesian tensors as indicated by the two free indices in each. AccordingIj for a set of rotated axes Xil having the transformation matrix [bij] with respect to the set of local unprimed axes Xi a t point PO as shown in Fig. 3-6(a), the components of L& and 1; are given by

and l!. 11 - - b. I P b. 34 G, 4 or l f = B * L . B , (3.76)

( b ) Fig. 3-6

Likewise, for the rotated axes xi' having the transformation matrix [aij] in Fig. 3-6(b), the components of E; and are given by

and C!. = a . a . € or E ' = A * E a A C (3.78) 1P 34 P 4

By analogy with the stress quadric described in Section 2.9, page 50, the Lagrangian and Eulerian linear strain quadrics may be given with reference to local Cartesian coordinates qi and (Si a t the points Po and P respectively as shown in Fig. 3-7. Thus the equation of the Lagrangian strain quadric is given by


and the equation of the Eulerian strain quadric is given by

Two important properties of the Lagrangian {Eulerian) linear strain quadric are:

1. The normal strain with respect to the original {final) length of a line element is inversely proportional to the distance squared from the origin of the quadric PO {P) to a point on its surface.

2. The relative displacement of the nedghboring particle located a t QO {Q) per unit original {final) length is parallel to the normal of the quadric surface a t the point of intersection with the line through POQO {PQ).

Additional insight into the nature of local deformations in the neighborhood of PO is provided by defining the strain ellipsoid a t that point. Thus for the undeformed continuum, the equation of the bounding surface of an infinitesimal sphere of radius R is given in terms of local material coordinates by (3.28) as

After deformation, the equation of the surface of the same material particles is given by (3.30) as

(dx) ' = Cijdxidxj = R2 or (dX)2 = d x . C . d x = R2 (3.82)

which describes an ellipsoid, known as the material strain ellipsoid. Therefore a spherical volume of the continuum in the undeformed state is changed into an ellipsoid a t Po by the deformation. By comparison, an infinitesimal spherical volume a t P in the deformed continuum began as an ellipsoidal volume element in the undeformed state. For a sphere of radius r a t P, the equations for these surfaces in terms of local coordinates are given by (3.32) for the sphere as

and by (3.34) for the ellipsoid as

The ellipsoid of (3.84) is called the spatial strain ellipsoid. Such strain ellipsoids as described here are frequently known as Cauchy strain ellipsoids.

3.13 PRINCIPAL STRAINS. STRAIN INVARIANTS. CUBICAL DILATATION The Lagrangian and Eulerian linear strain tensors are symmetric second-order Cartesian

tensors, and accordingly the determination of their principal directions and principal strain values follows the standard development presented in Section 1.19, page 20. Physically, a principal direction of the strain tensor is one for which the orientation of an element at a given point is not altered by a pure strain deformation. The principal strain value is simply the unit relative displacement (normal strain) that occurs in the principal direction.

For the Lagrangian strain tensor lij, the unit relative displacement vector is given by (3.477, which may be written

n

Calling 1;"' the normal strain in the direction of the unit vector ni, (3.85) yields for pure strain (Wij 0) the relation

A A

1;"' = Ljnj or 1'"' = 1.2 (3.86)


If the direction n, is a principal direction with a principal strain value 1, then

1:" = lni = /aijnj or 1';) = l$ = $102 (3.87)

Equating the right-hand sides of (3.86) and (3.87) leads to the relationship

which together with the condition nin+ = 1 on the unit vectors ni provide the necessary equations for determining the principal strain value 1 and its direction cosines ni. Nontrivial solutions of (3.88) exist if and only if the determinant of coefficients vanishes. Therefore

which upon expansion yields the characteristic equation of lij, the cubic

P - IL12 + IIJ - IIIL = O (3.90)

where IL = &i = tr L, 1 = ( Z i Z j - Zjij), , IIIL = I&jl = det L (3.91)

are the first, second and third Lagrangian strain invar.iants respectively. The roots of (3.90) are the principal strain values denoted by l(l), and z(31.

The first invariant of the Lagrangian strain tensor may be expressed in t e m s of the principal strains as

IL = Zii = Z(1) + Z(2) + Z(3) (3.92)

and has an important physical interpretation. To see this, consider a differential rectangular parallelepiped whose edges are parallel to the principal strain directions as shown in Fig. 3-8. The change in volume per unit original volume of this element is called the cubical dilatation and is given by

For small strain theory, the first-order approximation of this ratio is the sum

Do = l(l> + 1<2> + Z(3) = IL (3.94)

Fig. 3-8

CHAP. 31 DEFORMATION AND STRAIN 9 1

With regard to the Eulerian strain tensor eij and its associated unit relative displacement vector E:"', the principal directions and principal strain values C,,,, E,,,, C,,, are determined in exactly the same way as their Lagrangian counterparts. The Eulerian strain invariants may be expressed in terms of the principal strains as

I E = E(i, + + E<3,

IIE = '(1)'<2) '(2)'(3) + '(3)'(1)

IIIE = c (1 )E(2 , E ( 3 )

The cubical dilatation for the Eulerian description is given by

3.14 SPHERICAL AND DEVIATOR STRAIN TENSORS

The Lagrangian and Eulerian linear strain tensors may each be split into a spherz'cal and deviator tensor in the same manner in which the stress tensor decomposition was carried out in Chapter 2. As before, if Lagrangian and Eulerian deviator tensor components are denoted by dij and eij respectively, the resolution expressions are

Ikk I(tr L) lij = dij+6..- or L = LD+- i3 3 3

and 'kk l(tr E) ci. = e.. + 6.. - or E = ED + - L3 i 3 3 3

The deviator tensors are associated with shear deformation for which the cubical dilatation vanishes. Therefore it is not surprising that the first invariants di and eii of the deviator strain tensors are identically zero.

3.15 PLANE STRAIN. MOHR'S CIRCLES FOR STRAIN

When one and only one of the principal strains a t a point in a continuum is zero, a state of plane strain is said to exist a t that point. In the Eulerian description (the Lagrangian description follows exactly the same pattern), if x3 is taken as the direction of the zero principal strain, a state of plane strain parallel to the x1x2 plane exists and the linear strain tensor is given by

t , - (i: i: ;) E . . - or [ c ] = [; :i %] (3.99)

When xi and xz are also principal directions, the strain tensor has the form

o o o Cij = k' €;) 01 [Eij] = ['O) E;. i]

In many books on "Strength of Materials" and "Elasticity", plane strain is referred to as plane deformation since the deformation field is identical in a11 planes perpendicular to the direction of the zero principal strain. For plane strain perpendicular to the x3 axis, the displacement vector may be taken as a function of xl and xz only. The appropriate displacement components for this case of plane strain are designated by


U l = u1(x1, xz)

Uz = uz(x1, xz)

u3 = C (a constant, usually taken as zero)

Inserting these expressions into the definition of eij given by (3.43) produces the plane strain tensor in the same form shown in (3.99).

A graphical description of the state of strain a t a point is provided by the Mohr's circles for strain in a manner exactly like that presented in Chapter 2 for the Mohr's circles for stress. For this purpose the strain tensor is often displayed in the form

Here the yij (with i # j ) are the so-called "engineering" shear strain components, which are twice the tensorial shear strain components.

The state of strain a t an unloaded point on the ~ 1 2 1'

bounding surface of a continuum body is locally plane strain. Frequently in experimental studies involving strain measurements a t such a surface point, Mohr's strain circles are useful for reporting the observed data. Usually three normal strains are measured a t the given point by means of a strain €1

rosette, and the Mohr's circles diagram constructed from these. Corresponding to the plane stress Mohr's circles, a typical case of plane strain diagram is shown in Fig. 3-9. The principal normal strains are labeled as such in the diagram, and the maximum shear strain values are represented by points D and E. Fig. 3-9

3.16 COMPATIBILITY EQUATIONS FOR LINEAR STRAINS If the strain components rii are given explicitly as functions of the coordinates, the six

independent equations (3.43) 1 8%

cii = -(- 2 axj + 2) may be viewed as a system of six partia1 differential equations for determining the three displacement components ui. The system is over-deterrnined and will not, in general, possess a solution for an arbitrary choice of the strain components ci i . Therefore if the displacement components ui are to be single-valued and continuous, some conditions must be imposed upon the strain components. The necessary and sufficient conditions for such a displacement field are expressed by the equations

There are eighty-one equations in a11 in (3.103) but only six are distinct. These six written in explicit and symbolic form appear as

CHAP. 31 DEFORMATION AND STRAIN

5. - - 8x3 axi

d (dc23 '€31 - 6. - -+- - - - ~ X S 8x1 ax2 axs 8x1 8x2

Compatibility equations in terms of the Lagrangian linear strain tensor lij may also be written down by an obvious correspondence to the Eulerian form given above. For plane strain parallel to the ~ 1 x 2 plane, the six equations in (3.104) reduce to the single equation

where E is of the form given by (3.99).

Solved Problems

DISPLACEMENT AND DEFORMATION (Sec. 3.1-3.5)

3.1. With respect to superposed material axes Xi and spatial axes xi, the displacement field of a continuum body is given by xl = Xi, 2 2 = X2 + AX3, xs = X3 + AX2 where A is a constant. Determine the displacement vector components in both the material and spatial forms.

From (3.13) directly, the displacement components in material form are ul = xl -X l = 0, u2 = x2 - X2 = AX3, u3 = x3 - X3 = AX,. Inverting the given displacement relations to obtain X, = x,, X2 = (x, - Ax3)l(1 - AZ), X3 = (2, - Ax2)l(1 - AZ), the spatial components of u are U, = O, u, = A(x3 - Ax2)l(l - A2), ~3 = A(x2 - Ax3)l(1 - A2).

From these results i t is noted that the originally straight line of material particles expressed by X, = O, X2 + X3 = lI(1 + A ) occupies the location xl = 0, x2 + x3 = 1 after displacement. Likewise the particle line X1 = O, X2 = X3 becomes after displacement x1 = 0, x2 = x3. (Inter- pret the physical meaning of this.)

For the displacement field of Problem 3.1 determine the displaced location of the material particles which originally comprise (a) the plane circular surface XI = 0, XE + X : = 1/(1- A2), (b) the infinitesimal cube with edges along the coordinate axes of length dXi = d X . Sketch the displaced configurations for (a) and (b) if A = .S. (a) By the direct substitutions X2 = (x, - Ax3)l(l - A2) and X3 = (x3 - Ax2)l(1 - A2), the circular

surface becomes the elliptical surface (1 + A2)xl - 4Ax2x3 + (1 + A2)xi = (1 - A2). For

A = 4, this is bounded by the ellipse 5x22 - 8x223 + 5x3 = 3 which when referred to i ts prinzipal axes x i (at 45O with xi, i = 2,3) has the equation xi2 + 9 ~ ; ~ = 3. Fig. 3-10 below shows this displacement pattern.


Fig. 3-10 Fig. 3-11

( b ) From Problem 3.1, the displacements of the edges of the cube are readily calculated. For the edge X , = X,, X 2 = X 3 = O, u , = u , = u3 = O. For the edge X 1 = O = X2, X 3 = X3, u, = u3 = O , u2 = A X 3 and the particles on this edge are displaced in the X , direction propor- tionally to their distance from the origin. For the edge X , = X 3 = O , X , = X, , u, = u, = 0 , u3 = AX,. The initial and displaced positions of the cube are shown in Fig. 3-11.

33. For superposed material and spatial axes, the displacement vector of a body is given by u = 4x:ê1 + X2Xiê2 + Xixiê3. Determine the displaced location of the particle originally a t (1 ,0 ,2) .

The original position vector of the particle is X = ê, + 2êg. I ts displacement is u = 4ê1 + 4 ê 3 and since x = X + u, its final position vector is x = 5ê1 + 6ê3.

3.4. With respect to rectangular Cartesian material coordinates Xi, a displacement field is given by Ul = -AX2X3, U2 = A X l X 3 , U3 = 0 where A is a constant. Determine the displacement components for cylindrical spatial coordinates xi if the two systems have a common origin.

From the geometry of the a h s (Fig. 3-12) the A

transformation tensor apK = e p IK is

cosx, sin x , O

aPK (-7x2 0) and from the inverse form of (3.9) up = aPKUK. Thus since Cartesian and cylindrical coordinates are related through the equations X , = xl cos x,, X2 = x1 sin x,, X 3 = x,, equation (3.9) gives Fig. 3-12

ul = (-cos x2)AX,X3 + (sin x 2 ) A X l X 3

= (-cos x2)Ax3xl sin 2, + (sin x2)Ax3x1 cos x2 = O

u, = (sin x,)AX,X3 + (cos x,)AXlX3 = ( s i n ~ x 2 ) A x l x 3 + (cos2 x2)Ax1x3 = Ax1x3

U g = o This displacement is that of a circular shaft in torsion.


3.5. The Lagrangian description of a deformation is given by XI = XI + X3(e2- I), x2 = X2 + X3(e2 - e-2), x3 = e2X3 where e is a constant. Show that the Jacobian J does not vanish and determine the Eulerian equations describing this motion.

Inverting the equations, X1 = x1 + x3(e-2 - I) , X2 = x2 + x3(e-4 - I), X3 = e-2x3.

From (.9.16), J =

3.6. A displacement field is given by u = êl + x:x2ê2 + x2x3ê3. Determine independently the material deformation gradient F and the material displacement gradient J and verify (3.24), J = F - I.

From the given displacement vector u, J is found to be

1 O (e2 - 1)

O 1 (e2 - e-2)

O O (e2)

Since x = u + X, the displacement field may also be described by equations xl = X1(l + x:), x2 = X2(1 + x:), x3 = X3(l + x:) from which F is readily found to be

= e2 # 0.

Direct substitution of the calculated tensors F and J into (3.26) verifies that the equation is satisfied.

3.7. A continuum body undergoes the displacement u = (3x2 - 4X3)ê1 + (2x1 - X3)$2 + (4x2 - X1)ê3. Determine the displaced position of the vector joining particles A(1,0,3) and B(3,6 ,6) , assuming superposed material and spatial axes.

From (3.13), the spatial coordinates for this displacement are xl = Xl + 3X2 - 4X3, x2 = 2X1 + X2 - X3, s3 = -Xl + 4X2 + X3. Thus the displaced position of particle A is given by xl = -11, x2 = -1, x3 = 2; and of particle B, xl = -3, x2 = 6, x3 = 27. Therefore the displaced position of the vector joining A and B may be written V = 82, + 7ê2 + 25ê3.

3.8. For the displacement field of Problem 3.7 determine the displaced position of the position vector of particle C(2 ,6 ,3 ) which is parallel to the vector joining particles A and B. Show that the two vectors remain parallel after deformation.

By the analysis of Problem 3.7 the position vector of C becomes U = 82, + 7ê2 + 25ê3 which is clearly parallel to V. This is an example of so-called homogeneous deformation.

3.9. The general formulation of homogeneous deformation is given by the displacement field ui = AijXj where the Aij are constants'or a t most functions of time. Show that this deformation is such that (a) plane sections remain plane, (b) straight lines remain straight.

(a) From (3.13), xi = Xi + ui = Xi + AijXj = (a..+A..)X. i.3 13 3


According to (3.16) the inverse equations Xi = ( a i j + B i j ) x j exist provided the determinant Iaij + Aijl does not vanish. Assuming this is the case, the material plane piXi + cu = O becomes &(aij + B i j ) x j + cu = O which may be written in standard form a s the plane h j x j + a. = O where the coefficients = &(ai j + Bij ) .

( b ) A straight line may be considered a s the intersection of two planes. In the deformed geometry, planes remain plane a s proven and hence the intersection of two planes remains a s t raight line.

3.10. An infinitesimal homogeneous deformation ui = AijXj is one for which the coefficients Aij are so small that their products may be neglected in comparison to the coefficients themselves. Show that the total deformation resulting from two successive infinitesimal homogeneous deformations may be considered as the sum of the two individual deformations, and that the order of applying the displacements does not alter the final configuration.

Let xi = ( a i j + A i j ) X j and x l = ( a i j + Bi j ) x j be successive infinitesimal homogeneous displacements. Then xl = (a i j + Bij ) (Sjk + A j k ) X k = ( S i k + Bik + Aik + B i j A j k ) X k . Neglecting the higher order product terms B i j A j k , this becomes xl = ( S i k + Bik + A i k ) X k = (Oi lc + Cik )Xk which represents the infinitesimal homogeneous deformation

uE' = X E - X i = CikXk = (Bik + A i k ) X k = (A ik + Bik )Xk = tii + U ;

DEFORMATION AND STRAIN TENSORS (Sec. 3.6-3.9)

3.11. A continuum body undergoes the deformation xi = X i , x2 = X Z + AX3, x3 = X3 + AX2 where A is a constant. Compute the deformation tensor G and use this to determine the Lagrangian finite strain tensor LG.

From (3.35), G = F,* F and by (3.20) F i s given in matrix form a s

o o so tha t G i j ] = [i 1 l f 2

yAj Therefore from (3.37), L~ = S ( G - I) =

2 O 2 A A2

3.12. For the displacement field of Problem 3.11 calculate the squared length ( d ~ ) ~ of the edges OA and OB, and the diagonal OC after deformation for the small rectangle shown in Fig. 3-13.

Using G a s determined in Problem 3.11 in (3.34), the squared length of the diagonal OC is given in matrix form by

o ( d ~ ) ~ = [ O , d X 2 , dX3] O 1 + A 2 [: 2 A l:Aj[il Fig. 3-13

= ( 1 + A"(dX2)2 + 4 A d X 2 dX: , + ( 1 + A"(dX:,)"

Similarly for O A , (dx)' = ( 1 + A2)(dX'2)2; and for O B , (dx)z = ( 1 + A2)(dX3)2.


3.13. Calculate the change in squared length of the line elements of Problem 3.12 and check the result by use of (3.36) and the strain tensor LG found in Problem 3.11.

Directly from the results of Problem 3.12, the changes are:

(a) for OC, ( d ~ ) ~ - (dX)2 = , ( I + A Z ) ( ~ X ; + d x ; ) + 4 A d X , dX3 - ( d x ; + d x ; )

= A Z ( ~ X : + d x ; ) + 4 A d X , d X 3

( b ) for OB, (dx)2 - (dX)2 = ( 1 + A ' ) d x ; - d x ; = A2 d x ;

(c) for O A , (dx)2 - (dX)2 = ( 1 + A,) d x i - d ~ i = A2 d ~ i .

By equation (3.36), for OC

o (dx). - ( U ) 2 = O , d X 2 -4 L: A2 o'] [A;] = A Z ( ~ X ~ + d x ; ) + 4 A d X , dX3

O 2 A A2 d X g

The changes for O A and O B may also be confirmed in the same way.

3.14. For the displacement field of Problem 3.11 calculate the material displacement gradient J and use this tensor to determine the Lagrangian finite strain tensor LÇ. Compare with resillt of Problem 3.11.

From Problem 3.11 the displacement vector components a r e ul = O , u, = A X 3 , z(3 = A X , so tha t

O O A2

Thus from (3.40)

~ L G

O A O O A O O O A 2 O 2 A A2

the identical result obtained in Problem 3.11.

3.15. A displacement field is given by XI = X I + AX2, x2 = X Z + AX3, x s = X3 + A X l where A is a constant. Calculate the Lagrangian linear strain tensor L and the Eulerian linear strain tensor E. Compare L and E for the case when A is very small.

From (3.42), / 0 A O\ 10 O A \ / o A A\

Inverting the displacement equations gives

ul A(A2x l + x2 - A x 3 ) l ( l + A3), u2 = A ( - A x l + A2x2 + x3) / (1 + A3),

~3 = A ( x l - A x 2 + A2x3)/(1 + A3) from which by (3.43)

A2 1 - A A2 - A

2E = ( K + K,) = A 1 + A 3 :,) + h(-: t2 -i2)

When A is very small, A? and higher powers may be neglected with the result t h a t E reduces to L.


3.16. A displacement field is specified by u = X? X2ê1 + (X2 - x ; ) ê 2 + X; X3ê3. Determine the relative displacement vector du in the direction of the -XZ axis a t P ( 1 , 2 , -1). Determine the relative displacements uai - UP for Q1(1,1, -I), Q2(1,3/2, -I), Q3(1,7/4, -1) and Q4(1, 1518, -1) and compare their directions with the direction of du.

For the given u, the displacement gradient J in matrix form is

2XlX2 x; ~ a u i l ~ x i = [ : -i]

so that from (3.46) a t P in the -X2 direction,

- Next by direct calculation from u, up = 2 ê1 + ê2 - 4 ê3 and UQ, - êl - ê3. Thus A A

UQ, - up = -e, - e2 + 3 e,. Likewise, uQ2 - up = (-e1 - ê2 + 3.5 ê3)/2, ug3 - up = (- e, - e, + 3.75ê3)/4, uQ4 - up = (- ê, - ê2 + 3.875ê3)/8. I t is clear that as Qi approaches P the direction of the relative displacement of the two particles approaches the limiting direction of du.

3.17. For the displacement field of Problem 3.16 determine the unit relative displacement vector a t P (1 ,2 , -1) in the direction of Q(4,2,3).

The unit vector a t P in the direction of Q is $ = 3ê1/5 + 4ê3/5, so that from (3.47) and the matrix of J as calculated in Problem 3.16,

3.18. Under the restriction of small deformation theory, L = E. Accordingly for a displacement field given by u = (xl - ~ 3 ) ~ ê l + (x2 + ~ ~ ) ~ ê 2 - x1x2ê3, determine the linear strain tensor, the linear rotation tensor and the rotation vector at the point P(O,2, -1).

Here the displacement gradient is given in matrix form by

2(x1- x3) 0 -a21 - x3)

[auilaxj] = O 2(x2 + x3) 2(x2 + 23) 1 which a t the point P becomes r 2

Decomposing this matrix into its symmetric and antisymmetric components gives

[eijI + [wijI = -2 o -1 o

Therefore from (3.56) the rotation vector wi has components wl = -1, w2 = w3 = 0.


3.19. For the displacement field of Problem 3.18 determine the change in length per unit length (normal strain) in the direction of Y = (8êi - ê2 + 4ê3)/9 a t point P(O,2, -1).

From (3.59) and the s t rain tensor a t P a s computed in Problem 3.18, the normal s t rain a t P in the direction of Y is the matr ix product

3.20. Show that the change in the right angle between two orthogonal unit vectors G and í; in the undeformed configuration is given by G.2L.fi for small deformation theory.

Assuming small displacement gradients, the unit vectors in the deformed directions of $ and $ a r e given by (3.47) a s ($ + J * c ) and ( l i + J .fi) respectively. (The student should check equations (3.61) and (3.62) by this method.) Wri t ing J in the equivalent form $ * J c and dotting the two displaced uni t vectors gives (as i n (3.63)), cos e = sin ( ~ 1 2 . - e) = sin y,, = y,, or y,, = [$ + $ * J,] [ f i + J . C ] = $ . f i + G - ( ~ + ~ , ) . f i + Y . ~ , . ~ . f i . Here J,.J is of higher order fo r small displacement gradients and since Y I fi, 4. fi = O so t h a t finally by (3.42), y,, = $ 21. c.

3.21. Use the results of Problem 3.20 to compute the change in the right angle between G = (8ê1 - êz + 4ê3)/9 and i; = (4ê1 + 4ê2 - 7ê3)/9 a t the point P(O,2, -1) for the displacement field of Problem 3.18.

Since L = E for small deformation theory, the s t rain tensor eij = lij and so a t P

y,, = [8/9, -119, 4/91

-719

STRETCH AND ROTATION (Sec. 3.10-3.11)

3.22. For the shear deformation XI = Xi, xz = XZ + AX3, x 3 = X3 + AX2 of Problem 3.11 show that the stretch A,;, is unity (zero normal strain) for line elements parallel to the XI axis. For the diagonal directions OC and D B of the infinitesimal square OBCD (Fig. 3-14), compute A:;, and check the results by direct calculation from the displacement field.

From (3.66) and the matr ix of G a s determined in Problem 3.11, the squared stretch for & = êl is Fig. 3-14

Likewise for OC, & = ( ê 2 +&)I& and so


From the displacement equations the deformed location of C is x, = 0, x2 = dL + AdL, x, = dL + AdL. Thus (dx)2 = 2(1+ A)2(dL)2 and since dX = fi d ~ , the squared stretch (dx/dX)2 is (1 + A)2 a s calculated from (3.66).

2 Similarly, fo r DB, & = (-ê2 + ê 3 ) / f i and so A(m, = (1 - A ) 2 .

The stretch ratios n , ~ , and h(;, are equal only if 2 is the deformed direction of A. For the displacement field of Problem 3.22, calculate A(;, for & = ( $ 2 + & ) / f i and show that i t agrees with A : ~ , for the diagonal OC in Problem 3.22.

Inverting the displacement equations of Problem 3.22 one obtains

X1 = xl, X2 = (2 , - Ax3)/(l - A2), X3 = ($3 - Ax2)l(1 - A2)

from which the Cauchy deformation tensor C may be computed. Then using (3.67),

2 Thus A:;, = (1 - A2)2/(1- A)2 = (1 + A)2 which is identical with A(m, calculated for OC. The diagonal element OC does not change direction under the given shear deformation.

3.24. By a polar decomposition of the deformation gradient F for the shear deformation xl = X1, x2 = X2 + AX3, x3 = X B + AX2, determine the right stretch tensor S together with the rotation tensor R. Show that the principal values of S are the stretch ratios of the diagonals OC and DB determined in Problem 3.22.

In the polar decomposition of F, the stretch tensor S = 6; and from (3.73), R = FS-1. By

r 1 O O 1 (3.35), G = F, . F or here [Gij] = 1 0 1 + A2 2A I . The principal axes of G a r e given by

L 0 2A i t ~ 2 J

a 4 5 O rotation about X1 with the tensor in principal form [G~;] =

o o Therefore [Sij] = [ G ] =

Relative to the coordinate axes Xi, the decomposition is

In this example the deformation gradient F is i ts own stretch tensor S and R = i. This is the result of the coincidence of the principal axes of LG and EA fo r the given shear deformation.

3.25. An infinitesimal rigid body rotation is given by ul = - C X 2 + B X 3 , us = C X l - A X 3 , u3 = - B X l + AX2 where A, B , C are very small constants. Show that the stretch is zero (S = I) if terms involving squares and products of the constants are neglected.

F o r this displacement, rl + c- + B2 -A R -.c 1


Neglecting higher order terms, this becomes

[Gijl = O 0 1

STRAIN TRANSFORMATIONS AND PRINCIPAL STRAINS (Sec. 3.12-3.14)

3.26. For the shear deformation X I = XI, xz = X2 + f i X 3 , x3 = X3 + f i ~ 2 show that the principal directions of LG and Ea coincide as was asserted in Problem 3.24.

o From (3 .37) [Lij] = [! i :] which for principaI axes given by the transformation

o o

matrix [ a i j = [: 11. becomes [ L : = [i 1 ;fi . o -1 l f i 11fi

Likewise from (3 .39 , [Ed = which by the same transformation matr ix Iaij]

r 0 O O 1 is converted into the principal-axes form [E:] = . The student should verify these calculations. o -l+fi

3.27. Using the definition (3.37), show that the Lagrangian finite strain tensor Lij transforms as a second order Cartesian tensor under the coordinate transformations xi = bjixf and X( = bijXj.

1 axk axk By (3.37), Lij = - -- - Sij which by the stated transformation becomes 2 í ax, ax,

ax; axq d(b,,zL) ax; b .b .S (since bpkbqk = S p q ) mZ n~ p q ax:, ax:, a,; ax,

3.28. A certain homogeneous deformation field results in the finite strain tensor

. Determine the principal strains and their directions for this

deformation.

Being a symmetric second order Cartesian tensor, the principal strains a re the roots of


Thus L( , , = -2, L(2, = 2, Lo, = 8. The transformation matr ix fo r principal directions is

[aijl =

1 1 2 1 6

3.29. For the homogeneous deformation xi = fi XI, x, = 2 x 2 , xj = fi X3 - X2 determine the material strain ellipsoid resulting from deformation of the spherical surface X: + X: + X: = 1. Show that this ellipsoid has the form xS/A:,, + x;/Ak) + x i l ~ ~ , , = 1.

By (3.82), or alternatively by inverting the given displacement equations and substituting into X i x i = 1, the material s t rain ellipsoid is xS + xi + x3 + x2x3 = 3. This equation is put into the principal-axes form xS13 + x;/6 + x i l 2 = 1 by the transformation

o 11,h llfi

= [E From the deformation equations, the stretch tensor S = 6 i s given (calculation is similar to

tha t in Problem 3.24) a s

O 1

which by the transformation [aij] = is put into the principal form

with principal stretches A:~, = 3, A:,, = 6, A&) = 2. Note also t h a t the principal stretches may be calculated directly from (3.66) using [aij] above.

3.30. For the deformation of Problem 3.29, determine the spatial strain ellipsoid and show that i t is of the form A:,,x? + A?,,X; t A:,,X~ = 1.

By (3.84) the sphere xixi = 1 resulted from the ellipsoid X G X = 1, or

This ellipsoid is put into the principal-axes form 3 ~ : + 6x22 + 2 ~ : = 1 by the transformation


3.31. Verify by direct expansion that the second invariant IIL of the strain tensor may be expressed by

111 112 + 111 113 IIL =

122 123 I 1.1 122 1 / 131 133 1 1 132 133 1 Expansion of the given determinants results in IIL = l l1l2, + 1,,13, + 133111 - ( 1 f 2 + l i 3 + l i l ) .

I n comparison, direct expansion of the second equation of ( 3 . 9 1 ) yields

IIL = +[(111 + lZ2 + 133) l j j - ( l l j l l j + 12j12j + 13j13j)]

= &[(I11 + l22 + 133)(111 + l22 + l33) - ( l l l l l l + 112112 + 113113

+ l2l12l + l22l22 + l23l23 + 131131 + l32l32 + 133133)l

= 111122 + l22l33 f 133111 - (1S2 + 123 + 1321)

3.32. For the finite homogeneous deformation given by ui = AijXj where Aij are constants, determine an expression for the change of volume per unit original volume. If the Aij are very small, show that the result reduces to the cubical dilatation.

Consider a rectangular element of volume having original dimensions d X l , d X 2 , d X 3 along the coordinate axes. For the given deformation, xi = ( A i j + S i j ) X j . Thus by ( 3 . 8 3 ) the original volume d V o becomes a skewed parallelepiped having edge lengths d x i = ( A i c n ) + Si(,)) d X ( , , , n = 1,2,3 . From ( 1 . 1 0 9 ) this deformed element has the volume d V = e i j k ( A i l + S i 1 ) ( A j 2 4- S j 2 ) ( A k 3 + S k 3 ) d X 1 d X 2 d X 3 . Then

If the Aij a r e very small and their powers neglected,

A V / d V o = ~ ~ ~ ~ ( A ~ ~ 8 ~ ~ 8 ~ ~ + S i l A j 2 S k 3 + S i l S j 2 A k 3 + 8 i 1 8 j 2 8 k 3 ) - 1' = Ali + AZ2 + A33

F o r linear theory the cubical dilatation lii = auilaXi, which for ui = A i j X j is lii = A,, + A,, + A33,

3.33. A linear (small strain) deformation is specified by ul = 4x1 - xz + 3x3, u z = x1 + 7x2, us = -3x1 + 4x2 + 4x3. Determine the principal strains and the principal deviator strains e,,,, for this deformation.

Since cij is the symmetrical par t of the displacement gradient a u i l d x j , i t i s given here by

or in principal-axes form by = . Also, €,,I3 = 5 and so the

strain deviator is eij = i ) and i ts principal-axes form e$ = O -1 (1 u). Note

2 -1

t h a t e( , ) = c(,) - e k k / 3 .

PLANE STRAIN AND COMPATIBILITY (Sec. 3.15-3.16)

3.34. A 45" strain-rosette measures longitudinal strain along the axes shown in Fig. 3-15. At a point P, C , , = 5 x lOP4, C ; , = 4 x 10-4, = 7 x 10-4 inlin. Determine the shear strain C,, a t the point.

1 By ( 3 . 5 9 ) , with = (êl+ê2)/\/S a s the unit vector in the si

direction, Fig. 3-15


5 x 10-4 e, , llfi 11. I/., O] [ : 7 .:Op4 ]-[';I = 4 x 1 O p 4

12 x 10-4 + 2 ~ , , Therefore

2 = 4 x 10-4 or €,, = -2 x 10-4.

3.35. Construct the Mohr's circles for the case of plane strain 1 '"

and determine the maximum shear strain. Verify the result analytically.

With the given s tate of s t rain referred to the xi axes, the points B(eZZ = 5, €23 = & ) and D a r e established a s the diameter of the larger inner circle in Fig. 3-16. Since

= O is a principal value for plane strain, the other circles a r e drawn a s shown.

A rotation of 30' about the x , axis (equivalent to 60° in the Mohr's diagram) results in the principal s t rain axes with the principal s t rain tensor given by Fig. 3-16

Fig. 3-1 7 Fig. 3-18

Next a rotation of 45O about the x: axis (90° in the Mohr diagram) results in the xl axes and the associated strain tensor e / j given by

the first two rows of which represent the s tate of strain specified by point F in Fig. 3-16. Note tha t a rotation of -45O about xg would correspond to point E in Fig. 3-16.


3.36. The state of strain throughout a continuum is specified by

Are the compatibility equations for strain satisfied? Substituting directly into (3.104), a11 equati'ons a r e satisfied identically. The student should

car ry out the details.


3.37. Derive the indicial form of the Lagrangian finite strain tensor L, of (3.40) from its definition (3.37).

From (3.24), dx,/dXi = S i j + aui/dXi. Thus (3.37) may be written

Li$ = [( S k i f - " k ) ( S k j + - a?1i) - ax, axj

= Ldui au, auk auk

axi ax, axj 2 ax,+-+-- I 3.38. A displacement field is defined by xl = X I - CX2 + BX3, xz = C X i + Xz - AX3,

x3 = -BXl + AX2 + X3. Show that this displacement represents a rigid body rotation only if the constants A, B, C are very small. Determine the rotation vector w for the infinitesimal rigid body rotation.

1 -C B F o r the given displacements, F = and from (3.37),

If products of the constants a r e neglected, this s t rain tensor i s zero and the displacement reduces to a rigid body rotation. From (3.50), the rotation vector is

W = - 2

3.39. For the rigid body rotation represented by ul = 0.02X3, uz = -0.03X3, u3 = -0.02X1 + 0.03X2, determine the relative displacement of Q(3,0.1,4) with respect to P(3, o, 4 ) .

A A A A From the displacement equations, UQ = .O8 ê1 - .12êz - .O57 e, and up = .O8 el - .12 e2 - .O6 e3.

Hence du = UQ - UI> = -.003ê3. The same result is obtained by (3.51), with w = .03ê1 + .02ê,:

A A A

e1 e2 e3 â/dXl a/ax, alax,

-CXz + BX3 C X l - AX3 -BXl + AX,

d u =

= Aê, + s ê , + cê3

A A A

e1 e2 e3

.O3 .O2 O

o .1 o = -.003ê:,


3.40. For a state o£ plane strain parallel to the ~ 2 x 3 axes, determine expressions for the normal strain r;, and the shear strain r;, when the primed and unprimed axes are oriented as shown in Fig. 3-19.

By equation (3.59), o

"42 = ~ o , c o s e s i n e l [ ~ €J[:;:] - - E,, cos2 e + 2e23 sin e cos O i- sin2 o

- '22 + €33 '22 - €83 - - 2

+ - 2

eos 2e + sin 20

Similarly from (3.65) and Problem 3.20,

o 4 = [o, cos e, sin e, [. em €7 [-si! e]

E23 E33 COS 0

= -E, , sin O cos O + €23 C O S ~ 0 - €23 sin2 e + 9, sin e cos e

"22 - "33 = €23 COS20 - - 2

sin 20

Fig. 3-19 Fig. 3-20

3.41. For a homogeneous deformation the small strain tensor is given by

What is the change in the 90" angle ADC depicted by the small tetrahedron OABC in Fig. 3-20 if OA = OB = OC, and D is the midpoint of A B ?

The unit vectors Y and a t D a r e given by f; = (êl - ê2)/\/S and 2 = (2 ê3 - ê2 -êl)/& . From the result of Problem 3.20,


3.42. At a point the strain tensor is given by rij - - i ) and in principal form

. Calculate the strain invariants for each of these tensors

and show their equivalence.

By (3.95) and Problem 3.31, IE = 5 + 4 + 4 = 13, IE* = 6 + 4 + 3 = 13. Likewise IIE = 19 + 19 + 16 = 54, 11,. = 24 + 18 + 12 = 54. Finally IIIE = 5(16) - 4 - 4 = 72, 111,. = (6)(4)(3) = 72. The student should check these calculations.

3.43. For the displacement field xl = X I + AX3, xz = Xz - AX3, x3 = X3 - A X l + AX2, determine the finite strain tensor LG. Show that if A is very small the displacement represents a rigid body rotation.

Since u, = AX3, u, = -AX3, u3 = -AXl + AX,, by (3.40),

If A is small so t h a t A2 may be neglected, LG = 0; and by (3.50) the rotation vector w = A ê1 + A ê2.

3.44. Show that the displacement field ul = Axl + 3x2, u2 = 3x1 - Bxz, u3 = 5 gives a state of plane strain and determine the relationship between A and B for which the deformation is isochoric (constant volume deformation).

From the displacement equations, by (3.43), eii = 1 3 B O 1 which is of the form of

L 0 O 0 1 (3.99). From (3.96), the cubical dilatation is D eii = A - B, which is zero if A = B.

3.45. A so-called delta-rosette for measuring longitudinal surface strains has the shape of the equilateral i""/"; triangle A and records normal strains E ~ ~ , E ; , , E ; ; in the directions shown in Fig. 3-21. If E , , = a, r l l = b, e;; = c, determine E , , and E,, a t the point.

2 1

By (3.59) with L = E, fo r the x{ direction, Fig. 3-21

Also for the x;' direction

Solving simultaneously fo r e12 and ez2 yields e l z = ( b - c ) / 6 and E,Q = (-a + 2b + 2c)/3.


3.46. Derive equation (3.72) expressing the change in angle between the coordinate directions Xp and X3 under a finite deformation. Show that (3.72) reduces to (3.65) when displacement gradients are small.

Let y , ~ = ~ / 2 - 8 be the angle change a s shown in Fig. 3-5. Then sin yz3 = cos (r12 - 8) = A A n, n,, or by (3.33) and (3.34)

dx2 dx3 -.- - dX, . F, . F dX3

sin Yzn = - Idx21 ldxzl d d ~ , G dX2 d d ~ , G dX3

Now dividing the numerator and denominator of this equation by IdX21 and IdX31 and using (3.353 and (3.66) gives A

A A e, G e, - e, . G ê,

sin yz3 = - ,I- ,I- A(:,) A(&]

A A A A A A A Next from (3.37), e,. G -e3 = e, (21, + I) *e3 = ê2. 2~~ e3 + e, l e3 = 2Lz3 since e2 e3 = 0.

Also from (3.68), A(;2) = d m , etc., and so

sin y?, = 2L23

d - d m

If aui/dX, 4 1 this reduces to sin yz3 = duz/aX3 + du3/ax2 = 2123.

3.47. For the simple shear displacement XI = XI, x2 = XZ, x3 = X3 + 2 ~ z / f i , determine the direction of the line element in the XzX3 plane for which the normal strain is zero.

Let &i = m2ê, + m,ê, be the unit normal i n the direction of zero strain. Then from (3.66), since AZcm, = 1,

o

or 7mE + 4~ m2m3 + 3mi = 3. Also mi + rn; = 1, and solving simultaneously m2 = ffi12, m, = 7112, or m2 = O, m3 = f 1. Thus there is zero s t rain along the X3 axis and f o r the element a t 60' to the X3 axis.

A The student should verify this result by using the relation & 2LG m = O derived from (3.36).

Supplementary Problems 3.48. For the shear displacement of Problem 3.47, determine the equation of the ellipse into which the

circle X: + X: = 1 is deformed. Ans. x: + 9%; = 3

3.49. Determine the shear angle y,, fo r the deformation of Problem 3.47 (Fig. 3-22). Ans. y,, = sin-12/$?

3.50. Given the displacement field x, = X, + 2X3, x2 = Xz - 2X,, x, = X3 - 2X1 + 2X2, determine the Lagrangian and Eulerian finite strain tensors L, and E*.

2 -2 o 2 -2 o

x, Fig. 3-22


3.51. Determine the principal-axes form of the two tensors of Problem 3.50.

4 o 419 o Ans . 1; = ( c 4). E: = (: :) 419

3.52. For the displacement field of Problem 3.50 determine the deformation gradient i, and by a polar decomposition of F find the rotation tensor R and the right stretch tensor S.

1 2 -1 o 2

Ans. R = ? ( 3 ' 2 -i), s = i: : B', F = (-i : -:' -2 2

3.53. Show that the first invariant of 1, may be written in terms of the principal stretches as -

ILG - - 1 ) + (h:ê2) - 1 ) + (h2,, - 1)] /2 . Hint: See equation (3.68). (e31

-3 fi 3.54. The strain tensor a t a point is given by qj = (i -:). Determine the normal strain

in the direction of = $,I2 -$,I2 + ê 3 / f i and the shear strain between and fl = -ê1/2 + $212 + $,/\/S. Ans . e A = 6 , y,, = 0.

( V )

3.55. Determine the principal-axes form of eij given in Problem 3.54 and note that and f l of that problem are principal directions (hence y,, = 0) .

Ans . eij -

o -2

3.56. Draw the Mohr's circle for the state of strain giv'en in Problem 3.54 and determine the maximum shear strain value. Verify this result analytically. Ans. y,,, = 4

3.57. Using eij of Problem 3.54 and E: given in Problem 3.55, calculate the three strain invariants from each and compare the results. Ans. IE = 6 , IIE = -4, IIIE = -24.

3.58. For eij of Problem 3.54, determine the deviator tensor eij and calculate its principal values.

3.59. A displacement field is given by ul = 3 ~ ~ x 2 2 , u2 = 2x3xl, u3 = x i - x1x2. Determine the strain tensor eij and check whether or not the compatibility conditions are satisfied.

32: 3xlx2+ X 3 -z2/2

o 2112 2x3

3.60. For a delta-strain-rosette the normal strains are found to be those shown in Fig. 3-23. Determine e12 and ez2 a t the location.

Ans. ez2 = 1 X 10-4, e l z = -0.2885 X 10-4 r;l = 1 x 10-4 A= 10-4

3.61. For the displacement field x l = X l + AX,, x2 = X 2 , x3 = X 3 - A X 1 , calculate the volume change and show that i t is zero if A .I , = 2 x 10-4

is a very small constant. Fig. 3-23

Chapter 4

Motion and Flow

4.1 MOTION. FLOW. MATERIAL DERIVATIVE

Motion and flow are terms used to describe the instantaneous or continuing change in configuration of a continuum. Flow sometimes carries the connotation of a motion leading to a permanent deformation as, for example, in plasticity studies. In fluid flow, however, the word denotes continuing motion. As indicated by (3.14) and (3.15), the motion of a continuum may be expressed either in terms of material coordinates (Lagrangian description) by

~i = x ~ ( X I , X ~ , X ~ , t) = xi(X, t) or x = x(X, t)

or by the inverse of these equations in terms of the spatial coordinates (Eulerian description)

The necessary and sufficient condition for the inverse functions (4.2) to exist is that the Jacobian determinant

J = laxiIaXil (4.3)

should not vanish. Physically, the Lagrangian description fixes attention on specific particles of the continuum, whereas the Eulerian description concerns itself with a particular region of the space occupied by the continuum.

Since (4.1) and (4.2) are the inverses of one another, any physical property of the continuum that is expressed with respect to a specific particle (Lagrangian, or material description) may also be expressed with respect to the particular location in space occupied by the particle (Eulerian, or spatial description). For example, if the material description of the density p is given by

p = P(Xi,t) or p = p(X,t)

the spatial description is obtained by replacing X in this equation by the function given in (4.2). Thus the spatial description of the density is

P = p(Xi(x, t), t) = p*(~i, t) ~r p = p(X(x, t), t) = p*(x> t) (4.5)

where the symbol p* is used here to emphasize that the functional form of the Eulerian description is not necessarily the same as the Lagrangian form.

The time rate of change of any property of a continuum with respect to specific particles of the moving continuum is called the material derivative of that property. The material derivative (also known as the substantial, or comoving, or convective derivative) may be thought of as the time rate of change that would be measured by an observer traveling with the specific particles under study. The instantaneous position xi of a particle is itself a property of the particle. The material derivative of the particle's position is the instantaneous velocity of the particle. Therefore adopting the symbol dldt or the superpositioned dot ( O ) as representing the operation of material differentiation (some books use DIDt), the velocity vector is defined by

vi = dxi/dt = X i or V = dxldt = X (4.6)

CHAP. 41 MOTION AND FLOW 111

In general, if Pij,. . is any scalar, vector or tensor property of a continuum that may be expressed as a point function of the coordinates, and if the Lagrangian description is given by

the material derivative of the property is expressed by

The right-hand side of (4.8) is sometimes written ;jX7 t)] to emphasize that the X

X coordinates are held constant, i.e. the same particles are involved, in taking the derivative. When the property Pij.. . is expressed by the spatial description in the form

the material derivative is given by

where the second term on the right arises because the specific particles are changing position in space. The first term on the right of (4.10) gives the rate of change a t a particular Iocation and is accordingly called the local rate of change. This term is sometimes written [dP'.ir9 t)] to emphasize that x is held constant in this differentiation. The second term

X

on the right in (4.10) is called the convective rate of change since it expresses the contribution due to the motion of the particles in the variable field of the property.

From (4.6), the material derivative (4.10) may be written

which immediately suggests the introduction of the material derivative operator

which is used in taking the material derivati,ves of quantities expressed in spatial coordinates.

4.2 VELOCITY. ACCELERATION. INSTANTANEOUS VELOCITY FIELD

One definition of the velocity vector is given by (4.6) as vi = dxildt (or v = dxldt). An alternative definition of the same vector may be obtained from (3.11) which gives xi = ui + Xi (or x = u + X). Thus the velocity may be defined by

since X is independent of time. In (4.13), if the displacement is expressed in the Lagrangian form ui = ui(X, t), then

112 MOTION AND FLOW [CHAP. 4

If, on the other hand, the displacement is in the Eulerian form zci = u~(x , t), then

In (6.15) the velocity is given implicitly since i t appears as a factor of the second term on the right. The function

vi = vi(x, t ) or V = V ( X , t ) (4.1 6 )

is said to specify the instantaneous velocity field.

The material derivative of the velocity is the acceleration. If the velocity is given in the Lagrangian form (4.14), then

If the velocity is given in the Eulerian form (4.15), then

4.3 PATH LINES. STREAM LINES. STEADY MOTION

A path 2ine is the curve or path followed by a particle during motion or flow. A stream line is the curve whose tangent a t any point is in the direction of the velocity a t that point. The motion of a continuum is termed steady motion if the velocity field is independent of time so that av,lat = O . For steady motion, stream lines and path lines coincide.

4.4 RATE OF DEFORMATION. VORTICITY. NATURAL STRAIN INCREMENTS

The spatial gradient of the instantaneous velocity field defines the velocity gradient tensor, dvildxj (or Yij). This tensor may be decomposed into its symmetric and skew- symmetric parts according to

dvi y.. - = a X j

or Y = 9(vvx + Vx V ) + 4(vVx - Vx V ) = (D+V) (6.1 9 )

This decomposition is valid even if vi and avildxj are finite quantities. The symmetric tensor

is called the rate o f deformation tensor. Many other names are used for this tensor; among them rate o f strain, stretching, strain rate and velocity strain tensor. The skew-symmetric tensor

is called the vorticity or spin tensor.


The rate of deformation tensor is easily shown to be the material derivative o'f the Eulerian linear strain tensor. Thus if in the equation

the differentiations with respect to the coordinates and time are interchanged, i.e. if

(e) is replaced by , the equation takes the form d t axj dxj d t

By the same procedure the vorticity tensor may be shown to be equal to the material derivative of the Eulerian linear rotation tensor. The result is expressed by the equation

A rather interesting interpretation may be attached to (4.23) when that equation is rewritten in the form

deij = Dijd t or dE = D d t

The left hand side of (4.25) represents the components known as the natural s train increments , widely used in flow problems in the theory of plasticity (see Chapter 8).

4.5 PHYSICAL INTERPRETATION OF RATE OF DEFORMATION AND VORTICITY TENSORS

In Fig. 4-1 the velocities of the neighboring particles a t points P and Q in a moving continuum are given by vi and vi + dvi respectively. The relative velocity of the particle a t Q with respect to the one a t P is therefore

aVi dvi = -dxj or d v = vVX d x (4.26) dxj

in which the partia1 derivatives are to be evaluated a t P. In terms of Dij and Vij, (4.26) becomes

dvi = (Dij + Vi j ) dxj

O?.- d v = ( D + V ) . d x (4.27) Fig. 4-1

If the rate of deformation tensor is identically zero (Dij = O),

and the motion in the neighborhood of P is a rigid body rotation. For this reason a velocity field is said to be irrotational if the vorticity tensor vanishes everywhere within the field.

Associated with the vorticity tensor, the vector defined by

qi = cijk V k , j or q = v , x v (4.29)

is known as the vorticity vector. The symbolic form of (4.29) shows that the vorticity vector is the curl of the velocity field. The vector defined as one-half the vorticity vector,

MOTION AND FLOW [CHAP. 4

ai = q = i i k v k , j or a = &q = &Vx x v (4.30)

is called the rate of rotation vector. For a rigid body rotation such as that described by (4.28), the relative velocity of a neighboring particle separated from P by dxi is given as

dvi = ~,,,n, dx, or d v = 0 x dx (4.31)

The components of the rate of deformation tensor have the following physical interpretations. The diagonal elements of Dii are known as the stretching or rate of eztension components. Thus for pure deformation, from (S.27),

dv, = Dij dxj or dv = D d x (4.32)

and, since the rate of change of length of the line element dxi per unit instantaneous length is given by

dvi dxi - A

d i ( ~ ) = - dx = D..- - Dijvj or d'") = ü a $

a3 dx (4.33)

the rate of deformation in the direction of the unit v,ector V , is

d = di(V'~i = Dijvjvi or d = 3.D.D (4.34)

From (4.34), if vi is in the direction of a coordinate axis, say $2, A d = d22 or d = ê2 D . e2 = D22 (4.35)

The off-diagonal elements of Dij are shear rates, being a measure of the rate of change between directions a t right angles (See Problem 4.18).

Since Dij is a symmetric, second-order tensor, the concepts of principal axes, principal values, invariants, a rate of deformation quadric, and a rate of deformation deviator tensor may be associated with it. Also, equations of compatibility for the components of the rate of deformation tensor, analogous to those presented in Chapter 3 for the linear strain tensors may be developed.

4.6 MATERIAL DERIVATIVES OF VOLUME, AREA AND LINE ELEMENTS

In the motion from some initial configuration a t time t = O to the present configuration a t time t , the continuum particles which occupied the differential volume element dVo in the initial state now occupy the differential volume element dV. If the initial volume element is taken as the rectangular parallelepiped shown in Fig. 4-2, then by (1.10)

dV = dXiêi X dXzê2 dX3ê3 = dXi dX2dXa (4.36)

Due to the motion, tkis parallelepiped is moved and distorted, but because the motion is assumed continuous the volume element does not break up. In fact, because of the relationship (3.33), dxi - (axilaXj) dXj between the material and spatial line elements, the "line of particles" Fig. 4-2


that formed dX1 now form the differential line segment dx,'" = (dxilaXi) dX1. Similarly dX2 becomes d ~ : ~ ' = (axildxn) dX2 and dX3 becomes dx13' = (axiIdX3) dXa. Therefore the differential volume element dV is a skewed parallelepiped having edges dx?', dx:", dxi3) and a volume given by the box product

dV = dxC1) x dxC2) dxC3) = Éijk dxi(l) dxj2' dxk' (4.37)

But (4.37) is seen to be equal to

axi dX2 dX3 = J dVo dV = c.. --- dX ax, ax, ax3 (4.38)

where J = laxiIdXjl is the Jacobian defined by (4.3).

Using (4.38), it is now possible to obtain the material derivative of dV as

d since dVo is time independent, so that - (dVo) = O. The material derivative of the Jacobian dt

J rnay be shown to be (see Problem 4.28)

and hence (4.39) rnay be put into the form

For the initial configuration of a continuum, a differential element of area having the magnitude dSO rnay be represented in terms of its unit normal vector ni by the expression dSoni. For the current configuration of the continuum in motion, the particles initially making up the area d S O n now fill an area element represented by the vector dSnt or dSi. It rnay be shown that ax.

d S i = J L d X j or d S = J d X * X v x axi (4.42)

from which the material derivative of the element of area is

The material derivative of the squared length of the differential line element dxi rnay be calculated as follows,

However, since dxi = (axildXj) d X j ,

and (4.44) becomes

d dvi d - ( d x 2 ) = 2 - d x k d x i or z ( d x 2 ) = 2 d x . v X v . d x d t dxk (4.46)

The expression on the right-hand side in the indicial form of (4.46) is symmetric in i and k, and accordingly rnay be written


or, from (4.20), d d - (dx2) = 2 Dij dxidxj ~r - (dx2) = 2 d ~ ' D d~ dt dt (4.48)

4.7 MATERIAL DERIVATIVES OF VOLUME, SURFACE AND LINE INTEGRALS

Not a11 properties of a continuum may be defined for a specific particle as functions of the coordinates such as those given by (4.7) and (4.9). Some properties are defined as 'integrals over a finite portion of the continuum. In particular, let any scalar, vector or tensor property be represented by the volume integral

P 2j... (t) = J P < ; . . . ( x , ~ ) ~ v v

(4.49)

where V is the volume that the considered part of the continuum occupies a t time t. The material derivative of Pij. . . (t) is

and since the differentiation is with respect to a definite portion of the continuum (i.e. a specific mass system), the operations of differentiation and integration may be interchanged. Therefore

which, upon carrying out the differentiation and using (4.411, results in

Since the material derivative operator is given by (4.12) as dldt = d/& + v~aldxp, (4.52) may be put into the form

P;. . .(x, t) dV = a d t V

+ ~ X P -( vPPt . . ( x, t))] dV

By using Gauss' theorem (1.157), the second term of the right-hand integral of (4.53) may be converted to a surface integral, and the material derivative then given by

P: ( x , t ) d v = . '(xf t, dV + 1 v,[P:. . . (x, t)] dSp at (4.54)

This equation states that the rate of increase of the property Pij.. .(t) in that portion of the continuum instantaneously occupying V is equal to the sum of the amount of the property created within V plus the flux V,[P;. . .(x, t)] through the bounding surface S of V.

The procedure for determining the material derivatives of surface and line integrals is essentially the same as that used above for the volume integral. Thus for any tensorial property of a continuum represented by the surface integral

where S is the surface occupied by the considered part of the continuum a t time t, then, as bef ore,

& 1 Q: . . . (x, t) dSp = i & [Q: . . . (x, t) dSp] (4.56)


and, from (4.43), the differentiation in (4.56) yields

dQij ... ( t ) - - [dQ:i. ( x , t, + Q* (x, t ) ] d S p - [Q* -" d S p ] (4.57) d t s d t 8% s . . dXq v . . .

For properties expressed in line integral form such as .

R 13.. . ( t ) = I R ; ...( ~ , t ) d x , C

(4.58)

the material derivative is given by

& 1 R&. . ( x , t ) dxp = 1 2 [R:. . (r, t ) d x p ] (4.59)

Differentiating the right hand integral as indicated in (4.59) , and making use of (4.45) , results in the material derivative

d d [ R : . . . ( x , t )] - [ R i j . . . ( t ) ] = J

d t d t C d x , + 12 [ ~ ~ . . ( x , t ) ] dx , (4.60)

Solved Problems

9

MATERIAL DERIVATIVES. VELOCITY. ACCELERATION (Sec. 4.1-4.3)

Inverting the motion equations, X 1 = xle-t + x3(e-t - I), X 2 = x2 - x3(et- e-t), X3 = x3. Note that in each description when t = 0, xi = Xi.

4.1. The spatial (Eulerian) description of a continuum motion is given by X I = X1et + X3(et - I), x2 = X3(et - ec t ) + XZ, x3 = X3. Show that the Jacobian J does not vanish for this motion and determine the material (Lagrangian) description by inverting the displacement equations.

By (4.9) the Jacobian determinant is

4.2. A continuum motion is expressed by xl = X I , xz = et(X2 + X3)/2 + ec t (X2 - X3)/2, x3 = et(X2 + X3)/2 - e-t(X2 - X3)/2. Determine the velocity components in both their material and spatial forms.

J = laxi/aXjl =

From the second and third equations, X 2 + X 3 = e-t(x , + x,) and X , - X3 = et(x, - x,). Solving these simultaneously the inverse equatjons become X1 = x,, X , = e-t(x2+xj) /2 + et(xz - x3)/2, X3 = e-t(x2 + x3)/2 - et(x2 - x3)/2. Accordingly, the displacement components ui = zi - Xi may be written in either the Lagrangian form ul = O, u , = et(X2 + X3)/2 + e-t(X, - X3)/2 - X2, u3 = et(X2 + X3)/2 - e-t(X2 - X3)/2 - X3, or in the Eulerian form ul = O, u2 = x2 -

+ x3)/2 - et(x2 - x3)/2, u3 = x3 - e-t(x2 + x3)/2 + et(xz - x3)/2.

By (4.14), vi = aui/dt = üXi/ãt and the velocity components in Lagrangian form are v l = 0, v2 = et(X2 + X3)/2 - e - f ( X 2 - X3)/2, v , = et(Xz + X3)/2 + e-t(X2 - X3)/2. Using the relationships X z 4- X3 = e-t(x2 + x3) and X 2 - X3 = et(x2 - x3) these components reduce to v , = O, v , = x3, v3 = 2 2 .

et O e t - 1

O 1 e t -e - t

O 0 1

= et


Also, from (4.15), for the Eulerian case,

du~. ldt = v2 = C - ~ ( X ~ + x3)/2 - et(x, - x3)/2 + v2(2 - e-t - et)/2 + v3(-e-t + et) /2

du$dt = v3 = e-Yxz + x3)12 + et(xz - %,)I2 + v2(-e-t + et)/2 + v3(2 - e-t - et) /2

Solving these equations simultaneously for v , and v,, the result is as before v, = x,, v 3 = xs.

4.3. A velocity field is described by v1 = x ~ l ( l + t), vn = 2x2/(1+ t) , v3 = 3x3/(1+ t). Deter- mine the acceleration components for this motion.

4.4. Integrate the velocity equations of Problem 4.3 to obtain the displacement relations xi = xi(X, t ) and from these determine the acceleration components in Lagrangian form for the motion.

By (4.13), v , = dx l /d t = x l l ( l + t ) ; separating variables, dxl lx l = dt l (1 + t) which upon integration gives ln x , = ln ( 1 + t ) + ln C where C is a constant of integration. Since xl = X , when t = O, C = X , and so x1 = X l ( l + t). Similar integrations yield z2 = X 2 ( l + t ) 2 and x, = X3(1 + t)3. '

Thus from (4.14) and (4.17), v , = X1, v , = 2X2(1 + t ) , v , = 3X3(1 + t ) 2 and a, = O, a, = 2X,, a , = 6X3(1 + t).

4.5. The motion of a continuum is given by xl = A + (e-BAIX) sinA(A + w t ) , x2 = -B - (ecBVX) cos X(A + d) , x3 = X3. Show that the particle paths are circles and that the velocity magnitude is constant. Also determine the relationship between X l and X Z and the constants A and B.

By writing x , - A = (e-BVk) sin x ( A + ~ t ) , x , + B = (-e-BV/h) cos X(A + ~ t ) , then squaring and adding, t is eliminated and the path lines are the circles ( x l -A)2 + ( x , + B)2 = e-zBh/X. From (4.6), v l = oe-Bh cos h ( A + wt), v , = we-BA sin h ( A + wt), v , = O and v2 = v: + v; + v i = ~2e-zBA. Finally, when t = O, xi = Xi and so X , = A + (e-Bh/X) sin x A , X , = -B - ( e - B ~ x ) cos xA.

4.6. A velocity field is specified by the vector v = xStêl + x2t2ê2 + x1x3tê3. Determine the velocity and acceleration of the particle a t P(1,3,2) when t = 1.

A By direct substitution, v, = e, + 3 ê2 + 2 ê3. Using the vector form of (4.18) the acceleration field is given by

a = xSê, + 2x2tê2 + x1x3ê3 + ( x S t ê , + x2t2ê2 + x1x3tê3)

(2x1tê$, + x3tê,ê3 + t2ê$, + x1tê3ê3)

or a = ( x S + 2x1t2) ê, + (2x2t + %,e) ê2 + (51x3 + 2x x3t2)ê3

Thus ap = 3 ê l + 9 ê 2 + 6 ê , .

4.7. For the velocity field of Problem 4.3 determine the streamlines and path lines of the flow and show that they coincide.

At every point on a streamline the tangent is in the direction of the velocity. Hence for the differential tangent vector d x along the streamline, v X d x = O and accordingly the differential equations of the streamlines become dx l l v l = dx2/vz = dx3/v,. For the given flow these equations are d x l l x l = dx,/2x2 = dx3/3x3. Integrating and using the conditions xi = Xi when t = 0, the equations of the streamlines are ( x l / X l ) z = x,/X,, (x1/X1)3 = x3/X3, (x2/X2)3 = (x3/X3)2.

Integration of the velocity expressions dxi /dt = vi as was carried out in Problem 4.4 yields the displacement equations x l = X l ( l + t ) , x2 = X z ( l + t)*, x3 = X 3 ( l + t)3. Eliminating t from these equations gives the path lines which are identical with the streamlines presented above.

C H A P . 41 MOTION AND FLOW 119

4.8. The magnetic field strength of an electromagnetic continuum is given by A = ecAtlr where r2 = X : + X ; + xi and A is a constant. If the velocity field of the continuum is given by vi = Bx1x3t, v2 = B x ~ t2, v3 = B x ~ x ~ , determine the rate of change of magnetic intensity for the particle a t P(2, -1,2) when t = 1.

Since a(r-l)/axi = -xi/r3, equation (4.11) gives

i = -Ae-At/r - e-~ t (Bx:x , t + B X ~ t2 + Bx$x2)/r3

Thus for P a t t = 1, i, = -e-A(3A + B)/9.

4.9. A velocity field is given by vi = 4x3 - 3x2, v2 = 3x1, v3 = - 4 ~ ~ . Determine the acceleration components a t P(b, O, 0 ) and &(O, 4b, 3b) and note that the velocity field corresponds to a rigid body rotation of angular velocity 5 about the axis along A e = (4ê2 + 3ê3)/5.

From (4.18), a , = -25s1, a2 = -9x2 i 12x3, a , = 12x2 - 162,. Thus a t P(b, 0 , O ) , a = -25bêl which is a normal component of acceleration. Also, a t Q(O,4b, 3b) which is on the axis of rotation,

A a = O . Note that v = w X x = ( 4 ~ 2 + 3 ~ 3 ) ~ ( x l ~ ~ + ~ 2 ê 2 + x 3 ~ 3 ) = ( 4 ~ 3 - 3 x 2 ) ~ 1 + 3 ~ 1 ~ 2 - 4 ~ 1 e 3 .

RATE OF DEFORMATION, VORTICITY (Sec. 4.4-4.5)

4.10. A certain flow is given by vi = O, v2 = A ( X I X ~ - xi)ecBt, v3 = A ( x ~ - x1x3)ecBt where A and B are constants. Determine the velocity gradient dv i ldx j for this motion and from it compute the rate of deformation tensor D and the spin tensor V for the point P(1,0,3) when t = 0.

o By (4.19) a v i h x j = ( " x, which may be evaluated a t P when t = O

-23 22, -2,

and decomposed according to (4.20) and (4.21) as

4.11. For the motion xi = X l , x2 = X2 + - I ) , x3 = X 3 + X I ( ~ - ~ ~ - 1) compute the rate of deformation D and the vorticity tensor V. Compare D with de i j ld t , the rate of change of the Eulerian small strain tensor E.

Here the displacement components are u , = O , u2 = %,(e-2t- 1), u3 = x,(e-3t- 1 ) and from (4.14) the velocity components are v, = O , v2 = -2x1e-2t, v3 = -3x,e-3t. Decomposition of the velocity gradient avilaxj gives avi/asj = Dij + V i j . Thus

Likewise, decomposition of the displacement gradient gives aui/dxj = cij + uij. Thus


Comparing D with dEldt,

-e-% -3e-3t/2

ds,ldt = -e-2t O = Dij

-3e-3tI2 O ! O o

The student should show that dwijldt = V i j .

4.12. A vortex line is one whose tangent a t every point in a moving continuum is in the direction of the vorticity vector q. Show that the equations for vortex lines are dxilqi = dx21q2 = dx31q3.

Let d x be a differential distance vector in the direction of q. Then q X d x E O , or

(9, dx3 - q3 dx,) 2, 3- (93 dxl - d ~ 3 ) $2 (91 dxz - 9, dx i )ê3 E O

from which dxl lql = dx21q2 = dx31q3.

4.13. Show that for the velocity field v = ( A x s - Bx2)êl + (Bxl - Cx3)êz + (CXZ - Axl)ê3 the vortex lines are straight lines and determine their equations.

From (4.29), q = V , X V

lines are A dx3 = B dx,, B d x , tions of the vortex lines x3 = constants of integration.

= 2 ( C ê l + A 3, + sê3), and by Problem 4.12, the d.e. for the vortex = C dx,, C dxz = A dxl. Integrating these in turn yields the equa-

Bx21A + K1, x1 = Cx31B + K,, x z = A x J C + K , where the Ki are

4.14. Show that the velocity field of Problem 4.13 represents a rigid body rotation by showing that D = 0.

Calculating the velocity gradient avilaxj, i t is found to be antisymmetric. Thus avildxj = O -B A (-; r -f ) = V i j and Dij = O.

4.15. For the rigid body rotation v = 3x3ê1 - 4x3& + (4x2 - 3xl)ê3, determine the rate of rotation vector O and show that v = O X x.

From (4.30), 2iZ = q, or i2 = 4 ê 1 + 3ê2. This vector is along the axis of rotation. Thus

( 4 ê l f 3 ê 2 ) X ( ~ ~ ê ~ f ~ ~ ê ~ + ~ ~ ê ~ ) = 3x3ê1 - 4x3ê2 + ( 4 x 2 - 3 x 1 ) ê 3 = V

4.16. A steady velocity field is given by v = ( x ; - xix:)êl + (x: $2 + xz)ê2. Determine the unit relative velocity with respect to P(1,1,3) of the particles a t Q1(1,0,3), Q2(1,3/4,3), Q3(1, 7/8,3) and show that these values approach the relative velocity given by (4.26).

h h By direct calculation v , - v n , = -e1 + 2 e,, 4(vp - vQ2) = -7ê1/4 + 2 ê 2 and 8 ( v p - V Q ) = 3

-15$,/8 + 22,. The velocity gradient matrix is

3x1 - xp -2xix2


and a t P(l, 1,3) in the negative x2 direction,

Thus ( d v l d x ) ~ = -22, + 2ê2 which i s the value approached by the relative unit velocities v p - Vgi.

e2

4.17. For the steady velocity field v = 3x;x2êl + 2xix3ê2 + ~ 1 x 2 ~ : ê3, determine the rate of extension a t P(1,1,1) in the direction of $= (3êl- 4ê3)/5.

6xlx2 3%;

Here the velocity gradient is [8vi/8xi] = [ O 4x2x3 2:: ] and its symmetric part

x,x; xlx; 2x1x2x3

Thus from (4.34) for 2 = (3 ê1 - 4ê3)/5,

4.18. For the motion of Problem 4.17 determine the rate of shear a t P between the orthogonal directions 2 = (3êi - 4&)/5 and f i = (4&+ 3ê3)/5.

In analogy with the results of Problem 3.20 the shear rate Y,, is given by Y,, = c * 28 -2, or in matrix form

Y,, = [4/5, O , 3/51

4.19. A steady velocity field is given by v1 = 2x3, v2 = 2x3, 213 = O. Determine the principal directions and principal values (rates of extension) of the rate of deformation tensor for this motion.

0 0 2 O o o 1

Here avi/axiI = 0 = E : r] + [ o o ;] and for principal values

O 0 0 -1 -1

A o f D,,

Thus X 1 = + f i , A I I = O , X I , l = - f i

The transformation matrix to principal axis directions is

-112 -112 l / f i


with the rate of deformation matrix in principal form [D;] =

o -fi

4.20. Determine the maximum shear rate :,,,,, for the motion of Problem 4.19.

Analogous to principal shear strains of Chapter 3, the maximum shear rate is ?,,, = ( A 1 - A I I 1 ) / 2 = fi.

This result is also available by observing that the motion is a simple shearing parallel to the x l x 2 plane in the direction of the unit vector 2 = (êl +$,)I@. Thus, as before,

i.,,, = 7," = [0,0,11

I t is also worth noting that the maximum rate of extension for this motion occurs in the direction A

n = ( ê l + ê2 + f iê3)/2 as found in Problem 4.19. Thus Fig. 4-3

MATERIAL DERIVATIVES OF VOLUMES, AREAS, INTEGRALS, ETC. (Sec. 4.6-4.7) 4.21. Calculate the second material derivative of the scalar product of two line elements, i.e.

determine d2(dx2)ldt2. d(dxi) avi

From (4.451, 7 d(dx2) = - dxk; and i t is shown in (4.48) that - = 2Dii dxi dxj . Therefore 8 3 , d t

and by simple manipulation of the dummy indices,

4.22. Determine the material derivative pidSi of the flux of the vector property pi through the surface S. d t

BY (4.5%

4.23. Show that the transport theorem derived in Problem 4.22 may be written in symbolic notation as $A P * & ~ s = [$ + v(T7.p) + V x ( p x v ) ] . & d s

By a direct transcription into symbolic notation of the result in Problem 4.22,


Now use of the vector identity V X ( p X v) = p(V v) - v(V p) + (v V ) p - ( p V ) v (see Problem 1.65) gives it p - ~ d s = L ["z + v ( v o p ) + v x ( p x V ) ] ~ ~ n d s

4.24. Express Reynold's transport theorem as given by equations (4.53) and (4.54) in symbolic notation.

Let P*(x, t ) be any tensor function of the Eulerian coordinates and time. Then (4.53) is

.and by Gauss' divergence theorem this becomes (4.54),

4.25. If the function P*(x, t ) in Problem 4.24 is the scalar 1, the integral on the left is simply the instantaneous volume of a portion of the continuum. Determine the material derivative of this volume.

Using the vector form of (4.58) as given in Problem 4.24, $i dV = i V . v dV. Here

V v d V represents the rate of change of dV, and so V v is known as the cubical rate of dilatation. This relationship may also be established by a direct differentiation of (4.38). See Problem 4.43.


4.26. From the definition of the vorticity vector (4.29), q = curl v, show that qi = e i j k V k j and that 2Vii = ejik qk.

By (4.29), qi = ~ ~ ~ ~ v ~ , ~ = ~ ~ ~ ~ ( v ~ ~ , ~ ~ + and since = O (see, for example, Problem 1.50), qi = qjkV[k,jl = eiikVkj. From this result q,,qi = eirsYjkVkj = ( G r i S s k - S r k S s i ) V k j = 2Vsr.

av 4.27. Show that the acceleration a may be written a s a = - + q x v + +Vv2. dt

avi avi From (4.18), ai = - + v , - a t and so

axk

which, as the student should confirm, is the indicial form of the required equation.

4.28. Show that d(ln J)ldt = div v.

Let axi/aXp be written here as si,, so that J = eP~Rx1,P%2,Q%3,R and J becomes the sum of the three determinants, = f p Q R ( j S l , p ~ 2 , Q ~ 3 , R + x1,pjS2,Qx3,R + c ~ ~ , p x ~ , ~ j S ~ , ~ ) . NOW = vl,,x,,p, ek., and so J = E P Q R ( ~ ~ , ~ X ~ , P ~ Z , Q %3,R + ~ I . P ~ ~ , s ~ s , Q ~ . ~ , R + Z ~ , P X Z . Q ~ ~ , ~ ~ S , R ) . Of the nine 3 x 3 beterminants resulting from summation o? s in this expression, the three non-vanishing ones yield J = v,, J + v2,2 J + v3,3 J = v,,, J. Thus J = J V v and so d(ln J) ld t = div v.


4.29. Show that for steady motion (avilat = 0) of a continuum the streamlines and pathlines coincide.

As shown in Problem 4.7, a t a given instant t streamlines are the solutions of the differential equations dx l l v l = dx21v2 = dx31v3. Pathlines are solutions of the differential equations dxildt = vi(x , t). If vi = vi(x) , these equations become d t = dx,lv, = dxZ/v2 = dx3/v3 which coincide with the streamline differential equations.

4.30. For the steady velocity field v1 = x l x z + x i , v2 = -x; - ~ 1 x 2 , v3 = O determine expressions for the principal values of the rate of deformation tensor D a t an arbitrary point P(x1, X Z , x3).

Principal values d(i , are solutions of

2x,x2 - d -2: + x i , O

-x; + z; -2x1x2 - d O = O = -d[-4xix; + d2 - (4 - x;)2]

o o -d

Thus d,,, = O, do, = -(x: + x i ) , do, = (x ; + 2;). Note here that dI = (x: + x i ) , d I I = O, d I I I = -(x; + 22).

4.31. Prove equation (4.43) by taking the material derivative of d S i in its cross product form d S i = cijk d x j 2 ) d x k 3 ) .

axi axi a%, axk Using (S.&?), dSi = ~ ~ ~ ~ ( a x ~ / a X ~ ) dXz(axklaX3) dX3 and - dSi = qjk ax, d X 2 dX3 = axi ax ax. 8 x 1

J d X z dX3. Thus 2 dSi = 6, dSi = d S p = - J d X z d X 3 and by Problem 4.28, axP ax, a x ~

4.32. Use the results of Problems 4.27 and 4.23 to show that the material rate of change

of the vorticity flux - q * 6 dS equals the flux of the curl of the acceleration a. d"t L

Taking the curl of the acceleration as given in Problem 4.27,

V x a = V x + V x ( q x v) + V x ~ ( ~ 2 1 . 2 ) at

or V x a = aqlat + V x ( q x v ) = dqldt + q ( V v ) - ( q V ) v

since q = V X v and V X V(v212) = O. Thus if q is substituted for p in Problem 4.23,

$L , * n d ~ = 1 [$ + q ( V * v ) - ( q * V ) v ] . R d S = L ( V x a ) * R d S


4.33. For the vorticity qi show that a qi d V = [€,,ak + qjvi - qivi] d s j . a t

From Problem 4.32 the identity V X a = aqlat + V X (q X v) may be written in indicial form as aqi/at = eijkak, - E ~ ~ ~ ( E ~ ~ ~ ~ ~ v , . ) , ~ . Thus

i $ d~ = S [Eijkak,i - ( . ,srn~rn~r),sl dv v

and by the divergence theorem of Gauss (1.157),

Supplementary Problems A continuum motion is given by x1 = Xlet + X3(et- 1 ) , x2 = X 2 + X,(et - e-t), x3 = X3. Show that J does not vanish for this motion and obtain the velocity components. Ans. v , = ( X , + X3)et, v2 = X,(et + e-t), v3 = O or v , = x1 - x3, v2 = x3(et -i e-t), v3 = O

A velocity field is specified in Lagrangian form by v , = -X2e-t, v , = -X3, v3 = 2t. Determine the acceleration components in Eulerian form. Ans. a , = e-t(x2 + t x3 - e), a, = O, a3 = 2

Show that the velocity field vi = eijkbjxk + ci where b, and ci are constant vectors, represents a rigid body rotation and determine the vorticity vector for this motion.

Ans. qi = bixLi - b, = 2bi

Show that for the flow vi = xi l ( l + t ) the streamlines and path lines coincide.

The electrical field strength in a region containing a fluid flow is given by h = ( A cos 3t) lr where r2 = x ; + x i and A is a constant. The velocity field of the fluid is v , = X S X , + 22, v , = -x; - xixi, v , = O. Determine dXldt a t P(x,, x,, x,). Ans. dhldt = (-3A sin 3t) lr

Show that for the velocity field v , = x l x 2 + 22, v , = -x; - x i x i , v3 = O the streamlines are circular.

i

For the continuum motion x , = X,, x , = et(X2 + X3)/2 + e- t (X2 - X,)/2, x3 = et(X2 + X,)/2 - c ~ - ~ ( X , - X3)12, show that Dij = deijldt a t t = O. Compare these tensors a t t = 0.5.

For the velocity field v , = s; x, + si, v , = - (x t + x,x;), v3 = 0, determine the principal axes and principal values of ü a t P ( l , 2,3).

5 0 o 31- 11- o Am. D: = (: o); a = ( O

o -5 1 3 o

For the velocity field of Problem 4.41 determine the rate of extension in the direction A - V - (3, - 2 3 , + 2ê3) /3 a t P ( l , 2,3). What is the maximum shear rate a t P?

h

Ans. d"' = -2419, Ymax = 5

Show that d(axilaXj)ldt = v~,,x, ,~ and use this to derive (4.41) of the text directly from (4.38).

Prove the identity ~pq,(v,vr,,),q = qPvq,, +-vqqP,, - qqvP,, where vi is the velocity and qi the vorticity. Also show that v,, jvi,i = Dij Dij - qiqi/2.

Prove that the material derivative of the total vorticity is given by

Chapter 5

Fundamental Laws of Continuum Mechanics

5.1 CONSERVATION OF MASS. CONTINUITY EQUATION

Associated with every material continuum there is the property known as mass. The amount of mass in that portion of the continuum occupying the spatial volume V a t time t is given by the integral

(5.1)

in which p(x, t ) is a continuous function of the coordinates called the m a s density. The law of conservation of mass requires that the mass of a specific portion of the continuum remain constant, and hence that the material derivative of (5.1) be zero. Therefore from (4.52) with P:. . .(x, t ) = p(x, t ) , the rate of change of m in (5.1) is

Since this equation holds for an arbitrary volume V , the integrand must vanish, or

This equation is called the continuity equation; using the material derivative operator it may be put into the alternative form

For an incompressible continuum the mass density of each particle is independent of time, so that dpldt = O and (5.3) yields the result

v,,, = O or divv = O (5.5)

The velocity field v (x , t ) of an incompressible continuum can therefore be expressed by the equation

vi = c ~ ~ ~ s ~ , ~ or V = V x s (5.6)

in which s(x, t ) is called the vector potential of v.

The continuity equation may also be expressed in the Lagrangian, or material form. The conservation of mass requires that

where the integrals are taken over the same particles, i.e. V is the volume now occupied by the material which occupied V0 a t time t = O . Using (4.1) and (4.38), the right hand integral in (5.7) may be converted so that

CHAP. 51 FUNDAMENTAL LAWS OF CONTINUUM MECHANICS 127

Since this relationship must hold for any volume VO, it follows that

Po = PJ (5.9)

which implies that the product pJ is independent of time since V is arbitrary, or that

d ~ ( P J ) = 0 (5.10)

Equation (5.10) is the Lagrangian differential form of the continuity equation.

5.2 LINEAR MOMENTUM PRINCIPLE. EQUATIONS OF MOTION. EQUILIBRIUM EQUATIONS

A moving continuum which occupies the volume V a t time t is shown in Fig. 5-1. Body forces bi per unit mass are given. On the differential element dS of the bounding surface, the stress vector is tin'. The velocity field Vi = duildt is prescribed throughout the region occupied by the continuum. For this situation, the total linear momentum of the mass system within V is given by

Fig. 5-1

Based upon Newton's second law, the principle o f linear momentum states that the time rate of change of an arbitrary portion, of a continuum is equal to the resultant force acting upon the considered portion. Therefore if the interna1 forces between particles of the continuum in Fig. 5-1 obey Newton's third law of action and reaction, the momentum principle for this mass system is expressed by

A

Upon substituting tin' = ajini into the first integral of (5.12) and converting the resulting surface integral by the divergence theorem of Gauss, (5.12) becomes

In calculating the material derivative in (5.13), the continuity equation in the form given by (5.10) may be used. Thus

Replacing the right hand side of (5.13) by the right hand side of (5.14) and collecting terms results in the linear momentum principle in integral form,

128 FUNDAMENTAL LAWS OF CONTINUUM MECHANICS [CHAP. 5

Since the volume V is arbitrary, the integrand of (5.15) must vanish. The resulting equations,

are known as the equations of motion.

The important case of static equilibrium, in which the acceleration components vanish, is given a t once from (5.16) as

u ~ ~ , ~ + pbi = O or V, 1 + pb = O (5.1 7')

These are the equilibrium equations, used extensively in solid mechanics.

5.3 MOMENT OF MOMENTUM (ANGULAR MOMENTUM) PRINCIPLE

The moment of momentum is, as the name implies, simply the moment of linear momentum with respect to some point. Thus for the continuum shown in Fig. 5-1, the total moment of momentum or angular momentum as it is often called, with respect to the origin, is

~ ( t ) = J ~,,x,pv, d~ or v

N = ( X X ~ V ) d v (5.1 8)

in which xj is the position vector of the volume element dV. The moment of momentum principle states that the time rate of change of the angular momentum of any portion of a continuum with respect to an arbitrary point is equal to the resultant moment (with respect to that point) of the body and surface forces acting on the considered portion of the continuum. Accordingly, for the continuum of Fig. 5-1, the moment of momentum principle is expressed in integral form by

Equation (5.19) is valid for those continua in which the forces between particles are equal, opposite and collinear, and in which distributed moments are absent.

The moment of momentum principle does not furnish any new differential equation of h

motion. If the substitution tp ' = uPknp is made in (5.19), and the symmetry of the stress tensor assumed, the equation is satisfied identically by using the relationship given in (5.16). If stress symmetry is not assumed, such symmetry may be shown to follow directly from

h

(5.19), which upon substitution of t p ' = upknp, reduces to

Since the volume V is arbitrary, ... ,,, .V. ,, = O or 1, = O

which by expansion demonstrates that uik = ukj.

5.4 CONSERVATION OF ENERGY. FIRST LAW OF THERMODYNAMICS. ENERGY EQUATION

If mechanical quantities only are considered, the principle of conservation of energy for the continuum of Fig. 5-1 may be derived directly from the equation of motion given by (5.16). To accomplish this, the scalar product between (5.16) and the velocity vi is first computed, and the result integrated over the volume V. Thus


But vivi L pv,;,dV = i cLdV = dK d t

which represents the time rate of change of the kinetic energy K in the continuum. Also, v i ~ j i , j = ( v , ~ ~ , ) , ~ - vi,j uji and by (4.19) = Dij + V,,, so that (5.22) may be written

since Vijuj i O. Finally, converting the first integral on the right hand side of (5.24) to a surface integral by the divergence theorem of Gauss, and making use of the identity ti"' = ~ . . n . , the energy equation for a continuum appears in the form

31 3

This equation relates the time rate of change of total mechanical energy of the continuum on the left side to the rate of work done by the surface and body forces on the right hand side of the equation. The integral on the left side is known as the time rate of change of internal mechanical energy, and written dUldt . Therefore (5.25) may be written briefly as

where dWld t represents the rate of work, and the special symbol d is used to indicate that this quantity is not an exact differential.

If both mechanical and non-mechanical energies are to be considered, the principle of conservation of energy in its most general form must be used. In this form the conservation principle states that the t ime rate o f change o f t he kinetic plus the internal energy i s equal t o t he s u m o f the rate o f work plus a11 other energies supplied to, or removed f r o m the continuum per uni t t ime. Such energies supplied may include thermal energy, chemical energy, or electromagnetic energy. In the following, only mechanical and thermal energies are considered, and the energy principle takes on the form of the well-known first law o f thermodynamics.

For a thermomechanical continuum it is customary to express the time rate of change of internal energy by the integral expression

where u is called the specijic internal energy. (The symbol u for specific energy is so well established in the literature that it is used in the energy equations of this chapter since there appears to be only a negligible chance that it will be mistaken in what follows for the magnitude of the displacement vector ui.) Also, if the vector s is defined as the heat fEux per unit area per unit time by conduction, and z is taken as the radiant heat constant per unit mass per unit time, the rate of increase of total heat into the continuum is given by

Therefore the energy principle for a thermomechanical continuum is given by


or, in terms of the energy integrals, as

Converting the surface integrals in (5.30) to volume integrals by the divergence theorem of Gauss, and again using the fact that V is arbitrary, leads to the local form of the energy equation:

or

Within the arbitrarily small volume element for which the local energy equation (5.31) is valid, the balance of momentum given by (5.16) must also hold. Therefore by taking the scalar product between (5.16) and the velocity p'Uivi = vioji,j + pvibi and, after some simple manipulations, subtracting this product from (5.31),, the result is the reduced, but h i g h l ~ useful form of the local energy equation,

This equation expresses the rate of change of internal energy as the sum of the stress power plus the heat added to the continuum.

5.5 EQUATIONS OF STATE. ENTROPY. SECOND LAW OF THERMODYNAMICS

The complete characterization of a thermodynamic system (here, a continuum) is said to describe the state of the system. This description is specified, in general, by severa1 thermodynamic and kinematic quantities called state variables. A change with time of the state variables characterizes a thermodynamic process. The state variables used to describe a given system are usually not a11 independent. Functional relationships exist among the state variables and these relationships are expressed by the so-called equations of state. Any state variable which may be expressed as a single-valued function of a set of other state variables is known as a state function.

As presented in the previous section, the first law of thermodynamics postulates the interconvertibility of mechanical and thermal energy. The relationship expressing conversion of heat and work into kinetic and internal energies during a thermodynamic process is set forth in the energy equation. The first law, however, leaves unanswered the question of the extent to which the conversion process is reversible or irreversible. A11 real processes are irreversible, but the reversible process is a very useful hypothesis since energy dissipation may be assumed negligible in many situations. The basic criterion for irreversibility is given by the second law of thermodynamics through its statement on the limitations of entropy production.

The second law of thermodynamics postulates the existence of two distinct state functions; T the absolute temperature, and S the entropy, with certain following properties. T is a positive quantity which is a function of the empirical temperature O, only. The entropy is an extensive property, i.e. the total entropy in the system is the sum of the entropies of its parts. In continuum mechanics the specific entropy (per unit mass), or

entropy density is denoted by s, so that the total entropy L is given by L = 1 ps dV. The - .

entropy of a system can change either by interactions that occur with the surroundings, or by changes that take place within the system. Thus

ds = + ds(i) (5.33)


where d s is the increase in specific entropy, dsce) is the increase due to interaction with the exterior, and ds'" is the internal increase. The change dsCi) is never negative. I t is zero for a reversible process, and positive for an irreversible process. Therefore

d d i ) > O (irreversible process) (5.34)

ds( i ) = O (reversible process) (5.35)

In a reversible process, if d q ( ~ ) denotes the heat supplied per unit mass to the system, the change dsCe) is given by

dS(e) = - dq(Ri (reversible process) T (5.36)

5.6 THE CLAUSIUS-DUHEM INEQUALITY. DISSIPATION FUNCTION

According to the second law, the time rate of change of total entropy L in a continuum occupying a volume V is never less than the sum of the entropy influx through the continuum surface plus the entropy produced internally 'by body sources. Mathematically, this entropy principle may be expressed in integral form as the Clausius-Duhem inequality,

where e is the local entropy source per unit mass. The equality in (5.37) holds for reversible processes; the inequality applies to irreversible processes.

The Clausius-Duhem inequality is valid for arbitrary choice of volume V so that trans- forming the surface integral in (5.37) by the divergence theorem of Gauss, the local form of the in ternal entropy production rate y, per unit mass, is given by

This inequality must be satisfied for every process and for any assignment of state variables. For this reason it plays an important role in imposing restrictions upon the so-called constitutive equations discussed in the following section.

In much of continuum mechanics, it is often assumed (based upon statistical mechanics of irreversible processes) that the stress tensor may be split into two parts according to the scheme, a,. = a!?) + u ' D '

i j (5.39)

where u::) is a conservative stress tensor, and is a dissipative stress tensor. With this assumption the energy equation (5.32) may be written with the use of (4.25) as

1 In this equation, - o;?) iij is the rate of energy dissipated per unit mass by the stress, and

0

dqld t is the rate Af heat influx per unit mass into the continuum. If the continuum undergoes a reversible process, there will be no energy dissipation, and furthermore, dqldt = d q ( ~ ) l d t , so that (5.40) and (5.36) may be combined to yield

Therefore in the irreversible process described by (5.40), the entropy production rate may be expressed by inserting (5.41). Thus


The scalar ;ij is called the dissipation function. For an irreversible, adiabatic process (dq = O), dsldt > O by the second law, so from (5.42) it follows that the dissipation function is positive definite, since both p and T are always positive.

5.7 CONSTITUTIVE EQUATIONS. THERMOMECHANICAL AND MECHANICAL CONTINUA

In the preceding sections of this chapter, severa1 equations have been developed that must hold for every process or motion that a continuum may undergo. For a thermomechanical continuum in which the mechanical and thermal phenomena are coupled, the basic equations are

(a) the equation of continuity, (5.4)

(b) the equation of motion, (5.16)

(c) the energy equation, (5.32)

Assuming that body forces bi and the distributed heat sources z are prescribed, (5.43), (5.44) and (5.45) consist of five independent equations involving fourteen unknown functions of time and position. The unknowns are the density p, the three velocity components vi, (or, alternatively, the displacement components ui), the six independent stress components oij, the three components of the heat flux vector ci, and the specific interna1 energy u. In addition, the Clausius-Duhem inequality (5.38)

which governs entropy production, must hold. This introduces two additional unknowns: the entropy density s, and T, the absolute temperature. Therefore eleven additional equations must be supplied to make the system determinate. Of these, six will be in the form known as constitutive equations, which characterize the particular physical properties of the continuum under study. Of the remaining five, three will be in the form of temperature- heat conduction relations, and two will appear as thermodynamic equations of state; for example, perhaps as the caloric equation of state and the entropic equation of state. Specific formulation of the thermomechanical continuum problem is given in a subsequent chapter.

I t should be pointed out that the function of the constitutive equations is to establish a mathematical relationship among the statical, kinematical and thermal variables, which will describe the behavior of the material when subjected to applied mechanical or thermal forces. Since real materials respond in an extremely complicated fashion under various loadings, constitutive equations do not attempt to encompass a11 the observed phenomena related to a particular material, but, rather, to define certain ideal materials, such as the ideal elastic solid or the ideal viscous fluid. Such idealizations or material models as they are sometimes called, are very useful in that they portray reasonably well over a definite range of loads and temperatures the behavior of real substances.


In many situations the interaction of mechanical and thermal processes may be neglected. The resulting analysis is known as the uncoupled thermoelastic theory of continua. Under this assumption the purely mechanical processes are governed by (5.43) and (5.44) since the energy equation (5.45) for this case is essentially a first integral of the equation of motion. The system of equations formed by (5.43) and (5.44) consists of four equations involving ten unknowns. Six constitutive equations are required to make the system determinate. In the uncoupled theory, the constitutive equations contain only the statical (stresses) and kinematic (velocities, displacements, strains) variables and are often referred to as stress-strain relations. Also, in the uncoupled theory, the temperature field is usually regarded as known, or a t most, the heat-conduction problem must be solved separately and independently from the mechanical problem. . In isothermal problems the temperature is assumed uniform and the problem is purely mechanical.

Solved Problems

CONTINUITY EQUATION (Sec. 5.1)

5.1.' An irrotational motion of a continuum is described in Chapter 4 as one for which the vorticity vanishes identically. Determine the form of the continuity equation for such motions.

B y (1.29), curl v = O when q - 0, and so v becomes the gradient of a scalar field +(xi, t ) (see Problem 1.50). Thus vi = +,i and (5.3) is now dpldt + p+,kk = O or dpldt + pV2+ = 0 .

5.2. If P:?, .(x, t) represents any scalar, vector or tensor property per unit mass of a continuum so that P:. . . (x, t) = pp". . (x, t) show that

since by (5 J), dpldt + pvk, = 0 .

5.3. Show that the material form d(pJ)ldt = O of the continuity equation and the spatial form dpldt + pv , , , = O are equivalent.

Differentiating, d(pJ ) ld t = (dp ld t )J + p d J / d t = O and from Problem 4.28, dJ ld t = J v k , , so that d(pJ ) ld t = J(dp ld t + P V , , ~ ) = 0 .

5.4. Show that the velocity field vi = Axil?, where xixi = r2 and A is an arbitrary constant, satisfies the continuity equation for an incompressible flow.

From (5.5) v k , k = O for incompressible flow. Here

and so u k S k = (3 - 3) /$ = O to satisfy the continuity equation.


5.5. For the velocity field vi = x i l ( l + t), show that pxlx,x3 = p,,X1X2X3.

Here vk,k = 3/(1 + t ) and integrating (5.3) yields ln p = -1n (1 + t)3 + ln C where C is a constant of integration. Since p = po when t = 0, this equation becomes p = pd(l + t)3. Next by integrating the velocity field dxi/xi = d t / ( l + t ) (no sum on 9, xi = X i / ( l + t ) and hence Pxlx2x3 = ~ 0 ~ 1 ~ 2 ~ 3 .

LINEAR AND ANGULAR MOMENTUM. EQUATIONS OF MOTION (Sec. 5.2-5.3) A

5.6. Show by a direct expansion of each side that the identity eijkujke, = 1, used in (5.20) and (5.21) is valid.

By (1.15) and (2.8),

L, = a l S l x ê1 + olGl x $2 + a l g l x ê3 + . . + u33ê3 x $3

- - (023 - 032) ê1 + ('731 - 013) $2 + ( ~ 1 2 - ~ 2 1 ) 23

Also, expanding eijkajk gives identical results, (az3 - a32) for i = 1, (031 - ul3) for i = 2, (al2 - oZ1) for i = 3.

5.7. If distributed body moments m i per unit volume act throughout a continuum, show that the equations of motion (5.16) remain valid but the stress tensor can no longer be assumed symmetric.

Since (5.16) is derived on the basis of force equilibrium, i t is not affected. Now, however, (5.19) acquires an additional term so that

which reduces to (see Problem 2.9) (Yjkajk +mil dV = O, and because V i s arbitrary, eijkajk + mi = O for this case.

5.8. The momentum principle in differential form (the so-called local or "in the small" form) is expressed by the equation d ( p v , ) l d t = pbi + (ui j - ~ V ~ V ~ ) , ~ . Show that the equation of motion (5.16) follows from this equation.

Carrying out the indicated differentiation and rearranging the terms in the resulting equation yields

~ , ( a ~ / a t + p, jv j + pvj, j) + p(avi/Jt + vivi, j ) pbi aij, j

The first term on the left is zero by (5.4) and the second term is pai. Thus pai = pbi + aij,j which is (5.16).

5.9. Show that (5.19) reduces to (5.20). h

Substituting oPknp for tp' in (5.19) and applying the divergence theorem (1.157) to the resulting surface integral gives

Using the results of Problem 5.2, the indicated differentiations here lead to

The term in parentheses is zero by (5.16), also zj,, = S j p and cijkvjvk = O, so that finally s, €i jkUjk .v = o.


5.10. For a rigid body rotation about a point, vi = cijkojxk. Show that for this velocity (5.19) reduces to the well-known momentum principle of rigid body dynamics.

The left hand side of (5.19) is the total moment Mi of a11 surface and body forces relative to the origin. Thus for vi = ei jkwjXk,

d = [O. J" ~ ( 4 P q x q -

where Zip = p(8ipzqzq - 44 dV is the moment of inertia tensor.

ENERGY. ENTROPY. DISSIPATION FUNCTION (Sec. 5.4-5.6)

5.11. Show that for a rigid body rotation with vi = C~,W~X,, the kinetic energy integral o£ (5.23) reduces to the familiar form given in rigid body dynamics.

From (5.2?)),

K = i p Y d v = i peiikUj%k€ipq~pZq d v

= i i ~ ~ p w j ( 8 j p ~ k ~ - 6iqSkp)xkzq dV

- - w i w ~ wiwpzip 2- i p(6ipzqzq - zPz$ dV - 2

o. l .0 , In symbolic notation note that K = -

2

5.12. At a certain point in a continuum the rate of deformation and stress tensors are given by

and uii = - -: i 4 2 5

Determine the value h of the stress power Dijuij a t the point. Multiplying each element of Dij by its counterpart jn oij and adding, h = 4 + O - 4 + O - 6 +

14 - 4 + 14 + 40 = 58.

5.13. If uij = -pSij where p is a positive constant, show that the stress power may be p dp expressed by the equation D i j u i j = - - . P dt

By (4.19), Dij = viBj - Vij; and since Vi ju i j = O, i t follows that Dijoij = = -pvi,i. From the continuity equation (5.3), vi,i = -(llp)(dpldt) and so Dijq j = (plp)(dpldt) for oij = -paij.

5.14. Determine the form of the energy equation if uij = (-p + h* Dkk)8,, 4- 2 , ~ * D,, and the heat conduction obeys the Fourier law ci = -kTai.

From (5.92), du - (-p+ X*Dkk)SijDij + 2p*DijDij + kTSii + z

p d t - - P do - -- + (h* + 2 / ~ * ) ( 1 , ) ~ - 4/~*11, + kT,ii + z

P dt

where I, and 11, are the first and second invariants respectively of the rate of deformation tensor.


5.15. If uij = - p s i j , determine an equation for the rate of change of specific entropy during a reversible thermodynamic process.

ds du p d p Here oij = o::' and (5.41) gives T - = - + -- upon use of the result in Problem 5.13. d t dt p2 d t

5.16. For the stress having = /3DikDkj, determine the dissipation function in terms of the invariants of the rate of deformation tensor D.

Here by (4.25), o:;) iij = PDsDkjDij which is the trace of D~ (see page 16) and may be evaluated by using the principal axis values D( , , , D (2 ) , DC3). Thus by (1.138) the trace

DiiD,Dki = D:,) i- D:Z) + ~ ( 3 )

= ( D ( l ) + + D ( 3 ) ) 3 - 3 ( D ( 1 ) + D ( 2 ) + D ( 3 ) ) ( D ( 1 ) D ( 2 ) + D ( 2 ) D ( 3 > + D ( 3 ) D ( 1 ) )

+ 3D,l,D(2,D,3)

Therefore o:;' g j = p[1: - 31DIID + 3111D].

CONSTITUTIVE EQUATIONS (Sec. 5.7)

5.17. For the constitutive equations uij = Ki,,DPq show that because of the symmetry of the stress and rate of deformation tensors the fourth order tensor Kijp, has a t most 36 distinct components. Display the components in a 6 x 6 array.

Since oij = uji, KiTpq = Kjipq; and since Dij = Dji, KijPq = Kijqp. If Kijpq is considered as the outer product of two symmetric tensors AijBpq = Kijpq, i t is clear that since both Aij and Bij have six independent components, Kijpq will have a t most 36 distinct components.

The usual arrangement followed in displaying the components of Kib, is

5.18. If the continuum having the constitutive relations uij = KijpqDpq of Problem 5.17 is assumed isotropic so that Kijpq has the same array of components in any rectangular Cartesian system of axes, show that by a cyclic labeling of the coordinate axes the 36 components may be reduced to 26.

The coordinate directions may be labeled in six different ways as shown in Fig. 5-2. Isotropy of Kijpq then requires x1 or x2 that K,,, , = Kl13, = KZZ3, = K2,,, = K3,,, = K3322 and that K l q I 2 = 1(1313 = K2323 = K2121 = K3131 = K3232 which reduces the 36 components to 26. By suitable reflections and rotations 53 of the coordinate axes these 26 components may be reduced to 2 for the case of isotropy. Fig. 5-2

5.19. For isotropy Kijpq may be represented by Kijpq = A*8ij8pq + , ~ * ( 8 ~ ~ 8 ~ , + 8iq8jp). Use this to develop the constitutive equation uij = KijpqDpq in terms of A* and P* .


5.20. Show that the constitutive equation of Problem 5.19 may be split into the equivalent equations aii = (3A* + 2p*)Dii and sij = 2p*DG where sij and DC are the deviator tensors of stress and rate of deformation, respectively.

Substituting U i j = Sij + Sijukk/3 and Dij = D;j + SijDkk/3 int0 uij = h*SijDkk + 2@*Dij 0f Problem 5.19 resuits in the equation sij + aijukk/3 = h*SijDkk + 28*(Dii + SijDkk/3). From this when i f j, sij = 2@*D& and hence ukk = ('&h* + 2@*)Dkk.


581. Show that = ( ~ , , a ~ , ~ + qjvi,j)lp where p is the density, ai the acceleration and

qi the vorticity vector.

dd, (f) = 4 - But 4 = rijkak,i + qivi,i - qivjj (see Problem By direct differentiation - P p 2 '

4.32); and by the continuity equation (5.4, e = -pvci. Thus

5.22. A two dimensional incompressible flow is given by v1 = A(xS - xi)l.P, v2 = A ( 2 x l x z ) l l A , v3 = 0, where r2 = X: + x;. Show that the continuity equation is satisfied by this motion.

By (5.5), vi,i = O for incompressible flow. Here v l , , = A[-4x1(xS - xi)/18 + 2x1/r<l] and v e j 2 = A [2x1/1A - 8x1xi/r8]. Adding, v,, , + v,, , = 0.

5.24. In a two dimensional incompressible steady flow, v1 = -Ax21r2 where r2 = x: + xi . Determine v2 if v2 = O a t xl = O for a11 x2. Show that the motion is irrotational and that the streamlines are circles.

5.23. Show that the flow of Problem 5.22 is irrotational.

By (4.29), curl v = 0 for irrotational flow. Thus

From (5.5), viBi = O or v,,, = -v2,, = 2Axlx2/.P for this incompressible flow. Integrating with respect to x2 and imposing the given conditions on v 2 yields v2 = Axllr2.

curlv =

For an irrotational motion, curl v = O. Here

curl v = A [(x: - xe)/ la + (-E: + x i )I+] ê3 = 0

A A A

e1 e2 e3

alasl ala%, alax,

A ( X ; - xi)/1.4 2Ax1x2/r<l O

From Problem 4.7, page 118, the equations for streamlines are dxl lvl = dx21vz. Here these

equations are xl dxl + x2 dx2 = O which integrate directly into the circles x; + x i = constant.

= A[2x2/r<l - 8 x ~ x 2 / r ~ + 2x2/r<l + 4x2(x: - xg)/r6] ê3 = O

5.25. For a continuum whose constitutive equations are a, = (-p + A* Dkk)Sij + 2p*Dij, determine the equations of motion in terms of the velocity vi.

From (5.16), p$i = pbi + uijSj or here p6i = pbi - p,jSij + X*D,,,~S, + 28*DilBj. By definition, 2Dij = v i s j t vjBi so that Dkk = v k j k and 2Diisj = vi,jj + viaij. Therefore

FUNDAMENTAL LAWS OF CONTINUUM MECHANICS [CHAP. 5

pCi = pbi - p,i + (X* + p * ) ~ j , ~ j + p*vi,jj

In symbolic notation this equation is

p; = p b - V p + (A* + p*)V(V V ) + p*V2v

5.26. If the continuum of Problem 5.25 is considered incompressible, show that the divergente of the vorticity vanisbes and give the form of the equations of motion for this case.

By (4.29), qi = ei jkvk , and div q = eijkvk, ji = O since eijk is antisymmetric and vk, ji is symmetric in i and j . Thus for V v = O the equations of motion become p'Ui = pbi - p,i + p*vi,jj or in Gibbs notation p; = p b - V p + p*VZv.

5.27. Determine the material rate of change of the kinetic energy of the continuum which occupies the volume V and give the meaning of the resulting integrals. ,.

By (5.23). dK/d t = J pvi'UidV. Also the total stress power of the surface forces is V . L viti'' d S which may be written viuijni d S and by the divergence theorem (1.157) and the

equations of motion (5.16) expressed as the volume integrals r

( p(v$i - bivi) dV . Thus -

This sum of integrals represents the rate of work done by the body forces, the interna1 stresses and the surface stresses, respectively.

5.28. A continuum for which = h*Dk,8,, + 2p*Dij undergoes an incompressible irrotational flow with a velocity potential + such that v = grad +. Determine the dissipation function ;ij.

Here o::) aij = u : ~ ) D ~ ~ = (?,*Dkksij + 2p*Dij)Dij = 2p*DijDij since Dkk = v k , = O for incompres-

sible flow. Also since vi = +,i, the scalar DijDij = +,ij+,ij and so u : y 3 D ' ~ i j = 2p*+,ij+,ij.

Because the motion is incompressible and irrotational, +,li = O and 2+,u+,ij = (+,i+,i),jj = V2(V+)2. I t is also interesting to note that

which for +,ii = O reduces to 4+,ij+,ij. Thus +i:' gi j = p * ~ z ( v + ) z = p*v4(+z)/2.

5.29. For a continuum with uij = -p8. . the specific e n t h a l p y h = u + p l p . Show that the 23'.

energy equation may be written h = +Ip + Ti using this definition for the enthalpy.

From (5.41), = -pSijDijlp .+ T S for the given stress; and by the result of Problem 5.13 and

the definition of h, ii = a - +Ip - p;/pZ = -p;/pZ + T B . Canceling and rearranging, i = ;IP + TS.

5.30. If the continuum of Problem 5.25 undergoes an incompressible flow, determine the equation of motion in terms of the vorticity q in the absence of body forces and assuming constant density.

For incompressible flow, V v = 0 ; and if b 0, the equation of motion in Problem 5.25 reduces to pCi = -pZi + P*vi , j j . Taking the cross product V X with this equation for p = constant gives c ~ ~ ~ B ~ , ~ = - ~ ~ ~ ~ p , ~ ~ l ~ + (P* /P)~pq iv i , j jq . But ~ ~ ~ ~ p , ~ ~ = O and by (4.29) the result is 4, = ( ~ * / p ) q , ~ ~ . In symbolic notation, dqldt = (p*lp)V2q.


Supplementary Problems

5.31. Show that for the rate of rotation vector a, P

5.32. Show that the flow represented by v , = - ~ X , X ~ ~ / T ~ , v , = (x: - x~)x, l+, v , = %,/r2 where r2 = x; + xg, satisfies conditions for an incompressible flow. 1s this motion irrotational?

5.33. In terms of Cartesian coordinates x , y, z the continuity equation is

aplat + a ( p ~ , ) l a ~ + a ( p ~ , ) l a ~ + a ( ~ v , ) l a ~ = O

Show that in terms of cylindrical coordinates r , 6 , z this equation becomes

r(aplat) + d(rpv,)lar + a(pve)la6 + r(d(pv,)ldz) = O

5.34. Show that the flow v, = (1 - rz) cos elrz, ve = (1 + r,) sin #/r,, v , = O satisfies the continuity equation in cylindrical coordinates when the density p is a constant.

5.35. I f P i j , . , (x , t ) is an arbitrary scalar, vector or tensor function, show that

5.36. I f a continuum is subjected to a body momeht per unit mass h in addition to body force b, and a

couple stress g'"^ in addition to the stress t'"', the angular momentum balance may be written

where m is distributed angular momentum per unit mass. If G = g'L', show that the local form of this relation is p dmldt = h + V G + Z,.

5.37. If a continuum has the constitutive equation oii = -pGij + PDij + aDikDkj, show that uii = 3(-p - 2~~11~13) . Assume incompressibility, Dii = 0.

5.38. For a continuum having uij = -pSij, show that du = T ds - p dv where in this problem v = l I p , the specific volume.

5.39. I f T dsldt = -vi,ilp and the specific free energy is defined by * = u - Ts , show that the energy equation may be written p dqldt + ps dTldt = oiiDij.

5.40. For a thermomechanical continuum having the constitutive equation

q j = hekkSij f 2peii - (3h f zp)a Gij(T - To)

where To is a reference temperature, show that ekk = 3a(T - To) when oij = sij = oij - ukkSij/3.

Linear Elasticity

6.1 GENERALIZED HOOKE'S LAW. STRAIN ENERGY FUNCTION

In classical linear elasticity theory it is assumed that displacements and displacement gradients are sufficiently small that no distinction need be made between the Lagrangian and Eulerian descriptions. Accordingly in terms of the displacement vector ui, the linear strain tensor is given by the equivalent expressions

L = E = ~ ( u v , + Vxu) =. &(uVx + Vxu) = 4(uV + Vu)

In the following it is further assumed that the deformation processes are adiabati; (no heat loss or gain) and isothermal (constant temperature) unless specifically stated otherwise.

The constitutive equations for a linear elastic solid relate the stress and strain tensors through the expression -

uii = Cijkmekm or I: = C : E (6-2)

which is known as the generalized Hooke's law. In (6.2) the tensor of elastic constants Cijkm has 81 components. However, due to the symmetry of both the stress and strain tensors, there are a t most 36 distinct elastic constants. For the purpose of writing Hooke's law in terms of these 36 components, the double indexed system of stress and strain components is often replaced by a single indexed system having a range of 6. Thus in the notation - - -

'=H - '=I u23 - u32 - O4

and

- €33 - €3 2c12 = 2~~~ = E6

Hooke's law may be written -

uK - CKMcM ( K , M = 1 ,2 ,3 ,4 ,5 ,6 )

where CKM represents the 36 elastic constants, and where upper case Latin subscripts are used to emphasize the range of 6 on these indices.

When thermal effects are neglected, the energy balance equation (5.32) may be written

CHAP. 61 LINEAR ELASTICITY 141

The interna1 energy in this case is purely mechanical and is called the strain energy (per unit mass). From (6.6),

1 du = - u..d,,,

P 23 (6.7)

and if u is considered a function of the nine strain components, u = u(c,,), its differential is given by

au du = -de.. a,,, '3

Comparing (6.7) and (6.8), it is observed that

The strain energy density u* (per unit volume) is defined as

u* = t'u

and since p may be considered a constant in the small strain theory, u* has the property that

Furthermore, the zero state of strain energy may be chosen arbitrarily; and since the stress must vanish with the strains, the simplest form of strain energy function that leads to a linear stress-strain relation is the quadratic form

U* = -&Cijkmcijckm (6.12)

From (6.2), this equation may be written

U* = +uijeij or U* = -&Z : E (6.18)

In the single indexed system of symbols, (6.12) becomes

U* = -&CKMcKeM (6.14)

in which CKM = CMK. Because of this symmetry on CKM, the number of independent elastic constants is a t most 21 if a strain energy function exists.

6 2 ISOTROPY. ANISOTROPY. ELASTIC SYMMETRY If the elastic properties are independent of the reference system used to describe it, a

material is said to be elasticauy isotropic. A material that is not isotropic is called anko- tropic. Since the elastic properties of a Hookean solid are expressed through the coefficients CKM, a general anisotropic body will have an elastic-constant matrix of the form

When a strain energy function exists for the body, CKM = CMK, and the 36 constants in (6.15) are reduced to 21.

142 LINEAR ELASTICITY [CHAP. 6

A plane of elastic symmetry exists a t a point where the elastic constants have the same values for every pair of coordinate systems which are the reflected images of one another with respect to the plane. The axes of such coordinate systems are referred to as "equivalent elastic directions." If the ~ 1 x 2 plane is one of elastic symmetry, the constants CKM are invariant under the coordinate transformation

X ; = XI, xá = ~ 2 , ~3 = -x3 (6.16)

as shown in Fig. 6-1. The transformation matrix of (6.16) is given by Fig. 6-1

r 1 O 01

Inserting the values of (6.17) into the transformation laws for the linear stress and strain tensors, (2.27) and (3.78) respectively, the elastic matrix for a material having ~1x2 as a plane of symmetry is

C11 C12 c13 0 O

C21 C22 C23 0 0

The 20 constants in (6.18) are reduced to 13 when a strain energy function exists.

If a material possesses three mutually perpendicular planes of elastic symmetry, the material is called orthotropic and its elastic matrix is of the form

having 12 independent constants, or 9 if CKM = CMK.

An axis of elastic symmetry of order N exists a t a point when there are sets of equivalent elastic directions which can be superimposed by a rotation through an angle of 2nlN about the axis. Certain cases of axial and plane elastic symmetry are equivalent.

6.3 ISOTROPIC MEDIA. ELASTIC CONSTANTS

Bodies which are elastically equivalent in a11 directions possess complete symmetry and are termed isotropic. Every plane and every axis is one of elastic symmetry in this case.

J


For isotropy, the number of independent elastic constants reduces to 2, and the elastic matrix is symmetric regardless of the existence of a strain energy function. Choosing as the two independent constants the well-known Lamé constants, A and p, the matrix (6.19) reduces to the isotropic elastic form

In terms of A and p, Hooke's law (6.2) for an isotropic body is written

where C = c,, = IE. This equation may be readily inverted to express the strains in terms of the stresses as

- A 1 - A C . . = + -oij ~r E = 1

2 ~ ( 3 A + 2 ~ ) S i j u k k I @ + - E

2P 2 ~ ( 3 h +2P) 2P

where o = u,, = I,, the symbol traditionally used in elasticity for the first stress ihvariant.

For a simple uniaxial state of stress in the XI direction, engineering constants E and V

may be introduced through the relationships u,, = Ec,, and c,, = C,, = -vc,,. The constant E is known as Young's modulus, and V is called Poisson's ratio. In terms of these elastic constants Hooke's law for isotropic bodies becomes

E ( u.. = - ) E

13 l + v or E = - ( 1 + V E +:ic) 1 - 2~

or, when inverted, - l + v

c . . - E U i j - E L f 3 i j u kk 001" E = - l + v x -Li@

E E

From a consideration of a uniform hydrostatic pressure state of stress, it is possible to define the bulk modulus,

which relates the pressure to the cubical dilatation of a bpdy so loaded. For a so-called state of pure shear, the shear modulus G relates the shear components of stress and strain. G is actually equal to p and the expression

- E p - G = ----

2(1+ v )

may be proven without difficulty.

6.4 ELASTOSTATIC PROBLEMS. ELASTODYNAMIC PROBLEMS

In an elastostatic problem of a homogeneous isotropic body, certain field equations, namely,

(a) Equilibrium equations,

~ . . . + ~ b , = 0 31.1 or V . t + p b = O (6.27')


(b ) Hooke's law, uij = X a i j c k k + 2peij or 2 = hlc + ~ , . L E (6.28)

(c) Strain-displacement relations,

c i j = 4 ( ~ , ~ + u ~ , ~ ) or E = &(uV + Vu) (6.29)

must be satisfied a t a11 interior points of the body. Also, prescribed conditions on stress andlor displacements must be satisfied on the bounding surface of the body.

The boundary value problems of elasticity are usually classified according to boundary conditions into problems for which

(1) displacements are prescribed everywhere on the boundary,

(2) stresses (surface tractions) are prescribed everywhere on the boundary,

(3) displacements are prescribed over a portion of the boundary, stresses are prescribed over the remaining part.

For a11 three categories the body forces are assumed to be given throughout the continuum.

For those problems in which bound'ary displacement components are given everywhere by an equation of the form

ui = gi(X) or u = g(X)

the strain-displacement relations (6.29) may be substituted into Hooke's law (6.28) and the result in turn substituted into (6.27) to produce the governing equations,

which are called the Navier-Cauchy equations. The solution of this type of problem is therefore given in the form of the displacement vector Ui, satisfying (6.31) throughout the continuum and fulfilling (6.30) on the boundary.

For those problems in which surface tractions are prescribed everywhere on the boundary by equations of the form

the equations of compatibility (3.104) may be combined with Hooke's law (6.24) and the equilibrium equation (6.27) to produce the governing equations,

which are called the Beltrami-Michell equations of compatibility. The solution for this type of problem is given by specifying the stress tensor which satisfies (6.33) throughout the continuum and fulfills (6.32) on the boundary.

For those problems having "mixed" boundary conditions, the system of equations (6.27), (6.28) and (6.29) must be solved. The solution gives the stress and displacement fields throughout the continuum. The stress components must satisfy (6.32) over some portion of the boundary, while the displacements satisfy (6.30) over the remainder of the boundary.

In the formulation of elastodynamics problems, the equilibrium equations (6.27) must be replaced by the equations of motion (5.16)


and initial conditions as well as boundary conditions must be specified. In terms of the displacement field ui, the governing equation here, analogous to (6.31) in the elastostatic case is ..

p ~ ~ , ~ ~ + ( h + p ) ~ ~ , ~ ~ + ~ b ~ = p ~ ~ 01 p V 2 ~ + ( A + p ) V V * ~ + p b = p Ü (6.35)

Solutions of (6.35) appear in the form ui = u~(x, t) and must satisfy not only initial conditions on the motion, usually expressed by equations such as

ui = ui(x,O) and Ui = Ui(x,O) (6.36)

but also boundary conditions, either on the displacements,

ui = gi(x, t) or u = g(x, t)

or on the surface tractions,

6.5 THEOREM OF SUPERPOSITION. UNIQUENESS OF SOLUTIONS. ST. VENANT PRINCIPLE

Because the equations of linear elasticity are linear equations, the principle of superposition may be used to obtain additional solutions from those previously established. If, for example, ,ujl) represent a solution to the system (6.27), (6.28) and (6.29) with body forces b:" , and o;;), ut2) represent a solution for body forces b12', then <rij = a;:) + d2) ij 9

ui = u,'" + ui2) represent a solution to the system for body forces bi = bil' + b'2). i

The uniqueness of a solution to the general elastostatic problem of elasticity may be established by use of the superposition principle, together with the law of conservation of energy. A proof of uniqueness is included among the exercises that follow.

St. Venant's principle is a statement regarding the differences that occur in the stresses and strains a t some interior location of an elastic body, due to two separate but statically equivalent systems of surface tractions, being applied to some portion of the boundary. The principle asserts that, for locations sufficiently remote from the area of application of the loadings, the differences are negligible. This assumption is often of great assistance in solving practical problems.

6.6 TWO-DIMENSIONAL ELASTICITY. PLANE STRESS AND PLANE STRAIN

Many problems in elasticity may be treated satisfactorily by a two-dimensional, or plane theory of elasticity. There are two general types of problems involved in this plane analysis. Although these two types may be defined by setting down certain restrictions and assumptions on the stress and displacement fields, they are often introduced descrip- tively in terms of their physical prototypes. In plane stress problems, the geometry of the body is essentially that of a plate with one dimension much smaller than the others. The loads are applied uniformly over the thickness of the plate and act in the plane of the plate as shown in Fig. 6-2(a) below. In plane strain problems, the geometry of the body is essentially that of a prismatic cylinder with one dimension much larger than the others. The loads are uniformly distributed with respect to the large dimension and act perpendicular to it as shown in Fig. 6-2(b) below.


('J) Fig. 6-2

For the plane stress problem of Fig. 6-2(a) the stress components a,,, a,,, a,, are taken a s zero everywhere, and the remaining components are taken as functions of xl and 2 2 only,

na8 = (xl> xZ) (a, P = 1 ~ 2 ) (6.39)

Accordingly, the field equations for plane stress are

a a in which V = - êl + - $2 and

8x1 ax,

(6.43)

Due to the particular form of the strain tensor in the plane stress case, the six compatibility equations (3.104) may be reduced with reasonable accuracy for very thin plates to the single equation

In terms of the displacement components u,, the field equations may be combined to give the governing equation

d 2 d 2 where V 2 = - + - dx: 8 ~ ; '

For the plane strain problem of Fig. 6-2(b) the displacement component us is taken as zero, and the remaining components considered as functions of X I and X n only,


In this case, the field equations may be written

(a) u, , , ,+pb, = O or ~ . I : + p b = 0

(b) u,,3 - - XSa,'uv + ~ P E ~ , or 2: = Xlr + ZpE

- vu,, = X

O33 - 2(X + /L)

in which I: = [i: i: :) and E = (i: i) O33

From (6.47), (6.48), (6.49), the appropriate Navier equation for plane strain is

As in the case of plane stress, the compatibility equations for plane strain reduce to the single equation (6.44).

If the forces applied to the edge of the plate in Fig. 6-2(a) are not uniform across the . thickness, but are symmetrical with respect to the middle plane of the plate, a state of generalized plane stress is said to exist. In formulating problems for this case, the field variables a,,, r,, and u, must be replaced by stress, strain and displacement variables averaged across the thickness of the plate. In terms of such averaged field variables, the generalized plane stress formulation is essentially the same as the plane strain case if X is replaced by

2XP - A' = --- - VE

X + 2/.L 1 - v2 (6.52)

A case of generalized plane strain is sometimes mentioned in elasticity books when r,,

is taken as a constant other than zero in (6.50).

6.7 AIRY'S STRESS FUNCTION

If body forces are absent or are constant, the solution of plane elastostatic problems (plane strain or generalized plane stress problems) is often obtained through the use of the Airy stress function. Even if body forces must be taken into account, the superposition principie allows for their contribution to the solution to be introduced as a particular integral of the linear differential field equations.

For plane elastostatic problems in the absence of body forces, the equilibrium equations reduce to

- O or 8-2: = O u a 4 . ~ - (6.53)

and the compatibility equation (6.44) may be expressed in terms of stress components as

The stress components are now given as partia1 derivatives of the Airy stress function = +(xl, XP) in accordance with the equations

- - - ull - 4,229 u12 - -4,129 u22 - +,I1 (6.55)


The equilibrium equations (6.53) are satisfied identically, and the compatibility condition (6.54) becomes the biharmonic equation

Functions which satisfy (6.56) are called biharmonic functions. By considering biharmonic functions with single-valued second partia1 derivatives, numerous solutions to plane elastostatic problems may be constructed, which satisfy automatically both equilibrium and compatibility. Of course these solutions must be tailored to fit whatever boundary conditions are prescribed.

6.8 TWO-DIMENSIONAL ELASTOSTATIC PROBLEMS IN POLAR COORDINATES

Body geometry often deems it convenient to formulate two-dimensional elastostatic problems in terms of polar coordinates r and O. Thus for transformation equations

x1 = r cos 8, x2 =. r sin O (6.57)

the stress components shown in Fig. 6-3 are found to lead to equilibrium equations in the form

in which R and Q represent body forces per unit volume in the directions shown.

'c' Fig. 6-3

Taking the Airy stress function now as @ = @(r, O), the stress components are given by

The compatibility condition again leads to the biharmonic equation

7 ' (V2@) = V 4 @ = O

a2 i a i a2 but, in polar form, V 2 = - + -- + -- ar2 r ar r2 ao2 '


6.9 HYPERELASTICITY. HYPOELASTICITY Modern continuum studies have led to constitutive equations which define materials

that are elastic in a special sense. In this regard a material is said to be hyperelastic if it possesses a strain energy function U such that the material derivative of this function is equal to the stress power per unit volume. Thus the constitutive equation is of the form

in which Dij is the rate of deformation tensor. In a second classification, a material is said to be hypoelastic if the stress rate is a homogeneous linear function of the rate of deformation. In this case the constitutive equation is written

= KijkrnDkm

in which the stress rate U: is defined as

d u; = (cij) - uiq Vqi - ujq Vqi

where V i j is the vorticity tensor.

6.10 LINEAR THERMOELASTICITY If thermal effects are taken into account, the components of the linear strain tensor r ,

may be considered to be the sum c , , = ,(SI + € ( T )

i j i j (6.67)

in which r::' is the contribution from the stress field and rr is the contribution from the temperature field. Due to a change from some reference temperature TO to the temperature T , the strain components of an elementary volume of an unconstrained isotropic body are given by

€ C T ) = a(T - TO)Sij i j (6.68)

where a denotes the linear coefficient of thermal expansion. Inserting (6.68), together with Hooke's law (6.22), into (6.67) yields

which is known as the Duhamel-Neumann relations. Equation (6.69) may be inverted to give the thermoelastic constitutive equations

uij = XSi j rkk + 2pe, - (3X + 2,)aSij(T - To) (6.70)

Heat conduction in an isotropic elastic solid is governed by the well-known Fourier law of heat conduction,

ci = -kTai (6.71)

where the scalar k, the thermal conductivity of the body, must be positive to assure a positive rate of entropy production. If now the specific heat a t constant deformation c(") is introduced through the equation

- c ~ , ~ = p ~ ( " ) T (6.72)

and the interna1 energy is assumed to be a function of the strain components cij and the temperature T, the energy equation (5.45) may be expressed in the form


kTBii = p~'"' !i' + (3h + 2,.~)aT,,&

which is known as the coupled heat equation.

The system of equations that formulate the general therrnoelastic problem for an isotropic body consists of

(a) equations of motion .. ~ , , ~ + p b ~ = u ~ or V o Z + p b = Ü

(b ) thermoelastic constitutive equations

uii = Xai j rkk + BPcij - (3A + 2 , . ~ ) a 8 ~ ~ ( T - To)

I: = Xlr + ~ P E - (3A + 2 ~ ) a l ( T - To)

(c) strain-displacement relations

( d ) coupled heat equation

This system must be solved for the stress, displacement and temperature fields, subject to appropriate initial and boundary conditions. In addition, the compatibility equations must be satisfied.

There is a large collection of problems in which both the inertia and coupling effects may be neglected. For these cases the general thermoelastic problem decomposes into two separate problems which must be solved consecutively, but independently. Thus for the uncoupled, quasi-static, thermoelastic problem the basic equations are the

(a) heat conduction equation

(b ) equilibrium equations

a i j , j + p b i = O ~r V 0 E + p b = O

(c) thermoelastic stress-strain equations

aij = X S i j c k k + 2 p i j - (3A + 2 ~ ) a S ~ ~ (T - TO)

E Ale + ~ P E - (3X + 2,.L)aI(T - To)

( d ) strain-displacement relations


Solved Problems

HOOKE'S LAW. STRAIN ENERGY. ISOTROPY (Sec. 6.1-6.3)

6.1. Show that the strain energy density u* for an isotropic Hookean solid may be expressed in terms of the strain tensor by u* = X(tr E)V2 + /.LE: E, and in terms of the stress tensor by u* = [(I + v)Z : I: - ~ ( t r Z)2]/2E.

Inserting (6.21) into (6.13), u* = ( X S i j e k k + 2peij) e i j / = Xeiiyjl2 + peijyj which in symbolic notation is u* = X(tr E)2/2 + pE : E.

Inserting (6.24) into (6.13), u* = oij[( l + v)oij - vSijukk]/2E = [(I + v)oijoij - voiiojj]12E which in symbolic notation is u* = [ (1 + v)L : L - ~ ( t r L)2]/2E.

6.2. Separating the stress and strain tensors into their spherical and deviator components, express the strain energy density v,* as the sum of a dilatation energy density u&) and distortion energy density u&).

Inserting (3.98) and (2.70) into (6.13),

u* = &(sij + okksij/3)(eij + e,,Sij/3) = +(sijeij + oiiejj/3 + siieii/3 + u,ejj/3)

and since eii = sii = O this reduceS to U* = u'lS> + uTD, = oiiejj/6 + sijeij/2.

6.3. Assuming a state of uniform compressive stress uij = -pai,, develop the formulas for the bulk modulus (ratio of pressure to volume change) given in (6.25).

With uij = -pSii, (6.24) becomes eij = [ ( I + V)(-pSii) + vSij(3p)]lE and so yi = [-3p(l+ V) + 9pvIIE. Thus K = -p/eii = E / 3 ( 1 - 2v). Likewise from (6.21), oii = (31 + 2p) eii = -3p so that K = (3h + 2p)/3.

6.4. Express u:,, and u&, of Problem 6.2 in terms of the engineering constants K and G and the strain components.

From a result in Problem 6.3, uii = 3Keii and so

u : ~ ) = uiiejj/6 = Keiiejj/2 = K(IE)2/2

From (6.21) and (2.70), oij = ASij ekk + 2peii = sij + okk Sij/3 and since oii = (3X + 2p) eii i t foiiows that sij = 2p(eij - ekk Sijl3). Thus

u : ~ ) = zP(e i j - ~ ~ ~ 8 ~ ~ / 3 ) ( ~ ~ ~ - = p ( ~ i j ~ i i - eiieji/3)

Note that the dilatation energy density u:,) appears as a function of K only, whereas the distortion energy u'lD, is in terms of p (or G), the shear modu?us.

6.5. In general, u* may be expressed in the quadratic form u* = c ~ ~ z ~ E ~ in which the C&, are not necessarily symmetrical. Show that this equation may be written in the form of (6.14) and that au*la~, = a,.

Write the quadratic form as

U* = & c Z M e K c M + - & c K ~ ~ ~ E ~ = &CKMeKeM + + ~ : ~ e ~ e ~ = &(CzM + cMK)eKeM = *CKMeKeM

where CKM = CMK.

Thus the derivative au*laeR is now

au*/aeR = ~ C K M ( ~ K , R ' M + ~ K ~ M , R ) = & C K M ( ~ K R E M + E K ~ M R ) = J(CRMEM + CKREK) = C R M ~ M = "R


Show that for an orthotropic elastic continuum (three orthogonal planes of elastic symmetry) the elastic coefficient matrix is as given in (6.19), page 142.

Let the zlx2 (or equivalently, z;xh) plane be a plane of elastic symmetry (Fig. 6-4). Then UK = CKMcM and also u a = CKMcM. The transformation matrix between xi and zl is

Fig. 6-4

and from (2.27) and (J.78), o;( = UK, = cK for K = 1,2,3,6 whereas UL = -UK, e g = -cK

for K = 4,5. Thus, for example, from c; = CIMcM,

These two expressions for U; = o, are equal only if C14 = C15 = O. Likewise, from o; = o,, 03 = u3, U; = -v4, U; = -u5, U; = o, i t is found that C% = C25 = = C35 = C64 = C65 C41 =

= c43 = = = C53 = C5@ = 0.

If Z2x3 (or XZX:) is a second plane of elastic symmetry such that a; = CKMcM, the transformation array is

I1 - and now from (2.27) and (5.78), 02 = UK, cK - -€K for K = 1,2,3,4 whereas O; = -UK, ) I - c~ - -cK for K = 5,6. Now C16 = C2, = C3, = C45 = C54 = = Cs2 = Cs3 = O and the elastic

coefficient matrix attains the form (6.19). The student should verify that elastic symmetry with respect to the (third) xlx3 plane is identically satisfied by this array.

6.7. Give the details of the reduction of the orthotropic elastic matrix (6.19) to the isotropic matrix (6.20).

For isotropy, elastic properties are the same with respect to a11 Cartesian coordinate axes. In particular, for the rotated zl axes shown in Fig. 6-5, the method of Problem 6.6 results in the matrix (6.19) being further simplified by the conditions Cll = CZ2 = C33, C44 = C55 = and C12 = = Cl3 = = Cz3 = c32.

Fig. 6-5 Fig. 6-6


Finally, for the axes x:'obtained by a 45O rotation about x3 as in Fig. 6-6 above, the transformation matrix is

[aijl =

so that o[ = (o, - u1)/2 = (Cll - Clz)(az - 4 1 2 and e; = e, - e,. But o[ = C44~[ and so 2C4, = C,, - C,,. Thus defining p = C44 and X = C1,, (6.20) is obtained.

6.8. Give the details of the inversion of (6.21) to obtain (6.22).

From (6.21) with i = j, qi = (3X + 2P)Yi and so 2 p i j = uij - XSij~kk/(3X + 2p) or eij = oij/2p - X8ii~kk/2p(3X + 2p).

6.9. Express the engineering constants V and E in terms of the Lamé constants h and p.

From (6.25), E l ( 1 - 2 ~ ) = 3X + 2p; and from (6.26), El ( l+ v ) = 2p. Thus (3X + 2p)(1 - 2v) = 2p(l + v ) from which v = X/2(X + p). Now by (6.26), E = 2p(l + v) = p(3X f 2p)/(X + p).

6.10. Determine the elastic coefficient matrix for a continuum having an axis of elastic symmetry of order N = 4. Assume CKM = CMK.

Let x3 be the axis of elastic symmetry. A rotation e = 2a/4 = r12 of the axes about x3 produces equivalent elastic directions for N = 4. The transformation matrix is

[aiil = o o

and by (2.27) and (3.78), o; = o,, o; = o,, 04 = o,, o: = -05, o; = u4, u; = and e: = e2, E; = cl, = '3,

1 - e4 - -e5, 4 = e4, e: = -eg. Thus, for example, from o3 = 83, C34 = C35 = = 0, C3, = C3,. Likewise, from the remaining five stress relations, the elastic matrix becomes

[CKMI =

with seven independent constants.

Fig. 6-7

ELASTOSTATICS. ELASTODYNAMICS (Sec. 6.4-6.5)

6.11. Derive the Navier equations (6.31).

Replacing the strain components in (6.28) by the equivalent expressions in terms of displacements yields oij = AGijuk,, + lr(ui,j + u ~ , ~ ) . Thus oij,j = Au,,,, + $(ui, i j + ujPij). Substituting this into the equilibrium equations (6.27) and rearranging terms gives pu , j j + ( X + lr)uj, ji + pbi = 0.


6.12. Show that if V4Fi = 0, the displacement ui = ( A + 2p)Fi,jiIp(A + p) - Fj, j i l p is a solution of the Navier equation (6.31) for zero body forces.

Differentiating the assumed soiution, the terms @ui,jj = ( A 2 p ) F i , k k j j / ( A + P ) - F k , k i j j and ( A + = ( A + 2 p ) F j , k k j i / P - (A + p ) F k , k j j i / P are readily calculated. Inserting these into ( 6 . 3 1 ) gives

( A + 2 G ) F i , k k j j / ( A + P ) - [ P - ( A + 2 P ) f ( A + P) IF j , jkki = O

provided Fi,kkjj = V 4 F i = 0.

6.13. If body forces may be neglected, show that (6.35) is satisfied by ui = +,i + c ~ ~ ~ $ ~ , ~ provided + and $i each satisfies the familiar three dimensional wave equation.

Substituting the assumed ui into ( 6 . 9 5 ) yields . . . . ~ ( @ , i k k + eijk$'k,jqq) + ( A + ~ ) ( @ , j j i + €jpq$'q.pji) = P( @,i + eijk$'k,j)

Since ypq\l.q,pji = 0, this equation may be written . . ( ( A f 2 1 ~ ) @ , k k - P ? ) , i + ' i jk(~$'k ,qq - ~ $ ' k ) , j = O

which is satisfied when V2@ = p ? / ( A + 2 ~ ) and V 2 $ ' k = p Y k / f l .

6.14. Writing c2V2+ = where c2 = ( A + 2p) /p for the wave equation derived in Problem

6.13, show that + = g(r + ct) + h(r - ct) is a solution with g and h arbitrary func- r tions of their arguments and r2 = X i x i .

; C (r2:) since @ = + ( r , t ) . ~ h u s Here i t is convenient to use the spherical form V 2 --

r 2 ( a @ l a r ) = r ( g ' + h ' ) - ( g + h ) where primes denote derivatives with respect to the arguments of g and h. Then V 2 4 = (g" + h " ) / r . Also 4 = ( g ' c - h r c ) / r and ;b' = c2(g1' + h l ' ) l r . Therefore c 2 V 2 @ = ;6. for the given @.

6.15. Derive the Beltrami-Michell equations (6.33) and determine the form they take when body forces are conservative, i.e. when p b i = +,i.

Substituting ( 6 . 2 4 ) into ( 3 . 1 0 3 ) yields

where 8 = I= = uii. Only six of the eighty-one equations represented here are independent. Thus setting m = k and using ( 6 . 2 7 ) gives

from which = - ( i + ~ ) p b ~ , ~ / ( l - v ) . inserting this expression for O , k k into the previous equation leads to ( 6 . 9 3 ) .

If p b i = @ , i , then p ( b i , j + b j , i ) = 2 @ , i j and p b k , , = @ , k k = V 2 @ so that ( 6 . 3 3 ) becomes

TWO-DIMENSIONAL ELASTICITY (Sec. 6.6-6.8)

6.16. For plane stress parallel to the x1x2 plane, develop the stress-strain relations in terms of A and Show that these equations correspond to those given as (6.41).

Here u~~ = nl3 = uz3 = O so that ( 6 . 2 1 ) yields e13 = ep3 = O and €33 = -A(e11+ e Z 2 ) / ( A + 2 ~ ) . Thus ( 6 . 2 1 ) reduces to o,@ = 2ApS ,Bey , / (X + 2 ~ ) + with a , P, y = 1 ,2 from which o,, = 2 ~ ( 3 A + 2 p ) ~ , , / ( ~ + 2 ~ ) so that the equation may be inverted to give

Also, €33 = -Ae,,/(A + 2 @ ) = -Aoyy/2P(3) , + 2 p ) = - voyy /E


6.17. For plane strain parallel to x1x2, develop the stress-strain relations in terms of V and E. Show that these equations correspond to those given as (6.48).

Here u3 - O so that eg3 = O and (6.24) gives 033 = v(oll + o,,) = Ao,,/2(A + p). Thus (6.24)

becomes = (1 + v)o,plE - v ( l + v)SapoYylE from which e,, = ( 1 + v ) ( l - 2v)o,,lE. Finally, inverting,

o,p = vES,peyyl(l+ v ) ( l - 2") + Ee,pl( l + v) = AS,peyy + 2pe,p

6.18. Develop the Navier equation for plane stress (6.45) and show that i t is equivalent to the corresponding equation for plane strain (6.51) if A' = 2 A p / ( A + 2p) is substituted for A.

Inverting (6.41) and using (6.42) leads to a,p = E ( U , , ~ + up,,)/2(1 + v) + 2vES,puy,y/2(1 - v2). Differentiating with respect to x p and substituting into (6.40) gives

Eu,, pp/2(1+ v) + Eup,p,/Z(l- v) + pb, = hV2ua + p(3A + 2 p ) ~ ~ , p , / ( h + 2p) + pb, = O

Thus since p(3X + 2p)/(A + 2p) = (2Ap/(A + 2p) + p) = (A' + p), (6.45) and (6.51) have the same form for the given substitution.

6.19. Determine the necessary relationship between the constants A and B if + = AxSxi + Bx2 is to serve as an Airy stress function.

By (6.56), + must be biharmonic or +,1111 + 2+,1122 + +,2222 = O + 24Ax2 + 120Bx2 = 0 , which is satisfied when A = -5B.

6.20. Show that + = xi is suitable for use as an Airy stress function

and determine the stress components in the region x t > O, -c < xz < C.

Since 8 4 + is identically zero, + is a valid stress function. The stress components as given by (6.55) are oll = +,22 = -3Fx1x2/2c3 + P/2c, uiz = = -3F(c2 - xi ) /4c3, 022 = +,il = 0. These stresses are those of a cantilever beam subjected to a transverse end load F and an axial pull P (Fig. 6-8).

Fig. 6-8

6.21. In Problem 2.36 it was shown that the equilibrium equations were satisfied in the absence of body forces by uij = c ~ ~ ~ c ~ ~ ~ + ~ ~ , ~ ~ . Show that Airy's stress function is represented by the case +,, = +(xl , x,) with +,, = +,, = +,, = +,, = +,, 0.

Since +,, is the only non-vanishing component, oij = eipq ejmn+qn,pm becomes q j = eip3ejrn3 +33,pm which may be written oap = E , ~ ~ E ~ ~ ~ + ~ ~ , y t . Thus since +33 = +, ",B = ( S a ~ S y ( ; - S Y B ) @ , ~ ( ; = 6,p+,yy - +,,p . The stress components are therefore 011 = +,li + +,22 - +,li = +,22, 0 1 2 = -+,12,

"22 = 9.11 + +,22 - +,22 = +, 11.


6.22. In polar coordinates (r, O ) the Airy stress function CJ = BB is used in the solution of a disk of radius a subjected to a central moment M. Determine the stress components and the value of the constant B.

From (6.60) and (6.61), o(,,, = o(oe) = O. From (6.62) o(,,, = Blr2. Equilibrium of moments about the center of

2 r 2 r

the disk requires M = ~ ( , ~ ) d de = Jo' B de = 2nB. Thus B = M 1 2 ~ . Fig. 6-9

LINEAR THERMOELASTICITY (Sec. 6.10)

6.23. Carry out the inversion of (6.69) to obtain the thermoelastic constitutive equations (6.70).

From (6.69) with i = j, aii = (3X + - 3a(T - To)) . Solving (6.69) for oij gives

uij = 2peii + XSijokk/(3~ + 2p) - 2aa8ij(T - To)

= 2pyj + XSij(ekk - 3a(T - T o ) ) - 2paSij(T - To)

= 2peij + ASijekk - (3X + 2p)c18~~(T - T O ) I

6.24. Develop the thermoelastic energy equation (6.73) by use of the free energy f = u - Ts. Assuming the free energy to be a fynction of the strains and temperature, f = f(e,, T ) and

substituting into (5.41) p& aijS + pTs where dots indicate time derivatives, the result is (oij - pôf/dsij) iij - p(s + dfldT)T = O. Since the terms in parentheses are independent of strain and temperature rates, i t follows that uij = p d f l a ~ ~ ~ and s = -dfldT. From (5.38) for a reversible

isothermal process, = p ~ S = pT - i.. + - T . At constant deformation, Z i j = 0 and (a:j dT as "> comparing this equation with (6.72) gives c(") = T(ds/dT) or from above, since ds/dT = -a2flaT2, c(") = - d Z f l d T 2 . Also, from above, p(d2fld~ijdT) = dSjldT and so combining ( 5 3 8 ) with (6.71),

-ci,i = kTBii = pT -iij + - T . Finally from (6.70), doil/dT = (3X+ so that (YT c:) )

kT,ii = pc(")T + (3X + 2p)aT0eii which is (6.73).

6.25. Use (6.13) and (6.70) to develop the strain energy density for a thermoelastic solid.

Substituting (6.70) directly into (6.13),

U* = X8ij~kk€ii/2 + peijeij - (3X + 2p)aSij(T - To)eij/2

= Xeiiejj12 + peijeij - (3X + 2p)a(T - To)eii12


6.26. Show that the distortion energy density u:,, may be expressed in terrns of princigal stress values by the equation uy,, = [ ( a , - + (a2 - u J 2 + (a3 - ~ ~ ) ~ ] / 1 2 G .

\ From Problem 6.2, u:,) = sijeij/2 = siisij/4G which in terms of stress components becomes

u f D , = (uij-8ijukk/3)(aij- 8ijop,/3)/4G = (oijoij-oiioji/3)/4G

In terms of principal stresses this is

ufD) = [o: + o; + o: - (a1 + o2 + o3)(u1 + 4 2 + 03)/3]/4G

= [2(u; + a: + o: - olo2 - u2~3 - a3u1)/3]/4G

= [(vl - u2)2 + (v2 - u3)2 + (03 - u1)2]/12G


6.27. Use the results of Problem 6.1 to show that for an elastic material d u * l d ~ , ~ = uij and ~ u * I ~ u ~ , = É i j .

From Problem 6.1, u* = Xeiiejjl2 + prijsij and so

Ju*larPq = X/2[eii(aejjla~pq) + ejj(âciilòepq)] + Z p ~ ~ ~ ( a e ~ ~ l a $ , )

= X/2[eii8jp8jq + ~ ~ ~ 8 ~ ~ 8 ~ ~ ] + 2 / J ~ ~ ~ 8 ~ ~ 8 ~ ~ = X/2[€iispq $ €jjspq] f 2,Urpq - - heiiSpq + 2pePq = upq

Likewise from Problem 6.1, u* = [(I + v)uijuii - ~ u ~ ~ u ~ ~ ] / ~ E and so

au"la,, = [2(1+ u)uijSipSjq - v(uiiôpq + ujjôW]/2E = [ ( I + v)uPq - u8pquii]lE = epq

6.28. Express the strain energy density u* as a function of the strain invariants.

From Problem 6.1, u* = Xeiiqj/2 + priirij; and since by comparison with (3.91), I, = cii and 11, = (riirjj - 4jY j ) /2 , i t follows that

6.29. When a circular shaft of length L and radius a is subjected to end couples as shown in Fig. 6-10, the nonzero stress components are a13 = -G<Yx~, u~~ = G a x l where a is the angle of twist per unit length. Determine expressions for the strain energy density and the total strain energy in the shaft.

From Problem 6.1, u* = [ ( I + v) I : I - v(tr I )2] /2E. Here t r I O and I : I = 2G2a2r2 where r 2 = x: + x:. Thus u* = Ga2r2/2. The total strain energy is given by Fig. 6-10

Note that since T = LPT f ~ a ( % : + x:)r dr de = Gaa4n-12, U = TaLI2, the externa1 work.

6.30. Show that for a continuum having an axis of elastic symmetry of order N = 2, the elastic properties (Hooke's law and strain energy density) are of the same form as a continuum having one plane of elastic symmetry.

Here a rotation of axes e = 2 ~ 1 N = 2 ~ 1 2 = n produces equivalent elastic directions. But this is precisely the same situation a s the reflection about a plane of elastic symmetry.

6.31. Show that (6.19) with C11 = C22 = C33, C44 = Cs5 = C66 and C12 = C13 = C23 may be reduced to (6.20) by an arbitrary rotation í3 of axes about 2 3

(Fig. 6-11). The transformation between xi and xf axes is

cos 0 sin 0 0

and from (2.27), Fig. 6-11

158 L I N E A R E L A S T I C I T Y [CHAP. 6

ai2 = (-sin e cos e)ull + (cos2 e - sin2 e)o12 + (s in 6 cos @)oz2

or i n single index notation, = (-sin e cos e)ul + (cos2 8 - sin2 e)a6 + (s in e cos e)u2

Likewise from (3.78) and (6.4),

e; = (-2 sin e cos e ) € , + (cos2 e - sin2 @)e6 + ( 2 sin e cos @)e2

Bu t a; = C44€; for an isotropic body and so here u2 - ai = 2C44(€2- cl) . Finally f rom (6.19) wi th the given conditions, ul = Cllel i- C12(e2 + el) and q = CtlY + C12(el + e,) and so o2 - ul = (Cll - C12)(e2 - el). Therefore (Cll - C12) = 2C44 and wi th C* = p, G12 = h, Cll = h + 2p as given i n (6.20).

A

6.32. For an elastic body in equilibrium under body forces bi and surface forces ti"', show that the total strain energy is equal to one-half the work done by the externa1 forces acting through their displacements ui.

I t is required t o show tha t i pbiui dV + t:"ui dS = 2 u* dV. Consider first the surface A Jv

integral wi th ti("' = ujiuj and convert b y Gauss' theorem. T h u s

L uijuinj dS = l, (cijui), dV = L (aij, jui + uijui, j ) dV

But a i jub j = uij(cij + oij) = uijeij, and f rom equilibrium oij,j = -pbi T h u s

4 t$'ui dS = - Sv pbiui dV + 2 aiieij/2 dV

and the theorem is proved.

S 6.33. Use the result of Problem 6.32 to establish uniqueness of the elastostatic solution of a

linear elastic body by assuming two solutions O:), u:') and O:;) , ui2).

For linear elasticity superposition holds, so uij = b:;) - a:;', ui = uil' - ~ 1 ~ ) would also be

a solution for which bi = O. T h u s for this "difference" solution t:")uidS = 2 u* dV from L A Jv Problem 6.32. SinceAthe two assumed solutions sat is fy boundary conditions, the l e f t hand integral is zero here since tln' = tu' i - ti2' on the boundary for equation (6.32) and ui = u:l' - u'i2' on

the boundary for equation (6.5'0). Thus rs* dV = O and since u* is positive definite this occurs

e.. Jv

only i f eij = e:!) - i 2 ) 0 , or e::) = e!?). I f the strains are equal for the two assumed solutions, the stresses are also equal b y Hooke's law and the displacements are equal t o within a rigid body displacement. T h u s uniqueness is established.

6.34. The Navier equations (6.31) may be put in the form pui ,s + uj.ji + pbi = O

which for the incompressible case ( V = 4) are clearly indeterminate. Use the equilibrium equations for this situation to S ~ O W that pUi,jj + @,i13 + pbi = 0.

From equation (6.24), eii = ( 1 - 2v)aiilE; and for v = Q, eii uiBi = O. T h u s f rom (6.24),

2cij, = u . .. + u . .. = 2(1 + v)aij, jlE - 2v8ijukk,j/E 1.33 3,13

But uj,ji = O and E = 3G when v = 3, so tha t ui, j j = -pbilG - ukkJ3G or pV2ui + e j i / 3 + pbi = 0.

Supplementary Problems - 6.35. Prove tha t the principal axes o f the stress and strain tensors coincide for a homogeneous isotropic

elastic body.

6.36. Develop the expression for the strain energy density u* for an orthotropic elastic medium. Use equations (6.14) and (6.1 9).

U* = ( C l l ~ l f 2C12e2 + 2 C 1 3 ~ 5 ) ~ 1 / 2 f (C22e2 f 2c23e4)"2/2 + C3363 f C44'42 + Cs5e: f C66ei.


6.37. Determine the form of the strain energy density for the case of ( a ) plane stress, (b ) plane strain.

Ans. (a) u* = [o;, + - 2volla22 + 2(1 + v)oy2]12~

( b ) u* = ( a + ~/2)(e?1+ egz) + Xe11'22 + 2aef2

2a 6.38. Determine the value of c for which ul = A sin- ( x l c t ) , u2 = u3 = O is a solution of equation I

( 6 3 5 ) when body forces are zero. Ans. c = d w

6.39. Show that the distortion energy density uyD, = (uijoii - oiiojj/3)/4G and the dilatation energy density * urS, = oiiajjl18K.

6.40. Show that 1/ (1+ V ) = 2(X + & ( 3 ~ + 2p) and vl(1- V ) = X/(h + 2p).

6.41. For plane strain parallel to x,x,, show that b, O and that b, and b2 are functions of x , and x, only.

6.42. Use the transformation laws for stress and strain to show that the elastic constants Cijk,,, are the components of a fourth order Cartesian tensor so that Ci;km = aipaiqakTa,,,sCpqrs.

6.43. Show that the Airy stress function + = 2%; + 12x;x; - 6x2 satisfies the biharmonic equation V4+ = O and determine the stress components assuming plane strain.

2: - 3x2 -2x,x,

Ans. <rij = 24

o 2v(x; - x; )

6.44. Determine the strains associated with the stresses of Problem 6.43 and show that the compatibility equation (6.44) is satisfied.

x; - 32; - 24%; - 2;) -2x1x2

Am. eij = 24 (%) ( -2xlx2 x; + x; - 24%: - x ; ) O

o o

6.45. For an elastic body having an axis of elastic symmetry of order N = 6, show that C2- = C,,, C,, = CG, Ce6 = 2(Cll - CI2) and that C13 and C33 are the only remaining nonzero coefficients.

6.46. Show that for an elastic continuum with conservative body forces such that pba = V+ = $,,, the compatibility condition (6.44) may be written V20aa = Q2+/(1 - V ) for plane strain, or 8 2 0 a a = ( 1 + v)V2+ for plane stress.

6.47. If V4Fi = 0, show that ui = 2(1- v)V2Fi/G - FjBji /G is a solution of the Navier equation (6.31) when bi O (see Problem 6.12). If F = B(x2ê l - x lê2) / r where r 2 = xixi, determine the stress components. Ans. o,, = -oz2 = 6QGx1x2/r5, = 0, u12 = 3QG(x; - 4 ) / ~ 5 , u13 = -oz3 = 3QGx2x3/r5, where

Q = 4B(1- v)lG.

6.48. In polar coordinates an Airy stress function is given by @ = Cr2(cos 2e - cos 2a) where C and a are constants. Determine C if use = O, ora = T when e = a, and = O , a,, = -T when e = -a. Ans. C = r / ( 2 sin 2a)

6.49. Show that in plane strain thermoelastic problems 0 3 ~ = v(oll + oZ2) - a E ( T - To) and that = hSapeaa + 2 p n B - Sap(3X 2p)a(T - To). In plane stress thermoelasticity show that

€33 = -v(ul1 + aZ2)lE + a(T - To) and eap = ( 1 + v)aaBIE - vSaPoauIE + Sap ( T - To)a

6.50. In terms of the Airy stress function + = +(xl , x2), show that for plane strain thermoelasticity the compatibility equation (6.44) may be expressed as V4+ = -aEV2(T - T o ) / ( l - v ) and that for plane stress as V 4 + = -aEV2(T - To).

Chapter 7

Fluids

7.1 FLUID PRESSURE. VISCOUS STRESS TENSOR. BAROTROPIC FLOW

In any fluid a t rest the stress vector tin) on an arbitrary surface element is collinear with the normal íi of the surface and equal in magnitude for every direction a t a given point. Thus A

A tin) = ~ . . n . = -poni or t(^) = 1-íi = -p,n 23 3 (r*1)

in which po is the stress magnitude, or hydrostat ic pressure. The negative sign indicates a compressive stress for a positive value of the pressure. Here every direction is a principal direction, and from (7.1)

a, , = -p o s i j or 1 = -p I O

which represents a spherical state of stress often referred to as hydrostatic pressure. From (7.2), the shear stress components are observed to be zero in a fluid a t rest.

For a fluid in motion, the shear stress components are usually not zero, and it is customary in this case to resolve the stress tensor according to the equation

where rii is called the viscous s tress tensor and p is the pressure. A11 real fluids are both compressible and viscous. However, these characteristics vary

widely in different fluids so that it is often possible to neglect their effects in certain situations without significant loss of accuracy in calculations based upon such assumptions. Accordingly, an inviscid , or so-called perfect fluid is one for which rij is taken identically zero even when motion is present. VZscous fluids on the other hand are those for which rij must be considered. For a compressible fluid, the pressure p is essentially the same as the pressure associated with classical thermodynamics. From (7.3), the mean normal stress is given by

For a fluid a t rest, r , vanishes and p reduces to p, which in this case is equal to the negative of the mean normal stress. For an incompressible fluid, the thermodynamic pressure is not defined separately from the mechanical conditions so that p must be considered as an independent mechanical variable in such fluids.

In a compressible fluid, the pressure p, the density p and the absolute temperature T are related through a kinetic equation of state having the form

An example of such an equation of state is the well-known ideal gas law .

CHAP. 71 FLUIDS 161

where R is the gas constant. If the changes of state of a fluid obey an equation of state that does not contain the temperature, i.e. p = P ( ~ ) , such changes are termed barotropic. An isothermal process for a perfect gas is an example of a special case which obeys the barotropic assumption.

7.2 CONSTITUTIVE EQUATIONS. STOKESIAN FLUIDS. NEWTONIAN FLUIDS The viscous stress components of the stress tensor for a fluid are associated with the

dissipation of energy. In developing constitutive relations for fluids, i t is generally assumed that the viscous stress tensor T~~ is a function of the rate of deformation tensor Dij . If the functional relationship is a nonlinear one, as expressed symbolically by

the fluid is called a Stokesian fluid. When the function is a linear one of the form

where the constants Ki,, are called viscosity coeficients, the fluid is known as a Newtonian fluid. Some authors classify fluids simply as Newtonian and non-Newtonian.

Following a procedure very much the same as that carried out for the generalized Hooke's law of an elastic media in Chapter 6, the constitutive equations for an isotropic homogeneous Newtonian fluid may be determined from (7.7) and (7.3). The final form is

where A* and p* are viscosity coefficients of the fluid. From (7.9)' the mean normal stress is given by

where K* = Q(3A* + 2p*) is called the coeficient of bulk viscosity. The condition that

is known as Stokes' condition, and guarantees that the pressure p is defined as the average of the normal stresses for a compressible fluid a t rest. In this way the thermodynamic pressure is defined in terms of the mechanical stresses.

In terms of the deviator components sij = aii - 6ijokk/3 and D i = Dii - 6iiD,k/3, equation (7.9) above may be rewritten in the form

Therefore in view of the relationship (7.10), equation (7.12) may be expressed by the pair of equations

sij = 2p*D,; or S = 2p*Df (7.13)

the first of which relates the shear effects in the fluid and the second gives the volumetric relationship.

162 FLUIDS [CHAP. 7

7.3 BASIC EQUATIONS FOR NEWTONIAN FLUIDS. NAVIER-STOKES-DUHEM EQUATIONS

In Eulerian form, the basic equations required to formulate the problem of motion for a Newtonian fluid are

I

(a) the continuity equation (5.3),

= O or ; + p ( V x 0 v ) = O (7.1 5)

(b) the equations of motion (5:16),

( c ) the energy equation (5.32),

(d) the constitutive equations ( 7 4 ,

( e ) the kinetic equation of state (7.5), P = ~ ( p , T )

If thermal effects are considered, as they very often must be in fluids problems, the additional equations

( f ) the Fourier law of heat conduction (6.71),

( g ) the caloric equation of state, 9-4 = ~ ( p , T)

are required. The system of equations (7.15) through (7.21) represents sixteen equations in sixteen unknowns and is therefore determinate.

If (7.18) above is substituted into (7.16) and the definition 20 , = (vLj + is used, the equations that result from the combination are the Navier-Stokes-Duhem equations o f motion,

pGi = pbi - pSi + (A* + p * ) ~ , , , ~ + p * ~ ~ , ~ ~

or pV = pb - VP + (A* + p * ) v ( v V ) + p*V2v

When the flow is incompressible ( v , = O), (7.22) reduce to the Navier-Stokes equations for incompressible flow,

If Stokes condition is assumed (A* = -pp*), (7.22) reduce to the Navier-Stokes equations for compressible flow

PVi = Pbi - P*i + *!J*vj,ji + p*vcjj or (7.24)

PV = pb - V p + +p.*v(v V ) + p*V2v

CHAP. 71 FLUIDS 163

The Navier-Stokes equations (7.23), together with the continuity equation (7.15) form a complete set of four equations in four unknowns: the pressure p and the three velocity components vi. In any given problem, the solutions of this set of equations must satisfy boundary and initial conditions on traction and velocity components. For a viscous fluid, the appropriate boundary conditions a t a fixed surface require both the normal and tangentiaI components of velocity to vanish. This condition results from the experimentally established fact that a fluid adheres to and obtains the velocity of the boundary. For an inviscid fluid, only the normal velocity component is required to vanish on a fixed surface.

If the Navier-Stokes equations are put into dimensionless form, severa1 ratios of the normalizing parameters appear. One of the most significant and commonly used ratios is the Reynolds number N,,, which expresses the ratio of inertia to viscous forces. Thus if a flow is characterized by a certain length L, velocity V and density p, the Reynolds number is

N,,, = VL/v (7.25)

where V = p*lP is called the kinematic viscosity. , For very large Reynolds numbers, the viscous contribution to the shear stress terms of the momentum equations may be neglected. In turbulent flow, the apparent stresses act on the time mean flow in a manner similar to the viscous stress effects in a laminar flow. If turbulence is not present, inertia effects outweigh viscous effects and the fluid behaves as though it were inviscid. The ability of a flow to support turbulent motions is related to the Reynolds number. It is only in the case of laminar flow that the constitutive relations (7.18) apply to real fluids.

7.4 STEADY FLOW. HYDROSTATICS. IRROTATIONAL FLOW

The motion of a fluid is referred to as a steady flow if the velocity components are independent of time. For this situation, the derivative avilat is zero, and hence the material derivative of the velocity

reduces to the simple form i i = V . v . . J 2 . 3 O r + = V V . V X V

A steady flow in which the velocity is zero everywhere, causes the Navier-Stokes equations (7.22) to reduce to

pbi = P, , or p b = VxP (7.28)

which describes the hydrostatic equilibrium situation. If the barotropic condition p = p ( p ) is assumed, a pressure function

P(p) = spg (7.29) pa P

may be defined. Furthermore, if the body force may be prescribed by a potential function

equations (7.28) take on the form

A flow in which the spin, or vorticity tensor (4.21),

164 FLUIDS [CHAP. 7

vanishes everywhere is called an irrotational fiow. The vorticity vector qi is related to the vorticity tensor by the equation

q, = cijkVkj 01: q = V" (7.33)

and therefore also vanishes for irrotational flow. Furthermore,

Qi = z i jkvk , j or q = V x v (7.34)

and since V x v = O is necessary and sufficient for a velocity potentkl 4 to exist, the velocity vector for irrotational flow may be expressed by

v . = - 4 , or v = -V+ (7.35)

7.5 PERFECT FLUIDS. BERNOULLI EQUATION. CIRCULATION

If the viscosity coefficients h* and p* are zero, 'the resulting fluid is called an inviscid or perfect (frictionless) ftuid and the Navier-Stokes-Duhem equations (7.22) reduce to the form

which is known as the Euler equation of motion. For a barotropic fluid with conservative body forces, (7.29) and (7.30) may be introduced so that (7.36) becomes

;i = - (n+P), or V = -v(a+P) (7.37)

For steady flow (7.37) may be written

v . v . . = -(n+P),i or v . V v = -V(n+P) 3 t i 3 (7.38)

If the Euler equation (7.37) is integrated along a streamline, the result is the well- known Bernoulli equation in the form (see Problem 7.17)

For steady motion, avilat = O and C(t) becomes the Bernoulli constant C which is, in general, different along different streamlines. If the flow is irrotational as well, a single constant C holds everywhere in the field of flow.

When the only body force present is gravity, the potential n = gh where g is the gravitational constant and h is the elevation above some reference level. Thus with h, = Plg defined as the pressure head, and v2/2g = hv defined as the velocity head, Bernoulli's equation requires the total head along any streamline to be constant. For incompressible fluids (liquids), the equation takes the form

h + h, + h" = h + plpg + v2/2g = constant (7.40)

By definition, the velocitz~ circulation around a closed path of fluid particles is given by the line integral

r, = f v i d x i or r, = f vedx (7.41)

From Stokes theorem (1.153) or (1.154), page 23, the line integral (7.41) may be converted to the surface integral

where íi is the unit normal to the surface S enclosed by the path. If the flow is irrotational, V x v = O and the circulation is zero. In this case the integrand of (7.42) is the perfect differential d+ = -v. dx with the velocity potential.

CHAP. 71 FLUIDS 165

The material derivative drcldt of the circulation may be determined by using (4.60) which when applied to (7.41) gives

For a barotropic, inviscid fluid with conservative body forces the circulation may be shown to be a constant. This is known as Kelvin's theorem of constant circulation.

7.6 POTENTIAL FLOW. PLANE POTENTIAL FLOW

The term potential flow is often used to denote an irrotational flow since the condition of irrotationality, V x v = O, is necessary and sufficient for the existence of the velocity potential + of (7.35). For a compressible irrotational flow, the Euler equation and the continuity equation may be linearized and combined as is done in acoustics to yield the governing wave equation .. + = or ; = c2V2+ (7.44)

where c is the velocity of sound in the fluid. For a steady irrotational flow of a compressible barotropic fluid, the Euler equation and continuity equation may be combined to give

which is the so-called gas dynamical equation.

For incompressible potential flow the continuity equation attains the form

and solutions of this Laplace equation provide the velocity components through the definition (7.35). Boundary conditions on velocity must also be satisfied. On a fixed boundary, for example, d+ldn = O. An important feature of this formulation rests in the fact that the Laplace equation is linear so that superposition of solutions is possible.

In a two-dimensional incompressible flow parallel to the x,x, plane, v, = 0, and the continuity equation becomes

v,,, = o or B'V = o (7.47)

where, as usual in this book, Greek subscripts have a range of two. By (7.47), regardless of whether the flow is irrotational or not, i t is possible to introduce the stream function + = +(x,, x,) such that

If the plane flow is, indeed, irrotational so that

then from (7.48) and (7.49) the stream function and velocity potential are seen to satisfy the Cauchy-Riemann conditions

4.1 = q . 2 and +,, = -+,i (7.50)

By eliminating + and + in turn from (7.50) i t is easily shown that

166 FLUIDS [CHAP. 7

Thus both + and + are harmonic functions when the flow is irrotational. Furthermore, the complex potential

@(4 = + ( x 1 9 x,) + i+(x1, $2) (7.53)

is an analytic function of the complex variable, z = x , + i x , so that its derivative dqldz defines the c o m p l e x v e l o c i t y

dqldz = -vl + iv, (7.54)

Solved Problems

FUNDAMENTALS OF FLUIDS. NEWTONIAN FLUIDS (Sec. 7.1-7.3)

7.1. Show that the deviator sij for the stress tensor a , of ( 7 . 3 ) is equal to t , , the deviator of T i j of ( 7 . 3 ) .

From ( 7 4 , aii = -3p + rii and so here

7.2. Determine the mean normal stress o i i / 3 for an incompressible Stokesian (nonlinear) fluid for which T , = a D i j + pDikDkj where a and p are constants.

From ( 7 4 , aij = -pSij + aDij + /3DikDkj and so ai< = -3p + aDii + /3DikDki. But Dik = Dki and Dii = = O for an incompressible fluid so that

4 3 = -p + /3DijDij13 = -p - 2/311D/3

where 11, is the second invariant of the rate of deformation tensor.

7.3. Frictionless adiabatic, or isentropic flow of an ideal gas, is a barotropic flow for which p = cpk where C and k are constants with k = c(*)/c("), the ratio of specific heat a t constant pressure to that a t constant volume. Determine the temperature-density and temperature-pressure relationships for such a flow.

Inserting p = Cpk into equation (7.6), the temperature-density relationship is # - l / T = RIC, a constant. Also, since p = (plC)l/k here, (7.6) yields the temperature-pressure relationship as p(k-l)/k/T = R/Cl/k, a constant.

7.4. Determine the constitutive equation for a Newtonian fluid with zero bulk viscosity, i.e. with K* = 0.

If K* O , A* = -2 /~*/3 by (7.11) and SO (7.9) becomes aii = -pSij - (2/l*/3)SijDkk + 2/~*Dij which is expressed in terms of the rate of deformation deviator by

If the deviator stress si j is introduced, this constitutive relation is given by the two equations S i j = ~ / L * D ; and aii = -3p.

CHAP. 71 FLUIDS 167

7.5. Determine an expression for the "stress power" cijDij of a Newtonian fluid having equation (7.9) as its constitutive relation.

From (7.9) and the stress power definition,

o..D.. u 23 = -pSijDij + h*SijDkkDij + 2p*DijDij = -pDii + h*DiiDjj + 2p*DijDii

In symbolic notation, this expression is written

I : D = -p(tr D ) + A*(tr D)2 + 2p*D : D

In terms of D& the expression is

In symbolic notation, : D = -p(tr D ) + ~ * ( t r D ) 2 + 2p* D' : D'

7.6. Determine the conditions under which the mean normal pressure p,,, = -aii/3 is equal to the thermodynamic pressure p fo r a Newtonian fluid.

With the constitutive equations in the form (7.13) and (7.14), the latter equation gives K Dii. Thus p(m, = p when K* = O (by (7.11) when A* = -Qa*) or when Dii = 0. P ( , > - - P = - *

7.7. Verify the Navier-Stokes-Duhem equations of motion (7.22) fo r a Newtonian fluid and determine the form of the energy equation (7.17) for this fluid if the heat conduction follows the Fourier law (7.20).

Since Dii = vi,i, equation (7.18) may be written aij = -pSii + h*Sijvk,k + p*(vt j + v j J . Thus

o.. i13 3 . = -pj jSi j + X * S i j ~ k , k j + p*(vi,jj + vj,ij) = -P,i (h* + p*)vj,ji + p*vi,jj

and with this expression inserted into (7.16) a direct verification of (7.22) is complete.

Substituting the above equation for qj together with (7.20) into the energy equation (7.17), the result is

P; = [-psij + h*6ijvk,k + + v ~ , ~ ) ] ( v ~ , + vj , i ) /2 - kT,ii + p Z

which reduces to

7.8. Determine the traction force T, acting on the closed surface S which surrounds the volume V of a Newtonian fluid for which the bulk viscosity is zero.

A d s The element of traction is dTi = t i n ' d S and

the total traction force is T i = tin' dS which L A because of the stress principle is Ti = ojinj d s . From Problem 7.4, this becomes i

T i = ( - p ~ i j + 2p*D;)9dS /------ for a zero bulk modulus fluid; and upon application of Gauss' theorem,

Fig. 7-1

168 FLUIDS [CHAP. 7

7.9. In an axisymmetric flow along the x3 axis the velocity is taken as a function of x3 and r where r2 = X : + x2. If the velocity is expressed by v = qê, + v3ê3 where 6, is the unit radial vector, determine the form of the continuity equation.

Equation (5.4) gives the continuity equation in symbolic notation as aplat + V (pv) = 0. 1 d(rpg) i a(Pv3) Here the cylindrical form of the operator V may be used to give V (pv) = --

Inserting this into (5.4) and simplifying, the continuity equation becomes r ar 8x3 '

7.10. In a two-dimensional flow parallel to the X I X Z plane, v3 and dldxa are zero. Determine the Navier-Stokes equations for an incompressible fluid and the form of the continuity equation for this case.

From (7.23) with i = 3, pb, = p,, and when i = 1,2, p'U, = pb, - P,, + P * V , , ~ ~ . The continuity equation (7.15) reduces to v,,, = 0. 4

If body forces were zero and v , = v l ( x l , x,, t) , v , = O , p = p(xl , x,, t ) the necessary equations would be pGl = -dp/axl + a * ( d z ~ l / d ~ : + d2v1/dxi) and dvl/axl = 0.

HYDROSTATICS. STEADY AND IRROTATIONAL FLOW (Sec. 7.4)

7.11. Assuming a i r is an ideal gas whose temperature varies linearly with altitude as T = TO - axg where TO is ground leve1 temperature and X B measures height above the earth, determine the a i r pressure in the atmosphere as a function of x3 under .

hydrostatic conditions.

From (7.6) in this case, p = P R ( T , - CYX,); and from (7.28) with the body force b3 = -g, the gravitational constant, dp/dx3 = - ~ g = -pg/R(T o - ax3). Separating variables and integrating yields ln p = (g /Ra) ln ( T o - CYX,) + ln C where C is a constant of integration. Thus p = C(To - ax3)glRff and if p = po when x3 = O, C = po To-g/Rff and so p = po(l - a ~ 3 / T o ) g / R f f .

7.12. A barotropic fluid having the equation of state p = hpk where h and k are constants is a t rest in a gravity field in the x3 direction. Determine the pressure in the fluid with respect to x3 and po, the pressure a t x3 = 0.

From (7.28), dp/dx3 = -pg, dp/dxl = dpldx, = O. Note that pressure in x , and x z directions is constant in the absence of body forces b1 and bz. Since here p = (p/h)Uk, p-llkdp = -gh-1fkdx3 and integration gives ( k / ( k - l))pck-1)lk = + c. But p = po when x3 = O so that C = ( k / ( k - 1 ) ) ( k - l ) / k . Therefore x3 = (kpol(k - l ) g p o ) ( l - (p /p0) (k - l ) f k ) where po = ( p o / ~ ) ~ ~ k .

po

7.13. A large container filled with an incompressible liquid is accelerated a t a constant rate a = a2êz + a3ê3 in a gravity field which is parallel to the x3 direction. Determine the slope of the free surface of the liquid.

From (7.28), dp ldx , = 0 , dpldx, = pa, and dp/dx3 = -p(g -a3) . Integrating, p = pa,x, + f(x,) and p = -p(g - a,)%, + h(x2) where f and h are arbitrary functions of their arguments. In general, therefore, p = pazx2 - p(g - a,)%, + po where po is the pressure a t the origin of coordinates on the free surface. Since p = po everywhere I

on the free surface, the equation of that surface is x2/x3 = (9 - a3)/az. Fig. 7-2

CHAP. 71 FLUIDS 169

7.14. If a fluid motion is very slow so that higher order terms in the velocity are negligible, a limiting case known as creeping flow results. For this case show that in a steady incompressible flow with zero body forces the pressure is a harmonic function, i.e. v 2 p = o.

For incompressible flow the Navier-Stokes equations (7.23) are

p(avilat + v j v i , = pbi - P , ~ + p*vi, j j

and for creeping flow these linearize to the form

Hence for steady flow with zero body forces, p,i = Taking the divergence of this equation yields pJii = P * ~ i , i j j ; and since the continuity equation for incompressible flow is viai = O, i t follows that here p. , = V 2 p = 0 .

7.15. Express the continuity equation and the Navier-Stokes-Duhem equations in terms of the velocity potential 4 for an irrotational motion.

By (7.35), v i = -@, i so that from (7.15) the continuity equation becomes i - pV2@ = O . Also with v i = -+,i, (7.22) becomes

- P d , i = Pbi - P,i - (h* p*)+,jji - p*@,iii

or -p(a$,i/at + @ , k @ , i k ) = pbi - P,i - (h* + 2p*)$,jji

In symbolic notation this equation is written

- p v ( a + / a t + ( v + ) 2 / 2 ) = pb - v p - (A* + z p * ) v ( v 2 + )

7.16. Deterinine the pressure function P(p) for a barotropic fluid having the equation of state p = X p k where X and k are constants.

From the definition (7.29),

Also since d p = hkpk-1 dp, the same result may be obtained from

PERFECT FLUIDS. BERNOULLI EQUATION. CIRCULATION (Sec. 7.5) 7.17. Derive equation (7.39) by integrating Euler's equation (7.37) along a streamline.

Let dxi be an increment of displacement along a streamline. Taking the scalar product of this increment with (7.37) and integrating gives

Since dxi = d a and P,i dxi = d P the last two terms integrate a t once. Also, along a streamline, dxi = (v iIv) d s where ds is the increment of distance. Thus in the second integral,

v j v i , dxi = v j v i , ( v i / v ) ds = viv i , ( v j / v ) ds = v iv i , dx j = v i dvi

Therefore S v j v i , dxi = v . dvi = +vivi = +v2, and (7.39) is aehieved. s

170 FLUIDS [CHAP. 7

7.18. The barotropic fluid of Problem 7.16 flows from a large closed tank through a thin smooth pipe. If the pressure in the tank is N times the atmospheric pressure, determine the speed of the emerging fluid.

Applying Bernoulli's equation for steady flow between point A , a t rest in fluid of the tank and point B, in the emerging free stream, (739) assumes the form aA + PA + & v i = aB + PB + &vi . But v, = 0, and if gravity is assumed negligible this equation becomes (see Problem 7.16),

Since p B I P A = (pBlpA)-lJk = N-llk the result may be written

7.19. Show that for a barotropic, inviscid fluid with conservative body forces the rate of change of the circulation is zero (Kelvin's theorem).

From (7.M) :, = (Gi dxi + vi dvi) and by (7.37), Gi = -(a + P),i for the case a t hand. Thus § i., = $ (-qi dxi - Pai dxi + vi dvJ = - f (& + dP - d(vz12)) = - f d(Q -i- P - P / 2 ) = 0 , the inte-

grand being a perfect differential.

7.20. Determine the circulation around the square xl = k1, x2 = 1+1, x3 = O (see Fig. 7-3) for the two-dimensional f l o ~ v = ( X I + x ~ ) ~ ~ + (x; - x ~ ) ~ ~ . rh

Using the symbolic form of (7.42) with n̂ = ê3 and v x v = (2x1 - 1) ê3,

1

= s:. Sl (2x1 - 1) dx, dx, = -4

The same result is obtained from (7.41) where Fig. 7-3

r, = $ v - d x

1 - 1 = ( i - x 2 ) d x 2 + ~ - l ( x l + l ) d x l + (1 - x,) dx, + ( x , - 1) dxl = -4

with the integration proceeding counterclockwise from A.

POTENTIAL FLOW. PLANE POTENTIAL FLOW (Sec. 7.6)

7.21. Give the derivation of the gas dynamical equation (7.45) and express this equation in terms of the velocity potential +.

For a steady flow the continuity equation (5.4) becomes pjivi + pvi,i = O and the Euler equation (7.36) becomes pvjvi,j + pIi = O if body forces are neglected. For a barotropic fluid, p = p(p) and so dp = ( a p l a ~ ~ ) ( a ~ ~ l a ~ ) d ~ ; or rearranging, p,i = ( d ~ l d p ) p , ~ = c ~ P , ~ where c is the local velocity of sound. Inserting this into the Euler equation and multiplying by vi gives pvivjvi,j + c2vip,i = 0. From the continuity equation c ~ v ~ ~ , ~ = -C%. 1 , t . = -c2S..v. 21 ~3 . and so (c2Sij - vivj)vi,j = O. In terms of +,i = -vi this becomes (c2Sij - @,i@, j)+,ij = 0.

7.22. Show that the function + = A(-x: - x2 + 2 x i ) satisfies the Laplace equation and determine the resulting velocity components.

Substituting into (7.46) gives -2A - 2A + 4 A = O. From (7.3~9, v , = 2Ax,, v , = 2AxZ, v3 = -4Ax3. Also, by the analysis of Problem 4.7 the streamlines in the x , plane are represented by x i x3 = constant; in the x, plane by xSx3 = constant. Thus the flow is in along the x3 axis against the x,x, plane (fixed wall).

CHAP. 71 FLUIDS 171

7.23. Show that the stream function +(xI, x2) is constant along any streamline.

From (7.48) and the differential equation of a streamline, dxl/vl = dx21v2 (see Problem 4.7), - d ~ , l + , ~ = or +,ldxl + = d+ = O. Thus + = a constant along any streamline.

7.24. Verify that + = A(xl - x t ) is a valid velocity x2 potential and describe the flow field.

For the given +, (7.46) is satisfied identically by 2A - 2A = 0; and from (7.49), v l = -2Axl, v2 = +2Ax2. The streamlines are determined by integrating dxl/xl = -dx2/x2 to give the rectangular hyperbolas xlx2 = C (Fig. 7-4). The equipotential lines A ( X ; - 2;) = C1 form an orthogonal set of rectangular hyperbolas with the streamlines. Finally from (7.50), + = -2Az1x2 + CO and is seen to be constant along the streamlines a s was asserted in Prob- lem 7.23. Fig. 7-4

7.25. A velocity potential is given by + = Axl + Bx1Ir2 where r2 = x: + x:. Determine the stream function + for this flow.

From (7.50), = - G , ~ = 2Bx1z2/+ so that by integrating, + = -Bx21r2 + f (x2) where f ( z2) is an arbitrary function of z,. Differentiating, +,2 = -B(X; - s i ) / + + f'(x2). But from (7.50),

+,2 = = A + B(-z; + z8)/+. Thus f '(x2) = A and f (x2) = Axz + C. Finally then $ = Axz - Bx2/r2 + C.

7.26. Differentiate the complex potential a(z) = Alx to obtain the velocity components.

Here dWdz = -AI22 = -A/(xl + ix2)2 which after some algebra becomes d<P/dz = -A(%; - x;)/+ + i2Ax1x2/+. Thus

v1 = A(%; - si)+ and v2. = 2Ax1x2l.P

Note that since @ = A l z = A(%, - iz2)/r2, + = Axllr2 and JI = -Ax21r2. Also note that

v , = = A ( x ; - X : ) I + and v2 = -9.2 = 2 A ~ ~ x ~ 1 7 . 4

MISCELLANEOUS PROBLEMS 7.27. Derive the one-dimensional continuity equation for the flow of an inviscid incompres-

sible fiuid through a stream tube.

Let V be the volume between arbitrary cross sections A and B of the stream tube shown in Fig. 7-5. In integral form, for this volume (5.2) becomes

Sv V * v dV = O since p is constant here. Convert-

ing by Gauss' theorem, n o v dS = O where "ns ntA L A the outward unit normal to the surface S enclosing V . A Since n i. v on the lateral surface, the integration reduces to Fig. 7-5 " "

A

The velocity is assumed unifonn and perpendicular over SA and S,; and since va = - v B n ~ ,

v A iA dS - vB iB dS = O or v A S A = v B S B = a constant.

172 FLUIDS [CHAP.' 7

7.28. The stress tensor a t a given point for a Newtonian fluid with zero bulk viscosity is -6 2 -1

c = ( 2 -9 ). Determine -1 4 -3

From (7.14), for this fluid p = -uii/3 = 6. Then from ( 7 4 , . -6 2 -1 6 O 2 -1

oi; + 6 8 or -: : ) + = (4 -: :) 7.29. Show that aij and T~~ of (7.3) have the same principal axes.

When written out, (7.8) becomes .oll = -p + T ~ I , u 2 ~ = -p + 722, u33 = -p + T33, u12 = 712 ,

u23 = u13 = 7 1 ~ . For principal directions x: of qj, uTz = = uT3 = O and by the last three equations of (7.39, o; = 7; = O for i # j . Thus Z: are principal axes for T~~ also.

7.30. A dissipation potential Q, is often defined for a Newtonian fluid by the relationship a, = ( K / ~ ) D ; ~ D ~ ~ + p,*D;D;. Show that d a D l d D i j = r i j .

Here d+,laDpq = ( ~ / 2 ) [ D ~ ~ ( a D ~ ; l a D ~ , ) + (aDiilaDpq)Djj] + 25[D, ' , (a~&3~, , ) ] . But aDiilaDpq = Sip S i , = S p q and d ~ , ( ~ / a ~ , , = 6 , a j 4 - S i j Spq/3 so that

d@~laD, , = K D ~ ~ S ~ ~ + 2p*(Dij - 8ijDkk/3)(8ip8jq - SijSpq/3) = K D ~ ~ ~ ~ ~ + eP*(Dpq - SpqDii/3)

Finally since K = A* + 2p*/3, a+,/a~,, = ?,*sPqoii+ 2 p * ~ p q = Tpq

. 7.31. Determine the pressure-density relationship for the ideal gas discussed in Problem 7.11. At s3 = O, p = po and p = po. The ideal gas law (7.6) is here p = pR(To - as3) so that

po = poRT0; and from the pressure elevation relationship p = p,(l - a ~ ~ / T ~ ) g / R a of Problem 7.11, p l p o = (T/To)(g/Ra-i ) . Thus writing p = po(l - as31To)g/Ra in the form p/po = (T/To)g/Ra, i t is seen that T / T o = (plpo) Ralo and so P/Po = (p/po)(l-Ra/g).

7.32. For a barotropic inviscid fluid with conservative body forces show that the material d

derivative of the total vorticity, a L qi dV = viqi dSi.

From (4.54) and the results of Problem 4.33, $ i qi dV =.L (ciik ak + qivi) dSj. But here

ak = -(R + P) , , from (7.37); and by the divergence theorem (1.157),

since the integrand is zero (product of a symmetric and antisymmetric tensor). Hence

7.33. For an incompressible Newtonian fluid moving inside a closed rigid container a t rest,

show that the time rate of change of kinetic energy of the fluid is -,.L* ing zero body forces. q is the magnitude of the vorticity vector.

CHAP. 'i] FLUIDS 173

From Problem 5.27, the time rate of change of kinetic energy of a continuum is

In this problem the first and third integrals are zero; and for a Newtonian fluid by (7.18),

But incompressibility means vili = Dii = O and so

7.34. Show that for a perfect fluid with negligible body forces the rate of change of cir-

culation bc may be given by - ~ ~ ~ ~ ( í / ~ ) , j p , a s i .

From (1.43). bc = $ V i d z , + $ v i dvi; and since d ( f v 2 ) = O, the second integral i s zero.

From (7.36) with bi = 0, Gi = - ~ , ~ l p and so now d

bc = - 6 ( P . ~ ~ I P ) d.. = - i eijk(p, k/p),ini d s

where (7.42) has been used in converting to the surface integral. Differentiating as indicated,

h = -L e i jk [ ( l l~ ) , jp ,k + P , L ~ P ] @i = - L ei jk( lI~) . j ~ , k dSi


7.35. The constitutive equation for an isotropic fluid is given by aij = -paij + Ki jpqDpq with Kihq constants independent of the coordinates. Show that the principal axes of stress and rate of deformation coincide.

7.36. Show that ( l l p ) ( d p l d t ) = O i s a condition for ,-uii/3 = p for a Newtonian fluid.

7.37. Show that the constitutive relations for a Newtonian fluid with zero bulk viscosity may be expressed by the pair of equations sij = 2r*Dii and -oii = 3p.

7.38. Show that in terms of the vorticity vector q the Navier-Stokes equations may be written \I = b - V p l p - "*V x q where v* = r*lP is the ,kinematic viscosity. Show that for irrotational motion this equation reduces to (7.36).

7.39. If a fluid moves radially with the velocity v = v(r, t) where r2 = xixi, show that the equation of a a continuity is A + v 2 + 2 (r%) = 0. at ar r2 ar

7.40. A liquid rotates as a rigid body with constant angular velocity o about the vertical s3 axis. If gravity is the only body force, show that plp - o2r212 + gz3 = constant.

174 FLUIDS [CHAP. 7

7.41. For an ideal gas under isothermal conditions (constant temperature = To), show that p l p , = pIpo = e-(glRTozs) where po and po are the density and pressure a t x3 = 0.

7.42. Show that if body forces are conservative so that bi = -nJi, the Navier-Stokes-Duhem equations for the irrotational motion of a barotropic fluid may be integrated to yield - p ( a @ l a t +- (V@)V2) + pQ + P + (i* + 2@*)V2+ = f ( t ) . (See Problem 7.15.)

7.43. Show that the velocity and vorticity for an inviscid flow having conservative body forces and constant density satisfy the relation Gi - qjvi,j = O. For steady flow of the same fluid, show that

- vjqi,j - qjvi,j-

7.44. For a barotropic fluid having p = p(p) and P(p) defined by (7..29), show that grad P = grad plp.

7.45. Show that the Bernoulli equation (7.39) for steady motion of an ideal gas takes the form (a) Q + p ln (p lp ) + v2/2 = constant, for isothermal flow, ( b ) + (klk - l ) ( p lp ) + v212 = constant, for isentropic flow.

7.46. Show that the velocity field vl = -2x1x2x3/T1, v2 = (x: - x;)x3/r4, v3 = x2/r2 where r2 = x: + x; + $32 is a possible flow for an incompressible fluid. 1s the motion irrotational? Ans. Yes

7.47. If the velocity potential +(z) = + + i$ is an analytic function of the complex variable z = x , + ix2 = rei0 show that in polar coordinates 3 = '9 1 a6 and -- = -*

ar r ae r ae ar '

7.48. If body forces are zero, show that for irrotational potential flow +,ii = v* = ,~*lp is the kinematic viscosity.

Chapter 8

8.1 BASIC CONCEPTS AND DEFINITIONS

Elastic deformations, which were considered in Chapter 6, are characterized by complete recovery to the undeformed configuration upon remova1 of the applied loads. Also, elastic deformations depend solely upon the stress magnitude and not upon the straining or loading history. Any deformational response of a continuum to applied loads, or to environmental changes, that does not obey the constitutive laws of classical elasticity may be spoken of as an inelastic deformation. In particular, irreversible deformations which result from the mechanism of slip, or from dislocations a t the atomic level, and which thereby lead to permanent dimensional changes are known as plastic deformations. Such deformations occur only a t stress intensities above a certain threshold value known as the elastic limit, or yield stress, which is denoted here by o,.

In the theory of plasticity, the primary concerns are with the mathematical formulation of stress-strain relationships suitable for the phenomenological description of plastic deformations, and with the establishment of appropriate yield criteria for predicting the onset of plastic behavior. By contrast, the study of plastic deformation from the microscopic point of view resides in the realm of solid state physics.

The phrase plastic flow is used extensively in plasticity to designate an on-going plastic deformation. However, unlike a fluid flow, such a continuing plastic flow may be related to the amount of deformation as well as the rate of deformation. Indeed, a solid in the "plastic" state can sustain shear stresses even when a t rest.

Many of the basic concepts of plasticity may be introduced in an elementary A O B way by consideration of the stress-strain diagram for a simple one-dimensional tension (or compression) test of some hypo- thetical material as shown by Fig. 8-1. In this plot, u is the nominal stress (force1 original area), whereas the strain r may ou

- -

represent either the conventional (engineering) strain defined here by

e = (L- Lo)/Lo (8.1) - e

where L is the current specimen length and LO the original length, or the natural (logarithmic) strain defined by Fig. 8-1

For small strains, these two measures of strain are very nearly equal as seen by (8.2) and it is often permissible to neglect the difference.

176 PLASTICITY [CHAP. 8

The yield point P, corresponding to the yield stress o,, separates the stress-strain curve of Fig. 8-1 into an elastic range and a plastic range. Unfortunately, the yield point is not always well-defined. It is sometimes taken a t the proportional limit, which lies a t the upper end of the linear portion of the curve. It may also be chosen as the point J, known as Johnson's apparent elastic limit, and defined as that point where the slope of the curve attains 50% of its initial value. Various offset methods are also used to define the yield point, one such being the stress value a t 0.2 per cent permanent strain.

In the initial elastic range, which may be linear or nonlinear, an increase in load causes the stress-strain-state-point to move upward along the curve, and a decrease in load, or unloading causes the point to move downward along the same path. Thus a one-to-one stress-strain relationship exists in the elastic range.

In the plastic range, however, unloading from a point such as B in Fig. 8-1 results in the state point following the path BC which is essentially parallel with the linear elastic portion of the curve. At C, where the stress reaches zero, the permanent plastic strain cP remains. The recoverable elastic strain from B is labeled rE in Fig. 8-1. A reloading from C back to B would follow very closely the path BC but with a rounding a t B, and with a small hysteresis loop resulting from the energy loss in the unloading-reloading cycle. Upon a return to B a load increase is required to cause further deformation, a condition referred to as work hardening, or strain hardening. It is clear therefore that in the plastic range the stress depends upon the entire loading, or strain history of the material.

Although it is recognized that temperature will have a definite influence upon the plastic behavior of a real material, it is customary in much of plasticity to assume isothermal conditions and consider temperature as a parameter. Likewise, it is common practice in traditional plasticity to neglect any effect that rate of loading would have upon the stress- strain curve. Accordingly, plastic deformations are assumed to be time-independent and separate from such phenomena as creep and relaxation.

8 2 IDEALIZED PLASTIC BEHAVIOR Much of the three-dimensional theory for analyzing plastic behavior may be looked

upon as a generalization of certain idealizations of the one-dimensional stress-strain curve of Fig. 8-1. The four most commonly used of these idealized stress-strain diagrams are shown in Fig. 8-2 below, along with a simple mechanical model of each. In the models the displacement of the mass depicts the plastic deformation, and the force F plays the role of stress.

In Fig. 8-2(a), elastic response and work-hardening are missing entirely, whereas in (b), elastic response prior to yield is included but work-hardening is not. In the absence of work-hardening the plastic response is termed perfectly plastic. Representations (a) and (b) are especially useful in studying contained plastic deformation, where large deformations are prohibited. In Fig. 8-2(c), elastic response is omitted and the work-hardening is assumed to be linear. This representation, as well as (a), has been used extensively in analyzing uncontained plastic flow.

The stress-strai,n curves of Fig. 8-2 appear in the context of tension curves. The compression curve for a previously unworked specimen (no history of plastic deformation) is taken as the reflection with respect to the origin of the tension curve. However, if a stress reversal (tension to compression, or vice versa) is carried out with a real material that has been work-hardened, a definite lowering of the yield stress is observed in the second type of loading. This phenomenon is known as the Bauschinger effect, and will be neglected in this book.

CHAP. 81 PLASTICITY 177

L - E

Rough

(a) Rigid- Perfectly Plastic

E F

c Rough

(b) Elastic - Perfectly Plastic

I ......................................

Rough

(c) Rigid-Linear Work Hardening

Rough

(d) Elastic - Linear Work Hardening

Fig. 8-2

8.3 YIELD CONDITIONS. TRESCA AND VON MISES CRITERIA A ?jield condition is essentially a generalization to a three-dimensional state of stress of

the yield stress concept in one dimensional loading. Briefly, the yield condition is a mathematical relationship among the stress components a t a point that must be satisfied for the onset of plastic behavior a t the point. In general, the yield condition may be expressed by the equation

f(.,,) = c, (8.3')

where C, is known as the yield constant, or as is sometimes done by the equation

in which f , (ai j ) is called the yield function.


For an isotropic material the yield condition must be independent of direction and may therefore be expressed as a function of the stress invariants, or alternatively, as a symmetric function of the principal stresses. Thus (8.3) may appear as

Furthermore, experiment indicates that yielding is unaffected by moderate hydrostatic stress so that it is possible to present the yield condition as a function of the stress deviator invariants in the form

f3(IIxD, 1111~) = o (8.6)

Of the numerous yield conditions which have been proposed, two are reasonably simple mathematically and yet accurate enough to be highly useful for the initial yield of isotropic materials. These are:

(1) Tresca yield condition (Maximum Shear Theory)

This condition asserts that yielding occurs when the maximum shear stress reaches the prescribed value C,. Mathematically, the condition is expressed in its simplest form when given in terms of principal stresses. Thus for U, > a,, > a,,,, the Tresca yield condition is given from (2.54b) as

+(aI - uIII) = Cy (a constant) (8.7)

To relate the yield constant C, to the yield stress in simple tension u,, the maximum shear in simple tension a t yielding is observed (by the Mohr's circles of Fig. .8-3(a), for example) to be ~ ~ 1 2 . Therefore when referred to the yield stress in simple tension, Tresca's yield condition becomes

u~ - '-'III = u~ (8.8)

The yield point for a state of stress that is so-called pure shear may also be used as a referente stress in establishing the yield constant C,. Thus if the pure shear yield point value is k, the yield constant C, equals k (again the Mohr's circles clearly show this result, as in Fig. 8-3(b), and the Tresca yield criterion is written in the form

(a) Simple Tension ( 6 ) Pure Shear

Fig. 8-3

(2) von Mises yield condition (Distortion Energy Theory)

This condition asserts that yielding occurs when the second deviator stress invariant attains a specified value. Mathematically, the von Mises yield condition states


which is usually written in terms of the principal stresses as

(aI - uII)2 + (gII - uIII)2 + (vIII - u1)2 = 6cy (8.11)

With reference to the yield stress in simple tension, it is easily shown that (8.11) becomes

(oI - uIJ2 + (cII - aIII)2 f (vIII - a1)2 = 2 ~ ; (8.12)

Also, with respect to the pure shear yield value k, von Mises condition (8.11) appears in the form

(aI - uIJ2 + (rII - uIII)2 f bIII - uI)2 = 6k2 (8.13)

There are severa1 variations for presenting (8.12) and (8.13) when stress components other than the principal stresses are employed.

8.4 STRESS SPACE. THE II-PLANE. YIELD SURFACE

A stress space is established by using stress magnitude as the measure of distance along the coordinate axes. In the Haigh-Westergaard stress space of Fig. 8-4 the coordinate axes are associated with the principal stresses. Every point in this space corresponds to a state of stress, and the position vector of any such point P(u,, oII, uIII) may be resolved into a component OA along the line 0 2 , which makes equal angles with the coordinate axes, and a component OB in the plane (known as the n-plane) which is perpendicular to OZ and passes through the origin. The component along OZ, for which U, = a,, = uIII, represents hydrostatic stress, so that the component in the n-plane represents the deviator portion of the stress state. I t is easily shown that the equation of the n-plane is given by

a, + ali + a,,, = 0 (8.1 4) Fig. 8-4

In stress space, the yield condition (8.5), f2(a,, u,,,olII) = C,, defines a surface, the so-called yield surface. Since the yield conditions are independent of hydrostatic stress, such yield surfaces are general cylinders having their generators parallel to 02. Stress points that lie inside the cylindrical yield surface represent elastic stress states, those which lie on the yield surface represent incipient plastic stress states. The intersection of the yield surface with the R-plane is called the yield curve.

In a true view of the n-plane, looking along OZ toward the origin 0, the principal stress axes appear symmetrically placed 120" apart as shown in Fig. 8-5(a) below. The yield curves for the Tresca and von Mises yield conditions appear in the n-plane as shown in Fig. 8-5!b) and (c) below. In Fig. 8-5(b), these curves are drawn with reference to (8.7) and (8.11), using the yield stress in simple tension as the basis. For this situation, the von Mises circle of radius m3 U, is seen to circumscribe the regular Tresca hexagon. In Fig. 8-5(c), the two yield curves are based upon the yield stress k in pure shear. Here the von Mises circle is inscribed in the Tresca hexagon.


( b )

Fig. 8-5

The location in the II-plane of the projection of an arbitrary stress point P(u,, a,,, o,,,) is straightforward since each of the stress space axes makes cos-I @ with the II-plane. Thus the projected deviatoric components are ( m v I , d2/3uII, moIII). The inverse problem of determining the stress components for an arbitrary point in the n-plane is not unique since the hydrostatic stress component may have any value.

8.5 POST-YIELD BEHAVIOR. ISOTROPIC AND KINEMATIC HARDENING Continued loading after initial yield is reached leads to plastic deformation which may

be accompanied by changes in the yield surface. For an assumed perfectly plastic material the yield surface does not change during plastic deformation and the initial yield condition remains valid. This corresponds to the one-dimensional perfectly plastic case depicted by Fig. 8-2(a). For a strain hardening material, however, plastic deformation is generally accompanied by changes in the yield surface. To account for such changes it is necessary that the yield function f,(oij) of (8.4) be generalized to define subsequent yield surfaces beyond the initial one. A generalization is effected by introduction of the loading function

which depende not only upon the stresses, but also upon the plastic strains r; and the work- hardening characteristics represented by the parameter K. Equation (8.15) defines a loading surface in the sense that f: = O is the yield surface, f: < 0 is a surface in the (elastic) region inside the yield surface and f r > 0, being outside the yield surface, has no meaning.

Differentiating (8.15) by the chain rule of calculus,

Thus with f: = O and (af :Iaoij) doij < 0, unloading is said to occur; with f: = O and (af :/aoij) doij = 0, neutra1 loading occurs; and with f i = O and (af:la~,,) doij > 0, loading occurs. The manner in which the plastic strains c; enter into the function (8.15) when loading occurs is defined by the hardening rules, two especially simple cases of which are described in what follows.

The assumption of kotropic hardening under loading conditions postulates that the yield surface simply increases in size and maintains its original shape. Thus in the II-plane the yield curves for von Mises and Tresca conditions are the concentric circles and regular hexagons shown in Fig. 8-6 below.


( b ) Tresca Hexagons Fig. 8-6

(a) Mises Circles

In kinematic hardening, the initial yield surface is translated to a new location in stress space without change in size or shape. Thus (8.4) defining an initial yield surface is replaced by

f i (oij - a,) = O (8.1 7)

where the aij are coordinates of the center of the new yield surface. If linear hardening is assumed,

Xi. = c;: (8.1 8)

where c is a constant. In a one-dimensional case, the Tresca yield curve would be translated as shown in Fig. 8-7. Fig. 8-7

8.6 PLASTIC STRESS-STRAIN EQUATIONS. PLASTIC POTENTIAL THEORY

Once plastic deformation is initiated, the constitutive equations of elasticity are no longer valid. Because plastic strains depend upon the entire loading history of the material, plastic stress-strain relations very often are given in terms of strain increments - the so-called incremental theories. By neglecting the elastic portion and by assuming that the principal axes of strain increment coincide with the principal stress axes, the Levy-Mises equations relate the total strain increments to the deviatoric stress components through the equations

dcij = sij dA

Here the proportionality factor dX appears in differential form to emphasize that incremental strains are being related to finite stress components. The factor dA may change during loading and is therefore a scalar multiplier and not a fixed constant. Equations (8.19) represent the flow rule for a rigid-perfectly plastic material.

If the strain increment is split into elastic and plastic portions according to

and the plastic strain increments related to the stress deviator components by

the resulting equations are known as the Prandtl-Reuss equations. Equations (8.21) represent the flow rule for an elastic-perfectly plastic material. They provide a relationship between the plastic strain increments and the current stress deviators but do not specify the strain increment magnitudes.


The name plastic potential function is given to that function of the stress components g(u,) for which dg de; = - dA (8.22)

8%

For a so-called stable plastic material such a function exists and is identical to the yield function. Moreover when the yield function f l (u i j ) = IIr,, (8.22) produces the Prandtl- Reuss equations (8.21).

8.7 EQUIVALENT STRESS. EQUIVALENT PLASTIC STRAIN INCREMENT

With regard to the mathematical formulation of strain hardening rules, i t is useful to define the equivalent or efSective stress a,, as

This expression may be written in compact form as

In a similar fashion, the equivalent or effective plastic strain increment de:, is defined by

which may be written compactly in the form

In terms of the equivalent stress and strain increments defined by (8.24) and (8.25) respectively, dA of (8.21 ) becomes

3 de;, dh = -- 2 OEQ

8.8 PLASTIC WORK. STRAIN-HARDENING HYPOTHESES

The rate a t which the stresses do work, or the stress power as i t is called, has been given in (5.32) as uijDij per unit volume. From (4.25), den = D,dt, so that the work increment per unit volume may be written

dW = ali dcij (8.2 8)

and using (8.20) this may be split into

For a plastically incompressible material, the plastic work increment becomes

Furthermore, if the same material obeys the Prandtl-Reuss equations (8.21), the plastic work increment may be expressed as

dWP = a,,dcEQ (8.31)

and (8.21) rewritten in the form

de; = 3 dWP -- S . . 2 a;,


There are two widely considered hypotheses proposed for computing the current yield stress under isotropic strain hardening plastic flow. One, known as the work-hardening hypothesis, assumes that the current yield surface depends only upon the total plastic work done. Thus with the total plastic work given as the integral

the yield criterion may be expressed symbolically by the equation

for which the precise functional form must be determined experimentally. A second hardening hypothesis, known as the strain-hardening hypothesis, assumes that the hardening is a function of the amount of plastic strain. In terms of the total equivalent strain

this hardening rule is expressed symbolically by the equation

f , (aij) = (8.36)

for which the functional form is determined from a uniaxial stress-strain test of the material. For the Mises yield criterion, the hardening rules (8.34) and (8.36) may be shown to be equivalent.

8.9 TOTAL DEFORMATION THEORY

In contrast to the incrernental theory of plastic strain as embodied in the stress-strain increment equations (8.19) and (8.21), the so-called total deformation theory of Hencky relates stress and total strain. The equations take the form

In terms of equivalent stress and strain, the parameter + may be expressed as

3 (P = -- 2 um

where here L& = d Z 5 7 3 so that 3 cEe €P = --

23 2 UEQ

8.10 ELASTOPLASTIC PROBLEMS

Situations in which both elastic and plastic strains of approximately the same order exist in a body under load are usually referred to as elasto-plastic problems. A number of well-known examples of such problems occur in beam theory, torsion o£ shafts and thick- walled tubes and spheres subjected to pressure. In general, the governing equations for the elastic region, the plastic region and the elastic-plastic interface are these:

(a) Elastic region

1. Equilibrium equations (2.23), page 49

2. Stress-strain relations (6.23) or (6.24), page 143

3. Boundary conditions on stress or displacement

4. Compatibility conditions


(b) Plastic region

1. Equilibrium equations (2.23), page 49

2. Stress-strain increment relations (8.21)

3. Yield condition (8.8) or (8.11)

4 . Boundary conditions on plastic boundary when such exists

(c) Elastic-plastic interf ace

1. Continuity conditions on stress and displacement

8.11 ELEMENTARY SLIP LINE THEORY FOR PLANE PLASTIC STRAIN

In unrestricted plastic flow such as occurs in metal-forming processes, i t is often possible to neglect elastic strains and consider the material to be rigid-perfectly plastic. If the flow may be further assumed to be a case of plane strain, the resulting velocity field may be studied using slip line theory.

Taking the X l X 2 plane as the plane of flow, the stress tensor is given in the form

and since elastic strains are neglected, the plastic strain-rate tensor applicable to the situation is . .

'li '12 0 (8.42)

In (8.41) and (8.42) the variables are functions of X I and xz only, and also

where v, are the velocity components.

For the assumed plane strain condition, de,, = 0; and so from the Prandtl-Reuss equations (8.21), the stress V,, is given by

u33 = +(all + nz2) (8.44)

Adopting the standard slip-line notation o,, = -p, and ~((r , , - ~ , , ) ~ / 4 + ( u , , ) ~ = k, the principal stress values of (8.41) are found to be

The principal stress directions are given with respect to the xlx2 axes as shown in Fig. 8-8, where tan 28 = 2ul2/(al1 - u,,).

As was shown in Section 2.11, the maximum shear directions are a t 45" with respect to the principal stress directions. In Fig. 8-8, the maximum shear directions are des- 21

ignated as the a and p directions. From the Fig. 8-8


geometry of this diagram, O = r14 + + so that 1 tan2+ = - - . tan28 (8.46)

and for a given stress field in a plastic flow, two families of curves along the directions of maximum shear a t every point may be established. These curves are called shear lines, or slip lines.

For a small curvilinear element bounded by the two pairs of slip lines shown in Fig. 8-9,

u,, = -p - k sin2+

qZz = -p + k sin 2+

U12 = FC COS2+

and from the equilibrium equations i t may be shown that

p + 2k+ = C, a constant along an a line

p - 2k+ = C, a constant along a ,8 line

Fig. 8-9 Fig. 8-10

With respect to the velocity components, Fig. 8-10 shows that relative to the a and /i' lines,

v, = v, cos + - v, sin + (8.49)

vz = v, sin + + v, cos + For an isotropic material, the principal axes of stress and plastic strain-rate coincide.

Therefore if x, and x, are slip-line directions, ;,, and i,, are zero along the slip-lines so that

(& (v, cos + - v, sin +) S*=. = o

These equations lead to the relationships

dv, - v, d+ = O on a lines (8.52)

dv, + v, d+ = O on /i' lines (8.53)

Finally, for statically determinate problems, the slip line field may be found from (8.48), and using this slip line field, the velocity field may be determined from (8.52) and (8.53).


8.5. Convert the von Mises yield condition (8.10) to its principal stress form as given in (8.11).

From (2.72), - 1 1 ~ ~ = -(slsI1 + SIISIII + sII1s1); and by (2.71), SI = o1 - UM, etc., where

UM = (01 + oI1 + uIII)/3. Hence

Thus (01 - QII )2 + (011 - 0111)~ + (oIII - = 6Cy

8.6. With the rectangular coordinate system OXYZ oriented so that the XY plane coincides with the n-plane and the a,,, axis lies in the YOZ plane (see Fig. 8-13 and 8-4), show that the Mises yield surface intersects the n-plane in the Mises circle of Fig. 8-5(b).

Fig. 8-13

The table of transformation coefficients between the two sets of axes is readily determined to be a s shown above. Therefore

,,I = -x/fi - Y/& + z/fi, o,, = x/fi - Y/& + z /h , QIII = 2 ~ 1 6 + 216

and (8.12) becomes (-fix)z + ( x / h - 3 ~ / f i ) 2 + (x/fi+ 3~/1&)2 = 20.;

which simplifies to the Mises yield circle 3X2 + 3Y2 = 20; of Fig. 8-5(b).

8.7. Using the transformation equations of Problem 8.6, show that (8.14), a, + a,, + u,,, = 0, is the equation of the n-plane.

Substituting into (8.14) the o's of Problem 8.6, a1 + o,, + aII1 = fi Z = O, or Z = O which is the XY plane @-plane).

8.8. For a biaxial state of stress with a,, = O, determine the yield loci for the Mises and Tresca conditions and compare them by a plot in the two-dimensional u,/u, vs. a I I I /~ ,

space. From (8.12) with o,, = 0, the Mises yield condi-

tion becomes 0: - o,o,11 + o;,, = o;

which is the ellipse

( ~ I / U Y ) ~ - ( Q I U I I I I ~ ) f (UIII/"Y)~ = 1 Fig. 8-14


with axes a t 45' in the plot. Likewise, from (8.8) and the companion equations a111 - a,, = <ry,

uI1 - u1 = o,, the Tresca yield condition results in the line segments AB and ED with equations (oI /oy) - (o I I I /oy ) = 21, DC and FA with equations oI,,/uy = f 1, and BC and EF with equations oI/oy = TI, respectively.

8.9. The von Mises yield condition is referred to in Section 8.3 as the Distortion Energy Theory. Show that if the distortion energy per unit volume u,dD, is set equal to the yield constant C, the result is the Mises criterion as given by (8.12).

From Problem 6.26, uTD) is given in terms of the principal stresses by

and for a uniaxial yield situation where o, = oy, oI1 = o I I I = O , uTD) = ~ $ 1 6 ~ . Thus C, = a$/6G

and, as before, the Mises yield condition is expressed by (8.12).

PLASTIC DEFORMATION. STRAIN-HARDENING (Sec. 8.4-8.8)

8.10. Show that the Prandtl-Reuss equations (8.21) imply that principal axes of plastic strain increments coincide with principal stress axes and express the equations in terms of the principal stresses.

From the form of (8.21), when referred to a coordinate system in which the shear stresses are zero, the plastic shear strain increments are seen to be zero also. In the principal axes system,

P P P P (8.21) becomes de;lsI = deII lsII = d e I I I / ~ I I I = dX. Thus deI = (oI - aM)dX, deII = ( o I I - oM)dh, etc., and by subtracting,

P de: - de: - de: - dcFII de:[ - deI

- - - = dX " I - "11 "11 - "111 "111 - " I

8.11. For the case of plastic plane strain with r,, = O, de,, = O and u,, = 0, show that the Levy-Mises equations (8.19) lead to the conclasion that the Tresca and Mises yield conditions (when related to pure shear yield stress k) are identical.

Here (8.19) becomes dell = (2õ11 - u3,) dh/3, dez2 = -(oll + dh/3, O = 203, - 011. Thus in the absence of shear stresses, C , = u l l , ull = 03, = uI1/2, ( r I I 1 = O = az2. Then from (8.9) the Tresca yield condition is o, - a,,, = o,, = 2k. Also, from (8.19) for this case, Mises condition becomes (ul1/2)2 + (-ul1/2)2 + (-ul1)2 = 6k2 or oS1 = 4k2 and oll = 2k.

8.12. Show that the Prandtl-Reuss equations imply equality of the Lode variable p (see Problem 8.3) and v = ( 2 d 4 - dc; - d r g I ) l ( d r ; - d&).

From equations (8.21),

8.13. Writing 111, = sijsij/2, show that ~ I I z , / ~ u ~ ~ = sij.

Here a I I I /aopq = ( a ~ , ~ l a ~ ~ , ) ~ , ~ where = a(oij - 6 i i ~ k k / 3 ) / a ~ p q = O i p S j q - ~ ~ ~ ~ ~ ~ l 3 . Thus D

a I I x D / a ~ p q = ( S i p S j q - SijSpq/3)sij = spq since sii = I I D = 0.


8.14. Show that when the plastic potential function g(aii) = IIr,, the plastic potential equations (8.22) become the Prandtl-Reuss equations.

The proof follows directly from the result of Problem 8.13, since aglaa, = sij in this case and (8.22) reduce to (8.21).

8.15. Expand (8.24) to show that the equivalent stress a,, may be written in the form of (8.23).

From equation (8.24),

ogQ = 3siisij/2 = 3(aij - Sijupp/3)(aij - Sijaqq/3)/2 = (3aiiuij - uiioii)/2

Expanding this gives

[3(all + ai2 + 033 + 6(aS2 + 0223 + ail ) - (<r l l + a22 + ~ ~ ~ ) 2 ] / 2

- [2(uli 4 2 + o33 - a11cz2 - 0 2 2 ~ 3 3 - "33ul1) 6(a:2 + ai3 + ) ] /2 - - [(ali - ~ 2 2 ) ~ + (o22 - 033)' f (a33 - ~ 1 1 ) ~ + 6(oS2 + c i3 + uil )]I2

which confirms (8.23).

8.16. In plastic potential theory the plastic strain increment vector is normal to the loading (yield) surface a t a regular point. If [N,, N,, N3] are direction numbers of the normal to the yield surface f , (a i j ) , show that drp l s , = d~;ls , , = d ~ ~ ~ l s ~ ~ ~ under the Mises yield condition and flow law.

The condition of normality is expressed by N = grad f l which requires N l l ( a f l l a ~ I ) = N2/ (d f l /doI I ) = N31(afllauIII) for the Mises case where f l = (aI - aII)2 + (o I I - o I I I ) 2 + ( a I I I - oJ2 - 2a; = O. Here af1/aaI = 2(2u1 - a,, - a I I I ) = 6s,, etc., and since the plastic strain increment vector

P is along the normal i t follows that deF/sI = deII /sII = d e g I / s I I I .

8.17. Determine the plastic strain increment ratios for (a) simple tension with a,, = a,,

( b ) biaxial stress with V,, = -~ , / f i , uZ2 = a Y / f i , u~~ = a12 = u~~ = u13 = O, ( C ) pure shear with V,, = ~ , / f l . (a) Here all = a1 = a,, aI1 = 'aII I = O and s1 = 2ay/3, s I I = s I I I = -oy/3. Thus from Problem

8.16, depl2 = -degl l = -deEI/ l .

(b) Here o1 = a y / f i , aI1 = O , a I I I = - a y / f i and S I = o y / d , S I I = 0, S I I I = - u y / d . Thus

deF/ l = -degI / l and the third term is omitted since i t is usually understood in the theory that if the denominator is zero the numerator will be zero too.

8.18. Determine the plastic. work increment dWP and the equivalent plastic strain increment de;, for the biaxial stress state a,, = - ~ ~ / f i , u~~ = uy/\/3, = (rl2 = = a,, = O if plastic deformation is controlled so that de: = C, a constant.

In principal-axis form, (8.30) becomes d W P = o, de; + aI1 de: + a,,, de;,; and for the stress

state given, Problem 8.17 shows that de; = -deEI, de: = 0 ; hence

dWP = -u,c/\/~ ( a y / f i ) ( - c ) = - 2 c a y / d

From (8.25), = {2[(dep - deg)2 + (de: - degI)2 + - dep)21}112/3


8.19. Verify (8.32) by showing that for a Prandtl-Reuss material the plastic work increment is dWP = o,, d ~ ; , as given in (8.31).

From (8.30), dWP = sijsijdh for a Prandtl-Reuss material satisfying (8.21). But from (8.27), P dh = 3 deEQ/2UEQ for such a material and so dWP = ( 3 ~ ~ ~ 8 ~ ~ / 2 ) ( d e ; ~ / u ~ ~ ) which because of the

definition (8.24) gives dWP = UEQ &EQ. Thus de:, = dWP/uEQ here and ( 832 ) follows directly from (8.21).

8.20. For a material obeying the Mises yield condition, the equivalent stress a,, may be taken as the yield function in the hardening rules (8.36) and (8.36). Show that in this case uEQFr = H' where Fr and Hr are the derivatives of the hardening functions with respect to their respective arguments.

Here (8.34) becomes U E Q = F(WP) and so doEQ = F' dWP. Likewise (8.36) is given here by OEQ = H(€;,) and so daEQ = H' de;,. Thus F' dWP = H' de;Q; and since from (8.31) (or Problem 8.19) dWP = UEQ i t follows a t once that uEQF1 = H'.

TOTAL DEFORMATION THEORY (Sec. 8.9) 8.21. The Hencky total deformation theory may be represented through the equations

C. t j = C ; + C ; with C ; = e; + 6i j~kk/3 = (sij/2)G + 6,,(1 - 2v )okk /3E and C; = +sij. Show that these equations are equivalent to (8.37) and (8.38).

The equation e; = QSij implies e: = O so that €5 = e; = +vij, and from eij = E: + €5 i t fol-

lows that here yj = e:. From the same equation, eij + Sijekk/3 = e: + sijefk/3 + e; which reduces

to eij = e: + e: = (Q + * ~ ) s ~ ~ , (8.37). ~ i s o from eii = e:, eii = ( 1 - 2v)Ukk/E, (8.38).

8.22. Verify that the Hencky parameter 4 may be expressed as given in (8.39).

Squaring and adding the components in the equation e: = @sij of Problem 8.21 gives e; €5 = +2sijsij or Q = 4-luEQ which when multiplied on each side by 213 becomes

ELASTOPLASTIC PROBLEMS (Sec. 8.10)

8.23. An elastic-perfectly plastic rectangular beam is loaded steadily in pure bending. Using simple beam theory, determine the end moments M for which the remaining elastic core extends from -a to a as shown in Fig. 8-15.

Fig. 8-15

Here the only nonzero stress is the bending stress ull. In the elastic portion of the beam (-a < x , < a), ull = Ecll = E x d R where R is the radius of curvature and E is Young's modulus. In the plastic portion, ull = uy. Thus

where uy = EaIR, the stress co?dition a t the elastic-plastic interface, has been used. From the result obtained, M = 2bc2uy/3 a t first yield (when a = c) , and M = bczu, for the fully-plastic beam (when a = 0).


8.24. Determine the moment for a beam loaded as in Problem 8.23 if the material is a piecewise linear hardening material for which a, , = a, + A(€,, - u Y / E ) after yield.

The stress distribution for this beam is shown in Fig. 8-16. Again cl l = x2/R and so

or using oy = Ea/R as in Problem 8.23, 1 2 3

M e2boy(l - AIE) + 2c3bA/3R + ~ u ; R ~ ( A / E - 1)/3E2 Fig. 8-16

8.25. An elastic-perfectly plastic circular shaft of radius c is twisted by end torques T as shown in Fig. 8-17. Determine the torque for which an inner elastic core of radius a remains.

The shear stress u12 is given here by alz = krla for O 6 r 6 a, and by o,, - k for a 5 r 6 e where k is the 1

yield stress of the material in shear. Thus

(k+/a) dr + 2a kr2 dr = S.' ? q c 3 - 3 3 a /4)

Therefore the torque a t first yield is T1 = ak~3 /2 when a = c; and for the fully plastic condition, T, = 2ake3/3 = 4T1/3 when a = 0. Fig. 8-17

A thick spherical shell of the dimensions shown in Fig. 8-18 is subjected to an increasing pressure po. Using the Mises yield condition, determine the pressure a t which first yield occurs.

Because of symmetry of loading the principal stresses a re the spherical components o(%%) = o1 = U I I , o(,r) = 0111. Thus the Mises yield condition (8.12) becomes o(%%) -o( , , ) = UY. The elastic stress components may be slíown to be

Therefore ay = 3b3p0/2r3(b3/a3 - 1) and po = 2ay(l - a3/b3)/3 a t first yield which occurs a t the inner radius a. Fig. 8-18

SLIP LINE THEORY (Sec. 8 .11 )

8.27. Verify directly the principal stress values (8.45) for the stress tensor ( 8 . 41 ) with = ( a l i + a, , ) /2 as given in ( 8 . 44 ) .

The principal stress values a re found from the determinant equation (2.37) which here becomes

Expanding by the third column,


The roots of this equation a r e clearly o = -p and o = &(o11 + oZ2) 2 d i ( o l l + ~ 2 2 ) ~ + = -P k .

8.28. Using the condition that the yield stress in shear k is constant, combine (8.47) with the equilibrium equations and integrate to prove (8.48).

From the equilibrium equations aall/dxl + ao121axz = O and do12/ax1 + aa22/a~2 = O which a r e valid here, (8.47) yields

-aplaxl - k(2 cos 2g)(aglax1) + k(-2 sin 2g)(agldx,) = O

and -k(2 sin 2g)(ag/axl) - aplax, -I k(2 cos 2g)(a+lax2) = O

If x1 is along a n a line and x, along a ,L3 line, @ = O and these equations become -dplax, - 2k(dglaxl) = O along the a line, -aplaxz + 2k(agldx2) = O along the ,L3 line. Integrating directly, p + 2kg = C1 on the a line, p - 2kg = Cz on the line.

8.29. I n the frictionless extrusion through a square die causing a fifty per cent reduction, the centered fan region is composed of straight radial ,8 lines and circular a lines as shown in Fig. 8-19. Determine the velocity components along these slip lines in terms of the approach velocity U and the polar coordinates r and O.

Fig. 8-19

Along the s t raight ,L3 lines, d@ = 0; and from (8.53), dv2 = O o r v2 = constant. , From the normal velocity continuity along BC, the constant here must be U cose and so v2 = U cose. Along the circular a lines, dg = de; and from (8.52),

0

= S-,,, U cos e de = U(sin e + l/fi)


8.30. Show that the vonSMises yield condition may be expressed in terms of the octahedral shear stress uoCt (see Problem 2.22) by ao,, = f i ~ , / 3 .

In terms of principal stresses 3o,,, = d ( o I - o I I ) 2 + (o I I - u I I I ) 2 + ( o I I I - uI)2 (Problem 2.22)

and so 9oOCt = (oI - oI I )2 + ( o I I - a I I I ) 2 + ( o I I I - ar), = 2u$ i n agreement with (8.12).

8.31. Show that equation (8.13) for Mises yield condition may be written as

sS + s:, + s;,, = 2k2

From (2.71), o1 = S I + U M , etc., and so (8.13) a t once becomes

( S I - SI I ) ' + ( S I I - ~ 1 1 1 ) ' + ( S I I I - ~ 1 ) ~ = 6k2

Expanding and rearranging, this may be written s: + sSI + sSIr - ( s I + s I I + s I I I ) 2 / 3 = 21~2. But sI + sl, + s,,, = IpD O and the required equation follows.

C H A P . 81 PLASTICITY 193

8.32. At what value of the Lode parameter p = (2u1 , - U , - u, , , ) / (u , - u I I I ) are the Tresca and Mises yield conditions identical?

From the definition of p, o,, = ( O , + o I I I ) / 2 + p(o, - o I I 1 ) / 2 which when substituted into the Mises yield condition (8.12) gives after some algebra (see Problem 8.42) o, - U I I , = 2 o y / m . Tresca's yield condition, equation (8.8), is o, - C,,, = uP Thus when /I = 1 the two are identical. When a,, = o,, p = 1 which is sometimes called a cylindrical state of stress.

For the state of stress where u and T are constants, determine the

yield condition according to Tresca and von Mises criteria.

The principal stresses here are readily shown to be a1 = o + 7 , o,, = u, = u - r. Thus from (8.8), the Tresca condition V , - O,,, = UY gives 27 = uy. From (8.12) the Mises condition gives 7 = cy/*. Note that in each case yielding depends on 7, not on c , i.e. yielding is independent of hydrostatic stress.

8.34. Show that the Prandtl-Reuss equations imply incompressible plastic deformation and write the equations in terms of actual stresses.

P From (8.21), deii = sii dh = O since sii = IXD - O and the incompressibility condition dez = O

is attained. In terrns of stresses, dez = (oij - Siiukk/3) d ~ . ~ h u s de:, = (2/3)[ul i - (oz2 + q 3 ) / 2 ] d ~ , etc., for the normal components and de:, = o, , dA, etc., for the shear components.

8.35. Using the von Mises yield condition, show that in the U-plane the deviator stress components at yield are

' S, = [ - 2 ~ , cos (9 - ~/6) ] /3 , s,, = [2uY cos (9 + ~/6)]/3, s,,, = ( 2 u y sin $)I3

where 9 = tan-I Y / X in the notation of Problem 8.6.

The radius of the Mises yield circle is &%ry so that by definition X = m o y cose,

Y = m o y sin e a t yield. From the transformation table given in Problem 8.6 together with

o, = s , + UM, etc., the equations s , - s , , = - f i ~ = - ( 2 / f i ) a y cose and S I -i - 2~111 = -6 Y = -20y sin e are obtained. Also, in the n-plane, sI + s I I + s I I1 = O . Solving these three equations simultaneously yields the desired expressions, as the student should verify.

8.36. An elastic-perfectly plastic, incompressible material is loaded in plane strain between rigid pIates so that u,, = O and C, , = O (Fig. 8-20). Use Mises yield condition to determine the loading stress CT,, a t first yield, and the accompanying strain C , , .

The elastic stress-strain equation

Ec33 = o33 - v(oi i+ 0 2 2 )

reduces here to o,, = vu,,. Thus the principal stresses are o, = O, a,, = -vull, u I I 1 = - c l1 ; and by (8.12) we have

(vo1,)2 + ('Tll(l - v ) ) , + ( - u , ~ ) ~ = 20;

from whieh a l i = - u v / d ~ (cornpressive) a t A, yield. Likewise, from Eell = o l l - + u3R) we

see that here = -ay ( l - V Z ) / E ~ - a t yield. Fig. 8-20


8.37. An elastic-perfectly plastic rectangular beam is loaded in pure bending until fully plastic. Deter- mine the residual stress in the 6 r l . ) q-J _L beam upon removal of the bending k b - 4 m0ment M. Fig. 8-21

For the fully plastic condition, the moment is (see Problem 8.23) M = bc20y. This moment would cause an elastic stress having a = Mel1 = 3oy/2 a t the extreme fibers, since I = 26~313. Thus remova1 of M is equivalent to applying a corresponding negative elastic stress which results in the residual stress shown in Fig. 8-22.

fully plastic

3 0 ~ 1 2 oy/2

+ x - - negative residual elastic stress

Fig. 8-22

8.38. A thick-walled cylindrical tube of the dimensions shown in Fig. 8-23 is subjected to an interna1 pressure p , . Determine the value of p, a t first yield if the ends of the tube are closed. Assume (a) von Mises and ( b ) Tresca's yield conditions.

Fig. 8-23 Fig. 8-24

The cylindrical stress components (Fig. 8-24) are principal stresses and for the elastic analysis may be shown to be o(,,) = -pi(b2/r2 - 1) /Q , o(eo) = pi(b2/r2 + l ) / Q , o(zL) = pilQ where Q = (b2Ia2 - 1) .

( a ) Here Mises yield condition is

(a(, , ) - o ( 8 0 ) ) ~ + (o(e,) - ~ ( z z ) ) ~ + (O( , , ) - = 20; or pf b41+ = Q20;/3

The maximum stress is a t r = a , and a t first yield pi = ( o y / f i ) ( l - a2/b2) .

(b) For the Tresca yield condition, o(oe) - o(,,) = oy since o,, is the intermediate principal stress. Thus 2pibz/r2 = Qoy and now a t r = a , pi = (oy /2 ) (1 - a2/b2) a t first yield.

CHAP. 81 PLASTICITY

Supplementary Problems A one-dimensional stress-strain law is given by o = Ken where K and n are constants and E is true strain. Show that the maximum load occurs a t e = n.

Rework Problem 8.4 using the yield stress in shear k in place of uy in the Mises and Tresca yield conditions. Ans. Mises: ( o l f i k)2 + (rlk)2 = 1; Tresca: ( ~ / 2 k ) 2 + ( ~ / k ) 2 = 1

Making use of the material presented in Problem 8.6, verify the geometry of Fig. 8-5(c). .

From the definition of Lode's parameter p (see Problem 8.3) and the Mises yield condition, show that o, - u,,, = 2 o y / ~ ~ .

In the 11-plane where s = tan-1 Y I X with X and Y defined in Problem 8.6, show that p = -6 tan 0 .

Show that the invariants of the deviator stress IrID = sijsij/2 and IIIED = sijsjkski/3 may be written

= (s; + s;, + s:,,)/2 and III,, = ( s i + s;, + si,,)/3 respectively.

Show that von Mises yield condition may be written in the form

(011 - ~ 2 2 ) ~ (o22 - ~ 3 3 ) ~ 4- (a33 - 011)2 + 6(a:2 + ui3 + m i l ) = 6k2

Following the procedure of Problem 8.17, determine the plastic strain increment ratios for (a) biaxial tension with ul1 = o,, = uy, ( b ) tension-torsion with oll = oy/2, u12 = oY/2.

Ans . (a) deFl = dei, = -der3 12 ( b ) deTl /2 = -dei2 = -der3 = ddeL/3

Verify the following equivalent expressions for the effective plastic strain increment degQ and note that in each case degQ = deFl for uniaxial tension ull.

(a ) deeQ [(derl )2 + (de; )2 + (de; )2 + 2(dcr2 )2 + 2(de& )2 + 2(de& )2]1/2

( b ) de:, = (\/Sj3)[(der1 - dei2)2 + (de; - ds[3)2 + (dei3 - deTl)2 + 6 ( d ~ f , ) ~ + + 6(de&)2]112

A thin-walled elastic-perfectly plastic tube is loaded in combined tension-torsion. An axial stress o = uy/2 is developed first and maintained constant while the shear stress T is steadily increased from zero. At what value of r will yielding first occur according to the Mises condition? Ans . 7 = uy/2

I x9

The beam of triangular cross section shown in Fig. 8-25 is subjected to pure bending. Determine the location of the neutra1 axis (a distance b from top) of the beam when fully plastic. Ans. b = h l f i

Show that the stress tensor (8.41) becomes - 2 h ---4

Fig. 8-25

\ 0 o - P /

when referred to the axes rotated about x3 by an angle O

in Fig. 8-8.

8.51. A centered fan of a circle arcs and P radii includes an angle of 30° as shown in Fig. 8-26. The pressure on A B is k. Determine the pressure on AC. Ans. p = k(1 + a/3) Fig. 8-26

Linear Viscoeiasticity

9.1 LINEAR VISCOELASTIC BEHAVIOR

Elastic solids and viscous fluids differ widely in their deformational characteristics. Elastically deformed bodies return to a natural or undeformed state upon remova1 of applied loads. Viscous fluids, however, possess no tendency a t a11 for deformational recovery. Also, elastic stress is related directly to deformation whereas stress in a viscous fluid depends (except for the hydrostatic component) upon rate of deformation.

Material behavior which incorporates a blend of both elastic and viscous characteristics is referred to as viscoelastic behavior. The elastic (Hookean) solid and viscous (Newtonian) fluid represent opposite endpoints of a wide spectrum of viscoelastic behavior. Although viscoelastic materials are temperature sensitive, the discussion which follows is restricted to isothermal conditions and temperature enters the equations only as a parameter.

9.2 SIMPLE VISCOELASTIC MODELS

Linear viscoelasticity may be introduced conveniently from a one-dimensional viewpoint through a discussion of mechanical models which portray the deformational response of various viscoelastic materials. The mechanical elements of such models are the massless linear spring with spring constant G, and the viscous dashpot having a viscosity constant 7. As shown in Fig. 9-1, the force across the spring u is related to its elongation r by

a = Ge (9.1)

and the analogous equation for the dashpot is given by

a = q e

where : = deldt. The models are given more generality and dimensional effects removed by referring to u as stress and r a s strain, thereby putting these quantities on a per unit basis.

(a) Linear Spring ( b ) Viscous Dashpot Fig. 9-1

CHAP. 91 LINEAR VISCOELASTICITY 197

The Maxwell model in viscoelasticity is the combination of a spring and dashpot in series as shown by Fig. 9-2(a). The Kelvin or Voigt model is the parallel arrangement shown in Fig. 9-2(b). The stress-strain relation (actually involving rates also) for the Maxwell model is

u u - + - = ; ' G 7

(9.3)

and for the Kelvin model is u = Ge+rl;

These equations are essentially one-dimensional viscoelastic constitutive equations. I t is helpful to write them in operator form by use of the linear dijferential t i m e operator d, = a/dt. Thus (9.3) becomes

{a,/G + l/7)u = {a,), (9.5)

and (9.4) becomes u = {G + qat)€

with the appropriate operators enclosed by parentheses.

(a) Maxwell ( b ) Kelvin Fig. 9-2

The simple Maxwell and Kelvin models are not adequate to completely represent the behavior of real materials. More complicated models afford a greater flexibility in por- traying the response of actual materials. A three-parameter model constructed from two springs and one dashpot, and known as the standard linear solid is shown in Fig. 9-3(a). A three-parameter viscous model consisting of two dashpots and one spring is shown in Fig. 9-3(b). I t should be remarked that from the point of view of the form of their constitutive equations a Maxwell unit in parallel with a spring is analogous to the standard linear solid of Fig. 9-3(a), and a Maxwell unit in parallel with a dashpot is analogous to the viscous model of Fig. 9-3(b).

G9 G ,

(a) Standard Linear Solid ( b ) Three-parameter Viscous Model Fig. 9-3

A four-parameter model consisting of two springs and two dashpots may be regarded as a Maxwell unit in series with a Kelvin unit as illustrated in Fig. 9-4 below. Severa1 equivalent forms of this model exist. The four-parameter model is capable of a11 three of the basic viscoelastic response patterns. Thus it incorporates "instantaneous elastic re-

198 LINEAR VISCOELASTICITY [CHAP. 9

sponse" because of the free spring G,, "viscous flow" because of the free dashpot v,, and, finally, "delayed elastic response" from the Kelvin unit.

Fig. 9-4

The stress-strain equation for any of the three or four parameter models is of the general form

p2a + pl& + pou = q2Y+ ql; + Q o C (9.7)

where the p,'s and qi's are coefficients made up of combinations of the G's and ?'s, and depend upon the specific arrangement of the elements in the model. In operator form, (9.7) is written

{p2a: + plat + p0}u = {q2a: + qlat + qo)€ (9-8)

9.3 GENERALIZED MODELS. LINEAR DIFFERENTIAL OPERATOR EQUATION

The generalized Kelvin model consists of a sequence of Kelvin units arranged in series as depicted by Fig. 9-5. The total strain of this model is equal to the sum of the individual Kelvin unit strains. Thus in operator form the constitutive equation is, by (9.6),

Fig. 9-5

Similarly, a sequence of Maxwell units in parallel as shown in Fig. 9-6 is called a generalized Maxwell model. Here the total stress is the resultant of the stresses across each unit; and so from t9.5),

t 0

Fig. 9-6


For specific models, (9.9) and (9.10) result in equations of the form

p O u + p l ~ + p 2 a + . . . = qOc+ql;+q2;.+ . . .

which may be expressed compactly by

This linear diflerential operator equation may be written symbolically as

{P>u = { & } c (9.13)

where the operators { P ) and {Q) are defined by m ai

{P) = C p.- i=O 2dta' i=O

9.4 CREEP AND RELAXATION

The two basic experiments of viscoelasticity are the creep and relaxation tests. These tests may be performed as one-dimensional tension (compression) tests or as simple shear tests. The creep experiment consists of instantaneously subjecting a viscoelastic specimen to a stress U , and maintaining the stress constant thereafter while measuring the strain (creep response) as a function of time. In the relaxation experiment an instantaneous strain e0 is imposed and maintained on the specimen while measuring the stress (relaxation) as a function of time. Mathematically, the creep and relaxation loadings are expressed in terms of the unit step function [U( t - t,)], defined by

and shown in Fig. 9-7. 1 For the creep loading, 1 t 1 t

0 = ",[u(t)l ( 9 .I 6 ) Fig. 9-7

where [U( t ) ] represents the unit step function applied a t time t , = O. The creep response of a Kelvin material is determined by solving the differential equation

which results from the introduction of (9.16) into (9.4). Here T = rllG is called the retarda- t ion time. For any continuous function of time f ( t ) , it may be shown that with t' as the variable of integration,

f ( t ' )[U(tr - t,)] dt' = [U( t - t,)] St f (t') dt' (9.1 8 ) t i

by means of which (9.17) may be integrated to yield the Kelvin creep response

"o .(t) = - ( 1 - e-t17)[U(t)] G (9.19)

The creep loading, together with the creep response for the Kelvin and Maxwell models (materials) is shown in Fig. 9-8 below.


The stress relaxation which occurs in a Maxwell material upon application of the strain

uo

is given by the solution of the differential equation

1 ,,

-

*

obtained by inserting the time derivative of (9.20) into (9.3). Here [ S ( t ) ] = d [ U ( t ) ] l d t is a singularity function called the unit impulse funct ion, or Dirac delta funct ion. By definition,

t t

(a) Creep Loading ( b ) Creep Response

Fig. 9-8

This function is zero everywhere except a t t = t , where i t is said to have an indeterminate spike. For a continuous function f ( t ) , i t may be shown that when t > t,,

with the help of which (9.21) may be integrated to give the Maxwell stress relaxation

The stress relaxation for a Kelvin material is given directly by inserting ; = c o [ S ( t ) ] into (9.4) to yield

d t ) = GcOIU(t)l + V E ~ [ ~ ( ~ ) I (9.25)

The delta function in (9.25) indicates that i t would require an infinite stress to produce an instantaneous finite strain in a Kelvin body.

9.5 CREEP FUNCTION. RELAXATION FUNCTION. HEREDITARY INTEGRALS

The creep response of any material (model) to the creep loading u = u0[U( t ) ] may be written in the form

~ ( t ) = *( t )u , (9.26)

where q ( t ) is known as the c r e e p f z~~zct ion. For example, the creep function for the generalized Kelvin model of Fig. 9-5 is determined from (9.19) to be

Y

@(t ) = J , ( 1 - e t"l)[ú'(t)] (9.27) 1 - 1


where Ji = l/Gi is called the compliance. If the number of Kelvin units increases indefinitely so that N + m in such a way that the finite set of constants ( r , Ji) may be replaced by the continuous compliance function J(T), the Kelvin creep function becomes

The function J(T) is referred to as the "distribution of retardation times", or retardation spectrum.

In analogy with the creep response, the stress relaxation for any model subjected to the strain c = co[U(t)] may be written in the form

where +(t) is called the relaxation function. For the generalized Maxwell model of Fig. 9-6, the relaxation function is determined from (9.24) as

Here, as N + .o the function G(T) replaces the set of constants (G, T ~ ) and the relaxation function is defined by

+(t) = Sm G(T)~-"' dr (9.31)

The function G(7) is known as the "distribution of relaxation times", or relaxation spectrum.

In linear viscoelasticity, the superposition principle is valid. Thus the total "effect" of a sum of "causes" is equal to the sum of the "effects" of each of the "causes". Accord- ingly, if the stepped stress history of Fig. 9-9(a) is applied to a material for which the creep function is ~k(t), the creep response will be

Therefore the arbitrary stress history u = ~ ( t ) of Fig. 9-9(b) may be analyzed as an infinity of step loadings, each of magnitude da and the creep response given by the superposition integral

~ ( t ) = Jt - m $p 9( t - t') dt'

Such integrals are often referred to as hereditary integrals since the strain a t any time is seen to depend upon the entire stress history.

(a, Fig. 9-9

"+ gdt' "3

-v - - - - - - "

"1

"o ' --------------- I I I

I I

-- -- ---- ---- /e I I

I I I I I I

tl t2 t 3 t t' t' + dt' t -


For a material initially "dead", i.e. completely free of stress and strain a t time zero, the lower limit in (9.33) may be replaced by zero and the creep response expressed as

c(t) = lt d"(t3 qr(t - t') dt' dt'

Furthermore, if the stress loading involves a step discontinuity of magnitude o, a t t = 0, (9.34) is usually written in the form

Following similar arguments as above, the stress as a function of time may be represented through a superposition integral involving the strain history e(t) and the relaxation function +(t). In analogy with (9.33) the stress is given by

.(t) = Jm g +(t - t') dtf

and with regard to a material that is "dead" a t t = 0, the integrals comparable to (9.34) and (9.35) are respectively

.(t) = Xt%+( t - t ' ) dt' (9.37)

and

Since either the creep integral (9.34) or the relaxation integral (9.37) may be used to specify the viscoelastic characteristics of a given material, it follows that some relationship must exist between the creep function +(t) and the relaxation function +(t). Such a relationship is not easily determined in general, but using the Laplace transform definition

it is possible to show that the transforms G(s ) and $(t) are related by the equation

where s is the transform parameter.

9.6 COMPLEX MODULI AND COMPLIANCES

If a linearly viscoelastic test specimen is subjected to a one-dimensional (tensile or shear) stress loading o = U, sin wt, the resulting steady state strain will be c = E, sin (ot - a), a sinusoidal response of the same frequency but out of phase with the stress by the lag angle 6. The stress and strain for this situation may be presented graphically by the vertical projections of the constant magnitude vectors rotating a t a constant angular velocity w as shown in Fig. 9-10 below.

The ratios of the stress and strain amplitudes define the absolute dynamic modulus ao/%, and the absolute dynamic compliance C ~ / U , . In addition, the in-phase and out-of-phase components of the stress and strain rotating vectors of Fig. 9-10(a) are used to define

(a) the storage modulus

CHAP. 91 LINEAR VISCOELASTICITY

( b ) the loss modulus

(c ) the storage compliance

(d) the loss compliance

o0 sin S G, = -

€0

c , cos S J , = -

"o

c, sin 6 J , = -

,o

Fig. 9-10

A generalization of the above description of viscoelastic behavior is achieved by expressing the stress in complex form as

,* , eiwt o (9.41)

and the resulting strain also in complex form as = c e i ( w t - 6 )

o

From (9.41) and (9.42) the complex modulus G*(io) is defined as the complex quantity

V*/€* = G*(iw) = (o O / r o )eis = G , + iG, (9.43)

whose real part is the storage modulus and whose imaginary part is the loss modulus. Similarly, the complex compliance is defined as

e*/,* = J*(iw) = (Eoluo)e-i6 (9 4 4 )

= J , - U,

where the real part is the storage compliance and the imaginary part the negative G1

of the loss compliance. In Fig. 9-11 the vector diagrams of G* and J* are shown. Note that G* = 1/J*. Fig. 9-11

9.7 THREE DIMENSIONAL THEORY In developing the three dimensional theory of linear viscoelasticity, it is customary to

consider separately viscoelastic behavior under conditions of so-called pure shear and pure dilatation. Thus distortional and volumetric effects are prescribed independently, and subsequently combined to provide a general theory. Mathematically, this is handled by resolving the stress and strain tensors into their deviatoric and spherical parts, for each of which viscoelastic constitutive relations are then written. The stress tensor decomposition is given by (2.70) as


and the small strain tensor by (3.98) as

eij = e,, + Sijrkk/3 (9.46)

Using the notation of these equations, the three dimensional generalization of the viscoelastic constitutive equation (9.13) in differential operator form is written by the combination

and {M)uii = 3{N)eii (9.47b)

where { P ) , {Q), { M ) and { N ) are operators of the form (9.14) and the numerical factors are inserted for convenience. Since practically a11 materials respond elastically to moderate hydrostatic loading, the dilatational operators { M ) and { N ) are usually taken as constants and (9.47) modified to read

{P>s i j = 2{Q)e i j ( 9 . 4 8 ~ )

uii = 3K9, (9.48b) where K is the elastic bulk modulus.

Following the same general rule of separation for distortional and volumetric behavior, the three-dimensional viscoelastic constitutive relations in creep integral form are given by

and in the relaxation integral form by

The extension to three-dimensions of the complex modulus formulation of viscoelastic behavior requires the introduction of the complex bulk modulus K*. Again, writing shear and dilatatíon equations separately, the appropriate equations are of the form

9.8 VISCOELASTIC STRESS ANALYSIS. CORRESPONDENCE PRINCIPLE

The stress analysis problem for an isotropic viscoelastic continuum body which occupies a volume V and has the bounding surface S as shown in Fig. 9-12, is formulated as follows: Let body forces bi be given throughout V and let the surface tractions t$'(xk, t ) be prescribed over the portion SI of S, and the surface displacements gi(xk, t ) be prescribed over the portion S, of S. Then the governing field equations take the form of: Fig. 9-12


1. Equations of motion (or of equilibrium)

2. Strain-displacement equations 2fij = (ui, j + uj,i)

or strain-rate-velocity equations 2Gij = (v, ,~ + V,~)

3. Boundary conditions A

uij (x,, t) n, (x,) = tin) (x,, t) on S ,

4. Initial conditions ui (x,, 0) = UO

5. Constitutive equations

(a) Linear differential operator form (9.48)

(b) Hereditary integral form (9.49) or (9.50)

(c) CompIex modulus form (9.51)

If the body geometry and loading conditions are sufficiently simple, and if the material behavior may be represented by one of the simpler models, the field equations above may be integrated directly (see Problem 9.22). For more general conditions, however, it is conventional to seek a solution through the use of the correspondente pr.inciple. Thjs principle emerges from the analogous form between the governing field equations of elasticity and the Laplace transforms with respect to time of the basic viscoelastic field equations given above. A comparison of the pertinent equations for quasi-static isothermal problems is afforded by the following table in which barred quantities indicate Laplace transforms in accordance with the definition

f(xk9 S ) = xw f ( ~ , > t)e-st dt

Elastic Transformed Viscoelastic - 1. a,j3i + bi = O 1. üij,i + b, = O

From this table it is observed that when G in the elastic equations is replaced by &lP, the two sets of equations have the same form. Accordingly, if in the solution of the "corresponding elastic problem" G is replaced by &lP for the viscoelastic material involved, the result


is the Laplace transform of the viscoelastic solution. Inversion of the transformed solution yields the viscoelastic solution.

The correspondence principle may also be stated for problems other than quasi-static . problems. Furihermore, the form of the constitutive equations need not be the linear differential operator form but may appear as in (9.49), (9.50) or (9.51). The particular problem under study will dictate the appropriate form in which the principle should be used.

Solved Problems VISCOELASTIC MODELS (Sec. 9.1-9.3)

9.1. Verify the stress-strain relations for the Maxwell and Kelvin models given by (9.3) and (9.4) respectively.

In the Maxwell model of Fig. 9-2(a) the total strain is the sum of the strain in the spring plus the strain of the dashpot. Thus e = es + ED and also ; = Z s + G D . Since the stress across each element is o, (9 .1) and (9.2) may be used to give ; = O/G + a/?.

, In the Kelvin model of Fig. 9-2(b), o = os + a~ and directly from (9.1) and (9.2), o = v ; + Ge.

9.2. Use the operator form of the Kelvin model stress-strain relation to obtain the stress- strain law for the standard linear solid of Fig. 9-3(a).

Here the total strain is the sum of the strain in the spring plus the strain in the Kelvin unit. Thus E = es + e~ or in operator form e = olG1 + o/{G2 + q2at). From this

and so GIGze + G l s z ; = (G1 + Gg)a + 72;.

9.3. Determine the stress-strain equation for the four parameter model of Fig. 9-4. Let

T,I~ + and compare with the result of Problem 9.2.

Here the total strain e = e~ + e~ which in operator form is

e = o/{Gp + vzat) + {at + l / ~ l ) o / G l { a t )

Expanding the operators and collecting terms gives

As 7 , -+ m this becomes Ü + (G1 + G2)&l'lg = G1'E + G1G2;Iq2 which is equivalent to the result of Problem 9.2.

9.4. Treating the model in Fig. 9-13 as a special case of the generalized Max- well model, determine its stress-strain equation. <I - O

.. -

Writing (9.10) for N = 2 in the form

o = G l ; / { a t + 1 / ~ ~ ) + Gq; / {a t + 1 1 ~ ~ )

and operating as indicated gives Fig. 9-13


{at 4- 1/r2)(6 + "/r1) = Gl{at + 1/.r2); + G2{at + 1/7,);

which when expanded and rearranged becomes .. 0 (71 72);/r1r2 + = (G1 + G2) 7 + (G1/72 + G2/r1);

9.5. The model shown in Fig. 9-14 may be considered as a degenerate form of the generalized Maxwell model with

- m for the case N = 3. Using these values in ' 7 1 = '12 - (9.10) develop the stress-strain equation for this model.

Here (9.10) becomes o = vl; + G2;/{at) + ;/{at/G3 + l / q 3 1 or

{dt/G3 + l / v 3 ) 8 = {atlG3 + 1/73) (~17+ G2:) + ; 'I 1 93

Application of the operators gives

t which may also be written

713& + G30 = B ~ T ~ ~ Y + (G213 + G391 G393) : + G2G3e Fig. 9-14

CREEP AND RELAXATION (Sec. 9.4)

9.6. Determine the Kelvin and Maxwell creep response equations by direct integration of (9.1 7') and (9.21) respectively.

Using the integrating factor et/7, (9.17) becomes eetlr = - et'/7[U(t1)] dt' which by formula (9.18) yields n. i'

eet/7 = ( ~ ~ [ U ( t ) ] / ~ ) [ r e t ' / ~ ] ~ = (uO/G)(et/7- l ) [ U ( t ) ] or e = (ao/G)(l - e-t/T)[U(t)]

Use of et/' as the integrating factor in (9.21) gives oet/7 = GeO et1/T[8(t')] dt'; and by formula (9.23), i'

aet17 = Geo[U(t)] or a = Geoe- t /~[U( t ) ]

9.7. Determine the creep response of the standard linear solid of Fig. 9-3(a).

Since e = es + EK for this model the creep response from (9.1) and (9.19) is simply

e(t) = [ l /Gl + (1/G2) ( 1 - e- t /7~)]ao[U( t ) ]

The same result may be obtained by setting q2 = m in the generalized Kelvin (N = 2) response 2

c = 2 J i ( l - e- t / r , )~o[U( t ) ] or by integrating directly the standard solid stress-strain law. The i=l

student should carry out the details.

9.8. The creep-recovery experiment consists of a A 0

creep loading which is maintained for a period of time and then instantaneously removed. Determine the creep-recovery response of the 00

standard solid (Fig. 9-3(a)) for the loading shown in Fig. 9-15.

From Problem 9.7 the response while the load is

. I

I I I 6

272 t

on ( t < 272) is

e = oo[l/Gl + ( l / G z ) ( l - e - t / 7 ~ ) ] Fig. 9-15


At t = 2r2 the load is removed and o becomes zero a t the same time that the "elasticJJ deformation uo/G1 is recovered. For t > 272 the response is governed by the equation + €/r2 = O which is the stress-strain law for the model with u = O (see Problem 9.2). The solution of this differential equation is C = C e - T / r ~ where C is a constant and !i' = t - 2 ~ ~ . A t T = 0, C = C = oO(l - e-2)/G2 and so

e = aO(l - e-2)e-T1r2/G2 = uo(e2 - l ) e - t / r z G 2 / for t > 2 r 2

9.9. The special model shown in Fig. 9-16 is elongated a t a constant rate ; = ~, / t , as indicated in Fig. 9-17. Determine the stress in the model under this straining.

Fig. 9-16 Fig. 9-17

From Problem 9.5 the stress-strain law for the model is + 017 = qZ+ 3G; + GE/T and so here O + U / T = 3Gso/tl + Ge0t/7tl. Integrating this yields o = a0(39 + G t - ?) + Ce-t/r where C is the constant of integration. When t = 0, o = q~o/ t l and so C = - l )~Oltl . Thus o = eo(2q + Gt - qe-tlr)/tl. Note that the same result is obtained by integrating

3 ~ ~ t ' i r dt' + 2 X t d t f t l

9.10. Determine by a direct integration of the stress-strain law for the standard linear solid its stress relaxation under the strain E = e , [U( t ) ] .

Writing the stress-strain law (see Problem 9.2) as 6 + (G1 + G2)uIq2 = ~ ~ G ~ ( [ 8 ( t ) ] + G1G2[U(t)]lT2) for the case a t hand and employing the integrating factor e(Gi+Gz)tlnz i t is seen that

Integrating this equation with the help of (9.18) and (9.23),

u = eOGl(G2 + Gle-(Gi+Gz)tlnz)[U(t)]/(G1 + G2)

CREEP AND RELAXATION FUNCTIONS. HEREDITARY INTEGRALS (Sec. 9.5)

9.11. Determine the relaxation function +(t) for the three parameter model shown in Fig. 9-18.

The stress-strain relation for this model i s -o-- - - u

6 + u/r2 = (G1 + G2); + G1G2c/q2

and so with = eo[U(t)] and ; = eo[6(t)] use of the integrating factor etlrz gives Fig. 9-18

Thus by use of (9.18) and (9.23), o = eo(Gl + G2e-t/rz) = eo$(t). Note that this result may also be obtained by putting q1 + m in (9.30) for the generalized Maxwell model.

CHAP. 91 LINEAR VISCOELA'STICITY 209

9.12. Using the relaxation function +(t) for the model of Problem 9.11, determine the creep function by means of (9.40).

The Laplace transform of $ ( t ) = G1 + G2e-t/Tr is 5 ( s ) = Gl/s + Gz/(s + 1 1 ~ ~ ) (see any standard table of Laplace transforms). Thus from (9.40),

which may be inverted easily by a Laplace transform table to give

li/ = l / G l - [G2/G1(G1 + G2)]e-Git/(Gi+G2)T8

This result may be readily verified by integration of the model's stress-strain equation under creep loading.

9.13. If a ramp type stress followed by a sus- tained constant stress U , (Fig. 9-19) is applied to a Kelvin material, determine the resulting strain. Assume ~ , / t , = A.

The stress may be expressed as t . t

u = ht[U( t ) ] - h(t - t l ) [ U ( t - t,)]

which when introduced into (9.4) leads to

- I

Fig. 9-19

Integrating with the aid of (9.18) gives

E = (hlG)(( t i- ~ ( e - ~ l ~ - l ) ) [ U ( t ) ] - ((t - t l) + ~ ( e ( t i - t ) / ~ - l ) ) [ U ( t - t l ) ] )

which reduces as t -+ m to E = x ~ , I G = u,/G.

9.14. Using the creep integral (9.34) together with the Kelvin creep function, verify the result of Problem 9.13.

For the Kelvin body, $(t) = (1 - e-tI7)/G and (9.34) becomes

.(t) = S-tm $ ( [ U ( t l ) ] + t f [ s ( t l ) ] - [u(tf - t l)] - ( t f - t l )[s( t f - t l)])(l - e- ( t - t l ) /T) dt'

which by (9.18) and (9.23) reduces to

X t = - [ [ ~ ( t ) ] St ( 1 - e - ( t - t l ) l ~ ) dt' - [ U ( t - t l)] $ (1 - e-( t - t f ) lr) d t f

G o tl I A straightforward evaluation of these integrals confirms the result presented in Problem 9.13.

9.15. By a direct application of the superposition principle, determine the response of a Kelvin material to the stress loading shown in Fig. 9-20.

Fig. 9-20 Fig. 9-21

210 L I N E A R V I S C O E L A S T I C I T Y [CHAP. 9

The stress may be represented as a sequence of ramp loadings as shown in Fig. 9-21 above. From Problem 9.13, E = (h /G)[ t + r(e- t l f - I ) ] [U( t ) ] for this stress loading. In the présent case therefore

~ ( t ) = ( i / G ) [ ( t + r(e-tlT - l ) ) [ U ( t ) ] - ( ( t - t i ) + r ( e - ( t - t l ) l ~ - l ) ) [ U ( t - t i )]

- (( t - 2tl) + ~ ( e - ( t - z t ~ ) / ~ - l ) ) [ U ( t - 2t l ) ] + ( ( t - 3tl) + ~(e-( t -3t1)I7 - l ) ) [ U ( t - 3 t l ) ] l

Note that as t + m, E + 0.

COMPLEX MODULI AND COMPLIANCES (Sec. 9.6) 9.16. Determine the complex modulus G* and the lag angle S for the Maxwell material of

Fig. 9-2.

Writing (9.3) as 6 + U/T = G ; and inserting (9.41) and (9.42) gives iwooeiwt 4- o0eiot/r = Giw~,ei(wt-6) from which ~ ~ e i ~ l ~ ~ = G* = G i w ~ l ( l + i w ~ ) , or in standard form

From Fig. 9-11, tan 8 = GdG1 = Gor/Gw2r2 = 1 1 ~ ~ .

9.17. Show that the result of Problem 9.16 may also be obtained by simply replacing the operator d, by i. in equation (9.5) and defining u / e = G*.

After the suggested substitution (9.5) becomes (iolG + l / q ) o = &e from which

U / E = Giol(io + 117) = G ~ w T / ( ~ + ~ W T )

as before.

9.18. Use equation (9.10) for the generalized Maxwell model to illustrate the rule that "for models in parallel, the complex moduli add".

From Problem 9.17 the complex modulus for the Maxwell model may be written G* = a/€ = Giwrl(l+ iwr). , Thus writing (9.10) as

the generalized Maxwell complex modulus becomes

9.19. Verify the relationship J l = l / G l ( l + tan2 8) between the storage modulus and compliance.

From (9.43) and (9.44), J* = 1/G* and so J1 - iJ2 = l / ( G l + iG2) = (G1 - ~G,)I(G? + 62). Thus

J1 = G ~ / ( G : + 02) = l / G 1 ( l + (G2/G1)2) = l / G 1 ( l + tan2 6)

9.20. Show that the energy dissipated per cycle is related directly to the loss compliance

J2 by evaluating the integral u d ~ over one cycle. S I

For the stress and strain vectors of Fig. 9-10, the integral odr evaluated over one cycle is S ds 2 ~ 1 0

d t = l (oo sin U ~ ) E ~ W cos (ut - 6) d t

- - ooEo. J2'I* sin w t (cos u t cos a + sin u t sin a ) d t o

2nlo sin 2ot dt + J , lbiu (sin2 ot ) d t ] = o:sJ2


THREE DIMENSIONAL THEORY. VISCOELASTIC STRESS ANALYSIS (Sec. 9.7-9.8)

9.21. Combine ( 9 . 4 8 ~ ) and (9 .483) to obtain the viscoelastic constitutive relation uij = ai j {R}ekk + {S)cij and determine the form of the operators {R} and {SI .

Writing ( 9 . 4 8 ~ ) a s {P)(aii - oijakk/3) = 2{Q)(eij - oijekk/3) and replacing ukk here by the right hand side of (9.48b), the result after some simple manipulations is

aij oij{(3KP - 2Q)/3p)ekk + {2Q/P)eij

9.22. A bar made of Kelvin material is pulled in tension so that a,, = a o [ U ( t ) ] , a,, = a33 = -

ai, - a,, = a,, = O where ao is constant. Determine the strain ell for this loading.

From (9.48b), 3cii = oo[U( t ) ] lK for this case; and from ( 9 . 4 8 ~ ) with i = j = 1, {P)(u l l - a11/3) = { 2 Q ) ( ~ l l - eii/Q). But from (9.61, { P ) 1 and { Q ) = {G + Tat) for a Kelvin material; so that now

200[U(t)I/3 = 2{G + ~ a ~ l ( ~ ~ ~ - aO[U(t)]lgK)

Solving this differential equation yields

€11 = aO(3K + e-t1T) [U( t ) ] l9KG + aoe-tlT[U(t)]/9K

As t + m, €11 -> (3K + G)aO19KG = aJE.

9.23. A block of Kelvin material is held in a container with rigid .walls so that e, = e,, = O when the stress o,, = -ao[U( t ) ] is applied. Determine c,,

and the retaining stress components V,, and u3, for this situation.

Here qi = ell and a,, = so that (9.48b) becomes ai1 i- 2u2, = 3Kell and (9.48a) gives 2(u11 - u2,)/3 = 2G{1+ ~ a ~ ) ( 2 ~ ~ ~ / 3 ) for a Kelvin body. Combining these relations yields the differential equation Fig. 9-22

which upon integration gives

Inserting this result into ( 9 . 4 8 ~ ) for i = j = 2 gives

a,, = (+ao/2 - 9Kao(l - e-(4G+3K)t/4G~)/(8G + GK))[U( t ) ]

9.24. The radial stress component in an elastic half- space under a concentrated load a t the origin may be expressed as

a,,,, = ( P 1 2 ~ ) [(I - 2 v ) a ( r , 2 ) - P(r, 41 where a and p are known functions. Determine the radial stress for a Kelvin viscoelastic half- space by means of the correspondence principle when P = Po[U(t)]. Fig. 9-23


The viscoelastic operator for the term (1 - 2 ~ ) is {3Q) l {3KP + Q ) so that for a Kelvin body the transformed viscoelastic solution becomes

which may be inverted with help of partia1 fractions and transform tables to give the viscoelastic stress

9.25. The correspondence principie may be used to obtdin displacements as well as stresses. The x displacement of the surface of the elastic half space in Problem 9.24 is given by W ( z = ~ ) = P(l- v 2 ) l E T r . Determine the viscoelastic displacement of the surface for the viscoelastic material of that problem.

The viscoelastic operator corresponding to (1 - v2)lE is { 3 K + 4Q)/4Q(3K + Q ) which for the Kelvin body causes the transformed displacement to be

After considerable manipulation and inverting, the result is

Note that when t = O, w ( , = o ) = O and when t + m, w ( , = ~ ) -+ P o ( l - u z ) / E ~ r , the elastic deflection.

9.26. A simply supported uniformly loaded beam is assumed to be made of a Maxwell material. , Determine the bending stress a , , and the deflection w(x,, t) if the load is p = p o [ U ( t ) ] . Fig. 9-24

The bending stress for a simply supported elastic beam does not depend upon material properties, so the elastic and viscoelastic bending stress here are the same. The elastic deflection of the beam is w ( x l ) = pon(xI)/24EZ where a(xJ is a known function. For a Maxwell body, {P) = {a, + 117) and { Q ) = {Ga,), so that the transformed deflection is

which when inverted gives

When t = O, w ( x , , O ) = pw(x1)/24EZ, the elastic deflection.

9.27. Show that as t += .o the stress a,, in Problem 9.23 approaches ao (material behaves as a fluid) if the material is considered incompressible ( V = 112).

From Problem 9.23,

0221t+m = -oO(9K - (4G + 3K)) /2(4G + 3 K ) = -oo(3K - 2G)l(3K + 4G)

which may be written in terms of v as 02,1t,, = -voo/(l - v). Thus for v 112, o,,lt+, = -o*



9.28. Determine the constitutive relation for the 81

Kelvin-Maxwell type model shown in Fig. .

9-25 and deduce from the result the Kelvin and Maxwell stress-strain laws. c<t-

Here

a = U M + aK = ;/{dt/G1 + l / v l } + {Gz + q2at}r 82

which upon application of the time operators I I becomes

6 + a/rl = q2T + (G1 G2 q 2 / ~ 1 ) ( G 2 / ~ 1 ) ~ Fig. 9-25

In this equation if q2 = 0 (spring in parallel with Maxwell), g + d r l = (Gl + G2); + (G21~1)~ . Further, if G, = 0, the Maxwell law 6 + U / T ~ = G1; results. Likewise, if G2 is taken zero first (dashpot in parallel with Maxwell), 6 + a/rl = q2Y+ (Gl i- q2/r1);; and when q , = 0, this also reduces to the Maxwell law.

If the four-parameter constitutive relation is rewritten

1118 + Glu = 7118~Y + ( G l ~ l + G281+ G172); + GlG2e

and q1 set equal to zero, the result is the Kelvin law a = 8,; + Gze. Likewise, if Gl = O the reduced equation is 8 = q,T+ G,;, again representing the Kelvin model.

9.29. Use the superposition principie to obtain the creep recovery response for the standard linear solid of Fig. 9-3(a) and compare the result with that obtained in Problem 9.8.

With the stress loading expressed by

U = ao[U(t)] - ao[U(t - 2r2)]

(see Fig. 9-26), the strain may be written a t once from the result of Problem 9.7 as Fig. 9-26

E = ao(l/Gl + ( 1 - e - t ' ~ ~ ) / G ~ ) [ U ( t ) l - aO(l/G1 -i ( 1 - e-( t - z ~ z ) ~ ~ ) l G ~ ) [ U ( t - 2r2)]

At times t > 272 both step functions equal unity and

- E - a0(-e-t/7, + e-( t - 2 7 2 ) / ~ ~ ) / G ~ = aO(e2 - 1)e-t/r2/G2

which agrees with the result in Problem 9.8.

9.30. Determine the stress in the model of Prob- lem 9.9 when subjected to the strain history shown in Fig. 9-27. Show that eventually the "free" spring in the model carries the entire stress.

From Problem 9.9 and the superposition prin- I

ciple the stress is Fig. 9-27

- - ~ 0 ( 7 ( 2 - e-t/7) + G t ) [ U ( t ) ] l t l - eo(7(2 - e-( t - t i ) /7) + G(t - t l ) ) [ U ( t - t l ) ] / t l

For times t > t 1 the stress is o = cotl(etl/T - 1)e-t/r/tl f Gco, and as t + m this reduces to o = Gço.

LINEAR VISCOELASTICITY [CHAP. 9

9.31. The "logarithmic retardation spectrum" L is defined in terms of the retardation spectrum J by L(ln r ) = r J ( r ) . From this definition determine the creep function +(t) in terms of L(ln r ) .

Let ln r = h so that eX = T and thus drldh = e" T , or dr = ~ d ( l n T ) . From this, (9.28)

defining g ( t ) becomes q ( t ) = L(ln ~ ) ( l - e-t")d(ln r). In the same way, if H(ln r) = TG(T) LW defines the logarithmic relaxation spectrum, $ ( t ) in (9.81) may be written

9.32. For the Maxwell model of Fig. 9-2(a) determine the storage and loss moduli, Gl and G,, as functions of ln o r and sketch the shape of these functions.

From Problem 9.16,

G* = G ( w ~ T ~ + L r ) / ( 1 + ~ 2 ~ 2 )

for a Maxwell material. Thus

G1 = Gw2~2/( l+ ~ 2 ~ 2 ) = Ge2X/(l+ e2h)

where 1 = l n w ~ . For A = O, G1 = G/2; for h = m, G1 = G ; and for h = -a, Gl = O. Like- wise G, = GeX/(l+ e2X) and for A = O, G, = G/2; for h = I m , G2 = O. The shape of the curves for these functions is as shown in Fig. 9-28. Fig. 9-28

9.33. Determine the viscoelastic operator form of the elastic constant V (Poisson's ratio) using the constitutive relations (9.48).

Under a uniaxial tension o,, = o,, (9.48b) gives si/3 = od9K so that ( 9 . 4 8 ~ ) for i = j = 1 yields cll = (3KP + Q)ooI{9KQ). In the same way ( 9 . 4 8 ~ ) for i = j = 2 yields cz2 = (2Q - 3PK)oo/{18KQ). Thus in operator form, V = -c2,/c11 = (3PK - 2Q)/{6KP + 2Q).

9.34. A cylindrical viscoelastic body is inserted into a rigid snug-fitting container (Fig. 9-29) so that

= O (no radial strain). The body is elastic in dilatation and has the creep function +, = A + T Bt + CeM where A, B, C, A are constants. If h

;33 = ;OIU(t)], determine a,,(t).

Here oii = 3Keii and by the symmetry of the problem, 2all + o33 = 3Ke3,. Also from ( 9 . 5 0 ~ ) with i = j = 1,

011 - 413 = -it 2 h(t - t') dt'. Soiving for o,, from

these two relations we obtain Fig. 9-29

osr = KE,, + 3 it 2 +,(t - t f ) dt'

The relaxation function +, may be found with the help of (9.40). The result is

$, = [(rl - h)e+it - (r, - h)eTzt]/(rl - r,)

where r,, , = [ A h - B * ~ ( A A + B)' + 4BCh ] / 2 ( A + C). Thus finally


t - ~ ) ~ r l ( t - t ' ) - - h)eVz(t-tl) 2

rs3 = K b t + i. I IU(tl)] dt' (r1 - ~ 2 )

which upon integration gives

033 = (K;Ot + 2Z0[-(r1 - X ) ( I - e~it)/3r, + (r, - X ) ( I - erzt)/3r2]/(rl - r , ) ) [ ~ ( t ) ]

9.35. The "creep buckling" of a viscoelastic column may be analyzed within the linear theory through the correspondence principle. Determine the deflection w(x,, t ) of a Kelvin pinned-end column by this method.

Fig. 9-30

The elastic column formula is d2wldx: + Pow/EZ = 0, and for a Kelvin material E may be replaced by the operator {E + 78,) so that for the viscoelastic column { E + q a t ) ( d 2 ~ l d ~ : ) + P,wlZ = O. Assuming the deflection in a product form w(xl, t) = W(xl)e(t) , the operator leads to the differential equation

(Ee + qè)(d2wldx;) + PoWeIZ = O from which + [l + P, W / ~ ~ ( d ~ ~ / d z ; ) ] e l ~ = O

where T = 7IE. But the elastic buckling load is P, = - ~ Z ( d 2 ~ l d x : ) l ~ and so + (1 - P,lP,)el~ = O which integrates easily to yield e = e ( p ~ / ~ ~ - l ) t / ~ . Finally then the "creep buckling" deflection is w = We(Po/p~-l)t/T.

9.36. Formulate the steady-state vibration problem for a viscoelastic beam assuming' the constitutive relations are those given by (9.48). ,

The free vibrations of an elastic beam are governed by the equation E Z ( ~ ~ W / ~ X ; ) + pA(a2~lat2) = O. From (9.48) the viscoelastic operator for E is {9KQl(3KP + Q ) ) , and if the deflection w(xl , t ) = W(xl)e( t ) the resulting viscoelastic differential equation may be split into the space equation d4Wldx~ - k4W = O and the time equation (3KP + Q)(d2eldt2) + (k4ZlpA){9KQ)(~) = 0. The solution W i of the space equation represents the ith mode shape, and from the time equation for

N k = ki the solution ei = 2 Aijehilt where N depends upon the degree of the operator. The total

j=1 m N

solution therefore is w(xl , t ) = Wi(xl)AijeAijt in which the Aij are complex. i=1 j=l


9.37. Determine the constitutive equation for the four parameter mode1 shown in Fig. 9-31.

Ans. 8 + (Gilv2 + G2/v2 + Gl/7l); + ( G l G z / ~ l ~ z ) u = Glf .+ (G1G2/72)'

Fig. 9-31


9.38. Determine the creep response of the standard linear solid by direct integration of E + € 1 ~ ~ = oo[U(t)](Gl + G2)/G1v2 + uo[8(t)]/Gl. (See Problem 9.7.)

9.39. Deduce the Kelvin and Maxwell stress-strain laws from the results established in Problem 9.5 for the four parameter model of that problem. (Hint. Let G3 = 0 , etc.)

9.40. Use equation (9.40) to obtain $( t ) if +(t) = a(b/t)m with m < 1. (Hint. Take m = 1 - k ; then +(t) =

abmtk-1.) A m . $ ( t ) =

9.41. Determine the creep and relaxation functions for the model shown in Fig. 9-32.

Ans. $ ( t ) = 1/G2 - Gle-Czt/(Gl+c2)71/G2(G1 + G2) L"""d '11

+(t) = G2 + Gle-t/Tl Fig. 9-32

9.42. Determine G* for the model shown in Fig. 9-33.

G1(l + 722 0,) + G2a272 o(G2r2 + q3(1 + 7; a2)) Ans. G* = + i

1 + 027; 1 + ~ ~ 7 2

Fig. 9-33 Fig. 9-34

9.43. In the model of Problem 9.42 let G1 = G, = G and q, = q3 = q and determine the stress history of the resulting model when it is subjeeted to the strain sequence shown in Fig. 9-34.

€0 Ans. u = - (G(2t - t,) + ~ ( 4 - (1 + eti'r)e-tlT)) for t l < t < 2tl tl

9.44. - A viscoelastic block having the constitutive equation 8 + ao = P ; + ya where a, P, y are constants is loaded under conditions such that o,, = -oo[U(t)], 9, = O , q3 = O (see Fig. 9-35). Assuming oii = 3Kqi, determine ~ ~ ~ ( t ) , ~ ~ ~ ( 0 ) and ~ ~ ~ ( 0 0 ) .

Ans. o33 = 3aK - 2y 3K - 2P - 3aK - 2y ) e - ~ t ] where h = (3aK + y)/(3K + p). 2(3aK + y)h

+ (2 (3K + P ) 2(3aK + y)h

Fig. 9-35

wo sin (7x11) Po[U(t)l P,[U(t)l

Fig. 9-36

9.45. A pinned-end viscoelastic column is a Maxwell material for which 8 + olr = E;. The initial shape of the column is w = w o sin ( ~ z l l ) when the load Po [U( t ) ] is applied (see Fig. 9-36). Deter- mine the subsequent deflection w ( x l , t ) as a function of P,, the elastic buckling load. Ans. w ( x l , t ) = w0 sin ( a ~ ~ / l ) e - t ~ ( ' - P ~ / P o ) ~

I N D E X

Absolute dynamic, compliance, 202 modulus, 202

Acceleration, 111 Addition and subtraction,

of matrices, 17 of (Cartesian) tensors, 1 5 of vectors, 2

Adiabatic process, 140 Airy stress function, 147 Almansi s t rain tensor, 82 Angle change, 86 Angular momentum, 128 Anisotropy, 44, 141 Antisymmetric,

dyadic, 5 matrix, 19 tensor, 19

Axis of elastic symmetry, 142

Barotropic change, 161 Base vectors, 6 Basis, 6, 9

orthonormal, 6 Bauschinger effect, 176 Beltrami-Michell equations, 144 Bernoulli equation, 164 Biharmonic equation, 148 Body forces, 45 Boundary conditions, 144, 146 Bulk,

modulus, 143 viscosity, 161

Caloric equation of state, 132 Cartesian,

coordinates, 6 tensors, 1, 12, 13

Cauchy, deformation tensor, 81 strain ellipsoids, 89 stress principle, 45 stress quadric, 50

Cauchy-Riemann conditions, 165 Circulation, 164

Kelvin's theorem of, 165 Clausius-Duhem inequality, 131 Column matrix, 17 Compatibility equations, 92, 114 Complex,

niodulus, 202 potential, 166

Compliance, 201 Coniponent, 1, 7 Compressible fluid, 165 Configuration, 77 Conformable niatrices, 17

Conjugate dyadic, 4 Conservation of,

energy, 128 .mas , 126

Constitutive equations, 132 Continuity equation, 126 Continuum concept, 44 Contraction, 16 Contravariant tensor, 11 Convective,

derivative, 110 ra te of change, 111

Conventional stress and strain, 175 Coordinate transformation, 11 Correspondence principle, 205 Couple-stress vector, 45 Coupled heat equation, 150 Covariant tensor, 12 Creep,

function, 200 test, 199

Cross product, 3, 5, 16 Cubical dilatation, 90 Curvi1ine:ir coordinates, 7 Cylindrical coordinates, 8

Dashpot, 196 Decomposition,

polar, 87 velocity gradient, 112

Deformation, 77 gradients, 80 inelastic, 175 plane, 91 plastic, 175 tensors, 81 total (deformation) theory, 183

De1 operator, 22 Density, 44, 126

entropy, 130 strain energy, 141

Derivative, material, 110, 114 of tensors, 22 of vectors, 22

Deviatoric, s t rain tensor, 91 stress tensor, 57

Diagonal matrix, 17 Dilatation, 90 Direction cosines, 7 Displacement, 78, 83

gradient, 80 relative, 83

Dissipation function, 132 Dissipative stress tensor, 131 Distortion energy theory, 178

INDEX

Divergence theorem (of Gauss), 23 Dot product of,

dyads, 5 vectors, 3

Duhamel-Neumann relations, 149 Dyadics, 1, 4

antisymmetric, 4 conjugate of, 4 symmetric, 4

Dyads, 4 nonion form, 7

Dynamic moduli, 203

Effective, plastic strain increment, 182 stress, 182

Elastic, constants, 141 limit, 175 symmetry, 142

Elasticity, 140 Elastodynamics, 143 Elastoplastic problems, 183 Elastostatics, 143 Energy,

kinetic, 129 strain, 141 thermal, 129

Engineering stress and strain, 175 Entropy, 130

specific, 130 '

e-o identity, 39 Equations of

equilibrium, 48, 128 motion, 128 state, 130

Equivalent, plastic strain increment, 182 stress, 182

Euclidean space, 11 Eulerian,

coordinates, 78 description, 79 finite strain tensor, 82 linear strain tensor, 83

Field equations, elastic, 143, 147 viscoelastic, 205

Finite strain tensor, 81, 82 First law of thermodynamics, 129 Flow, 77, 110

creeping, 169 irrotational, 164 plastic, 175 potential, 175 rule, 181 steady, 163

Fluid, inviscid, 160 perfect, 160,164 pressure, 160

Forces, body, 45 surface, 45

Fourier heat law, 149 Fundamental metric tensor, 13

Gas, dynamical equation, 165 law, 160

Gauss's theorem, 23 Generalized,

Booke's law, 140 Kelvin model, 198 Maxwell model, 198 plane strain, 147 plane stress, 147

Gibb's notation, 2 Gradient,

deformation, 80 displacement, 80

Green's deformation tensor, 81 finite strain tensor, 82

Hamilton-Cayley equation, 21 Hardening,

isotropic, 180 ,

kinematic, 181 strain, 176, 183 work, 176, 183

Harmonic functions, 166 Heat,

conduction law, 149 flux, 129 radiant, 129

Hencky equations, 183 Hereditary integrals, 201 Homogeneous,

deformation, 95 material, 44

Hookean solid, 196 Hydrostatic pressure, 160 Hydrostatics, 163 Hyperelastic material, 149 Hypoelastic material, 149 Hysteresis, 176

Ideal, gas, 160 materiais, 132

Idealized stress-strain curves, 177 Idemfactor, 5 Identity matrix, 18 Incompressible flow, 126 Incremental theories, 181 Indeterminate vector product, 4 Indices, 9 Indicial notation, 8 Inelastic deformation, 175 Inertia forces, 48 Infinitesimal strain, 83 Initial conditions, 145 Inner product, 16

INDEX

Integral theorems, 23 Interna1 energy, 129 Invariants, 21

of ra te of deformation, 114 of strain, 89, 90 of stress, 51

Irreversible process, 130 Irrotational flow, 113, 164 Isothermal process, 140 Isotropy, 44, 142

Jacobian, 11, 79 Johnson's apparent elastic-limit, 176

Kelvin (material) model, 197 Kinematic,

hardening, 181 viscosity, 163

Kinetic energy, 129 Kronecker delta, 13

Lagrangian, description, 79 finite s t rain tensor, 82 infinitesimal s t rain tensor, 83

Lamé constants, 143 Laplace,

equation, 165 transform, 202, 205

Left stretch tensor, 87 Levy-Mises equations, 181 Line integrais, 23 Linear,

momentum, 127 rotation tensor, 83 therinoelasticity, 149 vector operator, 8 viscoelasticity, 196

Local rate of change, 111 Logarithmic strain, 175

Mass, 126 Material,

coordinates, 78 derivative, 110, 114-117 description of motion, 79

Matrices, 17-19 Maximum,

normal stress, 52 shearing stress, 52, 53

Maxwell (material) model, 197 Metric tensor, 13 Mises yield condition, 178 Modulus,

bulk, 143 complex, 203 loss, 203 shear, 143 storage, 202

Mohr's circles, f o r strain, 91 for stress, 54-56

Moment of momentum, 128

Momentum principie, 127 Motion, 110

steady, 112 Multiplication of,

matrices, 18 tensors, 15 vectors, 3

Navier-Cauchy equations, 144 Navier-Stokes equations, 162 Navier-Stokes-Duhem equations, 162 Newtonian viscous fluid, 161, 196 Normal stress components, 47

Octahedral, plane, 59 shear stress, 192

Orthogonal, tensor, 87 transformation, 13

Othogonality conditions, 13, 14 Orthotropic, 142 Outer product, 1 5

Parallelogram law of addition, 2 Particle, 77 Pa th lines, 112 Perfectly plastic, 176 Permutation symbol, 16 r-plane, 179 Plane,

deformation, 91 elasticity, 145 strain, 91, 145 stress, 56-57, 145

Plastic, deformation, 175 flow, 175 potential theory, 181 range, 176 strain increment, 182

Point, 77 Poisson's ratio, 143 Polar,

decomposition, 87 equilibrium equations, 148

Position vector, 77 Post-yield behavior, 180 Potential,

flow, 165 plastic, 182

Prandtl-Reuss equations, 181 Pressure,

fluid, 160 function, 163

Principal, axes, 20 strain values, 89 stress values, 51, 58

Proper transformation, 17 Proportiona1 limit, 177 Pure shear, 178

INDEX

Quadric of, strain, 88, 89 stress, 50

Quasistatic viscoelastic problem, 205

Range convention, 8 Rate of deformation tensor, 112 Rate of rotation vector, 114 Reciproca] basis, 28 Rectangular Cartesian coordinates, 6 Kelative displacement, 83 Relaxation,

function, 201 spectrum, 201 test, 199

Retardation, spectrum, 201 time, 199

Reversible process, 130 Reynold's transport theorem, 123 Right stretch tensor, 87 Rigid displacement, 82 Rotation tensor,

finite, 87 infinitesimal, 83, 84

Rotation vector, 83

St. Venant's principle, 145 Scalar, 1

field, 22 of a dyadic, 4 triple product, 4

Second law of thermodynamics, 130 Shear,

modulus, 143 strain components, 86 stress components, 52

Slip line theory, 184 Small deformation theory, 82 Spatial coordinates, 78 Specific,

entro&, 130 heat, 149

Spherical, coordinates, 8 tensor, 57, 91

Spin tensor, 112 Standard linear solid, 197 State of stress, 46 Stokes' condition, 161 Stokes' theorem, 23 Stokesian fluid, 161 Strain,

deviator, 91 ellipsoid, 88 energy, 141 hardening, 176 natural, 112, 175 plane, 91 rate, 112 shearing, 86 spherical, 91 transformation laws, 88

Stream function, 165 Stream lines, 112 Stress,

components, 47 conservative, 131 deviator, 57 effective, 182 ellipsoid, 52 function, 147 invariants, 51 Mohr's circles for, 54-56 normal, 47 plane, 56 power, 130 principie, 51 quadric, 50 shear, 47, 52 spherical, 57 symmetry, 48 tensor, 46 transformation laws, 49 vector, 45

Stretch, ratio, 86 tensor, 87

Summation convention, 8, 10 Superposition theorem, 145 Surface forces, 45 Symbolic notation, 2, 10 Symmetry,

elastic, 142 tensor, 19

Temperature, 130 Tension test, 175 Tensor,

Cartesian, 1, 12, 13 components of, 1 contravariant, 11 covariant, 12 ,

deformation, 81 derivative of, 22 fields, 22 general, 1, 11 metric, 12 multiplication, 1 5 powers of, 21 rank, 1 stretch, 87 transformation laws, 1

Tetrahedron of stress, 47 Thermal equation of state, 132 Thermodynamic process, 130 Thermoelasticity, 133, 149 Transformation,

laws, 13, 88 of tensors, 1 orthogonal, 13, 14

Tresca yield condition, 178 Triangle rule, 2 Triple vector product, 4

Tw-o-dimensional elastostatics, in polar form, 148 in rectangular form, 145

Uncoupled thermoelasticity, 133, 150 Uniqueness theorem, 145, 158 Unit,

dyadic, 5 relative displacements, 83 triads, 7 vector, 3

Vector, addition, 2 displacement, 78 dual, 16 field, 22 of a dyadic, 4 position, 12, 77 potential, 126 products, 3, 4, 16 rotation, 83 traction, 46 transformation law, 14 vorticity, 113

INDEX 221

Velocity, 110, 111 complex, 166 potential, 164 strain, 112

Viscoelastic stress analysis, 204-206 Viscoelasticity, 196 Viscous stress tensor, 160 Voigt model, 197 Vorticity,

tensor, 112 vector, 113

Work, hardening, 176 plastic, 182

Yield, condition, 177 curve, 179 surface, 179

Young's modulus, 143

Zero, matrix, 17 order tensor, 1 vector, 2

35737861 Schaum Continuum Mechanics Mase

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