SCHAUM'S O U T L I N E O F

THEORY AND PROBLEMS

CONTINUUM MECHANICS

GEORGE E. MASE, Ph.D.Professor o f Mechanics Michigan State University

SCHAUM9S OUTLINE SERIESMcGRAW-HILL BOOK COMPANYNezo York, St. Louis, San Francisco, London, Sydney, Toronto, Mexico, and Panama

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ISBN 07-04M63-4

PrefaceBecause 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 a s a text f o r an introductory course in Continuum Mechanics. Otherwise, the book should prove especially useful as a supplementary reference for a number of courses f o r 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 a n 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 a n initial chapter on the mathematics relevant to the study, the theory portion contains additional chapters on the Analysis of Stress, Deformation 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 f o r 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

CONTENTSChapter

11.1 1.2 1.3 1.4 1.5 1.6 1.7

Page

MATHEMATICAL FOUNDATIONSTensors and Continuum Mechanics .................................... General Tensors . Cartesian Tensors Tensor Rank ...................... Vectors and Scalars .................................................. Vector Addition . Multiplication of a Vector by a Scalar .................. Dot and Cross Products of Vectors .................................... Dyads and Dyadics .................................................... Coordinate Systems Base Vectors Unit Vector Triads .................. Linear Vector Functions Dyadics a s Linear Vector Operators .......... Indicial Notation Range and Summation Conventions Summation Convention Used with Symbolic Convention .................. Coordinate Transformation . General Tensors .......................... The Metric Tensor. Cartesian Tensors .................................. Transformation Laws f o r Cartesian Tensors . The Kronecker Delta Orthogonality Conditions ........................................ Addition of Cartesian Tensors Multiplication by a Scalar ................ Tensor Multiplication .................................................. Vector Cross Product . Permutation Symbol Dual Vectors .............. Matrices Matrix Representation of Cartesian Tensors .................. Symmetry of Dyadics. Matrices and Tensors ............................ Principal Values and Principal Directions of Symmetric Second-Order Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Powers of Second-Order Tensors Hamilton-Cayley Equation ............ Tensor Fields Derivatives of Tensors .................................. Line Integrals. Stokes' Theorem ...................................... The Divergence Theorem of Gauss ......................................1 1 2 2 3

.

4G

.

1.8

1.9 1.10 1.11 1.12 1.13

.

.

.

88 10 11 12 13 15 15

..................

.

.

.

.

1617 19 20 21 22 23 23

.

.

Chapter

2

ANALYSIS OF STRESSThe Continuum Concept ................................................ Homogeneity Isotropy Mass-Density .................................. Body Forces . Surface Forces .......................................... Cauchy's Stress Principle . The Stress Vector ............................ State of Stress a t a Point . Stress Tensor ................................ The Stress Tensor-Stress Vector Relationship .......................... Force and Moment. Equilibrium . Stress Tensor Symmetry . . . . . . . . . . . . . . Stress Transformation Laws .......................................... Stress Quadric of Cauchy ..............................................44 44 4545

.

.

46 4748 49 50 51 52 54 56 57

Principal Stresses Stress Invariants Stress Ellipsoid . . . . . . . . . . . . . . . . . . Maximum and Minimum Shear Stress Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . Mohr's Circles f o r Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Plane Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Deviator and Spherical Stress Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

.

.

CONTENTS

Chapter

33.1 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 STRAINParticles and Points ................................................ Coritinuum Configuration . Deformation and Flow Concepts . . . . . . . . . . . . . . Position Vector . Displacement Vector .................................. Lagrangian and Eulerian Descriptions .................................. Deformation Gradients. Displacement Gradients ........................ Deformation Tensors . Finite Strain Tensors ............................ Small Deformation Theory . Infinitesimal Strain Tensors ................ Relative Displacements . Linear Rotation Tensor . Rotation Veztor ........ Interpretation of the Linear Strain Tensors ............................ Stretch Ratio . Finite Strain Interpretation ............................ Stretch Tensors . Rotation Tensor ...................................... Transformation Properties of Strain Tensors ............................ Principal Strains. Strain Invariants . Cubical Dilatation ................ Spherical and Deviator Strain Tensors .................................. Plane Strain . Mohr's Circles f o r Strain ................................ Compatibility Equations for Linear Strains ..............................77

77 77 79 80 81 82 83 84 86 87 88 89 91 91 92

Chapter

44.1 4.2 4.3 4.4 4.5 4.6 4.7

MOTION AND FLOWMotion . Flow . Material Derivative .................................... Velocity. Acceleration . Instantaneous Velocity Field .................... P a t h Lines. Stream Lines. Steady Motion .............................. Rate of Deformation . Vorticity . Natural Strain ........................ Physical Interpretation of Rate of Deformation and Vorticity Tensors . . . . . . Material Derivatives of Volume. Area and Line Elements .................. Material Derivatives of Volume, Surface and Line Integrals ..............110 111 112 112 113 114 116

Chapter

55.1 5.2

FUNDAMENTAL LAWS OF CONTINUUM MECHANICSConservation of Mass . Continuity Equation ............................ Linear Momentum Principle . Equations of Motion . Equilibrium Equations .............................................. Moment of Momentum (Angular Momentum) Principle .................... Conservation of Energy . F i r s t Law of Thermodynamics. Energy Equation .................................................. Equations of State. Entropy . Second Law of Thermodynamics . . . . . . . . . . . . The Clausius-Duhem Inequality . Dissipation Function .................... Constitutive Equations . Thermomechanical and Mechanical Continua . . . . . .126 127 128 128 130 131 132

Chapter

66.1 6.2 6.3 6.4 6.5 6.6

LINEAR ELASTICITYGeneralized Hooke's Law . Strain Energy Function ...................... Isotropy . Anisotropy . Elastic Symmetry ................................ Isotropic Media . Elastic Constants .................................... Elastostatic Problems . Elastodynamic Problems ........................ Theorem of Superposition. Uniqueness of Solutions . S t. Venant's Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Two-Dimensional Elasticity. Plane Stress and Plane Strain . . . . . . . . . . . . . .140 141 142 143 145 145

CONTENTS

Page6.7 6.8 6.9 6.10

Airy's Stress Function ................................................ Two-Dimensional Elastostatic Problems in Polar Coordinates .............. Hyperelasticity . Hypoelasticity ........................................ Linear Thermoelasticity ................................................

147 148 149 149

Chapter

77.1 7.2 7.3 7.4 7.5 7.6

FLUIDSFluid Pressure . Viscous Stress Tensor Barotropic Flow ................ Constitutive Equations Stokesian Fluids Newtonian Fluids ............ Basic Equations f o r Newtonian Fluids Navier-Stokes-Duhem Equations .... Steady Flow Hydrostatics Irrotational Flow .......................... Perfect Fluids Bernoulli Equation . Circulation ........................ Potential Flow Plane Potential Flow ..................................

.

.

160 161 162 163 164 165

.

. .

.

.

.

Chapter

88.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 $8.9 8.10 8.11

PLASTICITYBasic Concepts and Definitions ........................................ Idealized Plastic Behavior .............................................. Yield Conditions . Tresca and von Mises Criteria ........................ Stress Space The 1I.Plane Yield Surfaces ............................ Post-Yield Behavior Isotropic and Kinematic Hardening ................ Plastic Stress-Strain Equations Plastic Potential Theory ................ Equivalent Stress Equivalent Plastic Strain Increment .................. Plastic Work Strain Hardening Hypotheses ............................ Total Deformation Theory .............................................. Elastoplastic Problems ................................................ Elementary Slip Line Theory f o r Plane Plastic Strain ....................175 176 177 173 180 181 182 182 183 183 184

.

.

.

.

.

.

Chapter

99.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8

VISCOELASTICITYLinear Viscoelastic Behavior ............................................ Siniple Viscoelastic Models ............................................ Generalized Models Linear Differential Operator Equation .............. Creep and Relaxation .................................................. Creep Function . Relaxation Function . Hereditary lntegrals . . . . . . . . . . . . . . Complex Moduli and Compliances ...................................... Three Dimensional Theory .............................................. Viscoelastic Stress Analysis Correspondence Principle ..................196 196 198 199 200 202 203 204

.

.

Chapter 1Mathematical Foundations1.1 TENSORS AND CONTINUUM MECHANICSContinuum mechanics deals with physical quantities which a r e independent of any particular coordinate system that may be used to describe them. A t the same time, these physicaI quantities a r e very often specified most conveniently by referring to a n appropriate system of coordinates. Mathematically, such quantities a r e represented by tensors. As a mathematical entity, a tensor has a n existence independent of any coordinate system. Yet i t may be specified in a particular coordinate system by a certain set of quantities, known a s its components. Specifying the components of a tensor in one coordinate system determines the components in any other system. Indeed, the l w o f a transformation of the components of a tensor is used here as a means f o r definicg the tensor. Precise statements of the definitions of various kinds of tensors a r e given a t the point of their introduction in the material that follows. The physical laws of continuum mechanics a r e expressed by tensor equations. Because tensor transformations a r e linear and homogeneous, such tensor equations, if they are valid in one coordinate system, a r e valid in any other coordinate system. This invarz'unce of tensor equations under a coordinate transformation is one of the principal reasons f o r 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 a r e known a s general tensors. When attention is restricted to transformations from one homogeneous coordinate system t o another, the tensors involved are referred to a s 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 r a n k , o r 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 a r e represented by scalars. Tensors of order one have three coordinate components in physicaI space and a r e known as vectors. Quantities possessing both magnitude and direction a r e represented by vectors. Second-order tensors correspond to dyadics. Severa1 important quantities in continuum mechanics a r e represented by tensors of rank two. Higher order tensors such as triudics, or tensors of order three, and tetradics, o r tensors of order four a r e 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 n 1 or zero vector is one having zero length and an u1 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 a s 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 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

(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)

a-b

= -b+a

= d

Fig. 1-2

CHAP. 1 1

MATHEMATICAL FOUNDATIONS

3

Multiplication of a vector by a scalar produces in general a new vector having the same direction a s 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. Multiplication 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 A,,,, > ~ ( I I I , .

For principal axes labeled OxTx2x8, the transformation from Oxixzx3 axes is given by the elements of the table$1

x2

x3

~12 ;

all = n(l)aZ1 n(2) =a31

a12= n(1) 2 aZ2 n(2) =

al, = ni1)(L*,

= n(32)

E;

= ni3)

=

a,, = n 3 )

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

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

A comparison of (1.135) and (1.137) indicates that Tij and a11 its integer powers have thesame principal axes. Since each of the principal values satisfies (1.133), and because of the diagonal matrix form of T ngiven by (1.138), the tensor itself will satisfy (1.133). Thus

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,

22

MATHEMATICAL FOUNDATIONS

[CHAP. 1

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 42) .1 tensor fields of various orders are represented in indicial and symbolic notation a s follows,

(a) scalar field: + = +(xi, t) or vi = vi(x, t) or (b) vector field: (c) second-order tensor field: Tij = Tij(x, t) or

$J =V

+(x, t) = V(X, t)

(1.143) (1 ,144) (1.145)

T = T(x, t)

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

i

aaXi

=

A

eidi

(1 46) .I

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

avi (b) - = vi,i axi (c)

(e)

2*

aTij

=

Ti!,

k

avi a2Tij - = vi,j ax ( f ) a s r d z ,km From these examples i t is seen t h a t 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 f o r reference.

divv = V - v curl v = V x v

orOr

divt = visi ~ijkajvk= 'ijkvk,i

(1.148) (1.I49)

CHAP. 11

MATHEMATICAL FOUNDATIONS

23

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

XF-~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,Fidxi (zi)* Fidxi (1.I 52)

J

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. . I n 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 incorporate a tensor field of any order. Thus for the arbitrary tensor field Tijk... the theorem is written Tijk . . . p dV = Tijknp dS (1.157)

24

MATHEMATICAL FOUNDATIONS

[CHAP. 1

Solved ProblemsALGEBRA 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 = 7A

+A

v = VIV = (217)i

+ (317)j

+

A

- (617)kA

A

( b ) The vector extending from P to Q is

u =

(O-I)?+A A

(2-O);+ (1-3)k -2kA

= -i+2ju = d(-1)2Thus or

+ (2)2+ (-2)2A

= 3A

Fig. 1-12directed from P to Q directed from Q to P

u = -(1/3) i

6

=

+ (213); - (213)k (1/3)? - (213)3 + ( 2 / 3 ) k

1.2.

Prove that the vector v = a? b c k is normal to the plane whose equation is a x by cz = A.Let P ( x l , yl, zl) and Q(x2,y2, z2) be any two points in the plane. Then a s 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 -*- - u [ ( x z - x IA ~ + v 2 - y 1 ) A ) A ( j u ( 2 , - 2,) k ] [a? b3 cc] 1 ( a s 2 by2+ cz2- axl - byl - cz,) = X - X = O = -

+ i+ + +

Iz

+ +

+ +

+

+

U

+ +

+

+

Z

U

Since u is any vector in the plane, v is I to the plane.

Fig. 1-15

13 ..

If r = x? + y 3 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,

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

A

A

A

A

A

A

= x2+y2+z2-ax-by-czAdding d2/4 = (a2

= O

+ 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 [ ( bX c ) X r]. By direct expansion of the cross product in brackets, a X [ ( bX 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 -

CHAP. 11 Thus -(b

MATHEMATICAL FOUNDATIONS 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 ) bx c

25

+

(b.r)c

X

a

+ ( c * r ) aX 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 . linear dependence or independence of the basis

Check the

u = 3i+j-2k

A

A

A

v = 4 i - j -kA

A

A

A

w = i -2j +kThe vectors a, b and c are linearly dependent if there exist constants h, p and v, not a11 zero, vc = O. The component scalar equations of this vector equation are such that ha pb

A

A

+ +

+ pbY + ha, + pb, + vc,ha,VC,

= O = O

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

a y by cY a, b, c, which is equivalent to a b X c = O.

=

O

For the proposed basis u, v, w,

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

+ w.

16 ..

Show that any dyadic of N terms may be reduced to a dyadic of three terms in a 3 form having the base vectors l, 2, as (a) antecedents, (b) consequents.

+ a2b2+ - - - + aNbN= aibi (i = 1,2, . . .,N ) . ( a ) In terms of base vectors, ai = aliel + azie2 + a3ie3 = a j i e jLet D = alblA A

A

A

and so D = a-2.b. = .(a..b.)= $.c. 3 1 3 ~ 3 t t 1 1 3

with j = 1,2,3.

( b ) Likewise setting bi = bjij i t follows that D = aibjij = (bjiai)j= g j 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.D

Then

v = al(b, v)=(V.

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

v . D,

1.8.

Prove that ( D , D ) , = D, .D.From (1.71), = D i j i j and D , = D j i i j . ThereforeD,.DA A A A = D.....D p q pq = DjiDpq(ej'ep)eieq jt I1 A

and

(o, o),

= DjiDpq( e j e,) e,ei

A

A A

= Dpqeq( e , e j ) e i D j i = D,,$,

A

A

A

A

.~

A A

~ = D, D e ~

.

~

e

~

26

MATHEMATICAL FOUNDATIONS

[CHAP. 1

1.9.

Show that (D x v), = -v(D X v), =

X

Dc.V)' ' '

D X v = al(bl X V)

+ a,(b2 X + + aN(bNx V) - + (bNX v)aN (bl X v)al + (b, x v)a2 +-(V

=A A

X b,)al - (v x b,)a, -

..- - (v X bN)aN =

-v X D,A

1.10.

If D = a ii b j j + c k k and r is the position vector r = x i that r D r = 1 represents the ellipsoid ax2 by2 cx2 = 1.AA AA

+

+

+

+ y j + xk,A A

show

r - ~ . r (s?+ =A

y3+z k )A A A A

(a??+A

b33+ ckk)A A

(s?+

zk)

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

11 . Forthedyadics D = 3 ? ? + 2 j j - j k + s k k .1 and F = 4 i k - t - 6 3 3 - 3 k j + k k , pute and compare the double dot products D :F and D F.

..

com-

ab

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

+ +

D : F = 12

+ 5 = 17.

Also, from

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??AA

+2 ; :

.

AA

3 k + 5 k k ) . (4;: + :6 :A A A A

-3k3A A

+kk)

= 12ik+12j j + 3 j jAA AA AA

-

jk -15kj+5kkAA AA

AA

Similarly,

H

= (4ik+6jj-3kj+kk).(3ii+2jj-jk+5kk) = 20ik+123!-63k.-6k3+8kkAA

AA

AA

1.13.

Show A directly from theAnonion form of the dyadic D that D = (Do?)? 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,II

+ (D-?)? +

= (D,,iA

+ D,,j + D,,k) i + (D,, i + D,,; + Dryk)j + (Dxzi + D,,? + D,$); = d l i + d 2 j +d,k = (o-?)?+ ( D * 3 ) 3 + ( ~ - 2 ) kA A A A A A A A A A A

Also now

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

A

e

A

A

6

A

= D,,

j * d l = ; . ~ . i = D,,,

6

j * d 2 = 3 . ~ . j = D,,,

A

etc.

1.14.

For an antisymmetric dyadic A and the arbitrary vector b, show that 2 b . A = A, x b.From Problem 1.6(a), A = &cl orA

2 = A

+ &c2 + $,c3; and because it is antisymmetric, ( elel + e2c2+ e3c3 - elel - c2e2 - c3e3)A A A

2 = (A - A,) A

A

A

= (lcl - elel

A

and so

2b A = [(b $,)cl - (b c,) ,]

=

+ [(b.2)c2- (b . 2]+ [(b 3)c3 [ ( , x c 1 ) x b + ( z X c 2 ) X b+ ( 3 X ~ 3 ) X b ] (AvXb) =C,)

.

.

+ e2c2- c2e2 + esc3 - c3e3)A A A A

-

(b c3)$31

1.15.

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

+

AA

A

A

A

Since u X v = ( 2 i + k ) X 5 j = 1 0 k - 5 i ,D * ( U X V= ( 6 i i + 3 i j + 4 k k ) - ( - 5 i + l O k ) )AA AA A A A A

A

A

A

A

A

= -30?+40kAA AA AA A A

Next, snd

D

~

= (6??+3?;+4kk)~(2i+k) UAA AA AA

A

A

= -6ik+8kj-6ij+3iiAA A

AA

(DXu)*v = ( 3 i i - 6 i j - 6 i k + 8 k j ) .

5 j = -30i+40k

CHAP. 1 1

MATHEMATICAL FOUNDATIONS

27

Considering the dyadicA A A A A A

D =

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

as a linear vector operator, determine the vector r' produced when D operates on r = 4?+23+5k.

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.

+ +A A

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

uv

= f(i) = i = f(j) = jA A

A

A

A

+ iXr + j XrA A A A A

= i -zj+yk = xi

A

A

+jA A

A

-xkA

A

w = f(k) = k + k X r = - y i + x j + k Then and = ui+vjA A

A

A

+ wk

A

= (i -zj+yk)i+(xi

A

A

A

t j -xk)j +(-yi+xj+k)kA

A

A

A

A

A

A

a = D b = (b,

+ b,z

- b,y) i

A

+ (-b,z + b, + b,x) j + (b,yA

- b,xA

+ b,) k

A

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

+bXr

= b,i -i-b,3

A

+ b,k + (b,z - b,y) i + (b,x - bxz) j + (bxy- b,x)kA A

A

1.18. Express the unit triad 4, in terms of , , A A A i, j, k and confirm that the curvilinear triad A is right-handed by showing that x = e,. , ,By direct projection from Fig. 1-15,A

e,

=

(COS

g cos e) iA

A

+ (cos g sin e) jA

A

A

- (sin g ) kA

A

e, e,

= (- sin e) i

A

=

+ (cos e) j (sin g cos e) i + (sin g sin e) j + (cos @) kA A

and soA

Fig. 1-15

ee+

A

1

k- sin

x ,

=

cos g cos e- sin eA

cos g sin e cos e

@

OA

=

(sin @ cos e ) i

+

(sin g sin e) j

+

[(cosze

+ sin2 e) cos g] kA

=

2,.

1.19. Resolve the dyadic D = 3: ; and antisymmetric parts.Let D = EE

+ 4:: + 6 3 1+ 7;: + 10 ko;f 2 k3ThenA A A A A A A A A A

into its symmetric

+F

where E = E, and F = -F,.

= (1/2)(D D,) = (1/2)(6 i iA A

+

+ 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 A A A A A A A A A A A A A A A A

A A

A A

= 3ii+3ij+7ik+3ji+7jj+jk+7ki+kjA A A A

= E,A A

F

= (1/2)(~-o,) = ( 1 / 2 ) ( 4 i k - 4 k i + 6 j i - 6 i j + 1 0 k i = -3i j - 3 i k + 3 j i A A

10ik+2kj-2jk)

jk+3ki+kj

= -F,

28

MATHEMATICAL FOUNDATIONS

[CHAP. 1

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 rela-

tionships for constructing the reciprocal base vectors and carry out the calculations for the basis bl = 3 i + 4 3 , bz = - i + 2 ? + 2 k , bs = ? + $ + %A A A

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 b l , b2,b3, l / h = bi ba X b3 = 12 and so

bl = (b, x b3)/12 = ( j -k)/4 b2 = (b3x bl)/12 =

A

A

- i13 + j 14 + k / 1 2A A A

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 Bijj represents three sums:(1) For (2) For

(3) For

+ A,, + A,,. i = 1, Blll + B12, + B133. i = 2, BZll + BZz2 BZ33. + i = 3, B3,, + B322+ B333.R32> R33.

Rii represents the nine components Rll, R12,Ri3,R,,, Rz2,RZ3? ai Tij represents three sums:(1) For j = 1, al Tll

(2) For j = 2, (3) For j = 3,

+ a2 T2i + a3 T3i. a1T12+ a2TZ2f a3 T3,. a1T13+ a2T23 + a3T33. ++

aibjSijrepresents 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 ~ ~ 3 ~ 1 ~ ~ 3 ~ 2 S 3 2~ 3 ~ 3 ~ 3 3 2 3 3 1

+

+

+

122. Evaluate the following expressions involving the Kronecker delta

aij

for a range of

three on the indices.( a ) Sii = S l l

+ Sz2 +

= 33

+ 62jS2j + S3jS3j = 3 S i j S i k S j k = S l j s l k 8 j k + 82j62k8jk + 83j83k8jk = ( d ) Gijsjk = s i l S l k + + 8i383k = Sik(b) SijSij = S l j S l j(C)

Si2S2k

(e)

6ifiik =

S1#11,

+ 82jA2k + 83jA3kCijk

= Ajk

1.23.

For the permutation symbol( b ) ' i j k a j a k = 0.(a) First sum on i,

S~OW by

direct expansion that (a)

Cijk'kij

= 6,

.. k i.. = i ~ j

ljkCklj

+ ' 2 j k e k 2 j + '3jkek3j

CHAP. 11

MATHEMATICAL FOUNDATIONS Next sum on j. The nonzero terms are'ijk'kij

29

=

'12kEk12

+ '13kCk13 + '21kek21 f

'23kek23

+ '31kCk31 + '32kEk32= 6

Finally summing on k, the nonzero terms aree.. .. ~k k21

'

-

'123'312

= ( l ) ( l ) (-I)(-1)( b ) Summing on j and k in turn, Eijkajak = ' i l k a l a k = 'i12ala2From this expression, when i = 1, when i = 2, when i = 3, Note thateiikajak eljkajak 2jkajak 3jkajak

+

+ '132'213 + '213'321 + '231'123 + '312'231 + '321'132

+ (-I)(-1) + ( l ) ( l+ ( l ) ( l+ (-I)(-1) ) )

+

Ei2ka2ak

+

Ei3ka3ak

+ ' i 1 3 ~ 1 ~ 3 ' i 2 1 ~ 2 ~ 1 'i23a2a3 f +

f ' i 3 1 ~ 3 ~ 1 Ei32a3a2

+

= a2a3- a3a2 = O = a i a 3 - a3a1 = 0 =a1U2

- a2al =

0

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

1.24. Determine the componentfi f2 'ijkTjk

f2

for the vector expressions given below.=-T13

= = = = =

'2jkTjkC,, j b j

=

213~13

+ '231T31

+ T31

( b ) fif2

- cj,ibj

c2,1b1

fc2, 1 - ~ l , 2 ) ~ 1 ( ~ 2 . 3 ~ 3 , 2 ) ~ 3 Bijf; B21f:

+ ~ 2 , 2 +2~ 2 , 3 ~ 3 - ~ 1 , 2 ~~12- . 2 ~ 2 ~ 3 , 2 ~ 3 ~ + +B22f2

(C)

fi f2

=

+ B23f3*Dijxixj

1.25.

Expand and simplify where possible the expression (b) D i j = - D j i .Expanding, Dijxixj = Dljx1xj

for

(a)

Dij

=

Dji,

+ D2jx2xj+ Dsjx3xjD33(x3)2

=

D11x1x1

+ D 1 2 x 1 x 2 + 0 1 3 x 1 5 3 + D 2 1 ~ 2 ~ 1D 2 2 ~ 2 ~ 2 + + D23x2x3 + D31x3x1 + D32x3x2 + D33x3x3

(a) Dijxixj =

D11(x1)2

+

D22(x2)2

+

+ 2D12x1x2+ 2D23x2x3+ 2D13x1x3etc.

( b ) Dijxixj = O since Dll = -Dll,

D12 = -D2,,

1.26.

- 8 i q s j p for (a)i = 1, j = q = 2, p = 3 and for (b) i = q = 1, Sh0w that c i j k c k p q = 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,a11 values. Then'ijk'kpq

q =2

and note that since k is a summed index i t takes on-

-

'12kEk32 - E121E132

+ '122'232 + '123'332- S12S23 = Oeijkekpq

= Oand- SiqSjp =

and

SipSjq

q

aiqS,

=

813S22

(b) Introduce i = 1, j = 2, p = 2, S12S21 - S11S22 = -1.

= 1.

Then

= ~123'321 = -1

SipSjq

1.27. Show that the tensor Bik = cijkaj is skew-symmetric.Since by definition ofcijk

an interchange o f two indices causes a sign change, Bik = eijkaj = - ( e k j i a j ) = - ( B k i ) = -Bki

30

M A T H E M A T I C A L FOUNDATIONS

[CHAP. 1

1.28. If Bij is a skew-symmetric Cartesian tensor for which the vector show that Bpq= t p qbi . iMultiply t h e given equation b yepqi

bi = ( + ) E ~ ~ ~

and use t h e identity given i n Problem 1.26.

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

+

1.29. Determine directly the components of the metric tensor f o r spherical polar coordinates as shown in Fig. 1-7(b).W r i t e (1.87) a s gpq = axi azi

;j~q

and label t h e co@

ordinates as shown i n Fig. 1-16 (r = e,, Then x , = 8 , sin 0, cos 8, x , = e , sin e, sin e,

= e,, e = e,).

Hence

ax1 ao1

-

sin O , cos 8 ,

ax1 - as2 8x2 - 302 8x3 as2-

e , cos 0 , cos e3

Fig. 1-16 8x1 - -e, sin e, sin e, 863 8x2 as3 8x3 - = 8636,

8x2 861

-

sin e2 sin e,

0,

cos O , sin

e,

sin e2 cos e,

-el sin e,

o= 1

f r o m which g,,

axi axi = -as, asl

-

sinz e, cos2 8 ,

+ sin2 o, sin2 e, + cos2 e,

g33

=

axi axi -- ao3 ae,

eSsin2 e, sin2 e,

+ eSsin2 6 , cos2 e ,

= eSsin2 O ,

Also, gpq = O for p Z q. For example, axi axig12

=

aBi dB., .e

= =

(sin e, cos @,)(e,cos e, cos e,)

+ (sin O 2 sin e3)(e1cos e, sin e3) o

(COS

eZ)(elsin

e,)

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 = gpqde, de,. T h u s f o r t h e line element (de,, O , O ) , i t follows t h a t (ds), = g l l ( d e l ) 2 and d s = \/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( 1 ) For (del,O,O),( 2 ) For ( O , de,, O ) ,

MATHEMATICAL FOUNDATIONSds = del = d r ds = e, de, = r d+ ds =

31

( 3 ) For (O, O, de,),

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,, =

812

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) a s dxi = -dek, and since (A), cos plz h,dsz, =aek

66.

-

ax, a%, - -de, de,

ae, ae,

ax, + as, ax, de, de2 i- - ax3de1 de2 ae, ae2 ae, as,-

= giz de1 dez

Hence using the result of Problem 1.30,

tos

plz = g,,

del de2 - -= ds1ds2

812

66

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

+

+

From the definition (see Section 1.13) ai; = cos (si, j ) and Fig. 1-18, the table of dire8ion s cosiiies is51

22

x3

X;

cos e-sin

sin e cos 6

OO

x x3

e

ocos 6

o

1

Thus the transformation tensor is sin 6 cos 6O O

A

=

-sin e

Fig. 1-18By the transformation law for vectors (1.94),

v ; = a I j v j = v , cose v i = a Z j v j = -vl v j = a3jvj =V,

+ v , sin 6 sin e + v , cos 6

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.

32

MATHEMATICAL FOUNDATIONS

[CHAP. 1

The unit vector along the x axis is given by the first row of the table a s = (315)1- (415) 2 . Also from the table 2; = 3. For a right-handed primed system 2 = 3 Xi, or A t, = e [ ( 3 / 5 ) 1 (4/5)$2]x -

$I

2 = (-3/5)& - (416)$~and ,

the third row is

1 1x:

-415

1 -3/5 1

O

I.

1 3 . Let the angles between the primed and .4unprimed coordinate directions be given by the table shown on the right. Determine 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 aij-

1

112 llfi 112

-112

o1 require:

The orthogonality conditions aiiaik = S i ,

-112

1.

For j = k = 1 that a l l a l , a2,a,, the elements in the first column.

+

+ a3,a3, = 1

which is seen to be the sum of squares of

2.3.

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

+

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 3 . Show that the sum x A + pBij represents the components of a second-order tensor .5 ~ ~ 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 and Bii = apiaqjBiq. Hence q XAij

+ @Bii =

~

(

~

~

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

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

1 3 . S ~ O that .6 W

(Pijk

+

Pjki

+

Pjik) XiXjXk

= 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 dummyvariables. Thus replacing i, j, k in the second and third terms by p, q , r , the sum becomesPijkxixjxk

+ Pqrpxpxqxr + Pqprxpxqx,

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

+ Pijk

+

Pijkxjxixk = 3Pijk xixjxk

1.37.

If Bij is skew-symmetric and Aij is symmetric, show that AijBij = 0.Since A . = A l,t. and B . . = -B.. A..B.. -A..B.. or A..B..+ A..B.. = A..B..+ A P P BPP = O. 31' 13 t3 I1 I1 1J 13 31 31 L3 i 1 Since a11 indices are dummy indices, ApPBpq AijBij and so 2AijBij= O, or AijBij= 0. =

CHAP. 1 1

MATHEMATICAL FOUNDATIONS

33

1.38.

Show t h a t the quadratic form Dijxixjis unchanged if Dij is replaced by its symmetric part D(ij).Resolving Dij into its symmetric and anti-symmetric parts, Dij = D(;j) Then D,ijlxixj =

+ DLijl = S ( D i j+ Dji) + &Dij Dji) #Dij + Dji)xixj = S(Dijxixj+ Dpqxqxp) = Dijxixj-

1.39.

Use indicial notation to prove the vector identities (1) a x (b x c) = ( a *c)b - ( a . b)c,w p = ePqiaqeijkbjck (1) Let v = b X c. Then vi = eijkbjck; and if a X v = w, then

(2) a x b a = O

.

- ( 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 skewsymmetric 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 cijkA1iA2jA~k.From (1.52) and (1.109) the box product [abc] rnay be written a1 a2C2

a3C3

x

= a.bXc

=

[abc] =

eijkaibjck =

blC1

b2 b3

If now the substitutions ai = A l i , b, = A Z i and ci = A s i are introduced,X = eijkaibjck = eijkAliAZjA3k

This result may also be obtained by direct expansion of the determinant. An equivalent expression for the determinant is eijkAilAj2Ak3.

1.41.

If the vector

vi

is given in terms of base vectors a, b, c by vi =

aa i i

+ pbi +

yCi,

34

MATHEMATICAL FOUNDATIONS

[CHAP. 1

By Cramer's rule,

a

=

I v3a1 a2

b3 bl b2

c3 c1 c2

I

and by (1.52) and (1.I 0 9 ) , a = eiikvibjck'pqrapbq~r'

Likewise /3 =

EijkaivjCk epqrapbqcr

cijkaibjvk = ~pqrapbqcr'

MATRICES AND MATRIX METHODS

(Sec. 1.17-1.20)

1.42. For the vectors a = 3:+42, b = 23-62 and the dyadic D = 3 ? ? + 2 ? k compute by matrix multiplication the products a D, D b and a . D b. 4 3 - 5k^

5

3,D

Let a

= v; then [ v l ,v2, v3] = [3,0,4] O -4

[ ]-6

.

= 9 , -20,6].

Let D b = w; thenO -5

.

-10

Let a

D

b = v . b = h; then

[h] = [9, -20,6]

(

2

1=

[-761.

1.43.

Determine the principal directions and principal values of the second-order Cartesian tensor T whose matrix representation is 3 -1 o -1 3 O [Tij] =

OFrom (1.132), for principal values h,

0

1

which results in the cubic equation h3 - 7h2 values are h ( 1 , = 1, = 2, h ( 3 )= 4.Ao,

+ 14h - 8 = ( h - l ) ( h- 2 ) ( h- 4 ) = O

whose principal

Next let n:l) be the components of the unit normal in the principal direction associated with 24'' = 0, from which n:" = = 0 ; and from nini = 1, nB1)= *I.

= 1. Then the first two equations of (1.231) give 2n:" - nil' = O and -n:"

nll)

+

For Forn33'

h t 2 ) = 2,

ny' = ni2' = %1/\/2,= 0,

(1.131) yields ni2' = 0 , -ny' since nini = 1 and ni2' = 0. -ni3'

nl2'

+ nf'

= O,

and

-ny'

= O.

Thus Thus

h,,, = 4, (1.131) yields n : 3 ) = -nF' = TI/@.

- n:'

= 0, -ni3' - ns' = O, and

3 n F ) = O.

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

CHAP. 11

MATHEMATICAL FOUNDATIONS

35

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 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.

+

A

n(2)

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

A

e 1

e2

e3

i

i1

-116

o o

=

(4+ 4)3

= e3

A

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

Fig. 1-19

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).

36

MATHEMATICAL FOUNDATIONS

[CHAP. 1

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

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. directions to be perpendicular.

Hence n:2'n:3' = 0 , the condition for the two

1.47.

Compute the principal values of axes coincide with those of T.3 -1 [Ti.

( T ) 2 of

Problem 1.43 and verify that its principal3 -1

=

[-i i !][-i i

!] [-:10 -6

=

1;

The characteristic equation for this matrix is10 - h -6 -6 10 - h

O Ol - h

= ( 1 - h)[(10- h)2 - 361

= ( 1 - X)(h - 4)(h- 16) = O

onini = 1,

O

from which h(,) = 1,gn"'1

= 4, h ( 3 )= 16.- Gn(l'2

Substituting these into (1.131) and using the condition

= = O =

For h ( , ) = 1,

-6n:"6n(2) 1

$ .

9n;"2

n21'

= 0'

n(l) 3

=21

- Cn(2) = O

For A ( 2 , = 4,

-Gn?'

+ 6nk2'-Sn'2'3

= 0-

=

ni2'= k l f i , nF) = 0

- 0

For

= 16,-15ni3'

= O

which are the same a s 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 fiwhen

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

CHAP. 1 1

MATHEMATICAL FOUNDATIONS

37

with the transformation matrix being

[aijl =

[

llfi0

lldzllfi

llh11,h

Therefore

fi

=

;" 3oo

-216

and using [aij; to relate this to the original axes by the

transformation

fi = ~ A, the matrix ~ f i

equation is

CARTESIAN TENSOR CALCULUS

(Sec. 1.21-1.23)

1 4 . For the function x = Aijxixj where Aij is constant, show that dhldxi = (Aij Aji)xj .9 and d2X/dxidxj = Aij + Aji. Simplify these derivatives for the case Aij = Aji.Consider - = axk

+

ax

A.. -X .%3axk 3

axi

+ A,.%.a%, ' a> 'axk

Since

-z

axi

Jxk

Sik, i t i~ seen that Jxk

ax = A k j z j+ Aikzi =

(Aaxk

+ A jk)xj.

a2A 8% Continuing the differentiation, -- ( A k j + A . ) ax, axk lk axp

- A k P + A P k .If A . . = A . 23 3i9

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

1.50.

Use indicial notation to prove the vector identities (a) V

X

V+ = O, ( b ) V V

X a

= 0., ~

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

+

+,

( b ) V V x e = h = ( ~ ~ ~= ~ a ~ = O since eijkak,ji ,

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 = (2/7)&- (3/7)&- (6/7)3 or & = (261- 32 - 63)/7.

n

+

The required derivative is - = V h n = X,ini.an

ax

A

Thus

381.52.

MATHEMATICAL FOUNDATIONS

[CHAP. 1

Ifxk,

Aij

is a second-order Cartesian tensor, show that its derivative with respect to namely Aij,k, is a third-order Cartesian tensor.F o r the Cartesian coordinate systems xi and xl, xi = ajixi and xildxi = ajt. Hence

which is the transformation law f o r a third-order Cartesian tensor.

1.53.

If r2= Xixi and f ( r ) is a n arbitrary function of r, show that ( a ) V ( f ( r )= f'(r)xlr, ) and ( b ) V 2 ( f ( r = f l r ( r ) Zf'lr, where primes denote derivatives with respect to r. ))

+

a f ar ( a ) The components of V f a r e simply f,+ Thus f,i = - - ; and since ar axi ar 2. a f ar follows t h a t - = 3. Thus f,i = - - = f'xi/r. dxj r ar axi

axj

= 2r ax, = 2 ~2 . .~x . i t -a

ar

Use the divergence theorem of Gauss toS ~ O W that

L

Xinj

d S = V6ij

where nj d S

represents the surface element of S , thebounding surf ace of the volume V shown

in Fig. 1-20. xi is the position vector of n d S , and n its outward normal. j i

= SijV

Fig. 1-20

1.55.

If the vector b = V x v, S ~ O W that a scalar function of the coordinates.Since b = VX

i

hbisdS =

i

h,ibi d V

where h = h ( x i ) is

v, bi = eijkvk, and so

i

hbini d S

=

i

rijkhvk, d s jni

=

& Li

bi d ~

since hrijkvk, = O ii

CHAP. 11

MATHEMATICAL FOUNDATIONS

39

MISCELLANEOUS PROBLEMS 1.56. For the arbitrary vectors a and b, show thatA = ( a x b) (a x b)

+ (a.b)2 =

( ~ b ) ~

Interchange the dot and cross in the first term. Thenh

= a - b x (axb) = =

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

+ (a.b)(a0b) + ( a . b)(a0b) ( a . a ) ( b - b ) - ( b . a ) ( a W b+ ( a . b ) ( a . b ) )(ab)2

since the second and third terms cancel.

1.57. If

ii = o x u and V

=

w

x v, show that

d (u X v) = dt-

c X a

(u x v . )

(a) In symbolic notation, d z(uxv)

= h X v + ux;=(V.

= (mXu)Xv+ uX(mXv)U)OI@)V

m)u -

(V

+

(U

v)a a X (U

(UV)

@)V

= (V' ,)U ( b ) I n indicial notation, let

(U

=

X

d

(u X

V)

=d

W.

Then

wi = and since Aj = e j p q w p u q andW i.

( E ~ ~ ~= ~ v ~ ) u eijkGjwk

+

eijkuj;k

ck = ekmnamVn,+ eijk'kmnUjamvn- S i n S j m - SimGjn('ijkEkmn

y k E3pqapUqwk .. ,

-~ink'kmj)~~jamwn

and using the result of Problem 1.59(a) below,wi =(SimSjn (SijSmn

+ SijSmn)~jamwn~irnk'~j~Ujarnw~

=

-8in8jm)uj~m~n E

which is the indicial form of

m

X (u X v).

8mp

8mq 8nq 8rq

Sms 8ns 8rs

1.58.

Establish the identity

cPqsrmnr

=

Snp 8rp

.A11 -412 A13 A23 A33

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

det A = AZl Az2A31 A32

.

An interchange of rows

A,, A,,A31

A,, A,, A,,

A23

A12

A,,A,,A31

A13

A,,

=

A,,

AZ3A33

=

-detA

A,,

A,,

and for an arbitrary number of row changes,Aml Anl Ar1 Am, Ana Am3 An3 4 3

-

em,,

det A

A,

40or column changes,

MATHEMATICAL FOUNDATIONS

[CHAP. 1

A,,-42,

A,,-42, -43,

A I S

-4% A,,

-

epqS det A

A,,

Hence f o r a n arbitrary row and column interchange sequence, A,,Anp

Am,Anq

Ams Ans Ars-

'mnr'pqs

det A

A,

Ar,

When Aij = Sij, det A = 1 and the identity is established.

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

+

+ Sms(SnpSrq

-

SnqSrp)

( a ) Identifying s with m yields, 'pqs'snr

= Ssp(8nqsrs - SnsSrq)-

f

S s q ( 8 n s s r p - SnpSrs)-

+ S s s ( S n p ~ r q- SnqSrp)

srpsnq

- SpnSrq + SqnSrp

SnpSqr

+ 3SnpSrq - 3Snqsrp

- S n p a r q - SnqSrp

( b ) Identifying q with n in ( a ) , ~pqs'sqr - Sqparq

- aqqsrp

= S p r - 3 S p r = -2Spr

1.60.

If the dyadic B is skew-symmetric B = -B,,Writing B = bll

show that B , x a = 2a B.

+ b22+ b33 (see Problem 1.6), B, = bl x 1 + bz X z + b3 X 3 + bzz+ b 3 3 ) - a .( l b l + 2b2 3 b 3 ) += 2a.B

and

B , X ~

= (blXl)xa+ (b2x2)xa+ (b3x3)xa A = ( a b l ) , - ( a . Al ) b l e ( a . b , ) z - ( a e z ) b 2 -t ( a b 3 ) 3 - ( a 3 ) b 3

+

= a (bll-

a.0 - a.0,

1 6 . Use the Hamilton-Cayley equation to obtain .1 Check the result directly by squaring ( B ) 2 .The characteristic equation f o r B is given by

(B)4

f o r the tensor B =

By the Hamilton-Cayley theorem the tensor satisfies its 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

+

+

+

CHAP. 1 1

MATHEMATICAL FOUNDATIONS

41

Checking by direct matrix multiplication of (B)2,

(23)4

=

O 26

are invariant under the coordinate trans1.62. Prove that (a) Aii, (b) AijAij, ( c ) c i j k ~ k j p A t p formation represented by (1.1 03), i.e. show t h a t Aii = ~ : i ,etc.(a) By (1.103), Aij = aPiaqjAP,. Hence Aii = a,,aqiA;, = S,,A:,= A:, = Aii.

(b) AijAij = a p i a q j A ~ q a m i a n j =~ n ~ ~ s ~ ~ A PApqApq = A!.A!. A B = ~ A & ~ 21 13(C)

cijk'kjpAip = ijk'kjpamianp~hn ( 8 i j s j p- ~ipsjj)amianpAkn =-

- ( S m n - &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 ~ i j i but that the product TijSijof 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 = eijk T L j k l since ) cijkT(jk) O (cijk is anti-symmetric in j and k, T ( j k ,symmetric in j and k). = For the product T i j S 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.D = lIijij and E = ~,,,,. By (1.31), D : E = D i j ~ , , ( i p)(j2,). By (1.35), = DijEpq g j $,)(Si 3 = DijE,, (j A A A Now interchanging the ( , ) %)( ei e,) since E,, = E,,. A A role of dummy indices p and q in this last expression, D E = q)( e,). ei

Write

D..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 ~ orVp

+

=

'pqi'ijk(Spjsqk

(ajbk),q = epqi'ijk(aj,qbk ajbk,q)

=

- Spksqj)(aj,q

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

-

bk

+ ~ j ~ k , q= )

bq

- aq,q bp

+

- aqbp,q

+ a(V

b) - a Vb.

1.66. By means of the divergente theorem of Gauss show t h a t

& 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 the volume integralq p j ,

(rqPirijkajxk),, and since a is constant, the last expression becomes dVp

s,

cqPinpcijkajxkdS. By (1.157) this becomes

d

=:

(Sqjspk

- s q k s p ~ a j x k , p dv

=

(aqxp,p - apxq,p) 'V

L

(aqS p p - a, a,,) dV =

(3aq- a,) dV = Za,V

42

MATHEMATICAL FOUNDATIONS

[CHAP. 1

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. I n matrix form, by (1.117),

i-:

L 0 0 1 ] L 0 0 1 J

:l[u-i 01I

r: : 01L 0 0 1 JA A A A A A A

Fig. 1-21

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

= =

A

=Hence

+ kk) X (v,i+v,j +v,k) i(v,k-v,j) + j(-v,k+v,i) + k(v,j-vyi) (vXi)i + (vXj)j + (vXk)k = v X IA A A A A A

(i i + j jA

A

A

A

A

A

A

A

A

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

Supplementary Problems1.69.

Show t h a t u = i j - k and v = i - j a r e perpendicular to one another. Ans. w = ( - l / f i ) ( ? + 2c) t h a t u, v, & forms a right-handed triad.

A

+

A

A

A

A

3+

Determine w so

h

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 vector and a is a constant vector.

X

x = O, (c) a . V x = a where x is the position

1.72.

Determine the principal values of the symmetric p a r t of the tensor Ans. h ( , , = -15,h ( Z )=

Tij =-3-18

5 , h,,) = 10

CHAP. 11

MATHEMATICAL FOUNDATIONS

43

1.73.

F o r the symmetric tensor of the principal axes.

Tij =

determine the principal values and the directions

Ans.

h c l , = 2 , A ( z , = 7, A,,, = 1 2 ,

1.74.

Given the arbitrary vector v and a n y unit vector , show t h a t v may be resolved into a component parallel and a component perpendicular to 2, i.e. v = ( v * ) X ( v X 3 ) .

+

1.75.

If

V v = O, V x v = 6 and V X w = -c,

.

show t h a 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 show t h a t ~ det (AB) ~= ~det A det 0. ~ ~ A ~ A ,

1.79.

Verify t h a t (a) O3,,vp= v3, ( b ) 8siAji = Aj3,

(C)

Sijeijk

= 0 , ( d ) ai2aj3Aij = A23.

1.80.

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

( a ) Show t h a t ' t h e orthogonality conditions aijaik= S j k and apqaSq = ( c ) W h a t is the equation of the plane xl - x,

a,,

a r e satisfied.

( b ) W h a t a r e the primed coordinates of the point having position vector x = 2 ,

- 3?

+ 3x3 = 12fiX;

in the primed system? = 1

Ans. ( b ) ( 2 / 5 f i , 1115, - 2 1 5 f i )

( c ) fiX ;

- X; -

1.81.

Show t h a t the volume V enclosed by the surface S may be given a s V = Qx ia the ponition vector and n the unit normal to the surfaee. Hint: Write V = (116) and use (1.157). '

h

1S

(xixi),ini dS

Analysis of Stress2.1 THE CONTINUUM CONCEPTThe molecular nature of the structure of matter is well established. --- numeroIn -investigations of material behavior,-- however, the individual molecule is of no conce- _and ----only tKe35havior of & material as a whole is-deemed important. For these cases the e . 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 - _- quantities such as stress and displacement are expressed field __ _ functions of the space coordinates and time.--_-_r

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 - the same in a11 directions - a is - at , _- --- -- -_ point. A material is called anisotropic with gspect to those . properties which are directional -- --- a t a pi$t. --____I_--- - - - A

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 thereforeP(av)

-

anlAV(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 = -M d= lim -*"-O

AV

d!'

(2.2)Fig. 2-1

Mass-density

p

is a scalar quantity.

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 a s 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 d f i / d S as AS approaches zero a t the point P- . , while a t the same time the moment of afiabout the point P vanishes in-thgimiting process. The resulting vector d f i l d S (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.__h

46

ANALYSIS O F STRESS

[CHAP. 2

Mathematically the stress vector is defined bydfi

dS

Or

f (n)

A

=

as-o

1im

.

-

Af

AS

=

df

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 ASchosen 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 V upon the material outside is the vector -t:"'. Thus by Newton's law of action and reaction, = - t' (2.5) The stress v e c t o r is very often referred to as the t r a c t i o n v e c t o r .A

"'h

2.5 STATE O F STRESS AT A POINT.A

STRESS TENSOR

At an arbitrary point P in a continuum, Cauchy's stress principie associates a stress vector t:"' with each unit normal vector ni, representing the orientation of a n infinitesimal surface element having P a s a n 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 s t a t e 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 f o r 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 a s 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,

t:ei)

A

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.

THE STRESS TENSOR - STRESS VECTOR RELATIONSHIP The relationship between the stress tensor 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. Designating 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. 2-72.6h

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 toA

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

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

rt$iwhich is written explicitlyA A A

=

Ln1k]

[vki]

1

'=12

u13

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

orL t ( ^ ' d ~ + L ~ b = ~O dA

(2.16 )

Replacing tin' here by ujinjand 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 thatajijj

+ pbi =

O

or

V I

+ pb

= O

(2.18)

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 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 toh

For the arbitrary volume V, (2.20) requires Equation (2.21) represents the equationsu,, =

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 ~ ~ bi = O p , ~ 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 and Px;x:x~ Fig. ~X~ of 2-9 be related to one another by the table of direction cosines

Fig. 2-9

50

ANALYSIS OF STRESS

[CHAP. 2

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 components of the stress vector tln)referred to the unprimed axes are related to the primed axes components tl'"' by the equation t;(P) = a..t(n) or t ' ( n ) - A . (2.26 ) JA AA

A

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 writtenA

[ti:")]= [aii] [t$]and the stress tensor transformation as

(2.28)

[a,] [apq] [aqjI Explicitly, the matrix multiplications in (2.28) and (2.29) are given respectively byLaijI

(2.29)

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 P61g253shown in Fig. 2-10. The equationuijgigj = fk2

(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 n is the unit normal in the direction of r. i At the point P the normal component aNni of the stress vector tin' has a magnitudeA

' 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, resulting quadric the

uij[icj=

-iaNr2

(2.34)

CHAP. 21

ANALYSIS O F STRESS

51

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 ELLIPSOIDAt the point P for which the stress tensor components are aij, the equation (2.12), t:"' = uiini, associates with each direction ni a stress vector t,'"'. Those directions f o r which tiA and ni are collinear as shown in Fig. 2-11 are called principal stress directions. For a principal stress direction,A A

tin) =

A

h

&&i

or

t'"' -

an

A

(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(aij- Sija)ni=

O

or

(1- lu) n = 0

.

A

(2.36)

Fig. 2-11

In the three equations (2.36), there are four unknowns, namely, the three direction cosinesni and the principal stress valueO.

luij - Sijal, must

For solutions of (2.36) other than the trivial one nj = 0, the determinant of coefficients, vanish. Explicitly,all

-

"12 u22

u13 u23O33

lu.. - S 7. . 11

1

I

~

= 0

or

a~~

-

=

0

(2.,?7)

as 1

a32

-

which upon expansion yields the cubic polynomial in a,

where

I, 11,

= aii = tr 1 = +(aiiujj a. .c..) - 21 lilaijl

(2.39)

111, =

= det

are known respectively as the first, second and third stress invariants. 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

CHAP. 21

ANALYSIS O F STRESS

53

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)

t(n) =3

h

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 (2.51) F = U: - Anini 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{uTI - 2uI1(uIn; ~ , , n ; ~,,,n;)

+ + + A} = O n3{uSII 2 ~ , , , ( ~ , n ; ~ , , n + ~,,,n;)+ A = O + ; }

(2.523) (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 n1=&1, nz=O, nl = O , ni = O , na = 11, nn = O , n3=0; n3 = 0; na = ? I ; forwhich~,=O for which for whicha,U,

(2.53~) (2.533) (2.53~)

=O=O

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 bynl = 0, ni = *l/fi, nl = 11/fi, n2 = - I + n z = 0, nz = e l l f i , , n3 = +l/fi; n3 = *1/fi; n3 = 0; for which for which for whichU,

= (a,, - uIII)/2

(2.54a) (2.54b) (2.54c)

a,U,

= (oIII- uI)/2 = (aI - uII)/2

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.

54

ANALYSIS O F STRESS

[CHAP. 2

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 wellknown 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

> O I I > u~~~ (2.55) For this arrangement the stress vector tln) has normal and shear components whose magnitudes satisfy the equations U , = u1nS uI1n; uI1,n; +

Fig. 2-14

+

(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 , - u I I 1 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 In Fig. 2-15, this circle is labeled CI. plane that are on or exterior to the circle

(V,,

os)

Fig. 2-15

56

ANALYSIS OF STRESS

[CHAP. 2

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 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.h

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

CHAP. 21

ANALYSIS OF STRESS

57

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 and having the equation u,,xS 2 u l z ~ , ~ ,u,,x~ = -Ck2

XIXZ

plane(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 given in (2.67). Point A on the circle represents the stress R = j/[(u1, - uzz)/2]2 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