Schaum Continuum Mechanics Mase
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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 SERIES McGRAW-HILL BOOK COMPANY Nezo York, St. Louis, San Francisco, London, Sydney, Toronto, Mexico, and Panama
Ctrpyright @ 1!)70 1)s Illv(:raw-Hill. lric*. A11 Rights Rcserved. Printed in the United States of America. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior wrltten permission of the publisher.
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 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
CONTENTS 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 .................................... 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 ......................................
.
.
.
.
.
..................
.
.
.
.
2
1 2 2 3
4 G
8 8 10 11 12
.
.
Chapter
1
13 15 15
16 17 19 20 21 22 23 23
ANALYSIS OF STRESS The 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 ..............................................
.
.
.
.
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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
44 44 45 45
46 47 48 49 50 51 52 54 56 57
CONTENTS
Chapter
DEFORMATION AND STRAIN
3.1
3.13 3.14 3.15 3.16
Particles 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 ..............................
4
MOTION AND FLOW
4.1 4.2 4.3 4.4 4.5 4.6 4.7
Motion . 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 ..............
5
FUNDAMENTAL LAWS OF CONTINUUM MECHANICS
5.1 5.2
Conservation 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 . . . . . .
3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10 3.11 3.12
Chapter
Chapter
Chapter
Page
3
6 6.1 6.2 6.3 6.4 6.5 6.6
77
77 77 79 80 81 82 83 84 86 87 88 89 91 91 92
110 111 112 112 113 114 116
126 127 128 128 130 131 132
LINEAR ELASTICITY Generalized 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
Page 6.7 6.8 6.9 6.10
Chapter
7
8
PLASTICITY
7.4 7.5
.
.
.
.
. .
.
8.11
Basic 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 ....................
9
VISCOELASTICITY
9.1
Linear 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 ..................
8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 $8.9 8.10
Chapter
.
7.6
7.2 7.3
9.2 9.3 9.4 9.5 9.6 9.7 9.8
147 148 149 149
FLUIDS Fluid 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 ..................................
7.1
Chapter
Airy's Stress Function ................................................ Two-Dimensional Elastostatic Problems in Polar Coordinates .............. Hyperelasticity . Hypoelasticity ........................................ Linear Thermoelasticity ................................................
.
.
.
.
.
.
.
.
160 161 162 163 164 165
175 176 177 173 180 181 182 182 183 183 184
196 196 198 199 200 202 203 204
Chapter 1 Mathematical Foundations 1.1 TENSORS AND CONTINUUM MECHANICS Continuum 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 law o f 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 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 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 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
Fig. 1-2
(1.3)
CHAP. 11
3
MATHEMATICAL FOUNDATIONS
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
all = n(l)
a12= n(1) 2
al, = ni1)
2;
aZ1= n(2)
aZ2= n(2)
E;
a31
~1
= ni3)
=
(L*,
= n(32)
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 the same 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 (1.142) 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
+(x, t) = V(X,t)
(1.143) (1,144)
T = T(x, t)
(1.145)
$J = V
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
=
êi
a
=
aXi
(1.I46)
eidi
A
Frequently, partia1 differentiation with respect to the variable xi is represented by the comma-subscript convention as illustrated by the following examples.
avi (b) axi = vi,i
(e)
aTij
ã2*
=
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. (c)
Severa1 important differential operators appear often in continuum mechanics and are given here f o r reference.
divv = V - v
or
divt = visi
(1.148)
curl v = V x v
Or
~ijkajvk= 'ijkvk,i
(1.I49)
CHAP. 11
23
MATHEMATICAL FOUNDATIONS
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
-
~
(1.151)
X
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. 1-11
Fig. 1-10
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
[CHAP. 1
MATHEMATICAL FOUNDATIONS
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
+ (317)j
A
v = VIV = (217)i
A
- (617)k
( b ) The vector extending from P to Q is
u =
A
(O-I)?+ A
= -i+2j u = d(-1)2
(2-O);+ (1-3)k
A
A
-2k
+ (2)2+ (-2)2 A
Thus or
1.2.
= 3
Fig. 1-12
+ (213); - (213)k (1/3)? - (213)3 + ( 2 / 3 ) k A
u = -(1/3) i
directed from P to Q
6
directed from Q to P
=
+ i+ + +
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 A A u *-v - 1 -[(xz-xI)~+ u A (v2-y1)j u ( 2 , - 2,) k ] [a? b3 cc] 1 -X = - ( a s 2 by2+ cz2- axl - byl - cz,) = X= O
+ +
+ +
+
U
+ +
+
Iz
+
+
Z
U
Since u is any vector in the plane, v is I to the plane.
1.3.
+
Fig. 1-15
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, A
A
A
A
A
(r-d)*r = [(x-a)i+(y-b)j+(z-c)k]*[xi+yj+zk]
= x2+y2+z2-ax-by-cz
A
= 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
Adding d2/4 = (a2
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
25
MATHEMATICAL FOUNDATIONS
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 ) bx c
(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 A
A
A
A
A
A
A
Check the
A
u = 3i+j-2k
v = 4 i - j -k A
w = i -2j +k The 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
+ +
+ 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
1.6.
+ w.
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.
+ a2b2+ - - - + aNbN= aibi (i = 1,2, . . .,N ) . ( a ) In terms of base vectors, ai = aliel + azie2 + a3ie3 = a j i e j Let D = albl
A
A
A
A
and so D = a-2.b. 3 1 3 = ~ ê.(a..b.) 3 t t = $.c. 1 1 3
with j = 1,2,3.
( b ) Likewise setting bi = bjiêj i t follows that D = aibjiêj = (bjiai)êj= g j ê j where j = 1 , 2 , 3 .
1.7.
For the arbitrary dyadic D and vector v, show that D - v = v . D,. Let D = albl
+ a2b2+ . - - + aNbN. D
+ a,(b2 v ) + . . . + a N ( b N v ) b,)a, + ( v .b2)a2+ . . + ( v . b N ) a N =
v = al(b, v) =
1.8.
Then
(V.
v . D,
Prove that ( D , D ) , = D, .D. From (1.71),ü = D i j ê i ê j and D , = D j i ê i ê j . Therefore D,.D
and
(o, o),
A
A A A A = D..ê.ê..D êê jt I p q p q = DjiDpq(ej'ep)eieq 1
A
A A
= DjiDpq( e j e,) e,ei
A
A
A
A
= Dpqeq( e , e j ) e i D j i = D,,ê$,
.~
A A
.
~ = D~, D e
~
e
~
26
1.9.
Show that (D x v), = -v
X
Dc.
+ a,(b2 X + + aN(bNx V) - + (bNX v)aN (bl X v)al + (b, x v)a2 +
D X v = al(bl X V)
(D X v), =
=
+
V)
X b,)al - (v x b,)a, -
-(V
' ' '
..- - (v X bN)aN =
-v X D,
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. A A
1.10.
[CHAP. 1
MATHEMATICAL FOUNDATIONS
AA
A
AA
+
+
y3+z k )
r - ~ . r= (s?+ A
A
(a??+
A
b33+ ckk) A
A
(s?+
+ y j + xk, A
A
show
zk)
A
= ( z i + y j + z k ) * ( a x i + b y j + c z k ) = as2+by2+cz2 = 1 A A
AA
AA
AA
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 , pute and compare the double dot products D :F and D F.
..
ab
1.12.
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.
com-
Also, from
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,
+ 2;:
= (3??
O
-
AA
3 k + 5 k k ) . (4;: + :6:
AA
A A
A A
= 12ik+12j j + 3 j j AA
Similarly,
AA
A A
jk -15kj+5kk
-
AA
+kk)
-3k3
AA
AA
AA
AA
AA
= (4ik+6jj-3kj+kk).(3ii+2jj-jk+5kk)
H
AA
= 20ik+123!-63k.-6k3+8kk
1.13.
Show directly from theAnonion form ofA 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.
+ (D-?)? +
Writing D in nonion form and regrouping terms, II
+ 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 ) k
= (D,,i
A
A
A
A
A
A
A
A
A
A
A
e
A
A
A
A
6
i - d l = i * ( ~ - i =) i . ( D , , i + D Y , ~ +D,,k)
Also now
A
= D,,
A
6
j * d 2 = 3 . ~ . j = D,,,
j * d l = ; . ~ . i = D,,,
1.14.
A
etc.
For an antisymmetric dyadic A and the arbitrary vector b, show that 2 b . A = A, x b.
+ &c2 + $,c3; and because it is antisymmetric, ( elel + e2c2+ e3c3 - elel - c2e2 - c3e3)
From Problem 1.6(a), A = &cl or
A
2A =
A
A
A
= (êlcl - elel
and so
.
2A = (A - A,)
A
+ e2c2- c2e2 + esc3 - c3e3) A
A
A
A
+ [(b.ê2)c2- (b . ê2]+ [(b ê3)c3 [ ( ê , x c 1 ) x b + ( ê z X c 2 ) X b+ ( ê 3 X ~ 3 ) X b ]= (AvXb)
2b A = [(b $,)cl - (b c,) ê,]
=
C,)
-
(b c3)$31
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
1.15.
.
A
A
+
AA
A
A
A
A
A
A
A
A
Since u X v = ( 2 i + k ) X 5 j = 1 0 k - 5 i , AA
AA
A A
A
A
D * ( U X V= ) (6ii+3ij+4kk)-(-5i+lOk) A
Next,
D
~
=U ( 6 ? ? + 3 ? ; + 4 k k ) ~ ( 2 i + k ) AA
snd
A
AA
AA
AA
= -30?+40k AA
AA
= -6ik+8kj-6ij+3ii AA
(DXu)*v = ( 3 i i - 6 i j - 6 i k + 8 k j ) .
A
A
A
5 j = -30i+40k
AA
CHAP. 11
27
MATHEMATICAL FOUNDATIONS
Considering the dyadic A A
A A
A A
31:-4:;+2ji+jj+kk
D =
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.
+ +
In accordance with (1.59) and (1.60), construct the vectors A
A
A
A
A
A
u
= f(i) = i
v
= f(j) = j
+ iXr + j Xr A
A
A
= i -zj+yk
A
+j
A
A
= xi
A
A
A
-xk
A
A
w = f(k) = k + k X r = - y i + x j + k A
A
ü = ui+vj
Then
+ wk
A
A
a = D b = (b,
and
A
A
A
= (i -zj+yk)i+(xi
+ b,z
A
- b,y) i
A
A
A
A
A
A
A
A
A
t j -xk)j +(-yi+xj+k)k
+ (-b,z + b, + b,x) j + (b,y A
- b,x
+ b,) k
A
AS a check the same result may be obtained by direct expansion of the vector function, a = b
+bXr
+ b,k + (b,z - b,y) i + (b,x - bxz) j + (bxy- b,x)k
A
A
= b,i -i-b,3
A
A
A
1.18. Express the unit triad ê4,ê, ê, in terms of A A A i, j, k and confirm that the curvilinear triad A is right-handed by showing that ê, x ê, = e,.
A
By direct projection from Fig. 1-15, A
A
e,
=
A
e,
= (- sin e) i
A
=
e,
(COS
g cos e) i
+ (cos g sin e) j
A
A
- (sin g ) k
+ (cos e) j (sin g cos e) i + (sin g sin e) j + (cos @) k A
A
A
A
A
Fig. 1-15
and so
e A
e+
x ê,
=
A
- sin e
=
E
+F
- sin
cos e
+
A
(sin @ cos e ) i
@
O
+
A
(sin g sin e) j
+ sin2 e) cos g] k A
[(cosze
+ 4:: + 6 3 1+ 7;: + 10 ko;f 2 k3
1.19. Resolve the dyadic D = 3;: and antisymmetric parts. Let D = E
k
1
cos g sin e
cos g cos e
where E = E, and
+
= -F,.
F
=
into its symmetric
Then
+ 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
= (1/2)(D D,) = (1/2)(6 i i 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+kj A A
F
2,.
A A
A A
A A
A A
= (1/2)(~-o,) = ( 1 / 2 ) ( 4 i k - 4 k i + 6 j i - 6 i j + 1 0 k i A A
A A
A A
= -3i j - 3 i k + 3 j i -
A A
A A
A A
jk+3ki+kj
A A
A A
= E, A A
10ik+2kj-2jk)
= -F,
28
[CHAP. 1
MATHEMATICAL FOUNDATIONS
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 A
A
bl = (b, x b3)/12 = ( j -k)/4
- i13 + j 14 + k / 1 2 A
b2 = (b3x bl)/12 =
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.
+ A,, + A,,. i = 1, Blll + B12, + B133. i = 2, BZll + BZz2+ BZ33. i = 3, B3,, + B322+ B333.
Ai, repi.esents the single sum Aii = All Bijj represents three sums:
(1) For (2) For
(3) For
Rii represents the nine components Rll, R12,Ri3,R,,, Rz2,RZ3? ai Tij represents three sums:
R32>R33.
+ a2 T2i + a3 T3i. a1T12+ a2TZ2f a3 T3,. a1T13+ a2T23 + a3T33.
(1) For j = 1, al Tll
(2) For j = 2, (3) For j = 3,
+
+
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 ~ ~2 33 ~ 1 ~ ~ 3 31~ 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
+ Sz2 +
= 3
+ 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)
3
Si2S2k
(e)
1.23.
6ifiik =
S1#11,
+ 82jA2k + 83jA3k
For the permutation symbol
Cijk
= Ajk
S ~ O Wby
direct expansion that (a)
( b ) ' i j k a j a k = 0. (a) First sum on i,
= €.. i ~ k i.. j
€ljkCklj
+ ' 2 j k e k 2 j + '3jkek3j
Cijk'kij
= 6,
CHAP. 11
29
MATHEMATICAL FOUNDATIONS 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) ( b ) Summing on j and k in turn, Eijkajak = ' i l k a l a k = 'i12ala2
+
+ (-I)(-1) + ( l ) ( l+) ( l ) ( l+) (-I)(-1)
Ei2ka2ak
+
= 6
Ei3ka3ak
+ ' i 1 3 ~ 1 ~f3 ' i 2 1 ~ 2 ~+1 'i23a2a3
+
f ' i 3 1 ~ 3 ~ 1 Ei32a3a2
From this expression, when i = 1,
eljkajak
when i = 2,
€2jkajak
= a2a3- a3a2 = O = a i a 3 - a3a1 = 0
when i = 3,
€3jkajak
=
Note that
eiikajak
f2
( b ) fi f2
(C)
1.25.
0
is the indicial form of the vector a crossed into itself, and so a X a = 0.
1.24. Determine the component fi
- a2al =
a1U2
for the vector expressions given below.
f2
'ijkTjk
=
'2jkTjk
=
C,, j b j
= =
c2,1b1
=
+ '231T31
=
€213~13
+ T31
-T13
- cj,ibj
+~ 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
fi
=
Bijf;
f2
=
B21f:
+
+ B23f3*
B22f2
Expand and simplify where possible the expression (b) D i j = - D j i .
Dijxixj
for
(a)
Dij
=
Dji,
+ D2jx2xj+ Dsjx3xj
Expanding, Dijxixj = Dljx1xj
=
(a) Dijxixj =
D11(x1)2
+
+ 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
D11x1x1
D22(x2)2
+
D33(x3)2
( b ) Dijxixj = O since Dll = -Dll,
1.26.
+ 2D12x1x2+ 2D23x2x3+ 2D13x1x3
D12 = -D2,,
etc.
- 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
and
-
q =2
and note that since k is a summed index i t takes on -
'12kEk32 - E121E132
SipSjq
(b) Introduce i = 1, j = 2, p = 2, S12S21 - S11S22 = -1.
-
aiqS,
=
q
= 1.
Then
+ '122'232 + '123'332
813S22
= O
- S12S23 = O
eijkekpq
= ~123'321 = -1
and
1.27. Show that the tensor Bik = cijkaj is skew-symmetric. Since by definition of
an interchange o f two indices causes a sign change, Bik = eijkaj = - ( e k j i a j ) = - ( B k i ) = -Bki
cijk
SipSjq
- SiqSjp =
30
[CHAP. 1
M A T H E M A T I C A L FOUNDATIONS
1.28. If Bij is a skew-symmetric Cartesian tensor for which the vector show that Bpq= t p qbi i . Multiply t h e given equation b y
epqi
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,
@
= e,, e = e,).
x , = e , sin e, sin e,
Hence
ax1 -
-
-
sin O , cos 8 ,
ax1 - as2
e , cos 0 , cos e3
sin e2 sin e,
8x2 - 302
0,
8x3 as2
-el sin e,
ao1
8x2 861
f r o m which g,,
g33
axi axi = -as, asl
=
-
axi axi -- ao3 ae, -
-
sinz e, cos2 8 ,
cos O , sin
e,
Fig. 1-16 8x1 - -e, sin e, sin e, 863 -
8x2 as3
8x3 = 863
6,
o
+ sin2 o, sin2 e, + cos2 e,
eSsin2 e, sin2 e,
+ eSsin2 6 , cos2 e ,
sin e2 cos e,
= 1
= eSsin2 O ,
Also, gpq = O for p Z q. For example, axi axi g12
=
aBi. dB.,
=
(sin e, cos @,)(e,cos e, cos e,)
+ (sin O 2 sin e3)(e1cos e, sin e3) -
e
=
(COS
eZ)(elsin
o
T h u s for spherical coordinates, (ds)2 = (de1),
+ (el)2(de2)"
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),
(e1 sin de^)'.
e,)
CHAP. 11
31
MATHEMATICAL FOUNDATIONS ( 1 ) For (del,O,O),
ds = del = d r
( 2 ) For ( O , de,, O ) ,
ds = e, de, = r d+
( 3 ) For (O, O, de,),
ds =
de, = r sin + de
e , sin 8 ,
1.31. If the angle between the line elements represented by (doi, O, O) and (O, don,O) is 812
denoted by p,,, show that cosp,, =
66.
Let ds, = 6de, 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
ax, a%, - -de, de,
-
ae, ae,
ax, ax, ax3 + as, - de, de2 i- - -de1 de2 ae, ae2 ae, as, -
Hence using the result of Problem 1.30,
tos
plz = g,,
del de2 - -= ds1ds2
= giz de1 dez
812
66
1.32. A primed set of Cartesian axes O X ~ X $isXobtained ~ 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,s j ) and Fig. 1-18, the table of dire8ion cosiiies is
X;
51
22
x3
cos e
sin e
O
e
cos 6
O
o
1
-sin
xó
o
x3
Thus the transformation tensor is
A
=
cos 6
sin 6
O
-sin e
cos 6
O
Fig. 1-18 By the transformation law for vectors (1.94),
+ v , sin 6 sin e + v , cos 6
v ; = a I j v j = v , cose v i = a Z j v j = -vl v j = a3jvj =
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.
V,
32
MATHEMATICAL FOUNDATIONS
[CHAP. 1
$I
The unit vector along the xí axis is given by the first row of the table a s = (315)ê1- (415)ê 2 . Also from the table 2; = ê3. For a right-handed primed system 23 = Xêi, or Ae t, =
2, = (-3/5)& - (416)$~and
[ ( 3 / 5 ) ê 1- (4/5)$2]x
the third row is
1 1 x:
-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. 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
o
-
+
llfi
1
The orthogonality conditions aiiaik = S i ,
-112
112
-112
require:
+ a3,a3, = 1
1.
For j = k = 1 that a l l a l , a2,a,, 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. sum of squares of elements of any column to be unity.
which is seen to be the sum of squares of
+
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.
1.35. 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 and q Bii = apiaqjBiq. Hence XAij
+ @Bii =
~
(
~
+ ay(apia,iBg,) ~ ~ ~ = k a p~i a ,)i ( ~ ~+ k qyBP,)
~
which demonstrates that the sum transforms a s a second-order Cartesian tensor.
1.36. S ~ O that 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 dummy variables. Thus replacing i, j, k in the second and third terms by p, q , r , the sum becomes Pijkxixjxk
+ Pqrpxpxqxr + Pqprxpxqx,
Now change dummy indices in these same terms again so that the form is Pijkxixixk
1.37.
+ Pijk
+
Pijkxjxixk = 3Pijkxixjxk
If Bij is skew-symmetric and Aij is symmetric, show that AijBij = 0. Since A . = A l,t. and B . . = -B..31' A..B.. -A..B.. or A..B.. + A..B.. = A..B.. + A P P BPP = O. 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. 11
1.38.
33
MATHEMATICAL FOUNDATIONS
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,
+ DLijl = S ( D i j+ Dji) + &Dij Dji) #Dij + Dji)xixj = S(Dijxixj+ Dpqxqxp) = Dijxixj
Dij = D(;j) Then
1.39.
D,ijlxixj =
-
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 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
x
= a.bXc
=
[abc] =
eijkaibjck =
a1
a2
a3
bl
b2 b3
C1
C2
C3
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 =
aiai
+ pbi +
yCi,
34
MATHEMATICAL FOUNDATIONS
By Cramer's rule,
Likewise /3 =
a
=
I v3
b3
c3
a1
bl
c1
a2
b2
c2
EijkaivjCk epqrapbqcr
I
[CHAP. 1
and by (1.52) and (1.I 0 9 ) , a = eiikvibjck 'pqrapbq~r'
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,
Let a
D
[ ]
= v; then [ v l ,v2, v3] = [3,0,4] O -4
.
= 9 , -20,6].
.
Let D b = w; then O -5
Let a
1.43.
D
b = v . b = h; then
-10
-6
[h] = [9, -20,6]
(
2
1=
[-761.
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] =
O
0
1
From (1.132), for principal values h,
which results in the cubic equation h3 - 7h2 values are h ( 1 , = 1, = 2, h ( 3 )= 4.
+ 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. Ao,
= 1. Then the first two equations of (1.231) give 2n:" - nil' = O and -n:"
nll)
For
h t 2 ) = 2,
nl2'
(1.131) yields ni2' = 0 , -ny' since nini = 1 and ni2' = 0.
+ nf'
= O,
and
+
= O.
Thus
3 n F ) = O.
Thus
-ny'
ny' = ni2' = %1/\/2, For n33'
= 0,
h,,, = 4, (1.131) yields n : 3 ) = -nF' = TI/@.
-ni3'
- n:'
= 0, -ni3' - ns' = O, and
The principal axes x: may be referred to the original axes xi through the table of direction cosines
CHAP. 11
35
MATHEMATICAL FOUNDATIONS
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.
+
Finally for the system to be right-handed, ~ ( 3 = ) n ( l ) . ~h~~
A
n(2)
A
A
A
e1
e2
e3
i
-116
i 1
o o
=
(4+ 4)ê3
A
= e3
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 r e along the principal directions with x: being a right-handed system, x;" a left-handed system.
1.45.
Fig. 1-19
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 ~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. directions to be perpendicular.
1.47.
Compute the principal values of axes coincide with those of T.
( T ) 2 of
Problem 1.43 and verify that its principal
[-i i !][-i i 3 -1
=
[Ti.
Hence n:2'n:3' = 0 , the condition for the two
3 -1
!] [-: 10 -6
=
1;
The characteristic equation for this matrix is 10 - h
-6
O
-6
10 - h
O
o
O
l - h
from which h(,) = 1,
= ( 1 - h)[(10- h)2 - 361
= 4, h ( 3 )= 16.
= ( 1 - X)(h - 4)(h- 16) = O
Substituting these into (1.131) and using the condition
nini = 1, gn"'
For h ( , ) = 1,
1
-6n:" 6n(2) 1
For A ( 2 , = 4,
-Gn?'
- Gn(l' 2
$.
9n;"
= O
=
n21'
=
ni2'= k l f i , nF) = 0
= 0'
n(l) 3
=21
- Cn(2) = O 2
+ 6nk2' -Sn'2'
3
For
=
= 0 -
- 0
= 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. 11
37
MATHEMATICAL FOUNDATIONS
with the transformation matrix being
[aijl =
;" ó 3 o
Therefore
fi
transformation
=
[
llfi
lldz
llh
llfi
11,h
0
-216
o
and using [aij; to relate this to the original axes by the
fi = ~ A,~ the matrix f i
CARTESIAN TENSOR CALCULUS
equation is
(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.
ax
Consider - = axk
(A
+ A jk)xj.
axi
a%, + A,.%. a> 'axk '
A.. -X . %3axk 3
Since
axi
-z
Jxk
ax = A k j z j+ Aikzi =
Sik, i t i~ seen that Jxk
a2A 8% Continuing the differentiation, -- ( A k j + A . ) ax, axk lk axp
- A k P + A P k .If A 23. . = A3.i 9
- = 2Akjxj and -- 2APk. as, axk
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 $,i and so v = V X V + has components v i = ~ ~ ~= ~~i j k +a, k j .~But eijk is anti-symmetric in j and k, whereas + , k jis symmetric in j and k; hence the product E ~ ~ vanishes. ~ + , ~ ~ The same result may be found by computing individually the components of v. For example, by expansion vl = '123+,23~ ~ 3 ~ = + ($,23 , ~ 232) = 0.
+
( b ) V V x e = h = ( ~ ~ ~= ~eijkak,ji a ~ =, O since
1.51.
+,
ak,ij
ak,ji
+
and
~ i j k= -'jik.
~ (~ ~ 3 )in~ the direction Determine the derivative of the function X = ( X I ) ~ 2 x 1 of the unit normal = (2/7)&- (3/7)&- (6/7)ê3 or & = (261- 3ê2 - 6ê3)/7.
n
ax
A
The required derivative is - = V h n = X,ini. an
Thus
,
~
38 1.52.
[CHAP.1
MATHEMATICAL FOUNDATIONS
If xk,
is a second-order Cartesian tensor, show that its derivative with respect to namely Aij,k, is a third-order Cartesian tensor.
Aij
F o r the Cartesian coordinate systems xi and xl, xi = ajixi and âxildxi = 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
ar
axj
= 2r ax, -- = 2 ~ . . x . i t 2 ~ a
Use the divergence theorem of Gauss to S ~ O W that
L
Xinj
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
1.55.
Fig. 1-20
If the vector b = V x v, S ~ O Wthat a scalar function of the coordinates. Since b = V
X
v, bi = eijkvk, and so
i
hbini d S
i
=
i
=
& Li
hbisdS =
i
h,ibi d V
rijkhvk,jni d s
bi d ~
since hrijkvk,ii = O
where h = h ( x i ) is
CHAP. 11
39
MATHEMATICAL FOUNDATIONS
MISCELLANEOUS PROBLEMS 1.56. For the arbitrary vectors a and b, show that A = ( a x b) (a x b)
+ (a.b)2 =
( ~ b ) ~
Interchange the dot and cross in the first term. Then
=
+ (a.b)(a0b) + ( a . b)(a0b) ( a . a ) ( b - b ) - ( b . a ) ( a W b+ ) (a.b)(a.b)
=
(ab)2
= a - b x (axb)
h
= a [(b. b)a - (b.a)b]
since the second and third terms cancel.
1.57. If
ii = o x u and V
=
w
(a) In symbolic notation, d z(uxv)
d
(V
U)OI
= (V' ,)U -
(U
@)V
(u X
V)
wi =
d
+
v)a -
(U
=
a X (U
(U
X
@)V
V)
Then
W.
( E ~ ~ ~=u ~eijkGjwk v ~ )
+
eijkuj;k
ck = ekmnamVn,
and since Aj = e j p q w p u q and
€y. k. E3pqapUqwk ,
W i.
=
(u x v).
= (mXu)Xv+ uX(mXv)
m)u -
(V.
ca X
-
= h X v + ux; =
( b ) I n indicial notation, let
d (u X v) = dt
x v, show that
+ eijk'kmnUjamvn
('ijkEkmn
-~ink'kmj)~~jamwn
and using the result of Problem 1.59(a) below, (SimSjn
- S i n S j m - SimGjn
=
(SijSmn
- 8 i n 8 j m ) u j ~ m ~ nE
which is the indicial form of
1.58.
Establish the identity
+ SijSmn)~jamwn
wi =
m
cPqsrmnr
~irnk'~j~Ujarnw~
X (u X v).
=
8mp
8mq
Sms
Snp
8nq
8ns
8rp
8rq
8rs
. A11
Let the determinant of Aij be given by or columns causes a sign change. Thus
A,, A,,
A,, A,,
A23
A31
A,,
A,,
A31
=
A,,
-412
det A = AZl Az2 A32
A12
A,,
A13
A,,
A,,
AZ3
A,,
A31
A33
A13 A23
Am,
Am3
Anl
Ana
An3
Ar1
A,
4
3
-
em,,
An interchange of rows
A33
=
and for an arbitrary number of row changes, Aml
.
det A
-detA
40
MATHEMATICAL FOUNDATIONS
[CHAP. 1
or column changes,
A,,
A,,
A I S
-42,
-42,
-4%
A,,
-43,
A,,
-
epqS det A
Hence f o r a n arbitrary row and column interchange sequence, A,,
Am,
Ams
Anp
Anq
Ans
A,
Ar,
Ars
-
'mnr'pqs
det A
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) -
srpsnq
f
S s q ( 8 n s s r p - SnpSrs)
- SpnSrq + SqnSrp
-
SnpSqr
+ S s s ( S n p ~ r q- SnqSrp)
+ 3SnpSrq - 3Snqsrp
-
- S n p a r q - SnqSrp
( b ) Identifying q with n in ( a ) , ~pqs'sqr - Sqparq
1.60.
If the dyadic B is skew-symmetric Writing B = blêl
and
B , X ~
- aqqsrp
B = -B,,
= S p r - 3 S p r = -2Spr
show that B , x a = 2a B.
+ b2ê2+ b3ê3 (see Problem 1.6), B, = bl x ê1 + bz X êz + b3 X ê3
= (blXêl)xa+ (b2xê2)xa+ (b3xê3)xa A = (a bl)ê, - (a.A el)bl ( a . b , ) ê z - ( a e z ) b 2 -t ( a b 3 ) ê 3 - ( a ê 3 ) b 3
+
= a (blêl -
+ bzêz+ b 3 ê 3 ) - a .( ê l b l + ê2b2+ ê 3 b 3 )
a.0 - a.0,
= 2a.B
1.61. Use the Hamilton-Cayley equation to obtain Check the result directly by squaring ( B ) 2 .
(B)4
f o r the tensor B =
The characteristic equation f o r B is given by
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. 11
41
MATHEMATICAL FOUNDATIONS
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 pare invariant under the coordinate transformation represented by (1.103), 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 = A ~ nB ~ ~ s ~ ~ A= PApqApq ~ A & =~ A!.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 ibut 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 = lIijêiêj and E = ~,,ê,ê,. By (1.31), D : E = D i j ~ , , ( ê i êp)(êj2,). By (1.35), = DijEpq( g j $,)(Si 3,) = DijE,, (êj A%)(Aei Ae,) since E,, = E,,. Now interchanging the A A role of dummy indices p and q in this last expression, D E = êq)(ei e,).
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 ~or~ ~ a ~ a ~ b ~ Vp
=
'pqi'ijk
(ajbk),q = epqi'ijk(aj,qbk ajbk,q)
=
(Spjsqk
- Spksqj)(aj,q
-
Hence v = b V a - b ( V a)
+ a(V
bk
+ ~ j ~ k , q=)
bq
- aq,q bp
+
- aqbp,q
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
p
cqPinpcijkajxkdS.By (1.157) this becomes
(rqPirijkajxk),, dV and since a is constant, the last expression becomes
the volume integral q
s,
j ,
p
d
=
(Sqjspk
L
:
- s q k s p ~ a j x k , p dv
(aqS p p - a, a,,) dV =
=
(aqxp,p - apxq,p) 'V
(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-:
r: : 01
:l[u-i 01
L 0 0 1 ] L 0 0 1 J
Fig. 1-21
L 0 0 1 J
1.68. Show that (I x v) D = v x D. I
Xv
= =
= 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 I A A
A A
A A
A
A
A
(i i + j j
A
A
A
A
A
A
A
A A
A
A
A
A
A
( I X v ) - D = V X I * D = V X D .
Supplementary Problems A
+
A
A
A
A
h
1.69.
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.
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.
1.72.
Determine the principal values of the symmetric p a r t of the tensor Ans. h ( , , = -15,
h ( Z )=
3+
5 , h,,) = 10
X
Determine w so
x = O, (c) a . V x = a where x is the position
Tij = -3
-18
CHAP. 11
1.73.
43
MATHEMATICAL FOUNDATIONS
F o r the symmetric tensor of the principal axes.
Ans.
Tij =
determine the principal values and the directions
h c l , = 2 , A ( z , = 7, A,,, = 1 2 ,
1.74.
Given the arbitrary vector v and a n y unit 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
1.76.
Check the result of Problem 1.48 by direct rnultiplication to show t h a t
1.77.
Determine the square root of the tensor
+
.
V v = O, V x v = 6 and V
*(&+I)
X
w = -c,
#&-I)
B
show t h a t
V2v = y.
fifi = T.
=
0
det A = E ~ ~ ~ A show ~ ~ t hAa t ~ det ~ (AB) A ~= ~det, A det 0.
1.78.
Using the result of Problem 1.40,
1.79.
Verify t h a t (a) O3,,vp= v3, ( b ) 8siAji = Aj3,
1.80.
Let the axes 0 x ~ x be ~ xrelated ~ to Oxlx,x3 by the table
(C)
Sijeijk
= 0 , ( d ) ai2aj3Aij = A23.
( a ) Show t h a t ' t h e orthogonality conditions aijaik= S j k and apqaSq=
a,,
a r e satisfied.
( b ) W h a t a r e the primed coordinates of the point having position vector x = 2 ê , ( c ) W h a t is the equation of the plane xl - x,
Ans. ( b ) ( 2 / 5 f i , 1115, - 2 1 5 f i )
1.81.
( c ) fiX ;
+ 3x3 = 1
- X; -
2fi
X;
- ê3?
in the primed system? = 1
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 unit normal to the surfaee. Hint: Write V = (116) and use (1.157). '
h
1 S
(xixi),inidS
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 - -fi11 the space it occupies. This continuum concept of matter is the and to completely -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. of t_hecontinuum viewpoint as- the description of --Adofion -------- --- -basis for the mathematical - . -. and displacement are expressed material behavior means that -field _- quantities __ _ _ such as stress -_ functions of the space coordinates and time. --_-_r
2.2 HOMOGENEITY. ISOTROPY. MASS-DENSITY
A homogeneoousmaterial is-tne having-id~nticalpropertiesat a11 poi&. With respect a11 directions at a to some property, a material is isotropic if that property is the same in - , _- --- -- -_ point. A material is called anisotropic with gspect to those properties which are directional -- . --- a t a pÒi$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 therefore P(av)
-
anl 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, dM = lim -al)I = *"-O
Mass-density
p
AV
d!'
is a scalar quantity.
(2.2) 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 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 by dfi
dS
A
f (n)
Or
=
.
1im
as-o
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.
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. A
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
47
ANALYSIS O F STRESS
Fig. 2-5
The nine stress vector components,
Fig. 2-6 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.
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. 2-7 2.6
h
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" =.u j i , (2.11) becomes
Equation (2.12) is also often expressed in the matrix form
rt$i
=
Ln1k]
[vki]
which is written explicitly 1 A
A
'=12
u13
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
(2.16 )
or L t ( ^ ' d ~ + L ~ b =d ~O A
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
49
ANALYSIS OF STRESS
CHAP. 21
Since the volume V is arbitrary, the integrand in (2.17) must vanish, so that
+ pbi =
ajijj
or
O
V I
+ pb
(2.18)
= O
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 to h
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 ~ ~p b i, = ~ 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 and ~ XPx;x:x~ ~ of Fig. 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 ) J A
A
A
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 written A
[ti:")]= [aii][t$]
(2.28)
and the stress tensor transformation as
[a,] [apq][aqjI Explicitly, the matrix multiplications in (2.28) and (2.29) are given respectively by LaijI
(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 equation uijgigj = 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 ni is the unit normal in the direction of r. At the point P the normal component aNni of the stress vector tin' has a magnitude A
' 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)
51
ANALYSIS O F STRESS
CHAP. 21
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$' acting on dS a t P is parallel to the normal of the tangent plane of the Cauchy quadric a t the point identified by r.
2.10 PRINCIPAL STRESSES. STRESS INVARIANTS. STRESS ELLIPSOID At the point P for which the stress tensor components are aij, the equation (2.12), t:"' = uiini, associates with each direction ni a stress vector t,'"'. Those directions f o r which tiA and ni are collinear as shown in Fig. 2-11 are called principal stress directions. For a principal stress direction, A
A
h
A
tin) =
t'"' -
or
&&i
A
(2.35)
an
in which a, the magnitude of the stress vector, is called a principal stress zjalue. Substituting (2.35) into (2.12) and making use of the identities ni = Siinj and uij = ajii results in the equations (aij- Sija)ni=
O
.
A
(1- lu) n = 0
or
(2.36)
Fig. 2-11
In the three equations (2.36), there are four unknowns, namely, the three direction cosines ni and the principal stress value
O.
For solutions of (2.36) other than the trivial one nj = 0, the determinant of coefficients, vanish. Explicitly,
luij - Sijal, must
all
lu.. - S.. 11 7
1
I
~
= 0
or
-
a~~
"12 u22
as 1
-
a32
u13 u23 O33
=
0
(2.,?7)
-
which upon expansion yields the cubic polynomial in a,
where
I,
= aii = tr 1
11,
= +(aiiujj - a.21.c..) li
111, =
laijl
(2.39)
= det
are known respectively as the first, second and third stress invariants. The three roots of (2.38), a,,,, a,,,, a(,) are the three principal stress values. Associated with each principal stress a,,,, there is a principal stress direction for which the direction cosines n,'k'are solutions of the equations
53
ANALYSIS O F STRESS
CHAP. 21
This resolution is shown in Fig. 2-13 where the axes are chosen in the principal stress directions and it is assumed the principal stresses are ordered according to U, > u,, > u,,,. Hence from (2.12), the components of t$) are ti.^' = u1n1 =
(2.48)
'-'II~z
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 (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
+ + + A} = O n3{uSII- 2 ~ , , , ( ~ , n+; ~ , , n + ; ~,,,n;)+ A} = O
n2{uTI - 2uI1(uIn; ~ , , n ; ~,,,n;)
(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,
n3=0;
forwhich~,=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 n3 = +l/fi; -
for which
U,
= (a,, - uIII)/2
(2.54a)
n z = 0,
n3 = *1/fi;
for which
a,
= (oIII- uI)/2
(2.54b)
nz = e l l f i ,
n3 = 0;
for which
U,
= (aI - uII)/2
(2.54c)
nl = 0,
n2 = + I
ni = *l/fi, nl = 11/fi,
,
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
(V,,
os)
plane that are on or exterior to the circle
In Fig. 2-15, this circle is labeled CI.
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
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.
2.13
=O Fig. 2-18
CHAP. 21
57
ANALYSIS OF STRESS
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 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
58
ANALYSIS O F STRESS
[CHAP. 2
where U , = - p = ukk/3 is the mean normal stress, and the second (the deviator stress tensor) has the form
This decomposition is expressed by the equations uij
+ Sij
= sijUkk/3
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 stress values are S(k)
-
u(k)
(2.71)
- a,
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) 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. A
It is required to show that
t:^.*'n, =
h
A
Frorn (2.12) tln*'ni = ojin;ni, and b y (2.22) oJ.i . = 0 . . so that A
o..n*n. 31
1
I
= (o..n.)n* = t:")n; 1J 1 1
Fig. 2-20
2.2.
59
ANALYSIS O F STRESS
CHAP. 21
The stress tensor values a t a point P are given by the array 7 o -2
o
-2
4
Determine the traction (stress) vector on the plane a t P whose unit normal is (113)$3.
+
íi = (213)êi - (213)$2
From (2.12), t';' =
n. 1.
The multiplication is best carried out in the matrix form of (2.13): 7
o
7 A
A
-2
-
A
[ t p ) ,t?', t!j")] = [ 2 / 3 , -213, 1/31 h
10A tCn' = 4 e , - 3 e,.
Thus
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 ) tin'
a
,
= ( 4 2, - - Ae,)
A
( b ) jtin'( = d 1 6 ( c ) Since
+ 10019
tia' 6
(%2, -
ê,
+ Qê3) =
4419
= 5.2
= t : ; ' [ cos O ,
cos O = ( 4 4 / 9 ) / 5 . 2 = 0.94 and e = 20'. ,
2.4.
A
Then
Let S be the sum in question.
s which from (2.7) becomes S
2.5.
,
The stress vectors acting on the three coordinate planes are given by t l e l ' , tlez' and t i ê 3 ) . Show that the sum of the squares of the magnitudes of these vectors is independent of the orientation of the coordinate planes. A
A
A
A
A
A
= t;ei)t:ei)+ t(.edt!ez)+ t!edt(ed
uliuli
t
t
a
i
+ uziu2i + u3iu3i= ujiuji, an invariant.
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
I n 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 -0/2
-012
-012
60 2.6.
[CHAP. 2
ANALYSIS O F STRESS
The stress tensor a t point P is given by the array t : =
(1 -
5
3 -
53
i)
1
Determine the stress vector on the plane passing through P and parallel to the plane ABC shown in Fig. 2 - 2 1 . B
+
The equation of the plane ABC is 3x1 6x2+ 22, = 12, and the unit normal to the plane is therefore (see Problem 1.2) A n = 3ê1 + d7 ê2 + 2s 7 3
52
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 by x 3 the array 3xix2 5x;
I: = 2x3
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.
+
A t P the stress components a r e given by
The unit normal to the surface a t P is determined from grad + = V+ = ~ ( 2 ; 23 - 4 ) . Thus V 4 = 2x2ê2 223ê3 and so
+
+
Therefore the unit normal a t P is
=
ê 2
,h -ê,. 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 $,I2
+ 3 ê2+ d ê 3 .
Fig. 2-22
,
61
ANALYSIS O F STRESS
CHAP. 21
EQUILIBRIUM EQUATIONS (Sec. 2.7) 2.8. For the distribution of the state of stress given in Problem 2.7, what form must the body force components have if the equilibrium equations (2.24) are to be satisfied everywhere. Computing (2.24) directly from i given in Problem 2.7, 3x2
+ 10x2 -k 0 + pbl O+O+2+pb2 O+O+O+pb3
= 0 = O = O
These equations are satisfied when bl = -13x21p, b2 = -2Ip,
2.9.
b3 = 0.
Derive (2.20) from (2.19), page 49. Starting with equation (2.19),
substitute t:" = oiini in the surface integral and convert the result to a volume integral by (1.157): ( E ~ ~ d~ s x =~ o ~(eijkxjopk),p ~ ) ~ dV ~
Carrying out the indicated differentiation in this volume integral and combining with the first volume integral gives E i i k ( ~ j , p o p+ k ~ j ( ~ p k+ , p pbk)} dV = O But from equilibrium equations,
u
+ pbk ~
~
=, 0 ;~ and since xi,, -
this volume integral
reduces to (2.20),
\
STRESS TRANSFORMATIONS (Sec. 2.8) 2.10. The state of stress a t a point is given with respect to the Cartesian axes the array
0
~
~
byx
I
Determine the stress tensor Z' for the rotated axes OxíxSxl related to the unprimed axes by the transformation tensor
llfi 1
llfi
112
The stress transformation law is given by (2.27) a s ali = aiPaiqopq O r 1' = A A,. The detailed calculation is best carried out by the matrix multiplication [ali] = [aip][opq] [aqi] given in (2.29). Thus
l l f i -1lfi [o$]
= 1
112
o
-112
2
~
62 2.11.
ANALYSIS O F STRESS
[CHAP. 2
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 = uijnin 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
(i ; :) o o
Uij
=
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
CHAP. 21
63
ANALYSIS O F STRESS
l/,h
l l f i -1lfi
1 1 6
1 1 6
llfi
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 -=O ( b ) uniaxial tension o,, = U ; o,, - V~~ - u13= = a,, = o,, = O ( C ) simple shear o,, = o,, = T ; a,, - a,, = - o;, - T; - 03, = 0. (d) plane stress with a,, = u = o,,; From (2.32),the quadric surface is given in symbolic notation by the equation { * I (rl2
[i E I!]![
J
= *k2.
Thus using the matrix form,
a )
[{I,
= ~ 1 :+
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 as representing the operation of material differentiation (some books use DIDt), the dot velocity vector is defined by vi = dxi/dt = X i or V = dxldt = X (4.6) ( O )
111
MOTION AND FLOW
CHAP. 41
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
;jX7t)]
X
to emphasize that the
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 t)] to emphasize that x is held constant in this differentiation. The second term
[dP'.ir9 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.
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 4.2
+
+
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 ~ ( xt), , then
In (6.15) the velocity is given implicitly since i t appears a s a factor of the second term on the right. The function vi = vi(x,t ) or V = V(X, t) (4.16 ) is said to specify the instantaneous velocity field. The material derivative of the velocity is the acceleration. the Lagrangian form (4.14), then
If the velocity is given in
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 o r 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 skewsymmetric parts according to y..
dvi
-
=
aXj
or
Y =
9(vvx+ V xV ) + 4(vVx - VxV )
= (D+V)
(6.19 )
This decomposition is valid even if vi and avildxj are finite quantities. The symmetric tensor
is called the rate of deformation tensor. Many other names are used for this tensor; among them rate of strain, stretching, strain rate and velocity strain tensor. The skew-symmetric tensor
is called the vorticity or spin tensor.
113
MOTION AND FLOW
CHAP. 41
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 dxj
d t axj
dt
, the equation takes the form
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 = Dijdt or dE = D d t
The left hand side of (4.25) represents the components known as the natural strain 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 + Vij)dxj O?.-
dv = (D+V).dx
(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,xv (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
ai = q
= iikvk,j
or
a
[CHAP. 4
= &q = &Vxx 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,
d v = 0 x dx
or
(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
dv = D d x
or
(4.32)
and, since the rate of change of length of the line element dxi per unit instantaneous length is given by dvi dxi d i ( ~=) = D..- Dijvj or d'") = ü (4.33) dx a3 dx A
a $
the rate of deformation in the direction of the unit v,ector V , is
d = di(V'~i = Dijvjvi From (4.34), if
vi
d = 3.D.D
or
(4.34)
is in the direction of a coordinate axis, say $2,
d = d22
or
d =
ê2
D
.e2 = D22 A
(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
CHAP. 41
115
MOTION AND FLOW
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 dV =
c..
axi
--- dX ax, ax, ax3
d X 2d X 3 = J dVo
(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 dSOrnay 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 nnow fill an area element represented by the vector dSnt or dSi. It rnay be shown that ax. dSi= J L d X j or dS= JdX*Xvx (4.42) axi 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 -(dx2)=2-dxkdxi dt dxk
or
d z(dx2)
=
2dx.vXv.dx
(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
116
MOTION AND FLOW
or, from (4.20),
d - (dx2) = 2 Dij dxidxj dt
~r
[CHAP. 4
d - (dx2) = 2 d ~D ' d~ 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
P2j... (t) =
Jv P < ; . . . ( x , ~ ) ~ 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/& may be put into the form dt
P;. . .(x, t) dV
+ v~aldxp, (4.52)
+ ~-(aX vPPt P . . (x, t))] dV
= V
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 =
. '(xft, dV at
+
1
v,[P:.
. .(x, t)] dSp
(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,
& 1Q:
. . .(x, t) dSp
=
i& [Q:
. . .(x, t) dSp]
(4.56)
CHAP. 41
117
MOTION AND FLOW
and, from (4.43), the differentiation in (4.56) yields dQij ... ( t ) dt
-
s
[dQ:i. ( x ,t, + Q* (x,t ) ]d S p dt 8% v . . .
For properties expressed in line integral form such as
R 13.. .( t ) =
IR; 12
[Q*
s
-"d S p ]
. . dXq
.
...(~ , t ) d x ,
(4.58)
C
the material derivative is given by
&1
R&. . ( x , t )d x p
[R:. . (r,t )d x p ]
=
(4.57)
(4.59)
Differentiating the right hand integral as indicated in (4.59), and making use of (4.45), results in the material derivative d -[R dt
ij...(t)]
=
J
C
d[R:. . .(x, t)] dx, dt
+
12
[ ~ ~ . . ( x , dt x) ,]
(4.60)
Solved Problems 9
MATERIAL DERIVATIVES. VELOCITY. ACCELERATION (Sec. 4.1-4.3) 4.1. The spatial (Eulerian) description of a continuum motion is given by X I = X1et X3(et - I), x2 = X3(et - e c t ) X Z , 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
J
= laxi/aXjl
=
et O O 1
et-1 et-e-t
=
et
1
O 0
+
Inverting the motion equations, X 1 = xle-t x3(e-t - I), X 2 = x2 - x3(et- e-t), Note that in each description when t = 0, xi = Xi.
4.2.
+
X3 = x3.
+
A continuum motion is expressed by xl = X I , xz = et(X2 X3)/2 e c t ( X 2- X3)/2, x3 = et(X2 X3)/2 - e-t(X2 - X3)/2. Determine the velocity components in both their material and spatial forms.
+
+
+
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 X 1 = 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 .
+
+
+
118
MOTION AND FLOW
[CHAP. 4
Also, from (4.15), for the Eulerian case,
du~.ldt = v2
=
= v3
du$dt
+ x3)/2 - et(x, - x3)/2 + v2(2 - e-t - et)/2 + v3(-e-t
C - ~ ( X ~
= e-Yxz
+ x3)12 + et(xz - %,)I2 + v2(-e-t + et)/2 + v3(2 - e-t
Solving these equations simultaneously for v , and v,, the result is a s before v, = x,,
+ et)/2 - et)/2
v 3 = xs.
4.3.
A velocity field is described by v1 = x ~ l ( l + t), vn = 2x2/(1+ t), v3 = 3x3/(1+ t). Determine 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 , = d x l / d t = x l l ( l t ) ; separating variables, dxllxl = d t 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 , = X 1 , v , = 2X2(1 t ) , v , = 3X3(1 a , = 6X3(1 t).
+
4.5.
+
+
t)2
+
and a, = O, a, = 2X,,
+
+
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 ( x , B)2 = e-zBh/X. and adding, t is eliminated and the path lines are the circles ( x l -A)2 From (4.6), v l = oe-Bh cos h ( A wt), v , = we-BA sin h ( A wt), v , = O and v2 = v: v; vi = ~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, tion field is given by a = xSê,
+ 3 ê2+ 2 ê3.
+ 2x2tê2 + x1x3ê3 +
Using the vector form of (4.18) the accelera-
+ x2t2ê2 + x1x3tê3) + x3tê,ê3 + t2ê$, + x1tê3ê3) (xS + 2x1t2)ê, + (2x2t + %,e)ê2+ (51x3 + 2x (xStê,
(2x1tê$,
or Thus
4.7.
a =
x3t2)ê3
ap = 3 ê l + 9 ê 2 + 6 ê , .
For the velocity field of Problem 4.3 determine the streamlines and path lines of the flow and show that they coincide. A t 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 d x 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 a s 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
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 , the rate of change of is given by vi = Bx1x3t, v2 = B x t2, ~ v3 = B x ~ x ~determine magnetic intensity for the particle a t P(2, -1,2) when t = 1.
+ +
Since a(r-l)/axi = -xi/r3,
i= Thus for P a t t = 1,
4.9.
119
MOTION AND FLOW
equation (4.11) gives - e-~t(Bx:x,t
-Ae-At/r
i, = -e-A(3A
+ B X ~ + Bx$x2)/r3 t2
+ B)/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 = (4ê2 3ê3)/5.
A
e
+
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
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)ecBtwhere d v i l d x j for this motion and Determine the velocity gradient A and B are constants. from it compute the rate of deformation tensor D and the spin tensor V for the point P(1,0,3) when t = 0. By (4.19)
avihxj =
(
"
-23
o which may be evaluated a t P when t = O
x, 22,
-2,
and decomposed according to (4.20) and (4.21) a s
4.11.
+
+
For the motion xi = X l , x2 = X2 ~ compute ~ the - I ) , x3 = X 3 X I ( ~-- 1) rate of deformation D and the vorticity tensor V. Compare D with d e i j l d 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
120
MOTION AND FLOW Comparing
D
[CHAP. 4
with dEldt,
!
=
ds,ldt
-e-%
-e-2t
O
-3e-3tI2 O
O
-3e-3t/2
=
Dij
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 (9, dx3 - q3 dx,)
2, 3-
(93 dxl
-
(91 dxz - 9, d x i ) ê 3
d ~ 3 $2 )
E
E
O , or
O
from which dxllql = dx21q2 = dx31q3.
4.13.
+
Show that for the velocity field v = ( A x s - Bx2)êl (Bxl - Cx3)êz the vortex lines are straight lines and determine their equations.
+
+ ( C X Z- Axl)ê3
+
From (4.29), q = V , X V = 2 ( C ê l A 3, sê3),and by Problem 4.12, the d.e. for the vortex lines are A dx3 = B dx,, B d x , = C dx,, C d x z = A dxl. Integrating these in turn yields the equations of the vortex lines x3 = Bx21A K1, x 1 = Cx31B K,, x z = A x J C K , where the Ki are constants of integration.
+
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. A
(-; r )
Thus avildxj =
O -B
-f
4.15.
= V i j and Dij = O.
For the rigid body rotation v = 3x3ê1- 4x3& rotation vector O and show that v = O X x. From (4.30), 2iZ = q, or i2 = 4 ê 1 ( 4 ê l f 3 ê 2 )X (
4.16.
~
~
ê
~
+ 3ê2. f
+ (4x2- 3xl)ê3, determine the
rate of
This vector is along the axis of rotation.
+
~ = ~ ê3x3ê1 ~ + - 4x3ê2 ~ ~ ê ( 4~x 2)- 3 x 1 ) ê 3 =
+
Thus
V
+
A steady velocity field is given by v = ( x ; - xix:)êl (x: $2 xz)ê2. Determine the unit relative velocity with respect to P(1,1,3) of the particles a t Q1(1,0,3),Q2(1,3/4,3), Q3(1,7/8,3) and show that these values approach the relative velocity given by (4.26). h
+
h
By direct calculation v , - v n , = -e1 2 e,, -15$,/8 22,. The velocity gradient matrix is
+
4(vp - vQ2)= -7ê1/4
3x1 - xp
-2xix2
+ 2ê2
and
8 ( v p- V Q3) =
CHAP. 41
121
MOTION AND FLOW
and a t P(l,1,3) in the negative x2 direction,
Thus vp
4.17.
( d v l d x ) ~= -22,
- Vgi.
e2
+ 2ê2
which i s the value approached by the relative unit velocities
+
[
6xlx2
[8vi/8xi] =
Here the velocity gradient is
Thus from (4.34) for
4.18.
+
~ : determine the rate For the steady velocity field v = 3x;x2êl 2xix3ê2 ~ 1 x 2 ê3, of extension a t P(1,1,1) in the direction of $ = (3êl- 4ê3)/5.
O x,x;
]
3%;
4x2x3 2:: xlx; 2x1x2x3
and its symmetric part
2 = ( 3ê1- 4ê3)/5,
For the motion of Problem 4.17 determine the rate of shear a t P between the orthogonal directions 2 = (3êi - 4&)/5 and fi = (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, 4.19.
=
[4/5,O, 3/51
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 Here
avi/axiI
=
2
0
0
E : r] O
=
0 O
A of
0
+
[ ;] o o o o
-1
-1
D,,
Thus X 1 = + f i ,
AII=O,
XI,l=-fi
The transformation matrix to principal axis directions is -112
-112
l/fi
1
and for principal values
122
MOTION AND FLOW
with the rate of deformation matrix in principal form
[CHAP. 4
[D;] =
o 4.20.
Determine the maximum shear rate :,
-fi
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, a s 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 fiê3)/2 a s found in Problem 4.19. Thus
+ +
Fig. 4-3
MATERIAL DERIVATIVES OF VOLUMES, AREAS, INTEGRALS, ETC. (Sec. 4.6-4.7) Calculate the second material derivative of the scalar product of two line elements, i.e. determine d2(dx2)ldt2. d(dxi) avi d(dx2) From (4.451, 7 = -d x k ; and i t is shown in (4.48) that - = 2Dii dxi dxj. Therefore
4.21.
dt
83,
and by simple manipulation of the dummy indices,
4.22.
Determine the material derivative dt through the surface S.
pidSi of the flux of the vector property pi
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,
CHAP. 41
123
MOTION AND FLOW
Now use of the vector identity V X ( p X v) = p(V v) - v(V p) 1.65) gives
it
4.24.
p - ~ d s=
L ["z+
v(vop)
+ (v
V ) p - ( p V ) v (see Problem
+ v x ( p x V ) ]~ ~ n d s
Express Reynold's transport theorem a s 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) a s given in Problem 4.24,
$i i dV =
V . v dV.
Here
V v d V represents the rate of change of dV, and so V v is known a s the cubical rate of dilatation. This relationship may also be established by a direct differentiation of (4.38). See Problem 4.43.
MISCELLANEOUS PROBLEMS 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 = ejikqk.
+
By (4.29), qi = ~ ~= ~ ~ ~ ~ ~ v (and ~v since ~, ~ ~ , = ~ O ~(see, for example, Problem = (GriSsk - S r k S s i ) V k j = 2Vsr. 1.50), qi = qjkV[k,jl = eiikVkj. From this result q,,qi = eirsYjkVkj
4.27.
av Show that the acceleration a may be written a s a = dt avi From (4.18), ai = at
avi + v , axk -
+ q x v + +Vv2.
and so
which, a s the student should confirm, is the indicial form of the required equation.
4.28.
Show that d(ln J)ldt = div v. and J becomes the sum of Let axi/aXp be written here a s si,, so that J = eP~Rx1,P%2,Q%3,R the three determinants, = f p Q R ( j S l , p ~ 2 , Q ~ 3 ,x1,pjS2,Qx3,R R c ~ ~ , p x ~ , ~ j NOW S ~ , ~ ) . = vl,,x,,p, ek., and so J = E P Q R ( ~ ~ , ~ X%3,R ~ , P+~~ZI, .QP ~ ~ , s ~ s , QZ ~~ , .P~X,ZR. Q ~ ~ , ~ ~ S , Of R ) . the nine 3 x 3 beterminants resulting from summation o? s in this expression, the three non-vanishing ones yield J = v,, J v2,2J v3,3J = v,,, J. Thus J = J V v and so d(ln J ) l d t = div v.
+
+
+
+
+
124 4.29.
MOTION AND FLOW
[CHAP. 4
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 d x 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 -x;
+ z;
-2:
,
-2x1x2 - d
o
+
O
O
o
Thus d,,, = O, do, = -(x: d I I I= -(x; 22).
4.31.
+xi
=
O
=
-d[-4xix;
+ d2 - (4- x;)2]
-d
+xi),
do, = ( x ;
+
2;).
Note here that
d I = (x:
+xi),
d I I = O,
Prove equation (4.43) by taking the material derivative of d S i in its cross product form d S i = cijkd x j 2 )d x k 3 ) .
axi axi a%, axk dX2dX3 = Using (S.&?), dSi = ~ ~ ~ ~ ( a xdXz(axklaX3) ~ / a X ~ ) d X 3 and axi - dSi = qjk ax ax. dSi = 6, dSi = d S p = 8x1 J d X z d X 3 and by Problem 4.28, J d X z dX3. Thus 2 axP ax, a x ~
ax,
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 a s given in Problem 4.27,
V V
or since q = V
X
$L
x
a
x a
x at + V x ( q x v) + V x ~ ( ~ 2 1 . 2 ) V x ( q x v ) = dqldt + q ( V v ) - ( q V ) v
= V
= aqlat
+
v and V X V(v212) = O.
,*nd~ =
1[$ +
Thus if q is substituted for p in Problem 4.23,
q ( V * v )- ( q * V ) v ] . R d S
=
L
(Vxa)*RdS
CHAP. 41
4.33.
125
MOTION AND FLOW
a
For the vorticity qi show that
qi d V =
at
From Problem 4.32 the identity V X a = aqlat .~ ~ ~ a s aqi/at = eijkak, - E ~ ~ ~ ~~ v , (. ) , E~Thus
i$
=
d~
S
+ qjvi- qivi]d s j .
[€,,ak
+ V X (q X v)
may be written in indicial form
[Eijkak,i- (.,srn~rn~r),sl
v
dv
and by the divergence theorem of Gauss (1.157),
Supplementary Problems A continuum motion is given by x 1 = 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, v 2 = X,(et e-t), v3 = O or v , = x1 - x3, v2 = x3(et -e-t), i v3 = O
+
+
A velocity field is specified in Lagrangian form by v , = -X2e-t, v , = -X3, v3 = 2t. Determine Ans. a , = e-t(x2 t x 3 - e), a, = O, a3 = 2 the acceleration components in Eulerian form.
+
+
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 = xil(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 = x; x i and A is a constant. The velocity field of the fluid is v , = 2 2 , v , = -x; Ans. dhldt = (-3A sin 3t)lr xixi, v , = O. Determine dXldt a t P(x,, x,, x,).
+
r2
XSX, +
Show that for the velocity field v , = x l x 2 circular.
+
22,
v , = -x;
- xixi,
v3 = O
the streamlines are
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.
+ si,
For the velocity field v , = s;x, and principal values of ü a t P ( l , 2,3).
(: 5
Am.
D:
=
0
o o);
o
-5
v , = -(xt
(
+ x,x;),
31-
a
=
1
11-
v3
= 0, determine the principal axes
o
O 3
o
For the velocity field of Problem 4.41 determine the rate of extension in the direction V - (3,- 2 3 , 2 ê 3 ) / 3 a t P ( l , 2,3). What is the maximum shear rate a t P?
+
A-
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 ~ ~ ~ ors ~ V, = ~ V xs (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 (5.9)
Po = PJ
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 differentialform 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
Since the volume V is arbitrary, the integrand of (5.15) must vanish. equations,
[CHAP. 5
The resulting
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 (5.17')
+
+
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 or N = ( X X ~ Vd ) v (5.18) ~ ( t=) v ~,,x,pv, d~
J
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 motion. If the substitution t p ' = uPknpis 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 (5.19), which upon substitution of t p ' = upknp, reduces to h
h
Since the volume V is arbitrary,
,,, ,
... .V.
= O
which by expansion demonstrates that
or
1, = O
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
CHAP. 51
FUNDAMENTAL LAWS OF CONTINUUM MECHANICS
L
129
i
dK vivi dt cLdV = which represents the time rate of change of the kinetic energy K in the continuum. Also, = Dij V,,, so that (5.22) may be written v i ~ j i ,=j ( v , ~ ~ , )vi,j , ~ uji and by (4.19)
But
pv,;,dV
=
+
since Vijuji 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"' = ~31. . n3 . ,the energy equation for a continuum appears in the form
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 a s
where d W l d 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 time rate o f change of t h e kinetic plus the internal energy i s equal t o t h e 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 unit time. 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 of 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 a s 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
130
[CHAP. 5
FUNDAMENTAL LAWS OF CONTINUUM MECHANICS
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
1
entropy density is denoted by s, so that the total entropy L is given by L = 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)
CHAP. 51
FUNDAMENTAL LAWS OF CONTINUUM MECHANICS
131
where d s is the increase in specific entropy, d s c e )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 ddi)
>
O
ds(i) = O
(irreversible process)
(5.34)
(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) (5.36)
T
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 transforming the surface integral in (5.37) by the divergence theorem of Gauss, the local form of the internal 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 'iDj ' (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) a s
1 In this equation, - o;?) iij is the rate of energy dissipated per unit mass by the stress, and 0
d q l d 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
132
FUNDAMENTAL LAWS O F CONTINUUM MECHANICS
[CHAP. 5
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 temperatureheat 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.
CHAP. 51
133
FUNDAMENTAL LAWS O F CONTINUUM MECHANICS
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.
+ pvk,
= 0.
Show that the material form d(pJ)ldt = O of the continuity equation and the spatial form dpldt p v , , , = O are equivalent.
+
Differentiating, d ( p J ) l d t = ( d p l d t ) J that d ( p J ) l d t = J ( d p l d t P V , , ~ )= 0 .
+
5.4.
+
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 ( 5J), dpldt
5.3.
+
+ p dJ/dt
= O and from Problem 4.28, d J l d t = J v k , , so
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.
134 5.5.
[CHAP. 5
FUNDAMENTAL LAWS O F CONTINUUM MECHANICS
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) 5.6. Show by a direct expansion of each side that the identity eijkujke,= 1, used in (5.20) and (5.21) is valid. A
By (1.15) and (2.8),
L, = -
alSl (023
x ê1 + olGl x $2
- 0 3 2 ) ê1
+ a l g l x ê3 + . . + u33ê3 x
+ ('731 - 013) $2 + ( ~ 1 -2 ~ 2 123)
Also, expanding eijkajkgives identical results, (az3- a32) for i = 1 , for i = 3.
5.7.
$3
- ul3) for i = 2,
(031
(al2- oZ1)
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 a n additional term so that
which reduces to (see Problem 2.9) mi = O for this case.
5.8.
+
( Y j k a j k+mil dV = O, and because V i s arbitrary, eijkajk
The momentum principle in differential form (the so-called local or "in the small" ) , ~ . that the equaform) is expressed by the equation d ( p v , ) l d t = pbi + ( u i j - ~ V ~ V ~ Show tion 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,jvj
+ pvj, + j)
p(avi/Jt
+ vivi,
j)
The first term on the left is zero by (5.4) and the second term is pai. is (5.16).
5.9.
pbi
aij, j
Thus pai = pbi
+ aij,j
which
Show that (5.19) reduces to (5.20). h
Substituting oPknpfor 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
s,
The term in parentheses is zero by (5.16), also €ijkUjk
.v
= o.
zj,,
= S j p and
cijkvjvk = O, so that finally
CHAP. 51
5.10.
135
FUNDAMENTAL LAWS O F CONTINUUM MECHANICS
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 = eijkwjXk,
=
where Zip =
d
[O.
J"~ ( 4 P q x q-
p(8ipzqzq- 44 d V 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.
i
From (5.2?)),
K
= i p Y d v =
ii
-
wiw~ 2-
In symbolic notation note that K =
=
peiikUj%k€ipq~pZq dv
~ ~ p w j ( 8 j p-~6ki q~S k p ) x k z q dV
i
wiwpzip 2
p(6ipzqzq - z P z $ d V
o.l.0, 2
5.12. At a certain point in a continuum the rate of deformation and stress tensors are
given by
-:i
and 4
2
uii
5
-
=
Determine the value h of the stress power Dijuij a t the point. 14
5.13.
Multiplying each element of Dij by its counterpart jn oij and adding, h = 4 14 40 = 58.
-4
+ +
+O -4 +O -6 +
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 = v i B j- Vij; and since V i j u i j= O, i t follows that Dijoij = = -pvi,i. From the continuity equation (5.3), vi,i = -(llp)(dpldt) and so D i j q j = (plp)(dpldt) for oij = -paij.
5.14.
Determine the form of the energy equation if uij = (-p heat conduction obeys the Fourier law ci = -kTai. From (5.92), du
p d t
+
+ h* Dkk)8,, 4- 2 , ~D,, *
+ +
and the
+z +z
- (-p+ X*Dkk)SijDij 2p*DijDij kTSii P do - -(h* 2 / ~ * ) ( 1 , )~ 4/~*11, kT,ii P dt
+
+
where I, and 11, are the first and second invariants respectively of the rate of deformation tensor.
136 5.15.
If uij = - p s i j , determine an equation for the rate of change of specific entropy during a reversible thermodynamic process. Here oij '= ::o
5.16.
[CHAP. 5
FUNDAMENTAL LAWS O F CONTINUUM MECHANICS
ds
du
and (5.41) gives T - = dt dt
p dp + -dt p2
upon use of the result in Problem 5.13.
= /3DikDkj, determine the dissipation function in terms of For the stress having 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) =
(D(l)
+
+
~ ( 3 )
+ 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:;'
gj = 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 a s the outer product of two symmetric tensors AijBpq= Kijpq, i t is clear that since both Aij and Bij have six independent components, Kijpqwill 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 = KijpqDpqof Problem 5.17 is assumed isotropic so that Kijpqhas 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 a s shown in Fig. 5-2. Isotropy of Kijpqthen requires 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 of the coordinate axes these 26 components may be reduced to 2 for the case of isotropy.
5.19.
x1 or x2
53
Fig. 5-2
+
+
Use For isotropy Kijpqmay be represented by Kijpq= A*8ij8pq , ~ * ( 8 ~ ~ 8iq8jp). 8~, this to develop the constitutive equation uij = KijpqDpqin terms of A* and P * .
5.20.
137
FUNDAMENTAL LAWS O F CONTINUUM MECHANICS
CHAP. 51
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 = S i j 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.
+
+
+
+
MISCELLANEOUS PROBLEMS
+
= ( ~ , , a ~ , ~ qjvi,j)lpwhere p is the density, ai the acceleration and
581. Show that
qi the vorticity vector. By direct differentiation
dd, (f) = 4 -
-
P
4.32); and by the continuity equation (5.4,
e
4 = rijkak,i + qivi,i - qivjj
But
p2' = -pvci.
(see Problem
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
+
=p ~ = S pT - i.. dT - T . A t constant deformation, Zij = 0 and 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),
isothermal process,
-ci,i
= kTBii = pT -iij
(YT
kT,ii = pc(")T
+ )T . :c )
Finally from (6.70), doil/dT = (3X+
so that
+ (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
= Xeiiejj12
-
(3X
+ 2p)aSij(T- To)eij/2
+ peijeij - (3X + 2p)a(T - To)eii12
MISCELLANEOUS PROBLEMS 6.26. Show that the distortion energy density u:,, may be expressed in terrns of princigal (a2 - u J 2 ( a 3- ~ ~ ) ~ ] / 1 2 G . stress values by the equation uy,, = [ ( a , -
+
\
+
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
+ o ; + o: - (a1+ o2 + o3)(u1+ + 03)/3]/4G = [2(u; + a: + o: - olo2 - u 2 ~ 3- a3u1)/3]/4G = [(vl- u2)2 + (v2- u3)2 + - u1)2]/12G
ufD) = [o:
42
(03
CHAP. 61
157
LINEAR ELASTICITY
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
+ + + ~ ~ ~+ 82 / ~J ~~ ~ 8=~ 8X/2[€iispq ~ ~ ~~ 8] $ ~ ~
Ju*larPq = X/2[eii(aejjla~pq)ejj(âciilòepq)] Z p ~ ~ ~ ( a e ~ ~ l a $ , )
= X/2[eii8jp8jq
€ j j s p q ]f
+ 2pePq =
- heiiSpq
2,Urpq
upq
+ v)uijuii - ~ u ~ ~ u and ~ ~so] / ~ E + ujjôW]/2E = [ ( I + v)uPq- u8pquii]lE =
Likewise from Problem 6.1, u* = [(I
au"la,,
= [2(1+ u)uijSipSjq- v(uiiôpq
epq
6.28. Express the strain energy density u* as a function of the strain invariants. From Problem 6.1, u* = Xeiiqj/2 11, = (riirjj- 4 j Y j ) / 2 , i t follows that
+ priirij;
and since by comparison with (3.91), I, = cii and
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 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 . when referred to the yield stress in simple tension, example) to be ~ ~ 1 2Therefore Tresca's yield condition becomes (8.8) u~ - -'I'II = u~ 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
( 6 ) Pure Shear
(a) Simple Tension
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
179
PLASTICITY
CHAP. 81
which is usually written in terms of the principal stresses as
+
(aI - uII)2
(gII
- uIII)2+ (vIII- u1)2 = 6 c y
(8.11)
With reference to the yield stress in simple tension, it is easily shown that (8.11) becomes (oI - uIJ2
+ (cII- aIII)2f
(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 (8.13) (aI - uIJ2 + (rII - uIII)2f bIII - uI)2 = 6k2 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
(8.14)
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 a s 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 3U, is seen to circumscribe the regular Tresca hexagon. In Fig. 8-5(c), circle of radius m 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.
180
PLASTICITY
[CHAP. 8
(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 workhardening 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.
CHAP. 81
181
PLASTICITY
(a) Mises Circles
( b ) Tresca Hexagons
Fig. 8-6
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 (8.17) f i (oij- a,) = O where the aij are coordinates of the center of the new yield surface. If linear hardening is assumed,
Xi. = c;:
(8.18)
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 = sijdA 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.
182
PLASTICITY
[CHAP. 8
The name plastic potential function is given to that function of the stress components g(u,) for which dg dA de; = (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 PrandtlReuss 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 effectiveplastic 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 uijDijper unit volume. From (4.25), den = D,dt, so that the work increment per unit volume may be written (8.2 8) dW = ali dcij 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 d W P = a,,dcEQ (8.31) and (8.21) rewritten in the form de;
=
3 dWP S.. 2 a;, --
CHAP. 81
183
PLASTICITY
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 a s 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
I n terms of equivalent stress and strain, the parameter (P
=
+ may be expressed as
3 -2 um
where here
L&
= d Z 5 7 3 so that €P 23
=
3 cEe --
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 thickwalled 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
184
PLASTICITY
[CHAP. 8
( 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
XI
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 (8.44) u33 = +(all+ nz2) Adopting the standard slip-line notation o,, = -p, principal stress values of (8.41) are found to be
and
~((r,,
-~ , , ) ~ + / 4( u , , ) ~=
k, the
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 designated as the a and p directions. From the
21
Fig. 8-8
CHAP. 81
PLASTICITY
geometry of this diagram, O = r14
++
185
so that
tan2+ =
1
.- 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
U12
= FC COS2+
+ k sin 2+
and from the equilibrium equations i t may be shown that p
+ 2k+
= C, a constant along a n
p - 2k+ =
a
line
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
/i' lines,
v, = v, cos +
-
vz = v, sin +
+ v, cos +
a
and
v, sin + (8.49)
For a n 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 t h a t
(& (v, cos + - v, sin +)S*=. =
o
These equations lead to the relationships dv, - v, d+ = O dv,
+ v, d+
= O
on a lines
(8.52)
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).
CHAP. 81 8.5.
PLASTICITY
187
Convert the von Mises yield condition (8.10) to its principal stress form a s given in (8.11). UM
From (2.72), - 1 1 ~=~-(slsI1 = (01 oI1 uIII)/3. Hence
+ +
Thus
8.6.
+
(01 - QII )2
SIISIII
+ sII1s1);
and by (2.71),
+ (011 - 0111)~+ (oIII -
SI
= o1 - UM, etc., where
= 6Cy
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
+ ( x / h - 3 ~ / f i )+ (x/fi+ 3~/1&)2 = which simplifies to the Mises yield circle 3X2 + 3Y2 = 20; of Fig. 8-5(b). (-fix)z
8.7.
2
Using the transformation equations of Problem 8.6, show that (8.14), a, is the equation of the n-plane. Substituting into (8.14) the o's of Problem 8.6, a1 is the XY plane @-plane).
8.8.
20.;
+ o,, + aII1 = fiZ
+ a,, + u,,,
= O, or Z = O which
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 condition becomes 0: - o,o,11 o;,, = o;
+
which is the ellipse (~I/UY -) (~Q I U I I I If~ )(UIII/"Y)~ = 1
= 0,
Fig. 8-14
188
PLASTICITY
[CHAP. 8
with axes a t 45' in the plot. Likewise, from (8.8) and the companion equations a111 - a,, = dt Transformed Viscoelastic
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
206
LINEAR VISCOELASTICITY
[CHAP. 9
is the Laplace transform of the viscoelastic solution. Inversion of the transformed solution yields the viscoelastic solution. The correspondence principle may also be stated f o r problems other than quasi-static
. problems. Furihermore, the form of the constitutive equations need not be the linear differential operator form but may appear a s 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 f o r 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 GD. Since the stress across each the strain of the dashpot. Thus e = es ED and also ;= Zs element is o, ( 9 . 1 ) and (9.2) may be used to give ;= O/G a/?.
+
,
9.2.
In the Kelvin model of Fig. 9-2(b), o = os
+
+
+ a~
and directly from (9.1) and (9.2), o = v ;
+ Ge.
Use the operator form of the Kelvin model stress-strain relation to obtain the stressstrain 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. e~ or in operator form e = olG1 o / { G 2 q2at). From this Thus E = es
+
+
and so GIGze
9.3.
+
+ G l s z ; = ( G 1+ Gg)a + 72;.
Determine the stress-strain equation for the four parameter model of Fig. 9-4. Let and compare with the result of Problem 9.2. T,I~ + Here the total strain
e
= e~ e
+ e~
which in operator form is
= o/{Gp + vzat)
+ {at + l / ~ l ) o / G l { a t )
Expanding the operators and collecting terms gives
As 7 , -+ m this becomes Ü Problem 9.2.
9.4.
+ ( G 1+ G2)&l'lg= G1'E+ G1G2;Iq2 which
Treating the model in Fig. 9-13 a s a special case of the generalized Maxwell model, determine its stress-strain equation.
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 here O U / T = 3Gso/tl Ge0t/7tl. Integrating this is the constant of integration. When t = 0, o eo(2q G t - qe-tlr)/tl. Note that the same result is
+
+
+
+
3 ~ ~ t 'dt' i r+2 tl
9.10.
+
model is 017 = q Z + 3G; GE/T and so yields o = a0(39 G t - ?) Ce-t/r where C = q ~ o / t land so C = -l ) ~ O l t lThus . o = obtained by integrating
Xt
+
+
dtf
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) a s 6 (G1 G2)uIq2= ~ ~ G ~ ( [ 8 (G1G2[U(t)]lT2) t)] 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 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
6 + u/r2 = (G1+ G2); + G1G2c/q2 and so with = eo[U(t)] and ;= eo[6(t)] use of the integrating factor etlrz gives
-o--
(Sec. 9.5)
-u
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
9.12.
209
LINEAR VISCOELA'STICITY
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 ) = G l / s ard table of Laplace transforms). Thus from (9.40),
+ G z / ( s+ 1 1 ~ (see ~ ) any stand-
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 sustained constant stress U , (Fig. 9-19) is applied to a Kelvin material, determine , A. the resulting strain. Assume ~ , / t= The stress may be expressed a s u
t- I.
= h t [ U ( t ) ]- h ( t - t l ) [ U ( t- t,)]
t
Fig. 9-19
which when introduced into (9.4) leads to
Integrating with the aid of (9.18) gives E
= ( h l G ) ( ( ti~ ( e -~ l l) )~[ U ( t ) ]- ( ( t- t l )
which reduces a s t
-+ m
to
E
+ ~ ( e ( t i - t ) / ~l )-) [ U ( -t t l ) ] )
= 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$
+ 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'
([U(tl)]
which by (9.18) and (9.23) reduces to =
St
$
t X [ [ ~ ( t ) ] ( 1 - e - ( t - t l ) l ~dt' ) - [ U ( t- tl)] (1 - e - ( t - t f ) l r ) 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
LINEAR VISCOELASTICITY
[CHAP. 9
The stress may be represented a s a sequence of ramp loadings a s shown in Fig. 9-21 above. From Problem 9.13, E = ( h / G ) [ t r(e-tlf - 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 ) -((t Note that a s t
+ r ( e - ( t - t l ) l ~ - l ) ) [ U ( t- ti)]
- 2tl) + ~ ( e - ( t - z t ~-) /l ~) ) [ U ( t 2 t l ) ] + ( ( t- 3tl) + ~(e-(t-3t1)I7- l ) ) [ U ( t 3tl)]l
+ 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) a s 6 + U/T= G ; and inserting (9.41) and (9.42) gives iwooeiwt4- o0eiot/r = Giw~,ei(wt-6) from which ~ ~ e= G* i ~= G l i w~ ~ l~( l +i w ~ ) , or in standard form From Fig. 9-11, tan 8 = G d G 1 = 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*.
+ l / q ) o = &e G ~ w T+ /(~
After the suggested substitution (9.5) becomes (iolG
U / E = Giol(io
+ 117) =
from which
~ W T )
a s 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) a s the generalized Maxwell complex modulus becomes
9.19.
Verify the relationship compliance.
Jl = l / G l ( l+ tan28)
between the storage modulus and
+
+
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
S
u d over ~ one cycle. I
For the stress and strain vectors of Fig. 9-10, the integral
ds
dt
=
l
2 ~ 1 0
(oo sin U
- ooEo. J2'I*
cos (ut - 6 ) d t
~ ) E ~ W
sin w t (cos u t cos a
o 2nlo
S
odr evaluated over one cycle is
sin 2ot
dt
+ J,
+ sin u t sin a ) d t
lbiu
(sin2 o t ) d t ]
= o:sJ2
CHAP. 91
211
LINEAR VISCOELASTICITY
THREE DIMENSIONAL THEORY. VISCOELASTIC STRESS ANALYSIS (Sec. 9.7-9.8) 9.21. Combine ( 9 . 4 8 ~ )and ( 9 . 4 8 3 ) to obtain the viscoelastic constitutive relation a i j { R } e k k {S)cij and determine the form of the operators { R } and { S I .
+
uij =
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 = o o [ U ( t ) ] l Kfor 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 ~- aO[U(t)]lgK) l ( ~ ~ ~
Solving this differential equation yields €11
As t
+
m,
€11 ->
(3K
= aO(3K +
+ aoe-tlT[U(t)]/9K
e-t1T)[ U ( t ) ] l 9 K G
+ G)aO19KG= a J E .
9.23. A block of Kelvin material is held in a container with rigid .walls so that e, = e,, = O when the stress o,, = - a o [ 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 i2u2, = 3Kell and (9.48a) gives 2(u11- u2,)/3 = 2G{1+ ~ a ~ ) ( 2 ~for ~ ~a/ Kelvin 3) 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~)/(8GG K ) ) [ U ( t ) ]
9.24. The radial stress component in an elastic halfspace 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 halfspace by means of the correspondence principle when P = Po[U(t)].
Fig. 9-23
212
LINEAR VISCOELASTICITY
[CHAP. 9
The viscoelastic operator for the term ( 1 - 2 ~ is) { 3 Q ) l { 3 K P the transformed viscoelastic solution becomes
+ Q ) so that for a
Kelvin body
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 a s 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 Kelvin body causes the transformed displacement to be
+ 4Q)/4Q(3K + Q ) which
for the
After considerable manipulation and inverting, the result is
Note that when t = O, w ( , = o ) = O and when t
9.26.
+
m,
w ( , = ~ -+ ) P o ( l - u z ) / E ~ r ,the elastic deflection.
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 ( x J 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
which may be written in terms of
v
as
+ 3K))/2(4G+ 3 K ) 02,1t,,
= -voo/(l
= -oo(3K - 2G)l(3K -
v).
Thus for v
+ 4G)
112, o,,lt+,
= -o*
CHAP. 91
213
LINEAR VISCOELASTICITY
MISCELLANEOUS PROBLEMS 9.28. Determine the constitutive relation for the Kelvin-Maxwell type model shown in Fig. 9-25 and deduce from the result the Kelvin and Maxwell stress-strain laws.
81
.
c
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