35737861-Schaum-Continuum-Mechanics-Mase.pdf

SCHAUM'S O U T L I N E O FTHEORY 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, Deforma- tion and Strain, Motion and Flow, and Fundamental Continuum Laws. Applications are treated in the final four chapters on Elasticity, Fluids, Plasticity and Viscoelasticity. At the end of each chapter a collection of solved problems together with severa1 exercises 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 .. . .. .. .. .5 Dot and Cross Products of Vectors .. .... .. .... .... .. . .... ...... . .. ...6 Dyads and Dyadics ...... .... ..... .... ....... ... General Tensors .. Permutation Symbol Dual Vectors . 17 Symmetry of Dyadics.. .. . .. ...... ... .. . ... .. Multiplication of a Vector by a Scalar . .... ........... ... 44 .. Stress Tensor Symmetry .. . 44 Body Forces .... ......... ...... ...... ......... .... ..... 48 Stress Transformation Laws ...... ...... 45 State of Stress a t a Point .. . .. ......... ..... .... . ... ...... 46 The Stress Tensor-Stress Vector Relationship . ...... 54 Plane Stress .. Surface Forces .. .. . Orthogonality Conditions .. ...... ..... ....... . . ... .. Principal Stresses Stress Invariants Stress Ellipsoid ......... 52 Mohr's Circles f o r Stress .. .... .. ..... . ................. .. .......... .. .. .. .. ... . ......... ... 13 . ..7 .... The Stress Vector ............... 11 1... .......... .. 22 Line Integrals... 10 1........ ...... .9 . .. . 45 Cauchy's Stress Principle ... ........ ..... .. 49 Stress Quadric of Cauchy .... ...... .. ...... . .. 19 Principal Values and Principal Directions of Symmetric Second-Order Tensors .. .. . ....... ... . 15 Tensor Multiplication ..... ....... 57 . ............. . ... ....... ... .. 4 1.. .. . 2 1... 15 Vector Cross Product ..... ... .... 23 The Divergence Theorem of Gauss ................... . .8 ... .. ............ .. 1 1.. 23 Chapter 2 ANALYSIS OF STRESS The Continuum Concept ...... G 1....... ................. . .. ........... . ... ............. 8 1.. 56 Deviator and Spherical Stress Tensors . Matrices Matrix Representation of Cartesian Tensors ... .... ...... . ..... Addition of Cartesian Tensors Multiplication by a Scalar ..... 50 ....... 3 1......4 Vector Addition . ... 47 Force and Moment........ . Stokes' Theorem ..... .. . ............ .... 1 1.11 Coordinate Transformation . 20 . . .. The Kronecker Delta ..... . Matrices and Tensors ....... ......2 ......... ..10 Summation Convention Used with Symbolic Convention .......... .. . . . ..... 2 1... .... .. ........ General Tensors ... .. ....... ... .... . Linear Vector Functions Dyadics a s Linear Vector Operators .. .... .... ...... ..... ........ ... . ..... CONTENTS Page Chapter 1 MATHEMATICAL FOUNDATIONS 1. 12 1..... . . .. ..... ... 8 1........ . . .. . .. . ... .. .................. ... .... ... . .... .... . ......... ....... . . .1 Tensors and Continuum Mechanics ....... .. . ... ........... ............ .... . ............... ....... Homogeneity Isotropy Mass-Density ......13 Transformation Laws f o r Cartesian Tensors .. . . .. .... .. Equilibrium . ...... ....... . ...... .....3 Vectors and Scalars ..... Coordinate Systems Base Vectors Unit Vector Triads ..... ...... ...... ... .... . 51 Maximum and Minimum Shear Stress Values ...... ............. ....... . .... ..... ...... ........ Cartesian Tensors ............. . Indicial Notation Range and Summation Conventions ........ . .. . .... . ......... Stress Tensor .. .. 16 . ..... .. .... 21 .... ..... ...... Tensor Fields Derivatives of Tensors ... Powers of Second-Order Tensors Hamilton-Cayley Equation .. . . Cartesian Tensors Tensor Rank .... .. ..12 The Metric Tensor..... ..... ... ......... Linear Rotation Tensor ............... ....... 112 4. . Rotation Tensor .. Flow ...... 142 6......... ....14 Spherical and Deviator Strain Tensors . .. Energy Equation .. ........ Natural Strain .. ... .5 Theorem of Superposition.11 Stretch Tensors ..... . Displacement Gradients .. Strain Energy Function . 112 4............. 128 Equations of State.......... . 140 6.. Finite Strain Interpretation .......10 Stretch Ratio . Deformation and Flow Concepts .... ........... . ... .......... . ... ........... ......... Instantaneous Velocity Field . ..... Anisotropy .... 82 3....... ....... Infinitesimal Strain Tensors .. ...3 Position Vector .4 Lagrangian and Eulerian Descriptions ..... .6 Two-Dimensional Elasticity....... . 84 3...2 Linear Momentum Principle .... Elastic Symmetry .15 Plane Strain ... .. 130 The Clausius-Duhem Inequality ................ .. Elastic Constants ... Equations of Motion .... . 143 6. ... 110 4......7 Small Deformation Theory .. .. .....2 Velocity......... .... ........ Elastodynamic Problems ..... Vorticity ............ 145 6...5 Physical Interpretation of Rate of Deformation and Vorticity Tensors ..5 Deformation Gradients... 114 4........ ....... ...... . . 91 3.. Thermomechanical and Mechanical Continua .. 111 4...... 91 3.... .... .. 92 Chapter 4 MOTION AND FLOW 4. ......1 Motion . Dissipation Function ... Surface and Line Integrals ........ ... CONTENTS Page Chapter 3 DEFORMATION AND STRAIN 3.. 145 ....4 Elastostatic Problems .....7 Material Derivatives of Volume...2 Coritinuum Configuration . .............. ........9 Interpretation of the Linear Strain Tensors ... ............. . ........... .. ...... .1 Generalized Hooke's Law . Displacement Vector ... .8 Relative Displacements ...... .................. ... 87 3. 80 3.. .......... ................ Area and Line Elements ............ Stream Lines......... . 127 Moment of Momentum (Angular Momentum) Principle .... 86 3..... 77 3...12 Transformation Properties of Strain Tensors ...... 81 3...... Continuity Equation .......... 113 4.... 88 3. . Rotation Veztor ......... Venant's Principle . Cubical Dilatation . .... ..... ... 83 3........... . ..... .............6 Deformation Tensors ..... ... ... 77 3.. .... .. Equilibrium Equations . S t............. ................. Mohr's Circles f o r Strain ....... . Acceleration . .................. .. .... ..2 Isotropy ... 89 3........................... 141 6.. 116 Chapter 5 FUNDAMENTAL LAWS OF CONTINUUM MECHANICS 5... .3 P a t h Lines....... ....... ...... ........ 132 Chapter 6 LINEAR ELASTICITY 6.......... Entropy ....4 Rate of Deformation ..1 Particles and Points ....... 131 Constitutive Equations . ... . 126 5... Finite Strain Tensors .................. . Plane Stress and Plane Strain ...... F i r s t Law of Thermodynamics. .. 79 3.......... .................. 77 3.... Steady Motion ...... Strain Invariants ..........3 Isotropic Media ... ......... .......................13 Principal Strains... .... 128 Conservation of Energy ................ ...16 Compatibility Equations for Linear Strains ............ Second Law of Thermodynamics ..... Material Derivative .. Uniqueness of Solutions ... . ........... ...1 Conservation of Mass . ....... .... ..6 Material Derivatives of Volume... ... .... .... 161 7..................11 Elementary Slip Line Theory f o r Plane Plastic Strain ... 204 ...4 Creep and Relaxation ......... Potential Flow Plane Potential Flow ..3 Yield Conditions .....................................6 .................. 184 Chapter 9 VISCOELASTICITY 9......... 149 Chapter 7 FLUIDS 7....7 Airy's Stress Function .... Steady Flow Hydrostatics Irrotational Flow ..................... ...........7 ... Relaxation Function ...... Viscous Stress Tensor Barotropic Flow ........ 181 8......1 ...8 Two-Dimensional Elastostatic Problems in Polar Coordinates ................7 Three Dimensional Theory ..... Viscoelastic Stress Analysis Correspondence Principle .....6 ...................................................... ............... 165 Chapter 8 PLASTICITY 8.... . .3 ..........10 Linear Thermoelasticity ............ 160 7............ .............. 182 8......... 196 9...................... .... ....... 182 $8.... 198 9.9 Total Deformation Theory .. Stress Space The 1I.9 Hyperelasticity .........8 ....... Post-Yield Behavior Isotropic and Kinematic Hardening . ........5 ............................ 149 6...........1 Linear Viscoelastic Behavior .............4 ........... . Perfect Fluids Bernoulli Equation ..... CONTENTS Page 6..... 183 8....Plane Yield Surfaces ..2 Idealized Plastic Behavior ...8 ............... 199 9........ . ... Tresca and von Mises Criteria ..... Hereditary lntegrals ...... 175 8. Circulation .. Fluid Pressure ... 180 8. 147 6................................................ Plastic Stress-Strain Equations Plastic Potential Theory ... 176 8........... .......... Generalized Models Linear Differential Operator Equation ........... Equivalent Stress Equivalent Plastic Strain Increment ........ .............................5 Creep Function ........... Hypoelasticity ..... 202 9.........5 ......................... ......... 177 8...1 Basic Concepts and Definitions ........ 164 7............ 183 8.. .........................2 Siniple Viscoelastic Models ..... 163 7.. 200 9... 196 9............................... 148 6....................... Constitutive Equations Stokesian Fluids Newtonian Fluids ............... 203 9......6 Complex Moduli and Compliances ....... 173 8......... 162 7......................... Basic Equations f o r Newtonian Fluids Navier-Stokes-Duhem Equations ..2 ......10 Elastoplastic Problems ...............3 ......4 . ....... Plastic Work Strain Hardening Hypotheses ........ . the tensors defined a r e known a s general tensors. such tensor equations. Yet i t may be specified in a particular coordinate system by a certain set of quantities. Tensors of order zero are called scalars. Higher order tensors such as triudics. if they are valid in one coordinate system. or tensors of order three. TENSOR RANK.2 GENERAL TENSORS. In dealing with general coordinate transformations between arbitrary curvilinear coordinate systems. the word "tensor" in this book means "Cartesian tensor" unless specifically stated otherwise. Because tensor transformations a r e linear and homogeneous. Physical quantities having magnitude only a r e represented by scalars. the tensors involved are referred to a s Cartesian tensors. and tetradics. the law o f transformation of the components of a tensor is used here as a means f o r definicg the tensor. where N is the order of the tensor. When attention is restricted to transformations from one homogeneous coordinate system t o another. Tensors of order one have three coordinate components in physicaI space and a r e known as vectors. As a mathematical entity. Since much of the theory of continuum mechanics may be developed in terms of Cartesian tensors. a r e valid in any other coordinate system. o r tensors of order four a r e also defined and appear often in the mathematics of continuum mechanics. Second-order tensors correspond to dyadics. A t the same time. Tensors may be classified by r a n k . such quantities a r e represented by tensors. Thus in a three-dimensional Euclidean space such as ordinary physical space. This same classification is also reflected in the number of components a given tensor possesses in an n-dimensional space. . Specifying the components of a tensor in one coordinate system determines the components in any other system. The physical laws of continuum mechanics a r e expressed by tensor equations. Mathematically. according to the particular form of the transformation law they obey. a tensor has a n existence independent of any coordinate system. o r order. these physicaI quantities a r e very often specified most conveniently by referring to a n appropriate system of coordinates. Severa1 important quantities in continuum mechanics a r e represented by tensors of rank two. 1. known a s its components. 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. Accordingly a tensor of order zero is specified in any coordinate system in three-dimensional space by one component. Indeed. the number of components of a tensor is 3N. CARTESIAN TENSORS. Quantities possessing both magnitude and direction a r e represented by vectors. 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. 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. which possess magnitude only are represented by tensors of order zero which are called scalars.3) Fig. arbitrary vectors a and b are shown along with the unit vector ê and the pair of equal vectors c and d. 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. A. vectors are designated by bold-faced letters such a s a. 1. Pictorially. Unit vectors are further distinguished by a caret placed over the bold-faced letter. for example. 1-2(c). MULTIPLICATION OF A VECTOR BY A SCALAR Vector addition obeys the parallelogram law.2 MATHEMATICAL FOUNDATIONS [CHAP. or for emphasis it may be denoted by the vector symbol between vertical bars as Ia]. Thus a-b = -b+a = d (1-2) The operations of vector addition and subtraction are commutative and associative as illustrated in Fig. b. Algebraically.4 VECTOR ADDITION. such as mass and energy. 1-2 . 1-2(a). in Fig.1) Vector subtraction is accomplished by addition of the negative vector as shown. 1 1. Fig. Those physical quantities. The graphical construction for the addition of a and b by the parallelogram law is shown in Fig. The nu11 or zero vector is one having zero length and an unspecified direction. etc. In Fig. etc. Such directed line segments are the geometrical representations of first-order tensors and are called vectors. a vector is simply an arrow pointing in the appropriate direction and having a length proportional to the magnitude of the vector. In the symbolic. which defines the vector sum of two vectors as the diagonal of a parallelogram having the component vectors as adjacent sides. the addition process is expressed by the vector equation a+b = b f a = c (1. may be represented in a three-dimensional space by directed line segments that obey the parallelogram law of addition. Scalars are denoted by italic letters such as a.3 VECTORS AND SCALARS Certain physical quantities. Equal vectors have the same direction and equal magnitudes. or Gibbs notation. 1-1. such as force and velocity. which possess both magnitude and direction. for which the appropriate equations are (a+b) + g = a + (b+g) = h (1. b. The negative of a vector is that vector having the same magnitude but opposite direction. 1-1 The magnitude of an arbitrary vector a is written simply as a. 1-2(b) where the triangle rule is used. A unit vector is a vector of unit length. CHAP. Exceptions are multiplication by zero to produce the nu11 vector. 1-4(a).7) 1. Thus m(nb) = (mn)b = n(mb) (1. Fig. 1-4(b). and ê is a unit vector perpendicular to their plane such t h a t a right-handed rotation about ê through the angle 0 carries a into b. the result is a unit vector in the direction of the original vector. 11 MATHEMATICAL FOUNDATIONS 3 Multiplication of a vector by a scalar produces in general a new vector having the same direction a s the original but a different length.b = boa = a b c o s 8 in which '6 is the smaller angle between the two vectors a s shown in Fig. depending upon the numerical value of m. 1-3 Multiplication of a vector by a scalar is a.4) In the important case of a vector multiplied by the reciproca1 of its magnitude. 1-4 The cross or vector product of a into b is the vector v given by in which O is the angle less than 180" between the vectors a and b. The cross product is not commutative.ssociative and distributive.5 DOT AND CROSS PRODUCTS OF VECTORS The dot or scalar product of two vectors a and b is the scalar = a . . shown shade'd in Fig. The magnitude of v is equal to the area of the parallelogram having a and b as adjacent sides. O<m<l Fig. The dot product of a with a unit vector ê gives the projection of a in the direction of ê. and multiplication by unity which does not change a vector. Multi- plication of the vector b by the scalar m results in one of the three possible cases shown ip Fig. 1-3. This relationship is expressed by the equation A b = blb (1. never unique.11). A dyadic D corresponds to a tensor of order two and may always be represented as a finite sum of dyads D = albi azbz + + .(a b)c = w (1.(bxc) = ( a x b ) * c= a.(bxc)= h (1. defined by writing the vectors in juxtaposition as ab is called a dyad. In symbolic notation.b l + az-bz + +a ~ * b ~ (1. the second is called the consequent.19) (ha)b = a(hb) = hab (1 .12) the antecedents and consequents are interchanged.12). the resulting dyadic is called the conjugate dyadic of D and is written Dc = blal + bzaz + +b ~ a ~ (1 . If in each dyad of (1. 1.4 MATHEMATICAL FOUNDATIONS [CHAP. D. a ~ b ~ + (1.16) (a + b)c = ac + bc (1. ab # ba.vXb~ (1. however.l7) (a + b)(c + d) = ac + ad + bc + bd (1. The first vector in a dyad is known as the antecedent. (h + p)ab = Aab + pab (1.18) and if h and p are any scalars.e. i. the result is a scalar known as the scalar o f t h e dyadic D and is written Ds = a i . the parentheses are unnecessary and may be deleted as shown.15) I t can be shown that Dc. c as coterminous edges. The following identity is frequently useful in expressing the product of a crossed into b x c.6 DYADS AND DYADICS The indeterminate vector product of a and b.12) is replaced by the cross product of the two vectors. a x (b x c) = (a c)b ..l4) If each dyad of D in (1.20) .I 3) If each dyad of D in (1. The vector triple product is a cross product of two vectors. the result is called the vector o f t h e dyadic D and is written Dv = a 1 x b l + a z x b z + +a. The magnitude h of the scalar triple product is equal to the volume of the parallelepiped having a. the product vector w is observed to lie in the plane of b and c. since the cross operation must be carried out first.11) From (1.12) which is. a. The indeterminate product is not in general commutative. b. and Dv are independent of the representation (1. Also.12) is replaced by the dot product of the two vectors. The indeterminate vector product obeys the distributive laws + a(b c) = ab + ac (1. dyadics are denoted by bold-faced sans-serif letters as above. This product is sometimes written [abc] and called the box product. one of which is itself a cross product. one of which is a cross product. 1 The scalar triple product is a dot product of two vectors.10) the dot and cross operation may be interchanged in this product.10) As indicated by (1. E or D.CHAP. Double dot and cross products are also defined for the dyads ab and cd as follows.+ aNb~) (cidi + czdz + .34) From these definitions. êS3 + (1-24) where 21.v = E-v (1. ascalar (1.a1)bl+(v.26) DXv = al(b1 X v) + az(bz X v) + +a ~ ( Xbv)~ = G (1. ê2.28). or idemfactor I. ( b ~ C N ) ~ N ~ N = G (1-29) The dyadics D and E are said to be reciproca1 of each other if E-D = D-E = I (130) Fór reciprocal dyadics. ê3 constitute any orthonormal basis for three-dimensional Euclidean space (see Section 1.23) The unit dyadic. c)ad (1.32) ab cd = (a. Also.38) . either v * D = v..22) it is called the prefactor.v = v .33) ab c cd = (a x c)(bx d) = uw. a vector (1.Dc) = G + H (1. For the arbitrary dyadic D the decomposition is D = ?j(D + Dc) + ?j(D. a dyad (1. if D = D.. the notation E = D-I and D = E-' is often used.+C N ~ N ) = (bi ci)a1dl + (bi cz)ald~+ . is the dyadic which can be represented as + I = êl& i%.d) = A. and anti-self-conjugate. The cross products v x D and D x v are the dyadics defined respectively by + v x D = (v X a1)bl (vX az)bz + + (vX a ~ ) =b ~F (1. c d = (b*c)(a. or symmetric. a scalar (1.21) D is called the postfactor. if Every dyadic may be expressed uniquely as the sum of a symmetric and anti-symmetric dyadic. c)(b x d) = g. a vector (1. ab : cd = (a c)(b d) = A.25) for a11 vectors v.az)bz+ + ( v * ~ N ) ~ N= i.27) The dot product of the dyads ab and cd is the dyad defined by a b *cd = (b. . 11 MATHEMATICAL FOUNDATIONS 5 If v is any vector. the dot product of any two dyadics D and E is the dyadic D E = (albl+ azbz + . or anti-symmetric.i (1.21) In (1. double dot and cross products of dyadics may be readily developed. The dyadic I is characterized by the property 1.35) A dyadic D is said to be self-conjugate.31) ab cd = (a x c)(b d) = h.28) From (1. the dot products v D and D v are the vectors defined respectively by v ' D = (v.I = v (1. . Two dyadics D and E are equal if and only if for every vector v. some authors use the double dot product defined by a b . and in (1.7). D = G* + H*. .(D. G* = G and H* = H. noncoplanar vectors of the system. 1-5.39) and H.k = 1 Such a set of base vectors is often called an orthonormal bash. b. Fig. The reference system of coordinate axes provides units for measuring vector magnitudes and assigns directions in space by which the orientation of vectors may be determined.. k along the coordinate axes as shown in Fig. but in certain situations a particular choice may be advantageous. kxi = j (1. the vector v shown in Fig.43) Base vectors are by hypothesis linearly independent.) = +(D. 1-6 below may be expressed by in which the Cartesian components . j x k = i. the equation + + ha pb vc = O (1.e.42) in turn yields respectively the desired equalities. The well-known rectangular Cartesian coordinate system is often represented by the mutually perpendicular axes.). + (D. BASE VECTORS. p. The choice of coordinate system is arbitrary. For base vectors a.44) is satisfied only if h = p = V = O. + D) = G (symmetric) (1.) = +(D. j. k. 1-5 The most frequent choice of base vectors for the rectangular Cartesian system is the A A A set of unit vectors i. i. 1 for which G. UNIT VECTOR TRIADS A vector may be defined with respect to a particular coordinate system by specifying the components of the vector in that system. nonzero.45) A A A A A A and i. V the vector v is given by V = ha pb vc + + (1. which are called base vectors.7 COORDINATE SYSTEMS. j .D) = -H (anti-symmetric) (1. A set of base vectors in a given coordinate system is said to constitute a bash for that system. A A A In terms of the unit triad i. for which $ $ A A A C A A h i x j = k. Oxyz shown in Fig. = *(Dc .6 MATHEMATICAL FOUNDATIONS [CHAP.41) and (1. = &(D.j = k. Any vector v in this system may be expressed as a linear combination of three arbitrary. j. These base vectors constitute a right-handed unit vector triad. 1.40) Uniqueness is established by assuming a second decomposition.). c and suitably chosen scalar coefficients h. 1-5. Then and the conjugate of this equation is G*-H* = G-H Adding and subtracting (1. The triple scalar product may also be represented in component form by the determinant a. b. .k)*(b.53) Because of the nine terms involved. It is possible to put any dyadic into nonion form.i +a. 11 MATHEMATICAL FOUNDATIONS 7 are the projections of v onto the coordinate axes. a. (1. b y +~ ayb. However.b. the base vectors here do not all have fixed directions and a r e therefore.53) is known a s the nonion form of the dyad ab.j+b~) + + = axbx ayby a.CHAP. (1. [abc] = bx b~ b~ C. functions of position.b. a.O.S$ + a .b.49) Fig.ê+) of base vectors illustrated in the figure a r e associated with these systems.i k A A 9" + ayb.~+a.kk A A (1. j . 60.bykj + a.? b. the cross product a x b is A + a x b = (ayb. . 1-6 For the same two vectors. I n Cartesian component form.b.b. The nonion form of the idemfactor in A A A terms of the unit triad i.7) by ê" = VIU = (COSa)? + (COS P ) + (COS y ) k (1...b.48) Since v is arbitrary. k is given by A A A A A A I = ii+jj+kk (1. .b. Unit triads ( $ R . 1-7 below a r e also widely used. O..i i+ axbyij + a. curvilinear coordinate systems such as the cylindrical (R. i t follows that any unit vector will have the direction cosines of that vector as its Cartesian components.aybx)k A This result is often presented in the determinant form in which the elements a r e treated as ordinary numbers.54) In addition to the rectangular Cartesian coordinate system already discussed.kT + a.$k ~ + a.?+a. Cy c.) and (ê.i+b. systems C$) shown in Fig.by)i (a.a. a. ê.k)(b. In Cartesian component form the dot product of a and b is given by A A A a*b = (a..) j A + (a. The unit vector in the direction of v is given according to (1.k) + + A A = a. ês. the dyad ab is given by c + ab = (a. in general.b. z ) and spherical ( r .a. 1 (a) Cylindrical (b) Spherical Fig. 1-7 1. either subscripts or superscripts. In (1.i b. if f is linear. ' i j k .8 MATHEMATICAL FOUNDATIONS [CHAP. are appended to the generic o r kernel letter representing the tensor quantity of interest. bj. This functional relationship is expressed by the equation a = f(b) (1. R'>q .59) A A . f ( j ) = v . so that now which is recognized a s a dyadic-vector dot product and may be written a = D-b (1. may be written a = b. b. letter indices. DYADICS AS LINEAR VECTOR OPERATORS A vector a is said to be a function of a second vector b if a is determined whenever b is given. and indeed the tensor itself.f(k) (1. In this notation.9 INDICIAL NOTATION. and f o r any scalar A.k) + + (1.58) which.) + b.55) The function f is said to be linear when the conditions f(b + c) = f(b) + f(c) (1. f ( k ) = w. 1. This demonstrates that any linear vector function f may be expressed a s a dyadic-vector product.? I n (1.61) A + + A A where D = u i v j wk. may be represented clearly and concisely by the use of the indicial notation.56) are satisfied for a11 vectors b and c.8 LINEAR VECTOR FUNCTIONS. Typical examples illustrating use of indices are the tensor symbols ai.. equation (1.59) let f ( i ) = u . Tij. Writing b in Cartesian component form.61) the dyadic D serves a s a linear vector operator which operates on the argument vector b to produce the image vector a.55) becomes c a = f(b.f(?) + b. RANGE AND SUMMATION CONVENTIONS The components of a tensor of any order.f(. pij. which represents a vector in physical three-space.62) . ai is sometimes interpreted to represent the ith component of the vector or indeed to represent the vector itself. or tensor of zero order. B?. the symbol Aij represents nine components (of the second-order tensor (dyadic) A). When an index occurs unrepeated in a term. in one or the other of the two forms. that index is understood to take on the values 1. Aijip. The tensor Aij is often presented explicitly by giving the nine components in a square array enclosed by large parentheses as Ai1 A12 A13 Aij = ( 1. In general. Also. represents a scalar. . Also.N where N is a specified integer that determines the range of the index.CHAP. correctly written tensor equations have the same letters as free indices in every term. are also recognized as first-order tensor quantities: aijbj. Unrepeated indices are known as free indices. Second-order tensor quantities may also appear in various f o m s as. Tensors of first order are denoted by kernel letters bearing one free index. Accordingly the symbol aiis understood to represent the three components a1. a symbol such as X which has no indices attached. that index is understood to take on a11 the values of its range. 11 MATHEMATICAL FOUNDATIONS 9 In the "mixed" form. Under the rules of indicial notation.3.. the summation convention must be suspended. since their replacement by any other letter not appearing as a free index does not change the meaning of the term in which they occur. a letter index may occur either once or twice in a given term. no index occurs more than twice in a properly written term. Also. and the resulting terms summed. Thus the arbitrary vector a is represented by a symbol having a single subscript or superscript.. Fikk.az. the dot shows that j is the second index. . third-order tensors are expressed by symbols with three free indices. and so any vector in this space is completely specified by its three components. is 1.2. for example. The following terms. having only one free index.e. The tensorial rank of a given term is equal to the number of free indices appearing in that term. i. If it is absolutely necessary to use some index more than twice to satisfactorily express a certain quantity. When an index appears twice in a term.2.a3. Thus the arbitrary dyadic D will appear in one of the three possible f o m s In the "mixed" form. 8ifikvk By a logical continuation of the above scheme. rijkUjZ)k Second-order tensors are denoted by symbols having two free indices. noncoplanar vectors. The number and location of the free indices reveal directly the exact tensorial character of the quantity expressed in the indicial notation. where both subscripts and superscripts appear.jk. the dot shows that j is the second index. For a range of three on both indices. In ordinary physical space a basis is composed of three. Therefore the range on the index of ai. In this so-called summation convention. repeated indices are often referred to as dummy indices. 1-8 .10 MATHEMATICAL FOUNDATIONS [CHAP.10 SUMMATION CONVENTION USED WITH SYMBOLIC NOTATION The summation convention is very often employed in connection with the representation of vectors and tensors by indexed base vectors written in the symbolic notation.66) would represent nine equations. the four equations Ali = BiiCiiDii + BliC12D12 + B12CilD21 + B12Ci2D22 Ai2 = BiiC2iDii + BiiC22Di2 + BIZCZID~I + Bi2CmD22 (1. Thus if the rectangular Cartesian axes and unit base vectors of Fig. tensor character is not given by the free indices rule as it is in true indicial notation. In such a "combination" style of notation. Fig. The notation here is essentially symbolic. 1-8. 1 I n the same way. The usefulness of the indicial notation in presenting systems of equations in compact form is illustrated by the following two typical examples. each having nine terms on the right-hand side. the indicial equation . for a range of N. Applying the summation convention to (1. v3 are the rectangular Cartesian components of v.67) A21 + B ~ I C I ~+DB22CiiD21 = BSICIIDI~ I~ + Bz2C12D22 F o r a range of three on both i and j. as) or ai = (1. the equation may be written in the abbreviated form A V = Viei (1. in expanded form. Aij = BipCjqDpq (1. v2. a n nth order tensor will have Nn components.64) represents in expanded form the three equations F o r a range of two on i and j.68) in which vi. F o r a range of three on both i and j the indicial equation Xi = C i j 2 j (1.69) where i is a summed index.66) represents. but with the added feature of the summation convention. the arbitrary vector v may be written V = viêi + v2ê2 + v3ê3 (1. az. (1. the components of a first-order tensor (vector) in three-space may be displayed explicitly by a row or column arrangement of the form ai = (ai. 1-5 are relabeled a s shown by Fig. 1.63) a3 I n general.68). the nonion form of the arbitrary dyadic D may be expressed in compact notation by 1.72) possesses a unique inverse set of the form The coordinate systems represented by xi and Bi in (1.B3) in the Oi system.70) I t is essential that the sequence of the base vectors be preserved in this expression. x2.O3 in the same space. continuous. 02. 02.72) and (1. I] MATHEMATICAL FOUNDATIONS 11 Second-order tensors may also be represented by summation on indexed base vectors. bj are the components of the tensor in the xj coordinate system. Accordingly the dyad ab given in nonion form by (1. x2.x3 in a three-dimensional Euclidean space. In general . In (1. (1. while bti are the components in the Oi system. In similar fashion. and let Bi represent any other coordinate system B1. If the Jacobian does not vanish.x3) in the xi system a new set of coordinates (O1. the differential vector dOi is given by This equation is a prototype of the equation which defines the class of tensors known as contravariant vectors.75) are completely general and may be any curvilinear or Cartesian systems. The determinant or.11 COORDINATE TRANSFORMATIONS. The functions Bi relating the two sets of variables (coordinates) are assumed to be single-valued. The letter superscripts are indices a s already noted. according to the equation where the partia1 derivatives are evaluated at P. in compact form. Here the numerical superscripts are labels and not exponents. From (l.72).77). Powers of x may be expressed by use of parentheses as in ( x ) ~or ( x ) ~ . I n general. differentiable functions.53) may be written ab = (aiêi)(b$j) = aibjêiêj (1. a set of quantities bi associated with a point P are said to be the components of a c o n t r a v a ~ a n ttensor of order one if they transform. The coordinate transformation equations assign to any point (xl.CHAP. GENERAL TENSORS Let xi represent the arbitrary system of coordinates xl. is called the Jacobian of the transformation. under a coordinate transformation. In general tensor theory.02. cylindrical or spherical coordinates) in the same space. I n general. x2.g. and let 8' represent any system of rectangular or curvilinear coordinates (e.12 THE METRIC TENSOR.e3) relating the systems. b: are the covariant components in the ei system. I t is for this reason that the coordinates are labeled xi here rather than Xi. By a logical extension of the tensor concept expressed in (l. a set of quantities bi are said to be the components of a covariant . The vector x having Cartesian components xi is called the position vector of the arbitrary point P(xl. but it must be noted that it is only the differentials dxi.77). bi the components in the xi system. x2. the distance differential is . 1. Second-order covariant tensors obey the transformation law Covariant tensors of higher order and mixed tensors. CARTESIAN TENSORS Let xi represent a system of rectangular Cartesian coordinates in a Euclidean three- space. covariant tensors are recognized by the use of subscripts as indices. 1 tensor theory. The prototype of the covariant vector is the partia1 derivative of a scalar function of the coordinates. contravariant tensors are recognized by the use of superscripts as indices. fourth and higher orders are defined in a similar manner.x3) is such a function. and not the coordinates themselves. The square of the differential element of distance between neighboring points P(x) and Q(x +dx) is given by From the coordinate transformation xi = xi(ol.temor of order one if they transform according to the equation In (1.12 MATHEMATICAL FOUNDATIONS [CHAP. Thus if + = +(xl. which have tensor character. The word contravariant is used above to distinguish that class of tensors from the class known as covariant tensors.x3) referred to the rectangular Cartesian axes. such as are defined in the obvious way.80). the definition of contravariant tensors of order two requires the tensor components to obey the transformation law Contravariant tensors of third. the partia1 derivatives appearing in general tensor definitions. aij = cos (xl. this is the case for transformation laws between systems of rectangular Cartesian coordinates with a common origin. the tensors so defined are called Cartesian tensors. in the transformation laws defining Cartesian tensors. or by a reflection of axes in one of the coordinate planes.CHAP. For Cartesian tensors there is no distinction between contravariant and covariant components apd therefore it is customary to use subscripts exclusively in expressions representing Cartesian tensors. = 1 if p = q .. then where S. 1-9 or alternatively by the transformation tensor . is the Kronecker delta (see Section 1. i. such as (1. TRANSFORMATION LAWS FOR CARTESIAN TENSORS.83) is called a system of homogeneous coordinates. xj).80) and (1. = O if p # q and S. If Oi represents a rectangular Cartesian system. Coordinate transformations between systems of homogeneous coordinates are orthogonal transformations..xtx$ represent two rectangular Cartesian coordinate systems with a common origin a t an arbitrary point O as shown in Fig.13. 1-9.83) becomes where the second-order tensor g. ORTHOGONALITY CONDITIONS Let the axes Oxixzxs and Ox. If the symbol aij denotes the cosine of the angle between the ith primed and jth unprimed coordinate axes. are replaced by constants.81). 11 MATHEMATICAL FOUNDATIONS 13 and therefore (1. Any system of coordinates for which the squared differential element of distance takes the form of (1. the relative orientation of the individual axes of each system with respect to the other is conveniently given by the table Fig. The primed system may be imagined to be obtained from the unprimed by a rotation of the axes about the origin. As will be shown next. = (dxi/dOp)(dxi/d$q) is called the metric tensor. say the x" system. In particular.13) defined by S. THE KRONECKER DELTA. and when attention is restricted to such transformations.. 1. or by a combination of these.e. or fundamental tensor of the space. in the unprimed system by the equation v = .95) may be written In expanded form. whose value depends upon the subscripts j and k . is said to be an orthogonal transformation. 1-9 may be expresse(.14 MATHEMATICAL FOUNDATIONS [CHAP.91) by its equivalent form (1. 1 From this definition of aij. expressed by (1.Vi I (1. (1.90) reveals that the vector components in the primed and unprimed systems are related by the equations v 3.ê. however. - .95) must reduce to the identity vj = vj.98). whose coefficients satisfy (1.94) may also be combined to produce vi = &ijakjvk from which the orthogonality conditions appear in the alternative form aijakj = s i k (1. tjazkVk .93) or (1.. By an appropriate choice of dummy indices. the unit vector along the x: axis is given according to (1. Therefore the coefficient aijaik.80) and (1.93) is the transformation law for first-order Cartesian tensors. (1.94) may be combined to produce the equation v j = a. (1. defined b'r may be used to represent quantities such as aijaik. the arbitrary vector v shown in Fig.97) consists of nine equations which are known as the orthogonality or orthonormality conditions on the direction cosines aij.a.93) and (1.93) The expression (1.98) A linear transformation such as (1.97) or (1. (1. must equal 1 or O according to whether the numerical values of j and Ic are the same or different. Finally.77). (1.91) Replacing in (1.94). Thus with the help of the Kronecker delta the conditions on the coefficient in (1.95) Since the vector v is arbitrary. %. and as such is seen to be a special case of the general form of first-order tensor transformations.90) and in the primed system by V = vi ei (1.94).89) yields the result Comparing (1.93) and (1.. . By interchanging the roles of the primed and unprimed base vectors in the above development. In (1.. the free index appears as the first index.92) with (1.48) and the summation convention by An obvious generalization of this equation gives the arbitrary unit base vector as In component form. The Kronecker delta.93) the free index on aij appears as the second index. 2). Coordinate axes rotations and reflections of the axes in a coordinate plane both Iead to orthogonal transformations. (1. the inverse of (1.93) is found to be I t is important to note that in (1. 8ijbj = 8iibi + 8i2b2 f 8i3b3 = bi and.CHAP. For the scalar multiplier A. MULTIPLICATION BY A SCALAR Cartesian tensors of the same order may be added (or subtracted) component by component in accordance with the rule The sum is a tensor of the same order as those added. which is given by (1.15 TENSOR MULTIPLICATION The outer product of two tensors of arbitrary order is the tensor whose components are formed by multiplying each component of one of the tensors by every component of the other. .14 ADDITION OF CARTESIAN TENSORS. likewise. . the dyad UiVj has components in the primed coordinate system given by U I V ~ = ( U i p ~ ) ( ~ j q=~ qUipUjqUpVq ) (I. (IJ0-4) 1. since. Note that like indices appear in the same sequence in each term. This process produces a tensor having an order which is the sum of the orders of the factor tensors. = aipajqalcm. any second-order Cartesian tensor Tij obeys the transf ormation Iaw TC = aipajqTpq With the help of the orthogonality conditions i t is a simple calculation to invert (1. .106) 1. outer products are formed by simply setting down the factor tensors in juxtaposition.) .. typical examples written in both indicial and symbolic notation are bi = xai or b = Xa (1. Typical examples of outer products are (a) aibj = Tij (C) DijTkrn = Qiikrn (b) viFjk = Nijk (d) €ijkVm = @ijkm As indicated by the above examples.102).101). for example. According to the transformation law (1. 8ijFik = 8ijFik +8 2 j F 2 k $. 11 MATHEMATICAL FOUNDATIONS 15 The Kronecker delta is sometimes called the substitution operator. 83jF3k = F j k It is clear from this property that the Kronecker delta is the indicial counterpart to the symbolic idemfactor I. Multiplication of every component of a tensor by a given scalar produces a new tensor of the same order. (Note that a dyad is formed from two vectors by this very procedure.I 01) In an obvious generalization of (1..94).. Tpqm. thereby giving the transformation rule from primed components to unprimed components: The transformation laws for first and second-order Cartesian tensors generalize for an Nth order Cartesian tensor to T~&.54). 1 Contraction of a tensor with respect to two free indices is the operation of assigning to both indices the same letter subscript.. va3. 3 (i. PERMUTATION SYMBOL. contracted to Eij Ejm Emq. aiEjk aiEik = fk a. j . Eij Ekm Eij Ejm = Bim E . in both the indicial and symbolic notations.e.e. This useful tensor is defined by 1 if the values of i . involving one index from each tensor. the third-order tensor eij. or ( E ) 3 1. Eij Fkm contracted to Eij Fij. 2 .E = f aiEji = hj E-a = h 3. if I-1 they appear in sequence as in the arrangement 1 2 3 1 2 ) . k are not a permutation of 1 . Contraction produces a tensor having an order two less than the original. Eij Fkm EijFjm = Gim E. or E : F 2.. k are an even permutation of 1 .e. they appear in sequence as in the arrangement 3 2 13 2). Two such examples are 1. 2 . E = (E)2 Multiple contractions of fourth-order and higher tensors are sometimes useful.. = bi Eijai = cj Eiiak = dk (C) Contractions of EijFkm EijFim = Gim EijFkk = Pij EijFki = Hik EijFjm = Qim EiiFkm = Kkm EijFkj = Rik An Znner product of two tensors is the result of a contraction. ( a ) Contractions of Tij and uivj + + Tii = Til TZZ T33 UiVi = UiVi+ + U2V2 U3V3 ( b ) Contractions of Eijak E. if two or more of the indices have the same value). DUAL VECTORS In order to express the cross product a x b in the indicial notation. known a s the permutation symbol or alternating tensor. if 'ijk . Typical examples of contraction are the following. 3 (i. EijEkmE.16 MATHEMATICAL FOUNDATIONS [CHAP. j.16 VECTOR CROSS PRODUCT. must be introduced. I O if the values of i . Outer Product Inner Product Indicial Notation Symbolic Notation 1. if the values of i . performed on the outer product of the two tensors. 3 (i. Severa1 inner products important to continuum mechanics are listed here for reference.F = G 4. j. 2 . . thereby changing these indices to dummy indices. aibj aibi a*b 2. k are an odd permutation of 1 . the cross product a x b = c is written in indicial notation by Using this relationship. the M x N matrix 4 . is called the transpose matrix of 4. the first subscript denotes the row. used to represent the typical element of a matrix. The N X M matrix CAT. An M x 1matrix. A square matrix with zeros everywhere except on the main diagonal (from A11 to ANN)is called a diagonal matrix. A matrix having only zeros as elements is called the zero matrix. Such tensors are called pseudo-tensors.c <.17 MATRICES. The product of two matrices. cA%. . I t is worthwhile to note that cijk obeys the tensor transformation law for third order Cartesian tensors as long as the transformation is a proper one (det aij = 1) such as arises from a rotation of axes. the "vector of the dyadic Tu. is defined only if the matrices are conformable.e. Multiplication of the matrix [Aij] by a scalar h results in the matrix [AAij]. by the kernel letter of the matrix in script. a minus sign must be inserted into the transformation law for cijk. If the nonzero elements of a diagonal matrix are a11 unity. c = h may be written Since the same box product is given in the form of a determinant by (1. it is not surprising that the permutation symbol is frequently used to express the value of a 3 X 3 determinant.15). MATRIX REPRESENTATION OF CARTESIAN TENSORS A rectangular array of elements.CHAP. The product of an M X P matrix multiplied into a P X N matrix is an M X N matrix. v. 11 MATHEMATICAL FOUNDATIONS 17 From this definition. the box product a x b . i.52). For example. enclosed by square brackets and subject to certain laws of combination. The matrix itself is designated by enclosing the typical element symbol in square brackets. k ik (1. is called a row matrix. written [alk]. A 1x N matrix. ~ T.ix. formed by interchanging rows and columns of the M x N matrix 4 . An M x N matrix is one having M (horizontal) rows and N (vertical) columns of elements. is called a matrix. is called a square matrix.g. In the symbol Aij. Matrices having the same number of rows and columns may be added (or subtracted) element by element. e.or [Aij] is the array given by A matrix for which M = N. as defined by (1. written [akl].110) which is observed to be the indicial equivalent of T. The dual vector of an arbitrary second-order Cartesian tensor Tij is defined by 2 .. the second subscript the column occupied by the element. If the transformation is improper (det aij = -I). 1. Matrix multiplication is usually denoted by simply setting down the matrix symbols in juxtaposition as in . . if the prefactor matrix cA has the same number of columns as the postfactor matrix 23 has rows. is called a column matrix. the matrix is called the unit or identity mat. a reflection in one of the coordinate planes whereby a right-handed coordinate system is transformed into a left-handed one. or alternatively. the Kronecker delta. if O4 is orthogonal.116) I t is clear.ith row and kth column of the product matrix. Accordingly. the determinant of the square array remaining after the row and column of Aij are deleted. As suggested by the fact that any dyadic may be expressed in the nonion form (1.is defined by * = Aij (-l)i+jMij (1. that 4 is the matrix representation of 6. (a) Vector dot product a b = b *a = A [aij][bjl] = [A] . Although every Cartesian tensor of order two or less (dyadics. 1 Matrix multiplication is not. commutative: cA% z LBd .18 MATHEMATICAL FOUNDATIONS [CHAP. Severa1 such products occur repeatedly in continuum mechanics and are recorded here in the various notations for reference and comparison. and these products summed to give the element in the . A first-order tensor (vector) may be represented by either a 1 x 3 row matrix.114) it may be shown that where 4 is the identity matrix. since the components of a second-order tensor may be displayed in the square array (1. If both matrices in the product 4 % = C are 3 x 3 matrices representing second-order tensors. the unit dyadic. Expansion of (1. vectors. The adjoint matrix of O4 is obtained by replacing each element by its cofactor and then interchanging rows and columns. Any matrix cA for which the condition cAT = cA-I is satisfied is called an orthogonal matrix. the multiplication is equivalent to the inner product expressed in indicial notation by where the range is three. A square matrix cA whose determinant \Aij\is zero is called a singular matriz. or by a 3 x 1 column matrix. The cofactor of the element Aij of the square matrix 4 . and so named because of the property 4 c A = CAJ = (1. Thus From the inverse matrix definition (1.113) in which Mij is the minor of Aij.53). If a square matrix e4 = [Aij] is non-singular. and of I.e..I18) duplicates the "row by column" multiplication of matrices wherein the elements of the ith row of the prefactor matrix are multiplied in turn by the elements of the kth column of the postfactor matrix. and.62). denoted here by AT~. 3 x 3 matrices. i t possesses a unique inverse matrix 4-' which is defined as the adjoint matrix of O4 divided by the determinant of 04. i. of course. equivalently.. in general. having ones on the principal diagonal and zeros elsewhere. scalars) may be represented by a matrix. not every matrix represents a tensor. it proves extremely useful to represent second-order tensors (dyadics) by square. 18 SYMMETRY OF DYADICS. - .D.. a3 . a square matrix cA is symmetric if it is equal to its transpose CAT.38) is or.. or skew-symmetric. Since the interchange of indices of a second-order tensor is equivalent to the interchange of rows and columns in its matrix representation.CHAP. Consequently a symmetric 3 x 3 matrix has only six independent components as illustrated by Ali Ai2 (1. tja3. Ci [Eij][ajl] [cii] 1.36) (or (1. 3% (1. Similarly the second-order tensor Dij is symmetm'c if D. MATRICES AND TENSORS According to (1.. 11 MATHEMATICAL FOUNDATIONS 19 ( b ) Vector-dyadic dot product a*E = b a& = CB (c) Dyadic-vector dot product E*a = c &a = c E. Consequently a 3 x 3 anti-symmetric matrix CB has zeros on the main diagonal. in an equivalent abbreviated form often employed. if Therefore the decomposition of Dij analogous to (1.37)). and therefore only three independent components as illustrated by . and square brackets on the indices denote the anti-symmetric part.126) A23 A33 An anti-symmetric matrix is one that equals the negative of its transpose. where parentheses around the indices denote the symmetric part of Dii. a dyadic D is said to be symmetric (anti-symmetric) if it is equal to (the negative of) its conjugate D.. .I22) and is anti-symmetric. IIIT = O (I . the system to be solved is Note first that for every A. A3 . that is. the trivial solution ni = O satisfies the equations. defined a t some point in space. namely.19 PRINCIPAL VALUES AND PRINCIPAL DIRECTIONS OF SYMMETRIC SECOND-ORDER TENSORS In the following analysis. . With the help of the identity ni = Sijnj. ]Tij. an arbitrary tensor is said to be symmetric with respect to a pair of indices if the value of the typical component is unchanged by interchanging these two indices. ni and h.ITX' + IITA.129) can be put in the form which represents a system of three equations for the four unknowns. Also. In general. Examples of symmetry properties in higher-order tensors are (a) Rijkm = Rikjm (symmetric in k and j ) ( b ) cijk = -c kii (anti-symmetric in k and i) (c) Gijkm = Gjimk (symmetric in i and j. For (1. In expanded form. The purpose here.132) Expansion of this determinant leads to a cubic polynomial in h. from the homogeneity of the system (1.133) which is known as the characteristic equation of Tij. equivalently. For this case. there is associated with each direction (specified by the unit normal ni) a t that point. This simplifies the mathematics somewhat.131) to have a non-trivial solution. (1. the inner product may be expressed as a scalar multiple of ni.ASijl = O (1. or principal axis of Tij. only symmetric tensors with real components are considered. however. and for which the scalar coefficients. and this condition is imposed from now on.20 MATHEMATICAL FOUNDATIONS [CHAP. For every symmetric tensor Tij.131) it follows that no loss of generality is incurred by restricting attention to solutions for which nini = 1. a vector giveii by the inner p r oduct Here Tij may be envisioned as a linear vector operator which produces the vector vi conjugate to the direction ni. associated with each principal direction. (1. k and m) (d) Pijk = Pikj = P k j i = pjik (syrnmetric in a11 indices) 1. is to obtain non-trivial solutions. A tensor is anti- symmetric in a pair of indices if an interchange of these indices leads to a change of sign without a change of absolute value in the component. and the direction ni is called a principal direction. and since the important tensors of continuum mechanics are usually symmetric there is little sacrifice in this restriction. the determinant of coefficients must be zero.130) or. 1 Symmetry properties may be extended to tensors of higher order than two. If the direction is one for which vi is parallel to ni. any direction is a principal direction.138). > ~ ( I I I . For principal axes labeled OxTx2x8..h ( ~A(.. . are called the first. and if these values are distinct.134) IIIT = ITijl = det Tij (1. aZ1= n(2) aZ2= n(2) (L*.. ) .h(II).4 is the identity rnatrix. HAMILTON-CAYLEY EQUATION By direct matrix multiplication. Since each of the principal values satisfies (1. both the tensor array and its matrix appear in diagonal form. the cube as TikTkrnTmj. = n 3 ) E. the tensor has a diagonal form which is independent of the choice of X ( I ) and htz) axes.) and to display the ordering as in Ao. etc. the principal values are real. > A. labeled X(I)..135) and (1. Thus in which . The three roots of the cubic (1. . the three principal directions are mutually orthogonal. Therefore with Tii written in the diagonal form (1. CHAP.h(.20 POWERS OF SECOND-ORDER TENSORS. I] MATHEMATICAL FOUNDATIONS 21 IT = Tii = t r Tij (trace of Tij) (1.).136) . the tensor itself will satisfy (1. 1. it is customary to write thern as h ( . and because of the diagonal matrix form of T ngiven by (1. of Tij. second and third invariants.133).139) by I produces the equation. in which nlj) are the direction cosines of the jth principal direction. the square of the tensor Tij is given as the inner product TikTkj. = ni1) 2. Matrix multiplication of each term in (1. When referred to principal axes. respectively. the nth power of the tensor is given by A comparison of (1.. Thus If X(I) = h(z). once the principal axis associated with XG) has been established.137). are called the principal values of Tij.133). If a11 principal values are equal. This equation is called the Hamilton-Cayley equation.133). .. For a symmetric tensor with real components. = n(32) a31 = ni3) = a.137) indicates that Tij and a11 its integer powers have the same principal axes. the transformation from Oxixzx3 axes is given by the elements of the table $1 x2 x3 ~1 all = n(l) a12= n(1) 2 al. If the principal values are ordered. t) or $J = +(x.148) curl v = V x v Or ~ijkajvk= 'ijkvk.v or divt = visi (1. t) or T = T(x.142) tensor fields of various orders are represented in indicial and symbolic notation a s follows. t) or V = V(X.143) (b) vector field: vi = vi(x. DERIVATIVES OF TENSORS A tensor field assigns a tensor T(x. for which the position vector of an arbitrary point is A x = xiei (1. T and 8. If the components are functions of x only. In symbolic notation. or briefly in indicial form by a. t) to every pair (x.. (a) scalar field: + = +(xi.j ax ( f ) a s r d z . 1 Combining (1. t) are continuous (or differentiable) functions of x and t. the tensor field is said to be steady.i (1.. t) (1. 1. Severa1 important differential operators appear often in continuum mechanics and are given here f o r reference. The tensor field is said to be continuous (or differentiable) if the components of T(x. t) (1.I49) . avi aTij (b) - axi = vi. partia1 differentiation with respect to the variable xi is represented by the comma-subscript convention as illustrated by the following examples.t) (1. divv = V .145) Coordinate differentiation of tensor components with respect to xi is expressed by the differential operator aldxi. t) where the position vector x varies over a particular region of space and t varies over a particular interval of time.i (e) ã2* = Ti!. indicating a n operator of tensor rank one.140) and (1.144) (c) second-order tensor field: Tij = Tij(x.= vi.21 TENSOR FIELDS.139) by direct substitution. Continuation of this procedure yields the positive powers of 7 as linear combinations of T2.22 MATHEMATICAL FOUNDATIONS [CHAP. the corresponding symbol is the well-known differential vector operator V. and a tensor of order one lower if i becomes a dummy index ((b) above) in the derivative.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). With respect to a rectangular Cartesian coordinate system. k avi a2Tij - (c).I46) Frequently. pronounced de1 and written explicitly V = êi a aXi = eidi A (1. F = F(x). where 2 is the outward unit normal to the bounding surface S .CHAP.I 52) Fig.integral of F aIong C. p dV = Tijknp dS (1. In the indiciaI notation.151) becomes (z~)s S. and dS is the differential element of surface as shown by the figure.153) is written 1. in which 6 is the unit normal on the positive side of S . ~ X (1. (1.155) is written i vi.23 THE DIVERGENCE THEOREM OF GAUSS The divergence theorem of Gauss relates a volume integral to a surface integral. Explicitly. 11 MATHEMATICAL FOUNDATIONS 23 1. . If the differential tangent vector to the curve a t the arbitrary point P is dx.22 LINE INTEGRALS..156) The divergence theorem of Gauss as expressed by (1.157) . (1. may be expressed in terms of an integral over any two-sided surface S which has C as its boundary. the theorem is written Tijk .151) taken along the curve from A to B is known as the line . 1-10 Fig. . In the indicial notation. STOKESy THEOREM In a given region of space the vector function of position. 1-10.i dV = L vini dS (1.. 1-11 Stokes' theorem says that the line integral of F taken around a closed'reducible curve C. In its traditional form the theorem says that for the vector field v = v(x). is defined a t every point of the piecewise smooth curve C shown in Fig.156) may be generalized to incor- porate a tensor field of any order. the integral XF-~X = L ~ F . 1-11. . (1. as pictured in Fig.Fidxi J (zi)* Fidxi (1. I n the indicial notation. of the volume V in which the vector field is defined. Thus for the arbitrary tensor field Tijk. 3. Expanding the indicated dot product.4. Consider the product a X [ ( bX c ) X r]. A 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 = O Adding d2/4 = (a2 + b2 + c2)/4 to each side of this equation gives the desired equation ( x .) k ] [a? + b3 cc] + U 1 + = .6g. 1-15 1. setting b X c = v . l ) .24 MATHEMATICAL FOUNDATIONS [CHAP.( a .( c r)b] = .axl . Prove that [a b X c]r = ( a -r)b x c + (b r)c x a + (c r)a x b.y l ) j ( 2 . zl) and Q(x2.1-1. A A u u -[(xz-xI)~+ (v2-y1)j A +( 2 . Also.2. yl.(617)k A ( b ) The vector extending from P to Q is A u = (O-I)?+ (2-O).(c r)a X b . 1.2. z) and d = a? b j c k is a constant vector.y2.2.c/2)2 = (d/2)2 which is the equation of the sphere centered a t d/2 with radius d/2. .(213)k directed from P to Q or 6 = (1/3)? .d) r = O is the vector equation of a sphere.a/2)2 + ( y .0. + ( a ) Ivl = v = 1 / ( 2 ) ~ (3)2 (-6)2 = 7 + A v = VIV = (217)i A + (317)j A .x l ) i A + A + A (v2.(213)3 + ( 2 / 3 ) k directed from Q to P 1. + If r = x? + y 3 z c is the vector extending from the origin to the arbitrary point + + P ( x .( b r)c X a . -X = O U Z Since u is any vector in the plane.2 . + + Let P ( x l .8) Determine in rectangular Cartesian form the unit vector which is (a) parallel to the vector + v = 2? 3 3 . Prove that the vector v = a? b + i+ c k is normal to the plane whose equation is a x by cz = A. a x [ ( b x c ) x r ] = a x ( v x r ) = ( a W r ) b x c. a X [ ( bX c ) X r] = a X [ ( b r)c . .) k.1 .cz.byl . 1 Solved Problems ALGEBRA OF VECTORS AND DYADICS (Sec. 1. ( b ) along the line joining points P(1.+ (1-3)k A A A = -i+2j -2k u = d(-1)2 + (2)2+ (-2)2 = 3 Fig. z2) be any two points in Iz + + the plane. show that (r .b/2)2 + ( z . 1-12 Thus A u = -(1/3) i A + (213). The projection of v in the direction of u is u *-v . . Then a s l byl czl = X and ax2 by2 cz2 = + + and the vector joining these points is u = (x2.) = X. By direct expansion of the cross product in brackets.( a s 2 by2+ cz2.3) and Q(O. v is I to the plane. Fig. b x c ) r . y. ( b ) Likewise setting bi = bjiêj i t follows that D = aibjiêj = (bjiai)êj= g j ê j where j = 1 . D. b. p and v provided the determinant of coefficients vanishes. and indeed v = u + w.3. . ê2.j -k A A A w = i -2j +k The vectors a. b. D.CHAP.ê. D e ~ e ~ .v = v . 2 .+ aNbN. b N ) a N = v . 3 .) 3 1 1 with j = 1.ê.D. . .2.ê$. which is equivalent to a b X c = O.D = D. = O This set has a nonzero solution for h.8. not a11 zero. Let D = albl + a2b2+ ..b. For the proposed basis u.71). Therefore D. ~ = D~... Show that if the vectors a. v.ê3 as (a) antecedents.5. . Prove that ( D .D jt I 1 êê A A A A p q p q = DjiDpq(ej'ep)eieq and (o. .N ) .b.(a. w are linearly dependent. ( a ) In terms of base vectors.6. 3 t t = $. = O ha. a y by cY = O a.ei A A A A A A = Dpqeq( e .~ A A .b2)a2+ . show that D .( c r ) a X b = ( a r ) b X c . (b) consequents.. Check the linear dependence or independence of the basis A A A u = 3i+j-2k A A A v = 4 i .. . Let D = albl + a2b2+ . b and c are linearly dependent if there exist constants h. v) + a.) e. a b x c = O . . + a N ( b N v ) = (V. e j ) e i D j i = D. = D j i ê i ê j . = D. Hence the vectors u.. ~ ê. such that ha pb + + vc = O.. + vc.. 11 MATHEMATICAL FOUNDATIONS 25 Thus -(b r)c x a .2. w. 1. ai = aliel + azie2 + a3ie3 = a j i e j A A A A 3 1 3 = and so D = a-2. c. . b and c are linearly dependent. For the arbitrary dyadic D and vector v. . c. b. The component scalar equations of this vector equation are ha.)a. a. + pb. + ( v .ü = D i j ê i ê j and D .7. A A = DjiDpq( e j e.r)c X a + ( c * r ) aX b This identity is useful in specifying the displacement of a rigid body in terms of three arbitrary points of the body.. + ( v . 1. v. From (1. Show that any dyadic of N terms may be reduced to a dyadic of three terms in a form having the base vectors êl. + pbY + VC.c. o). 1.+ aNbN= aibi (i = 1. D ) . 1. Then D v = al(b.(b2 v ) + .( a b X c)r and so ( a * b x c)r = ( a * r ) bx c + (b. p and v.. i + D Y . 3 k + 5 k k ) .(b .) ê.6 i j . i + D..10. j = D..$). show by direct calculation + AA AA A A A 1. A A show that r D r = 1 represents the ellipsoid ax2 by2 cx2 = 1.ê2)c2... ( u x v ) = ( D x u ) . 2b A = [(b $.14. i = D.elel .. From the definition ab : cd = (a c)(b d) i t is seen that D : F = 12 + 5 = 17. H = (4ik+6jj-3kj+kk)..j k + s k k and F = 4 i k . that D . from ab cd = (b e)(a d) i t follows that D F = 12 3 5 = 20. D ~ =U ( 6 ? ? + 3 ? . = (bl X v)al + (b. show that 2 b . Also. . (b c3)$31 = [ ( ê . etc.15. From the definition ab cd = (b c)ad. A A A A A A A A = d l i + d 2 j +d. i = Dz.: + :6: -3k3 +kk) AA AA A A A A AA A A = 12ik+12j j + 3 j j ..13. 5 j = -30i+40k .. 1.? + D. ~ +D. ~ .t . . j * d 2 = 3 . A A A A A Since u X v = ( 2 i + k ) X 5 j = 1 0 k .)al .d l = i * ( ~ . x v)a2 + .. etc.5 i . For an antisymmetric dyadic A and the arbitrary vector b. = -v X Dc. + 3 ij 4 k k and u = 2 i + k .. compute and compare the double dot products D :F and D F..(b c.c2e2 + esc3 .j + D.. + Dryk)j + (Dxzi + D. + + 1. + 4 k k ) ~ ( 2 i + k ) = -6ik+8kj-6ij+3ii AA AA AA AA A A A snd (DXu)*v = ( 3 i i . and because it is antisymmetric.A.c3e3) A A A A A A 2A = = (êlcl .(v X bN)aN = -v X D. jk -15kj+5kk AA AA AA AA AA AA AA Similarly. r= (s?+ y3+z k ) (a??+ b33+ ckk) (s?+ zk) A A A A A A = ( z i + y j + z k ) * ( a x i + b y j + c z k ) = as2+by2+cz2 = 1 A A AA AA AA A A 1. i . Determine the dyadics G = D .. A A 6 j * d l = . From Problem 1. .i + D. 2A = (A .2 ) k A A A Also now A A e A i .12. A = A. + [(b. x b. A A + If D = a ii b j j + c k k and r is the position vector r = x i AA AA A + y j + xk.6(a).] .k) i + (D. x c 1 ) x b + ( ê z X c 2 ) X b+ ( ê 3 X ~ 3 ) X b ]= (AvXb) If D = 6.elel A + e2c2. Show that (D x v).6 i k + 8 k j ) . AA AA A A A A D * ( U X V= ) (6ii+3ij+4kk)-(-5i+lOk) = -30?+40k A A AA AA AA AA Next.(v x b.9.) or ( elel + e2c2+ e3c3 .26 MATHEMATICAL FOUNDATIONS [CHAP. II = (D. 1. O = (3?? + 2. ~ ..-6k3+8kk 1..)cl .i =) i .c3e3) A A A A and so . Forthedyadics D = 3 ? ? + 2 j j .. .)a..~ . D X v = al(bl X V) + a.c2e2 .) . .. (4. ( D . 1 1.k) A 6 A = D.k = (o-?)?+ ( D * 3 ) 3 + ( ~ .D .: . Show A A directly from A theAnonion form A ofA the dyadic D that D = (Do?)? + (D-?)? + ( D 0 k ) k and also i w D . v = 5 j. .11.3 k j + k k .(3ii+2jj-jk+5kk) AA = 20ik+123!-63k. Writing D in nonion form and regrouping terms. A = &cl + &c2 + $.c3.6 3 3 . ~ .+ (bNX v)aN = -(V X b. F and H = F * D if D and F are the dyadics given in Problem 1. + + r .11. ê2]+ [(b ê3)c3 C.j = D...(b2 X + + aN(bNx V) V) ' ' ' (D X v). z .ê. 10ik+2kj-2jk) A A A A A A A A A A A A = -3i j .y A .bxz) j + (bxy. Resolve the dyadic D = 3. = cos g cos e cos g sin e . 1-14 1. 1-15 e A 1 k A e+ x ê.6 i j + 1 0 k i .. A e. = e.b.(sin g ) k A A e. By direct projection from Fig.: + 4:: + 6 3 1+ 7. Let D = E +F where E = E.) = ( 1 / 2 ) ( 4 i k .f 2 k3 into its symmetric and antisymmetric parts.i -i-b.CHAP. determine the vector r' produced when D operates on r = 4?+23+5k.) A A A A A A A A A A A A A A A A = 3ii+3ij+7ik+3ji+7jj+jk+7ki+kj = E. 1-15.. and F = -F.x) j + (b. + b.sin e) i + (cos e) j A A (sin g cos e) i + (sin g sin e) j + (cos @) k A A A A e. 1.3 i k + 3 j i .) = (1/2)(6 i i A A + 4 i k + 4 k i + 6 j i + 6 i j + 14 j j A A A A A A A A A A + ioct + i o t k + 2 2 3 + 2.60)..b.+2ji+jj+kk as a linear vector operator. construct the vectors u = f(i) = i A A + iXr A A = i -zj+yk A A v = f(j) = j A A + j Xr A = xi A +j A -xk A A A A A A A w = f(k) = k + k X r = . A A A A A A A A A A A A A A A A F = (1/2)(~-o. + b.18.sin e cos e O = (sin @ cos e ) i A + (sin g sin e) j A + [(cosze + sin2 e) cos g] k A = 2. 11 MATHEMATICAL FOUNDATIONS 27 Considering the dyadic A A A A A A D = 31:-4:.4 k i + 6 j i .17.z .19. + + In accordance with (1. a = b +bXr A = b.y i + x j + k A A A A Then ü = ui+vj A A + wk A A = (i -zj+yk)i+(xi A A A A A A t j -xk)j +(-yi+xj+k)k A and a = D b = (b.. . = (COS g cos e) i A + (cos g sin e) j A .z + b. Fig. 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.sin @ .) k A AS a check the same result may be obtained by direct expansion of the vector function.y) i + (b. jk+3ki+kj = -F.x . = and so Fig.3 + b. = (.k + (b. k and confirm that the curvilinear triad A is right-handed by showing that ê.x)k A A A A 1. Then E + = (1/2)(D D.y) i A + (-b.b. ê. Express the unit triad ê4. in terms of A A A A i.59) and (1.x + b. x ê.b.: + 10 ko. j. BZll + BZz2+ BZ33.aibjSij. a2.CARTESIAN TENSORS (Sec. Bijj represents three sums: (1) For i = 1. X a. (3) For i = 3. ( b ) ' i j k a j a k = 0. in general. a1 = O. aibjSij= albjSlj a2bjSzj + + a3bjS3j. Bijj. ( a ) Sii = S l l + Sz2 + = 3 (b) SijSij = S l j S l j + 62jS2j + S3jS3j = 3 (C) S i j S i k S j k = S l j s l k 8 j k + 82j62k8jk + 83j83k8jk = 3 ( d ) Gijsjk = s i l S l k + Si2S2k + 8i383k = Sik (e) 6ifiik = S1#11. Ai. Since al a1 = 1. = 1 and h = l / ( a l a.. a. a1 = X(a2 X a3). B3.. i ~ k i.. + 82jA2k + 83jA3k = Ajk 1. Now summing each of these three terms on j. a1T13+ a2T23 + a3T33. l / h = bi ba X b3 = 12 and so A A bl = (b. the set a'. i.esents the single sum Aii = All + A. b2.9-1. ai Tij. give the meaning of the following Cartesian tensor symbols: Aii. + B322+ B333. and a. j = €ljkCklj + ' 2 j k e k 2 j + '3jkek3j .) = l/[ala2a3]. + A.16) 1. Rij. repi. Summing first on i.Ri3. a2. a3 (not necessarily unit vectors).e. (2) For i = 2.RZ3? R32>R33. By definition. Evaluate the following expressions involving the Kronecker delta aij for a range of three on the indices. (2) For j = 2.. €. al Xa2 X a. al a1 = 1. bz = . a3 a1 = O. With respect to the set of base vectors ai. aibjSijrepresents a single sum of nine terms. 1. Blll + B12. Hence a1 is perpendicular to both a.a3 = aij... For the permutation symbol Cijk S ~ O Wby direct expansion that (a) Cijk'kij = 6. R12. Thus..23.. x b3)/12 = ( j -k)/4 .21.. + B133. a3 is said to be a reciprocal basis if ai.20.28 MATHEMATICAL FOUNDATIONS [CHAP. (a) First sum on i. For a range of three on the indices. X a. al Tll + a2 T2i + a3 T3i. a1T12+ a2TZ2f a3 T3.+ a2b2Sz2 +~ + 2 ~ 3 ~ ~2 33 ~ 1 ~ ~ + + 3 31~ 2 S 3 2~ 3 ~ 3 ~ 3 3 122.. 1 1.i + 2 ? + 2 k .i13 + j 14 + k / 1 2 A A A b2 = (b3x bl)/12 = INDICIAL NOTATION . aibjSij = alblSll + alb2S12+ alb3S13+ a2b1S2.. Therefore i t is parallel to a. Rii represents the nine components Rll.R. ai Tij represents three sums: (1) For j = 1. (3) For j = 3. Determine the necessary relationships for constructing the reciprocal base vectors and carry out the calculations for the basis bs = ? + $ + % A A A bl = 3 i + 4 3 . For the basis b l . Rz2.b3. 1.27.E121E132 + '122'232 + '123'332 = O and SipSjq . q =2 and note that since k is a summed index i t takes on a11 values.( B k i ) = -Bki .a3a2 = O when i = 2.24. '12kEk32 .59 that this identity holds for every choice of indices. 1. j = q = 2. The nonzero terms are 'ijk'kij = '12kEk12 + '13kCk13 + '21kek21 f '23kek23 + '31kCk31 + '32kEk32 Finally summing on k.CHAP. - 'ijk'kpq . 2 ~ 2 ~..SiqSjp = S12S21 . (It is shown in Problem 1. p = 3 and for (b) i = q = 1.~ l . j = 2.~ 3 . p = 2. 2 ~ 3 = + fc2. Then . 2 ) ~ 1 ( ~ 2 . j = 2. Expanding.1b1 +~ 2 .D j i . and so a X a = 0.. D12 = -D2. Since by definition of cijk an interchange o f two indices causes a sign change.25. j b j . f2 = B21f: + B22f2 + B23f3* 1. Sh0w that c i j k c k p q = . q = 1. 2 ) ~ 3 (C) fi = Bijf.S11S22 = -1. when i = 1.26. 3.ibj f2 = c2. Eijkajak = ' i l k a l a k + Ei2ka2ak + Ei3ka3ak = 'i12ala2 + ' i 1 3 ~ 1 ~f3 ' i 2 1 ~ 2 ~+1 'i23a2a3 + f ' i 3 1 ~ 3 ~ 1 Ei32a3a2 From this expression. 3 ~ 3 .) ( a ) Introduce i = 1. 2 ~ ~1 2. = 813S22 . Determine the component f2 for the vector expressions given below..a2al = 0 Note that eiikajak is the indicial form of the vector a crossed into itself. aiqS. j = p = 2. €3jkajak = a1U2 . the nonzero terms are e.' .8 i q s j p for (a)i = 1. €2jkajak = a i a 3 . Bik = eijkaj = . 1. eljkajak = a2a3. (b) D i j = .a3a1 = 0 when i = 3.3 .~ 1 .S12S23 = O (b) Introduce i = 1. p = 3. 1 . 11 MATHEMATICAL FOUNDATIONS 29 Next sum on j.. ~k k21 - . Expand and simplify where possible the expression Dijxixj for (a) Dij = Dji. Dijxixj = Dljx1xj + D2jx2xj+ Dsjx3xj = D11x1x1 + D 1 2 x 1 x 2 + 0 1 3 x 1 5 3 + D 2 1 ~ 2 ~+1D 2 2 ~ 2 ~ 2 + D23x2x3 + D31x3x1 + D32x3x2 + D33x3x3 (a) Dijxixj = D11(x1)2 + D22(x2)2 + D33(x3)2 + 2D12x1x2+ 2D23x2x3+ 2D13x1x3 ( b ) Dijxixj = O since Dll = -Dll.( e k j i a j ) = .cj. fi 'ijkTjk f2 = '2jkTjk = €213~13 + '231T31 = -T13 + T31 ( b ) fi = C. Then eijkekpq = ~123'321 = -1 and SipSjq . etc.. Show that the tensor Bik = cijkaj is skew-symmetric. '123'312 + '132'213 + '213'321 + '231'123 + '312'231 + '321'132 + = ( l ) ( l ) (-I)(-1) + (-I)(-1) + ( l ) ( l+) ( l ) ( l+) (-I)(-1) = 6 ( b ) Summing on j and k in turn. 2 + ~ 2~ 2 . 1 1.- ao3 ae. + cos2 e. 1.)(e.. T h u s f o r t h e line element (de. and for ( 0 . de.. (ds)2 = (de1). ds = G d e 2 .29. . 1-16 Hence ax1 . 861 302 as3 8x3 . cos2 8 .. . 8x1 - .. 1-16 (r = e. gpq = O for p Z q.) = o T h u s for spherical coordinates. dB. + (el)2(de2)" (e1 sin de^)'. sin e.cos e.j~q ordinates as shown i n Fig. Apply this result to the spherical coordinate system of Problem 1.. W r i t e (1.). sin e. x . sin e3) . axi axi = (sin e. . . cos e. e + (sin O 2 sin e3)(e1cos e.86) as (ds)2 = gpqde. show that Bpq= t p qbi Multiply t h e given equation b y epqi and use t h e identity given i n Problem 1.. . cos 8. If Bij is a skew-symmetric Cartesian tensor for which the vector bi = ( + ) E ~ ~ ~ i . sin e2 cos e. . + ~pqibi = & ~ ~ q i ~ i j k=~ j&k( a p j a q k .29.30. sinz e. e . de3). sin e. axi azi W r i t e (1. = o as2 863 f r o m which g. ds = de3.SpkSqj)Bjk = & ( B p q.. sin 0. sin O .30 M A T H E M A T I C A L FOUNDATIONS [CHAP. 6. 8x3 . Similarly for (0. -e. cos 0 . cos2 e . 8x2 - . O ) . = 8 . cos 8 . sin2 e. Fig. + sin2 o. axi axi = -- as. asl - . sin e. cos O . 8x2 . sin e. sin2 e. = 1 axi axi g33 = -. 1-17). 0. = e .. = g l l ( d e l ) 2 and d s = \/g. . Also. . cos @. eSsin2 e. (COS eZ)(elsin e. 1-7(b). 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). For example. Therefore (Fig.28.26.. Then x . O ) . e = e.. ax1 . i t follows t h a t (ds). O . @ = e.. de. = eSsin2 O . sin e2 sin e..B q p ) = &(Bpq B p q ) = B p q 1. . ao1 as2 863 8x2 . -el sin e. . de.87) a s gpq = and label t h e co- . + eSsin2 6 . 0 . Determine directly the components of the metric tensor f o r spherical polar coordinates as shown in Fig.) g12 = aBi. cos e3 . a%. ae2 ax. sin 8 . sin 6 v i = a Z j v j = -vl sin e + v . ae. be the length of line element represented by (del.31. O .O) is 812 denoted by p. v .. + as. = giz de1 dez del de2 812 Hence using the result of Problem 1. de. If the angle between the line elements represented by (doi. axis. Determine the entries for the bottom row of the table so that 0 x i x . don. x i is a right-handed system. O). 1-18 By the transformation law for vectors (1. O) and (O. ae. O. 1.). O..33.. de. Let ds. A primed set of Cartesian axes O X ~ X $isXobtained ~ by a rotation through an angle e about the X . Write (1. cos 6 v j = a3jvj = V.-de1 de2 ae. = r d+ ( 3 ) For (O. ax3 .= cos plz h. ds = e . . de. and give the primed components of the vector v = ~ $ 1 vzêz v&. . aek ax.-= ds1ds2 66 1.. tos plz = g..32. = 6de. de2 i. The table of direction cosines relating two sets of rectangular Cartesian axes is partially given as shown on the right. = cos (si. and since (A). ae. O ) .CHAP.O).. ds = del = d r ( 2 ) For ( O . de. . = 66.94).85) a s dxi = -dek.dsz.-de.13) ai. Determine the transformation coefficients aij relating the axes. = r sin + de 1. 11 MATHEMATICAL FOUNDATIONS 31 ( 1 ) For (del. the table of dire8ion cosiiies is 51 22 x3 X.de..s j ) and Fig. de. - ax. cos e sin e O xó -sin e cos 6 O x3 o o 1 Thus the transformation tensor is cos 6 sin 6 O A = -sin e cos 6 O Fig. as.O. cose + v . show that cosp. 1-18. . de. + + From the definition (see Section 1. 0 ) and dsz = G d 6 2 axi be that of ( O .30. = a I j v j = v . ds = e. 13 t3 -A.35.. = -B.36. 1.a3. show that AijBij = 0.B. S ~ O that W (Pijk + Pjki + Pjik) XiXjXk = 3Pijk xixjxk.. Since a11 indices are dummy indices.. Now change dummy indices in these same terms again so that the form is Pijkxixixk + Pijk + Pijkxjxixk = 3Pijkxixjxk 1. q . ApPBpq= AijBij and so 2AijBij= O. 1J 13 + A. Since A . o llfi .. I1 I1 or A.a3. 3. 31 31 = A. Aij = a p i a q j ~ p and q Bii = apiaqjBiq. 1 112 -112 The orthogonality conditions aiiaik = S i . = O which is seen to be the sum of products of corresponding elements of the second and third columns.. = A l. The coefficients aij are direction cosines and may be calculated directly from the table.) which demonstrates that the sum transforms a s a second-order Cartesian tensor.B. Thus . each tem of the sum is equivalent to the others. .B. For orthogonality conditions in the form ajiaki= Sjk.iBg.t.B.) k qyBP. and B . If Bij is skew-symmetric and Aij is symmetric. + For j = 2. or Ae t..31' A.(416)$~and the third row is 1 1 x: -415 1 -3/5 1 O I. 1 The unit vector along the xí axis is given by the first row of the table a s $I = (315)ê1. the rows are multiplied together instead of the columns. = [ ( 3 / 5 ) ê 1. Any two columns "multiplied together element by element and summed" to be zero.(415)ê 2 .... L3 i 1 + A P P BPP = O. Since a11 indices are dummy indices and the order of the variables xi is unimportant. The sum of squares of elements of any column to be unity. require: 1. Also from the table 2. This may be readily shown by introducing new dummy variables. Deter- mine the transformation coefficients aij and show that the orthogonality conditions are satisfied. By (1. j. k = 3 that a12a13+aZ2a23 a3. the sum becomes Pijkxixjxk + Pqrpxpxqxr + Pqprxpxqx. + + a3. = (-3/5)& . A11 of these conditions are satisfied by the above solution. . r .. = ê3. Show that the sum x A +~pBij ~ represents the components of a second-order tensor if Aij and Bij are known second-order tensors.a.108) and the statement of the problem. 2. Hence XAij + @Bii = ~ ( ~ ~ ~ ~ ~ = k a p~i a . Thus replacing i. 1 112 -112 aij ..)i ( ~ ~+ + ay(apia. or AijBij= 0. For a right-handed primed system 23 = Xêi.B.34.(4/5)$2]x 2. 1. For j = k = 1 that a l l a l .32 MATHEMATICAL FOUNDATIONS [CHAP. = 1 which is seen to be the sum of squares of the elements in the first column. k in the second and third terms by p.37. Let the angles between the primed and unprimed coordinate directions be given by the table shown on the right.. 1. a2. while (aiajbk) is symmetric in i and j . Resolving Dij into its symmetric and anti-symmetric parts.26) = aqbpcq .41.( S p j S q k . 1. If the vector vi is given in terms of base vectors a.aqbqcp = (aqcq)b. Show t h a t the quadratic form Dijxixjis unchanged if Dij is replaced by its symmetric part D(ij). Dij = D(. c by vi = aiai + pbi + yCi.(aqbq)cp Transcribing this expression into symbolic notation. 1. An equivalent expression for the determinant is eijkAilAj2Ak3.ajbk. .109) the box product [abc] rnay be written a1 a2 a3 x = a. Then vi = eijkbjck. and if h = v a .( a . then X = eiik(aiajbk). w = a X ( b X c ) = ( a . = A Z i and ci = A s i are introduced.39. Thus vi = eij. b.c ) b.bXc = [abc] = eijkaibjck = bl b2 b3 C1 C2 C3 If now the substitutions ai = A l i . and if a X v = w. But eijk is skew- symmetric in i and i. Hence the product eijkaiajbk vanishes a s may also be shown by direct expansion.40.j) + DLijl = S ( D i j+ Dji) + &Dij Dji)- Then D. .ijlxixj = #Dij + Dji)xixj = S(Dijxixj+ Dpqxqxp) = Dijxixj 1. From (1. b ) c ( 2 ) Let a X b = v.38. Show that the determinant may be expressed in the form cijkA1iA2jA~k. b.SpkSqj)aqbjck (see Problem 1. then w p = ePqiaqeijkbjck . Use indicial notation to prove the vector identities (1) a x (b x c) = ( a *c)b .52) and (1. 11 MATHEMATICAL FOUNDATIONS 33 1.CHAP. .( a . X = eijkaibjck = eijkAliAZjA3k This result may also be obtained by direct expansion of the determinant. (2) a x b a = O (1) Let v = b X c. b)c. then [h] = [9.43. -20. compute by matrix multiplication the products a D. n : 3 ) = -nF' = TI/@. . h ( 3 )= 4. -ny' + nf' = O. which results in the cubic equation h3 .v2. b = 23-62 and the dyadic D = 3 ? ? + 2 ? k - 4 3 .n:' = 0.l ) ( h. .42. since nini = 1 and ni2' = 0.52) and (1. D b and a . = 4.I 0 9 ) . D b. Let a D = v. (1. 1 By Cramer's rule. and -ny' = O. Determine the principal directions and principal values of the second-order Cartesian tensor T whose matrix representation is 3 -1 o [Tij] = -1 3 O O 0 1 From (1.0.4 ) = O whose principal values are h ( 1 . = 2. For h t 2 ) = 2.231) give 2n:" . + from which n:" = nll) = 0 . and 3 n F ) = O. b = h.. Thus n33' = 0.6] ( 2 1= [-761.ns' = O. a = I v3 b3 c3 I and by (1. Next let n:l) be the components of the unit normal in the principal direction associated with Ao. 1.132). Then the first two equations of (1.20) 1.nil' = O and -n:" 24'' = 0. The principal axes x: may be referred to the original axes xi through the table of direction cosines ..7h2 + 14h . Thus ny' = ni2' = %1/\/2.8 = ( h . nB1)= *I.131) yields ni2' . = 1.131) yields -ni3' . for principal values h. then O -5 -6 -10 Let a D b = v . -20. For h. -ni3' .6].5k^5 3. (1. a = eiikvibjck a1 bl c1 'pqrapbq~r' a2 b2 c2 EijkaivjCk cijkaibjvk Likewise /3 = epqrapbqcr = ~pqrapbqcr' MATRICES AND MATRIX METHODS (Sec. = 1. and from nini = 1.34 MATHEMATICAL FOUNDATIONS [CHAP.2 ) ( h. For the vectors a = 3:+42. then [ v l . nl2' = 0 .4] O - -4 [ ] = 9 . 1. v3] = [3.17-1. Let D b = w. Fig.CHAP.43.ajqTPq. (or in matrix symbols T* = cATcAT).45. there are two sets of principal axes. Finally for the system to be right-handed.43 form a right-handed set of orthogonal axes. Since the condition nini = 1 was used in determining the aii. Orthogonality requires that the conditions aijaik= 6 j k be satisfied. Show that the principal axes determined in Problem 1. x. As shown by the sketch both sets a r e along the principal directions with x: being a right-handed system. ~ ( 3 = A n(2) ) n ( l ) . 11 MATHEMATICAL FOUNDATIONS 35 from which the transformation matrix (tensor) may be written: 1. Show that the matrix of the tensor Tij of Problem 1. 1-19 1." a left-handed system. 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 1.44. x: and x:*. ~h~~ A A A e1 e2 e3 i i o = (4+ 4)ê3 A = e3 -116 1 o As indicated by the plus and minus values of aij in Problem 1.43 may be put into diagonal (principal) form by the transformation law T: = a. orthogonality is automatically satisfied for j = k.. .43. = [-i i !][-i i 3 -1 3 -1 !] [-: = 10 -6 1.0 For = 16.46. ~ ( ~ .Cn(2) = O 1 2 For A ( 2 .. = 4. nF) = 0 -Sn'2' - 3 . 1. 1 1. their difference is not zero. Prove that if the principal values h ( l ) .~n ': Multiplying ~ ) .131) and using the condition nini = 1..X)(h .361 = ( 1 . Substituting these into (1. the first of these equations by ni3' and the second Since Tij is symmetric. ) = 1. 2 9n.h -6 O -6 10 . the diagonal form of T is given by . Use the fact that (T)2 has the same principal directions as the symmetrical tensor T to obtain fiwhen First. the principal directions are mutually orthogonal. .) = 1. n ~and j .47. h ( 3 )= 16. the principal values and principal directions of T are determined." = O = n21' = 0' n(l) 3 =21 6n(2) .43 and verify that its principal axes coincide with those of T. [Ti.129) is satisfied.. The proof is made for h t 2 ) and h ( 3 ) . Hence n:2'n:3' = 0 .48. gn"' .. ) n : ~ ' n=: ~O' Since h(. = 4.Gn(l' = For h ( . -15ni3' = O which are the same a s the principal directions of 1.h)[(10. For each of these (1. -Gn?' + 6nk2' = 0 = ni2'= k l f i . Compute the principal values of ( T ) 2 of Problem 1. the condition for the two directions to be perpendicular.h)2 . 1.h O = ( 1 .36 MATHEMATICAL FOUNDATIONS [CHAP.4)(h. Z h(. The characteristic equation for this matrix is 10 . h ( 3 ) of a symmetric second-order tensor are a11 distinct. -6n:" 1 $..16) = O o O l . so that T i j n j 2 )= ~ ' ~ ~ ~= n~ ~ ~ ( .43. . 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. Following the procedure of Problem 1.h from which h(. k j . whereas + .23 32) = 0.= ax axi A.h Therefore fi = . + ~ (~ ~ 3 )in~ the direction of the unit normal n = (2/7)&.If A 23. axk lk axp . 1. Use indicial notation to prove the vector identities (a) V X V+ = O.ji and ~ i j k= -'jik.~ f i the matrix equation is CARTESIAN TENSOR CALCULUS (Sec.~But + . 11 MATHEMATICAL FOUNDATIONS 37 with the transformation matrix being [aijl = [ llfi 0 -216 lldz llfi llh 11. +.(6/7)ê3 or & = (261.23) 1.. -X .21-1. by expansion vl = '123+.ij ak.i 9 ax.A k P + A P k .i and so v = V X V + has components v i = ~ ~ ~= ~~i j k +a.= V h n = X.6ê3)/7.a%. For example.50. A Thus an .51. V + is written $. axk as. show that dhldxi = (Aij Aji)xj + and d2X/dxidxj = Aij + Aji. i t i~ seen that - ax = A k j z j+ Aikzi = axk %3axk 3 a> 'axk ' Jxk Jxk (A + A jk)xj. hence the product E ~ ~ vanishes. ~ ~ The same result may be found by computing individually the components of v.%.49.. = A3.(3/7)&. ( b ) V V x e = h = ( ~ ~ ~= ~eijkak. + A.147). Determine the derivative of the function X = ( X I ) ~ 2 x 1 . O since ak.= 2Akjxj and -. a2A 8% Continuing the differentiation. . ~ 2- + ($. ~ + . ~ eijk is anti-symmetric in j and k. Consider . ax The required derivative is . ( a ) By (1. For the function x = Aijxixj where Aij is constant.2APk. 1. .23~ ~ 3 ~ = + .3ê2 . to relate this to the original axes by the transformation fi = ~ A. Simplify these derivatives for the case Aij = Aji.ini. axk 1.ji a ~ =.CHAP. Since axi -z Sik. k jis symmetric in j and k. -." ó 3 o o and using [aij.( A k j + A . ( b ) V V X a = 0. ) - - . and ni its outward normal. and so i hbini d S = i rijkhvk. the bounding surf ace of the volume V shown in Fig.= f'xi/r. 1. i t ar axi axj 2 ~ a ar 2. and since ar = 2r ax. 1-20 1.. xi is the position vector of nj d S . -. i hbisdS = i h. Hence which is the transformation law f o r a third-order Cartesian tensor.jni d s = & Li bi d ~ since hrijkvk.38 MATHEMATICAL FOUNDATIONS [CHAP. dxj r ar axi Use the divergence theorem of Gauss to that S ~ O W L Xinj d S = V6ij where nj d S represents the surface element of S . show that its derivative with respect to Aij xk. namely Aij. xi = ajixi and âxildxi = ajt. F o r the Cartesian coordinate systems xi and xl.k.55. If r2= Xixi and f ( r ) is a n arbitrary function of r. If is a second-order Cartesian tensor. show that ( a ) V ( f ( r )= ) f'(r)xlr.1 1. a f ar follows t h a t . 1-20.i = .ibi d V where h = h ( x i ) is Since b = V X v. If the vector b = V x v. and ( b ) V 2 ( f ( r = a f ar ( a ) The components of V f a r e simply f.= 3.. S ~ O Wthat a scalar function of the coordinates. x .. is a third-order Cartesian tensor. bi = eijkvk. + ) ) f l r ( r ) Zf'lr.+ Thus f. where primes denote derivatives with respect to r. = SijV Fig.i = . .ii = O . Thus f.53.= 2 ~ .52. 8mp 8mq Sms 1.59(a) below.b)2 = ( ~ b ) ~ Interchange the dot and cross in the first term. Then wi = d ( E ~ ~ ~=u ~eijkGjwk v ~ ) + eijkuj. a ) ( a W b+) (a. show that A = ( a x b) (a x b) + (a.(b.. m)u . 8rp 8rq 8rs A11 -412 A13 Let the determinant of Aij be given by det A = AZl Az2 A23 .. Thus A31 A32 A33 A. A. AZ3 = -detA A31 A. A. (a) In symbolic notation.. Aml Am. b)a ..S i n S j m .8 i n 8 j m ) u j ~ m ~ nE ~irnk'~j~Ujarnw~ which is the indicial form of m X (u X v). d z(uxv) = h X v + ux. show that - dt (u X v) = ca X (u x v). A. 4 3 . = (mXu)Xv+ uX(mXv) = (V. A23 A12 A. wi = (SimSjn . €y. det A Ar1 A. (U @)V = (V' . W i. E3pqapUqwk . a ) ( b . b)(a0b) = ( a . If ii = o x u and V = w x v. (V U)OI + (U v)a .57. = A.58..b)(a.. em. 11 MATHEMATICAL FOUNDATIONS 39 MISCELLANEOUS PROBLEMS 1.SimGjn + SijSmn)~jamwn = (SijSmn . Am3 Anl Ana An3 - .. For the arbitrary vectors a and b. Then h = a .k and since Aj = e j p q w p u q and ck = ekmnamVn. A. k.56. A31 A33 and for an arbitrary number of row changes.CHAP.b x (axb) + (a.. d 1.b)(a0b) = a [(b. + eijk'kmnUjamvn ('ijkEkmn -~ink'kmj)~~jamwn and using the result of Problem 1.b ) .( b . A. Establish the identity cPqsrmnr = Snp 8nq 8ns . let (u X V) = W... A13 A.)U .. An interchange of rows or columns causes a sign change.b) = (ab)2 since the second and third terms cancel. A..a)b] + ( a . (U @)V = a X (U X V) d ( b ) I n indicial notation. = bl x ê1 + bz X êz + b3 X ê3 and B . Use the results of Problem 1. B.0 .SpnSrq + SqnSrp . -43.A el)bl + A ( a .aqqsrp = S p r ..60.a .0.S m p ( S n q S r s . A. .61.Onpsrs) + Sms(SnpSrq . A. A.SnqSrp ( b ) Identifying q with n in ( a ) . det A = 1 and the identity is established.3Snqsrp - ..3 S p r = -2Spr 1. If the dyadic B is skew-symmetric B = -B.SnsSrq) f S s q ( 8 n s s r p . A. 1 or column changes. ) ê z ... epqS det A A. S ~ S Q=~ . 1. Writing B = blêl + b2ê2+ b3ê3 (see Problem 1.6). Use the Hamilton-Cayley equation to obtain (B)4 f o r the tensor B = Check the result directly by squaring ( B ) 2 . 5 8 .SnpSrs) + S s s ( S n p ~ r q. SnpSqr + 3SnpSrq .B 1.( a ê 3 ) b 3 = a (blêl + bzêz+ b 3 ê 3 ) .( a e z ) b 2 -t ( a b 3 ) ê 3 .Sqparq .58 to prove (a) cpqscsnr = S P n S ~ r. = 2a. Am. Hence + . Expanding the determinant of the identity in Problem 1 . 'pqs'snr = Ssp(8nqsrs .( ê l b l + ê2b2+ ê 3 b 3 ) - . X ~ = (blXêl)xa+ (b2xê2)xa+ (b3xê3)xa = (a bl)ê.40 MATHEMATICAL FOUNDATIONS [CHAP.SnqSrp) - - srpsnq .2 ( ~ -) 6~8 91 = 0.Spraqn. -4% . -42. The characteristic equation f o r B is given by By the Hamilton-Cayley theorem the tensor satisfies its own characteristic equation. - 'pqs'mnr . a. x a = 2a B.s n s a r q ) + Omq(SnsSrp . Ams - Anp Anq Ans . Hence .2 6 ~ ~ . show that B .59. and multiplying this equation by B yields = 2(B)3 6 ( 8 ) 2 .(a. - ~pqs'sqr . b . Hence f o r a n arbitrary row and column interchange sequence.181. SnqSrp) ( a ) Identifying s with m yields.9 8 01 + + (B)4 = 1 0 ( B ) z 38 .S n p a r q .a.. Ar. Ars When Aij = Sij. A I S - -42.. 'mnr'pqs det A A. ( b ) ~ P . ( ê i êp)(êj2.) = DijE.~ m n ~ j j ) ~=k n~mjk%jnALn 1. . 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. cijkT(jk) For the product T i j S i j= TCii>Sij TLij1 + Si? Here TLii.31). 1. p d = (Sqjspk .q Hence v = b V a .apxq. Let V X (a x b ) = v. ( c ) c i j k ~ k j p A t pare invariant under the coordinate transformation represented by (1. i. = E.~ipsjj)amianpAkn - .103)..p . 21 13 (C) cijk'kjpAip = €ijk'kjpamianp~hn = ( 8 i j s j p. Hence Aii = a.. Use the indicial notation to prove the vector identity V x (a x b) = b Va . D : E = D i j ~ . Prove that (a) Aii.62.ê.s q k s p ~ a j x k .V .) since E.110) the dual vector of Tij is vi = cijkTjk. or vi = eijk(TCjk)T c j k l = + = O (cijk is anti-symmetric in j and k.e.b(V a) + a ( V * b ). = S. The position vector to any point in V is x. .).( S m n .p) 'V : L (aqS p p .Spksqj)(aj.. D E = êq)(ei e. 11 MATHEMATICAL FOUNDATIONS 41 Checking by direct matrix multiplication of (B)2.157) this becomes dV and since a is constant.E . = ~ ~ . By (1.Sij = TCij. a. + a(V b) . (23)4 = O 26 1.103).CHAP.q bp + . (êj A%)(Aei Ae. Aij = aPiaqjAP.) dV = Za.65.q) = (Spjsqk . T ( j k . = A:..).qbk ajbk. 1.aqiA. then v. 1. In the indicial notation the surface integral is the volume integral (rqPirijkajxk).A:.a. = DijEpq( g j $.symmetric in j and k). s. q=) bq .etc. and a is an arbitrary constant vector. By (1. Now interchanging the A A role of dummy indices p and q in this last expression. ) eijk T L j k l since By (1.a Vb.. (b) AijAij.35). cqPinpcijkajxkdS. E if E is a symmetric dyadic.) dV = (3aq.Sii = O and T.S.&mnsjj)Akn= (SmjSnj . (a) By (1. (b) AijAij = a p i a q j A ~ q a m i a n j = A ~ nB ~ ~ s ~ ~ A= PApqApq ~ A & =~ A!. 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. D.. the last expression becomes q p j .V b ..)(Si 3.b ( V a) .ê.64 Show that D : E is equal to D 0 . ~ c ~or~ ~ a ~ a ~ b ~ Vp = 'pqi'ijk (ajbk).q = epqi'ijk(aj.aq.a.66.aqbp.63...q bk + ~ j ~ k ..By (1. p dv = (aqxp... WriteD = lIijêiêj and E = ~. show t h a t Aii = ~ : i .A!.. = Aii.a .. 42 MATHEMATICAL FOUNDATIONS [CHAP. 1 1.67. For the reflection of axes shown in Fig. 1-21 show that the transformation is orthogonal. From the figure the transformation matrix is The orthogonality conditions aijaik = S j k or ajiaki= S j k a r e clearly satisfied. I n matrix form, by (1.117), i-: :l[u-i 01 L 0 0 1 ] L 0 0 1 J r: : 01 L 0 0 1 J Fig. 1-21 1.68. Show that (I x v) D = v x D. I Xv = A A + kk) X (v,i+v,j +v,k) (i i + j j A A A A A A A i(v,k-v,j) + j(-v,k+v,i) + k(v,j-vyi) A A A A A A A A = (vXi)i + (vXj)j + (vXk)k = v X I A A A A A A = Hence ( I X v ) - D = V X I * D = V X D . Supplementary Problems 1.69. A + A A A A Show t h a t u = i j - k and v = i - j a r e perpendicular to one another. h Determine w so t h a t u, v, & forms a right-handed triad. Ans. w = ( - l / f i ) ( ? + 2c) 3+ 1.70. Determine the transformation rnatrix between the u,v,W axes of Problem 1.69 and the coordinate directions. 1.71. Use indicial notation to prove ( a ) V x = 3, (b) V X x = O, (c) a . V x = a where x is the position vector and a is a constant vector. 1.72. Determine the principal values of the symmetric p a r t of the tensor Tij = Ans. h ( , , = -15, h ( Z )= 5 , h,,) = 10 -3 -18 CHAP. 11 MATHEMATICAL FOUNDATIONS 43 1.73. F o r the symmetric tensor Tij = determine the principal values and the directions of the principal axes. Ans. 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 . V v = O, V x v = 6 and V X w = -c, show t h a t V2v = y. 1.76. Check the result of Problem 1.48 by direct rnultiplication to show t h a t fifi = T. 1.77. Determine the square root of the tensor B = *(&+I) #&-I) 0 1.78. Using the result of Problem 1.40, det A = E ~ ~ ~ A show ~ ~ t hAa t ~ det A ~= ~det, A det 0. ~ (AB) 1.79. Verify t h a t (a) O3,,vp= v3, ( b ) 8siAji = Aj3, (C) Sijeijk = 0 , ( d ) ai2aj3Aij = A23. 1.80. Let the axes 0 x ~ x be ~ xrelated ~ to Oxlx,x3 by the table ( 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 , - ê3? ( c ) W h a t is the equation of the plane xl - x, + 3x3 = 1 in the primed system? Ans. ( b ) ( 2 / 5 f i , 1115, - 2 1 5 f i ) ( c ) fiX ; - X; - 2fi X; = 1 h 1.81. Show t h a t the volume V enclosed by the surface S may be given a s V = Q x ia the ponition vector and n the unit normal to the surfaee. Hint: Write V = (116) and use (1.157). ' 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 and to completely - -- fi11 the space it occupies. This continuum concept of matter is the -fundamental postulate of Continuum Mechanics. Within the limitations for which the continuum assumption is valid, this concept provides a framework for studying the behavior of solids, liquids and gases alike. --Adofion -- ------ of t_hecontinuum viewpoint -- -- material behavior means that -field as- the -- -basis for the mathematical __ _ _ such as stress _- quantities and - . -. - displacement description of are expressed -- --_-_r _ - functions of the space coordinates and time. 2.2 HOMOGENEITY. ISOTROPY. MASS-DENSITY A homogeneoousmaterial is-tne having-id~nticalpropertiesat a11 poi&. With respect ,to some _- property, a material ____I_--- is isotropic - --- -- -_ if that - - - - A property - -is the same in a11 - - directions at a point. A material -- is called -- anisotropic - - with gspect to those properties . which are directional -- - a t a pÒi$t. --- The concept of density is developed from the mass-volume ratio in the neighborhood of a point in the continuum. In Fig. 2-1 the mass in the small element of volume AV is denoted by AM. The average density of the material within AV is therefore - anl P(av) - 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, = lim -al)I = - dM - *"-O AV - d!' (2.2) Mass-density p is a scalar quantity. Fig. 2-1 CHAP. 21 ANALYSIS O F STRESS 2.3 BODY FORCES. SURFACE FORCES Forces are vector quantities which are best described by intuitive concepts such as push or pull. Those forces which act on a11 elements of volume of a continuum are known 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 __h 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. Fig. 2-4. the appropriate stress and normal vectors are shown in Fig.69) in terms of its Cartesian components as . representing the orientation of a n infinitesimal surface element having P a s a n interior point.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 . This may be accomplished by giving the stress vector on each of three mutually perpendicular planes a t P. the three separate diagrams in Fig. For some differently oriented surface element.t' - . The totality of a11 possible pairs of such vectors tiE. Coordinate transformation equations then serve to relate the stress vector on any other plane a t the point to the given three.'and ni a t P defines the s t a t e of stress a t that point. Each of the three coordinate-plane stress vectors may be written according to (1. = df dS as-o AS The notation tln' (or tt. 2.5 STATE O F STRESS AT A POINT. 2-4 are often combined into a single schematic representation a s shown in Fig. Cauchy's stress principie associates a stress A vector t:"' with each unit normal vector ni. Adopting planes perpendicular to the coordinate axes f o r the purpose of specifying the state of stress a t a point. Af . 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.)) 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). 2-3. Thus by Newton's law of action and reaction. "' h (2. = . STRESS TENSOR At an arbitrary point P in a continuum. This is illustrated in Fig. 2-5 below. The stress vector arising from the action across A S a t P of the material within A V upon the material outside is the vector -t:"'. the associated stress vector a t P will also be different.46 ANALYSIS O F STRESS [CHAP. 2-4 For convenience. 2 Mathematically the stress vector is defined by dfi Or A f (n) = 1im . having a different unit normal. . a.. the stress tensor components may be displayed with reference to the coordinate planes as shown in Fig. dS2 = dS n2 for face APC. os?. A stress component is positive when it acts in the pgsitive direction of the coordinate axes. The equivalent stress dyadic is designated by Z. 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. dS3 = dS n3 for face BPA or Fig.. 2-6. and the three faces are taken perpendicular to the coordinate planes as shown by Fig. The components perpendicular to the planes (all. 2-6 The nine stress vector components. 21 ANALYSIS O F STRESS 47 Fig. respectively.) are called normal stresses. 2-7.. Des- ignating the area of the base ABC as dS. 2.o. take the forms Pictorially..6 THE STRESS TENSOR . A t:ei) E u.a2. Those acting in (tangent to) the planes (al2. h sor uij a t a point P and the stress vector t. 2-7 . the areas of the faces are the projected areas.STRESS VECTOR RELATIONSHIP The relationship between the stress ten. dSl = dS nl for face CPB.... 2-6 are a11 positive. so that explicit component and matrix representations of the stress tensor.. The stress components shown in Fig. 2-5 Fig. The base of the tetrahedron is taken perpendicular to %.CHAP. a.'"' 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. a.) are called shear stresses. having its vertex a t P. are the components of a second-order Cartesian tensor known as the stress tensor.a.. 7 FORCE AND MOMENT EQUILIBRIUM.48 ANALYSIS OF STRESS [CHAP.14) is equivalent to the component equations 2. Summation of surface and body forces results in the integral relation. the average stress vectors approach the specific values appropriate to the designated directions a t P. (2. STRESS TENSOR SYMMETRY Equilibrium of an arbitrary volume V of a continuum. At the same time. 2 The average traction vectors -t:'&' on the faces and t:':' on the base.16 ) 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.tf'"' d s 2 . if present).10) reduces to A Cancelling the common factor d S and using the identity tje" =. if present) as shown in Fig.157).) d S . requires that the resultant force and moment acting on the volume be zero. 2-8 . 2-8.10) If now the linear dimensions of the tetrahedron are reduced in a constant ratio to one another. 1 ti" d~ + 1 pbi d V = O or (2.tf'"' d s 3 + pbf d V = 0 (2. equation (2. being an order higher in the small dimensions.16) becomes Fig. equation (2. together with the average body forces (including inertia forces. acting on the tetrahedron are shown in the figure.12) is also often expressed in the matrix form rt$i = Ln1k] [vki] which is written explicitly 1 '=12 u13 A A A The matrix form (2. subjected to a system of surface forces t:. Equilibrium of forces on the tetrahedron requires that tf(.u j i . tend to zero more rapidly than the surface forces.11) becomes Equation (2.' and body forces bi (including inertia forces. the body forces.9). Therefore by this limiting process and the substitution (2. tf'"' d s l . 22).. 21 ANALYSIS OF STRESS 49 Since the volume V is arbitrary. a.. or in a11 which shows that the stress tensor is szjmmetric.18) are often written + ~ O o ~ ~p b i. = V. a. the integrais of (2.21) represents the equations u. In the absence of distributed moments or couple-stresses. - . making h the substitution tin' = ujini.18) which are called the equilibrium equations..x:x~ coordinate systems P X I X and ~ of Fig. the equilibrium of moments about the origin requires that in which xi is the position vector of the elements of surface and volume. Again.. so that + pbi = ajijj O or V I + pb = O (2..CHAP.18).. In view of (2. applying the theorem of Gauss and using the result expressed in (2. 2-9 be related to one another by the table of direction cosines Fig. = which appear in expanded form as 2. the equilibrium equations (2.17) must vanish.. 2-9 .8 STRESS TRANSFORMATION LAWS At the point P let the rectangular Cartesian ~ XPx...20) requires Equation (2.19) are combined and reduced to For the arbitrary volume V. (2. the integrand in (2. a. = a. 29) Explicitly. let the stress tensor have the values uij when referred to directions parallel to the local Cartesian axes P61g253shown in Fig. the stress tensor components in the two systems are related by O.9 STRESS QUADRIC OF CAUCHY At the point P in a continuum. (2. At the point P the normal component aNni of the A stress vector tin' has a magnitude = t i l ) n i = t ( D ) . 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.26 ) J Likewise. = aiPajqap.34) .50 ANALYSIS OF STRESS [CHAP. (2.33) ' J ~ Fig.27) In matrix form. 2-10 Accordingly if the constant k2 of (2. the stress vector transformation is written A [ti:")]= [aii][t$] (2. or the transformation dyadic According to the transformation law for Cartesian tensors of order one (1.(P) = a. the A components of the stress vector tln)referred to the unprimed axes are related to the primed A axes components tl'"' by the equation A A t. 2 or by the equivalent alternatives.102) for second-order Cartesian tensors. The equation uijgigj = fk2 (a constant) (2. or 1' = A . n = ~ i j n i n j (2.28) and the stress tensor transformation as [a. X A.32) represents geometrically similar quadric surfaces having a common center a t P. where ni is the unit normal in the direction of r. by the transformation law (1.28) and (2.] [apq][aqjI LaijI (2. 2-10.t(n) or t ' ( n ) - .32) is set equal to uNr2..93).A. the matrix multiplications in (2.29) are given respectively by and 2. the transformation matrix [aijl.the resulting quadric uij[icj= -iaNr2 (2. . Furthermore i t may be..35) into (2. Those directions f o r which tiA and ni are collinear as shown in Fig..a.'k'are solutions of the equations .Sijal.c.?7) as 1 a32 O33 - which upon expansion yields the cubic polynomial in a. the three direction cosines ni and the principal stress value O. a(. For a principal stress direction. a.lu) n = 0 . STRESS INVARIANTS. Explicitly.36).38). the determinant of coefficients.39) 11.= i-k2/r2.) are the three principal stress values.e.) li 111.. A (2. From this definition i t follows that the magnitude O .. 2. must all ... second and third stress invariants.'"'. the magnitude of the stress vector. is inversely proportional to r2.12). STRESS ELLIPSOID At the point P for which the stress tensor com- A ponents are aij.36) Fig. there is a principal stress direction for which the direction cosines n. is called a principal stress zjalue...35) in which a. u23 = 0 (2. 7 1 I ~ = 0 or a~~ u22 .i.12) and making use of the identities ni = Siinj and uij = ajii results in the equations (aij. 2-11 In the three equations (2.21. = laijl = det are known respectively as the first. "12 u13 lu.CHAP. namely..36) other than the trivial one nj = 0. A sociates with each direction ni a stress vector t. 2-11 are called principal stress directions. the equation (2. U . For solutions of (2. a..S. as. A h tin) = &&i or t'"' - A an (2..Sija)ni= O or (1. of the normal stress component on the surface element dS perpendicular to the position vector r of a point on Cauchy's stress quadric.10 PRINCIPAL STRESSES. Associated with each principal stress a. shown that the stress vector t$' acting on dS a t P is parallel to the normal of the tangent plane of the Cauchy quadric a t the point identified by r. there are four unknowns. Substituting (2. The three roots of (2. 11 . = +(aiiujj . t:"' = uiini. where I. = aii = tr 1 (2. luij . 21 ANALYSIS O F STRESS 51 is called the stress quadric of Cauchy. vanish.. . since (2..48) h t(n) = 3 u111n3 and from (2. ( ~ .53~) The shear stress values in (2. the squared magnitude of the shear stress as a function of the direction cosines ni is given by Fig. > u. Equation (2. n3 = 0.53) are obviously minimum values.CHAP.n.523) n3{uSII.uIII)/2 (2.33). nn = O.51) is clearly a function of the direction cosines ni. nz=O.53~) nl = O. A second set of solutions to (2.. conjugate to the extremum values of shear stress. ~. for which U.2 + (2.. and the associated shear stresses from (2. = (aI . together with the condition nini = 1. ~ . n3 = *1/fi. the normal component magnitude is U.533) ni = O.2uI1(uIn. for which U. .n.48) and (2. n3 = +l/fi. for which a. may be solved for A and the direction cosines nl.51) in which the scalar A is called a Lagrangian multiplier. = (oIII.53) are recognized as principal stress directions.543).. Hence from (2.^' = u1n1 = '-'II~z (2.54a) ni = *l/fi. Setting these partials equal to zero yields the equations + + n2{uTI . = u. 2-13 The maximum and minimum values of U.54b) gives the maximum shear stress value. which is equal to half the difference of the largest and smallest principal stresses.)+ A} = O (2.50).. .50) by the method of Lagrangian rnultipliers. so that the conditions for stationary (maximum or minimum) values of F are given by JFldni = O.Anini (2.uI)/2 (2. n . na = 11. the maximum shear stress component acts in the plane which bisects the right angle between the directions of the xhaximum and minimum principal stresses. the directions given by (2. nz = e l l f i .47). ne. the components of t$) are ti. =O (2.uII)/2 (2..12). n .49) into (2. > u.. . n3 = 0.2 ~ . 2-13 where the axes are chosen in the principal stress directions and it is assumed the principal stresses are ordered according to U.. .52).n.49) Substituting (2..n.52) may be verified to be given by nl = 0. ~ . Also from (2. n3=0. ~.54b) nl = 11/fi. na. + ~ . ~.52~) which. na = ? I . are n1=&1. forwhich~. =O (2.. for which a. Furthermore. One set of solutions to (2. The procedure is to construct the function F = U: . . . .35) indicates that shear components vanish on principal planes.) + A} = O (2. = (a. for which U. n z = 0. n+.54c) Equation (2. 21 ANALYSIS O F STRESS 53 This resolution is shown in Fig. may be obtained from (2.=O (2. . n2 = - + I .. n + . 12 MOHR'S CIRCLES FOR STRESS A convenient two-dimensional graphical representation of the three-dimensional state of stress a t a point is provided by the well- known Mohr's stress circles. for which the U .54 ANALYSIS O F STRESS [CHAP..55). axis is the abscissa. In (2.n. 2-15. The principal stresses are assumed to be distinct and ordered according to > O I I > u~~~ (2.u I I 1> O from (2. (2. 2-15. and since ( n ~ )is ' non-negative. 2 2.. > O and U . results in the eauations These equations serve as the basis for Mohr's stress circles. 2-15 . axis is the ordinate. Fig. + = u1nS uI1n.58a). and the U . 2-14. shown in the "stress plane" of Fig.56) Combining these two expressions with the identity nini = 1 and solving for the direction cosines ni. os) plane that are on or exterior to the circle In Fig. the coordinate axes are again chosen in the principal stress directions a t P as shown by Fig. In developing these. 2-14 U . since U . . . the numerator of the right-hand side satisfies the relationship which represents stress points in the (V.u.+ uI1. this circle is labeled CI.55) For this arrangement the stress vector tln) has normal and shear components whose magnitudes satisfy the equations Fig. . stress vectors for Q located on BC will have components given by stress points on the circle CI in Fig. 2-16 are doubled in the stress space of Fig. n3 = ~ 0 ~ x 1=2 O on AB According to the first of these and the equation (2. 2-16 whereas the conjugate stress points O. 2-15. points g. h and f are located in Fig. CA and AB. Such a situation occurs a t an unloaded point on the free surface bounding a body. on the bounding circle arcs BC. 2-16 corresponds to the circle C2. The stress vector components O. 2-17. and O. of the stress vector tin) on the plane having the normal direction ni a t Q in Fig. 2-17 (arc AB subtends 90" in Fig. =O Fig. 2-17 and the appropriate pairs joined by circle arcs having their centers on the O. CA in Fig. the Mohr's stress circles will have one of the characterizations appearing in Fig. for an arbitrary location of Q rnay be determined by the construction shown in Fig. 2-16.56 ANALYSIS OF STRESS [CHAP. 2-17 2. Likewise. are 180" apart on C3). If the principal stresses are ordered.58a). 2-15.. The intersection of circle arcs ge and hf h represents the components and O. and AB to the circle C3 in Fig. and O. n2 = COSTIZ= O on CA. 2 and. axis. nl = cosx12 = O on BC. Fig. 2-18 . Thus point e may be located on C3 by drawing the radial line from the center of C3 a t the angle 2B.13 PLANE STRESS In the case where one and only one of the principal stresses is zero a state of plane stress is said to exist. 2-18. In the same way. Note that angles in the physical space of Fig. A single circle Mohr's diagram is able.. 2-19 2. 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. however. the single plane stress Mohr's circle has the equation The essential features in the construction of this circle are illustrated in Fig.. The circle is drawn by locating the center C a t = (all + O.65) The stress quadric for this plane stress is a cylinder with its base lying in the XIXZ plane and having the equation u.x~ = -Ck2 (2.14 DEVIATOR AND SPHERICAL STRESS TENSORS I t is very often useful to split the stress tensor uij into two component tensors. .. the stress matrix has the form ull @12 0 (2. to display the stress points for a11 those planes a t the point P which include the zero principal stress axis. As seen from Fig. In particular. and points E and D on the circle are points of maximum shear stress value.. the state of stress is termed plane stress parallel to the xlxz plane. For such planes.65). 2-18 this representation is necessarily incomplete since a11 three circles are required to show the complete stress picture. one of which (the spherical or hydrostatic stress tensor) has the form . Principal stress points U .uzz)/2]2 given in (2. are so labeled.xS + + 2 u l z ~ . For arbitrary choice of orientation of the orthogonal axes X I and xz in this case. 21 ANALYSIS OF STRESS 57 If the principal stresses are not ordered and the direction of the zero principal stress is taken as the X Q direction.)/:! and using the radius + R = j/[(u1. ~ . Fig. 2-19).67). if the coordinate axes are chosen in accordance with the stress representation given in (2.CHAP. 2-19. Point B on the circle represents the stress state on the top surface of the parallelepiped with normal nz.u. and a.66) Frequently in elementary books on Strength of Materials a state of plane stress is represented by a single Mohr's circle. Point A on the circle represents the stress state on the surface element whose normal is nl (the right-hand face of the rectangular parallelepiped shown in Fig. comparable to (2. = 0 .72).n.*'n.*)in the direction of ni.ni. . 31 1 I = (o. 2-20 . STRESS TENSOR (Sec.58 ANALYSIS O F STRESS [CHAP.^)in the direction of n: is equal to the component of t:^. Fig.)n* 1J 1 1 = t:")n.6) 2. 2. which accounts for its absence in (2.n*n. and b y (2.p = ukk/3 is the mean normal stress. 2 where U .1. = .38) for the stress tensor. and the second (the deviator stress tensor) has the form This decomposition is expressed by the equations uij = sijUkk/3 + Sij or 1 = UMI + tD The principal directions of the deviator stress tensor Sij are the same as those of the stress tensor uij. Show that the component of ti. STRESS VECTOR.. At the point P the stress vectors tin) A and t:"*' act on the respective surface elements ni AS and n: AS*.a. It is required to show that h t:^.22) oJ.i . Thus principal deviator stress values are - S(k) . = A Frorn (2.71) The characteristic equation for the deviator stress tensor. Solved Problems STATE OF STRESS AT A POINT.1-2.12) tln*'ni = ojin. u(k) . is identically zero. is the cubic I t is easily shown that the first invariant of the deviator stress tensor Ir.. (2. so that A o. Let S be the sum in question.5.3. ' [ cos O .edt!ez)+ t!edt(ed t t A a A i A which from (2. + Qê3) = A ( a ) tin' a = ( 4 2. . A I n matrix forrn. cos O = ( 4 4 / 9 ) / 5 . ( C ) the angle between tin)and G.12).t?'. b and c so that the stress vector on the octahedral plane (2 = (11fi)êl ( 1 l f i ) ê ~+ + (11fl )ê3)vanishes. Solving these equations.' = n. + From (2.7) becomes S uliuli + uziu2i + u3iu3i= ujiuji.2. .13): 7 7 o -2 - A A A [ t p ) . . 2. . . t'. The stress tensor values a t a point P are given by the array 7 o -2 -2 o 4 Determine the traction (stress) vector on the plane a t P whose unit normal is íi = (213)êi . Show that the sum of the squares of the magnitudes of these vectors is independent of the orientation of the coordinate planes. -213. The multiplication is best carried out in the matrix form of (2. 4419 + 10019 A ( b ) jtin'( = d 1 6 = 5. determine (a) the component perpendicular to the plane.2 ( c ) Since tia' 6 = t : . 21 ANALYSIS O F STRESS 59 2. The state of stress a t a point is given by the stress tensor where a. c are constants and u is some stress value..) (%2. The stress vectors acting on the three coordinate planes are given by t l e l ' . 1. Determine the constants a. (b) the magnitude of t i n ) . 1/31 h 10A Thus tCn' = 4 e . t!j")] = [ 2 / 3 . ê. 2. Then s A A A = t. Therefore the solution tensor is -012 -012 -0/2 -012 .4.3 e.1 / 2 . b. For the traction vector of Problem 2.2. .. A 2.(213)$2(113)$3. tin' = uijnj must be zero for the given stress tensor and normal vector.Ae.ei)t:ei)+ t(. tlez' and t i ê 3 ) .CHAP. a = b = c = .94 and e = 20'. an invariant. 2 = 0. and the unit normal to the plane is therefore (see Prob- lem 1. = 12.60 ANALYSIS O F STRESS [CHAP. - Thus 2.6. x3 = 4 a t P. 2-21 From (2.the stress vector may be determined by matrix multiplication. 23 . The stress tensor a t point P is given by the 53 array t : = (1. 1 .h -ê. 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 . I: = 2x3 Determine the stress vector acting a t the point P ( 2 . 2-22. 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. 2 2 This may also be seen in Fig. 2-22 . fi)of the plane that is tangent + to the cylindrical surface x . Fig. 2 . 5 - 3 i) 1 Determine the stress vector on the plane passing through P and parallel to the plane ABC shown in Fig. Thus V 4 = + 2x2ê2 223ê3 and so Therefore the unit normal a t P is = ê .2 1 .2) n = 3ê1 + d7 ê2 + 2s A 7 3 Fig.14). Finally the stress vector a t P on the plane I to R is given by or t':' = 5 $. B 52 + The equation of the plane ABC is 3x1 6x2+ 22.. 2 2.4 ) .I2 + 3 ê2+ d ê 3 .7. Computing (2.8) 2.20). For the distribution of the state of stress given in Problem 2.7. this volume integral reduces to (2.~ and since xi. substitute t:" = oiini in the surface integral and convert the result to a volume integral by (1. what form must the body force components have if the equilibrium equations (2. u ~+ pbk ~ - =.9.. 3x2 + 10x2 -k 0 + pbl = 0 O+O+2+pb2 = O O+O+O+pb3 = O These equations are satisfied when bl = -13x21p. b2 = -2Ip. \ STRESS TRANSFORMATIONS (Sec..p ~ Carrying out the indicated differentiation in this volume integral and combining with the first volume integral gives k ~ j ( ~ p k+ E i i k ( ~ j .24) are to be satisfied everywhere. p o p+ . .8. page 49. Starting with equation (2. 2.29). The de- [aqi] given in tailed calculation is best carried out by the matrix multiplication [ali] = [aip][opq] (2.27) a s ali = aiPaiqopq O r 1' = A A. The state of stress a t a point is given with respect to the Cartesian axes 0 ~ ~ byx 2 ~ the array I Determine the stress tensor Z' for the rotated axes OxíxSxl related to the unprimed axes by the transformation tensor llfi llfi 1 112 The stress transformation law is given by (2..20) from (2. b3 = 0.10.157): ~ ) ~ dV ( E ~ ~ d~ s x =~ o ~(eijkxjopk). 2.7.7) 2. Derive (2. Thus l l f i -1lfi [o$] = 1 112 -112 o . 0 .19). 21 ANALYSIS O F STRESS 61 EQUILIBRIUM EQUATIONS (Sec.19).24) directly from i given in Problem 2. 2. CHAP. p pbk)} dV = O But from equilibrium equations. the transformation matrix is determined by computing the missing entries in the table.a.11. U!. 2-23 Using the orthogonality equations aijaik = Si. For the unprimed axes in Fig. 2.. Therefore ("!.a. makes equal angles with the xi axes. Thus Fig. Show that the stress transformation law may be derived from (2. :) o o Determine the stress tensor for the primed axes specified as shown in the figure.33). - 23 IP 3 9 O and since the directions of the unprimed axes are arbitrary. 2-23.n!n! = o!.62 ANALYSIS O F STRESS [CHAP. the first row of the transformation table together with a. is known. n a . Since x..n!nj a3 1 = uijnin j and since by (1.94) nE = aijnj. n = E3 Z 3 21 IP P 54 q "N = "PqnPnq where new dummy indices have been introduced in the last term. 2 2. 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. i t is given with respect to an arbitrary set of primed or unprimed axes in the same form as ohr = o!. I t is left as an exercise for the student to show that .12. the stress tensor is given by Uij = (i . a. Since U N is a zero-order tensor. I t is first necessary to determine completely the transformation matrix A. u13= (rl2 =O - ( C ) simple shear o.. = o. [i E I!]![ Thus using the matrix form. = U . 05: = k 2 and so the quadric surface for plane stress is a general conic cylinder parallel to the zero principal axis. o...03. axis. . = o. = O ( b ) uniaxial tension o. + 2 ~ ( ~ {+. = o. a ) [{I. a.V~~ - -- ..531 [ [] O 0 0 = 05.. = ~ 1 :+ <+ = *k2 Hence the quadric surface for uniform tension is the sphere 5: + 522 + c3 2 = kk2lcr. = o. .a.CHAP.. - . ai i 2. = = a.. - .o. o. Hence the quadric surface for uniaxial tension i s a circular cylinder along the tension axis. 21 ANALYSIS O F STRESS 63 l/.the quadric surface is given in symbolic notation by the equation { * I J = *k2. = a .9) 2... = a. - . Show that the Cauchy stress quadric for a state of stress represented by = is an ellipsoid (the stress ellipsoid) when a. = u = o..32). = T ....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. - . Determine the Cauchy stress quadric a t P for the following states of stress: (a) uniform tension o. = u.... = 0. CAUCHY'S STRESS QUADRIC (Sec. and so the quadric surface for simple shear is a hyperbolic cylinder parallel to the 5.13. b and c are a11 of the same sign. 2.T.. The equation of the quadric is given by 53 - Therefore the quadric surface is the ellipsoid bcg +g a c +- ab - +k2 aL bc ' .14. (d) [Si?52.. = O (d) plane stress with a.. . From (2. Then from (2. let xB be associated with Then from (2. and let n. = 1. 2.*x3.64 ANALYSIS O F STRESS [CHAP. (3 + 2 ) n i 1 )+ n k l ) + n"' 3 = 0 Hence n:" = O. ni3)= . Zn:2' + n (22 ) + n ( 2 ) = 0 Finally. Then from (2.15 transforms the original stress tensor into the diagonal principal axes stress tensor- According to (2.1 1 6 . upon expansion. [Gj]= [ a i p ] [ a P q ] [ a Qwhich j].15.h -2 o o O 0 4 ." be the direction cosines of this axis.42). nil' = l l f i . show that the transformation tensor of direction cosines determined in Problem 2.42). n i l ) = -i/dZ.1 ) = O 1 2 -a The roots are the principal stress values = -2. The stress tensor a t a point P is given with respect to the axes 0 x 1 ~ 2 ~ ~ by the values Determine the principal stress values and the principal stress directions represented by the axes Oxtx.2 / f i .10-2.' n") 2 = -ni" and since nini = 1. 2 PRINCIPAL STRESSES (Sec. -n(3) + n(3) + = 0 2 so that ni3' = . Likewise. Let the x: axis be the direction of o ( l ) . nF) = -1 / f i 2-16.4 ) ( o. for the ~ r o b l e ma t hand becomeS o -.29). (a + 2)(0. = (ni1))2 112. let x: be associated with a ( 2 ) .42). = 4. Therefore nil)= 0.37) the principal stress values o are given by 3-0 1 1 1 -a 2 = O or.11) 2. From (9. h 1 onality conditions aijaik= Sjk. consider the axes determined in Problem 2. For example.37). Hence = O. xi and x2 are in the same horizontal plane) a t e also principal axes for the stress tensor of Problem 2.42) yield + u2] + 27% = [37 . From the orthog.17.42) yield which together with nini = 1 a r e insufficient to determine uniquely the first and second principal directions. and xy. Determine the principal stress values and principal directions for the stress tensor From (2. (2. for which the transformation matrix is llfi llfi llfi [aiil According to the transformation law (2. x.] is given by Show that the axes O x T x ~ x ~(where.17. x3 and x: are in the same vertical plane.h llfi a s is evident from Fig. For o(.29). 7 7 L 7-a ~ o(3) a T = O or (7 .0102 = 0. 21 ANALYSIS O F STRESS 65 2. For o ( 3 )= 37.12. the principal stress matrix [a*. 2-24 .h 11.CHAP.*. 2-24.) = = 0. <rc2) ~ = 0. (2.a)[-27a = 37. and therefore n:3' = n:' = ni3)= l l f i . Thus any pair of axes perpendicular to the n:3' direction and perpendicular to each other may serve a s principal axes. The transformation matrix [aii] relating the two sets of axes clearly has the known elements [aijl = 11. the remaining four elements are determined so that Fig. = a1 + o11 + = 3 + 8 + 9 = 20 0111 IIE = o1011 + + = 24 + 72 + 27 ~11~111 ~ 1 1 1 ~ 1 = 123 1111 = OIO~IOIII = (24)9 = 216 . f f / aiPikakj = ai~ajqupqairaksarsakmajnomn = (aipair)(ajqajn)(aksakm)~pq"rs"mn = aprsqnssmopqarsamn = (sprupq)(aqnamn)(8smurs) . There- 3 fore the right hand side may be written in form oij = 2 p=1 apiapjop. oij = apiaqjop. 2.39). In terms of principal values. The principal stress values of aij are o1 = 3.27). By the transformation law (2. arqoqmarm - .41). 1. 1111for the stress tensor Determine the principal stress values for this state of stress and show that the diagonal form of the stress tensor yields the same values for the stress invariants.apjy..27).aijoikokj 2.20. 2. p=1 From the transformation law for stress (2.66 ANALYSIS O F STRESS [CHAP. Prove that uijuikokjis an invariant of the stress tensor. Evaluate directly the invariants IE. there are only three terms on the right side of this equation. From (2. From (2. and in each p = q.19.40). are principal stresses.412a12 .21.031a31 = 36+ 48 + 48 .023423 .* but since o:. IE = oii = 6 + 6 f 8 = 20. IIIz = (oij(= 6(48) + 3(-24) = 216. a111= 9.9 = 123. Show that the principal stresses and the stress components oij for an arbitrary set of axes referred to the principal directions through the transformation coefficients aij are related by oij = 5 ap.111. oI1 = 8. From (2. 2 I -1lfi 11d5 1 1 fi16 116 11a lld3 O As before. IIE = ( 1 / 2 ) ( ~-~u~ ~0 ~ ~u ~ ~ ) = 01ia22 + O22O33 + 033011 . The stress tensor a t a point is given by uij = . t(n) = h(o1+ o ~ I+ o ~ ~ ~ ) Fig.4 1 2 = -12. = -15. Fig..12) the stress vector on the octahedral plane is 1 = - 6 + 011ê2 + ~ ~ ~ $ 3 ) and its normal component is . ~11)' -I. = 5. Determine the maximum shear stress a t the point and show that it acts in the plane which bisects the maximum and minimum stress planes. From (2. the so-called octahedral shear stress..5.22. 2-25 Therefore the shear component is 2. 21 ANALYSIS O F STRESS 67 2.x = n. 2-26 .23. The principal axes 0 x ~ x are ~ x related ~ to the axes of maximum shear 0 x .(011 . x ~ xby~ the transformation table below and are situated as shown in Fig. is given by UOCT = 3d(uI .VI)' With respect to the principal axes.38) the reader should verify that the principal stresses are o. From (2. 2-26. o. = 10. Show that the shear stress on this plane. The octahedral plane is the plane which makes equal angles with the principal stress directions.CHAP. the normal to the octahedral plane is given by Hence from (2.~ 1 1 1 ) ' + ( ~ 1 1 1.51b) the maximum shear value is os = (oIII. the points A ( . 2-27). x2 and x3 are in the plane of x: x.12-2. 1 2 ) on the plane I to x . 2.24.13) 2.2O shown.6 . 1 2 ) on the plane L to x2 and B ( 1 . Fig. 2-29 . 2 The stress tensor referred to the primed axes is thus given by 1 o llfi [a&] = -12. are located. 2-29. 2-27 MOHR'S CIRCLES (Sec. Fig. axes are coincident. The upper half of the symmetric Mohr's circles diagram is shown in Fig.O) represents the stress state on the plane L to x. Draw the Mohr's circles for the state of stress discussed in Problem 2. Label important points. The transformation table of direction cosines is from which a diagram of the relative orientation of the axes is made a s shown in Fig. 2-28 Fig. Point C(5.8O and fl = 53. From the angles a = 36.68 ANALYSIS O F STRESS [CHAP..5 O -2.5 The results here may be further clarified by showing the stresses on infinitesimal cubes a t the point whose sides are perpendicular to the coordinate axes (see Fig. as shown. Relate the axes 0 x 1 ~ (conjugate ~ ~ 3 to uij) to the principal axes OX?x2 x9 and locate on the Mohr's diagram the points giving the stress states on the coordinate planes of O X ~ X ~ X ~ .23. 2-28 with the maximum shear point P and principal stresses labeled. The x1 and 2. 2' and $ = cos-1113 = 70. UN = t ( : ) hn = -7019.CHAP. = 10. Accordingly tlie angles of Fig. 2-16 a r e given by 0 = . Check the results by the Mohr's diagram for this problem. and the principal axes a r e related to 0 x l x 2 x 3by the table Thus in the principal axes frame. 2-17 is a s shown in Fig. the stress vector on the plane of A n is given by the matrix product Thus t':) = -10ê1/3 .13) and the symmetry property of the stress tensor. a. and the Mohr's diagram comparable to Fig. and from @. = aijnj or Ili] [: i53uj O -315 415 213 = . 21 ANALYSIS O F STRESS 69 2. From (2.47).= 70.719.33).10ê3/3.5O. From (2. 2-30. oI1 = -5. CIII = -15. Fig.25.1 0 ê 2 .8 = cos-I 213 = 48. 2-30 . The state of stress a t a point referred to axes 0 ~ 1 x 2 is ~ sgiven by o Determine analytically the stress vector components on the plane whose unit normal + + is íi = (2/3)ê1 (1/3)ê2 (2/3)&. n. F o r oij the principal stress values a r e o. 26. The decomposition is . 2.5 = 0. Sketch the Mohr's circles for the three cases of plane stress depicted by the stresses on the small cube oriented along the coordinate axes shown in Fig. Thus uij = uMSij + ~ i j = o -2 -5 and sii = 4 + 1 -. Determine the maximum shear stress in each case. 2-32 SPHERICAL AND DEVIATOR STRESS (Sec. (b) Fig. 2-31 The Mohr's circles are shown in Fig.27. 2-31.14) 12 4 o 2. Split the stress tensor aij = into its spherical and deviator parts and show that the first invariant of the deviator is zero. Show that the deviator stress tensor is equivalent to the superposition of five simple shear states. 2-32. Fig.28. UM = akk/3 (12 + 9 + 3 ) / 3 = 8.70 ANALYSIS OF STRESS [CHAP. 2 2. 2. 29. = ajqaipaqp= aie . The same result is obtained by first calculating the principal stress values of oij and then using (2.31. = + + sh). the transformed tensor in any other coordinate system is also symmetric.(SI + + 12) = -*(*I + + s : ~ ~ ) 811 8111 SII MISCELLANEOUS PROBLEMS 2. or by the alternative form IIn. Show that the second invariant of the stress deviator is given in terms of its principal + + stress values by 111.7 = -6.-s o = (SI . Determine the principal deviator stress values for the stress tensor 10 -6 O 0 0 1 The deviator of oij is from the determinant sij = (--i - O -6 g) and its principal values may be determined Thus sI = 9.72). sS1 In terms of the principal deviator stresses the characteristic equation of the deviator stress tensor is given by the determinant SI-s o o o SI.s~~= s ~ ~ .71).30.. a1 = 16.a s the reader should show. sII1= 1. For qj. 21 ANALYSIS OF STRESS 71 where the last two tensors are seen to be equivalent to simple shear states by comparison of cases (a) and (b) in Problem 2. oII1 = 1 and hence sI = 16 . = (sIsI1 SIISIII SIIISI).s) = o O 0 SI11 .s = s3 + (81811 + SII~III+ S I I I ~ I )-~ ~ 1 ~ 1 1 ~ 1 1 1 Hence from (2. -sll ..s)(sIII.26. I I n D = (sIsII + sI1sIII+ sIIIsI). sI1 = 4 . From (2. o. o11 = 4.7 = 9.s)(sII. Since sI + + 2811181 IInD = *(~SISII 2~11~111 . + sII + sIII = 0.7 = -3.27).CHAP. Also note that since sii = 0. 2. 2. sIII = -6. sII = -3. Prove that for any symmetric tensor such as the stress tensor uij. = aipajqo. $qp.. and ni . Assume that the stress components uij are derivable from a symmetric tensor field +ij by the relationship uij = Q ~ ~ ~ c Show ~ ~ +that ~ in ~ the ..23) are satisfied. Next from (2.ni .apjnj + a.a12] = O. Determine the unit + normal ni for the plane upon which U . and since n: + + n: + ni = 1. so X = h*. -@22. Thus by the theory of equations one root (principal stress) is real.72 ANALYSIS O F STRESS [CHAP.21 - a22 = $11. Then . Let a ( .36) defining principal stress directions. + a I I I ) 2 / 4+ n3ai1.1 = -nf ..) . Prove that the principal stress values are real if 1 is real and symmetric. 2 From (2. = -2nS into this equation and solving for n.qp) $..33+ @33. ji . Then aii = 3X = 3h*. ul u I I I .$pp. n3 = 1 / 2 f i .2hni.o i 2 ) 2 4aii > 0. these equations may be combined to yield n.0 o 12 or ( a c 3 ) . p j + +jp.~.2 a 2.(oI + o I I I ) 2 / 4= (aI .1a. ) be this root and consider a set of primed axes a: of which xi is in the direction of ( r c 3 ) .@ji. = u.2XSipni - . the stress components are given by oij = Sij($qq.38) are a11 real. At the point P the principal stresses are such that 2 ~ .aijni.4[a.~. the remaining roots must be real.p.a ) .o)(aS2. of body forces the equilibrium equations (2..33). Using the results of Problem 1. Thus in analogy with (2. correspond to principal values... ~ absence ~ . <rij = Sijh -t sij = Si$* + sTi with sii = O and sTi = O.11 + $11.pi .pp . the direction cosines are found to be nl = 1 / 2 f i .pni 8 % ~ = aijSipnj + oijniS. Since the discriminant of the quadratic in square brackets + + D = (ail a i 2 ) 2 .51) construct the function H = oN . + n i ( a . The reader should apply these results to the stress tensor oij = 2.pp or explicitly 011 = $33.22 a31 = "13 . = .. $ p i .a I I I)2/16 Substituting n l = n.)/4.11 O23 = O32 . 2 2. . Use the method of Lagrangian multipliers to show that the extrema1 values (maximum and minimum) of the normal stress U .2XSiPni = 2(api. . Show that the stress tensor uij may be decomposed into a spherical and deviatoric part in one and only one way.( o i 2 ) 2 ] = (aíl .n.22 + #22.12. a I I I ) / 2 .23 - O33 = $22.33. Assume two decompositions.36. and U .33). 2. From (2. oN = nloI + 2 + ni(oI a I I I ) / 2 n. With respect to aí1-.47).pnj+ aijninj.o) [(ai1.58. a i = n. n2 = f i / 2 . cri2 o such axes the characteristic equation i s given by the determinant -o O = O o& a& 0 a(. .33 012 = a21 = [email protected] . U N = aijninj with qni = 1. + = (o. and from ASij + + sij = hSij s$ i t foilows that sij = s*. = (u. = n3.A n i n i for which aHlani = O. For real values of the stress components the stress invariants are real and hence the coefficients in (2.XSip)ni = O which is equivalent t o equation (2. [email protected]. 331 .( ~=~ 2 ~t ' n~) . the traction vectors on the unprimed coordinate + axes are t ' . ) 2.12).112 .3 = $33.CHAP. gives which by the transformation of the unit vectors becomes Likewise and so that .7). this becomes ~ Now~ since 2uPimi or 2 9 . Projecting these vectors onto the primed axes by the vector form of (2. Let the quadric surface be given in the form $ = oijlirj2 k2 = O.221 + $22.$11.g 2 2 . t ( ê z ) = -10 ê1 5ê2. 3 3 2 + $33.6) and the identity tte" = uij (2.131 .2 + "13. The normal a t any point is then V g or &$/ali = A Hence $. l ) = 15ê1 .2 f "33. Check the result by the transformation formula (2.1 + "32. 2 3 2 f g 2 2 .223 = O 2. "11.10ê. - o 20 / / r If new axes O x l x 2 x 3 are chosen by a rotation about the origin for which the trans- r315 O -4151 formation matrix is [aii] = I O 1 O I . = 0. t ' .g 1 1 . the normal to the Cauchy stress quadric a t the point whose position vector is r is parallel to the stress vector tln).3 = [email protected] + "23.. ~ ~ li = rni.38. In this way determine a. 1 3 3 = O "21.37.9.27).1 + "12. Show that. as is asserted in Section 2. = u ~ ~ oijriSjp + S = 2uPiTi. 1 1 3 + +11.$33.212 . 21 ANALYSIS O F STRESS 73 Substituting these values into qj.. A From (2.233 = O "31. At a point P the stress tensor referred to axes 0 ~ 1 ~ 2 x 3 is given by 15 -10 c. 3 ) = 20 ê3 which correspond to the rows of the stress tensor. determine the traction vectors on each L _1 of the primed coordinate planes by projecting the traction vectors of the original axes onto the primed directions. A t(G) = nlt(el) + A n2t'ez) + A n 3 t ( e 3 ) ..l f "22.3 = [email protected] + g 1 1 . 40.i with $ = xixi . + $êz $ê3. the unit normal to the plane 2 2 . The maximum shear stress value is given by - (2. From Problem 2. Fig. and because o1 = U M + SI. The state of stress throughout a body is given by t h e stress tensor Cx3 o.. o I I 1 ) / 3= 0. is related to the octahedral shear stress by the equation a.74 ANALYSIS O F STRESS [CHAP. the equilibrium equations are satisfied identically. = o -cx1 where C is an arbitrary constant.7) calculate the stress + vector on the plane 2x1 2x2 . = \/-311r.o I I I ) 2 ( o I I I. and on the sphere ( ~ 1 ) ( ~ ~ 2 ) ( ~ ~ 3 =) (9)2. In the matrix form of (2..2.SI)^ Also s . (d) The Mohr's circles are shown in Fig. hence O -4 -a o1 = fi.24) from aij. OII = 6111 + s I 1 .X Q = -7. Slrow that the second invariant of the deviator stress tensor IIp. 2 2.14) the stress vector a t P is ( c ) From (2.37).SIII)' + ( S I I I .12) the stress vector on the plane a t P is A A The normal to the sphere xixi = (9)2 a t P i s ni = $. a I I 1 = -m. ( b ) From Problem 1.54b) as os = ( o I I -I a I ) / 2= *&.6 5 ) = 0. + sII + sII1 = O and so (sl + s I I + s I I I ) 2= O or Hence 2. Thus from (2. <rOCT + + = $ d ( o I .. 2-33 . 2x2 . -4. (d) Sketch the Mohr's circles for the state of stress a t P. Since the mean normal stress a t P is = + (oI4 o.39.a I I ) 2 (o11 . (a) Substituting directly into (2.. etc. 2-33. (a) Show that the equilibrium equations are satisfied if body forces are zero. oI1 = O . ~ + + ( c ) Determine the principal stresses.81.x3 = -7 + is 3 1 + 2ê fi = 22 3 2 . ~11)' + ( S I I . the principal deviator stresses are the same a s the principal stresses. or n = 49 e 1 -. ~ O C T= SI . for principal stresses o. (b) At the point P(4. -o 7 o 7 -a -4 = o(a2 .a I ) 2 . maximum shear stresses and principal deviator stresses a t P ..22.1" 3e3.. ) l f i 2. 8111 = -39 2. ( a ) t':' = 11 ê14. = 4 . 2. oN = 3 2. S I = 31. . 2-34 Ans. = o..46.2 ê 2 + ê 3 ) / 6 . o. The stress tensor a t a point is . ( b ) o. and 30 -27 determine the principal deviator stresses. o l ~ = olI1= -1. Supplementary Problems A t point P the stress tensor is aij = [r 2 . oz2 = 1. The principal stresses a t point P are o. Ans. o. Determine the stress vector and its normal component on the octahedral plane a t P. 2-34. os = 37.. U N = 6315..715 2.41.44..43.2 so that the stress vector on some plane a t the point will be zero. and (b) oij = (: 1:) 1 2 1 Ans. . Give the unit normal for this traction-free plane. ( b ) t '" + = (21 2.42. = 8.41. = 3. Determine the normal and shear stress components on the plane BGFC of Problem 2. = 12. ( a ) a1 = 2. = 1 3 -10 4 Decompose the stress tensor oij = O 3:) into its spherical and deviator par. f 14 ê2 21 ê . Ans. Ans. SI. Ans. = -6. Show that the normal component of the stress vector on the octahedral plane is equal to one third the first invariant of the stress tensor.. Determine 0. Fig. = ( S I . 0 1 .12 ê2+ 9 ê 3 . t':' = (12êl+3ê2-6ê3)/fi.ven a s oij = 1 2 1 o o ozz 1 :) with unspecified.47. 21 ANALYSIS O F STRESS 75 2. ( b ) BGFC of the small parallelepiped shown in Fig. Determine the stress vector on the plane a t P parallel to plane ( a ) BGE. Determine the principal stress values for (a) oij = (i 1) 1 0 1 and show that both have the same principal directions. f.CHAP. ( d ) . -a. the stress field is given by the tensor o Determine ( a ) the body force distribution if the equilibrium equations are to be satisfied throughout the field.76 ANALYSIS O F STRESS [CHAP. = o -7 : 0 -27 2.58.27).2 2. ( a ) b.2&). = -4x.50. (c) the maximum shear stress a t P . page 50.. Sketch the Mohr's circles and determine the maximum shear stress for each of the following stress states: (a) uij Ans. ( b ) os = 3712 = i:: ) 0 0 (b) 0. = 7.5~3. Use the result given in Problem 1.16a/3 . to show that ~ ~ ~ is an~invariant. (d) the principal deviator stresses a t P . ~ ~ ~ ~ u ~ ~ u ~ 2. 0.49. In a continuum. 8a.-5a/3. ( a ) U . ( c ) 24.1 1 ~ 1 3 . together with the stress transformation law (2. ( b ) a.48. Ans. ( b ) the principal stress values a t the point P(a. page 39. . of a continuum.1 PARTICLES AND POINTS In the kinematics of continua. The word "particle" will denote a small vo1umetri. Fig. Deformation and Strain 3. To avoid misunderstanding. The emphasis - in deformation studies is on the initial and final configurations. the meaning of the word "point" must be clearly understood since it may be construed to refer either to a "point" in space.__-_I___ -CI to the particular sequence _____-of configurations . the word jlow is used to designate -A----. configuration.c 'element.2 CONTINUUM CONFIGURATION. Indeed. The identification of the particles of the continuum with the points of the space it occupies a t time t by reference to a suitable set of coordinate axes is said to specify the configuration of the continuum a t that instant. the term "point" _will he u e d exclusively t-o designate alõcation in fixed space.------------I----- _ _ L . No attention is given to -\intermedia~configurations --- -_ or -I -. ãnd a subsequent -(deformed) .-_ by which the deformation o c c u r s-X y contrast. . the . or "material point". For the present development i t is useful to refer the initial and final configurations to separate coordinate axes as in the figure. In brief. The term deformation refers to a change in the shape of the continuum between some ini'tial (undeformed~configuration --. or to a "point" of a continuum. 3-1 . 3-1 the undeformed configuratiofi of a material continuum a t time t = O is shown together with the deformed configuration of the same continuum a t a later time t = t .---- of motion-of a continuum. a particle is a small part of a material continuum.-continuigg-sme .. a continuum having a volume V and bounding surface S will occupy a certain region R of physical space.3 POSITION VECTOR. a point i s a place in space. a configuration history is inherent in flow investigations -for which the specification of a time-dependent velocity field is given. DISPLACEMENT VECTOR In Fig. 3. DEFORMATION AND FLOW CONCEPTS At any instant of time t. 3. I. the components of the displacement vector are observed from (3.I. from which U~ = (3.... in the initial configuration a representative particle of the continuum occupies a point POin space and has the position vector with respect to the rectangular Cartesian axes OXlX2X3. = I. (3.I P= SKP and A A ekDep= 8kp. which are defined by the dot products of unit vectors as A A A A - e. e.7) into ( 3 4 . and a. in which the components UK and uk are interrelated through the direction cosines A a. = a. 3-1 serves to locate the origin o with respect to O.. . Their use here is to emphasize the connection of certain expressions with the coordinates (XiXaX3). the orthoganality conditions between spatial and material axes take the form Tn Fig.. From the geometry of the figure. u = b+x-X (3. 3 Accordingly. are constants. but their use as sixmmation indices is restricted to this section. which are called the material coordinates.9) Since the direction cosines a. Here lower-case letters are used as subscripts to identify with the coordinates (x1x2x3)which give the current position of the particle and are frequently called the spatial coordinates. of the particle).1) and will appear as such in severa1 equations that follow. In the remainder of the book upper-case subscripts or superscripts serve as labels only.78 DEFORMATION AND STRAIN [CHAP.1O) . respectively. 3-1 the vector u joining the points Po and P (the initial and final' positions. Upper-case letters are used as indices in (3.5) A or alternatively as U = U..9) to obey the law of transformation of first-order Cartesian tensors. In the deformed configuration the particle originally a t PO is located a t the point P and has the position vector when referred to the rectangular Cartesian axes oxixnx3. From (1. is known as the displacement vector.. This vector may be expressed as A u = ukek (3.89) the unit vector êk is expressed in terms of the material base vectors IK as A A ek = ~ILKIK Therefore substituting (3.a. lnasmuch as Kronecker deltas are designated by the equations I K . The vector b in Fig. The relative orientation of the material axes OXlX2X3 and the spatial axes 0 x 1 ~ 2 ~is3 specified through direction cosines a.3) No summ:. as they should..tion is implied by the indices in these expressions since k and K are distinct A h indices. which results in the direction cosine symbols a. the particles of the continuum move along various paths in space. the material and spatial axes are assumed superimposed and hence only lower-case indices will be used.15) is a continuous one-to-one mapping with continuous partial derivatives. A necessary and sufficient condition for the inverse functions to exist is that the Jacobian should not vanish. the two mappings are the unique inverses of one another. with b = O. the Lagrangian description given by the equations has the inverse Eulerian formulation.14) is known as the Lagrangian formulation.13) in which only lower-case subscripts appear. In the remainder of this book.4 LAGRANGIAN AND EULERIAN DESCRIPTIONS When a continuum undergoes deformation (or flow).et .1) 1 . (3. As a simple example. unless specifically stated otherwise.10) becomes O X I X Z Xand u = x-X (3. = X . If.CHAP. This motion may be expressed by equations of the f orm which give the present location xi of the particle that occupied the point (XIX2X3) a t time t = O.14) may be interpreted as a mapping of the initial configuration into the current configuration. 3. so that (3.X Z .X. ~ 3 ) . I t is assumed that such a mapping is one-to-one and continuous. This description may be viewed as one which provides a tracing to its original position of the particle that now occupies the location ( x l . the motion or deformation is given through equations of the form in which the independent variables are the coordinates xi and t. Accordingly. The description of motion or deformation expressed by (3. becoming Kronecker deltas. Xl = -XI + x2(et.11) In Cartesian component form this equation is given by the general expression However. with continuous partial derivatives to whatever order is required. (3. .. for superimposed axes the unit triads of base vectors for the two systems are identical. on the other hand. Also. the description is known as the Eulerian formulation. If (3. 31 DEFORMATION AND STRAIN 79 Very often in continuum mechanics it is possible to consider the coordinate systems ~ oxixexs superimposed.$ (3.e-t .14). as was also assumed for (3.12) reduces to U. In symbolic notation.15) with respect to x j produces the tensor dXi/dxj which is called the spatial d e f o r m a t i o n gradient. Thus Partial differentiation of (3. these tensors are given in terms of the deformation gradients as the material gradient and the spatial gradient In the usual manner. or the spatial displacement gradient duiIdxj.13). = --$idXi is applied from the right (as shown explicitly in the equation). 3 3. From (3. DISPLACEMENT GRADIENTS Partial differentiation of (3.80 DEFORMATION AND STRAIN [CHAP. the matrix f o m s of J and K are respectively and . dxildXj is represented by the dyadic a in which the differential operator V. This tensor is represented by the dyadic having a matrix form The material and spatial deformation tensors are interrelated through the well-known chain rule for partia1 differentiation.5 DEFORMATION GRADIENTS. as a difference of coordinates.11) with respect to Xj produces the tensor dxildXj which is called the material d e f o r m a t i o n gradient. when it appears as the consequent of a dyad. which expresses u. The matrix form of F serves to further clarify this property of the operator V. Partial differentiation of the displacement vector ui with respect to the coordinates produces either the material displacement gradient a~Jax~. the sqiare of the differential element of length between P and Q is From (3.CHAP.28) From (3. 31 DEFORMATION AND STRAIN 81 3. G . 3-2 The square of the differential element of length between POand QOis ( d X ) 2 . '* Fig.= d X * d X = d X i d X i = Gij d X i d X j (3. d X (3.6 DEFORMATION TENSORS.28) may be written in which the second-order tensor is known as Cauchy's deformation tensor. The neighboring particles which occupy points POand &O before deformation.14) the distance differential here is axi d x i = --dXj or d x = F-dX dXj so that the squared length ( d ~in) (3.32) ~ may be written d ~ dk ~ k ( d ~ ) '= ---. 3-2 the initial (undeformed) and final (deformed) configurations of a continuum are referred to the superposed rectangular Cartesian coordinate axes OX1X2X3and ox1x2x3.dXi dXj = GijdXi d X j or (dx)' = d X . the distance differential d X i is seen to be dXi d X i = =dxj or dX = H-dx so that the squared length ( d X ) 2in (3.15). - In the deformed configuration. move to points P and Q respectively in the deformed configuration.34) dXi d X j in which the second-order tensor is known as Green's deformation tensor . FINITE STRAIN TENSORS In Fig. 24) is substituted into (3.34) and (3.. .37).27) respectively. or LG = &(Fc F . The fundamental measure of deformation is the difference (dx)' . ' (dx)' = dX ( F .26) and (3. F . If this difference is identically zero for a11 neighboring particles of a continuum. *' / Using (3. the result after some simple algebraic manipulations is the Lagrangian finite strain tensor in the form In the same manner.41) into (3. .41) may be written directly from (3.I) / is called the Lagrangian (or Green's) finite strain tensor.(dX)' for two neighboring particles of a continuum is used as the measure of deformation that occurs in the neighborhood of the particles between the initial and final configurations.82 DEFORMATION AND STRAIN [CHAP. ( d ~ ). the result is the Eulerian finite strain tensor in the form The matrix representations of (3. .25) is substituted into (3. 3. a rigid displacement is said to occur.28). .39).30). INFINITESIMAL STRAIN TENSORS The so-called small deformation theory of continuum mechanics has as its basic condition the requirement that the displacement gradients be small compared to unity. Using (3. the same difference may be expressed in the form in which the second-order tensor is called the Eulerian (or Almansi's) finite strain tensor.36) in which the second-order tensor L. the finite strain tensors in (3.I) dX = dX -2Lc dX (3.32) and (3. 3 The difference (dx)' .7 SMALL DEFORMATION THEORY. which may be expressed in terrns of the displacement gradients by inserting (3. If the displacement gradients are small.40) and (3.36) and (3.38) respectively.36) and (3.38) reduce to infinitesimal strain tensors. this difference may be expressed in the form 01.40) and (3. if dXi/dxj from (3.(dX)'. . and the resulting equations represent small deformations. An especially useful form of the Lagrangian and Eulerian finite strain tensors is that in which these tensors appear as functions of the displacement gradients. 'rhus if dxi/dXj from (3. The resulting tensor is the Lagrangian infinitesimal strain tensor.46) is the Lagrangian form of the relative displacement vector. The vector v dui = u i < Q ~ ) .44) if both the displacements and displacement gradients are sufficiently small. 3-3 Here the parentheses on the partia1 derivatives are to emphasize the requirement that the derivatives are to be evaluated a t point Po. ROTATION VECTOR In Fig. u) = S(J + Jc) Likewise for duiIdxi < 1 in (3. the relative displacement vector can be written as Fig. if the displacement gradient components duildXj are each small compared to unity. 3-2). Neglect. a Taylor series expansion for uiPO) U(PO) rnay be developed in the neighborhood of Po. d X . 3-3 the displacements of two neighboring Q particles are represented by the vectors u:"o' and uiQO' (see also Fig. there is very little difference in the material and spatial coordinates of a continuum particle. 31 DEFORMATION AND STRAIN 83 In (3.CHAP. Assuming suitable continuity conditions on the P displacement field. or L = E (3. the relative displacement vector dui rnay be written a s . It is also useful to define the u n i t relative displacement vector d ~ l d Xin which d X is the magnitude of the differential distance vector dXi. Accordingly the material gradient components duildXj and spatial gradient components dui/dxj are very nearly equal. the product terms rnay be dropped to yield the Eulerz'ccn infinitesimal strain t e m o r . LINEAR ROTATION TENSOR.8 RELATIVE DISPLACEMENTS. then Since the material displacement gradient duildXj rnay be decomposed uniquely into a symmetric and an antisymmetric part. the product terms are negligible and rnay be dropped. which is denoted by or i. Equation (3. = r i . Thus l i .U ! P ~ ) or 1 du = ~ ( Q o ). 3. which is denoted by If both the displacement gradients and the displacements themselves are small.41). so that the Eulerian and Lagrangian infinitesimal strain tensors rnay be taken as equal. These derivatives are actually the components of the material displacement gradient. = i(uvx+ V. Accordingly if vi is a unit vector in the direction of d X i so that d X i = V .40).u ( P o ) dx is called the relative displacement vector of the particle originally a t Qo with respect to the particle originally a t Po. po ing higher-order terms in this expansion. This infinitesimal rotation may be represented by the rotation vector in terms of which the relative displacement is given by the expression The development of the Lagrangian description of the relative displacement vector.54) is the Eulerian linear strain tensor eij.36) may be replaced by the linear Lagrangian strain tensor lij. and that expression may now be written .55). the linear rotation tensor and the linear rotation vector is paralleled completely by an analogous development for the Eulerian counterparts of these quantities. 3 The first term in the square brackets in (3.48) is recognized as the linear Lagrangian strain tensor lij.the relative displacement a t that point will be an infinitesimal rigid body rotation.84 DEFORMATION AND STRAIN [CHAP. Accordingly the Eulerian description of the relative displacement vector is given by and the unit relative displacement vector by Decomposition of the Eulerian displacement gradient auilaxj results in the expression The first term in the square brackets of (3. the finite Lagrangian strain tensor Lij in (3. The second term is known as the linear Lagrangian rotation tensor and is denoted by In a displacement for which the strain tensor lij is identically zero in the vicinity of point Po. The second term is the linear Eulerian rotation tensor and is denoted by From (3. the linear Eulerian rotation vector is defined by in terms of which the relative displacement is given by the expression 3.9 INTERPRETATION OF THE LINEAR STRAIN TENSORS For small deformation theory. the diagonal terms of the linear strain tensor represent normal strains in the coordinate directions. 3-4 Thus the normal strain for an element originally along the X z axis is seen to be the component 122. d X for small deformations. become after deformation the line elements P Q and P M with respect to the parallel set of local axes with origin a t P. d X i l d X = dX3ldX = O. The original right angle between the line elements becomes the angle O.d X ) ( d x d X ) = 2Ljd X i d X j (dx)' .59) is recognized as the change in length per unit original length of the differential element and is called the normal strain f o r the line element originally having direction cosines dXildX. 31 DEFORMATION A N D STRAIN 85 + (dx)' . Fig. From (3. In Fig. (3. located with respect to the set of local axes a t POas shown in Fig. I n general.58) Since d x . d x (3. respectively.59) is applied to the differential line element PoQo. therefore. Likewise for elements originally situated along the X i and X 3 axes. 3-4. 3-5 the line elements POQOand POMOoriginally along the X2 and Xs axes. Because POQO here lies along the X Z axis. dXzldX = 1 and therefore (3.(dX)' = ( d x . a first order approximation gives the unit vector a t P in the direction of Q as . 3-5 The physical interpretation of the off-diagonal terms of lij may be obtained by a consideration of the line elements originally located along two of the coordinate axes.59) becomes Fig. the result will be the normal strain f o r that element. this equation may be put into the form The left-hand side of (3.CHAP.( d X ) 2 = ( d -~d X ) ( d + ~ dX) = d x 21.46) and the assumption of small deformation theory. When (3.59) yields normal strain values l1l and 133 respectively. 86 DEFORMATION AND STRAIN [CHAP. 3 and, for the unit vector a t P in the direction of M, as Therefore or, neglecting the product temi which is of higher order, duz dua COS o = - + - = 2JZ3 aX3 aX2 Furthermore, taking the change in the right angle between the elements as y2, = 712 - O, and remembering that for the linear theory r,, is very small, i t follows that Y23 - sin yZ3 = sin ( ~ 1 - 2 O ) = cos O = 24, (3.65) Therefore the off-diagonal terms of the linear strain tensor represent one-half the angle change between two line elements originally a t right angles to one another. These strain components are called shearing strains, and because of the factor 2 in (3.65) these tensor components are equal to one-half the familiar "engineering" shearing strains. A development, essentially paralleling the one just presented for the interpretation of the components of lij, may also be made for the linear Eulerian strain tensor eii. The essential difference in the derivations rests in the choice of line elements, which in the Eulerian description must be those that lie along the coordinate axes after deformation. The diagonal terms of cij are the normal strains, and the off-diagonal terms the shearing strains. For those deformations in which the assumption lij = aii is valid, no distinction is made between the Eulerian and Lagrangian interpretations. 3.10 STRETCH RATIO. FINITE STRAIN INTERPRETATION An important measure of the extensional strain of a differential line element is the ratio dxldX, known as the stretch or stretch ratio. This quantity may be defined a t either the point POin the undeformed configuration or a t the point P in the deformed configuration. Thus from (3.34) the squared stretch a t point POfor the line element along the unit vector i% = dXldX, is given by Similarly, from (3.30) the reciproca1 of the squared stretch for the line element a t P along the unit vector íi= dxldx is given by For an element originally along the local X2 axis shown in Fig. 3-4, = êz and therefore dXildX = dXJdX = O, dXzldX = 1 so that (3.66) yields for such an element Similar results may be determined for A:;,, and A:;,). CHAP. 31 DEFORMATION AND STRAIN 87 For a n element parallel to the xz axis after deformation, (3.67) yields the result with similar expressions f o r the quantities 1 / ~and ~ 1~1 ~ : ,~ In ~ )general, . A,&, is not equal to A,;2, since the element originally along the X2 axis will not likely lie along the x2 axis after deformation. The stretch ratio provides a basis for interpretation of the finite strain tensors. Thus the change of length per unit of original length is and f o r the element POQOalong the X2 axis (of Fig. 3-4), the unit extension is therefore This result may also be derived directly from (3.36). For small deformation theory, (3.71) reduces to (3.60). Also, the unit extensions L ( 1 )and L(,) a r e given by analogous equations in terms of Ltt and L33 respectively. F o r the two differential line elements shown in Fig. 3-5, the change in angle y,, = ~ 1- 26 is given in terms of A , ; ~ ,and A , ; ~ ,by When deformations are small, (3.72) reduces to (3.65). 3.11 STRETCH TENSORS. ROTATION TENSOR The so-called polar decomposition of a n arbitrary, nonsingular, second-order tensor is given by the product of a positive symmetric second-order tensor with a n orthogonal second- order tensor. When such a multiplicative decomposition is applied to the deformation gradient F, the result may be written in which R is the orthogonal rotation tensor, and S and T are positive symmetric tensors known a s the right stretch tensor and left stretch tensor respectively. The interpretation of (3.73) is provided through the relationship dxi = (dxi/dXi)dXj given by (3.33). Inserting the inner products of (3.73) into (3.33) results in the equations From these expressions the deformation of dXi into dxi a s illustrated in Fig. 3-2 may be given either of two physical interpretations. In the first form of the right hand side of (3.74), the deformation consists of a sequential stretching (by S ) and rotation to be followed by a rigid body displacement to the point P. In the second form, a rigid body translation to P is followed by a rotation and finally the stretching (by T). The translation, of course, does not alter the vector components relative to the axes Xi and X i . 88 DEFORMATION AND STRAIN [CHAP. 3 3.12 TRANSFORMATION PROPERTIES OF STRAIN TENSORS The various strain tensors Lij, Eij, lij and cij defined respectively by (3.379, (3.39), (3.42) and (3.43) are a11 second-order Cartesian tensors as indicated by the two free indices in each. AccordingIj for a set of rotated axes Xil having the transformation matrix [bij]with respect to the set of local unprimed axes Xi a t point PO as shown in Fig. 3-6(a), the components of L&and 1; are given by and l!. 11 - b.I P b.34 G,4 or lf=B*L.B, (3.76) (b) Fig. 3-6 Likewise, for the rotated axes xi' having the transformation matrix [aij] in Fig. 3-6(b), the components of E; and are given by and C!. = a1P. a34. €P 4 or E'=A*EaAC (3.78) By analogy with the stress quadric described in Section 2.9, page 50, the Lagrangian and Eulerian linear strain quadrics may be given with reference to local Cartesian coordinates q i and (Si a t the points Po and P respectively as shown in Fig. 3-7. Thus the equation of the Lagrangian strain quadric is given by CHAP. 31 DEFORMATION AND STRAIN 89 and the equation of the Eulerian strain quadric is given by Two important properties of the Lagrangian {Eulerian) linear strain quadric are: 1. The normal strain with respect to the original {final) length of a line element is inversely proportional to the distance squared from the origin of the quadric PO {P) to a point on its surface. 2. The relative displacement of the nedghboring particle located a t QO {Q) per unit original {final) length is parallel to the normal of the quadric surface a t the point of intersection with the line through POQO{PQ). Additional insight into the nature of local deformations in the neighborhood of POis provided by defining the strain ellipsoid a t that point. Thus for the undeformed continuum, the equation of the bounding surface of an infinitesimal sphere of radius R is given in terms of local material coordinates by (3.28) as After deformation, the equation of the surface of the same material particles is given by (3.30) as (dx)' = Cijdxidxj = R2 or ( d X ) 2 = d x . C . d x = R2 (3.82) which describes an ellipsoid, known as the material strain ellipsoid. Therefore a spherical volume of the continuum in the undeformed state is changed into an ellipsoid a t Po by the deformation. By comparison, an infinitesimal spherical volume a t P in the deformed continuum began as an ellipsoidal volume element in the undeformed state. For a sphere of radius r a t P, the equations for these surfaces in terms of local coordinates are given by (3.32) for the sphere as and by (3.34) for the ellipsoid as The ellipsoid of (3.84) is called the spatial strain ellipsoid. Such strain ellipsoids as described here are frequently known as Cauchy strain ellipsoids. 3.13 PRINCIPAL STRAINS. STRAIN INVARIANTS. CUBICAL DILATATION The Lagrangian and Eulerian linear strain tensors are symmetric second-order Cartesian tensors, and accordingly the determination of their principal directions and principal strain values follows the standard development presented in Section 1.19, page 20. Physically, a principal direction of the strain tensor is one for which the orientation of an element at a given point is not altered by a pure strain deformation. The principal strain value is simply the unit relative displacement (normal strain) that occurs in the principal direction. For the Lagrangian strain tensor lij, the unit relative displacement vector is given by (3.477, which may be written n Calling 1;"' the normal strain in the direction of the unit vector ni, (3.85) yields for pure strain (Wij 0) the relation A A 1;"' = Ljnj or 1'"' = 1.2 (3.86) 87) leads to the relationship which together with the condition nin+= 1 on the unit vectors ni provide the necessary equations for determining the principal strain value 1 and its direction cosines ni. 3-8. the first-order approximation of this ratio is the sum Do = l(l>+ 1<2>+ Z(3) = IL (3. The roots of (3. second and third Lagrangian strain invar.94) Fig. . consider a differential rectangular parallelepiped whose edges are parallel to the principal strain directions a s shown in Fig. IIIL = O (3. then 1:" = lni = /aijnj or 1'.87) Equating the right-hand sides of (3. Therefore which upon expansion yields the characteristic equation of lij.92) and has an important physical interpretation. 1 = (ZiZj . IIIL = I&jl = det L (3. and z(31.86) and (3.90) where IL = &i = tr L. Nontrivial solutions of (3.91) are the first. is a principal direction with a principal strain value 1.88) exist if and only if the determinant of coefficients vanishes.90 DEFORMATION AND STRAIN [CHAP. the cubic P . The change in volume per unit original volume of this element is called the cubical dilatation and is given by For small strain theory. 3-8 .) = l$ = $102 (3.iants respectively. To see this.IL12 + IIJ .90) are the principal strain values denoted by l(l).Zjij). 3 If the direction n. The first invariant of the Lagrangian strain tensor may be expressed in t e m s of the principal strains as IL= Zii = Z(1) Z(2) Z(3) + + (3. = e..CHAP. IIE = '(1)'<2) '(2)'(3) + '(3)'(1) IIIE = c(1)E(2. C.. In the Eulerian description (the Lagrangian description follows exactly the same pattern).... For plane strain perpendicular to the x3 axis. are determined in exactly the same way as their Lagrangian counterparts.15 PLANE STRAIN. - . The appropriate displacement components for this case of plane strain are designated by .. if Lagrangian and Eulerian deviator tensor components are denoted by dij and eij respectively. :i %] (3. plane strain is referred to as plane deformation since the deformation field is identical in a11 planes perpendicular to the direction of the zero principal strain.E.) linear strain tensor is given by t. a state of plane strain is said to exist a t that point. The Eulerian strain invariants may be expressed in terms of the principal strains as IE = E(i. + + E<3... the resolution expressions are lij = dij+6. o In many books on "Strength of Materials" and "Elasticity". As before. E ( 3 ) The cubical dilatation for the Eulerian description is given by 3..... the principal directions and principal strain values C. or [c] = [. - i33 'kk or E = ED +- l(tr E) 3 The deviator tensors are associated with shear deformation for which the cubical dilatation vanishes.. Therefore it is not surprising that the first invariants di and eii of the deviator strain tensors are identically zero.) o o 01 [Eij] = ['O) i] E. the strain tensor has the form Cij = k' €. E. 31 DEFORMATION AND STRAIN 91 With regard to the Eulerian strain tensor eij and its associated unit relative displacement vector E:"'. MOHR'S CIRCLES FOR STRAIN When one and only one of the principal strains a t a point in a continuum is zero. a state of plane strain parallel to the x1x2plane exists and the (i: i: .14 SPHERICAL AND DEVIATOR STRAIN TENSORS The Lagrangian and Eulerian linear strain tensors may each be split into a spherz'cal and deviator tensor in the same manner in which the stress tensor decomposition was carried out in Chapter 2.99) When xi and xz are also principal directions. L3 + 6. 3. if x3 is taken as the direction of the zero principal strain.-Ikk or L = LD+- I(tr L) i3 3 3 and ci. the displacement vector may be taken as a function of xl and xz only. possess a solution for an arbitrary choice of the strain components cii.16 COMPATIBILITY EQUATIONS FOR LINEAR STRAINS If the strain components rii are given explicitly as functions of the coordinates. 3-9. The principal normal strains are labeled as such in the diagram. A graphical description of the state of strain a t a point is provided by the Mohr's circles for strain in a manner exactly like that presented in Chapter 2 for the Mohr's circles for stress. and the Mohr's circles diagram constructed from these.92 DEFORMATION AND STRAIN [CHAP. which are twice the tensorial shear strain components. The system is over-deterrnined and will not.43) cii -(- 1 8% = 2 axj + 2) may be viewed as a system of six partia1 differential equations for determining the three displacement components ui. in general. a typical case of plane strain diagram is shown in Fig. Corresponding to the plane stress Mohr's circles. 3-9 3. 3 Ul = u1(x1. Therefore if the displacement components ui are to be single-valued and continuous. Mohr's strain circles are useful for reporting the observed data.99). The necessary and sufficient conditions for such a displacement field are expressed by the equations There are eighty-one equations in a11 i n (3. some conditions must be imposed upon the strain components. Fig. and the maximum shear strain values are represented by points D and E. These six written in explicit and symbolic form appear as . Frequently in experimental studies involving strain measurements a t such a surface point. The state of strain a t an unloaded point on the ~ 1 2 '1 bounding surface of a continuum body is locally plane strain.103) but only six are distinct.xz) u3 = C (a constant. usually taken as zero) Inserting these expressions into the definition of eij given by (3.43) produces the plane strain tensor in the same form shown in (3. Usually three normal strains are measured a t the given point by means of a strain €1 rosette. the six independent equations (3.xz) Uz = uz(x1. For this purpose the strain tensor is often displayed in the form Here the yij (with i # j ) are the so-called "engineering" shear strain components. Ax2)l(l. u. 3-10 below shows this displacement pattern.+ .99). = A(x3.5) 3. Inverting the given displacement relations to obtain X.) For the displacement field of Problem 3.A2). X3 = (2. = O. the six equations in (3.A2). = O.Ax2)l(1.Ax2)l(1.3) has the equation xi2 +9 ~ = . (Inter- pret the physical meaning of this. Likewise the particle line X1 = O. . the circular surface becomes the elliptical surface (1+ A2)xl .A2). x2 x3 = 1 after displacement.X3 = AX.. = x. From (3.1-3. ~ Fig. From these results i t is noted that the originally straight line of material particles expressed + + by X. (a) By the direct substitutions X2 = (x.Ax3)l(1. Determine the displacement vector components in both the material and spatial forms.104) reduce to the single equation where E is of the form given by (3. 3. the displacement components in material form are ul = xl . With respect to superposed material axes Xi and spatial axes xi. .. the spatial components of u are U.8x223 + 5x3 = 3 which when referred to its prinzipal axes x i (at 45O with xi. . - - 8x3 axi d (dc23 - 6.A2). For plane strain parallel to the ~ 1 x 2plane. .Ax3)l(1. + XE X : = 1/(1.S.CHAP.AZ).. x s = X3 AX2 where + A is a constant. Solved Problems DISPLACEMENT AND DEFORMATION (Sec. 31 DEFORMATION AND STRAIN 5.A2) and X3 = (x3 . 3. u3 = x3 .-'€31 - ~ X S8x1 ax2 axs 8x1 8x2 Compatibility equations in terms of the Lagrangian linear strain tensor lij may also be written down by an obvious correspondence to the Eulerian form given above. . x2 = x3. .X l = 0. 2 2 = X2 AX3.1 determine the displaced location of the material particles which originally comprise (a) the plane circular surface XI = 0. ~3 = A(x2 . u2 = x2 . X2 X3 = lI(1+ A ) occupies the location xl = 0.4Ax2x3 + (1+ A2)xi = (1 . For A = 4. (b) the infinitesimal cube with edges along the coordinate axes of length d X i = d X .X2 = AX3. X2 = (x. X2 = X3 becomes after displacement x1 = 0.A2).AZ). this is bounded by the ellipse 5x22 .1. i = 2.13) directly. Sketch the displaced configurations for (a) and (b) if A = . the displacement + field of a continuum body is given by xl = Xi.Ax3)l(l . 3-10 Fig.. = u3 = O . + 2êg. = 0 . u3 = AX. 3-12 + ul = (-cos x2)AX. Its displacement is u = 4ê1 4 ê 3 + and since x = X + u. U 3 = 0 where A is a constant. 3-12) the A transformation tensor apK = e p IK is cosx. (sin x2)Ax3x1cos x2 = O u. The initial and displaced positions of the cube are shown in Fig. = (sin x. 2 ) . = u. Determine the displacement components for cylindrical spatial coordinates xi if the two systems have a common origin.X3 (sin x 2 ) A X l X 3 + = (-cos x2)Ax3xl sin 2 .Thus since Cartesian and cylindrical coordinates are related through the equations X . u . . direction propor- tionally to their distance from the origin. the displacements of the edges of the cube are readily calculated. For the edge X . u.. X . For superposed material and spatial axes.)AXlX3 + = ( s i n ~ x 2 ) A x l x 3 (cos2 x2)Ax1x3 = A x 1 x 3 Ug = o This displacement is that of a circular shaft in torsion. u2 = A X 3 and the particles on this edge are displaced in the X .. 0 .4. a displacement field is given by U l = . 3-11. From the geometry of the a h s (Fig. For the edge X 1 = O = X2. = xl cos x.. Determine the displaced location of the particle originally a t ( 1 . u. sin x .1. = X.9) gives Fig. O aPK (-7x2 0) and from the inverse form of (3. For the edge X .94 DEFORMATION AND STRAIN [CHAP. X 3 = x. + 3.)AX. = u3 = O. the displacement vector of a body is given + + by u = 4x:ê1 X 2 X i ê 2 Xixiê3. = X 3 = O.9) up = aPKUK. 3-11 ( b ) From Problem 3.X3 + (cos x. 33..A X 2 X 3 . With respect to rectangular Cartesian material coordinates Xi. U2 = A X l X 3 .. = u . X 2 = X 3 = O. its final position vector is x = 5ê1 6ê3. = X. X 2 = x1 sin x. The original position vector of the particle is X = ê. equation (3. 3 Fig. X 3 = X3. 1 O (e2.8.3) and B ( 3 . x3 = 2. Thus the displaced position of particle A is given by xl = -11. x3 = e2X3 where e is a constant. x3 = X3(l x:)+from which F is readily found to be Direct substitution of the calculated tensors F and J into (3.24). x3 = 27.X1)ê3.9. 3. 3. x2 = 2X1 + X2 .1) From (. and of particle B. + 7ê2 25ê3. J is found to be + Since x = u X. The Lagrangian description of a deformation is given by XI = XI X3(e2. the displacement field may also be described by equations xl = X1(l + x:).7 the position vector of C becomes U = 82. Show that field u this deformation is such that ( a ) plane sections remain plane. 7ê2 + + 25ê3 which is clearly parallel to V. Show that the Jacobian J does not vanish and determine the Eulerian equations describing this motion.4X3.13).CHAP. This is an example of so-called homogeneous deformation. (b) straight lines remain straight. J = F .9.7 determine the displaced position of the position vector of particle C ( 2 . Show that the two vectors remain parallel after deformation.3 13 3 . + From (3. X2 = x2 + x3(e-4 .e-2). xi = Xi + ui = Xi + AijXj = (a.)X. A continuum body undergoes the displacement u = ( 3 x 2 . s3= -Xl + + 4X2 X3.+A.7. assuming superposed material and spatial axes. 6 . + + x2 = X2 X3(e2. For the displacement field of Problem 3. J = O 1 (e2. Determine the displaced position of the vector joining particles A(1.13). 6 .5. + 3. x2 = 6.Determine inde. Therefore the displaced position of the vector joining A and B may be written V = 82. The general formulation of homogeneous deformation is given by the displacement i = AijXj where the Aij are constants'or a t most functions of time. A displacement field is given by u = + êl x:x2ê2 x2x3ê3. 3 ) which is parallel to the vector joining particles A and B.26) verifies that the equation is satisfied. 3.6. x2 = -1. By the analysis of Problem 3. i.+ pendently the material deformation gradient F and the material displacement gradient J and verify (3.I).. + x2 = X2(1 x:). X1 = x1 + x3(e-2 .I)..X3. the spatial coordinates for this displacement are xl = Xl 3X2 . O O (e2) Inverting the equations.X3)$2 + + (4x2. X3 = e-2x3.I.e-2) = e2 # 0. From the given displacement vector u. 31 DEFORMATION AND STRAIN 95 3. xl = -3.4X3)ê1 (2x1.0. 6 ) .I).16). (a) From (3. 96 DEFORMATION AND STRAIN [CHAP. 3 + According to (3.16) the inverse equations Xi = (aij B i j ) x j exist provided the determinant + Iaij Aijl does not vanish. Assuming this is the case, the material plane piXi cu = O becomes + + + &(aij B i j ) x j cu = O which may be written in standard form a s the plane h j x j a. = O + where the coefficients = &(aij Bij). + ( b ) A straight line may be considered a s the intersection of two planes. I n the deformed geometry, planes remain plane a s proven and hence the intersection of two planes remains a straight line. 3.10. An infinitesimal homogeneous deformation ui = AijXj is one f o r which the coefficients Aij are so small that their products may be neglected in comparison to the coefficients themselves. Show that the total deformation resulting from two successive infinitesimal homogeneous deformations may be considered as the sum of the two individual deformations, and that the order of applying the displacements does not alter the final configuration. + + Let xi = (aij A i j ) X j and x l = (aij B i j ) x j be successive infinitesimal homogeneous displace- + + + + + ments. Then xl = (aij Bij)(Sjk A j k ) X k = ( S i k Bik Aik B i j A j k ) X k . Neglecting the higher + + + order product terms B i j A j k , this becomes xl = ( S i k Bik A i k ) X k = (Oilc C i k ) X k which represents the infinitesimal homogeneous deformation uE' = XE - X i = CikXk = (Bik +Aik)Xk = (Aik + B i k ) X k = tii + U; DEFORMATION AND STRAIN TENSORS (Sec. 3.6-3.9) 3.11. A continuum body undergoes the deformation xi = X i , x2 = X Z AX3, x3 = X3 AX2 + + where A is a constant. Compute the deformation tensor G and use this to determine the Lagrangian finite strain tensor LG. From (3.35), G = F,* F and by (3.20) F i s given in matrix form a s so t h a t Gij] = [i 1l o f 2 o yAj Therefore from (3.37), L~ = S ( G - I) = 2 O 2A A2 3.12. For the displacement field of Problem 3.11 calculate the squared length ( d ~of) the ~ edges OA and OB, and the diagonal OC after deformation for the small rectangle shown in Fig. 3-13. Using G a s determined in Problem 3.11 in (3.34), the squared length of the diagonal OC is given in matrix form by o ( d ~ = ) ~ [O, dX2,dX3] O = [: l:Aj[il 1 +A2 2A + A"(dX2)2 + 4 A d X 2 dX:, + ( 1 + A"(dX:,)" (1 Fig. 3-13 Similarly for O A , (dx)' = ( 1 + A2)(dX'2)2; and for O B , (dx)z = ( 1 + A2)(dX3)2. CHAP. 31 DEFORMATION A N D S T R A I N 97 3.13. Calculate the change in squared length of the line elements of Problem 3.12 and check the result by use of (3.36) and the strain tensor LG found in Problem 3.11. Directly from the results of Problem 3.12, t h e changes are: ( a ) f o r OC, ( d ~ -) (dX)2 ~ = ,(I+A Z ) ( ~ X ; dx;) + + 4 A dX, dX3 - ( d x ; + d x ; ) = A Z ( ~ X : dx;) + + 4 A dX, dX3 ( b ) for OB, (dx)2 - (dX)2 = ( 1 + A ' ) d x ; - dx; = A2 d x ; (c) f o r O A , (dx)2 - (dX)2 = ( 1 + A,) d x i - d ~ = i A2 d ~ i . By equation (3.36), f o r OC (dx). - ( U ) 2 = O , dX2 -4 L:O o A2 2A o'][A;] A2 dXg + = A Z ( ~ X ~d x ; ) + 4 A dX, dX3 The changes f o r O A and O B may also be confirmed in the same way. 3.14. For the displacement field of Problem 3.11 calculate the material displacement gradient J and use this tensor to determine the Lagrangian finite strain tensor LÇ. Compare with resillt of Problem 3.11. F r o m Problem 3.11 the displacement vector components a r e ul = O , u, = A X 3 , z(3 =AX, so that O O A2 Thus f r o m (3.40) ~ L G O A O O A O O O A2 O 2A A2 t h e identical result obtained in Problem 3.11. + 3.15. A displacement field is given by X I = X I A X 2 , x2 = X Z AX3, x s = X3 A X l where + + A is a constant. Calculate the Lagrangian linear strain tensor L and the Eulerian linear strain tensor E. Compare L and E for the case when A is very small. F r o m (3.42), / 0 A O\ 10 O A \ /o A A\ Inverting the displacement equations gives ul A(A2xl + x2 - A x 3 ) l ( l + A3), u2 = A ( - A x l + A2x2 + x 3 ) / ( 1+ A3), ~3 = A(xl -Ax2 + A2x3)/(1+ A3) f r o m which by (3.43) -i2) A2 1 -A A2 - A + K,) :,) A 2E = (K = 1 +A3 + t2 h(-: When A i s very small, A? and higher powers may be neglected with the result t h a t E reduces to L. 98 DEFORMATION AND STRAIN [CHAP. 3 3.16. A displacement field is specified by u = X? X2ê1 (X2 - x ; ) ê 2 + X; X3ê3. Determine + the relative displacement vector du in the direction of the -XZ axis a t P ( 1 , 2 , -1). Determine the relative displacements uai - UP for Q1(1,1, -I), Q2(1,3/2, -I), Q3(1,7/4, -1) and Q4(1, 1518, -1) and compare their directions with the direction of du. For the given u, the displacement gradient J in matrix form is ~ a u i l ~ x i= so that from (3.46) a t P in the -X2 direction, [: 2XlX2 x; -i] + Next by direct calculation from u, up = 2 ê1 ê2- 4 ê3 and UQ, - - êl - ê3. Thus UQ, - up = -e, - e2 + 3 e,. Likewise, uQ2- up = - ê2(-e1 + 3.5 ê3)/2, ug3 - up = (- e, - e, A A + + 3.75ê3)/4, uQ4- up = (- ê, - ê2 3.875ê3)/8. I t is clear that a s Qi approaches P the direction of the relative displacement of the two particles approaches the limiting direction of du. 3.17. For the displacement field of Problem 3.16 determine the unit relative displacement vector a t P ( 1 , 2 , -1) in the direction of Q(4,2,3). The unit vector a t P in the direction of Q is $ = 3ê1/5 + 4ê3/5, so that from (3.47) and the matrix of J a s calculated in Problem 3.16, 3.18. Under the restriction of small deformation theory, L = E. Accordingly for a displacement field given by u = (xl - ~ 3 ) ~ ê (x2 l + + ~ ~ ) -~ x1x2ê3, ê 2 determine the linear strain tensor, the linear rotation tensor and the rotation vector at the point P(O,2, -1). Here the displacement gradient is given in matrix form by [auilaxj] = 2(x1- x3) O 0 2(x2+ x3) -a21 - x3) 2(x2 + 23) 1 which a t the point P becomes r2 Decomposing this matrix into its symmetric and antisymmetric components gives [eijI + [wijI = -2 o -1 o Therefore from (3.56) the rotation vector wi has components wl = -1, w2 = w3 = 0. CHAP. 31 DEFORMATION AND STRAIN 99 3.19. For the displacement field of Problem 3.18 determine the change in length per unit + length (normal strain) in the direction of Y = (8êi - ê2 4ê3)/9 a t point P(O,2, -1). From (3.59) and the strain tensor a t P a s computed in Problem 3.18, the normal strain a t P in the direction of Y is the matrix product 3.20. Show that the change in the right angle between two orthogonal unit vectors G and í; in the undeformed configuration is given by G.2L.fi for small deformation theory. Assuming small displacement gradients, the unit vectors in the deformed directions of $ and $ + + a r e given by (3.47) a s ($ J * c ) and (li J .fi) respectively. (The student should check equations (3.61) and (3.62) by this method.) Writing J in the equivalent form $ * J c and dotting the two displaced unit vectors gives (as i n (3.63)), cos e = sin ( ~ 1 2 . -e) = sin y,, = y,, or y,, = [$ $ * J,] + [ f i + J . C ] = $ . f i + G - ( ~ + ~ , ) . f i + Y . ~ , . ~ . f i . Here J,.J is of higher order f o r small displacement gradients and since Y Ifi, 4. fi = O so t h a t finally by (3.42), y,, = $ 21. c. 3.21. Use the results of Problem 3.20 to compute the change in the right angle between + + G = (8ê1- êz 4ê3)/9 and i; = (4ê1 4ê2- 7ê3)/9 a t the point P(O,2, -1) f o r the displacement field of Problem 3.18. Since L = E for small deformation theory, the strain tensor eij = lij and so a t P y,, = [8/9, -119, 4/91 -719 STRETCH AND ROTATION (Sec. 3.10-3.11) 3.22. For the shear deformation XI = Xi, xz = XZ AX3, + x 3 = X3 + AX2 of Problem 3.11 show t h a t the stretch A,;, is unity (zero normal strain) for line elements parallel to the XI axis. For the diagonal directions OC and D B of the infinitesimal square OBCD (Fig. 3-14), compute A:;, and check the results by direct calculation from the displacement field. From (3.66) and the matrix of G a s determined in Problem 3.11, the squared stretch f o r & = êl is Fig. 3-14 Likewise for OC, & = ( ê 2 +&)I& and so 67). An infinitesimal rigid body rotation is given by ul = . = dL AdL. calculate A(. . F o r this displacement. + = (1 . a r e equal only if 2 is the deformed direction of A. + + x.c 1 .22 one obtains X1 = xl.Ax2)l(1.] = o o Therefore [Sij] = [ G ] = Relative to the coordinate axes Xi. & = (-ê2+ ê 3 ) / f i and so A(m. The principal axes of G a r e given by L0 2A i t ~ 2 J a 4 5 O rotation about X1 with the tensor in principal form [G~.A X 3 . By a polar decomposition of the deformation gradient F f o r the shear deformation + + xl = X1. calculated for OC.B X l + A X 2 where A . 3. f o r DB.25. Thus (dx)2 = 2(1+ A)2(dL)2 and since dX = fid ~ the . determine the right stretch tensor S together with the rotation tensor R..22. R = FS-1. rl + c-+ B2 -A R -.A2)2/(1. Thus A:.and from (3. C are very small constants.C X 2 + B X 3 . ~ and F o r the displacement field of Problem 3. X2 = (2.35).A2). us = C X l .73). f o r & = ( $ 2 + & ) / f iand show that it agrees with A : ~ .A2) from which the Cauchy deformation tensor C may be computed.22. Show that the stretch is zero (S = I) if terms involving squares and products of the constants are neglected. for the diagonal OC in Problem 3. .66). The stretch ratios n . The 2 diagonal element OC does not change direction under the given shear deformation. By r1 O O 1 (3. 3 From the displacement equations the deformed location of C is x. 2 Similarly. x2 = dL AdL. u3 = .A ) 2 .22. the stretch tensor S = 6 .24. = (1 . Show that the principal values of S a r e the stretch ratios of the diagonals OC and DB determined in Problem 3.100 DEFORMATION AND STRAIN [CHAP. F or here [Gij] = 1 0 1 + A2 2A I. the decomposition is I n this example the deformation gradient F is its own stretch tensor S and R = i. x2 = X2 AX3. squared stretch (dx/dX)2 + is (1 A)2 a s calculated from (3. h(. 3. B . Inverting the displacement equations of Problem 3. Then using (3. x3 = X B AX2.A)2 = (1 A)2 which is identical with A(m. = 0.. G = F.Ax3)/(l. X3 = ($3 . In the polar decomposition of F. This is the result of the coincidence of the principal axes of LG and EA f o r the given shear deformation. .. 3.27. axk axk --. axq d(b. ax:. a. o 11fi becomes [L: = [i .b .12-3. b .24.. Using the definition (3. 3. ax. 31 DEFORMATION AND STRAIN 101 Neglecting higher order terms.fi 1 o .Sij which by the stated transformation becomes ax. 3 7 ) [Lij] = [! i ]: o which f o r principaI axes given by the transformation matrix [aij = [: o -1lfi 11. x3 = X3 f i ~ 2 show that the + principal directions of LG and Ea coincide a s was asserted in Problem 3. o -l+fi 3. 3 9 .28. the principal strains a r e the roots of . For the shear deformation X I = XI. A certain homogeneous deformation field results in the finite strain tensor .37). The student should verify these calculations. Determine the principal strains and their directions f o r this deformation.26. Being a symmetric second order Cartesian tensor. xz = X2 f i X 3 .zL) ax. Likewise from ( 3 . From ( 3 . By (3..ax.37). [Ed = which by the same transformation matrix Iaij] r 0 O O 1 is converted into the principal-axes form [E:] = . Lij = - 2 1 íax. this becomes [Gijl = O 0 1 STRAIN TRANSFORMATIONS AND PRINCIPAL STRAINS (Sec. show that the Lagrangian finite strain tensor Lij transforms a s a second order Cartesian tensor under the coordinate transformations xi = bjixf and X( = bijXj.CHAP.S (since bpkbqk = S p q ) m Z n~ p q ax:.14) + 3. For the homogeneous deformation xi = fiX I .x? + A?.. = 2. 3. 1. o r alternatively by inverting the given displacement equations and substituting into X i x i = 1.29.X.84) the sphere xixi = 1 resulted from the ellipsoid X G X = 1. = 8. L(2. = -2. = 6. .. For the deformation of Problem 3. = 3.66) using [aij] above. Show that this ellipsoid has the form xS/A:.102 DEFORMATION AND STRAIN [CHAP.X~ = 1.82).. o r This ellipsoid is put into the principal-axes form 3 ~ : 6x22 + +2 : 1 by the transformation ~ = .30. A:. A&) = 2. determine the spatial strain ellipsoid and show that i t is of the form A:. The transformation matrix f o r principal directions is [aijl = 1 1 216 3. 3 Thus L ( .29. + + By (3. xj = fiX3 . This equation is put into + the principal-axes form xS13 x. t A:. x. By (3. Note also t h a t the principal stretches may be calculated directly from (3..24) a s O 1 which by the transformation [aij] = is put into the principal form with principal stretches A:~../Ak) x i l ~ ~=. the material strain ellipsoid is xS + + + xi x3 x2x3 = 3./6 + x i l 2 = 1 by the transformation = E[ o 11.X2 determine the material strain ellipsoid resulting from deformation of the spherical surface + + X: X: X: = 1. Lo. x.h llfi From the deformation equations. the stretch tensor S = 6 i s given (calculation is similar to t h a t in Problem 3. = 2 x 2 . . 1. . Determine the shear strain C. + A. xi = ( A i j S i j ) X j .xz 3x3.2. = 5 and so the Note PLANE STRAIN AND COMPATIBILITY (Sec. A 45" strain-rosette measures longitudinal strain along the axes shown in Fig. If the Aij a r e very small.3.33. AV/dVo = ~ ~ ~ ~ + S( i l A Aj 2 S k 3~ + ~S i l S8j 2 A k~3 +~8i18j28k3) 8 ~ -~ 1' = Ali + AZ2+ A33 F o r linear theory the cubical dilatation lii = auilaXi..)) d X ( .. + A33. with = (êl+ê2)/\/S a s the unit vector in the si direction. At a point P. 8 3 ) the original volume d V o becomes a skewed parallelepiped having edge lengths d x i = ( A i c n ) Si(.. 3-15. a t the point. + + 133111 . show that the result reduces to the cubical dilatation. From+ + ( 1 . For the finite homogeneous deformation given by ui = AijXj where Aij a r e constants.I3 e$ = (1 O -1 u). ) = c(. C .( l l l l l l + 112112 + 113113 + l2l12l + l22l22 + l23l23 + 131131 + l32l32 + 133133)l = 111122 + l22l33 f 133111 .. d X 3 along the + coordinate axes. Fig.e k k / 3 . . Determine the principal strains and the principal deviator strains e.. for this deformation.16) 3.. u z = x1 7x2. which f o r ui = A i j X j is lii = A.34. 3.(1S2 + 123 + 1321) 3. 3. €.15-3.. = 7 x 10-4 inlin. . + + + + us = -3x1 4x2 4x3. i t i s given here by strain deviator is eij = or in principal-axes form by t h a t e ( .. d X 2 .32. direct expansion of the second equation of ( 3 .) .13. 9 1 ) yields IIL = +[(111 + lZ2 + 133)ljj . n = 1.CHAP. Also.S j 2 ) ( A k 3 Sk3) d X 1 d X 2 d X 3 . A linear (small strain) deformation is specified by ul = 4x1 . 3-15 . . 1 By ( 3 . Thus by ( 3 . F o r the given deformation. C . 2 -1 i) = and its principal-axes form .1 112 122 1 /+ 111 131 113 133 1 1 122 132 123 133 1 Expansion of the given determinants results in IIL = ll1l2. 5 9 ) . = 4 x 10-4. I n comparison. + Then If the Aij a r e very small and their powers neglected.31. Since cij is the symmetrical p a r t of the displacement gradient a u i l d x j . 31 DEFORMATION AND STRAIN 103 3. = 5 x lOP4. Verify by direct expansion that the second invariant IILof the strain tensor may be expressed by IIL = I 111 1. Consider a rectangular element of volume having original dimensions d X l .( l l j l l j + 12j12j + 13j13j)] = &[(I11 + l22 + 133)(111 + l22 + l33) .( 1 f 2 + l i 3 + l i l ) . 1 0 9 ) this deformed element has the volume d V = e i j k ( A i l S i 1 ) ( A j 2 4. determine an expression for the change of volume per unit original volume. .I llfi = 4 x 1Op4 Therefore 12 x 10-4 + 2 ~ . 3 11. Construct the Mohr's circles for the case of plane strain 1 '" and determine the maximum shear strain. the other circles a r e drawn a s shown. 3-16. 3-16. Verify the result analytically. Since = O is a principal value f o r plane strain./I O] [: 5 x 10-4 e..104 DEFORMATION AND STRAIN [CHAP. = -2 x 10-4. Note t h a t a rotation of -45O about xg would correspond to point E in Fig.:Op4 ]-['. €23 = & ) and D a r e established a s the diameter of the larger inner circle in Fig. . 3-1 7 Fig. 3-16. A rotation of 30' about the x . With the given state of strain referred to the xi axes. 3-16 Fig. 2 3.€ . 3-18 Next a rotation of 45O about the x: axis (90° in the Mohr diagram) results in the xl axes and the associated strain tensor e / j given by the first two rows of which represent the state of strain specified by point F in Fig.= 4 x 10-4 or . axis (equivalent to 60° in the Mohr's diagram) results in the principal strain axes with the principal strain tensor given by Fig.35. .. 7 . the points B ( e Z Z= 5. O8 ê1. F o r the rigid body rotation represented by ul = 0. C a r e very small.AX3 -BXl + AX.51). Hence du = UQ. 3.03X2. The same result is obtained by (3..: A A A e1 e2 e3 du = . + + x3 = . determine the relative displacement of Q(3.AX3..O8 el .36. 1 -C B F o r the given displacements. this strain tensor i s zero and the displacement reduces to a rigid body rotation. Thus (3. From (3. and up = . Determine the rotation vector w f o r the infinitesimal rigid body rotation.+-+-- au.37). A displacement field is defined by xl = X I . From (3. 31 DEFORMATION AND STRAIN 105 3.40) from its definition (3. Show that this displacement represents a rigid body rotation only if the constants A.02ê. xz = C X i + Xz .24). the rotation vector is A A A e1 e2 e3 W = - 2 â/dXl a/ax.39. uz = -0. u3 = -0. F = and from (3.4) with respect to P(3. The student should c a r r y out the details.02X3. Derive the indicial form of the Lagrangian finite strain tensor L.1 o . + cê3 -CXz + BX3 C X l . A A A A From the displacement equations.003ê:.37.1. o .O3 .03ê1 + .0.37) may be written Li$ = [(Ski f - "k) ax. The state of strain throughout a continuum is specified by Are the compatibility equations for strain satisfied? Substituting directly into (3.12êz . ( Skj + axj- a?1i) - = Ldui 2 ax. dx. MISCELLANEOUS PROBLEMS 3. If products of the constants a r e neglected.O57 e.37)..12 e2.03X3. o.CHAP./dXi = S i j + aui/dXi. axi ax..50). axj auk auk I 3.O2 O = -.B X l + AX2 X3. + sê. B.O6 e3. alax. = Aê.104).02X1 + 0. UQ = . with w = .38.UI>= -. of (3.003ê3. 4 ) .CX2 BX3. a11 equati'ons a r e satisfied identically. . + o em E23 €23 C O S ~0 €7 E33 [-si! .20. -E.20.41. and the shear strain r. 3-19. 3-20 3. cos e.. 4 = = [o. sin e cos e "22 .. sin O cos O [. E. . By equation (3.106 DEFORMATION A N D S T R A I N [CHAP.cosesinel[~ o €J[:.65) and Problem 3.-sin 20 2 Fig. For a homogeneous deformation the small strain tensor is given by What is the change in the 90" angle ADC depicted by the small tetrahedron OABC in Fig.:] - .ê2-êl)/& . For a state o£ plane strain parallel to the ~ 2 x 3axes. From the result of Problem 3. = (êl .€23 COS e] 0 sin2 e + 9. cos2 e + 2e23sin e cos O i.ê2)/\/S and 2 = (2 ê3. when the primed and unprimed axes are oriented as shown in Fig. and D is the midpoint of A B ? The unit vectors Y and a t D a r e given by f. 3 3.59). determine expressions for the normal strain r.40.-2 eos 2e + '22 €83 sin 20 2 Similarly from (3. "42 = ~o. 3-20 if OA = OB = OC.-'22 + €33 + . 3-19 Fig."33 = €23 COS20 . sin2 o - . sin e. If E .4 . Calculate the strain invariants for each of these tensors and show their equivalence.B. . + + 3.42.50)the rotation vector w = A ê1 A ê2. u. by (3.A X l AX2. x3 = X3 . determine E . 3.43.31. and by (3. . 111. From (3. . a t the point.c ) / 6 and E.43). 3-21 Also f o r the x. IE = 5 4 4 = 13. the directions shown in Fig. u3 = -AXl + AX. e. LG = 0.40).59)with L = E. + + By (3. f o r the x{ direction. = c. 3-21. . From the displacement equations. u2 = 3x1 . Finally IIIE = 5(16). = a. = 24 + + 18 12 = 54.. 21 By (3.. . . IE*= 6 4 3 = 13.4 = 72. = AX3. eii = 1 3 B O 1 which is of the form of L 0 O 01 (3. and E.' direction Solving simultaneously f o r e12 and ez2 yields e l z = ( b . Likewise IIE = + + 19 + + 19 16 = 54. A so-called delta-rosette for measuring longitudinal surface strains has the shape of the equilateral triangle A and records normal strains E ~ ~ .96)..99). + determine the finite strain tensor LG. At a point the strain tensor is given by rij . Show that the displacement field ul = A x l 3x2. 31 DEFORMATION A N D STRAIN 107 3.45. E .AX3.CHAP. 11. Fig. The student should check these calculations.44. in i"/". by (3.. + 3. = (6)(4)(3)= 72.Bxz. r l l = b. = -AX3..Q = (-a + 2b + 2c)/3. which is zero if A = B. x z = X z .95) and Problem 3. Since u . Show that if A is very small the displacement represents a rigid body rotation. u3 = 5 gives a state of plane strain and determine the relationship between A and B for which the deformation is isochoric (constant volume deformation). For the displacement field xl = X I AX3. i) and in principal form . E.the cubical dilatation is D eii = A . If A is small so t h a t A2 may be neglected. Let & =im2ê. F dX3 . Thus there is zero strain along the X3 axis and f o r the element a t 60' to the X3 axis. F o r the shear displacement of Problem 3. and solving simultaneously m2 = ffi12. A(:.34) sin Yzn = dx2 -. x: 9%.. = 2L23 d ..G dX2 d d ~ . Also mi + rn. x2 = Xz . 3 3. Ans. be the unit normal i n the direction of zero strain. x2 = XZ. A(.46. determine the Lagrangian and Eulerian finite strain tensors L.I. F. m3 = f 1. or by (3.36).47. .2) = d m .68).72) expressing the change in angle between the coordinate directions Xp and X3 under a finite deformation. Idx21 ldxzl d d ~ .2X. Determine the shear angle y. Show that (3. = sin-12/$? 3.. = 1. Also from (3.50.37). Fig.8) = A A n. A A l e3 A A A = 2Lz3 since e2 e3 = 0. Then sin yz3 = cos (r12 . e. Supplementary Problems 3. 4 1 this reduces to sin yz3 = duz/aX3 + du3/ax2 = 2123. G -e3= e.353 and (3. = 7112.A (21. G e.47.ê. .66) gives sin yz3 = A A e.2X1 2X2. Then from (3.. A The student should verify this result by using the relation & 2LG m = O derived from (3.) A(&] A Next from (3.. = X. = + X3 .66).2~~ e3 + e. Let y . + m. x. o r m2 = O.49. = 3+ 3. n. G ê. ~= ~ / -2 8 be the angle change a s shown in Fig.- dx3 - - dX.I. o or 7mE + + 4~ m2m3 3mi = 3. x3 = X3 + 2 ~ z / f i . y. Derive equation (3.47 (Fig. + Given the displacement field x. determine the equation of the ellipse into which the circle X: +X: = 1 is deformed. = 1. 2X3.65) when displacement gradients are small.d m If aui/dX. 3. For the simple shear displacement X I = XI. and so sin y?.72) reduces to (3.G dX3 Now dividing t h e numerator and denominator of this equation by IdX21 and IdX31 and using (3. determine the direction of the line element in the XzX3plane for which the normal strain is zero. 3-22). and E*.108 DEFORMATION AND STRAIN [CHAP. since AZcm. 3-5. etc. . 3-22 .48. f o r the deformation of Problem 3. 2 -2 o 2 -2 o x.. . + I) *e3= ê2. Ans. - - A e. m.33) and (3. Determine e12 and ez2 a t the location. 32: 3xlx2+ X3 -z2/2 o A= 2112 2x3 3.60.):. calculate the volume change and show that i t is zero if A . and by a polar i:: (-i :-:' decomposition of F find the rotation tensor R and the right stretch tensor S. Ans. R = ?3 ( 2 s = .1 ) ] / 2 . + x2 = X 2 . For the displacement field of Problem 3. determine the deviator tensor eij and calculate its principal values.55. Draw the Mohr's circle for the state of strain giv'en in Problem 3. 3-23.'B F = -2 2 3. ez2 = 1 X 10-4. A n s . Hint: See equation (3.57. Fig.A X 1 . 4 o E: = (: :) 419 o 419 ' -i). Show that the first invariant of 1.1 ) (h:ê2) + + .52. = 0). 1. IE= 6 . Using eij of Problem 3. 31 DEFORMATION AND STRAIN 109 3. (e31 3. Determine the strain tensor eij and check whether or not the compatibility conditions are satisfied. The strain tensor a t a point is given by in the direction of = $. 3.50 determine the deformation gradient i.I2 -$. =4 3.1 ) (h2.I2 ê3/fi + qj = (i. IIE = -4.. . For a delta-strain-rosette the normal strains are found to be those shown in Fig.50. A displacement field is given by ul = 3 ~ ~ x 2 2u2 . y.54 and E : given in Problem 3.55.54 and note that and fl of that problem are principal directions (hence y. e ( AV ) = 6 ./\/S. Ans. .2885 X 10-4 r.58.61. 3.CHAP. e l z = -0.I.54.59. 1 2 -1 o 2 Ans. 3-23 . For eij of Problem 3. = 0. IIIE = -24. For the displacement field x l = X l AX. x3 = X 3 . = 2x3xl.54 and determine the maximum shear strain value. Determine the principal-axes form of the two tensors of Problem 3.. calculate the three strain invariants from each and compare the results. 3..l = 1 x 10-4 10-4 3. y. Ans.56.54. = ( c 4). may be written in terms of the principal stretches as ILG - . = 2 x 10-4 is a very small constant..51.x1x2.53. u3 = x i .-3 fi and the shear strain between Determine the normal strain and fl = -ê1/2 + + $212 $. eij - o -2 3.68). 3. Determine the principal-axes form of eij given in Problem 3. Verify this result analytically. Ans. Ans. X ~ .1 MOTION. The material derivative (also known as the substantial. for example. t) = p*(~i.t) X ~=. The instantaneous position xi of a particle is itself a property of the particle. any physical property of the continuum that is expressed with respect to a specific particle (Lagrangian. t) = p*(x>t) (4. In fluid flow.15).2) to exist is that the Jacobian determinant J = laxiIaXil (4. if the material description of the density p is given by p = P(Xi. As indicated by (3. whereas the Eulerian description concerns itself with a particular region of the space occupied by the continuum.2) are the inverses of one another.6) . Physically. 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. FLOW.1) and (4. MATERIAL DERIVATIVE Motion and flow are terms used to describe the instantaneous or continuing change in configuration of a continuum.t). For example. t) or by the inverse of these equations in terms of the spatial coordinates (Eulerian description) The necessary and sufficient condition for the inverse functions (4. or comoving. xi(X.t) the spatial description is obtained by replacing X in this equation by the function given in (4. Thus the spatial description of the density is P = p(Xi(x.14) and (3. Therefore adopting the symbol dldt or the superpositioned dot as representing the operation of material differentiation (some books use DIDt). 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 motion of a continuum may be expressed either in terms of material coordinates (Lagrangian description) by ~i = x ~ ( X I .3) should not vanish. the Lagrangian description fixes attention on specific particles of the continuum. the ( O ) velocity vector is defined by vi = dxi/dt = X i or V = dxldt = X (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. t) or x = x(X. t). in plasticity studies. t) ~r p = p(X(x. Flow sometimes carries the connotation of a motion leading to a permanent deformation as. or spatial description). The material derivative of the particle's position is the instantaneous velocity of the particle. Since (4. Chapter 4 Motion and Flow 4. or material description) may also be expressed with respect to the particular location in space occupied by the particle (Eulerian.2).t) or p = p(X. the word denotes continuing motion. however. The second term X on the right in (4..11) which gives xi = ui Xi + + (or x = u X).8) is sometimes written .e.2 VELOCITY.ir9 Iocation and is accordingly called the local rate of change.6).13).6) as vi = dxildt (or v = dxldt). vector or tensor property of a continuum that may be In general. 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. the material derivative (4. When the property Pij.. if Pij. In (4. i. then .10) may be written which immediately suggests the introduction of the material derivative operator which is used in taking the material derivati. in taking the derivative. The first term on the right of (4. 4. From (4.ves of quantities expressed in spatial coordinates. if the displacement is expressed in the Lagrangian form ui = ui(X. 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 same particles are involved. INSTANTANEOUS VELOCITY FIELD One definition of the velocity vector is given by (4.jX7t)] X to emphasize that the X coordinates are held constant.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. An alternative definition of the same vector may be obtained from (3. ACCELERATION. 41 MOTION AND FLOW 111 . Thus the velocity may be defined by since X is independent of time.10) gives the rate of change a t a particular [dP'. This term is sometimes written t)] to emphasize that x is held constant in this differentiation. is any scalar.CHAP. t). For steady motion. The function vi = vi(x. If the velocity is given in the Lagrangian form (4. . 4.19 ) This decomposition is valid even if vi and avildxj are finite quantities.t ) or V = V(X. t) (4.VxV ) = (D+V) (6.. dvildxj (or Yij). STREAM LINES. then 4. VORTICITY.lat = O . STEADY MOTION A path 2ine is the curve or path followed by a particle during motion o r flow. among them rate of strain. the displacement is in the Eulerian form zci = u ~ ( xt). 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. then In (6.3 PATH LINES. strain rate and velocity strain tensor.15) the velocity is given implicitly since i t appears a s a factor of the second term on the right.4 RATE OF DEFORMATION. This tensor may be decomposed into its symmetric and skew- symmetric parts according to y. stretching. then If the velocity is given in the Eulerian form (4. stream lines and path lines coincide. on the other hand.112 MOTION AND FLOW [CHAP. The skew-symmetric tensor is called the vorticity or spin tensor.14). The material derivative of the velocity is the acceleration. -dvi = aXj or Y = 9(vvx+ V xV ) + 4(vVx . The symmetric tensor is called the rate of deformation tensor. 4 If. NATURAL STRAIN INCREMENTS The spatial gradient of the instantaneous velocity field defines the velocity gradient tensor. Many other names are used for this tensor.15).16 ) is said to specify the instantaneous velocity field. . 26) becomes dvi = (Dij + Vij)dxj O?. widely used in flow problems in the theory of plasticity (see Chapter 8). The vector defined as one-half the vorticity vector.29) is known as the vorticity vector. Thus if in the equation the differentiations with respect to the coordinates and time are interchanged.25) represents the components known as the natural strain increments.(4. 4-1 If the rate of deformation tensor is identically zero (Dij = O).29) shows that the vorticity vector is the curl of the velocity field.dx (4..23) when that equation is rewritten in the form deij = Dijdt or dE = D d t The left hand side of (4. if d t axj (e) is replaced by dxj dt . . i. 41 MOTION AND FLOW 113 The rate of deformation tensor is easily shown to be the material derivative o'f the Eulerian linear strain tensor.5 PHYSICAL INTERPRETATION OF RATE OF DEFORMATION AND VORTICITY TENSORS In Fig.26) dxj in which the partia1 derivatives are to be evaluated a t P. Associated with the vorticity tensor. and the motion in the neighborhood of P is a rigid body rotation. 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. For this reason a velocity field is said to be irrotational if the vorticity tensor vanishes everywhere within the field. 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. dv = (D+V). The symbolic form of (4.CHAP. The result is expressed by the equation A rather interesting interpretation may be attached to (4.xv (4. In terms of Dij and Vij. 4.27) Fig.e. j or q=v. the vector defined by qi = cijk V k . 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. if vi is in the direction of a coordinate axis.35) The off-diagonal elements of Dij are shear rates. tkis parallelepiped is moved and distorted. the continuum particles which occupied the differential volume element dVo in the initial state now occupy the differential volume element dV. the "line of particles" Fig. is d = di(V'~i = Dijvjvi or d = 3. If the initial volume element is taken as the rectangular parallelepiped shown in Fig. AREA AND LINE ELEMENTS In the motion from some initial configuration a t time t = O to the present configuration a t time t.dx. since the rate of change of length of the line element dxi per unit instantaneous length is given by dvi d i ( ~=) . invariants.31) The components of the rate of deformation tensor have the following physical interpretations. In fact.. 4-2. because of the relationship (3..D.32) and..27).33) dx a3 dx the rate of deformation in the direction of the unit v.33).30) is called the rate of rotation vector. .n. the concepts of principal axes.e2 = D22 A (4. second-order tensor. Since Dij is a symmetric. MOTION AND FLOW [CHAP. 4 ai = q = iikvk.6 MATERIAL DERIVATIVES OF VOLUME.34). a rate of deformation quadric. the relative velocity of a neighboring particle separated from P by dxi is given as dvi = ~. 4-2 .36) Due to the motion. Thus for pure deformation. = Dij dxj or dv = D d x (4. say $2. d = d22 or d = ê2 D . or d v = 0 x dx (4.. then by (1. principal values. 4. The diagonal elements of Dii are known as the stretching or rate of eztension components. from (S.ector V .D (4.28). A = D. For a rigid body rotation such as that described by (4. and a rate of deformation deviator tensor may be associated with it. dxi . dv. Also.10) dV = dXiêi X dXzê2 dX3ê3 = dXi dX2dXa (4. equations of compatibility for the components of the rate of deformation tensor.j or a = &q = &Vxx v (4.18).34) From (4.Dijvj or d'") = ü a $ (4. being a measure of the rate of change between directions a t right angles (See Problem 4. but because the motion is assumed continuous the volume element does not break up. dxi (axilaXj)dXj between the material and - spatial line elements. analogous to those presented in Chapter 3 for the linear strain tensors may be developed. the particles initially making up the area d S O nnow fill an area element represented by the vector dSnt or dSi. a differential element of area having the magnitude dSOrnay be represented in terms of its unit normal vector ni by the expression dSoni. it is now possible to obtain the material derivative of dV as d since dVo is time independent. It rnay be shown that ax..vXv. and accordingly rnay be written .37) is seen to be equal to dV = c. For the current configuration of the continuum in motion. Therefore the differential volume element dV is a skewed parallelepiped having edges dx?'. 41 MOTION AND FLOW 115 that formed dX1 now form the differential line segment dx. dxi3)and a volume given by the box product dV = dxC1)x dxC2) dxC3) = Éijk dxi(l)dxj2' dxk' (4.44) becomes d dvi d -(dx2)=2-dxkdxi or z(dx2) = 2dx.38) where J = laxiIdXjl is the Jacobian defined by (4.CHAP.'" = (dxilaXi)dX1.(dVo)= O.38).28) and hence (4. The material derivative of the Jacobian dt J rnay be shown to be (see Problem 4. However. Using (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.3).39) rnay be put into the form For the initial configuration of a continuum. dSi= J L d X j or dS= JdX*Xvx (4. so that . axi --.46) is symmetric in i and k.46) dt dxk The expression on the right-hand side in the indicial form of (4. ax. and (4.dX d X 2d X 3 = J dVo ax.37) But (4. Similarly dX2 becomes d ~ : = ~ (axildxn) ' dX2 and dX3 becomes dx13' = (axiIdX3)dXa. ax3 (4. dx:".dx (4. since dxi = (axildXj)d X j . (x. In particular. (t) = Jv P < . 4 or. . t) dSp] (4. ... . t) dV = V + ~-(aX vPPt P . t)] through the bounding surface S of V. d d . Therefore which.157).52) may be put into the form dt P. .. Some properties are defined as 'integrals over a finite portion of the continuum.(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. the operations of differentiation and integration may be interchanged.116 MOTION AND FLOW [CHAP.[P. . '(xft. (4. the second term of the right-hand integral of (4.[P:. . (x.49) where V is the volume that the considered part of the continuum occupies a t time t. .7) and (4. t)] dSp (4.20).(t) is and since the differentiation is with respect to a definite portion of the continuum (i. from (4. dV at + 1 v. t))] dV By using Gauss' theorem (1. ( x .e. 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 (4.9).(dx2) = 2 d ~D ' d~ (4. & 1Q: .411. a specific mass system).48) dt dt 4. let any scalar. and the material derivative then given by P: ( x . vector or tensor property be represented by the volume integral P2j.(x.12) as dldt = d/& + v~aldxp. upon carrying out the differentiation and using (4.7 MATERIAL DERIVATIVES OF VOLUME. The procedure for determining the material derivatives of surface and line integrals is essentially the same as that used above for the volume integral. .. t) dSp = i& [Q: .56) .. . as bef ore. . ~ ) ~ v (4. t ) d v = . ..(x. then. . . . results in Since the material derivative operator is given by (4.53) may be converted to a surface 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.54) This equation states that the rate of increase of the property Pij.(x.(x.(dx2) = 2 Dij dxidxj ~r . The material derivative of Pij. . xi = Xi. .e-t(X2 . X . Using the relationships + X z 4. . from (4.t ) ]d S p . = et(Xz X3)/2 e-t(X2 . t )d x p =12 [R:.x3)/2. x. . Note that in each description when t = 0. = et(X2 X3)/2 e-t(X. 4.e-t(X2 . Solving these simultaneously the inverse equatjons become X 1 = x. xz = et(X2 X3)/2 e c t ( X 2.59) Differentiating the right hand integral as indicated in (4.X3.e c t ) X Z . x2 = X3(et .X3 = et(x. t ) d x .43). 4. u3 = x3 . u2 = x2 - + + x3)/2 .56) yields dQij .).) and X . + 12 [ ~ ~ . + + + v2 = et(X2 X3)/2 . .] (4. .x3)/2. = e-t(x2+xj)/2 + + et(xz .et(x2.X3)/2. s .x. or in the Eulerian form ul = O.(~ .e-t(x2 x3)/2 et(xz.e .I). = O. . + + + x3 = et(X2 X3)/2 . results in the material derivative d -[R dt ij. Accordingly. + + From the second and third equations. . 41 MOTION AND FLOW 117 and.x3(et. [Q* -"d S p ] (4. + + + X2.t.14).X3 = et(x2.X3 = e-t(x2 x3) and X 2 .60) Solved Problems 9 MATERIAL DERIVATIVES.x3)/2.et(x2.X3)/2 ...(x. X3 = x3.CHAP.Xi may be written in either the Lagrangian form ul = O.x3)/2. Determine the velocity components in both their material and spatial forms.58) the material derivative is given by &1 R&. By (4. . ( x .X3)/2. u3 = et(X2 X3)/2 . . + By (4. + Q* (x.( t ) = IR. . VELOCITY..59).X3)/2.X3)/2.45). (4.3) 4. t)] dt dx.x3) these components reduce to v . The spatial (Eulerian) description of a continuum motion is given by X I = X1et + + X3(et . dt x) .1. the displacement components ui = zi . v3 = 2 2 . dXq For properties expressed in line integral form such as .e-t). A continuum motion is expressed by xl = X I . and making use of (4. (r. Show that the Jacobian J does not vanish for this motion and determine the material (Lagrangian) description by inverting the displacement equations.f ( X 2 . v . . X 2 X 3 = e-t(x.I). [dQ:i.2. X3 = e-t(x2 x3)/2 . vi = aui/dt = üXi/ãt and the velocity components in Lagrangian form are v l = 0. u . X 1 = xle-t + x3(e-t ..57) dt s dt 8% v . ( x . .9) the Jacobian determinant is et O et-1 J = laxi/aXjl = O 1 et-e-t = et O 0 1 Inverting the motion equations.X3)/2 . C . ( t ) - ..1-4.t )d x p ] (4. ... x3 = X3. v . the differentiation in (4. ACCELERATION (Sec.. R 13.(t)] = J C d[R:. X 2 = x2 . = x3. ( x . v l = oe-Bh cos h ( A wt). Hence for the differential tangent vector d x along the streamline. + + By (4. the equations of the streamlines are ( x l / X l ) z= x. A + 3 ê2+ 2 ê3. when + + t = O. Deter- A velocity field is described by v1 = x ~ l ( l + mine the acceleration components for this motion.4. x2 = -B . for the Eulerian case. = 6X3(1 t). = X 1 . + (2x2t + %. t).e-t . dxllxl = d t l ( 1 t ) which upon inte- + + gration gives ln x . + ' + Thus from (4.%. Finally.17). v . vi = + + + ~2e-zBA. + squaring and adding. 4. 4. = ln ( 1 t ) ln C where C is a constant of integration. = X3(1 t)3..ldt = v2 = + x3)/2 . v 3 = xs. or a = (xS + 2x1t2)ê. = we-BA sin h ( A wt). = 2X. x3 = X 3 ( l t)3. v. Eliminating t from these equations gives the path lines which are identical with the streamlines presented above.. + x3tê.et)/2 + v3(-e-t C . t ) and from these determine the acceleration components in Lagrangian form for the motion.~ ( X ~ + et)/2 du$dt = v3 = e-Yxz + x3)12 + et(xz . Using the vector form of (4.13). du~. v .18) the acceleration field is given by a = xSê. when t = O. For the given flow these equations are d x l l x l = dx. (x1/X1)3= x3/X3. 4.B ~ x cos ) xA. By writing x .3 to obtain the displacement relations xi = xi(X. A velocity field is specified by the vector v = xStêl x2t2ê2 x1x3tê3. Integrate the velocity equations of Problem 4. X . the result is a s before v.e)ê2+ (51x3 + 2x x3t2)ê3 Thus ap = 3 ê l + 9 ê 2 + 6 ê .4 yields + + + the displacement equations x l = X l ( l t ) . = x. = d x l / d t = x l l ( l t ) . = e. v . v X d x = O and accordingly the differential equations of the streamlines become d x l l v l = dx2/vz = dx3/v. = O and v2 = v: v. B)2 = e-zBh/X. v3 = 3x3/(1+ t). separating variables. and v. xi = Xi and so X .5.3. .et)/2 Solving these equations simultaneously for v .6). For the velocity field of Problem 4. v . + The motion of a continuum is given by xl = A (e-BAIX)sinA(A w t ) . x2 = X z ( l t)*. 4. = 2X2(1 t ) . t is eliminated and the path lines are the circles ( x l -A)2 + ( x . = 3X3(1 + t)2 and a. vn = 2x2/(1+ t). Integration of the velocity expressions dxi/dt = vi a s was carried out in Problem 4. Integrating and using the conditions xi = Xi when t = 0.ê3 + t2ê$.6.14) and (4. . . = -B ./2x2 = dx3/3x3. Determine+ + the velocity and acceleration of the particle a t P(1. By direct substitution. a.2) when t = 1. and so x1 = X l ( l t). x3 = X3.A = (e-BVk) sin x ( A ~ t ) x.. . = A (e-Bh/X) sin x A .x3)/2 + v2(2 . . + 2x2tê2 + x1x3ê3 + + x2t2ê2 + x1x3tê3) (xStê.15). + + + From (4. + + B = (-e-BV/h) cos X(A ~ t ) then . + x1tê3ê3) (2x1tê$. Show that the particle paths are circles and that the velocity magnitude is constant.3. = O. v . + + (ecBVX)cos X(A d ) .118 MOTION AND FLOW [CHAP..)I2 + v2(-e-t + et)/2 + v3(2 . Similar integrations yield z2 = X 2 ( l t ) 2 and x./X.7. + a . Also determine the relationship between X l and X Z and the constants A and B. (x2/X2)3= (x3/X3)2. A t every point on a streamline the tangent is in the direction of the velocity. 4.3 determine the streamlines and path lines of the flow and show that they coincide.. from (4. v . 4 Also.( e .et(x.e-t . Since xl = X . C = X . = O.3) when t = 0.(e-2t.11. a .4-4.4 ~ ~ Determine . If the velocity field of the continuum is given by vi = Bx1x3t.. ~ v3 = A ( x . = 12x2 . VORTICITY (Sec. A velocity field is given by vi = 4x3 . i. 4. A a = O . O. 41 MOTION AND FLOW 119 4. + + xi and A is a constant.11) gives i= -Ae-At/r .e-3t. u3 = x. Since a(r-l)/axi = -xi/r3. a = -25bêl which is a normal component of acceleration. 4b. Decomposition of the + velocity gradient avilaxj gives avi/asj = Dij V i j . 3b) and note that the velocity field corresponds to a rigid body rotation of angular velocity 5 about the axis along A e + = (4ê2 3ê3)/5. From (4. 22.t + B X ~ + Bx$x2)/r3 t2 Thus for P a t t = 1. v2 = -2x1e-2t.3 x 2 ) ~ 1 + 3 ~ 1 ~ 2 . Here the displacement components are u .4 RATE OF DEFORMATION. 3b) which is on the axis of rotation. Also. A certain flow is given by vi = O.14) the velocity components are v.= -e-A(3A + B)/9. By (4. The magnetic field strength of an electromagnetic continuum is given by A = ecAtlr where r2 = X : X.5) 4. the acceleration components a t P(b. 0 . -1. v2 = B x t2.(e-3t. = O. equation (4. Determine the velocity gradient d v i l d x j for this motion and " from it compute the rate of deformation tensor D and the spin tensor V for the point P(1.2) when t = 1. a .1) + ~ compute ~ the rate of deformation D and the vorticity tensor V. v2 = 3x1.19) avihxj = (-23 o x. Compare D with d e i j l d t .I ) .1 ) and from (4. 4. v3 = -3x.10. Thus a t P(b. = -25s1. For the motion xi = X l .20) and (4. v2 = A ( X I - X xi)ecBt.0. 0 ) and &(O. O ) . u2 = %.9. Thus Likewise.4b.21) a s 4.e-~t(Bx:x. v3 = . Note that v = w X x = ( 4 ~ 2 + 3 ~ 3 ) ~ ( x l ~ ~ + ~ 2 ê 2 + x 3 ~ 3 ) = ( 4 ~ 3 .1).C H A P . the rate of change of the Eulerian small strain tensor E. x3 = X 3 X I ( ~-. a t Q(O. the rate of change of magnetic intensity for the particle a t P(2.162. x2 = X2 + .18).8. a2 = -9x2 i12x3. Thus . ~ x1x3)ecBtwhere A and B are constants. ~ v3 = B x ~ x ~determine . -2. decomposition of the displacement gradient gives aui/dxj = cij + uij. which may be evaluated a t P when t = O and decomposed according to (4.3x2. xp -2xix2 . or (9.vQ2)= -7ê1/4 + 2ê2 and 8 ( v p. X V = 2 ( C ê l A 3.q3 dx. By direct calculation v .120 MOTION AND FLOW [CHAP. Show that the velocity field of Problem 4. Then q X d x E O . From (4.3). 2iZ = q. . sê3). determine the rate of rotation vector O and show that v = O X x..3xl)ê3. Let d x be a differential distance vector in the direction of q. or i2 = 4 ê 1 + 3ê2.29).Q2(1.7/8.1. Determine the + + unit relative velocity with respect to P(1. dx3 . i t is found to be antisymmetric. q = V .xix:)êl (x: $2 xz)ê2.V Q3) = -15$.) 2. + + From (4.3) and show that these values approach the relative velocity given by (4.15. 4. Calculating the velocity gradient avilaxj. ! -e-% -3e-3t/2 ds.4x3& + (4x2. 3.12. For the rigid body rotation v = 3x3ê1. = C dx.ldt = -e-2t O = Dij -3e-3tI2 O O o The student should show that dwijldt = V i j . . A steady velocity field is given by v = ( x . Q3(1.3) of the particles a t Q1(1. (93 dxl .v n . Integrating these in turn yields the equa- + + tions of the vortex lines x3 = Bx21A K1.30). Thus avildxj = (-.Cx3)êz + + ( C X Z.4x3ê2 + ~ ~ ê ( 4~x 2).14.0. C d x z = A dxl.. the d. 4. d ~ 3 $2 ) (91 dxz . d x i ) ê 3 E O from which dxllql = dx21q2 = dx31q3. Show that for the velocity field v = ( A x s . B d x .13. The velocity gradient matrix is 3x1 .Bx2)êl (Bxl .3).13 represents a rigid body rotation by showing that D = 0. h + h 4(vp . A vortex line is one whose tangent a t every point in a moving continuum is in the direction of the vorticity vector q./8 + 22.. = -e1 2 e.. Show that the equations for vortex lines are dxilqi = dx21q2 = dx31q3.9. 4. 4. x 1 = Cx31B K. 4 Comparing D with dEldt.16.Axl)ê3 the vortex lines are straight lines and determine their equations.3 x 1 ) ê 3 = V 4. where the Ki are + constants of integration.e.3/4. for the vortex lines are A dx3 = B dx. r ) O -B A -f = V i j and Dij = O.26). This vector is along the axis of rotation..and by Problem 4.12. Thus ( 4 ê l f 3 ê 2 )X ( ~ ~ ê ~ f ~ = ~ ê3x3ê1 ~ + . x z = A x J C K . 213 = O. 3/51 4.O. = c *28 -2. v2 = 2x3.l=-fi The transformation matrix to principal axis directions is -112 -112 l/fi .17. Here the velocity gradient is [8vi/8xi] = [ 6xlx2 O x. Determine the principal E : r] directions and principal values (rates of extension) of the rate of deformation tensor for this motion. Thus ( d v l d x ) ~= -22.CHAP. 3%. 4. + + ~ : determine the rate of extension a t P(1.3) in the negative x2 direction. 2x1x2x3 ] and its symmetric part Thus from (4. Here avi/axiI = 0 O 0 0 0 2 0 = O + [ . e2 4.4ê3)/5. or in matrix form Y.1.1) in the direction of $ = (3êl. 4x2x3 2:: xlx.Vgi. AII=O.1.4&)/5 and fi = (4&+ 3ê3)/5. + 2ê2 which i s the value approached by the relative unit velocities vp .] -1 o o o o -1 1 and for principal values A of D.34) for 2 = ( 3ê1. In analogy with the results of Problem 3. Thus X 1 = + f i . is given by Y.19. XI.17 determine the rate of shear a t P between the orthogonal directions 2 = (3êi . A steady velocity field is given byv1 = 2x3.18.4ê3)/5. = [4/5.x. 41 MOTION AND FLOW 121 and a t P(l. For the steady velocity field v = 3x. For the motion of Problem 4.20 the shear rate Y.x2êl 2xix3ê2 ~ 1 x 2 ê3.. 23. AREAS. . 4. & d s By a direct transcription into symbolic notation of the result in Problem 4.)[email protected]. 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+$. dt and by simple manipulation of the dummy indices. .p) + V x ( p x v ) ] . Determine the material derivative pidSi of the flux of the vector property pi dt through the surface S.22 may be written in symbolic notation as $A P * & ~ s = [$ + v(T7. Show that the transport theorem derived in Problem 4. i. 7 = -d x k .19. Determine the maximum shear rate :.20. Thus Fig. Analogous to principal shear strains of Chapter 3.A I I 1 ) / 2 = fi.451.5% 4.19.e. = 7.? = ( A 1 . 4.7) Calculate the second material derivative of the scalar product of two line elements.0.21. the maximum shear rate is..6-4. determine d2(dx2)ldt2.122 MOTION AND FLOW [CHAP. i." = [0. for the motion of Problem 4. (Sec.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. 4 with the rate of deformation matrix in principal form [D. 4-3 MATERIAL DERIVATIVES OF VOLUMES. 4.22. INTEGRALS.] = o -fi 4. d(dxi) avi d(dx2) From (4. Therefore 83. and i t is shown in (4.= 2Dii dxi dxj.48) that . . a s before. Thus. BY (4. ETC. 41 MOTION AND FLOW 123 Now use of the vector identity V X ( p X v) = p(V v) .( p V ) v (see Problem 1.~ d s= L ["z+ v(vop) + v x ( p x V ) ]~ ~ n d s 4. Using the vector form of (4. Let P*(x. Show that d(ln J)ldt = div v. t ) in Problem 4. avi avi From (4.54). Q ~ 3 .29).28.24. 1.Q%3. qi = ~ ~= ~ ~ ~ ~ + ~ v (and ~.R + ~ . the integral on the left is simply the instantaneous volume of a portion of the continuum.p.53) and (4.. J v2. p ~ 2 . is the indicial form of the required equation.24 is the scalar 1. 4.27.S r k S s i ) V k j = 2Vsr. From this result q.2J v3.29). See Problem 4.x.Qx3. s ~ s .50). + = f p Q R ( j S l . so that J = eP~Rx1. . Determine the material derivative of this volume. Show that the acceleration a may be written a s a = - dt + q x v + +Vv2. ~ ~ S . ~ ) . ~ j NOW S ~ . $i i dV = V . the three non-vanishing ones yield + + J = v.ZR. This relationship may also be established by a direct differentiation of (4. v dV.R R + c ~ ~ .58) a s given in Problem 4. axk . From the definition of the vorticity vector (4.38). ek. If the function P*(x.qi = eirsYjkVkj av 4.54) in symbolic notation. 4. a s the student should confirm. Thus J = J V v and so d(ln J ) l d t = div v. J.P~X. and so V v is known a s the cubical rate of dilatation.43. Here V v d V represents the rate of change of dV. MISCELLANEOUS PROBLEMS 4. Q ~ ~ .25. and so which. the nine 3 x 3 beterminants resulting from summation o? s in this expression.. show that qi = e i j k V k j and that 2Vii = ejikqk..x1. ~ X%3.65) gives it p ..jl = eiikVkj. p x ~ .24. .QP ~ ~ ..CHAP.3J = v. Then (4. ai = - at + v . By (4. ~ ~ .pjS2. .v(V p) + (v V ) p .R and J becomes the sum of the three determinants. = vl. Let axi/aXp be written here a s si. Express Reynold's transport theorem a s given by equations (4. and so J = E P Q R ( ~ ~ . for example..and by Gauss' divergence theorem this becomes (4.18). QZ ~~ .26. qi = qjkV[k... Of R ) .53) is .P%2. = ~v since ~ O ~(see. q = curl v. P+~~ZI. Problem = (GriSsk . t ) be any tensor function of the Eulerian coordinates and time. ). Prove equation (4. are solutions of 2x. dSi = d S p = - 8x1 J d X z dX3. v2 = -x. axi axi a%.v3 = O determine expressions for the principal values of the rate of deformation tensor D a t an arbitrary point P(x1.(4.&?). = O. do. dSi = 6. -2x1x2 . As shown in Problem 4. a x ~ 4.7. R d S = L (Vxa)*RdS . x3).dSi = qjk dX2dX3 = ax. $L .23. axP ax. 4. If vi = vi(x).lv. do. a t a given instant t streamlines are the solutions of the differential equations d x l l v l = dx21v2 = dx31v3. Thus 2 J d X z d X 3 and by Problem 4. V x a = V x at + V x ( q x v) + V x ~ ( ~ 2 1 . + 2. + d2 .23 to show that the material rate of change of the vorticity flux - d"t L q *6 dS equals the flux of the curl of the acceleration a.*nd~ = 1[$ + q ( V * v ).31. = ( x . 2 ) or V x a = aqlat + V x ( q x v ) = dqldt + q ( V v ) . d I I = O. dSi = ~ ~ ~ ~ ( a xdXz(axklaX3) ~ / a X ~ ) d X 3 and axi .124 MOTION AND FLOW [CHAP.27 and 4.~ 1 x 2 . ax ax..x.32. Thus if q is substituted for p in Problem 4. Principal values d(i.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 ) . Pathlines are solutions of the differential equations dxildt = vi(x. .29. t). these equations become d t = dx.d -2: +xi .d O = O = -d[-4xix.( q * V ) v ] . Note here that d I = (x: +xi). 22).)2] o o -d Thus d. 4. + For the steady velocity field v1 = x l x z x i .x2 . 4 4. + z. X Z . Taking the curl of the acceleration a s given in Problem 4. + d I I I= -(x.30. O -x. axk Using (S. Use the results of Problems 4. Show that for steady motion (avilat = 0) of a continuum the streamlines and pathlines coincide.28.27. = dxZ/v2= dx3/v3 which coincide with the streamline differential equations. = -(x: +xi).( q V ) v since q = V X v and V X V(v212) = O.. v 2 = X.e-t).q = qPvq.32 the identity V X a = aqlat + V X (q X v) may be written in indicial form a s aqi/at = eijkak. = e-t(x2 t x 3 .)/2. .)..qiqi/2. Am. Determine dXldt a t P(x. X3)et. For the vorticity qi show that a at qi d V = [€.srn~rn~r). ) . = -x. . where vi is the velocity and qi the vorticity. v . = s.5.. Ans.i = Dij Dij .157). v . v3 = 2t.~ ~ ~ i$ d~ = Sv [Eijkak.. Prove that the material derivative of the total vorticity is given by . v .38). 2.2 3 .)/2 .. Also show that v.X3)12..x.~ and use this to derive (4.. 41 MOTION AND FLOW 125 4.CHAP. D: = (:5 0 o o o).x. represents a rigid body rotation and determine the vorticity vector for this motion.1 ) .XSX.. = 2bi Show that for the flow vi = xil(l +t) the streamlines and path lines coincide.x3. a3 = 2 + + Show that the velocity field vi = eijkbjxk ci where b.qqvP. a. + c~-~(-X . x3 = X3. d"' = -2419. determine the principal axes and principal values of ü a t P ( l . qi = bixLi .(et .e).ak + qjvi. Ymax = 5 Show that d(axilaXj)ldt = v~. v3 = 0. v .(3.). v2 = x3(et -e-t).x. Show that J does not vanish for this motion and obtain the velocity components.3). + si.xixi. Supplementary Problems A continuum motion is given by x 1 = Xlet + X3(et. v . = ( X . a . Compare these tensors a t t = 0..). + xixi. For the velocity field v . +-vqqP. = x1 .(v. Determine the acceleration components in Eulerian form. = O. v3 = O or v . x3 = et(X2 X.qivi]d s j . = 2 2 . Prove the identity ~pq. . = -x. i + + For the continuum motion x ... jvi. . = -(xt + x.3).33. x. show that Dij = deijldt a t t = O.(. x2 = X 2 + X. + Ans. x. The electrical field strength in a region containing a fluid flow is given by h = ( A cos 3t)lr where r2 = x. Ans. = -X3. (.X. = X.. = et(X2 X3)/2 e-t(X2 . + x i and A is a constant. = O. and ci are constant vectors.E ~ ~ ~ ~~ v . Ans.41 determine the rate of extension in the direction V .41) of the text directly from (4.vr. 2. dhldt = (-3A sin 3t)lr Show that for the velocity field v . What is the maximum shear rate a t P? Ans.i. = x l x 2 + 22.sl dv and by the divergence theorem of Gauss (1.b. -5 a = ( 31- 1 O 3 11- o o For the velocity field of Problem 4. v3 = O the streamlines are circular. A- h + 2 ê 3 ) / 3 a t P ( l .(et + e-t). From Problem 4. E~Thus . The velocity field of the fluid is v .. = -X2e-t. i v3 = O A velocity field is specified in Lagrangian form by v . x. v .. The continuity equation may also be expressed in the Lagrangian. the rate of change of m in (5.e. Therefore from (4.t ) = p(x.. or material form. 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. t ) . or This equation is called the continuity equation.1 CONSERVATION OF MASS. CONTINUITY EQUATION Associated with every material continuum there is the property known as mass.the right hand integral in (5.38).7) may be converted so that . = O or divv = O (5. t ) is called the vector potential of v. .3) yields the result v..52) with P:. . i.1) is Since this equation holds for an arbitrary volume V . Using (4. The law of conservation of mass requires that the mass of a specific portion of the continuum remain constant. the integrand must vanish. t ) is a continuous function of the coordinates called the m a s density.6) in which s(x. Chapter 5 Fundamental Laws of Continuum Mechanics 5. t ) of an incompressible continuum can therefore be expressed by the equation vi = c ~ ~ ~ ors ~ V.1)and (4. 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. =~ V xs (5.1) in which p(x. The conservation of mass requires that where the integrals are taken over the same particles. V is the volume now occupied by the material which occupied V0 a t time t = O . and hence that the material derivative of (5.1) be zero.5) The velocity field v ( x .(x. 51 FUNDAMENTAL LAWS OF CONTINUUM MECHANICS 127 Since this relationship must hold for any volume VO. the stress vector is tin'. Therefore if the interna1 forces between particles of the continuum in Fig. of a continuum is equal to the resultant force acting upon the considered portion.10) is the Lagrangian differentialform of the continuity equation.13) by the right hand side of (5.2 LINEAR MOMENTUM PRINCIPLE. .12) becomes In calculating the material derivative in (5.9) which implies that the product pJ is independent of time since V is arbitrary. EQUATIONS OF MOTION.13). (5. 5-1 Based upon Newton's second law. Body forces bi per unit mass are given.10) may be used. 5-1.CHAP.it follows that Po = PJ (5. For this situation. 5.10) Equation (5. Thus Replacing the right hand side of (5. the principle o f linear momentum states that the time rate of change of an arbitrary portion.12) and converting the resulting surface integral by the divergence theorem of Gauss. On the differential element dS of the bounding surface. 5-1 obey Newton's third law of action and reaction. or that d ~ ( P J= ) 0 (5. the momentum principle for this mass system is expressed by A Upon substituting tin' = ajini into the first integral of (5. EQUILIBRIUM EQUATIONS A moving continuum which occupies the volume V a t time t is shown in Fig. The velocity field Vi = duildt is prescribed throughout the region occupied by the continuum. the continuity equation in the form given by (5.14) and collecting terms results in the linear momentum principle in integral form. the total linear momentum of the mass system within V is given by Fig. The moment of momentum principle does not furnish any new differential equation of h motion.4 CONSERVATION OF ENERGY. used extensively in solid mechanics.19) is valid for those continua in which the forces between particles are equal. with respect to the origin. are known as the equations of motion.128 FUNDAMENTAL LAWS OF CONTINUUM MECHANICS [CHAP.17') These are the equilibrium equations. . = O or 1. . the moment of momentum principle is expressed in integral form by Equation (5.19).18) in which xj is the position vector of the volume element dV.. simply the moment of linear momentum with respect to some point. 5. Thus for the continuum shown in Fig. The resulting equations. Thus .pv. 5-1 may be derived directly from the equation of motion given by (5. FIRST LAW OF THERMODYNAMICS. 5-1. reduces to Since the volume V is arbitrary. and in which distributed moments are absent..15) must vanish. If the substitution t p ' = uPknpis made in (5. The important case of static equilibrium. If stress symmetry is not assumed. as the name implies. the integrand of (5.19). . ENERGY EQUATION If mechanical quantities only are considered. 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. is given a t once from (5. d~ or N = ( X X ~ Vd ) v (5.3 MOMENT OF MOMENTUM (ANGULAR MOMENTUM) PRINCIPLE The moment of momentum is.V.16) and the velocity vi is first computed.16). in which the acceleration components vanish. 5-1.16). Accordingly. is J ~ ( t=) v ~. 5. the total moment of momentum or angular momentum as it is often called.. 5 Since the volume V is arbitrary. opposite and collinear.. and the result integrated over the volume V.x. which upon substitution of t p ' = upknp. the scalar product between (5. the principle of conservation of energy for the continuum of Fig. = O which by expansion demonstrates that uik = ukj.16) as + u ~ ~pbi. . To accomplish this. for the continuum of Fig. = ~ O or V. such symmetry may be shown to follow directly from h (5. 1 pb = O + (5. the equation is satisfied identically by using the relationship given in (5.. and the symmetry of the stress tensor assumed. the principle of conservation of energy in its most general form must be used. v i ~ j i . so that (5. If both mechanical and non-mechanical energies are to be considered. In the following.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. .24) to a surface integral by the divergence theorem of Gauss.. if the vector s is defined as the heat fEux per unit area per unit time by conduction. Finally. Also. and written dUldt. and z is taken a s the radiant heat constant per unit mass per unit time. or removed f r o m the continuum per unit time. or electromagnetic energy. 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.. only mechanical and thermal energies are considered..22) may be written since Vijuji O. 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. The integral on the left side is known as the time rate of change of internal mechanical energy. n3 .19) + = Dij V. )vi.25) may be written briefly a s where d W l d t represents the rate of work. 51 FUNDAMENTAL LAWS OF CONTINUUM MECHANICS 129 But L pv. Therefore (5. the rate of increase of total heat into the continuum is given by Therefore the energy principle for a thermomechanical continuum is given by . and making use of the identity ti"' = ~31.CHAP. (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. converting the first integral on the right hand side of (5. ~ uji and by (4. and the energy principle takes on the form of the well-known first law of thermodynamics. ~ ~ - .=j ( v . chemical energy.dV = i vivi cLdV = dK dt which represents the time rate of change of the kinetic energy K in the continuum.j . Such energies supplied may include thermal energy. and the special symbol d is used to indicate that this quantity is not an exact differential.. .) Also. T the absolute temperature. and S the entropy. leaves unanswered the question of the extent to which the conversion process is reversible or irreversible. leads to the local form of the energy equation: or Within the arbitrarily small volume element for which the local energy equation (5. Therefore by taking the scalar product between (5. The basic criterion for irreversibility is given by the second law of thermodynamics through its statement on the limitations of entropy production. a continuum) is said to describe the state of the system. in terms of the energy integrals. the balance of momentum given by (5.31). The first law. The state variables used to describe a given system are usually not a11 independent. or entropy density is denoted by s. The entropy of a system can change either by interactions that occur with the surroundings. The second law of thermodynamics postulates the existence of two distinct state functions.. so that the total entropy L is given by L = 1 .5 EQUATIONS OF STATE. In continuum mechanics the specific entropy (per unit mass). This equation expresses the rate of change of internal energy as the sum of the stress power plus the heat added to the continuum. with certain following properties. The relationship expressing conversion of heat and work into kinetic and internal energies during a thermodynamic process is set forth in the energy equation.16) must also hold. Functional relationships exist among the state variables and these relationships are expressed by the so-called equations of state. ps dV. the result is the reduced. subtracting this product from (5. by severa1 thermodynamic and kinematic quantities called state variables.130 FUNDAMENTAL LAWS OF CONTINUUM MECHANICS [CHAP. but the reversible process is a very useful hypothesis since energy dissipation may be assumed negligible in many situations. ENTROPY. however. in general.31) is valid. The entropy is an extensive property. 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. 5.j+ pvibi and. Thus ds = + ds(i) (5. This description is specified.e. 5 or. the total entropy in the system is the sum of the entropies of its parts.. as Converting the surface integrals in (5. after some simple manipulations.33) .16) and the velocity p'Uivi = vioji. or by changes that take place within the system. T is a positive quantity which is a function of the empirical temperature O. and again using the fact that V is arbitrary. SECOND LAW OF THERMODYNAMICS The complete characterization of a thermodynamic system (here. but h i g h l ~ useful form of the local energy equation.30) to volume integrals by the divergence theorem of Gauss. the first law of thermodynamics postulates the interconvertibility of mechanical and thermal energy. only. A11 real processes are irreversible. As presented in the previous section. i. A change with time of the state variables characterizes a thermodynamic process. 34) ds(i) = O (reversible process) (5.41).36) T 5. that (5. With this assumption the energy equation (5.37) by the divergence theorem of Gauss. this entropy principle may be expressed in integral form as the Clausius-Duhem inequality.35) In a reversible process. DISSIPATION FUNCTION According to the second law.25) a s 1 In this equation.32) may be written with the use of (4. .36)may be combined to yield Therefore in the irreversible process described by (5. 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.the entropy production rate may be expressed by inserting (5. The equality in (5. The Clausius-Duhem inequality is valid for arbitrary choice of volume V so that trans- forming the surface integral in (5. Mathematically.37) holds for reversible processes. I t is zero for a reversible process.39) where u::) is a conservative stress tensor. If the continuum undergoes a reversible process.40). d s c e )is the increase due to interaction with the exterior. and furthermore. and 0 d q l d t is the rate Af heat influx per unit mass into the continuum. For this reason it plays an important role in imposing restrictions upon the so-called constitutive equations discussed in the following section. Therefore ddi) > O (irreversible process) (5. In much of continuum mechanics. The change dsCi)is never negative. = a!?) + u 'iDj ' (5. the inequality applies to irreversible processes.. 51 FUNDAMENTAL LAWS OF CONTINUUM MECHANICS 131 where d s is the increase in specific entropy. 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. is given by This inequality must be satisfied for every process and for any assignment of state variables.o. and ds'" is the internal increase. a. per unit mass. and positive for an irreversible process. and is a dissipative stress tensor. there will be no energy dissipation.CHAP. dqldt = d q ( ~ ) l d t so . where e is the local entropy source per unit mass.?) iij is the rate of energy dissipated per unit mass by the stress. if d q ( ~denotes ) the heat supplied per unit mass to the system. the local form of the internal entropy production rate y.40) and (5.6 THE CLAUSIUS-DUHEM INEQUALITY. Thus . the change dsCe)is given by dS(e) = -dq(Ri (reversible process) (5. to define certain ideal materials. the three velocity components vi. the displacement components ui). Since real materials respond in an extremely complicated fashion under various loadings. the six independent stress components oij. (5. and the specific interna1 energy u. adiabatic process (dq = O). kinematical and thermal variables. the absolute temperature. This introduces two additional unknowns: the entropy density s. I t should be pointed out that the function of the constitutive equations is to establish a mathematical relationship among the statical. and T. In addition. the basic equations are (a) the equation of continuity. alternatively.4) (b) the equation of motion. Of these. . the three components of the heat flux vector ci. The unknowns are the density p. (5.ij is called the dissipation function. THERMOMECHANICAL AND MECHANICAL CONTINUA In the preceding sections of this chapter.43). three will be in the form of temperature- heat conduction relations.132 FUNDAMENTAL LAWS O F CONTINUUM MECHANICS [CHAP. and two will appear as thermodynamic equations of state. which characterize the particular physical properties of the continuum under study. but. 5 The scalar . for example. (5. constitutive equations do not attempt to encompass a11 the observed phenomena related to a particular material. Of the remaining five. are very useful in that they portray reasonably well over a definite range of loads and temperatures the behavior of real substances. severa1 equations have been developed that must hold for every process or motion that a continuum may undergo. Such idealizations or material models as they are sometimes called. perhaps as the caloric equation of state and the entropic equation of state.16) (c) the energy equation. since both p and T are always positive. Therefore eleven additional equations must be supplied to make the system determinate. six will be in the form known as constitutive equations. so from (5. For a thermomechanical continuum in which the mechanical and thermal phenomena are coupled. For an irreversible. 5.44) and (5. rather.42) it follows that the dissipation function is positive definite.38) which governs entropy production. the Clausius-Duhem inequality (5. such as the ideal elastic solid or the ideal viscous fluid. (or.32) Assuming that body forces bi and the distributed heat sources z are prescribed. Specific formulation of the thermomechanical continuum problem is given in a subsequent chapter. dsldt > O by the second law.7 CONSTITUTIVE EQUATIONS. which will describe the behavior of the material when subjected to applied mechanical or thermal forces. (5. (5.45) consist of five independent equations involving fourteen unknown functions of time and position. must hold. . .29). satisfies the continuity equation for an incompressible flow. . strains) variables and are often referred to as stress-strain relations.CHAP. .(x. t) = pp".kk = O or dpldt pV2+ = 0 .1) 5. Here and so u k S k= ( 3 . so + that d ( p J ) l d t = J ( d p l d t P V . Under this assumption the purely mechanical processes are governed by (5. d ( p J ) l d t = ( d p l d t ) J + p dJ/dt = O and from Problem 4.3)/$ = O to satisfy the continuity equation. .4. or a t most. Also. Show that the velocity field vi = Axil?. - B y (1.i and (5. t) represents any scalar.50). in the uncoupled theory.curl v = O when q 0.' An irrotational motion of a continuum is described in Chapter 4 as one for which the vorticity vanishes identically. where xixi = r2 and A is an arbitrary constant. 51 FUNDAMENTAL LAWS O F CONTINUUM MECHANICS 133 In many situations the interaction of mechanical and thermal processes may be neglected. the constitutive equations contain only the statical (stresses) and kinematic (velocities. dpldt + pvk.43) and (5. t) show that since by ( 5J). . + 5. and so v becomes the gradient of a scalar field +(xi.28. . From (5.44) since the energy equation (5. Thus vi = +. .(x. 5. k = O for incompressible flow.(x. If P:?. = 0.45) for this case is essentially a first integral of the equation of motion.5) v k . the temperature field is usually regarded as known. Solved Problems CONTINUITY EQUATION (Sec. d J l d t = J v k . 5. vector or tensor property per unit mass of a continuum so that P:. Differentiating. . . the heat-conduction problem must be solved separately and independently from the mechanical problem. Show that the material form d(pJ)ldt = O of the continuity equation and the spatial + form dpldt p v . 5. = O are equivalent. In isothermal problems the temperature is assumed uniform and the problem is purely mechanical. ~ )= 0 . t ) (see + Problem 1. The system of equations formed by (5. displacements.3)is now dpldt p+.2. Six constitutive equations are required to make the system determinate. Determine the form of the continuity equation for such motions. The resulting analysis is known as the uncoupled thermoelastic theory of continua. In the uncoupled theory.43) and (5.3.1.44) consists of four equations involving ten unknowns. 0 3 2 ) ê1 + ('731 .X1X2X3.2-5. LINEAR AND ANGULAR MOMENTUM. Now. If distributed body moments m i per unit volume act throughout a continuum. i t is not affected. . For the velocity field vi = x i l ( l + t). +j) p(avi/Jt + vivi. + u33ê3 x $3 - .a32) for i = 1 .ul3) for i = 2. Since p = po when t = 0. = alSl x ê1 + olGl x $2 + a l g l x ê3 + . j) pbi aij.6.19) and applying the divergence theorem (1.5.3) yields ln p = -1n (1 t)3 ln C where C is a + + constant of integration..19) acquires a n additional term so that which reduces to (see Problem 2.8. h Substituting oPknpfor tp' in (5.4) and the second term is pai. + Here vk. The term in parentheses is zero by (5.134 FUNDAMENTAL LAWS O F CONTINUUM MECHANICS [CHAP.013) $2 + ( ~ 1 -2 ~ 2 123) Also. show that the equations of motion (5.7. expanding eijkajkgives identical results. 5.= 1. By (1.20).16) remain valid but the stress tensor can no longer be assumed symmetric. Carrying out the indicated differentiation and rearranging the terms in the resulting equation yields ~. ) l d t = pbi + ( u i j . Show by a direct expansion of each side that the identity eijkujke.v = o. 5 5.9. (031 . 5.19) reduces to (5.3) A 5..16) is derived on the basis of force equilibrium. that the equation of motion (5. 5. = S j p and . Since (5.21) is valid.(a~/+ a t p. j The first term on the left is zero by (5. (5. the indicated differentiations here lead to cijkvjvk = O. L. and because V i s arbitrary.16).16).9) ( Y j k a j k+mil dV = O. 5.8).20) and (5.jvj + pvj. show that pxlx.oZ1) for i = 3. Show that (5. (023 . (al2.x3 = p.15) and (2. this equation becomes p = pd(l t)3. so that finally s. Thus pai = pbi + aij. EQUATIONS OF MOTION (Sec.k = 3 / ( 1 t ) and integrating (5. xi = X i / ( l t ) and hence + Pxlx2x3 =~ 0 ~ 1 ~ 2 ~ 3 .2. (az3.~ V ~ V ~ Show ) . used in (5.j which is (5.16) follows from this equation. also €ijkUjk .157) to the resulting surface integral gives Using the results of Problem 5. ~ . eijkajk + mi = O for this case. zj. however. Next + by integrating the velocity field dxi/xi = d t / ( l + t ) (no sum on 9. The momentum principle in differential form (the so-called local or "in the small" form) is expressed by the equation d ( p v . 19).--do + + (h* 2 / ~ * ) ( 1 . * and the heat conduction obeys the Fourier law ci = -kTai.2?)).(-p+ X*Dkk)SijDij 2p*DijDij - + kTSii + +z .CHAP. At a certain point in a continuum the rate of deformation and stress tensors are given by 4 2 5 and Determine the value h of the stress power Dijuij a t the point. 4. 5. h = 4 +O -4 +O -6 + 14 -4 + + 14 40 = 58.6) 5. ENTROPY. i t follows that Dijoij = = -pvi.l.W~X.19) is the total moment Mi of a11 surface and body forces relative to the origin. i p(6ipzqzq .. ~D.Vij. vi. uii = . kT. show that the stress power may be -.23) reduces to the familiar form given in rigid body dynamics. the kinetic energy integral o£ (5. are the first and second invariants respectively of the rate of deformation tensor.13. J"~ ( 4 P q x q- where Zip = p(8ipzqzq. and since V i j u i j= O. DISSIPATION FUNCTION (Sec.19) reduces to the well-known momentum principle of rigid body dynamics.44 d V is the moment of inertia tensor. In symbolic notation note that K = . du p d t .P . From (5. For a rigid body rotation about a point.12. Determine the form of the energy equation if uij = (-p + h* Dkk)8. From the continuity equation (5.0.ii + +z where I. 5.11. The left hand side of (5. -:i Multiplying each element of Dij by its counterpart jn oij and adding.i.. ENERGY..2 5. If uij = -pSij where p is a positive constant. )- P dt ~ 4/~*11. . Thus for vi = eijkwjXk.z P z $ d V wiwpzip - 2 o. and 11.92).4-5. Dij = v i B j. = d [O.14. From (5. p dp expressed by the equation D i j u i j = - P dt By (4. wiw~ 2.10.3). Show that for this velocity (5. 5.i = -(llp)(dpldt) and so D i j q j = (plp)(dpldt) for oij = -paij.2 . K = i p Y d v = i peiikUj%k€ipq~pZq dv = ii ~ ~ p w j ( 8 j p-~6ki q~S k p ) x k z q dV . Show that for a rigid body rotation with vi = C~. vi = cijkojxk. 51 FUNDAMENTAL LAWS O F CONTINUUM MECHANICS 135 5. 18. Thus by (1. = Kl13. If uij = .13.136 FUNDAMENTAL LAWS O F CONTINUUM MECHANICS [CHAP.19.Dki = D:. 5..) i. determine an equation for the rate of change of specific entropy during a reversible thermodynamic process. .DC3).l. 3(D(1) + D(2) + D(3))(D(1)D(2) + D(2)D(3> + D(3)D(1)) + 3D. 5-2 5.3) Therefore o:.41) gives T ..17 is assumed isotropic so that Kijpqhas the same array of components in any rectangular Cartesian system of axes. For the constitutive equations uij = Ki. Display the components in a 6 x 6 array..D.= - dt du + -- p dp p2 dt upon use of the result in Problem 5. determine the dissipation function in terms of the invariants of the rate of deformation tensor D. 8~. i t is clear that since both Aij and Bij have six independent components. . + Use + this to develop the constitutive equation uij = KijpqDpqin terms of A* and P * .has a t most 36 distinct components..= K2. ~ * ( 8 ~ ~ 8iq8jp)....138) the trace DiiD.15. If the continuum having the constitutive relations uij = KijpqDpqof Problem 5. show that by a cyclic labeling of the coordinate axes the 36 components may be reduced to 26. For isotropy Kijpqmay be represented by Kijpq= A*8ij8pq .. Since oij = uji. 5 5.D(2. By suitable reflections and rotations 53 of the coordinate axes these 26 components may be reduced to 2 for the case of isotropy. CONSTITUTIVE EQUATIONS (Sec. 5-2. = KZZ3. D ( 2 ) .DPq show that because of the symmetry of the stress and rate of deformation tensors the fourth order tensor Kijp. and since Dij = Dji. is 5.) iij = PDsDkjDij which is the trace of D~ (see page 16) and may be evaluated by using the principal axis values D ( . The coordinate directions may be labeled in six different ways a s shown in Fig. If Kijpq is considered a s the outer product of two symmetric tensors AijBpq= Kijpq.D:Z) + ~ ( 3 ) = (D(l) + + D(3))3 . For the stress having = /3DikDkj.31DIID+ 3111D]. Fig.17.' gj = p[1: . Isotropy of Kijpqthen requires x1 or x2 that K. 5. Kijpqwill have a t most 36 distinct components. The usual arrangement followed in displaying the components of Kib. Here by (4. o:.16. = K3. = K3322 and that K l q I 2= 1(1313= K2323= K2121= K3131= K3232 which reduces the 36 components to 26.25).p s i j . KiTpq= Kjipq. KijPq= Kijqp.7) 5. . Here oij '= ::o dt ds and (5. By (4. Show that the motion is irrotational and that the streamlines are circles.19 resuits in the equation sij aijukk/3 = h*SijDkk 28*(Dii SijDkk/3). . By direct differentiation dd.29). + MISCELLANEOUS PROBLEMS 581.CHAP. Therefore .22. the equations for streamlines are dxllvl = dx21vz.P.7. + 2Dij = v i s jt vjBi so that Dkk = v k j k and 2Diisj = vi.P for this incompressible flow.. = -v2. sij = 2@*D& and hence ukk = ('&h* 2@*)Dkk. Integrating with respect to x2 and imposing the given conditions on v 2 yields v2 = Axllr2. A two dimensional incompressible flow is given by v1 = A(xS ..xi)/18 + 2x1/r<l] and v e j 2 = A [2x1/1A.i . Here v l ..25. Show that the constitutive equation of Problem 5.22 is irrotational. Show that + = ( ~ .j)lpwhere p is the density. Thus A A A e1 e2 e3 curlv = alasl ala%. ~ qjvi. Adding.i + qivi. x i = constant.5).j SijDkk/3 int0 uij = h*SijDkk 2@*Dij 0f + + + Problem 5. A ( X . .. Thus 5.qivjj (see Problem 4. p$i = pbi uijSj or here p6i = pbi .32). vi.i = O for incompressible flow. 51 FUNDAMENTAL LAWS O F CONTINUUM MECHANICS 137 5. Determine v2 if v2= O a t xl = O for a11 x2. = A[-4x1(xS .xg)/r6]ê3 = O 5. . = (-p + A* Dkk)Sij+ 2p*Dij.jSij X*D. By (5. + From (5. a ~ . + + Substituting U i j = S i j Sijukk/3 and Dij = D. alax.p. (f) = 4 - - P p2' But 4 = rijkak.xi)l. where r2 = X: + x. v1 = -Ax21r2 where r2= x: + xi . For a n irrotational motion.xe)/la + (-E: + x i )I+] ê3 = 0 From Problem 4. For a continuum whose constitutive equations are a.. Show that the continuity equation is satisfied by this motion.5). page 118.jj viaij. and by the continuity equation (5.16). viBi= O or v. v3 = 0. .8 x ~ x 2 / r ~2x2/r<l + 4x2(x: . . respectively.. + + 28*DilBj. determine the equations of motion in terms of the velocity vi.~S.4. v.23.24.8x1xi/r8]. e = -pvci. = 2Axlx2/.xi)/1. curl v = O... Here curl v = A [(x: . From this + when i f j.+ v. In a two dimensional incompressible steady flow. By definition. From (5. 5. Show that the flow of Problem 5.4 2Ax1x2/r<l O + = A[2x2/r<l . v2 = A ( 2 x l x z ) l l A .= 0. curl v = 0 for irrotational flow. ai the acceleration and qi the vorticity vector. Here these + equations are xl dxl x2 dx2 = O which integrate directly into the circles x. + 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.20. the interna1 stresses and the surface stresses. 5.I t is also interesting to note that which for +. .jj + or in Gibbs notation p. By (5. +.29. Show that the + + energy equation may be written h = +Ip Ti using this definition for the enthalpy. For incompressible flow.41). dqldt = (p*lp)V2q.V p + (A* + p * ) V ( V V) + p*V2v 5.ij+./pZ + T B . = p b . the equation of motion in Problem 5.ij+. Taking the cross product V X with this equation for p = constant + gives c ~ ~ =~. . Determine the dissipation function .l j jBut q~...*Dkksij ~ + 2p*Dij)Dij= 2p*DijDij since Dkk = v k .i p*vi. Because the motion is incompressible and irrotational.157) and the equations of motion (5. Thus This sum of integrals represents the rate of work done by the body forces. From (5. ~ ~ = O ~ and ~by (4.p. result ~ ~is 4. Here o::) aij = u : ~ ) D = ~(?.23'. and if b 0.li = O and 2+. show that the divergente of the vorticity vanisbes and give the form of the equations of motion for this case.symbolic notation.ij and so u : y 3 D ' = ~ i j2p*+.ij+.p 8 .ij. By (4.jj In symbolic notation this equation is p.26. L viti'' d S which may be written V viuijni d S and by the divergence theorem (1. Determine the material rate of change of the kinetic energy of the continuum which occupies the volume V and give the meaning of the resulting integrals./pZ = -p.i.bivi) d V . If the continuum of Problem 5.B~ ~~ . and by the result of Problem 5. the scalar DijDij = +. dK/dt = J pvi'UidV.ii = O reduces to 4+. + 2p*Dij undergoes an incompressible irrotational flow with a velocity potential + such that v = grad +.i+. A continuum for which = h*Dk. 5 pCi = pbi .ij.8. ~In~ . = pb . Canceling and rearranging. 5. Also since vi = +.ji is symmetric in i and j . Also the total stress power of the surface forces is . = -pSijDijlp .25 reduces to pCi = -pZi + P * v i . Thus for V v = O the equations of motion become p'Ui = pbi . and div q = eijkvk. FUNDAMENTAL LAWS O F CONTINUUM MECHANICS [CHAP. Thus +i:' gij = p * ~ z ( v + ) = z p*v4(+z)/2.+Ip .27.25 undergoes an incompressible flow. V v = 0 .23).25 is considered incompressible.13 and the definition of h.jj = V2(V+)2.IP + TS. ii = a .29) p the .i). determine the equation of motion in terms of the vorticity q in the absence of body forces and assuming constant density.u+. the specific e n t h a l p y h = u p l p . +~ j p*vi. 5. For a continuum with uij = .p. . If the continuum of Problem 5.i + (X* + p * ) ~ j . i= .+ T S for the given stress. respectively.V p p*VZv. qi = e i j k v k .ij.ij = (+.30. = O for incompressible flow.29). = ( ~ * / p ) q . + 5.ji = O since eijk is antisymmetric and vk.16) expressed a s the volume integrals r - ( p(v$i . ) ~~p q~i v i .28.p. ~ ~ ~( Pp* / P. j j . z the continuity equation is aplat + a ( p ~ .To) when oij = sij = oij . Show that for the rate of rotation vector a.31. the angular momentum balance may be written where m i s distributed angular momentum per unit mass..(x.x~)x. Dii = 0. + xg. show that ekk = 3a(T .. .l+.40. = O satisfies the continuity equation in cylindrical coordinates when the density p is a constant. If G = g'L'. ) l a+~ a ( p ~ . z this equation becomes r(aplat) + d(rpv. v e = (1 r. + Show that the flow v. y. = (1. In terms of Cartesian coordinates x . Show that the flow represented by v . satisfies conditions for an incompressible flow. 5.(3h f zp)a Gij(T. vector or tensor function. the specific volume. v . (x: .Ts. X ~ ~ /v T = . ) l a=~ O Show that in terms of cylindrical coordinates r . incompressibility. P 5.) sin #/r.)lar + a(pve)la6+ r(d(pv. show that uii = 3(-p .ukkSij/3. For a thermomechanical continuum having the constitutive equation q j = hekkSijf 2peii . I f T dsldt = -vi. I f a continuum i s subjected to a body momeht per unit mass h in addition to body force b. ~ r2= x.+ + 5. 5.rz) cos elrz. If a continuum has the constitutive equation oii = -pGij + PDij + aDikDkj. and a couple stress g'"^ in addition to the stress t'"'.T o ) where T o is a reference temperature. I f P i j . = .35.38.33. show that the energy + equation may be written p dqldt ps dTldt = oiiDij. t ) is a n arbitrary scalar. show that du = T ds .)ldz) = O 5. 6 .37.. For a continuum having uij = -pSij. = %.34.2 ~ ~ 1 1 ~ 1 3Assume ).ilp and the specific free energy is defined by * = u . 5.CHAP. ) l a+~ a ( ~ v . show that 5.36. 51 FUNDAMENTAL LAWS O F CONTINUUM MECHANICS 139 Supplementary Problems 5./r2 where .~ X . v . 1s this motion irrotational? 5. .p dv where in this problem v = l I p . show that the local form of this relation is p dmldt = h V G Z.32. 5.39. Linear Elasticity 6. Accordingly in terms of the displacement vector ui. &(uVx+ Vxu) = 4(uV + Vu) In the following it is further assumed that the deformation processes are adiabati.M = 1 .O4 and - €33 . Vxu) = .1 GENERALIZED HOOKE'S LAW. However. .CKMcM ( K .2) the tensor of elastic constants Cijkm has 81 components. - '=H . STRAIN ENERGY FUNCTION In classical linear elasticity theory it is assumed that displacements and displacement gradients are sufficiently small that no distinction need be made between the Lagrangian and Eulerian descriptions. and where upper case Latin subscripts are used to emphasize the range of 6 on these indices. €3 2c12 = 2~~~= E6 Hooke's law may be written - uK . 2 . 3 . Thus in the notation . The constitutive equations for a linear elastic solid relate the stress and strain tensors through the expression - uii = Cijkmekm or I: = C : E (6-2) which is known a s the generalized Hooke's law. due to the symmetry of both the stress and strain tensors. When thermal effects are neglected.32) may be written . the double indexed system of stress and strain components is often replaced by a single indexed system having a range of 6. 4 .u32 . the energy balance equation (5. 5 . For the purpose of writing Hooke's law in terms of these 36 components. the linear strain tensor is given by the equivalent expressions L = E = ~ ( u v+ . In (6. (no heat loss or gain) and isothermal (constant temperature) unless specifically stated otherwise.'=I u23 . there are a t most 36 distinct elastic constants. 6 ) where CKMrepresents the 36 elastic constants. 7) P 23 and if u is considered a function of the nine strain components. From (6.d.15) are reduced to 21.).14) in which CKM= CMK. A material that is not isotropic is called anko- tropic. and the 36 constants in (6. the simplest form of strain energy function that leads to a linear stress-strain relation is the quadratic form U* = -&Cijkmcijckm (6. (6. a.18) In the single indexed system of symbols. 1 du = . ANISOTROPY. the zero state of strain energy may be chosen arbitrarily..12) becomes U* = -&CKMcKeM (6.2). u = u(c. Since the elastic properties of a Hookean solid are expressed through the coefficients CKM. 6 2 ISOTROPY. and since the stress must vanish with the strains. (6.12) From (6.. .u.a general anisotropic body will have an elastic-constant matrix of the form When a strain energy function exists for the body. ELASTIC SYMMETRY If the elastic properties are independent of the reference system used to describe it. u* has the property that Furthermore. CKM= CMK. this equation may be written U* = +uijeij or U* = -&Z: E (6..6). a material is said to be elasticauy isotropic. 61 LINEAR ELASTICITY 141 The interna1 energy in this case is purely mechanical and is called the strain energy (per unit mass).. it is observed that The strain energy density u* (per unit volume) is defined a s u* = tu' and since p may be considered a constant in the small strain theory.8).7) and (6. 3' Comparing (6.CHAP.the number of independent elastic constants is a t most 21 if a strain energy function exists.. its differential is given by au du = -de... Because of this symmetry on CKM. the constants CKMare invariant under the coordinate transformation X. ~3 = -x3 (6. An axis of elastic symmetry of order N exists a t a point when there are sets of equivalent elastic directions which can be superimposed by a rotation through an angle of 2nlN about the axis. Every plane and every axis is one of elastic symmetry in this case.142 LINEAR ELASTICITY [CHAP. The axes of such coordinate systems are referred to as "equivalent elastic directions. = XI.16) is given by Fig.78) respectively. 6 A plane of elastic symmetry exists a t a point where the elastic constants have the same values for every pair of coordinate systems which are the reflected images of one another with respect to the plane. 6-1 r 1 O 01 Inserting the values of (6. 6.16) a s shown in Fig.3 ISOTROPIC MEDIA.27) and (3. 6-1. J ." If the ~ 1 x plane 2 is one of elastic symmetry. the elastic matrix for a material having ~ 1 x 2as a plane of symmetry is C11 C12 c13 0 O C21 C22 C23 0 0 The 20 constants in (6.17) into the transformation laws for the linear stress and strain tensors. xá = ~ 2 .18) are reduced to 13 when a strain energy function exists. (2. If a material possesses three mutually perpendicular planes of elastic symmetry. Certain cases of axial and plane elastic symmetry are equivalent. or 9 if CKM= CMK. The transformation matrix of (6. ELASTIC CONSTANTS Bodies which are elastically equivalent in a11 directions possess complete symmetry and are termed isotropic. the material is called orthotropic and its elastic matrix is of the form having 12 independent constants. . 13 = - l E + v ( ) or E = ..4 ELASTOSTATIC PROBLEMS.19) reduces to the isotropic elastic form In terms of A and p. and the elastic matrix is symmetric regardless of the existence of a strain energy function. when inverted... the shear modulus G relates the shear components of stress and strain. the number of independent elastic constants reduces to 2.. For a so-called state of pure shear. E U i j Lf3 i j ukk E E From a consideration of a uniform hydrostatic pressure state of stress. . = I..27') .. G is actually equal to p and the expression . = -A + 2 ~ ( 3 A 2 ~S i j )u k k + 2P 1 -oij ~r E = -A 2 ~ ( 3 +2P) h I@+-E 1 2P where o = u. which relates the pressure to the cubical dilatation of a bpdy so loaded.2~ or. The constant E is known as Young's modulus. Hooke's law (6. For a simple uniaxial state of stress in the X I direction. 61 LINEAR ELASTICITY 143 For isotropy. l + v E = - l + v x -Li@ -E 001" c. engineering constants E and V may be introduced through the relationships u. certain field equations. and V is called Poisson's ratio. G = ---- 2(1+ v) may be proven without difficulty. = or 0 V .. namely... E p . In terms of these elastic constants Hooke's law for isotropic bodies becomes E u. .CHAP. ( a ) Equilibrium equations.1. . A and p.. Choosing as the two independent constants the well-known Lamé constants.. it is possible to define the bulk modulus. = IE. the matrix (6. = -vc. + ~ b . the symbol traditionally used in elasticity for the first stress ihvariant. t + p b = O (6. ~31. ELASTODYNAMIC PROBLEMS In an elastostatic problem of a homogeneous isotropic body. 6.1( + V E +:ic)1 . = Ec. = C. and c. This equation may be readily inverted to express the strains in terms of the stresses as C.2) for an isotropic body is written where C = c. satisfying (6.LE (6. which are called the Beltrami-Michell equations of compatibility.24) and the equilibrium equation (6. ~ or ) E = &(uV + Vu) (6. For those problems having "mixed" boundary conditions. For those problems in which bound'ary displacement components are given everywhere by an equation of the form ui = gi(X) or u = g(X) the strain-displacement relations (6. the equilibrium equations (6.28) and the result in turn substituted into (6.30) on the boundary.29) may be substituted into Hooke's law (6. The boundary value problems of elasticity are usually classified according to boundary conditions into problems for which (1) displacements are prescribed everywhere on the boundary. the system of equations (6.144 LINEAR ELASTICITY [CHAP. Also.104) may be combined with Hooke's law (6.u ~~ .33) throughout the continuum and fulfills (6. The solution of this type of problem is therefore given in the form of the displacement vector Ui.29) must be satisfied a t a11 interior points of the body.32) over some portion of the boundary.27) to produce the governing equations. cij = 4 ( ~+ . which are called the Navier-Cauchy equations.28) and (6.. prescribed conditions on stress andlor displacements must be satisfied on the bounding surface of the body.29) must be solved.27) to produce the governing equations. (2) stresses (surface tractions) are prescribed everywhere on the boundary. stresses are prescribed over the remaining part.32) on the boundary. The solution for this type of problem is given by specifying the stress tensor which satisfies (6. 6 ( b ) Hooke's law.16) . while the displacements satisfy (6. uij = X a i j c k k + 2peij or 2 = hlc + ~.27). For a11 three categories the body forces are assumed to be given throughout the continuum.31) throughout the continuum and fulfilling (6.28) (c) Strain-displacement relations.27) must be replaced by the equations of motion (5. The stress components must satisfy (6. (6. (3) displacements are prescribed over a portion of the boundary. In the formulation of elastodynamics problems.30) over the remainder of the boundary. The solution gives the stress and displacement fields throughout the continuum. For those problems in which surface tractions are prescribed everywhere on the boundary by equations of the form the equations of compatibility (3. analogous to (6. In plane stress problems. A proof of uniqueness is included among the exercises that follow.O) and Ui = Ui(x.29) with body forces b:" .ujl) represent a solution to the system (6.). The loads are applied uniformly over the thickness of the plate and act in the plane of the plate as shown in Fig. (6. then <rij = a. for locations sufficiently remote from the area of application of the loadings. There are two general types of problems involved in this plane analysis. the geometry of the body is essentially that of a prismatic cylinder with one dimension much larger than the others. and o. If. being applied to some portion of the boundary.35) Solutions of (6. 6. PLANE STRESS AND PLANE STRAIN Many problems in elasticity may be treated satisfactorily by a two-dimensional.5 THEOREM OF SUPERPOSITION. ~ ~ + ( h + p ) ~ ~ . In plane strain problems.27). The loads are uniformly distributed with respect to the large dimension and act perpendicular to it as shown in Fig. ui = gi(x. i The uniqueness of a solution to the general elastostatic problem of elasticity may be established by use of the superposition principle. In terms of the displacement field ui. Venant's principle is a statement regarding the differences that occur in the stresses and strains a t some interior location of an elastic body.36) but also boundary conditions.. ST. the principle of superposition may be used to obtain additional solutions from those previously established. either on the displacements.t) and must satisfy not only initial conditions on the motion. St.'" +ui2) represent a solution to the system for body forces bi = bil' + b'2). the geometry of the body is essentially that of a plate with one dimension much smaller than the others. the governing equation here.28) and (6. t) o r on the surface tractions. 6. . 6-2(a) below.35) appear in the form ui = u~(x. ~0~ 1+ p~V b 2~ ~=+ p( A ~ + ~ p ) V V * ~ + p b = p Ü(6. p ~ ~ ..CHAP. for example.6 TWO-DIMENSIONAL ELASTICITY.:) + d2) ij 9 ui = u. VENANT PRINCIPLE Because the equations of linear elasticity are linear equations. due to two separate but statically equivalent systems of surface tractions. UNIQUENESS OF SOLUTIONS. The principle asserts that. usually expressed by equations such a s ui = ui(x.O) (6. Although these two types may be defined by setting down certain restrictions and assumptions on the stress and displacement fields. t) or u = g(x. 6-2(b) below. ut2) represent a solution for body forces b12'. This assumption is often of great assistance in solving practical problems. they are often introduced descrip- tively in terms of their physical prototypes. together with the law of conservation of energy.31) in the elastostatic case is . 61 LINEAR ELASTICITY 145 and initial conditions as well as boundary conditions must be specified. . the differences are negligible. or plane theory of elasticity. na8 = (xl>xZ) (a.39) Accordingly. 6-2(b) the displacement component us is taken as zero. .104) may be reduced with reasonable accuracy for very thin plates to the single equation In terms of the displacement components u.. the six compatibility equations (3. a. the field equations may be combined to give the governing equation where V 2 = - d2 dx: + 8- d2 ~ .146 LINEAR ELASTICITY [CHAP. the field equations for plane stress are in which a a -êl + -$2 and V = 8x1 ax.. and the remaining components are taken as functions of xl and 2 2 only. and the remaining components considered as functions of X I and X n only. are taken a s zero everywhere..P =1~2) (6. ' For the plane strain problem of Fig.43) Due to the particular form of the strain tensor in the plane stress case.. a. 6 ('J) Fig. 6-2 For the plane stress problem of Fig.. 6-2(a) the stress components a. (6.. 6. the generalized plane stress formulation is essentially the same as the plane strain case if X is replaced by A' = -2XP -.CHAP.L + 1 . the field equations may be written (a) u. 6-2(a) are not uniform across the .'uv . (6..XSa. a state of generalized plane stress is said to exist.229 u12 . I : + p b= 0 (b) u.. = 2(X + /L) in which I: = [i: i: :) O33 and E = (i: i) From (6. the compatibility equations for plane strain reduce to the single equation (6. r.48). is taken as a constant other than zero in (6.-4. In formulating problems for this case.. XP) in accordance with the equations .. but are symmetrical with respect to the middle plane of the plate.44) may be expressed in terms of stress components as The stress components are now given as partia1 derivatives of the Airy stress function = +(xl. 61 LINEAR ELASTICITY 147 In this case.50). (6...+pb. the solution of plane elastostatic problems (plane strain or generalized plane stress problems) is often obtained through the use of the Airy stress function...49). the appropriate Navier equation for plane strain is As in the case of plane stress. must be replaced by stress.52) X 2/.129 u22 . = O or ~ . Even if body forces must be taken into account. . or 2: = Xlr + ZpE . vu. (6. .7 AIRY'S STRESS FUNCTION If body forces are absent or are constant.O or 8-2:= O (6. the superposition principie allows for their contribution to the solution to be introduced a s a particular integral of the linear differential field equations.I1 .3 ... For plane elastostatic problems in the absence of body forces. thickness. - VE (6. +.v2 A case of generalized plane strain is sometimes mentioned in elasticity books when r.~- . In terms of such averaged field variables.55) ull . the field variables a. +~ P E ~ . the equilibrium equations reduce to ua4.53) and the compatibility equation (6. strain and displacement variables averaged across the thickness of the plate. If the forces applied to the edge of the plate in Fig. X O33 . and u.4.44).47). 57) the stress components shown in Fig. 6. 6-3 Taking the Airy stress function now as @ = @(r. 'c' Fig. in polar form. 6 The equilibrium equations (6.54) becomes the biharmonic equation Functions which satisfy (6. x2 =.O). numerous solutions to plane elastostatic problems may be constructed. V2 = - a2 + -- i a i a2 + r2 -- ar2 r ar ao2 ' .148 LINEAR ELASTICITY [CHAP. the stress components are given by The compatibility condition again leads to the biharmonic equation 7 ' ( V 2 @ )= V 4 @= O but.53) are satisfied identically. 6-3 are found to lead to equilibrium equations in the form in which R and Q represent body forces per unit volume in the directions shown. and the compatibility condition (6. r sin O (6. Thus for transformation equations x1 = r cos 8.56) are called biharmonic functions. Of course these solutions must be tailored to fit whatever boundary conditions are prescribed. By considering biharmonic functions with single-valued second partia1 derivatives.8 TWO-DIMENSIONAL ELASTOSTATIC PROBLEMS IN POLAR COORDINATES Body geometry often deems it convenient to formulate two-dimensional elastostatic problems in terms of polar coordinates r and O. which satisfy automatically both equilibrium and compatibility. .22). If now the specific heat a t constant deformation c(")is introduced through the equation .uiqVqi . ci = -kTai (6.70) Heat conduction in an isotropic elastic solid is governed by the well-known Fourier law of heat conduction.T o ) (6. the strain components of an elementary volume of an unconstrained isotropic body are given by €CT) ij = a(T .71) where the scalar k. In this case the constitutive equation is written = KijkrnDkm in which the stress rate U: is defined as d u. the thermal conductivity of the body.(3X + 2.)aSij(T . Due to a change from some reference temperature T Oto the temperature T . In a second classification. ujqV q i where V i j is the vorticity tensor. HYPOELASTICITY Modern continuum studies have led to constitutive equations which define materials that are elastic in a special sense.72) and the interna1 energy is assumed to be a function of the strain components cij and the temperature T.. Thus the constitutive equation is of the form in which Dij is the rate of deformation tensor.c ~ .10 LINEAR THERMOELASTICITY If thermal effects are taken into account.45) may be expressed in the form . may be considered to be the sum c. Equation (6. together with Hooke's law (6.69) may be inverted to give the thermoelastic constitutive equations uij = XSijrkk + 2pe.67) yields which is known a s the Duhamel-Neumann relations.68).TO)Sij (6.(SI ij + €i(j T ) (6. the components of the linear strain tensor r. the energy equation (5.9 HYPERELASTICITY. = . Inserting (6.= ~ p~(")T (6. 61 LINEAR ELASTICITY 149 6.CHAP. a material is said to be hypoelastic if the stress rate is a homogeneous linear function of the rate of deformation.68) where a denotes the linear coefficient of thermal expansion. must be positive to assure a positive rate of entropy production. In this regard a material is said to be hyperelastic if it possesses a strain energy function U such that the material derivative of this function is equal to the stress power per unit volume. 6. = (cij). into (6.67) in which r::' is the contribution from the stress field and rr is the contribution from the temperature field. To) ( d ) strain-displacement relations .(3A+ 2 . . j + p b i =O ~r V0E+pb= O ( c ) thermoelastic stress-strain equations aij = XSijckk + 2 p i j . ~ . quasi-static. 6 kTBii= p~'"' !i' + (3h + 2. displacement and temperature fields. the compatibility equations must be satisfied. but independently. In addition. Thus for the uncoupled. The system of equations that formulate the general therrnoelastic problem for an isotropic body consists of ( a ) equations of motion .150 LINEAR ELASTICITY [CHAP.(3A + 2 ~ ) a (TS ~-~T O ) E Ale +~ P - E (3X + 2..To) ( c ) strain-displacement relations ( d ) coupled heat equation This system must be solved for the stress.& which is known as the coupled heat equation. ~ ) a 8 ~-~To) (T I: = Xlr + ~ P -E (3A + 2 ~ ) a l ( T . There is a large collection of problems in which both the inertia and coupling effects may be neglected. .L)aI(T. ~ + p b ~ = oru ~ V o Z + p b = Ü ( b ) thermoelastic constitutive equations uii = Xaijrkk + BPcij .. For these cases the general thermoelastic problem decomposes into two separate problems which must be solved consecutively.~)aT.. thermoelastic problem the basic equations are the (a) heat conduction equation ( b ) equilibrium equations a i j . subject to appropriate initial and boundary conditions.. 14) and that au*la~.Sij/3) + + = +(sijeij oiiejj/3 siieii/3 + u.21).vSijukk]/2E= [(I + v)oijoij. 6. of Problem 6. uii = 3Keii and so u : ~ )= uiiejj/6 = Keiiejj/2 = K(IE)2/2 From (6.70) into (6. are not necessarily symmetrical. Separating the stress and strain tensors into their spherical and deviator components. Show that this equation may be written in the form of (6. express the strain energy density v. Show that the strain energy density u* for an isotropic Hookean solid may be expressed in terms of the strain tensor by u* = X(tr E)V2 /. oii = ( 3 1 2p) eii = -3p + so that + K = (3h 2p)/3.25).24) into (6.= oiiejj/6 sijeij/2. u* = ( X S i j e k k + 2peij)e i j / = Xeiiyjl2 + peijyj which in symbolic notation is u* = X(tr E)2/2 pE : E.ekk Sijl3).2 in terms of the engineering constants K and G and the strain components.LE: E. develop the formulas for the bulk modulus (ratio of pressure to volume change) given in (6..3. r 6.CHAP.2. ( ~ i~j ~- i i eiieji/3) Note that the dilatation energy density u:.~ ( tL)2]/2E.voiiojj]12Ewhich in + symbolic notation is u* = [ ( 1 v)L : L .. ~ K ~ M .* as the sum of a dilatation energy density u&) and distortion energy density u&). u* may be expressed in the quadratic form u* = c ~ ~ in z which ~ E the~ C&.3) 6. r Inserting (6..~ ( tZ)2]/2E.ejj/3) + and since eii = sii = O this reduceS to U* = u'lS> uTD. oij = ASij ekk + 2peii = sij + okkSij/3 and since oii = (3X + 2p)eii i t foiiows that sij = 2p(eij . u* = o i j [ ( l v)oij .1. Thus u : ~ )= zP(eij . 6.) appears a s a function of K only. Thus the derivative au*laeR is now + + + C K R E K= au*/aeR = ~ C K M (R~' MK . + + u* = &(sij okksij/3)(eij e. STRAIN ENERGY. Assuming a state of uniform compressive stress uij = -pai.24) becomes eij = [ ( I V)(-pSii) + + vSij(3p)]lE and so yi = [ . Inserting (3.& c K ~ ~=~ E&CKMeKeM ~ ++ ~ : ~ e= + ~ e& (~C z M c M K ) e K e M = *CKMeKeM where C K M= C M K . Express u:. Write the quadratic form a s U* = & c Z M e K c M + .5. In general. Thus K = -p/eii = E / 3 ( 1 .13).= a. 6.3.21) and (2. Likewise from (6. + 6.98) and (2. the shear modu?us. 61 LINEAR ELASTICITY 151 Solved Problems HOOKE'S LAW.70).13).13).~ ~ ~ / 3 ) (=~ p ~ ~ ~ 8 .. (6.+ + Inserting (6. and u&. and in terms of + + the stress tensor by u* = [(I v)Z: I: . whereas the distortion energy u'lD.3 p ( l + V) + 9pvIIE. From a result in Problem 6.2v). ISOTROPY (Sec. is in terms of p (or G). With uij = -pSii.=R )& C K M ( ~ K R EEMK ~ M R ) = J ( C R M E M = ) CRM~M "R .4.21) into (6.1-6. for example. Let the zlx2 (or equivalently.6 results in the matrix (6.19) to the isotropic matrix (6. z.U.19).27) and (5. 6 Show that for an orthotropic elastic continuum (three orthogonal planes of elastic symmetry) the elastic coefficient matrix is as given in (6.5. 6-5 Fig.-cK for K = 5.( = UK. 6-4 and from (2.78). )I - c~ . 6-5. = -UK. Then UK = CKMcM and also u a = CKMcM.6. from o. Fig. 6-4). cK . = C45 = C54 = = Cs2 = Cs3 = O and the elastic coefficient matrix attains the form (6. Now C16 = C2. If Z2x3 (or XZX:) is a second plane of elastic symmetry such that a. Thus. In particular.. = cK for K = 1.4 whereas O.20). Likewise. = CKMcM. = o. page 142.27) and (J. U. the transformation array is I1 - and now from (2. are equal only if C14 = C15 = O. for the rotated zl axes shown in Fig. C44 = C55 = and C12 = = Cl3 = = Cz3 = c32. For isotropy.78). the method of Problem 6.-€K for K = 1. = o. U. 02 = UK.3. = o. The student should verify that elastic symmetry with respect to the (third) xlx3 plane is identically satisfied by this array. = -v4.152 LINEAR ELASTICITY [CHAP.6 whereas UL = -UK. i t is found that C% = C25 = = C35 = C64 = C65 C41 = = c43 = = = C53 = C5@= 0. The transformation matrix between xi and zl is Fig. 03 = u3.2. eg = -cK for K = 4. = -u5. o.xh) plane be a plane of elastic symmetry (Fig.2.3. These two expressions for U.19) being further simplified by the conditions Cll = CZ2= C33.7. Give the details of the reduction of the orthotropic elastic matrix (6. elastic properties are the same with respect to a11 Cartesian coordinate axes. = C3. 6.19). from c. 6-6 . = CIMcM. o.21) with i = j.2 ~ = ) 3X + 2p.i j ujPij). lr(ui.CHAP.-e5.8.31). 6-6 above. + 2p) From (6..11.j u ~ . for example.C. ELASTOSTATICS..ji + pbi = 0. Likewise. Replacing the strain components in (6. E l ( 1 . ~ )Thus . and from (6. C3.26). ... = e. ..20) is obtained. But o[ = C44~[ and so 2C4. Give the details of the inversion of (6.. = and e: = e2. . 6. Let x3 be the axis of elastic symmetry. e: = -eg.28) by the equivalent expressions in terms of displace- + + ments yields oij = AGijuk. = cl.27) and rearranging terms gives p u . 6-7 [CKMI = with seven independent constants. A rotation e = 2a/4 = r12 of the axes about x3 produces equivalent elastic directions for N = 4.Clz)(az.10. . qi = (3X + 2P)Yi and so 2 p i j = uij . o.27) and (3. 61 LINEAR ELASTICITY 153 Finally. u. Express the engineering constants V and E in terms of the Lamé constants h and p..78). 04 = o.25)..4 1 2 and e.. = '3. Thus defining p = C44 and X = C1.5) 6.. j j + + ( X lr)uj...Substituting this into the equilibrium equations (6. E l ( l + v ) = 2p.26). E. Determine the elastic coefficient matrix for a continuum having an axis of elastic symmetry of order N = 4. Thus. E = 2p(l + v) = p(3X f 2p)/(X + p).. = u4. = C3. 4 = e4. 6. o. the transformation matrix is [aijl = so that o[ = (o. = C. (6.9. Assume CKM= CMK. = o. 6.XSij~kk/(3X or eij = oij/2p - + X8ii~kk/2p(3X 2p). From (6.22). from the remaining five stress relations. 6. C34= C35= = 0.u1)/2 = (Cll . ELASTODYNAMICS (Sec.2v) = 2p(l + + v ) from which v = X/2(X p). for the axes x:'obtained by a 45O rotation about x3 a s in Fig. 1 - e4 . The transformation matrix is [aiil = o o and by (2. Now by (6. from o3 = 83.21)to obtain (6.j = Au. = o.e.4-6. o: = -05. the elastic matrix becomes Fig. oij. Thus (3X + 2p)(1. + + $(ui. Derive the Navier equations (6. . Inserting these into ( 6 . .F k . j + bj.b' = c2(g1' hl')lr. jkki = O provided Fi. 2 7 ) gives from which = -(i + ~ ) p b ~ . ~ h u s . show that + = g(r ct) r h(r . + eijk$'k.(r2:) since @ = + ( r .Bey.8) 6. If p b i = @ . 6. develop the stress-strain relations in terms of A and Show that these equations correspond to those given as (6. Substituting ( 6 . Here i t is convenient to use the spherical form V 2 -. k k j j / ( A P ) . Then V 2 4 = (g" + h")/r.j i l p + + is a solution of the Navier equation (6. y = 1 . 2 4 ) into ( 3 . .31) for zero body forces.pji= 0. i .i +c ~ ~ ~ $ provided + and $i each satisfies the familiar three dimensional wave equation. Differentiating the assumed soiution.154 LINEAR ELASTICITY [CHAP.jj= ( A 2 p ) F i . j j i + €jpq$'q.33) and determine the form they take when body forces are conservative. = P( @.35) is satisfied by ui = +. 6. when p b i = +. = + + 2 ~ ( 3 A 2 p ) ~ . +2p) = -voyy/E .15./(X 2 ~ ) + + with a .jqq) + ( A + ~ ) ( @ . Substituting the assumed ui into ( 6 . . Show that if V4Fi = 0.(A p ) F k . then p ( b i . 6. i + 'ijk(~$'k. 9 3 ) .( A + 2 P ) f ( A + P)IFj. + Thus ( 6 .13. Thus setting m = k and using ( 6 .C r 2 ( a @ l a r ) = r ( g ' + h ' ) . k k j j / ( A P ) .h r c ) / r and . inserting this expression for O . = @ .i. k k = V 2 @ so that ( 6 .i Since ypq\l.pji) ./(A +2@) = -Aoyy/2P(3). k k into the previous equation leads to ( 6 ..[ P . this equation may be written ( ( A f 2 1 ~ ) @ .12. 2 from which o. the terms @ui. 6 6. €33 = -Ae. ~-/ (v l) . 3 3 ) becomes TWO-DIMENSIONAL ELASTICITY (Sec.. k j j i / P are readily calculated.P ? ) . show that (6. P. the displacement ui = ( A 2p)Fi.6.jiIp(A p ) . k k. 3 1 ) gives + + ( A 2 G ) F i . Also 4 = ( g ' c .41). k i j j and + (A + + + = ( A 2 p ) F j . 9 5 ) yields + ~ ( @ .@ = 2ApS. 2 1 ) reduces to o. / ( ~ 2 ~ so ) that the equation may be inverted to give Also. Only six of the eighty-one equations represented here are independent. Writing c2V2+= where c2 = ( A + 2 p ) / p for the wave equation derived in Problem 6. k k j i / P .( g + h ) where primes denote derivatives with respect to the arguments of g and h.q.6-6. i k k eijk$'k.14.~ $ ' k ) j. 6. If body forces may be neglected.e. 1 0 3 ) yields where 8 = I= = uii. i. for the given @.ij and p b k . Therefore + c 2 V 2 @ = .. 2 1 ) yields e13 = ep3 = O and €33 = -A(e11+ e Z 2 ) / ( A 2 ~ ) .kkjj= V 4 F i = 0. = O = p?/(A which is satisfied when V 2 @ +2 ~ and ) V2$'k = pYk/fl.13.i) = 2@.. Here u~~ = nl3 = uz3 = O so that ( 6 . For plane stress parallel to the x1x2plane.j).ct) is a solution with g and h arbitrary func- + + tions of their arguments and r2= X i x i .qq . t ) .Fj.16. Derive the Beltrami-Michell equations (6. 42) leads to a..2222= O + 2 4 A x 2 + 120Bx2 = 0 .45) and show that i t is equivalent to the corresponding equation f o r plane strain (6. 6-8 6. / (+h 2p) + pb. = h V 2 u a + p(3A + 2 p ) ~ ~ . Since +.pm becomes q j = eip3ejrn3+33. The stress components are therefore 0 1 1 = +. Differentiating with respect to x p and substituting into (6. = ( 1 v ) ( l ./2(A + + + + = (1 v)o.. 11.x.20.yy . F o r plane strain parallel to x1x2.11 + +. 0 2 2 = +. Fig.51) if A' = 2 A p / ( A 2 p ) is substituted + f o r A.22.1111 2+... 0 1 2 = -+. -c < xz < C.v ( l v ) S a p o Y y l E from which e.puy.22 . ~ Thus + ~ ~ .v) + pb. 6.p = vES./Z(l.56). Determine the necessary relationship between the constants A and B if + = + AxSxi Bx2 is to serve a s a n Airy stress function.pm which may be written oap = E . 6-8).B = ( S a ~ S y ( . =( .17. is represented by the case +.il = 0.36 it was shown that the equilibrium equations were satisfied in the absence of body forces by uij = c ~ ~ ~ c Show ~ ~ ~ that+ Airy's ~ ~ . " 2 2 = 9.55) are oll = +. +. uiz = These stresses are those of a cantilever beam subjected to a transverse end load F and an axial pull P (Fig. ".41) and using (6. - E.~ 6. The stress components as given by (6.)/2(1 + + v) + 2vES.. (6.1122 + + +.. pp/2(1+ v) + Eup. 0.+.. +. ~ up.p .li + +.22 = +.24) Finally. Here u3 O so that eg3 = O and (6.21.p + 6. 6. Thus (6.peyy + 2pe. Develop the Navier equation for plane stress (6.v2). +. + p). +. SYB)@.plE . Show that these equations correspond to those given a s (6.12.+.p = E ( U . Inverting (6.) = Ao.lE..40) gives Eu. p . Show that += xi is suitable f o r use a s an Airy stress function and determine the stress components in the region x t > O.24) gives 033 = v(oll o. = O Thus since p(3X + 2p)/(A+ 2p) = (2Ap/(A+ 2p) + p) = (A' + p).2") Ee...2v)o. 61 LINEAR ELASTICITY 155 6..45) and (6. By (6.) with = = = = +.x i ) / 4 c 3 .18.51) have the same form for the given substitution. o. . + is a valid stress function. + must be biharmonic or +. = -3F(c2 . = +(xl..p+.since +33 = +.li = +.develop the stress-strain relations in terms of V and becomes inverting.. .48). which is satisfied when A = -5B. oij = eipqejmn+qn.CHAP.22 . In Problem 2..p..pl(l + v) = AS.stress ~ ~ function .19.peyyl(l+ v ) ( l .+. Since 84+ is identically zero.y/2(1 .is the only non-vanishing component.22 = -3Fx1x2/2c3 + P/2c. ~ ~ E ~ y t~. 73) by use of the free energy f = u . Show that the distortion energy density u:.. (a2 . Solving (6.73). the result is + (oij.T o ) ). ) Finally from (6. From (6. From (6.u 2 ~ 3. 6.41) p& + aijS pTs where dots indicate time derivatives.T o ) = 2peij + ASijekk . + \ From Problem 6. Carry out the inversion of (6.70) to develop the strain energy density for a thermoelastic solid.24. since ds/dT = -a2flaT2..61). 156 LINEAR ELASTICITY [CHAP. (3X + 2p)aSij(T. . may be expressed in terrns of princigal stress values by the equation uy.3a(T .69) with i = j. In polar coordinates (r.. 6.d Z f l d T 2 . =p ~ = (a:j S pT . Substituting (6. T ) and substituting into (5.69) to obtain the thermoelastic constitutive equations (6. 2r de) = ~(. = dSjldT and so combining ( 5 3 8 ) with (6.pôf/dsij)iij .u1)2]/12G (03 .a3u1)/3]/4G = [(vl. + o: . Since the terms in parentheses are independent of strain ~ ~ s = -dfldT.13).70). p(d2fld~ijdT) -ci. Zij = 0 and comparing this equation with (6.13) and (6.i = kTBii = pT -iij (YT + )- :c T .3a(T . Develop the thermoelastic energy equation (6.u2)2 + (v2.Ts.. Determine the stress components and the value of the constant B.T o ) = 2pyj + XSij(ekk . = (uij-8ijukk/3)(aij..70). O ) the Airy stress function CJ= BB is used in the solution of a disk of radius a subjected to a central moment M.To)eii12 MISCELLANEOUS PROBLEMS 6. 6 6.38) for a reversible isothermal process.i.TO) I 6. doil/dT = (3X+ so that kT..70) directly into (6.2.p(s d f l d T ) T = O..(3X + 2p)a(T .(3X + 2p)c18~~(T . 6-9 LINEAR THERMOELASTICITY (Sec. i t follows that uij = p d f l a ~ and From (5.. Fig.ii = pc(")T + (3X + 2p)aT0eii which is (6.u J 2 + ( a 3. = Blr2. Use (6.2paSij(T .62) o(. c(") = .(a1+ o2 + o3)(u1+ + 03)/3]/4G 42 = [2(u. A t constant deformation. u:.u3)2 + .25. + a: + o: .26.olo2 . Equilibrium of moments about the center of the disk requires M = Thus B = M 1 2 ~ .T o ) ) . and temperature rates. f = f(e. Assuming the free energy to be a fynction of the strains and temperature. = o(oe) = O.23.To)eij/2 = Xeiiejj12 + peijeij .10) 6.8ijop.69) for oij gives uij = 2peii + X S i j o k k / ( 3+~ 2p) . + U* = X8ij~kk€ii/2 peijeij .~ d Jo' 2r B de = 2nB.) = sijeij/2 = siisij/4G which in terms of stress components becomes u f D . Also.2aa8ij(T .71). From (6. dTas + "> . o(./3)/4G = (oijoij-oiioji/3)/4G In terms of principal stresses this is ufD) = [o: + o .60) and (6.~ ~ ) ~ ] / 1 2 G . from above. aii = (3X + .T .22. = [ ( a .72) gives c(") = T(ds/dT) or from above. The total strain energy is given by Fig. But this is precisely the same situation a s the reflection about a plane of elastic symmetry.1. the elastic properties (Hooke's law and strain energy density) are of the same form as a continuum having one plane of elastic symmetry. + From Problem 6. Express the strain energy density u* as a function of the strain invariants. 6-10 Note that since T = LPT + f ~ a ( % : x:)r dr de = Gaa4n-12.heiiSpq + 2pePq = upq Likewise from Problem 6. u* = [(I + v)uijuii . Show that (6.1.4 j Y j ) / 2 . = cii and 11.CHAP. The transformation between xi and xf axes is cos 0 sin 0 0 and from (2.31.30. Thus u* = Ga2r2/2. the externa1 work.~ u ~ ~ u and ~ ~so] / ~ E au"la. and since by comparison with (3.u~~= G a x l where a is the angle of twist per unit length.1 to show that for an elastic material d u * l d ~ = .u8pquii]lE = epq 6. . u* = Xeiiqj/2 + priirij. Use the results of Problem 6. 61 LINEAR ELASTICITY 157 6. 6-11 . From Problem 6. Here a rotation of axes e = 2 ~ 1 N= 2 ~ 1 2= n produces equivalent elastic directions. ~ uij and ~ u * I ~=u É~i j . Determine expressions for the strain energy density and the total strain energy in the shaft. + From Problem 6. 6-11). 6.1.Urpq - . Here t r I O and I : I = 2G2a2r2 where r 2 = x: +x:.v(tr I ) 2 ] / 2 E . the nonzero stress components are a13 = -G<Yx~. = (riirjj. 6.91). ) = X/2[eii8jp8jq + ~ ~ ~+ 82 / ~J ~~ ~ 8=~ 8X/2[€iispq ~ ~ ~~ 8] $ ~ ~ € j j s p q ]f 2.20) by an arbitrary rotation í3 of axes about 2 3 (Fig. 6-10.27).29. Show that for a continuum having an axis of elastic symmetry of order N = 2.28. i t follows that 6. u* = [ ( I v ) I :I . u* = Xeiiejjl2 prijsij and so + + Ju*larPq = X/2[eii(aejjla~pq)ejj(âciilòepq)] Z p ~ ~ ~ ( a e ~ ~ l a $ . C44 = Cs5= C66 and C12 = C13 = C23 may be reduced to (6..19) with C11 = C22 = C33.1. U = TaLI2.27. When a circular shaft of length L and radius a is subjected to end couples as shown in Fig. I.v(uiiôpq + ujjôW]/2E = [ ( I + v)uPq. = [2(1+ u)uijSipSjq. Fig. 31) may be p u t in the form pui.'. ul = Cllel i. Use the equilibrium equations for this situation to S ~ O Wthat pUi. so t h a t ui. i2) 0 .32.s + uj.i13 pbi = 0. Use the result of Problem 6. For linear elasticity superposition holds. I f the strains are equal for the t w o assumed solutions. T h u s L uijuinjd S = l. + (cos2 e . t h e l e f t hand integral Jv u* dV from is zero here since tln' = tu' i .el). + C12(e2 e l ) and q = CtlY C12(el e. t h e stresses are also equal b y Hooke's law and the displacements are equal t o within a rigid body displacement. .jj @. Finally f r o m (6. Consider first the surface integral w i t h ti("' = ujiuj and convert b y Gauss' theorem.33.jui + uijui. + + U* = ( C l l ~fl 2C12e2 2 C 1 3 ~ 5 ) ~f1 / 2(C22e2f 2c23e4)"2/2 C3363 f C44'42 Cs5e: f C66ei.32 to establish uniqueness of the elastostatic solution of a linear elastic body by assuming two solutions O : ) . ) .32) and ui = u:l' . For an elastic body in equilibrium under body forces bi and surface forces ti"'. T h u s for this "difference" solution LA t:")uidS = 2 Problem 6. u:') and O : ..sin2 @)e6 ( 2 sin e cos @)e2 + B u t a. Use equations (6.32. 6. eii = ( 1 . or e::) = e!?). 6. Develop the expression for the strain energy density u* for an orthotropic elastic medium. = u 1. + + From equation (6. ui = uil' . Prove t h a t the principal axes o f the stress and strain tensors coincide for a homogeneous isotropic elastic body.u'i2' on the boundary for equation (6.20). .ul = + (Cll . 6.4).35. ui2). The Navier equations (6. 2cij.14) and (6. I t is required t o show t h a t A i pbiui dV + Jv t:"ui d S = 2 u* dV.sin2 e)a6 + (sin e cos e)u2 Likewise from (3..cl). for a n isotropic body and so here u2 . 6. pbiui dV +2 S aiieij/2 dV and t h e theorem is proved.j = -pbi Thus 4 t$'ui d S Sv = .~ 1 would ~ ) also be a solution for which bi = O. SinceAthe t w o assumed solutions satisfy boundary conditions. Supplementary Problems . u 3. Therefore (Cll . and for v = Q.j/E B u t uj. e.ji= O and E = 3G when v = 3. show that the total strain energy is equal to one-half the work done by the externa1 forces acting through their displacements ui. 6 ai2 = (-sin e cos e)ull + (cos2 e .24).19) w i t h the given conditions. 158 LINEAR ELASTICITY [CHAP.) ..ti2' on the boundary for equation (6.33 .) + and so o2 .5'0). eii uiBi= O.2v)aiilE. dV = L (aij. = (-2 sin e cos e)€.ukkJ3G or pV2ui + e j i / 3+ pbi = 0.jlE .C12)= 2C44 and w i t h C* = p. j) dV B u t a i j u b j= uij(cij+ oij)= uijeij.j j = -pbilG . and f r o m equilibrium oij. G12 = h. T h u s uniqueness is established.C12)(e2.e.sin2 e)o12 + (sin 6 cos @)oz2 or i n single index notation.19).34.ai = 2C44(€2. (cijui). Thus rs* dV = O and since u* is positive definite this occurs Jv only i f eij = e:!) .2v8ijukk.ji + pbi = O which for the incompressible case ( V = 4) are clearly indeterminate. = 2(1 v)aij.a:. = C44€.13 + + .24). + . A 6. T h u s f r o m (6.78)and (6.36. = (-sin e cos e)ul + (cos2 8 . so uij = b:. Cll = h 2p as + given i n (6. Ans.To)a + 6.V ) for plane strain. = -oz2 = 6QGx1x2/r5. Ans.49.43 and show that the compatibility equation (6..44) may be written V20aa = Q2+/(1 . C = r / ( 2 sin 2a) 6.CHAP. the compatibility condition (6. show t h a t ui = 2(1. Show that the distortion energy density uyD.2 .43.24%.48. o. + + + . ) 6.v)V2Fi/G . In plane stress thermoelasticity show that €33 = -v(ul1 + + + aZ2)lE a(T .x.x .44) may be expressed as V4+ = . u 2 = u3 = O is a solution of equation I ( 6 3 5 ) when body forces are zero.km = aipaiqakTa. .FjBji/G is a solution of the Navier equation (6. 6.To). Determine the value of c for which ul = A sin. = -T when e = -a.u13 = -oz3 = 3QGx2x3/r5.31) when bi O (see Problem 6. If F = B ( x 2 ê l . V) = X/(h + 2p).a E V 2 ( T . determine the stress components. Ans. where Q = 4 B ( 1 ..37. show that for plane strain thermoelasticity the compatibility equation (6.v ) and that for plane stress as V 4 + = -aEV2(T . 6. 6. Show that for an elastic continuum with conservative body forces such that pba = V+ = $. u12 = 3QG(x.3x2 -2x.x l ê 2 ) / r where r 2 = xixi. In terms of the Airy stress function + = +(xl. ora = T when e = a.39.47.To). <rij = 24 o 2v(x.. C.oiiojj/3)/4G and the dilatation energy density * urS.2volla22 2(1 v ) o y 2 ] 1 2 ~ ( b ) u* = (a + ~/2)(e?1+egz) + Xe11'22 + 2aef2 2a 6. .12). 2: . or 820aa = (1 + v ) V 2 + for plane stress.. . + Show that the Airy stress function + = 2%. 6.C I 2 ) and that C13 and C33 are the only remaining nonzero coefficients.41...Sap(3X 2p)a(T . 12x. Determine the form of the strain energy density for the case of ( a ) plane stress. ( b ) plane strain.44. and = O .24%: .. = oiiajjl18K. show that C2. = CG.46..x . If V4Fi = 0.sCpqrs. ) o O 6. Show that in plane strain thermoelastic problems 0 3 = ~ v(oll + oZ2).. a.T o ) / ( l. and x. For an elastic body having an axis of elastic symmetry of order N = 6. and b2 are functions of x .. only.a E ( T . Show that 1 / ( 1 + V ) = 2(X + & ( 3 ~+ 2p) and vl(1.44) is satisfied.40. eij = 24 (%) ( x.4 ) / ~ 5 . Am. For plane strain parallel to x.32. 61 LINEAR ELASTICITY 159 6. Ce6 = 2(Cll . are the components of a fourth order Cartesian tensor so that Ci.x2). 6.( x l c t ) . Ans. -2x1x2 + x.. .. c = d w 6.T o ) and that = + hSapeaa 2 p n B . Ans. O and that b.= C. Determine C if use = O.T o ) and eap = ( 1 v)aaBIE.6x2 satisfies the biharmonic equation V4+ = O and determine the stress components assuming plane strain.vSaPoauIE Sap ( T . Determine the strains associated with the stresses of Problem 6. ( a ) u* = [o.45.x. . Use the transformation laws for stress and strain to show that the elastic constants Cijk. 6.= (uijoii. ) -2xlx2 o x.38. 6. = 0.x.v)lG. .50.cos 2a) where C and a are constants. In polar coordinates an Airy stress function is given by @ = Cr2(cos 2e . .. . show that b.42. or hydrostatic pressure. BAROTROPIC FLOW In any fluid a t rest the stress vector tin)on an arbitrary surface element is collinear with the normal íi of the surface and equal in magnitude for every direction a t a given point. the pressure p. However. Here every direction is a principal direction. = -p o s i j or 1 = -p O I which represents a spherical state of stress often referred to as hydrostatic pressure. VZscous fluids on the other hand are those for which rij must be considered.1) a . the shear stress components are observed to be zero in a fluid a t rest. an inviscid. the mean normal stress is given by For a fluid a t rest.n (r*1) in which po is the stress magnitude. r . and from (7. the pressure p is essentially the same as the pressure associated with classical thermodynamics. which in this case is equal to the negative of the mean normal stress. Chapter 7 Fluids 7.2). In a compressible fluid. For an incompressible fluid. .3 n=. and it is customary in this case to resolve the stress tensor according to the equation where rii is called the viscous stress tensor and p is the pressure. or so-called perfect fluid is one for which rij is taken identically zero even when motion is present.1 FLUID PRESSURE. VISCOUS STRESS TENSOR. -poni or t(^) = 1-íi = -p. From (7. these characteristics vary widely in different fluids so that it is often possible to neglect their effects in certain situations without significant loss of accuracy in calculations based upon such assumptions. The negative sign indicates a compressive stress for a positive value of the pressure. For a fluid in motion. . For a compressible fluid. Accordingly. Thus A A tin) = ~23. the shear stress components are usually not zero.3). . A11 real fluids are both compressible and viscous. the density p and the absolute temperature T are related through a kinetic equation of state having the form An example of such an equation of state is the well-known ideal gas law . vanishes and p reduces to p. the thermodynamic pressure is not defined separately from the mechanical conditions so that p must be considered as an independent mechanical variable in such fluids. From (7. In this way the thermodynamic pressure is defined in terms of the mechanical stresses. 7.6iiD. the constitutive equations for an isotropic homogeneous Newtonian fluid may be determined from (7.e. In developing constitutive relations for fluids. In terms of the deviator components sij = aii .13) the first of which relates the shear effects in the fluid and the second gives the volumetric relationship. NEWTONIAN FLUIDS The viscous stress components of the stress tensor for a fluid are associated with the dissipation of energy.9) above may be rewritten in the form Therefore in view of the relationship (7. If the functional relationship is a nonlinear one.. 6ijokk/3and D i = Dii . i. When the function is a linear one of the form where the constants Ki. STOKESIAN FLUIDS. If the changes of state of a fluid obey an equation of state that does not contain the temperature. Following a procedure very much the same as that carried out for the generalized Hooke's law of an elastic media in Chapter 6. The condition that is known as Stokes' condition.. .3). equation (7. 71 FLUIDS 161 where R is the gas constant. the fluid is known as a Newtonian fluid.9)' the mean normal stress is given by + where K* = Q(3A* 2p*) is called the coeficient of bulk viscosity.k/3. Some authors classify fluids simply as Newtonian and non-Newtonian. are called viscosity coeficients. as expressed symbolically by the fluid is called a Stokesian fluid. p = P ( ~ ) such .CHAP.10).2 CONSTITUTIVE EQUATIONS. and guarantees that the pressure p is defined as the average of the normal stresses for a compressible fluid a t rest. i t is generally assumed that the viscous stress tensor T~~ is a function of the rate of deformation tensor D i j . The final form is where A* and p* are viscosity coefficients of the fluid.equation (7. or S = 2p*Df (7. changes are termed barotropic.7) and (7. An isothermal process for a perfect gas is an example of a special case which obeys the barotropic assumption. From (7.12) may be expressed by the pair of equations sij = 2p*D. + (A* + p * ) ~ .24) + PV = pb .P*i + *!J*vj.18) above is substituted into (7.*v(v V ) p*V2v + . If (7.22) reduce to the Navier-Stokes equations for incompressible flow. ( c ) the energy equation (5. ( g ) the caloric equation of state. .3).+.162 FLUIDS [CHAP.15) through (7.16) and the definition 2 0 . P = ~ ( p T. are required. (7.5). NAVIER-STOKES-DUHEM EQUATIONS I n Eulerian form. 9-4 = ~ ( p T) . ~p * ~ ~ pGi = pbi . The system of equations (7.32).22) reduce to the Navier-Stokes equations for compressible flow PVi = Pbi . (7. as they very often must be in fluids problems.21) represents sixteen equations in sixteen unknowns and is therefore determinate.VP + (A* + p * ) v ( v + p*V2v V) When the flow is incompressible ( v . = ( v L j + is used. the basic equations required to formulate the problem of motion for a Newtonian fluid are I (a) the continuity equation (5.71).V p +p.3 BASIC EQUATIONS FOR NEWTONIAN FLUIDS. ~ ~ or pV = pb . ( e ) the kinetic equation of state (7. (d) the constitutive equations ( 7 4 . = O). = O or . the equations that result from the combination are the Navier-Stokes-Duhem equations o f motion.ji+ p*vcjj or (7.pSi .15) (b) the equations of motion (5:16). ) If thermal effects are considered. If Stokes condition is assumed (A* = -pp*). 7 7. + p ( V x 0 v )= O (7. the additional equations ( f ) the Fourier law of heat conduction (6. . Furthermore.CHAP. causes the Navier-Stokes equations (7. only the normal velocity component is required to vanish on a fixed surface. 71 FLUIDS 163 The Navier-Stokes equations (7.J v . IRROTATIONAL FLOW The motion of a fluid is referred to as a steady flow if the velocity components are independent of time. 7.28) which describes the hydrostatic equilibrium situation.4 STEADY FLOW. the appropriate boundary conditions a t a fixed surface require both the normal and tangentiaI components of velocity to vanish. the Reynolds number is N. velocity V and density p..22) to reduce to pbi = P. Thus if a flow is characterized by a certain length L.. if the body force may be prescribed by a potential function equations (7.15) form a complete set of four equations in four unknowns: the pressure p and the three velocity components vi. If the barotropic condition p = p(p) is assumed. This condition results from the experimentally established fact that a fluid adheres to and obtains the velocity of the boundary. a pressure function P(p) = spgpa P (7. It is only in the case of laminar flow that the constitutive relations (7. The ability of a flow to support turbulent motions is related to the Reynolds number. viscous contribution to the shear stress terms of the momentum equations may be neglected. = VL/v (7. which expresses the ratio of inertia to viscous forces.. One of the most significant and commonly used ratios is the Reynolds number N. HYDROSTATICS. For an inviscid fluid.28) take on the form A flow in which the spin. .. the apparent stresses act on the time mean flow in a manner similar to the viscous stress effects in a laminar flow. the derivative avilat is zero.25) where V = p*lPis called the kinematic viscosity.3 Or +=VV. the . For very large Reynolds numbers.29) may be defined. or pb = VxP (7.21).23). and hence the material derivative of the velocity reduces to the simple form i i = V . inertia effects outweigh viscous effects and the fluid behaves as though it were inviscid. In turbulent flow. For this situation. severa1 ratios of the normalizing parameters appear. If the Navier-Stokes equations are put into dimensionless form. or vorticity tensor (4.VXV A steady flow in which the velocity is zero everywhere. In any given problem. together with the continuity equation (7. 2. the solutions of this set of equations must satisfy boundary and initial conditions on traction and velocity components.18) apply to real fluids. For a viscous fluid. . If turbulence is not present. 22) reduce to the form which is known as the Euler equation of motion. If the flow is irrotational. Qi = or zijkvk. Furthermore.37) is integrated along a streamline. = f v i d x i or r. in general. . 7 vanishes everywhere is called an irrotational fiow. Bernoulli's equation requires the total head along any streamline to be constant. the line integral (7. a single constant C holds everywhere in the field of flow. .Vv = -V(n+P) (7.38) If the Euler equation (7.17) For steady motion. 'the resulting fluid is called an inviscid or perfect (frictionless) ftuid and the Navier-Stokes-Duhem equations (7.41) may be converted to the surface integral where íiis the unit normal to the surface S enclosed by the path. = -4. In this case the integrand of (7.154). = cijkVkj 01: q = V" (7. (7. vt i 3. = vedx f (7.42) is the perfect differential d+ = -v. = Plg defined as the pressure head. = -(n+P).i = -(n+P).33) and therefore also vanishes for irrotational flow. avilat = O and C(t) becomes the Bernoulli constant C which is. dx with the velocity potential. different along different streamlines. For incompressible fluids (liquids). page 23.i or v.164 FLUIDS [CHAP. + h" = h + plpg + v2/2g = constant (7.35) 7.36) becomes . the potential n = gh where g is the gravitational constant and h is the elevation above some reference level. or v = -V+ (7.34) and since V x v = O is necessary and sufficient for a velocity potentkl 4 to exist. For a barotropic fluid with conservative body forces. If the flow is irrotational as well. BERNOULLI EQUATION.5 PERFECT FLUIDS. the velocitz~circulation around a closed path of fluid particles is given by the line integral r. CIRCULATION If the viscosity coefficients h* and p* are zero.153) or (1.37) For steady flow (7. or V = -v(a+P) (7. the velocity vector for irrotational flow may be expressed by v. the equation takes the form h + h. V x v = O and the circulation is zero.j q = Vxv (7.29) and (7. The vorticity vector qi is related to the vorticity tensor by the equation q.40) By definition.30) may be introduced so that (7. When the only body force present is gravity. the result is the well- known Bernoulli equation in the form (see Problem 7.37) may be written v3. and v2/2g= hv defined as the velocity head. Thus with h.41) From Stokes theorem (1. . regardless of whether the flow is irrotational or not. PLANE POTENTIAL FLOW The term potential flow is often used to denote an irrotational flow since the condition of irrotationality.35). For a compressible irrotational flow.50) By eliminating + and + in turn from (7. For a steady irrotational flow of a compressible barotropic fluid. For incompressible potential flow the continuity equation attains the form and solutions of this Laplace equation provide the velocity components through the definition (7. Greek subscripts have a range of two. = 0. 71 FLUIDS 165 The material derivative drcldt of the circulation may be determined by using (4. On a fixed boundary.1 = q..= c2V2+ (7. = -+. An important feature of this formulation rests in the fact that the Laplace equation is linear so that superposition of solutions is possible. d+ldn = O.) such that If the plane flow is. plane.2 and +.6 POTENTIAL FLOW.41) gives For a barotropic..60) which when applied to (7.49) the stream function and velocity potential are seen to satisfy the Cauchy-Riemann conditions 4.48) and (7.CHAP. is necessary and sufficient for the existence of the velocity potential + of (7. = o or B'V = o (7.44) where c is the velocity of sound in the fluid.x. indeed. as usual in this book.x. By (7. In a two-dimensional incompressible flow parallel to the x. the Euler equation and continuity equation may be combined to give which is the so-called gas dynamical equation. 7.47).35). Boundary conditions on velocity must also be satisfied.i (7.. + = or . i t is possible to introduce the stream function + = +(x.50) i t is easily shown that .. inviscid fluid with conservative body forces the circulation may be shown to be a constant. V x v = O. and the continuity equation becomes v. v. irrotational so that then from (7. for example. the Euler equation and the continuity equation may be linearized and combined as is done in acoustics to yield the governing wave equation . This is known as Kelvin's theorem of constant circulation.47) where. 3 ) is equal to t. $2) (7. the complex potential @(4 = +(x19 x.2.l / T = RIC.1-7. with K * = 0. Frictionless adiabatic.54) Solved Problems FUNDAMENTALS OF FLUIDS.3. the temperature-density relationship is # . 7. a constant.166 FLUIDS [CHAP. Determine the constitutive equation for a Newtonian fluid with zero bulk viscosity.4. i. Inserting p = Cpk into equation (7. + ix. Determine the temperature-density and temperature-pressure relationships for such a flow. or isentropic flow of an ideal gas. If K* O . is a barotropic flow for which p = cpk where C and k are constants with k = c(*)/c(").(2/l*/3)SijDkk+ 2/~*Dij which is expressed in terms of the rate of deformation deviator by If the deviator stress sij is introduced..2 / ~ * / 3 by (7. + (7.e. = a D i j pDikDkj where a and p are constants. 7.11) and S O (7.and aii = -3p. aii = -3p + rii and so here 7. the ratio of specific heat a t constant pressure to that a t constant volume. z = x . aij = -pSij + + aDij /3DikDkj and so ai< = -3p + aDii + /3DikDki. 7. Show that the deviator sij for the stress tensor a . Furthermore. 7 Thus both + and + are harmonic functions when the flow is irrotational. is the second invariant of the rate of deformation tensor. . since p = (plC)l/k here. of ( 7 .2/311D/3 where 11.6) yields the temperature-pressure relationship a s p(k-l)/k/T = R/Cl/k. 3 ) . NEWTONIAN FLUIDS (Sec. a constant. From ( 7 4 . From ( 7 4 .3) 7. A* = . the deviator of T i j of ( 7 .53) is an analytic function of the complex variable.6). But Dik = Dki and Dii = = O for an incompressible fluid so that 4 3 = -p + /3DijDij13 = -p .) + i+(x1. so that its derivative dqldz defines the c o m p l e x v e l o c i t y dqldz = -vl iv. Also.9) becomes aii = -pSij . Determine the mean normal stress o i i / 3 for an incompressible Stokesian (nonlinear) + fluid for which T . (7.1. this constitutive relation is given by the two equations S i j = ~ / L * D . .. Ti = (-p~ij + 2p*D.CHAP.4. Determine the conditions under which the mean normal pressure p.14). 7-1 .8.jj and with this expression inserted into (7. > .18) may be written aij = -pSii + h*Sijvk. ~ ) ] ( v v~j ..16) a direct verification of (7. equation (7. Substituting the above equation for qjtogether with (7..k + p * ( v t j + v j J . From (7. this expression is written I : D = -p(tr D) + A*(tr D)2 + 2p*D : D In terms of D& the expression is In symbolic notation.P = -K *Dii.20). Determine the traction force T.i.p j j S i j i13 + X * S i j ~ k . and upon application of Gauss' theorem. = -aii/3 is equal to the thermodynamic pressure p f o r a Newtonian fluid. 71 FLUIDS 167 7.7. k+j p*(vi.acting on the closed surface S which surrounds the volume V of a Newtonian fluid f o r which the bulk viscosity is zero.5. this becomes i ojinjd s .17). /----- Fig.i (h* + p*)vj. the latter equation gives P ( .jj + vj. = [-psij h*6ijvk. Verify the Navier-Stokes-Duhem equations of motion (7..D.22) is complete. 7. i ) / 2.ij) = -P.)9dS for a zero bulk modulus fluid.9) a s its constitutive relation. + h*SijDkkDij+ 2p*DijDij = -pDii + h*DiiDjj + 2p*DijDii In symbolic notation.17) f o r this fluid if the heat conduction follows the Fourier law (7. : D = -p(tr D ) + ~ * ( tD)2r + 2p* D' : D' 7.ii + pZ + which reduces to 7.11) when A* = -Qa*) or when Dii = 0.6.22) f o r a Newtonian fluid and determine the form of the energy equation (7. Determine a n expression for the "stress power" cijDijof a Newtonian fluid having equation (7.9) and the stress power definition.13) and (7. 3. With the constitutive equations in the form (7.kT.k + + + v ~ . = p when K* = O (by (7. Since Dii = vi. u 23 = -pSijDij o. = . Thus o.20) into the energy equation (7. the result is P..ji + p*vi. Thus p(m. A ds The element of traction is dTi = t i n ' d S and the total traction force is T i = L A tin' dS which because of the stress principle is T i = From Problem 7. Determine the slope of the free surface of the liquid.4) and simplifying. = pb. In a two-dimensional flow parallel to the X I X Z plane.) + ln C where C is a constant of integration.11.l ) / k . is + the unit radial vector. . t). determine the form of the continuity equation.6) in this case.)%. = 0. HYDROSTATICS.l))pck-1)lk = + c. x. p = pa.4) 7. Therefore x3 = ( k p o l ( k. 7. + 1 d(rpg) i a(Pv3) Here the cylindrical form of the operator V may be used to give V (pv) = -- Inserting this into (5. d p / d x l = d p l d x .CYX. p = p ( x l . = p.) and p = -p(g .. A large container filled with an incompressible liquid is accelerated a t a constant rate a = a2êz+ a3ê3 in a gravity field which is parallel t o the x 3 direction. = v l ( x l . = O . But p = po when x3 = O so that C = ( k / ( k .15) reduces to v. po where po is the pressure a t the origin of coordinates on the free surface. From (7. ~ ~ . and dp/dx3 = -p(g .CYX. t ) the necessary equations would be pGl = -dp/axl + + a * ( d z ~ l / d ~ : d 2 v 1 / d x i ) and dvl/axl = 0. Since p = po everywhere I on the free surface. x. dp/dx3 = . Since here p = (p/h)Uk.23) with i = 3. d p l d x . A barotropic fluid having the equation of state p = hpk where h and k a r e constants is a t rest in a gravity field in the x3 direction. p = P R ( T .).28). the equation of that surface is x 2 / x 3 = (9. p-llkdp = -gh-1fkdx3 and integration gives ( k / ( k . 4 If body forces were zero and v .a ~ 3 / T o ) g / R f f . From (7. Integrating. Separating variables and integrating yields ln p = ( g / R a ) ln ( T o. Determine the Navier-Stokes equations for a n incompressible fluid and the form of the continuity equation for this case.1 ) ) ( k .~ g= . 2 . Equation (5. v3ê3 where 6. .168 FLUIDS [CHAP.The continuity equation (7. = pa..P. p = pazx2 - + p(g ..10.a 3 ) . Determine the pressure in the fluid with respect to x3 and po. and x z directions is constant in the absence of body forces b1 and bz. d p l d x .axg where T Ois ground leve1 temperature and X B measures height above the earth.x. 7-2 . p'U. pb. v . the continuity equation becomes r ar 8x3 ' 7. v3 and dldxa a r e zero. Thus p = C ( T o. .p g / R ( T o . If the velocity is expressed by v = qê.. Fig. C = po To-g/Rff and so p = po(l .12. + h ( x 2 ) where f and h are arbitrary functions of their arguments. From (7. STEADY AND IRROTATIONAL FLOW (Sec. 7 7.9. dp/dx3 = -pg.28) with the body force b3 = -g.( p / p 0 ) ( k . and from (7.. = O. therefore. po 7.a.a3)/az.ax3)glRff and if p = po when x3 = O. hydrostatic conditions.l ) f k ) where po = ( p o / ~ ) ~ ~ k . the pressure a t x3 = 0. the gravitational constant. In general. + P * V .ax3).13.a. determine the a i r pressure in the atmosphere a s a function of x3 under . In a n axisymmetric flow along the x3 axis the velocity is taken a s a function of x3 and r where r2 = X : + x2.)%.28). 7. + f(x.l ) g p o ) ( l. = 0 . From (7.4) gives the continuity equation in symbolic notation a saplat V (pv) = 0. Note that pressure in x . and when i = 1 . Assuming a i r is an ideal gas whose temperature varies linearly with altitude a s T = T O.. v i = . j j and for creeping flow these linearize to the form Hence for steady flow with zero body forces. Also with v i = -+. dxi = s v .22) becomes -Pd.23) are p(avilat + vjvi.29)..e. dvi = +vivi = +v2. Taking the scalar product of this increment with (7. and (7. From the definition (7. Express the continuity equation and the Navier-Stokes-Duhem equations in terms of the velocity potential 4 for an irrotational motion.i . For this case show that in a steady incompressible flow with zero body forces the pressure is a harmonic function.iii or -p(a$. Thus in the second integral.CHAP. For incompressible flow the Navier-Stokes equations ([email protected].@ .(A*+ z p * ) v ( v 2 + ) 7.jji .P.P. By (7. p. 71 FLUIDS 169 7. (7. i j jand . Deterinine the pressure function P(p) for a barotropic fluid having the equation of state p = X p k where X and k are constants.i.(h* p*)+. v 2 p = o. ( v i / v )ds = vivi. since the continuity equation for incompressible flow i s viai= O.39) is aehieved. a limiting case known a s creeping flow results.i dxi = d P the last two terms integrate a t once. If a fluid motion is very slow so that higher order terms in the velocity are negligible. CIRCULATION (Sec. ( v j / v )ds = v i v i .37) and integrating gives Since dxi = d a and P. ~ p*vi.15) the continuity equation becomes i. i t follows that here p.35). dxi = (viIv)d s where ds is the increment of distance. the same result may be obtained from PERFECT FLUIDS.5) 7. 7. v j v i .i = Taking the divergence of this equation yields pJii = P * ~ i .p*@.i . i so that from (7. along a streamline.(h* + 2p*)$. + = pbi .i = Pbi .15. .16.pV2@ = O.i/at + @. 7. Also since d p = hkpk-1 dp. i. d x j = v i dvi Therefore S v j v i . dxi = v j v i . Let dxi be an increment of displacement along a streamline.14.jji In symbolic notation this equation is written -pv(a+/at + (v+)2/2) = pb .37) along a streamline. BERNOULLI EQUATION.v p .39) by integrating Euler's equation (7. Derive equation (7. = V 2 p = 0 .ik) = pbi .P . Also. 7-3) for the two-dimensional + f l o ~v = ( X I x ~ (x. 7 7.7 the streamlines in the x .Paidxi + vi dvJ = . t. and if gravity is assumed negligible this equation becomes (see Problem 7. in the x . by the analysis of Problem 4.22.42) with ^n = ê3 and v x v = (2x1.j= O.ij= 0.19. Gi = -(a + P).l J k= N . plane are represented by x i x 3 = constant. Determine the circulation around the square xl = k1.j pIi = O if body forces are neglected. For a barotropic fluid. Show that for a barotropic.2 A 4 A = O.i = 0.= $ (-qidxi . Applying Bernoulli's equation for steady flow between point A .45) and express this equation in terms of the velocity potential +. . In terms of + .1 )ê3. p = p(p) and so dp = ( a p l a ~ ~ ) ( a ~ ~ or l a rearranging. + ( x .1) d x . PLANE POTENTIAL FLOW (Sec. Inserting this into the Euler equation and multiplying by vi gives pvivjvi.) dx.x. determine the speed of the emerging fluid.i for the case a t hand. Substituting + into (7. = 2Ax.l l k the result may be written 7. The barotropic fluid of Problem 7. = 21 ~3 . From (7. If the pressure in the tank is N times the atmospheric pressure.36) becomes pvjvi.3~9. v 3 = -4Ax3. The same result is obtained from (7.i@. From (7. ~ c is the local velocity of sound. x3 = O (see Fig. Using the symbolic form of (7.21.P / 2 ) = 0 . i = -vi this becomes (c2Sij.vivj)vi.x2 2 x i ) satisfies the Laplace equation and determine the resulting velocity components.i = ( d ~ l d p ) p=.170 FLUIDS [CHAP.j + c2vip. = § (Gi dxi + vi dvi) and by (7.i = O and the Euler equa- + tion (7. dx. p. v . x2 = 1+1. Thus i. ~c ~ P where . plane (fixed wall).@. j)+. + For a steady flow the continuity equation (5. 7-3 r.v. = $v-dx 1 -1 = (i-x2)dx2 + ~ . + + But v. Show that the function + = A(-x: . 7. in the emerging free stream. a t rest in fluid of the tank and + + point B.1) dxl = -4 with the integration proceeding counterclockwise from A.18.46) gives -2A ..16 flows from a large closed tank through a thin smooth pipe. POTENTIAL FLOW. ~ 1.f (& + dP .x. . From the continuity equation c ~ v=~-C%. Since pBIPA = ( p B l p A ) . + ) -~x ~~ ) ~ ~ .16). v .. the d(Q-i. = 0.41) where = -4 Fig.37).M) :. Give the derivation of the gas dynamical equation (7.. integrand being a perfect differential..l ( x l + l ) d x l+ (1. and so (c2Sij. ( 7 3 9 ) assumes the form aA PA & v i = aB PB & v i .d(vz12)) = . rh 7. = 2 A x Z .20.Sl 1 (2x1 . + 7. inviscid fluid with conservative body forces the rate of change of the circulation is zero (Kelvin's theorem). Thus the flow is in along the x3 axis against the x. ~)d~.f P . = s:. plane by xSx3 = constant. ~ -c2S.4) becomes pjivi pvi.6) 7. Also. 7-4).x t ) is a valid velocity x2 potential and describe the flow field. ntA A n Since i. Determine + the stream function for this flow. Verify that + = A(xl . (7. = = A ( x . Here dWdz = -AI22 = -A/(xl + ix2)2 which after some algebra becomes d<P/dz = -A(%. and since va = . Differentiating. Thus f'(x2) = A and f ( x 2 ) = A x z C. 71 FLUIDS 171 7. Convert- ing by Gauss' theorem. For the given +. Finally then $ = + A x z .Bx2/r2 C. = -d~. Thus + = a constant along any streamline. ~ = 2Bx1z2/+ so that by integrating. But from (7. + + z8)/+.x. The streamlines are determined by integrating dxl/xl = -dx2/x2 to give the rectangular hyperbolas xlx2 = C (Fig.2 = 2 A ~ ~ x ~ 1 7 .v B n ~ . From (7.24. + = -Bx21r2 f ( x 2 ) where f ( z 2 ) is an arbitrary function of z. 7. + = Axllr2 and JI = -Ax21r2. +.2) becomes Sv V * v dV = O since p is constant here. Show that the stream function +(xI.2 = + = A B(-z. Also note that v.CHAP. and from (7.G .26. Let V be the volume between arbitrary cross sections A and B of the stream tube shown in Fig. dxl/vl = dx21v2 (see Problem 4.50). -X:)I+ and v2 = -9.50). Thus v1 = A(%. Derive the one-dimensional continuity equation for the flow of an inviscid incompressible fiuid through a stream tube. + = -2Az1x2 CO and is seen to be constant along the streamlines a s was asserted in Prob- lem 7. 7-5 A The velocity is assumed unifonn and perpendicular over SA and S. + = . v l = -2Axl.l+. 4 MISCELLANEOUS PROBLEMS 7. = 2Ax1x2l. Finally + from (7.~ or +.s i ) / + + f'(x2).P Note that since @ = A l z = A(%.23.)/+ + i2Ax1x2/+.vB dS = O or v A S A = v B S B = a constant. . . 7-5. From (7..50)..2.23.48) and the differential equation of a streamline.x2) is constant along any streamline. LA n o v dS = O where " n s the outward unit normal to the surface S enclosing V .2 = -B(X. The equipotential lines A ( X . v on the lateral surface. .. In integral form.ldxl += d+ = O. Fig. 7-4 7.27.iz2)/r2. v2 = +2Ax2.49).46) is satisfied identically by 2 A ..7). +. .2 A = 0 . the integration reduces to " " Fig. vA iA iB dS . si)+ and v2. for this volume (5.) = C1 form a n orthogonal set of rectangular hyperbolas with the streamlines. Differentiate the complex potential a(z) = Alx to obtain the velocity components. 7.25. A velocity potential is given by + = Axl + Bx1Ir2 where r2 = x: + x:. from (7. $ i qi dV =. .. u33 = -p + T33.*sPqoii+ 2 p * ~ p q= Tpq .For principal directions x: of qj. aL From (4.30. aj4 .(l . ( ~ / a = + S ~ ~. The stress tensor a t a given point for a Newtonian fluid with zero bulk viscosity is c = -6 ( 2 -9 2 -1 -1 4 -3 Determine ). ~p. p l p o = (T/To)(g/Ra-i).157).u 2 ~ +722. since the integrand is zero (product of a symmetric and antisymmetric tensor). (4-: :) -6 2 -1 6 O 2 -1 oi.32. Thus Z: are principal axes for T~~ also.oll = -p = -p T ~ I . 172 FLUIDS [CHAP. i t is seen that T / T o = (plpo) Ralo and so P/Po= (p/po)(l-Ra/g). = 7. From (7. Hence 7. Then from ( 7 4 .. = S p q and d ~ . p = po and p = po.6) is here p = pR(To . (7. Here D~. = ( K / ~ ) D . Determine the pressure-density relationship for the ideal gas discussed in Problem 7. 7.11. A dissipation potential Q.33. u12 = 7 1 2 .a ~ ~ / T ~ ) g /ofR aProblem 7. Thus writing p = po(l . 6.11. ' .SpqDii/3) ~ ~ ~ Finally since K = A* + 2p*/3.L (ciik ak + qivi) dSj. show that the time rate of change of kinetic energy of the fluid is -.. . . ) ] .' 7 7. For an incompressible Newtonian fluid moving inside a closed rigid container a t rest. But aDiilaDpq = ~ . and from the pressure elevation relationship p = p.SijSpq/3) = K D + eP*(Dpq ~ ~. and by the divergence theorem (1. +6 8 or -:: )+ = 7..as31To)g/Ra in the form p/po = (T/To)g/Ra.31.. Show that aij and T~~ of (7. d qi dV = viqi dSi.29.14). q is the magnitude of the vorticity vector.laDpq = ( ~ / 2 ) [ D ~ ~ ( a D ~ .8ijDkk/3)(8ip8jq d@~laD.8) becomes .33. = K D ~ ~ 2p*(Dij .3) have the same principal axes.54) and the results of Problem 4. A t s3 = O. The ideal gas law (7. But here ak = -(R +P).37).D. 7.)+ d+. + u23 = u13 = 7 1 ~ . is often defined for a Newtonian fluid by the relationship a. = ?. uTz = = uT3 = O and by the last three equations of (7. for this fluid p = -uii/3 = 6. For a barotropic inviscid fluid with conservative body forces show that the material derivative of the total vorticity. a+. When written out./a~. D~~ + Show that d a D l d D i j = r i j . = O for i # j. 7.39.28. .as3) so that po = poRT0.S i j Spq/3 so that Sip Si. ( a ~ & 3 ~ .L* ing zero body forces.*D. l a (aDiilaDpq)Djj] + 2 5 [ D . o. 36) with bi = 0. Show that for irrotational motion this equation reduces to (7.35. and since d d ( f v 2 ) = O. t) where r2 = xixi. k P . But incompressibility means vili = Dii = O and so 7.18). 6 IP) = ( P . + $ v i dvi. ~ ) .37. L ~ P @i ] = . 7.o2r212 + gz3 = constant.36. Show that ( l l p ) ( d p l d t = ) O i s a condition for . Differentiating a s indicated. The constitutive equation for an isotropic fluid is given by aij = -paij + K i j p q D p q with Kihq constants independent of the coordinates.k/p). show that plp .-uii/3 = p for a Newtonian fluid. Show that the principal axes of stress and rate of deformation coincide. h = -L + e i j k [ ( l l ~ ) . Show that in terms of the vorticity vector q the Navier-Stokes equations may be written \I = b .40. 7.27. ~ ~ ~ j~p .V p l p .ini d s where (7. . If gravity is the only body force. From (1.~ . Gi = . the second integral i s zero.kinematic viscosity. bc = $ Vi d z .36).42) has been used in converting to the surface integral. 7."*V x q where v* = r*lP is the .CHAP. 'i] FLUIDS 173 From Problem 5. i eijk(p. and for a Newtonian fluid by (7. A liquid rotates a s a rigid body with constant angular velocity o about the vertical s3 axis.39. . Show that for a perfect fluid with negligible body forces the rate of change of circulation bc may be given by . 7. j p . show that the equation of continuity is A a a + 2 (r%) = 0. ( a ís i/.34. ~ l pand so now bc = . L eijk(lI~). ~ ~d.38. +v 2 at ar r2 ar 7. Show that the constitutive relations for a Newtonian fluid with zero bulk viscosity may be expressed by the pair of equations sij = 2r*Dii and -oii = 3p. the time rate of change of kinetic energy of a continuum is In this problem the first and third integrals are zero.j ~ .43). If a fluid moves radially with the velocity v = v(r.. dSi k Supplementary Problems 7. From (7. 7. ~ * l p is the kinematic viscosity. Yes + + i$ + 7.)x3/r4. Show that the Bernoulli equation (7..j .46. v 2 = (x: .43. 7 7.44. 7.show that grad P = grad plp. . For steady flow of the same fluid.) 7.p ( a @ l a t +- (V@)V2) + + + pQ P + (i* 2@*)V2+= f ( t ) .j = O. show that - vjqi.qjvi. for isentropic flow. for isothermal flow.x.j- 7.+ $32is a possible flow for an incompressible fluid. Show that the velocity and vorticity for an inviscid flow having conservative body forces and constant density satisfy the relation Gi . Show that the velocity field vl = -2x1x2x3/T1.39) for steady motion of an ideal gas takes the form + + (a)Q p ln ( p l p ) v2/2 = constant. ( b ) + + (klk .15.48. 1s the motion irrotational? Ans. 7. the Navier-Stokes-Duhem equations for the irrotational motion of a barotropic fluid may be integrated to yield . (See Problem 7.qjvi. ar ' 7. Show that if body forces are conservative so that bi = -nJi.47.42. show that p l p .45.ii = v* = . show that for irrotational potential flow +.29).41.l ) ( p l p ) v212 = constant. For a barotropic fluid having p = p(p) and P ( p ) defined by (7. v3 = x2/r2 where r2 = x: + x. If the velocity potential +(z) = ix2 = rei0 show that in polar coordinates 3ar = '9 and -- r ae 1 a6 = r ae -* is an analytic function of the complex variable z = x .174 FLUIDS [CHAP. = pIpo = e-(glRTozs) where po and po are the density and pressure a t x3 = 0. For an ideal gas under isothermal conditions (constant temperature = To). If body forces are zero. In this plot.2) and it is often permissible to neglect the difference. Chapter 8 8. that does not obey the constitutive laws of classical elasticity may be spoken of as an inelastic deformation. the primary concerns are with the mathematical formulation of stress-strain relationships suitable for the phenomenological description of plastic deformations.1 BASIC CONCEPTS AND DEFINITIONS Elastic deformations. In particular. whereas the strain r may ou -- represent either the conventional (engineering) strain defined here by e = (L. irreversible deformations which result from the mechanism of slip. are characterized by complete recovery to the undeformed configuration upon remova1 of the applied loads. Indeed. Many of the basic concepts of plasticity may be introduced in an elementary A O B way by consideration of the stress-strain diagram for a simple one-dimensional tension (or compression) test of some hypo- thetical material as shown by Fig. or yield stress. or from dislocations a t the atomic level. By contrast. or the natural (logarithmic) strain defined by Fig. u is the nominal stress (force1 original area). elastic deformations depend solely upon the stress magnitude and not upon the straining or loading history.Lo)/Lo (8. Any deformational response of a continuum to applied loads. Such deformations occur only a t stress intensities above a certain threshold value known as the elastic limit. and which thereby lead to permanent dimensional changes are known as plastic deformations. or to environmental changes. these two measures of strain are very nearly equal as seen by (8. 8-1 For small strains. . such a continuing plastic flow may be related to the amount of deformation as well as the rate of deformation. the study of plastic deformation from the microscopic point of view resides in the realm of solid state physics. a solid in the "plastic" state can sustain shear stresses even when a t rest. which is denoted here by o. unlike a fluid flow.. In the theory of plasticity. which were considered in Chapter 6. Also. The phrase plastic flow is used extensively in plasticity to designate an on-going plastic deformation. 8-1.1) -e where L is the current specimen length and LOthe original length. However. and with the establishment of appropriate yield criteria for predicting the onset of plastic behavior. Upon a return to B a load increase is required to cause further deformation. This phenomenon is known as the Bauschinger effect. elastic response is omitted and the work-hardening is assumed to be linear. The compression curve for a previously unworked specimen (no history of plastic deformation) is taken a s the reflection with respect to the origin of the tension curve. This representation. A reloading from C back to B would follow very closely the path BC but with a rounding a t B. . The stress-strai. and will be neglected in this book. and the force F plays the role of stress. however. and with a small hysteresis loop resulting from the energy loss in the unloading-reloading cycle. which lies a t the upper end of the linear portion of the curve. It is sometimes taken a t the proportional limit. has been used extensively in analyzing uncontained plastic flow. which may be linear or nonlinear. 8-2 appear in the context of tension curves. and a decrease in load. However. 8-1. 8-1 results in the state point following the path BC which is essentially parallel with the linear elastic portion of the curve. Likewise. 8 2 IDEALIZED PLASTIC BEHAVIOR Much of the three-dimensional theory for analyzing plastic behavior may be looked upon as a generalization of certain idealizations of the one-dimensional stress-strain curve of Fig. 8-2(a). Unfortunately. the permanent plastic strain cP remains. It may also be chosen as the point J. the yield point is not always well-defined. In the initial elastic range. separates the stress-strain curve of Fig. Accordingly. if a stress reversal (tension to compression. it is common practice in traditional plasticity to neglect any effect that rate of loading would have upon the stress- strain curve. In the plastic range. an increase in load causes the stress-strain-state-point to move upward along the curve. one such being the stress value a t 0. or vice versa) is carried out with a real material that has been work-hardened. whereas in (b). a definite lowering of the yield stress is observed in the second type of loading. a condition referred to as work hardening. or strain hardening. In the absence of work-hardening the plastic response is termed perfectly plastic. The four most commonly used of these idealized stress-strain diagrams are shown in Fig.176 PLASTICITY [CHAP. elastic response and work-hardening are missing entirely. or strain history of the material. where the stress reaches zero. 8-1 into an elastic range and a plastic range.2 per cent permanent strain. Representations (a) and (b) are especially useful in studying contained plastic deformation. known as Johnson's apparent elastic limit. The recoverable elastic strain from B is labeled rE in Fig. where large deformations are prohibited. In Fig. along with a simple mechanical model of each. Thus a one-to-one stress-strain relationship exists in the elastic range. or unloading causes the point to move downward along the same path. plastic deformations are assumed to be time-independent and separate from such phenomena as creep and relaxation. as well a s (a). 8-2(c). 8-1.n curves of Fig. In Fig. In the models the displacement of the mass depicts the plastic deformation. Various offset methods are also used to define the yield point. 8 The yield point P. and defined as that point where the slope of the curve attains 50% of its initial value. unloading from a point such a s B in Fig. At C. corresponding to the yield stress o. Although it is recognized that temperature will have a definite influence upon the plastic behavior of a real material. it is customary in much of plasticity to assume isothermal conditions and consider temperature as a parameter. It is clear therefore that in the plastic range the stress depends upon the entire loading. elastic response prior to yield is included but work-hardening is not. 8-2 below. (aij) is called the yield function.Perfectly Plastic . TRESCA AND VON MISES CRITERIA A ?jield condition is essentially a generalization to a three-dimensional state of stress of the yield stress concept in one dimensional loading.3 YIELD CONDITIONS.....Perfectly Plastic c (b) Elastic ... 8-2 8. or a s is sometimes done by the equation in which f.... the yield condition is a mathematical relationship among the stress components a t a point that must be satisfied f o r the onset of plastic behavior a t the point.. is known a s the yield constant... Rough Rough L F 177 I ...... Briefly........... (8.....) = c...CHAP... .Linear Work Hardening Fig. the yield condition may be expressed by the equation f(. Rough (c) Rigid-Linear Work Hardening Rough (d) Elastic .3') where C. 81 E PLASTICITY E (a) Rigid. I n general..... two are reasonably simple mathematically and yet accurate enough to be highly useful for the initial yield of isotropic materials.. for example) to be ~ ~ 1 2Therefore .3) may appear as Furthermore. Thus if the pure shear yield point value is k. These are: (1) Tresca yield condition (Maximum Shear Theory) This condition asserts that yielding occurs when the maximum shear stress reaches the prescribed value C... Thus for U. and the Tresca yield criterion is written in the form (a) Simple Tension ( 6 ) Pure Shear Fig. as in Fig. Thus (8. .8) The yield point for a state of stress that is so-called pure shear may also be used as a referente stress in establishing the yield constant C.8-3(a). the Tresca yield condition is given from (2..7) To relate the yield constant C. or alternatively. 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. the maximum shear in simple tension a t yielding is observed (by the Mohr's circles of Fig. > a. the condition is expressed in its simplest form when given in terms of principal stresses.. Mathematically. > a.178 PLASTICITY [CHAP. equals k (again the Mohr's circles clearly show this result..uIII) = Cy (a constant) (8. 8 For an isotropic material the yield condition must be independent of direction and may therefore be expressed as a function of the stress invariants. to the yield stress in simple tension u. as a symmetric function of the principal stresses. 8-3(b).6) Of the numerous yield conditions which have been proposed. experiment indicates that yielding is unaffected by moderate hydrostatic stress so that it is possible to present the yield condition as a function of the stress deviator invariants in the form f3(IIxD. the von Mises yield condition states .1111~)= o (8.54b) as +(aI. Tresca's yield condition becomes u~ .-'I'II = u~ (8.. the yield constant C. when referred to the yield stress in simple tension. Mathematically. . von Mises condition (8. f2(a.11) appears in the form (aI . the yield condition (8. 8-5!b) and (c) below.12) Also. it is easily shown that (8. 8-5(a) below. (8.12) and (8. those which lie on the yield surface represent incipient plastic stress states.13) There are severa1 variations for presenting (8. + ali + a. 81 PLASTICITY 179 which is usually written in terms of the principal stresses as (aI . such yield surfaces are general cylinders having their generators parallel to 02. Every point in this space corresponds to a state of stress. represents hydrostatic stress.. In a true view of the n-plane.. YIELD SURFACE A stress space is established by using stress magnitude as the measure of distance along the coordinate axes.7) and (8.. u.CHAP. The intersection of the yield surface with the R-plane is called the yield curve.5). the two yield curves are based upon the yield stress k in pure shear.a1)2 = 2~. = 0 (8. Stress points that lie inside the cylindrical yield surface represent elastic stress states. THE II-PLANE. for which U.13) when stress components other than the principal stresses are employed. which makes equal angles with the coordinate axes. the principal stress axes appear symmetrically placed 120" apart as shown in Fig. looking along OZ toward the origin 0. and a component OB in the plane (known as the n-plane) which is perpendicular to OZ and passes through the origin. these curves are drawn with reference to (8. is seen to circumscribe the regular Tresca hexagon.uI)2 = 6k2 (8. with respect to the pure shear yield value k. The yield curves for the Tresca and von Mises yield conditions appear in the n-plane a s shown in Fig.uIJ2 + (cII.. so that the component in the n-plane represents the deviator portion of the stress state.aIII)2f (vIII. and the position vector of any such point P(u. oII. For this situation. . 8-5(c).uIII)2f bIII .. 8-4 In stress space.4 STRESS SPACE.11) becomes (oI .uIII)2+ (vIII. = uIII.u1)2 = 6 c y (8. defines a surface. = a.11) With reference to the yield stress in simple tension. I t is easily shown that the equation of the n-plane is given by a. using the yield stress in simple tension as the basis. In Fig. the von Mises circle of radius m 3U. Here the von Mises circle is inscribed in the Tresca hexagon.uII)2 + (gII . the so-called yield surface.14) Fig. Since the yield conditions are independent of hydrostatic stress. In the Haigh-Westergaard stress space of Fig. The component along OZ.11).uIII) may be resolved into a component OA along the line 0 2 .uIJ2 + (rII . In Fig.. 8..olII)= C. 8-5(b). 8-4 the coordinate axes are associated with the principal stresses. two especially simple cases of which are described in what follows. neutra1 loading occurs. Equation (8.. The manner in which the plastic strains c.180 PLASTICITY [CHAP. 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. Thus the projected deviatoric components are ( m v I . moIII). . loading occurs. This corresponds to the one-dimensional perfectly plastic case depicted by Fig.. 8-5 The location in the II-plane of the projection of an arbitrary stress point P(u.15) when loading occurs is defined by the hardening rules. unloading is said to occur.4) be generalized to define subsequent yield surfaces beyond the initial one. but also upon the plastic strains r. f: < 0 is a surface in the (elastic) region inside the yield surface and f r > 0.15) by the chain rule of calculus. d2/3uII. To account for such changes it is necessary that the yield function f. with f: = O and (af :/aoij) doij = 0. o. The assumption of kotropic hardening under loading conditions postulates that the yield surface simply increases in size and maintains its original shape. and the work- hardening characteristics represented by the parameter K. For a strain hardening material. Differentiating (8.. 8-6 below.. Thus with f: = O and (af :Iaoij) doij < 0. 8. 8-2(a)... however. For an assumed perfectly plastic material the yield surface does not change during plastic deformation and the initial yield condition remains valid. a.15) defines a loading surface in the sense that f: = O is the yield surface. being outside the yield surface. 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. has no meaning.(oij) of (8.) is straightforward since each of the stress space axes makes cos-I @ with the II-plane.) doij > 0. A generalization is effected by introduction of the loading function which depende not only upon the stresses. 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 (b) Fig. plastic deformation is generally accompanied by changes in the yield surface.5 POST-YIELD BEHAVIOR. enter into the function (8. and with f i = O and (af:la~. 21) represent the flow rule for an elastic-perfectly plastic material. = c.the so-called incremental theories. The factor dA may change during loading and is therefore a scalar multiplier and not a fixed constant. Because plastic strains depend upon the entire loading history of the material. By neglecting the elastic portion and by assuming that the principal axes of strain increment coincide with the principal stress axes.19) represent the flow rule for a rigid-perfectly plastic material. Equations (8. Thus (8.6 PLASTIC STRESS-STRAIN EQUATIONS.18) where c is a constant. They provide a relationship between the plastic strain increments and the current stress deviators but do not specify the strain increment magnitudes. Equations (8. Fig. If linear hardening is assumed.17) where the aij are coordinates of the center of the new yield surface.a. plastic stress-strain relations very often are given in terms of strain increments . 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. the constitutive equations of elasticity are no longer valid.: (8. PLASTIC POTENTIAL THEORY Once plastic deformation is initiated.) = O (8.4) defining an initial yield surface is replaced by f i (oij. In a one-dimensional case. the Tresca yield curve would be translated as shown in Fig. .CHAP. 8-6 In kinematic hardening. the initial yield surface is translated to a new location in stress space without change in size or shape. 8-7. Xi. 8-7 8. 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. 81 PLASTICITY 181 (a) Mises Circles ( b ) Tresca Hexagons Fig. (8. 2 OEQ 8.) for which dg dA de. = .the plastic work increment may be expressed as d W P = a.22) 8% For a so-called stable plastic material such a function exists and is identical to the yield function..2 8) and using (8.21).182 PLASTICITY [CHAP. From (4. 8 The name plastic potential function is given to that function of the stress components g(u. .21). S.8 PLASTIC WORK. 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.25) respectively.25). Moreover when the yield function f l ( u i j )= IIr.24) and (8. as This expression may be written in compact form as In a similar fashion. is defined by which may be written compactly in the form In terms of the equivalent stress and strain increments defined by (8. = -..7 EQUIVALENT STRESS. dA of (8.21) rewritten in the form 3 dWP de. or the stress power as i t is called.3 de.20) this may be split into For a plastically incompressible material... STRAIN-HARDENING HYPOTHESES The rate a t which the stresses do work. (8.21) becomes dh = -. 8.32) as uijDijper unit volume. 2 a.. has been given in (5. so that the work increment per unit volume may be written dW = ali dcij (8. den = D.. if the same material obeys the Prandtl-Reuss equations (8.31) and (8.dt. the plastic work increment becomes Furthermore.22) produces the Prandtl- Reuss equations (8.dcEQ (8. the equivalent or effectiveplastic strain increment de:. 8. 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. One.21). 81 PLASTICITY 183 There are two widely considered hypotheses proposed for computing the current yield stress under isotropic strain hardening plastic flow. Equilibrium equations (2. (aij) = (8.19) and (8. page 143 3.CHAP. The equations take the form I n terms of equivalent stress and strain. Boundary conditions on stress or displacement 4. In terms of the total equivalent strain this hardening rule is expressed symbolically by the equation f . known as the work-hardening hypothesis.36) for which the functional form is determined from a uniaxial stress-strain test of the material.23). torsion o£ shafts and thick- walled tubes and spheres subjected to pressure. A second hardening hypothesis. the so-called total deformation theory of Hencky relates stress and total strain.34) and (8.36) may be shown to be equivalent. the governing equations for the elastic region. In general. assumes that the hardening is a function of the amount of plastic strain. Stress-strain relations (6. assumes that the current yield surface depends only upon the total plastic work done. For the Mises yield criterion.23) or (6. the plastic region and the elastic-plastic interface are these: (a) Elastic region 1. page 49 2 . A number of well-known examples of such problems occur in beam theory.24).9 TOTAL DEFORMATION THEORY In contrast to the incrernental theory of plastic strain as embodied in the stress-strain increment equations (8. the parameter + may be expressed as (P = 3 -- 2 um where here L& = d Z 5 7 3 so that €P = 3 cEe -- 23 2 UEQ 8. the hardening rules (8. known a s the strain-hardening hypothesis.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. Compatibility conditions . In Fig. = 0. and ~((r. = -p..21) 3.11.21).). . de.8) or (8. 8-8.42) In (8. From the Fig. the maximum shear directions are a t 45" with respect to the principal stress directions.42) the variables are functions of XI and xz only. . where tan 28 = 2ul2/(al1. ) ~= k. Stress-strain increment relations (8. Taking the X l X 2 plane as the plane of flow. Yield condition (8. 8 ( b ) Plastic region 1.u. -~ . .184 PLASTICITY [CHAP. the resulting velocity field may be studied using slip line theory. 21 ignated as the a and p directions. 'li '12 0 (8. 8-8 .23). the stress V . 8-8. the maximum shear directions are des. As was shown in Section 2. i t is often possible to neglect elastic strains and consider the material to be rigid-perfectly plastic. and also where v. ) ~ + / 4( u .44) Adopting the standard slip-line notation o. the plastic strain-rate tensor applicable to the situation is .41) and (8. the stress tensor is given in the form and since elastic strains are neglected. and so from the Prandtl-Reuss equations (8.. is given by u33 = +(all+ nz2) (8. For the assumed plane strain condition. If the flow may be further assumed to be a case of plane strain. page 49 2..11) 4 . Boundary conditions on plastic boundary when such exists (c) Elastic-plastic interf ace 1. the principal stress values of (8.. Equilibrium equations (2.are the velocity components.41) are found to be The principal stress directions are given with respect to the xlx2 axes as shown in Fig. 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. . v. the velocity field may be determined from (8. 8-9 Fig. sin + + v. . v. the slip line field may be found from (8.k sin2+ qZz = -p + k sin 2+ U12 = FC COS2+ and from the equilibrium equations i t may be shown that p + 2k+ = C. + v.52) dv. sin + (8. the principal axes of stress and plastic strain-rate coincide.CHAP. These curves are called shear lines.48).v. 8-10 With respect to the velocity components. d+ = O on a lines (8. u.. = o These equations lead to the relationships dv.2k+ = C. 8-9.. cos + For a n isotropic material. cos + .52)and (8.. Fig. are slip-line directions. 81 PLASTICITY 185 geometry of this diagram.. a constant along a n a line p .53). For a small curvilinear element bounded by the two pairs of slip lines shown in Fig. = -p . two families of curves along the directions of maximum shear a t every point may be established.49) vz = v. and using this slip line field. d+ = O on /i' lines (8. v. and i.. 8-10 shows that relative to the a and /i' lines.8 line Fig. Therefore if x. are zero along the slip-lines so t h a t (& (v. sin +)S*=. cos + .53) Finally. and x.46) and for a given stress field in a plastic flow. a constant along a . .- tan28 (8. . = v. or slip lines. O = r14 ++ so that 1 tan2+ = . for statically determinate problems. . o. a I I I / ~ . 8-13 and 8-4).o. = 0. = x/fi . = o..6.7..QII )2 + (011 ..10) to its principal stress form a s given in (8. 8-14 .Y/& + z/fi..12) becomes (-fix)z + ( x / h .I = -x/fi . Therefore . Hence Thus (01 . which simplifies to the Mises yield circle 3X2 + 3Y2 = 20. determine the yield loci for the Mises and Tresca conditions and compare them by a plot in the two-dimensional u.. where UM + + = (01 oI1 uIII)/3.. For a biaxial state of stress with a.11 + o. = 6Cy 8. 81 PLASTICITY 187 8. or Z = O which is the XY plane @-plane).11). show that (8. and by (2.14). With the rectangular coordinate system OXYZ oriented so that the XY plane coincides with the n-plane and the a.. of Fig. + aII1 = fiZ = O.3 ~ / f i )+ (x/fi+ 3~/1&)2 = 2 20.1 1 ~=~-(slsI1 + SIISIII + sII1s1). 8. SI = o1 .Y/& + z / h .. 8-5(b). 8-13 The table of transformation coefficients between the two sets of axes is readily determined to be a s shown above. From (8. 8.CHAP.0111)~+ (oIII . Convert the von Mises yield condition (8. axis lies in the YOZ plane (see Fig. 8-5(b). = 0.UM.. . a. + a. = O. is the equation of the n-plane./u.72). which is the ellipse = 1 -) (~Q I U I I I If~ )(UIII/"Y)~ (~I/UY Fig. space.5. From (2. a1 + o.8.71).. etc.12) with o... show that the Mises yield surface intersects the n-plane in the Mises circle of Fig. QIII = 2 ~ 1+ 6 216 and (8..14) the o's of Problem 8.6.o. Substituting into (8. + u.6. Using the transformation equations of Problem 8. the Mises yield condition becomes 0: . Fig. vs. . C ."11 "11 . = u l l .011. u ~sij..aM)dX.12).19) becomes dell = (2õ11. Also. ~ l a ~ ~where .SijSpq/3)sij= spq since sii = I I D = 0. DC and F A with equations oI.deI .de: . 8. = o y . In the principal axes system. show that the Levy-Mises equations (8.19) for this case.( o I I I / o y )= 21. From Problem 6. Writing 111.) dh/3. STRAIN-HARDENING (Sec. uTD)= ~ $ 1 6 ~Thus .dcFII .4-8. .13.. de. Likewise. . P de: . O = 203. Show that if the distortion energy per unit volume u. .12). .)~.u3.12.o M ) d h . PLASTIC DEFORMATION. show that ~ I I z . and BC and EF with equations o I / o y = T I . = a$/6G and. Mises condition becomes + (ul1/2)2 (-ul1/2)2 + (-ul1)2 = 6k2 or oS1 = 4k2 and oll = 2k. the Mises yield condition is expressed by (8. Then from (8. / ~ = ~ Here a I I I D /aopq = ( a ~ .6 i i ~ k k / 3 ) / a ~=p qOipSjq .a.. = O and u. Show that the Prandtl-Reuss equations imply equality of the Lode variable p (see Problem 8. = dX " I .21) imply that principal axes of plastic strain increments coincide with principal stress axes and express the equations in terms of the principal stresses. dez2 = -(oll + dh/3. 8. Here (8. as before. Thus deI = (oI .lsI = d e I I l s I I= d e I I I / ~ I=I I dX.26.3)and v = ( 2 d 4 . ( r I I 1 = O = az2.~ ~ ~ Thus ~ ~ ~ a I I x D / a ~ p=q ( S i p S j q . = o. For the case of plastic plane strain with r.. The von Mises yield condition is referred to in Section 8. etc.dc.9) the Tresca yield condition is o. = 2k.. . 8 with axes a t 45' in the plot. and by subtracting.d r g I ) l ( d r . = u I 1 / 2 .is set equal to the yield constant C. uTD)is given in terms of the principal stresses by and for a uniaxial yield situation where o. = <ry.10. P P P P (8. uI1 . Show that the Prandtl-Reuss equations (8. From equations (8. C . = sijsij/2./uy = f1. when referred to a coordinate system in which the shear stresses are zero.21) becomes de.11.8) 8. From the form of (8.u1 = o. . deII = ( o I I. the result is the Mises criterion as given by (8.de:[ . from (8..9. the Tresca yield condition results in the line segments AB and ED with equations ( o I / o y ). Thus in the absence of shear stresses.d&)." I 8.. = 0.de: .a. 8."111 "111 . = O.~ = a(oij . 8. oI1 = o I I I = O .21).dD. ull = 03. the plastic shear strain increments are seen to be zero also.21). respectively.. from (8.188 PLASTICITY [CHAP.8) and the companion equations a111 ..3 as the Distortion Energy Theory.19) lead to the conclasion that the Tresca and Mises yield conditions (when related to pure shear yield stress k) are identical. . Problem 8. u~~= a12 = u~~= u13= O. .22) reduce to (8.d e g I ) 2+ . In principal-axis form. (8.d e g l l = . = ...Sijupp/3)(aij .30) becomes d W P = o. = sij in this case and (8.033)' f (a33 . a constant.. The proof follows directly from the result of Problem 8.(<rll + a22 + ~ ~ ~ ) 2 ] / 2 .d e g I / l and the third term is omitted since i t is usually understood in the theory that if the denominator is zero the numerator will be zero too. hence dWP = -u. 81 PLASTICITY 189 8.. etc. Expand (8. S I I = 0.a. = d~. ( b ) biaxial stress with V. Thus from Problem 8... Determine the plastic.. = .CHAP. de..24).[2(uli 4 2 + o33 .uiioii)/2 Expanding this gives [3(all + ai2 + 033 + 6(aS2 + 0223 + ail ) .14. / f l . ( a i j ) . work increment dWP and the equivalent plastic strain increment de.u y / d .16. aI1de: + + a. show that d r p l s . the plastic potential equations (8..0 2 2 ~ 3 3. In plastic potential theory the plastic strain increment vector is normal to the loading (yield) surface a t a regular point.22) become the Prandtl-Reuss equations. (a) Here all = a1 = a. may be written in the form of (8..15. Thus d e F / l = . a I I I = . (C) pure shear with V. depl2 = .aII)2 ( o I I.ls.d e E I / l .25).23).24) to show that the equivalent stress a.~ + 1 1 ) ~6(oS2 + c i 3 + uil ) ] I 2 which confirms (8.Sijaqq/3)/2 = (3aiiuij . = {2[(dep.18. = -deEI.23). S I I I = . de.17 shows that de. Determine the plastic strain increment ratios for (a) simple tension with a. = = = (rl2 a. and for the stress state given. (b) Here o1 = a y / f i . 8.dep)21}112/3 .17.a11cz2 . 8.~ .c/\/~ (ay/fi)(-c) = -2cay/d From (8. The condition of normality is expressed by N = grad f l which requires N l l ( a f l l a ~ I )= + + N 2 / ( d f l / d o I I )= N31(afllauIII) for the Mises case where f l = (aI .deg)2 + (de: ... = a.~ + (o22 2 2 ) ~ .a I I I ) = 6s. o g Q = 3siisij/2 = 3(aij. Show that when the plastic potential function g(aii) = IIr. and since the plastic strain increment vector is along the normal i t follows that deF/sI = d ePI I / s I I= d e g I / s I I I .uZ2= a Y / f i . = ~ . since aglaa. If [N. a I 1 = O ..[(ali. = O if plastic deformation is controlled so that de: = C. = O. aI1 = ' a I I I= O and s1 = 2ay/3.o J 2 - 2a. s I I = s I I I = -oy/3. N3] are direction numbers of the normal to the yield surface f .o I I I ) 2 ( a I I I. = d ~ ~ under ~ l the s Mises ~ ~ yield ~ condition and flow law. 8.a y / f i and S I = o y / d . 8. From equation (8."33ul1) 6(a:2 + ai3 + )]/2 - ...~ ~ / fu~~ i =.21). uy/\/3. / f i .16.13. Here af1/aaI = 2(2u1.. de: = 0 ...N.. for the biaxial stress state a. ~ ~ .30).22.21).2 v ) o k k / 3 E and C . (8. = dWP/uEQ here and ( 8 3 2 ) follows directly from (8. ~ i s ofrom eii = e:.2v)Ukk/E. The Hencky total deformation theory may be represented through the equations t j = C. Verify that the Hencky parameter 4 may be expressed as given in (8. + C. 8-15.21 gives e.31) (or Problem 8. €5 = +2sijsij or Q = 4-luEQ which when multiplied on each side by 213 becomes ELASTOPLASTIC PROBLEMS (Sec.23. which reduces to eij = e: + + e: = ( Q * ~ ) s (8. = +vij. 8. = e.34) becomes U E Q = F ( W P ) and so doEQ= F' dWP. as given in (8.(1 .36). and since from (8. may be taken as the yield function in the hardening rules (8. d ~ .19) dWP = U E Q i t follows a t once that uEQF1= H'. From the result obtained. 8 8.Q.27). TOTAL DEFORMATION THEORY (Sec. definition (8.31).20. C. 8-15 Here the only nonzero stress is the bending stress ull. But from (8.36) and (8. . From (8.39). Verify (8.21. and from eij = E: + €5 i t follows that here yj = . < a ) . Fig.24) gives dWP = U E Q Thus de:. For a material obeying the Mises yield condition. d W P = sijsijdh for a Prandtl-Reuss material satisfying (8.e: + From the same equation.) and so daEQ= H' de. the equivalent stress a.19.9) 8.. 8. eii = ( 1 . 8.. Show that in this case uEQFr= H' where Fr and H r are the derivatives of the hardening functions with respect to their respective arguments. and M = bczu. ~ / ubecause ~ ~ ) of the &EQ.37). M = 2bc2uy/3 a t first yield (when a = c ) . Here (8. Using simple beam theory.190 PLASTICITY [CHAP.38). has been used. determine the end moments M for which the remaining elastic core extends from -a to a as shown in Fig.21). Show that these equations are equivalent to (8.. Thus F' dWP = H' de. P dh = 3 deEQ/2UEQ for such a material and so d W P = ( 3 ~ ~ ~ 8 ~ ~ / 2 ) ( dwhich e . for the fully-plastic beam (when a = 0). with C .. = +sij. An elastic-perfectly plastic rectangular beam is loaded steadily in pure bending. 8. = QSij implies e: = O so that €5 = e.32) by showing that for a Prandtl-Reuss material the plastic work increment is dWP = o.. The equation e.36) is given here by OEQ = H(€.10) 8. the stress co?dition a t the elastic-plastic interface.38). In the plastic portion. ull = uy. eij Sijekk/3 = e: + + sijefk/3 e. ull = Ecll = E x d R where R is the radius of curvature and E is Young's modulus. In the elastic portion of the beam (-a < x .37) and (8. . Thus where uy = E a I R . Squaring and adding the components in the equation e: = @sij of Problem 8. + 6ij~kk/= + 3 (sij/2)G 6.. Likewise (8. Determine the torque for which an inner elastic core of radius a remains.AIE) + 2c3bA/3R + ~ u . Thus 1 (k+/a) dr + 2a S. determine the pressure a t which first yield occurs. 8-17. = 2ake3/3 = 4T1/3 when a = 0.u Y / E ) after yield. Thus the Mises yield condition (8. 8 .37) which here becomes Expanding by the third column. 12 3 M e2boy(l. 4 4 ) . Using the Mises yield condition.45) for the stress tensor (8.1) and po = 2ay(l . O 6 r 6 a. 8-18 SLIP LINE THEORY (Sec. ..) = UY. 8-18 is subjected to an increasing pressure po. Fig.27.41) with + = ( a l i a . 8-17 A thick spherical shell of the dimensions shown in Fig. 8-16 8. Again c l l = x2/R and so or using oy = E a / R a s in Problem 8.24. A(€.. . 81 PLASTICITY 191 8.12) becomes o(%%) -o(. T .. .25.r) = 0111. Fig.23 if the material is a + piecewise linear hardening material for which a . = a .a3/b3)/3 a t first yield which occurs a t the inner radius a. The stress distribution for this beam is shown in Fig. and for the fully plastic condition. Verify directly the principal stress values (8. The principal stress values a r e found from the determinant equation (2. Determine the moment for a beam loaded as in Problem 8. ) / 2 as given in ( 8 . o(.'kr2 dr = ?qc3- 3 a3 / 4 ) Therefore the torque a t first yield is T 1 = a k ~ 3 / 2when a = c. An elastic-perfectly plastic circular shaft of radius c is twisted by end torques T as shown in Fig. 1 1 ) 8. Because of symmetry of loading the principal stresses a r e the spherical components o(%%) = o1 = U I I .CHAP. 8-16. The elastic stress components may be slíown to be Therefore ay = 3b3p0/2r3(b3/a3.. - The shear stress u12 is given here by alz = krla for k for a 5 r 6 e where k is the yield stress of the material in shear. R ~ ( A-/ E1)/3E2 Fig.23. and by o. uI)2 (Problem 2. (8..SII)' + (SII -~111)' + (SIII -~ 1 = ) ~ 6k2 Expanding and rearranging. 8-19. o1 = SI + UM.. -I If x 1 is along a n a line and x . Show that equation (8. along a .o I I ) 2 ( o I I. normal velocity continuity along B C ..L3 lines. the constant here must be U cose and so v2 = U cose..47) yields -aplaxl .30. Using the condition that the yield stress in shear k is constant. and from (8. and so (8. But sI + + sl. 8. 0 = S-.53). p . Fig..8 lines and circular a lines a s shown in Fig. -aplaxz + 2k(agldx2) = O along the .12).u I I I ) 2 ( o I I I. - 2k(dglaxl) = O along the a line.52).28. From the .192 PLASTICITY [CHAP.o I I ) 2 ( o I I.22) by ao. Determine the velocity components along these slip lines in terms of the approach velocity U and the polar coordinates r and O. 8 The roots of this equation a r e clearly o = -p + and o = &(o11 oZ2)2 d i ( o l l +~ + 2 2 ) ~ = -P k.k ( 2 cos 2g)(aglax1)+ k(-2 sin 2g)(agldx.29.47) with the equilibrium equations and integrate to prove (8. and from (8. Along the circular a lines. = 2k2 From (2.13) a t once becomes (SI .L3 line.31.2kg = C z on the line. U cos e de = U(sin e + l/fi) MISCELLANEOUS PROBLEMS 8. combine (8.a I I I ) 2 ( o I I I.. I n the frictionless extrusion through a square die causing a fifty per cent reduction.22) + + and so 9oOCt = (oI .) = O and k ( 2 cos 2g)(a+lax2) = O -k(2 sin 2g)(ag/axl). the centered f a n region is composed of straight radial .. + + I n terms of principal stresses 3o. etc. p + 2kg = C1 on the a line. 8. = f i ~ .48).71). From the equilibrium equations aall/dxl + ao121axz = O and do12/ax1 + a a 2 2 / a ~ 2= O which a r e valid here. 8. @ = O and these equations become -dplax. . this may be written s: sSI sSIr ..L3 line. 8-19 Along the straight .( s I + + + s I I + s I I I ) 2 / 3= 21~2.ar).aplax. s. dg = de. = 2u$ i n agreement with (8. / 3 . = d ( o I . Show that the vonSMisesyield condition may be expressed in terms of the octahedral shear stress uoCt (see Problem 2. Integrating directly.. + s.13) for Mises yield condition may be written as sS + s:. d@= 0. = IpD O and the required equation follows.. dv2 = O o r v 2 = constant. o I I I ) / 2 p(o. = O (Fig. u3R) we + A. The radius of the Mises yield circle is &%ry so that by definition X = m o y cose.V Z ) / E ~ . .12) we have + (vo1. 8-20). = O.. Likewise.f i=~ . = ( 2 u y sin $)I3 where 9 = tan-I Y / X in the notation of Problem 8.2 ~ . = 2 o y / m . for the shear components.U I I . for the normal components and de:. a t first yield. and the accompanying strain C .s.. = o. Solving these three equations simultaneously yields the desired expressions.. from Eell = o l l . Show that the Prandtl-Reuss equations imply incompressible plastic deformation and write the equations in terms of actual stresses. o. deii = sii dh = O since sii = IXD O and the incompressibility condition dez = O + is attained. From (8.v ) ) . Using the von Mises yield condition... = ( O . from whieh a l i = .. s.C H A P .)2 ('Tll(l. = vu.35. At what value of the Lode parameter p = ( 2 u 1 . . An elastic-perfectly plastic. The elastic stress-strain equation Ec33 = o33 . = u. 8. = UY gives 27 = uy. = s . 8-20 .8). and by (8.. = O and C.. show that in the U-plane the deviator stress components at yield are S. . a s the student should verify.36. dA. . = o... = [ . equation (8.6. Thus from (8. etc.42) o. Also.12) gives after some algebra (see Problem 8. + U M ..~ / 6 ) ] / 3 .. the equations s ... see that here = -ay(l .2~111= -6 + + Y = -20y sin e are obtained. etc. + = u . Note that in each case yielding depends on 7. . ~ )= ~ 20. The principal stresses here are readily shown to b e a1 = o 7 . d e z = (oij.u v / d ~(cornpressive) a t yield. i. When a.a t yield. s... p = 1 which is sometimes called a cylindrical state of stress. . For the state of stress where u and T are constants.. . incompressible material is loaded in plane strain between rigid pIates so that u.8). . Fig. P - From (8.( 2 / f i ) a y cose and S I -i . = uP Thus when /I = 1 the two are identical.(oz2 q 3 ) / 2 ]d ~ . Use Mises yield condition to determine the loading stress CT.not on c . In terrns of stresses.Siiukk/3)d ~ ..O. Thus the principal stresses are o.32.. sI s I I s I I 1= O . Tresca's yield condition.u I I I ) are the Tresca and Mises yield conditions identical? + + From the definition of p. .34. 81 PLASTICITY 193 8..e... = -vull. = . 8.21). u I I 1= . + ( . ) / ( u . From the transformation table given in Problem 8.6 together with o. o.U .u . = (2/3)[uli. determine the yield condition according to Tresca and von Mises criteria.~ h u sde:.. etc.v(oii+ 0 2 2 ) reduces here to o. a. ' + = [2uY cos (9 ~/6)]/3. yielding is independent of hydrostatic stress.c l 1 . ..cos (9 . in the n-plane. 8. the Tresca condition V .u .12) the Mises condition gives 7 = cy/*. is o.o I I 1 ) / 2 which when substituted into the Mises yield condition (8.C.r. Y = m o y sin e a t yield. . o(oe)..23) M = bc20y. since I = 26~313.a2/b2).~ + ( z z ) ) ~( O ( ./3 The maximum stress is a t r = a . ) _L k b . .1).) = o y since o. . An elastic-perfectly plastic rectangular beam is loaded in pure bending until fully plastic. 8-22 - residual stress 8. 8-24 The cylindrical stress components (Fig.4 m0ment M.) . pi = ( o y / 2 ) ( 1.) . (b) For the Tresca yield condition.a2/b2) a t first yield. Thus 2pibz/r2 = Qoy and now a t r = a . the moment is (see Problem 8.37. A thick-walled cylindrical tube of the dimensions shown in Fig. 8-23 is subjected to an interna1 pressure p . 8-23 Fig. Fig. This moment would cause a n elastic stress having a = Mel1 = 3oy/2 a t the extreme fibers. 8-24) are principal stresses and for the elastic analysis + ) pi(b2/r2 l ) / Q . . Deter- mine the residual stress in the beam upon removal of the bending 6 r l . o(zL)= pilQ where Q = (b2Ia2 . a t first yield if the ends of the tube are closed. Fig.. or pf b41+ = Q20.o ( 8 0 ) ) ~ (o(e. and a t first yield pi = ( o y / f i ) ( l ..) = -pi(b2/r2 .38. 30~12 oy/2 fully plastic + negative elastic Fig. = 20. Assume (a) von Mises and ( b ) Tresca's yield conditions. may be shown to be o(.1 ) / Q . o ( e o = ( a ) Here Mises yield condition is + (a(. Determine the value of p. ) .o(. 8-22. 8 q-J 8. x Thus remova1 of M is equivalent to applying a corresponding negative elastic stress which results in the residual stress shown in Fig. is the intermediate principal stress. 8-21 For the fully plastic condition.194 PLASTICITY [CHAP. 41) becomes 2h ---4 Fig. + )~ + 6(de&)2]112 A thin-walled elastic-perfectly plastic tube is loaded in combined tension-torsion. + (de.CHAP. ( a ) deeQ + (de. determine the plastic strain increment ratios for (a)biaxial tension with ul1 = o. show that o. The pressure on A B is k.6. Tresca: ( ~ / 2 k ) 2 ( ~ / k ) = 2 1 + Making use of the material presented in Problem 8. 8. b = h l f i Show that the stress tensor (8.ds[3)2 + (dei3 . Ans. ( b ) tension-torsion with oll = oy/2. 8-25 is subjected to pure bending. = -der3 12 ( b ) deTl /2 = -dei2 = -der3 = ddeL/3 Verify the following equivalent expressions for the effective plastic strain increment degQ and note that in each case degQ = deFl for uniaxial tension ull. A t what value of r will yielding first occur according to the Mises condition? A n s . Mises: ( o l f i k)2 (rlk)2 = 1. u12 = oY/2. p = k(1 a / 3 ) Fig.6. 8-25 \ 0 o-P/ when referred to the axes rotated about x3 by an angle O in Fig. (a) deFl = dei.. show that p = -6tan 0 . . + si. + 2(dcr2 + 2(de& + 2(de& [(derl)2 )2 )2 )2 )2 )2]1/2 ( b ) de:.. Show that von Mises yield condition may be written in the form (011 . 8-5(c).3) and the Mises yield condition.17. Determine the pressure on AC.. + Ans. + Ans.011)2 + 6(a:2 + ui3 + mil ) = 6k2 Following the procedure of Problem 8. From the definition of Lode's parameter p (see Problem 8. Rework Problem 8. 8-8..~ 3 3 4-) ~ (a33.. + s. Show that the maximum load occurs a t e = n. . 7 = uy/2 I x9 - The beam of triangular cross section shown in Fig. + s:.)/3 respectively. 81 PLASTICITY Supplementary Problems A one-dimensional stress-strain law is given by o = Ken where K and n are constants and E is true strain. = 2 o y / ~ ~ .51. . A centered fan of a circle arcs and P radii includes an angle of 30° a s shown in Fig. 8-26 . verify the geometry of Fig. 8-26. = (\/Sj3)[(der1. Determine the location of the neutra1 axis (a distance b from top) of the beam when fully plastic. In the 11-plane where s = tan-1 Y I X with X and Y defined in Problem 8.u. Show that the invariants of the deviator stress IrID = sijsij/2 and IIIED= sijsjkski/3 may be written = (s.4 using the yield stress in shear k in place of u y in the Mises and Tresca yield conditions...)/2 and III. = uy.~ 2 2 ) ~(o22 .deTl)2 + 6 ( d ~ f . An axial stress o = uy/2 is developed first and maintained constant while the shear stress T is steadily increased from zero. = (si + s.. Ans.dei2)2 + (de. possess no tendency a t a11 f o r deformational recovery. 9. (a) Linear Spring ( b ) Viscous Dashpot Fig. the discussion which follows is restricted to isothermal conditions and temperature enters the equations only a s a parameter. As shown in Fig. Viscous fluids. The models are given more generality and dimensional effects removed by referring to u as stress and r a s strain. 9-1. Linear Viscoeiasticity 9. and the viscous dashpot having a viscosity constant 7. The mechanical elements of such models are the massless linear spring with spring constant G. Also.2 SIMPLE VISCOELASTIC MODELS Linear viscoelasticity may be introduced conveniently from a one-dimensional viewpoint through a discussion of mechanical models which portray the deformational response of various viscoelastic materials. elastic stress is related directly to deformation whereas stress in a viscous fluid depends (except for the hydrostatic component) upon rate of deformation. thereby putting these quantities on a per unit basis. the force across the spring u is related to its elongation r by a = Ge (9. Elastically deformed bodies return to a natural or undeformed state upon remova1 of applied loads. The elastic (Hookean) solid and viscous (Newtonian) fluid represent opposite endpoints of a wide spectrum of viscoelastic behavior. 9-1 . however. Although viscoelastic materials are temperature sensitive. Material behavior which incorporates a blend of both elastic and viscous characteristics is referred to a s viscoelastic behavior.1) and the analogous equation f o r the dashpot is given by a = qe where := deldt.1 LINEAR VISCOELASTIC BEHAVIOR Elastic solids and viscous fluids differ widely in their deformational characteristics. 9-2(a).).CHAP. 9-3 A four-parameter model consisting of two springs and two dashpots may be regarded as a Maxwell unit in series with a Kelvin unit as illustrated in Fig. A three-parameter viscous model consisting of two dashpots and one spring is shown in Fig. These equations are essentially one-dimensional viscoelastic constitutive equations. 9-4 below. 9-3(b).5) and (9./G + l/7)u = {a. G9 G. 9-3(a). and a Maxwell unit in parallel with a dashpot is analogous to the viscous model of Fig. Thus it incorporates "instantaneous elastic re- . = a/dt.3) G 7 and for the Kelvin model is u = Ge+rl. 9-3(b). 9-2 The simple Maxwell and Kelvin models are not adequate to completely represent the behavior of real materials. (a) Maxwell ( b ) Kelvin Fig.3) becomes {a. ' (9.u = . I t should be remarked that from the point of view of the form of their constitutive equations a Maxwell unit in parallel with a spring is analogous to the standard linear solid of Fig. The Kelvin or Voigt model is the parallel arrangement shown in Fig. Severa1 equivalent forms of this model exist.4) becomes u = {G + qat)€ with the appropriate operators enclosed by parentheses. 91 LINEAR VISCOELASTICITY 197 The Maxwell model in viscoelasticity is the combination of a spring and dashpot in series as shown by Fig. More complicated models afford a greater flexibility in por- traying the response of actual materials. and known as the standard linear solid is shown in Fig. I t is helpful to write them in operator form by use of the linear dijferential t i m e operator d. 9-2(b). Thus (9. (a) Standard Linear Solid ( b ) Three-parameter Viscous Model Fig. The four-parameter model is capable of a11 three of the basic viscoelastic response patterns. A three-parameter model constructed from two springs and one dashpot. 9-3(a). The stress-strain relation (actually involving rates also) for the Maxwell model is -u+ . (9. 9 sponse" because of the free spring G. Q o C + (9. Fig. In operator form. "delayed elastic response" from the Kelvin unit. 9-6 . The total strain of this model is equal to the sum of the individual Kelvin unit strains. 9-5. 9-5 Similarly.. 9-4 The stress-strain equation for any of the three or four parameter models is of the general form + + p2a pl& pou = q2Y+ ql.'s and qi's are coefficients made up of combinations of the G's and ?'s. t 0 Fig. and so from t9. by (9.. (9.7) where the p.7) is written {p2a: + plat + p0}u = {q2a: + qlat + qo)€ (9-8) 9. Here the total stress is the resultant of the stresses across each unit. and.198 LINEAR VISCOELASTICITY [CHAP.6). 9-6 is called a generalized Maxwell model. finally.5). LINEAR DIFFERENTIAL OPERATOR EQUATION The generalized Kelvin model consists of a sequence of Kelvin units arranged in series as depicted by Fig. and depend upon the specific arrangement of the elements in the model. a sequence of Maxwell units in parallel as shown in Fig.3 GENERALIZED MODELS. "viscous flow" because of the free dashpot v. Fig. Thus in operator form the constitutive equation is. )].t.(t) = - G ( 1 . Here T = rllG is called the retardation time. = O. The creep response of a Kelvin material is determined by solving the differential equation which results from the introduction of (9. For any continuous function of time f ( t ) . 9-7 where [ U ( t ) ]represents the unit step function applied a t time t.13) where the operators { P ) and {Q) are defined by m ai CO p. = qOc+ql. In the relaxation experiment an instantaneous strain e0 is imposed and maintained on the specimen while measuring the stress (relaxation) as a function of time. These tests may be performed as one-dimensional tension (compression) tests or as simple shear tests.+ .16) into (9.-2dta' {P) = i= i=O 9.. and maintaining the stress constant thereafter while measuring the strain (creep response) as a function of time.4).17) may be integrated to yield the Kelvin creep response "o . 9-7. . The creep experiment consists of instantaneously subjecting a viscoelastic specimen to a stress U .e-t17)[U(t)] (9.[u(t)l ( 9 .4 CREEP AND RELAXATION The two basic experiments of viscoelasticity are the creep and relaxation tests.defined by and shown in Fig.+q2.I 6 ) Fig.19) The creep loading. 9-8 below.CHAP.it may be shown that with t' as the variable of integration. 1 t t 1 0 = ".. 1 For the creep loading.t.t. the creep and relaxation loadings are expressed in terms of the unit step function [ U ( t.18 ) by means of which (9.10) result in equations of the form p O u + p l ~ + p 2 a +.)] ti dt' (9.9) and (9. . 91 LINEAR VISCOELASTICITY 199 For specific models. f (t')[U(tr. together with the creep response for the Kelvin and Maxwell models (materials) is shown in Fig.)]dt' = Stf(t') [ U ( t. (9.. . which may be expressed compactly by This linear diflerential operator equation may be written symbolically as {P>u = {&}c (9. Mathematically. .= c o [ S ( t ) ] into (9. or Dirac delta f u n c t i o n . RELAXATION FUNCTION.200 LINEAR VISCOELASTICITY [CHAP. uo - * t t (a) Creep Loading ( b ) Creep Response Fig.26) where q ( t ) is known as the c r e e p f z ~ ~ z c t i o n . HEREDITARY INTEGRALS The creep response of any material (model) to the creep loading u = u 0 [ U ( t ) ] may be written in the form ~ ( t =) * ( t ) u . 9-8 The stress relaxation which occurs in a Maxwell material upon application of the strain is given by the solution of the differential equation obtained by inserting the time derivative of (9. ( 1 . (9.i t may be shown t h a t when t > t. where i t is said to have a n indeterminate spike.25) The delta function in (9. For a continuous function f ( t ) . with the help of which (9.For example. Here [ S ( t ) ] = d [ U ( t ) ] l d t is a singularity function called the unit i m p u l s e f u n c t i o n .3). 9-5 is determined from (9.5 CREEP FUNCTION. 9 1. By definition.e t"l)[ú'(t)] (9.27) 1-1 .25) indicates that it would require a n infinite stress to produce a n instantaneous finite strain in a Kelvin body.21) may be integrated to give the Maxwell stress relaxation The stress relaxation for a Kelvin material is given directly by inserting .19) to be Y @(t) = J . 9.20) into (9.4) to yield dt) = GcOIU(t)l+ V E~[~(~)I (9. This function is zero everywhere except a t t = t . the creep function f o r the generalized Kelvin model of Fig.. For the generalized Maxwell model of Fig. if the stepped stress history of Fig. Ji) may be replaced by the continuous compliance function J(T). t' I I I I + dt' t - . 91 LINEAR VISCOELASTICITY 201 where Ji = l/Gi is called the compliance. or relaxation spectrum.. the superposition principle is valid.the Kelvin creep function becomes The function J(T)is referred to as the "distribution of retardation times". T ~ )and the relaxation function is defined by +(t) = G(T)~-"'dr Sm (9.v . each of magnitude da and the creep response given by the superposition integral -m ~ ( t ) = Jt $p 9 ( t .31) The function G(7) is known as the "distribution of relaxation times".o the function G(T) replaces the set of constants (G. 9-9(a) is applied to a material for which the creep function is ~k(t)..----- " ---- I I I I t' (a. Thus the total "effect" of a sum of "causes" is equal to the sum of the "effects" of each of the "causes". 9-9(b) may be analyzed a s an infinity of step loadings. as N + . 9-6..the creep response will be Therefore the arbitrary stress history u = ~ ( t )of Fig. or retardation spectrum. the relaxation function is determined from (9.- t Fig. 9-9 /e " + gdt' -.24) as Here.t') dt' Such integrals are often referred to as hereditary integrals since the strain a t any time is seen to depend upon the entire stress history.. If the number of Kelvin units increases indefinitely so that N + m in such a way that the finite set of constants ( r . Accord- ingly.CHAP. In linear viscoelasticity. the stress relaxation for any model subjected to the strain c = co[U(t)] may be written in the form where +(t) is called the relaxation function. "o ' "1 I tl I t2 "3 --------------- I I I t3 . In analogy with the creep response.. it follows that some relationship must exist between the creep function +(t) and the relaxation function +(t).a). sin wt.andthe absolute dynamic compliance C ~ / U . i. but using the Laplace transform definition it is possible to show that the transforms G ( s ) and $(t) are related by the equation where s is the transform parameter.34) is usually written in the form Following similar arguments a s above. the in-phase and out-of-phase components of the stress and strain rotating vectors of Fig.(t) = J m g +(t . the integrals comparable to (9. 9. The stress and strain for this situation may be presented graphically by the vertical projections of the constant magnitude vectors rotating a t a constant angular velocity w as shown in Fig. the stress as a function of time may be represented through a superposition integral involving the strain history e(t) and the relaxation function +(t). sin (ot . a sinusoidal response of the same frequency but out of phase with the stress by the lag angle 6.t') dt' Furthermore. Such a relationship is not easily determined in general. (9. In analogy with (9. The ratios of the stress and strain amplitudes define the absolute dynamic modulus ao/%. completely free of stress and strain a t time zero.34) and (9. the lower limit in (9. if the stress loading involves a step discontinuity of magnitude o. 9 For a material initially "dead". 9-10 below.t') dtf and with regard to a material that is "dead" a t t = 0. 9-10(a) are used to define (a) the storage modulus . the resulting steady state strain will be c = E.6 COMPLEX MODULI AND COMPLIANCES If a linearly viscoelastic test specimen is subjected to a one-dimensional (tensile or shear) stress loading o = U.37) may be used to specify the viscoelastic characteristics of a given material.202 LINEAR VISCOELASTICITY [CHAP. In addition.37) and Since either the creep integral (9.e. . a t t = 0.33) the stress is given by .33) may be replaced by zero and the creep response expressed as c(t) = ltd"(t3 dt' qr(t .(t) = Xt%+(t-t') dt' (9.35) are respectively .34) or the relaxation integral (9. Note that G* = 1/J*. this is handled by resolving the stress and strain tensors into their deviatoric and spherical parts. Fig. . sin 6 (d) the loss compliance J.43) whose real part is the storage modulus and whose imaginary part is the loss modulus.U. 9-10 A generalization of the above description of viscoelastic behavior is achieved by expressing the stress in complex form as *. and subsequently combined to provide a general theory.41) and (9.70) as . it is customary to consider separately viscoelastic behavior under conditions of so-called pure shear and pure dilatation. . = - "o c.oeiwt (9. = - €0 c. The stress tensor decomposition is given by (2. where the real part is the storage compliance and the imaginary part the negative G1 of the loss compliance.7 THREE DIMENSIONAL THEORY In developing the three dimensional theory of linear viscoelasticity. 9-11 the vector diagrams of G* and J* are shown. In Fig. + iG. Similarly. = - . for each of which viscoelastic constitutive relations are then written. the complex compliance is defined as e*/.* = J*(iw) = (Eoluo)e-i6 (94 4 ) = J.o Fig. 9-11 9. Thus distortional and volumetric effects are prescribed independently.42) the complex modulus G*(io) is defined as the complex quantity V*/€* = G*(iw) = (oO/ r o )eis = G.41) and the resulting strain also in complex form as = c ei(wt-6) o From (9. cos S ( c ) the storage compliance J. 91 LINEAR VISCOELASTICITY o0 sin S ( b ) the loss modulus G. (9. Mathematically.CHAP. 46) Using the notation of these equations. Since practically a11 materials respond elastically to moderate hydrostatic loading. 9 and the small strain tensor by (3. (9. the three-dimensional viscoelastic constitutive relations in creep integral form are given by and in the relaxation integral form by The extension to three-dimensions of the complex modulus formulation of viscoelastic behavior requires the introduction of the complex bulk modulus K*. 9-12 . is formulated a s follows: Let body forces bi be given throughout V and let the surface tractions t$'(xk. { M ) and { N ) are operators of the form (9.48~) uii = 3K9.47) modified to read { P > s i j= 2 { Q ) e i j (9..47b) where { P ) . Again. t ) be prescribed over the portion SI of S.t ) be prescribed over the portion S.204 LINEAR VISCOELASTICITY [CHAP.8 VISCOELASTIC STRESS ANALYSIS.98) as eij = e.13) in differential operator form is written by the combination and {M)uii = 3{N)eii (9.48b) where K is the elastic bulk modulus. the appropriate equations are of the form 9. of S. Then the governing field equations take the form of: Fig. {Q). writing shear and dilatatíon equations separately. Following the same general rule of separation for distortional and volumetric behavior. 9-12. and the surface displacements gi(xk.14) and the numerical factors are inserted for convenience. the three dimensional generalization of the viscoelastic constitutive equation (9. CORRESPONDENCE PRINCIPLE The stress analysis problem for an isotropic viscoelastic continuum body which occupies a volume V and has the bounding surface S as shown in Fig. the dilatational operators { M ) and { N ) are usually taken as constants and (9. + Sijrkk/3 (9. i + -b..22). Strain-displacement equations 2fij = (ui. Constitutive equations (a) Linear differential operator form (9.50) (c) CompIex modulus form (9. and if the material behavior may be represented by one of the simpler models. it is conventional to seek a solution through the use of the correspondente pr. + bi a. t) n.i) or strain-rate-velocity equations + 2Gij = (v. A comparison of the pertinent equations for quasi-static isothermal problems is afforded by the following table in which barred quantities indicate Laplace transforms in accordance with the definition f(xk9S ) = xw f ( ~ .~) 3.CHAP. the two sets of equations have the same form. For more general conditions.) = tin)(x. the field equations above may be integrated directly (see Problem 9.inciple.~ V..t)e-st > dt Elastic Transformed Viscoelastic 1. if in the solution of the "corresponding elastic problem" G is replaced by &lP for the viscoelastic material involved. (x.49) or (9. 4.j3i = O 1. Equations of motion (or of equilibrium) 2. Initial conditions ui (x. 0) = UO 5. t) on S . however. üij. the result . = O From this table it is observed that when G in the elastic equations is replaced by &lP.. 91 LINEAR VISCOELASTICITY 205 1. Boundary conditions A uij (x.48) (b) Hereditary integral form (9. Thjs principle emerges from the analogous form between the governing field equations of elasticity and the Laplace transforms with respect to time of the basic viscoelastic field equations given above. Accordingly.51) If the body geometry and loading conditions are sufficiently simple..j + uj. 9-3(a). Determine the stress-strain equation for the four parameter model of Fig. determine its stress-strain equation.2) may be used to give . / { a t 1 / ~ ~ G) q . 9-13 a s a special case of the generalized Max- well model. problems. 9-13 .206 LINEAR VISCOELASTICITY [CHAP. 9-2(a) the total strain is the sum of the strain in the spring plus + the strain of the dashpot. 9.3) 9. Solved Problems VISCOELASTIC MODELS (Sec. -+ m this becomes Ü + ( G 1+ G2)&l'lg= G1'E+ G1G2. o = os + a~ and directly from (9.= Zs + GD. In the Kelvin model of Fig. Treating the model in Fig.I~ + and compare with the result of Problem 9. + . The particular problem under study will dictate the appropriate form in which the principle should be used. 9 is the Laplace transform of the viscoelastic solution. -.Iq2 which is equivalent to the result of Problem 9.. The correspondence principle may also be stated f o r problems other than quasi-static .1.10) for N = 2 in the form o + + + = G l .3) and (9. Thus E = es + e~ or in operator form e = olG1 + o / { G 2 q2at).= O/G a/?. ( 9 . Furihermore.2. From this + and so GIGze + G l s z . + Ge.2).50) or (9. the form of the constitutive equations need not be the linear differential operator form but may appear a s in (9. Since the stress across each element is o. = ( G 1+ Gg)a + 72. Thus e = es ED and also . 9. 9-4.1) and (9. <I ..O Writing (9. Here the total strain is the sum of the strain in the spring plus the strain in the Kelvin unit. o = v . Let T.4) respectively. 9-2(b).2.4. In the Maxwell model of Fig. Verify the stress-strain relations f o r the Maxwell and Kelvin models given by (9. Inversion of the transformed solution yields the viscoelastic solution.51).3. 1 ) and (9.1-9. / { a t 1 1 ~ ~ ) and operating as indicated gives Fig. 9. (9. 9. Here the total strain e = e~ + e~ which in operator form is e = o/{Gp + vzat) + {at + l / ~ l ) o / G l { a t ) Expanding the operators and collecting terms gives As 7 . Use the operator form of the Kelvin model stress-strain relation to obtain the stress- strain law for the standard linear solid of Fig.2.49). and by formula et1/T[8(t')] 9. 9-14 CREEP AND RELAXATION (Sec. ) ~ o [ U ( t ) ]or by integrating directly the standard solid stress-strain law.19) is simply e(t) = [ l / G l + (1/G2)( 1 . 9.17') and (9. 9-14 may be considered as a degenerate form of the generalized Maxwell model with . + G2. which when expanded and rearranged becomes . Determine the Kelvin and Maxwell creep response equations by direct integration of (9.17) becomes eetlr = - (9..e ./r1r2 + + = (G1 G2)7 + (G1/72+ G2/r1).5.t / 7 ~ ) ] a o [ U ( t ) ] The same result may be obtained by setting q2 = m in the generalized Kelvin (N = 2) response 2 c = 2 i=l J i ( l .e-t/T)[U(t)] (9./{at) + .10) becomes o = vl. Determine the creep-recovery response of the 00 . The model shown in Fig.).CHAP. Using these values in ' 7 1 = '12 - (9. 9-15. 0 (71 72).1) and (9.7. Using the integrating factor et/7.10) develop the stress-strain equation for this model.7 the response while the load is on ( t < 272) is e = oo[l/Gl + ( l / G z ) ( l.e .4) 9.18) yields eet/7 = ( ~ ~ [ U ( t ) ] / ~ ) [ r= e t '(uO/G)(et/7- /~]~ l)[U(t)] n. 9. 9-3(a)) for the loading I I shown in Fig. The creep-recovery experiment consists of a A0 creep loading which is maintained for a period of time and then instantaneously removed./{at/G3 + l / q 3 1 or {dt/G3 + l/v3)8 = {atlG3 + 1 / 7 3 ) ( ~ 1 7 +G2:) + . Determine the creep response of the standard linear solid of Fig.r2). 6 272 t From Problem 9.t / r . + G2{at+ 1/7.23). I I standard solid (Fig. 9. (9.6.21) respectively.m for the case N = 3. The student should carry out the details.1/r2)(6 + "/r1) = Gl{at + 1/. i' or et'/7[U(t1)]dt' which by formula e = (ao/G)(l.t / 7 ~ ) ] Fig.e . Here (9.8.21) gives oet/7 = GeO aet17 = Geo[U(t)] or a = Geoe-t/~[U(t)] i' dt'. 9-15 . Use of et/' a s the integrating factor in (9. 93 'I 1 Application of the operators gives which may also be written t 713& + G30 = B ~ T ~(G213+ ~ Y + G391 G393):+ G2G3e Fig. 9-3(a). Since e = es + EK for this model the creep response from (9. 91 LINEAR VISCOELASTICITY 207 {at 4. qe-tlr)/tl. 9-18 + Thus by use of (9. 9 A t t = 2r2 the load is removed and o becomes zero a t the same time that the "elasticJJ deformation uo/G1 is recovered. / t . Fig. 9-16 Fig. 9-17 From Problem 9. GE/T and so + + here O U / T = 3Gso/tl Ge0t/7tl.t / r z/G 2 for t > 2 r 2 9.23). Note that the same result is obtained by integrating i r+2 3 ~ ~ t 'dt' tl Xt dtf 9. [ U ( t ) ] . The stress-strain relation for this model i s -o-.18) and (9. o = q ~ o / t land so C = -l ) ~ O l t lThus .23).= eo[6(t)] use of the integrating factor etlrz gives Fig. C = C = oO(l.2). 9. For t > 272 the response is governed by the equation + €/r2 = O which is the stress-strain law for the model with u = O (see Problem 9. Determine by a direct integration of the stress-strain law for the standard linear solid its stress relaxation under the strain E = e .11.18) and (9.= ~ . Determine the stress in the model under this straining. Determine the relaxation function +(t)for the three parameter model shown in Fig. u + = eOGl(G2 Gle-(Gi+Gz)tlnz)[U(t)]/(G1 + G2) CREEP AND RELAXATION FUNCTIONS.l ) e .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 (9.5) 9.T / r ~where C is a constant and !i' = t . + G1G2c/q2 and so with = eo[U(t)] and .9. -u 6 + u/r2 = (G1+ G2).208 LINEAR VISCOELASTICITY [CHAP. The special model shown in Fig.e-2)e-T1r2/G2 = uo(e2.?) Ce-t/r where C is the constant of integration. o = + eo(2q G t . o = eo(Gl G2e-t/rz) = eo$(t). The solution of this differential equation is C = C e . Integrating this + + yields o = a0(39 G t . When t = 0. . + + + Writing the stress-strain law (see Problem 9.10. HEREDITARY INTEGRALS (Sec. T = 0.e-2)/G2 and so e = aO(l.30) for the generalized Maxwell model. Note that this result may also be obtained by putting q1 + m in (9.2 ~ A~ t . 9-17. as indicated in Fig.5 the stress-strain law for the model is + + 017 = q Z + 3G. 9-18. 9-16 is elongated a t a constant rate . t l ) l ~dt' ) .34) together with the Kelvin creep function. 9-21 .12.( t . ~ l l) )~[ U ( t ) ].23) reduces to = X - G o St [ [ ~ ( t ) ] ( 1 . determine the resulting strain. A. 9.(t) = S-tm$ ([U(tl)] + t f [ s ( t l )-] [u(tf.( t .t l ) [ U ( t.15. For the Kelvin body.18) gives E ~ ( e -- = ( h l G ) ( ( ti.t l ) [ s ( t-f t l ) ] ) (-l e . Using the creep integral (9. If a ramp type stress followed by a sus- tained constant stress U .t f ) l r ) d t f I A straightforward evaluation of these integrals confirms the result presented in Problem 9. $ ( t )= (1.18) and (9.[ U ( t. Thus from (9.14. I G= u. which may be inverted easily by a Laplace transform table to give li/ = + l / G l .CHAP.4) leads to Integrating with the aid of (9. Using the relaxation function +(t)for the model of Problem 9. verify the result of Problem 9. 9. (Fig.h ( t .t l ) + ~ ( e ( t i . The stress may be expressed a s t.e .( ( t.34) becomes . 9-19 which when introduced into (9.t ) / ~l )-) [ U ( -t t l ) ] ) which reduces a s t -+ m to E = x ~ .40).e-tI7)/G and (9. Assume ~ .t l ) / T )dt' which by (9.( t f.40). / t=./G.( t . 9-20 Fig. t u = h t [ U ( t ) ]. 9-20. determine the creep function by means of (9. By a direct application of the superposition principle.t l ) ].13. 91 LINEAR VISCOELA'STICITY 209 9.tl)] tl t $ (1 .)] Fig.I. Fig. 9.t.13. 9-19) is applied to a Kelvin material. determine the response of a Kelvin material to the stress loading shown in Fig.11.13.[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. + The Laplace transform of $ ( t ) = G1 G2e-t/Tr is 5 ( s ) = G l / s + G z / ( s+ 1 1 ~ (see ~ ) any standard table of Laplace transforms).e . Writing (9. J* = 1/G* and so J1 . 9-21 above. 9-11. + From Problem 9.5) becomes (iolG + l / q ) o = &e from which U / E = Giol(io + 117) = G ~ w T+ /(~ ~ W T ) a s before. Use equation (9. Thus writing (9.l ) ) [ U ( t.iJ2 = l / ( G l iG2) = (G1. In the présent case therefore + ) ( i / G ) [ ( t r(e-tlT . the complex moduli add". lbiu (sin2 o t ) d t ] = o:sJ2 .6) 9. t 3tl)]l Note that a s t + m.i in equation (9. 9.l ) ) [ U ( - .2tl) + ~ ( e .ei(wt-6) from which ~ ~ e= G* i ~= G l i w~ ~ l~( l +i w ~ ) . tan 8 = G d G 1 = Gor/Gw2r2 = 1 1 ~ ~ .17. E + 0. Determine the complex modulus G* and the lag angle S for the Maxwell material of Fig.o0eiot/r = Giw~. 9.( t .19. From Problem 9.13.42) gives iwooeiwt4. and inserting (9. 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. the integral S odr evaluated over one cycle is ds dt = l 2 ~ 1 0 (oo sin U cos (ut . Show that the result of Problem 9.16 may also be obtained by simply replacing the operator d. J2'I* sin w t (cos u t cos a + sin u t sin a ) d t o 2nlo sin 2ot dt + J.)I(G? 62). COMPLEX MODULI AND COMPLIANCES (Sec.ti)] -((t t 2 t l ) ] + ( ( t.~G. 9.6 ) d t ~ ) E ~ W .5)and defining u / e = G*. 9-2. After the suggested substitution (9.I ) ][ U ( t ) ] for this stress loading. From (9.41) and (9. by . 9-10.210 LINEAR VISCOELASTICITY [CHAP.3tl) + ~(e-(t-3t1)I7.18.43) and (9.10) a s the generalized Maxwell complex modulus becomes 9.ooEo.16.l ) ) [ U ( t ) - ~(t= ] ( ( t.t l ) l ~ . .10) for the generalized Maxwell model to illustrate the rule that "for models in parallel.44).( t .t i ) + r ( e .17 the complex modulus for the Maxwell model may be written G* = a/€ = Giwrl(l+ iwr). or in standard form From Fig. + + Thus + + J1 = G ~ / ( G : 02) = l / G 1 ( l (G2/G1)2) = l / G 1 ( l tan2 6) + 9. Verify the relationship Jl = l / G l ( l+ tan28) between the storage modulus and compliance.3) a s 6 + U/T= G .z t ~-) /l ~) ) [ U ( . 9 The stress may be represented a s a sequence of ramp loadings a s shown in Fig. E = ( h / G ) [ t r(e-tlf .20. I For the stress and strain vectors of Fig. 4 8 ~ )and ( 9 .)/3 = 2G{1+ ~ a ~ ) ( 2 ~for ~ ~a/ Kelvin 3) body. 3cii = o o [ U ( t ) ] l Kfor this case.. = a o [ U ( t ) ] . 9-22 which upon integration gives Inserting this result into ( 9 .. 91 LINEAR VISCOELASTICITY 211 THREE DIMENSIONAL THEORY.oijakk/3)= 2{Q)(eij. { P ) ( u l l. A block of Kelvin material is held in a container with rigid .. = so that (9. = e.. Combining these relations yields the differential equation Fig. Writing ( 9 . From (9.48b)..23. 2 ) . + = (+ao/2 . a.u2.48a) gives 2(u11. VISCOELASTIC STRESS ANALYSIS (Sec. = O when the stress o. 4 8 3 ) to obtain the viscoelastic constitutive relation uij = + a i j { R } e k k {S)cij and determine the form of the operators { R } and { S I .24.oijekk/3) and replacing ukk here by the right hand side of (9.2 v ) a ( r . . Determine the radial stress for a Kelvin viscoelastic half- space by means of the correspondence principle when P = Po[U(t)]. = a. €11 -> (3K + G)aO19KG= a J E . But from (9.48b). 4 8 ~a) s {P)(aii. = 3Kell and (9. the result after some simple manipulations is aij oij{(3KP.a11/3)= + { 2 Q ) ( ~ l l -eii/Q). The radial stress component in an elastic half- space under a concentrated load a t the origin may be expressed as a.22.9Kao(l . and from ( 9 .P(r... Determine c.. Here qi = ell and a. A bar made of Kelvin material is pulled in tension so that a. and the retaining stress components V. = a33 = - ai.CHAP. 4 8 ~ for ) i = j = 2 gives a. 9. and u3. = . Combine ( 9 .8) 9.21.48b) becomes ai1 i-2u2. { P ) 1 and { Q ) = {G Tat) for a Kelvin material.e-(4G+3K)t/4G~)/(8GG K ) ) [ U ( t ) ] 9. 9-23 . ) . Determine the strain ell for this loading. so that now 200[U(t)I/3 = 2{G +~ a ~.41 = ( P 1 2 ~[(I where a and p are known functions.walls so that e.aO[U(t)]lgK) l ( ~ ~ ~ Solving this differential equation yields €11 = aO(3K + + aoe-tlT[U(t)]/9K e-t1T)[ U ( t ) ] l 9 K G As t + m.61. Fig. 4 8 ~with ) i = j = 1.a.2Q)/3p)ekk + {2Q/P)eij 9. 9.a o [ U ( t ) ] is applied.. for this situation....7-9. = O where ao is constant.. 25. the result is Note that when t = O. and the deflection w(x. . w ( . The correspondence principie may be used to obtdin displacements a s well as stresses. The elastic deflection of the beam is w ( x l ) = pon(xI)/24EZ where a ( x J is a known function.27. = -o* . 9-24 The bending stress for a simply supported elastic beam does not depend upon material properties.23. w ( x .v 2 ) l E T r . w ( .o the stress a. 117) + and { Q ) = {Ga.2G)l(3K + 4G) which may be written in terms of v as 02. . From Problem 9.. The x displacement of the surface of the elastic half space in Problem 9.26.the elastic deflection. A simply supported uniformly loaded beam is assumed to be made of a Maxwell material.24 is given by W ( z = ~=) P(l. For a Maxwell body. 0221t+m = -oO(9K . {P) = {a. Show that as t += . The viscoelastic operator corresponding to ( 1 . Thus for v 112. Determine the bending stress a . Determine the viscoelastic displacement of the surface for the viscoelastic material of that problem. O ) = pw(x1)/24EZ.. 9. t) if the load is p = p o [ U ( t ) ] .(4G + 3K))/2(4G+ 3 K ) = -oo(3K . = -voo/(l . so that the transformed deflection is which when inverted gives When t = O.).. ) P o ( l . . = o ) = O and when t + m. o.212 LINEAR VISCOELASTICITY [CHAP. 9 The viscoelastic operator for the term ( 1 . the elastic deflection. v).v2)lE is { 3 K + 4Q)/4Q(3K + Q ) which for the Kelvin body causes the transformed displacement to be After considerable manipulation and inverting. so the elastic and viscoelastic bending stress here are the same..lt+.23 approaches ao (material behaves as a fluid) if the material is considered incompressible ( V = 112).2 ~ is) { 3 Q ) l { 3 K P + Q ) so that for a Kelvin body the transformed viscoelastic solution becomes which may be inverted with help of partia1 fractions and transform tables to give the viscoelastic stress 9.u z ) / E ~ r . Fig. in Problem 9. = ~ -+ 9.1t. again representing the Kelvin model. if Gl = O the + reduced equation is 8 = q. if G.. 9. 9-25 and deduce from the result the Kelvin and Maxwell stress-strain laws. With the stress loading expressed by U = ao[U(t)].7 a s Fig. . Gze. Determine the constitutive relation for the 81 Kelvin-Maxwell type model shown in Fig..9 and the superposition principle the stress is Fig.e . the Maxwell law 6 U / T= + ~ G1. 9-25 In this equation if q2 = 0 (spring in parallel with Maxwell). 9-3(a) and compare the result with that obtained in Problem 9. the strain may be written a t once from the result of Problem 9. Use the superposition principie to obtain the creep recovery response for the standard linear solid of Fig. and when q . = 0. 9-26 E = ao(l/Gl + ( 1 . 91 LINEAR VISCOELASTICITY 213 MISCELLANEOUS PROBLEMS 9. results.. Show that eventually the "free" spring in the model carries the entire stress.e-(t-ti)/7) + G ( t . + e-(t .2r2)] (see Fig.8. this also reduces to the Maxwell law. 9-26).t ' ~ ~ ) / G ~ ) [ U (-t ) laO(l/G1-(i1. c<t- Here a = UM + aK = . . if G2 is taken zero first (dashpot in parallel with Maxwell).z ~ z ) ~ ~ ) l G ~-) 2r2)] [U(t A t times t > 272 both step functions equal unity and E - . Likewise. + GlG2e and q1 set equal to zero.29.1)e-t/r2/G2 which agrees with the result in Problem 9.t l ) ) [ U (-t t l ) ] / t l For times t > t 1 the stress is o = cotl(etl/T. and as t + m this reduces to o = Gço.28.ao[U(t. Likewise.. = 0. ( G 2 1 ~ 1 ) ~ . 9.1)e-t/r/tl f Gco. If the four-parameter constitutive relation is rewritten 1118 + Glu + = 7118~Y ( G l ~ l +G281+ G172).2 7 2 ) / ~ ~ ) / =G ~ aO(e2.T+ G. the result is the Kelvin law a = 8. Determine the stress in the model of Prob- lem 9.30.~ 0 ( 7 (- 2 e-t/7) + G t ) [ U ( t ) ] l t -l eo(7(2. 9-27 - . 9-27./{dt/G1 + l / v l } + {Gz + q2at}r 82 which upon application of the time operators becomes I I 6 + a/rl = q2T + (G1 G2 q 2 / ~ 1 ) ( G 2 / ~ 1 ) ~ Fig.CHAP. I From Problem 9. 6 a/rl = q2Y+ (Gl i.9 when subjected to the strain history shown in Fig. + q2/r1).e-(t . g d r l = (Gl G2).8. a0(-e-t/7. + + + Further. 9 9. Fig.t f )dt' The relaxation function +. where r.33 = .(t).(r. Gl = O. The result is = [(rl. Let ln r = h so that eX = T and thus drldh = "e T. Like- wise G. may be found with the help of (9. + 9. For A = O.. h(t t') dt'.OIU(t)]. Soiving for o. = A + Bt CeM where A.. (9. r). = [ A h . +3 it2 +.81) may be written H(ln r ) = TG(T) 9.48b) gives si/3 = o d 9 K so that ( 9 .. $ ( t ) in (9. (9.34. B.h)e+it . G1 = G/2. G2 = O.3PK)oo/{18KQ).B * ~ ( A+AB)' + 4BCh ] / 2 ( A+ C).16.h)eTzt]/(rl. and for h = -a. if defines the logarithmic relaxation spectrum.31. 2all + o33 = 3Ke3. = G/2.) $. Thus in operator form.40). 4 8 ~ )for i = j = 1 + yields cll = ( 3 K P Q)ooI{9KQ). Thus finally .28) defining g ( t ) becomes q ( t ) = LW L(ln ~ ) ( le-t")d(ln . 9-28.. Here oii = 3Keii and by the symmetry of the problem. 5 0 ~ )with i = j = 1... A cylindrical viscoelastic body is inserted into a rigid snug-fitting container (Fig. = o. Thus G1 = Gw2~2/(l+~ 2 ~ = 2 ) Ge2X/(l+ e2h) where 1 = l n w ~ . LINEAR VISCOELASTICITY [CHAP. G.48).(t . Gl and G. 9-29) so that = O (no radial strain). Also from ( 9 . 9-29 osr = KE.. From this. or dr = ~ d ( l Tn) . 4 8 ~ )for i = j = 2 yields cz2 = (2Q . 9-28 9. = GeX/(l+ e2X) and for A = O. In the same way.determine a. For the Maxwell model of Fig. for h = I m . G1 = G . From this definition determine the creep function +(t)in terms of L(ln r ) .. C. + G* = G ( w ~ TL~r ) / ( 1 + ~ 2 ~ 2 ) for a Maxwell material. Determine the viscoelastic operator form of the elastic constant V (Poisson's ratio) using the constitutive relations (9. A are constants.32.. for h = m. from these two relations we obtain Fig. as functions of ln o r and sketch the shape of these functions. V = -c2.r. The shape of the curves for these functions is a s shown in Fig.2Q)/{6KP 2Q). The body is elastic in dilatation and has the creep function +. In the same way ( 9 ..413 = -it 2 . 011 . The "logarithmic retardation spectrum" L is defined in terms of the retardation spectrum J by L(ln r ) = r J ( r ) .33. From Problem 9./c11 = (3PK . Under a uniaxial tension o. 9-2(a) determine the storage and loss moduli. If + T h . i=1 j=l Supplementary Problems 9.) so that for the viscoelastic column { E q a t ) ( d 2 ~ l d ~ : ) + + P. the operator leads to the differential equation (Ee + qè)(d2wldx. The "creep buckling" of a viscoelastic column may be analyzed within the linear theory through the correspondence principle. + (G1G2/72)' Fig. Assuming the deflection in a product form w(xl. and from the time equation for N k = ki the solution ei = 2 Aijehilt where N depends upon the degree of the operator. t ) of a Kelvin pinned-end column by this method.37. Fig.erzt)/3r2]/(rl -r .t '- ) .l )Finally t / ~ . .k4W = O and the time equation (3KP Q)(d2eldt2) (k4ZlpA){9KQ)(~) + = 0. and for a Kelvin material E may be + replaced by the operator {E 78.48) the viscoelastic operator for E is {9KQl(3KP Q ) ) . Determine the deflection w(x. 9-31 .~ Z ( d 2 ~ l d x : ) land ~ so + (1 ..48).)el~ = O which integrates easily to yield e = e ( p ~ / ~ ~ . 9-31. W / ~ ~ ( d ~ ~ / d z =.X ) ( I . t) = W(xl)e(t).lP. (r.X ) ( I . The free vibrations of an elastic beam are governed by the equation E Z ( ~ ~ W / ~ X .P.~ ) ~ r l ( t . 9-30 The elastic column formula is d2wldx: + Pow/EZ = 0. ) + + p A ( a 2 ~ l a t 2=) O. Formulate the steady-state vibration problem for a viscoelastic beam assuming' the constitutive relations are those given by (9.Ot 2Z0[-(r1.t ) = Wi(xl)AijeAijt in which the Aij are complex.e~it)/3r.CHAP. Determine the constitutive equation for the four parameter mode1 shown in Fig.~ 2 2 ) I IU(tl)]dt' which upon integration gives 033 + + = (K.and if the deflection w ( x l . ) O] e l ~ where T = 7IE. = .35.t ) = W ( x l ) e ( t ) the resulting viscoelastic differential equation may be split into the space + equation d 4 W l d x ~. The total j=1 m N solution therefore is w ( x l . (r1 . h)eVz(t-tl) rs3 = Kbt + i. then the "creep buckling" deflection is w = We(Po/p~-l)t/T. From (9. 91 LINEAR VISCOELASTICITY 215 t . 8 + (Gilv2 + G2/v2+ Gl/7l). But the elastic buckling load is P. Ans. ) ) [ ~ ( t ) ] 9.wlZ = O.36. The solution W i of the space equation represents the ith mode shape. 9. + ( G l G z / ~ l ~ z=) u G l f . .)+ PoWeIZ = O from which + [l+ P. 44. w ( x l . The initial + shape of the column is w = w o sin ( ~ z l l when ) the load Po [ U ( t ) ]is applied (see Fig.Gle-Czt/(Gl+c2)71/G2(G1 G2) + L"""d '11 + +(t) = G2 Gle-t/Tl Fig.(G(2t. Use equation (9. u = . + i + 1 ~ ~ 7 2 + Fig.) 9.40. (Hint.43. the elastic buckling load.t. determine ~ ~ ~ ~( t~ ) ~.P ~ / P o ) ~ . A viscoelastic block having the constitutive equation 8 ao = P .t ~ ( ' . 9 9. 9. $(t) = 9. q3= O (see Fig. Ans. t ) = w0 sin ( a ~ ~ / l ) e .~ t ] where h = (3aK + y)/(3K + p).) tl + ~ ( -4( 1 + eti'r)e-tlT)) for t l < t < 2tl 9. 9-36 9.2(3aK 3aK -+2y)hy ) e . Deduce the Kelvin and Maxwell stress-strain laws from the results established in Problem 9.5 for the four parameter model of that problem.45.216 LINEAR VISCOELASTICITY [CHAP.42 let G1 = G. €0 Ans. G* = 1 027.42. $ ( t ) = 1/G2 . 9-33.41. 9-34. then +(t) = abmtk-1.. Deter- mine the subsequent deflection w ( x l .7.2 y 2(3aK y)h + + ( 3 K. 9-32 9.= O. G1(l 7220. 9-32. etc. (See Problem 9.) A m . 9-35).k .38. = q3 = q and determine the stress history of the resulting model when it is subjeeted to the strain sequence shown in Fig. Ans. . 9-34 9. Determine the creep response of the standard linear solid by direct integration of E +€ 1 =~ ~ + + oo[U(t)](Gl G2)/G1v2 uo[8(t)]/Gl. Determine G* for the model shown in Fig. = -oo[U(t)]. a2)) + + Ans.. 9-35 Fig. t ) a s a function of P. wo sin (7x11) Po[U(t)l P.P. Ans. In the model of Problem 9..[U(t)l Fig. (Hint.40) to obtain $ ( t ) if +(t) = a(b/t)m with m < 1. = G and q. Determine the creep and relaxation functions for the model shown in Fig.)+ G2a272 + o(G2r2 q3(1 7 .2P ( 23K + P) . o33 = 3aK . Assuming oii = 3Kqi.39. y are constants is loaded under conditions such that o. Take m = 1 . + + ya where a. 9-36). Let G3 = 0 . A pinned-end viscoelastic column is a Maxwell material for which 8 olr = E.) 9. and ( 0~ )~ ~ ( 0 0 ) . 9-33 Fig. INDEX Absolute dynamic, Conjugate dyadic, 4 compliance, 202 Conservation of, modulus, 202 energy, 128 Acceleration, 111 . m a s , 126 Addition and subtraction, Constitutive equations, 132 of matrices, 17 Continuity equation, 126 of (Cartesian) tensors, 1 5 Continuum concept, 44 of vectors, 2 Contraction, 1 6 Adiabatic process, 140 Contravariant tensor, 11 Airy stress function, 147 Convective, Almansi s t r a i n tensor, 82 derivative, 110 Angle change, 86 r a t e of change, 111 Angular momentum, 128 Conventional stress and strain, 175 Anisotropy, 44, 141 Coordinate transformation, 11 Antisymmetric, Correspondence principle, 205 dyadic, 5 Couple-stress vector, 45 matrix, 19 Coupled h e a t equation, 150 tensor, 19 Covariant tensor, 1 2 Axis of elastic symmetry, 142 Creep, function, 200 Barotropic change, 161 test, 199 Base vectors, 6 Cross product, 3, 5, 1 6 Basis, 6, 9 Cubical dilatation, 90 orthonormal, 6 Curvi1ine:ir coordinates, 7 Bauschinger effect, 176 Cylindrical coordinates, 8 Beltrami-Michell equations, 144 Bernoulli equation, 164 Dashpot, 196 Biharmonic equation, 148 Decomposition, Body forces, 45 polar, 87 Boundary conditions, 144, 146 velocity gradient, 112 Bulk, Deformation, 77 modulus, 143 gradients, 80 viscosity, 161 inelastic, 175 plane, 91 Caloric equation of state, 132 plastic, 175 Cartesian, tensors, 81 coordinates, 6 total (deformation) theory, 183 tensors, 1, 12, 1 3 De1 operator, 22 Cauchy, Density, 44, 126 deformation tensor, 8 1 entropy, 130 strain ellipsoids, 89 strain energy, 141 stress principle, 45 Derivative, stress quadric, 50 material, 110, 114 Cauchy-Riemann conditions, 165 of tensors, 22 Circulation, 164 of vectors, 22 Kelvin's theorem of, 165 Deviatoric, Clausius-Duhem inequality, 131 s t r a i n tensor, 91 Column matrix, 17 stress tensor, 57 Compatibility equations, 92, 114 Diagonal matrix, 17 Complex, Dilatation, 90 niodulus, 202 Direction cosines, 7 potential, 166 Displacement, 78, 83 Compliance, 201 gradient, 80 Coniponent, 1, 7 relative, 83 Compressible fluid, 165 Dissipation function, 132 Configuration, 77 Dissipative stress tensor, 131 Conformable niatrices, 17 Distortion energy theory, 178 INDEX Divergence theorem (of Gauss), 23 Forces, Dot product of, body, 45 dyads, 5 surface, 45 vectors, 3 Fourier heat law, 149 Duhamel-Neumann relations, 149 Fundamental metric tensor, 1 3 Dyadics, 1, 4 antisymmetric, 4 Gas, conjugate of, 4 dynamical equation, 165 symmetric, 4 law, 160 Dyads, 4 Gauss's theorem, 23 nonion form, 7 Generalized, Dynamic moduli, 203 Booke's law, 140 Kelvin model, 198 Effective, Maxwell model, 198 plastic strain increment, 182 plane strain, 147 stress, 182 plane stress, 147 Elastic, Gibb's notation, 2 constants, 141 Gradient, limit, 175 deformation, 80 symmetry, 142 displacement, 80 Elasticity, 140 Green's Elastodynamics, 143 deformation tensor, 81 Elastoplastic problems, 183 finite strain tensor, 82 Elastostatics, 143 Energy, Hamilton-Cayley equation, 21 kinetic, 129 Hardening, strain, 141 isotropic, 180 , thermal, 129 kinematic, 181 Engineering stress and strain, 175 strain, 176, 183 Entropy, 130 work, 176, 183 specific, 130 ' Harmonic functions, 166 e - o identity, 39 Heat, Equations of conduction law, 149 equilibrium, 48, 128 flux, 129 motion, 128 radiant, 129 state, 130 Hencky equations, 183 Equivalent, Hereditary integrals, 201 plastic strain increment, 182 Homogeneous, stress, 182 deformation, 95 Euclidean space, 11 material, 44 Eulerian, Hookean solid, 196 coordinates, 78 Hydrostatic pressure, 160 description, 79 Hydrostatics, 163 finite strain tensor, 82 Hyperelastic material, 149 linear strain tensor, 8 3 Hypoelastic material, 149 Hysteresis, 176 Field equations, Ideal, elastic, 143, 147 gas, 160 viscoelastic, 205 materiais, 132 Finite strain tensor, 81, 82 Idealized stress-strain curves, 177 First law of thermodynamics, 129 Idemfactor, 5 Flow, 77, 110 Identity matrix, 18 creeping, 169 Incompressible flow, 126 irrotational, 164 Incremental theories, 181 plastic, 175 Indeterminate vector product, 4 potential, 175 Indices, 9 rule, 181 Indicial notation, 8 steady, 163 Inelastic deformation, 175 Fluid, Inertia forces, 48 inviscid, 160 Infinitesimal strain, 83 perfect, 160,164 Initial conditions, 145 pressure, 160 Inner product, 16 INDEX Integral theorems, 23 Momentum principie, 127 Interna1 energy, 129 Motion, 110 Invariants, 21 steady, 112 of r a t e of deformation, 114 Multiplication of, of strain, 89, 90 matrices, 1 8 of stress, 51 tensors, 15 Irreversible process, 130 vectors, 3 Irrotational flow, 113, 164 Isothermal process, 140 Navier-Cauchy equations, 144 Isotropy, 44, 142 Navier-Stokes equations, 162 Navier-Stokes-Duhem equations, 162 Jacobian, 11, 79 Newtonian viscous fluid, 161, 196 Johnson's apparent elastic-limit, 176 Normal stress components, 47 Kelvin (material) model, 197 Octahedral, Kinematic, plane, 59 hardening, 181 shear stress, 192 viscosity, 163 Orthogonal, Kinetic energy, 129 tensor, 87 Kronecker delta, 1 3 transformation, 1 3 Othogonality conditions, 13, 14 Lagrangian, Orthotropic, 142 description, 79 Outer product, 1 5 finite strain tensor, 82 infinitesimal strain tensor, 83 Parallelogram law of addition, 2 Lamé constants, 143 Particle, 77 Laplace, P a t h lines, 112 equation, 165 Perfectly plastic, 176 transform, 202, 205 Permutation symbol, 16 Left stretch tensor, 87 r-plane, 179 Levy-Mises equations, 181 Plane, Line integrais, 23 deformation, 91 Linear, elasticity, 145 momentum, 127 strain, 91, 145 rotation tensor, 83 stress, 56-57, 145 therinoelasticity, 149 Plastic, vector operator, 8 deformation, 175 viscoelasticity, 196 flow, 175 Local rate of change, 111 potential theory, 181 Logarithmic strain, 175 range, 176 strain increment, 182 Mass, 126 Point, 77 Material, Poisson's ratio, 143 coordinates, 78 Polar, derivative, 110, 114-117 decomposition, 87 description of motion, 79 equilibrium equations, 148 Matrices, 17-19 Position vector, 77 Maximum, Post-yield behavior, 180 normal stress, 52 Potential, shearing stress, 52, 53 Maxwell (material) model, 197 flow, 165 plastic, 182 Metric tensor, 1 3 Mises yield condition, 178 Prandtl-Reuss equations, 181 Modulus, Pressure, bulk, 143 fluid, 160 complex, 203 function, 163 loss, 203 Principal, shear, 143 axes, 20 storage, 202 strain values, 89 Mohr's circles, stress values, 51, 58 f o r strain, 91 Proper transformation, 17 for stress, 54-56 Proportiona1 limit, 177 Moment of momentum, 128 Pure shear, 178 INDEX Quadric of, Stream function, 165 strain, 88, 89 Stream lines, 112 stress, 50 Stress, Quasistatic viscoelastic problem, 205 components, 47 conservative, 131 Range convention, 8 deviator, 57 Rate of deformation tensor, 112 effective, 182 Rate of rotation vector, 114 ellipsoid, 52 Reciproca] basis, 28 function, 147 Rectangular Cartesian coordinates, 6 invariants, 51 Kelative displacement, 83 Mohr's circles for, 54-56 Relaxation, normal, 47 function, 201 plane, 56 spectrum, 201 power, 130 test, 199 principie, 51 Retardation, quadric, 50 spectrum, 201 shear, 47, 52 time, 199 spherical, 57 Reversible process, 130 symmetry, 48 Reynold's transport theorem, 123 tensor, 46 Right stretch tensor, 87 transformation laws, 49 Rigid displacement, 82 vector, 45 Rotation tensor, Stretch, finite, 87 ratio, 86 infinitesimal, 83, 84 tensor, 87 Rotation vector, 83 Summation convention, 8, 10 Superposition theorem, 145 St. Venant's principle, 145 Surface forces, 45 Scalar, 1 Symbolic notation, 2, 10 field, 22 of a dyadic, 4 Symmetry, triple product, 4 elastic, 142 Second law of thermodynamics, 130 tensor, 19 Shear, modulus, 143 Temperature, 130 strain components, 86 Tension test, 175 stress components, 52 Tensor, Slip line theory, 184 Cartesian, 1, 12, 1 3 Small deformation theory, 82 components of, 1 Spatial coordinates, 78 contravariant, 11 Specific, covariant, 12 , entro&, 130 deformation, 81 heat, 149 derivative of, 22 Spherical, fields, 22 coordinates, 8 general, 1, 11 tensor, 57, 91 metric, 12 Spin tensor, 112 multiplication, 1 5 Standard linear solid, 197 powers of, 21 State of stress, 46 rank, 1 Stokes' condition, 161 stretch, 87 Stokes' theorem, 23 transformation laws, 1 Stokesian fluid, 161 Tetrahedron of stress, 47 Strain, Thermal equation of state, 132 deviator, 91 ellipsoid, 88 Thermodynamic process, 130 energy, 141 Thermoelasticity, 133, 149 hardening, 176 Transformation, natural, 112, 175 laws, 13, 88 plane, 91 of tensors, 1 rate, 112 orthogonal, 13, 14 shearing, 86 Tresca yield condition, 178 spherical, 91 Triangle rule, 2 transformation laws, 88 Triple vector product, 4 148 complex. 83 Vorticity. 3. 7 tensor. 3 vector. Velocity. 113 vector. INDEX 221 Tw-o-dimensional elastostatics. 12. hardening. 143 products. 4. 5 Voigt model. 78 dual. 17 transformation law. 4 curve. Vector. 14 order tensor. 166 in rectangular form. 126 Young's modulus. 150 Viscoelastic stress analysis. field. Viscous stress tensor. 46 matrix. 179 position. 77 surface. 112 Uncoupled thermoelasticity. 145. traction. 83 Zero. 1 vorticity. 113 Work. 176 addition. 196 Unit. 160 dyadic. 112 vector. 110. 204-206 Uniqueness theorem. 164 strain. 179 potential. 16 rotation. 177 of a dyadic. triads. 158 Viscoelasticity. 111 in polar form. 145 potential. 2 plastic. 22 condition. 133. 16 Yield. 197 relative displacements. 2 . 182 displacement. 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