continuum notes

DRAFTME185 Introduction to Continuum Mechanics Panayiotis Papadopoulos Department of Mechanical Engineering, University of Calif ornia, Berkeley Copyright c/ci∇cleco√y∇t2008 by Panayiotis Papadopoulos i DRAFTIntroduction This set of notes has been written as part of teaching ME185, a n elective senior-year undergraduate course on continuum mechanics in the Departm ent of Mechanical Engineering at the University of California, Berkeley. Berkeley, California P. P. August 2008 ii DRAFTContents 1 Introduction 1 1.1 Solids and fluids as continuous media . . . . . . . . . . . . . . . . . . . . . . 1 1.2 History of continuum mechanics . . . . . . . . . . . . . . . . . . . . . . . . . 2 2 Mathematical preliminaries 3 2.1 Elements of set theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 2.2 Vector spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 2.3 Points, vectors and tensors in the Euclidean 3-space . . . . . . . . . . . . . . 8 2.4 Vector and tensor calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 3 Kinematics of deformation 19 3.1 Bodies, configurations and motions . . . . . . . . . . . . . . . . . . . . . . . 19 3.2 The deformation gradient and other measures of deformat ion . . . . . . . . . 28 3.3 Velocity gradient and other measures of deformation rat e . . . . . . . . . . . 48 3.4 Superposed rigid-body motions . . . . . . . . . . . . . . . . . . . . . . . . . 53 4 Basic physical principles 61 4.1 The divergence and Stokes’ theorems . . . . . . . . . . . . . . . . . . . . . . 61 4.2 The Reynolds’ transport theorem . . . . . . . . . . . . . . . . . . . . . . . . 63 4.3 The localization theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 4.4 Mass and mass density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 4.5 The principle of mass conservation . . . . . . . . . . . . . . . . . . . . . . . 69 4.6 The principles of linear and angular momentum balance . . . . . . . . . . . . 70 4.7 Stress vector and stress tensor . . . . . . . . . . . . . . . . . . . . . . . . . . 73 iii DRAFT4.8 The transformation of mechanical fields under superpose d rigid-body motions 85 4.9 The Theorem of Mechanical Energy Balance . . . . . . . . . . . . . . . . . . 88 4.10 The principle of energy balance . . . . . . . . . . . . . . . . . . . . . . . . . 91 4.11 The Green-Naghdi-Rivlin theorem . . . . . . . . . . . . . . . . . . . . . . . . 95 5 Infinitesimal deformations 99 5.1 The Gˆ ateaux differential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 5.2 Consistent linearization of kinematic and kinetic vari ables . . . . . . . . . . 101 6 Mechanical constitutive theories 109 6.1 General requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 6.2 Inviscid fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 6.3 Viscous fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 15 6.4 Non-linearly elastic solid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 6.5 Linearly elastic solid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 6.6 Viscoelastic solid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 7 Boundary- and Initial/boundary-value Problems 135 7.1 Incompressible Newtonian viscous fluid . . . . . . . . . . . . . . . . . . . . . 135 7.1.1 Gravity-driven flow down an inclined plane . . . . . . . . . . . . . . . 135 7.1.2 Couette flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 7.1.3 Poiseuille flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 9 7.2 Compressible Newtonian viscous fluids . . . . . . . . . . . . . . . . . . . . . 140 7.2.1 Stokes’ First Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 7.2.2 Stokes’ Second Problem . . . . . . . . . . . . . . . . . . . . . . . . . 1 41 7.3 Linear elastic solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 7.3.1 Simple tension and simple shear . . . . . . . . . . . . . . . . . . . . . 143 7.3.2 Uniform hydrostatic pressure . . . . . . . . . . . . . . . . . . . . . . 144 7.3.3 Saint-Venant torsion of a circular cylinder . . . . . . . . . . . . . . . 144 7.4 Non-linearly elastic solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 7.4.1 Rivlin’s cube . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 7.5 Multiscale problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 iv DRAFT7.5.1 The virial theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 9 7.6 Expansion of the universe . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 Appendix A A.1 A.1 Cylindrical polar coordinate system . . . . . . . . . . . . . . . . . . . . . . . A.1 v vi DRAFTList of Figures 2.1Schematic depiction of a set . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 2.2Example of a set that does not form a linear space . . . . . . . . . . . . . . . 5 2.3Mapping between two sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 3.1A body Band its subset S.. . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 3.2Mapping of a body Bto its configuration at time t.. . . . . . . . . . . . . . . 20 3.3Mapping of a body Bto its reference configuration at time t0and its current configuration at time t.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 3.4Schematic depiction of referential and spatial mappings fo r the velocity v.. . 22 3.5Particle path of a particle which occupies Xin the reference configuration. . 25 3.6Stream line through point xat time t.. . . . . . . . . . . . . . . . . . . . . . 26 3.7Mapping of an infinitesimal material line elements dXfrom the reference to the current configuration. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 3.8Application of the inverse function theorem to the motion χat a fixed time t.30 3.9Interpretation of the right polar decomposition. . . . . . . . . . . . . . . . . . 35 3.10Interpretation of the left polar decomposition. . . . . . . . . . . . . . . . . . . 36 3.11Interpretation of the right polar decomposition relative t o the principal direc- tionsMAand associated principal stretches λA.. . . . . . . . . . . . . . . . 38 3.12Interpretation of the left polar decomposition relative to the principal directions RM iand associated principal stretches λi.. . . . . . . . . . . . . . . . . . . 39 3.13Geometric interpretation of the rotation tensor Rby its action on a vector x.42 3.14Spatially homogeneous deformation of a sphere. . . . . . . . . . . . . . . . . 44 3.15Image of a sphere under homogeneous deformation. . . . . . . . . . . . . . . 46 vii DRAFT3.16Mapping of an infinitesimal material volume element dVto its image dvin the current configuration. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 3.17Mapping of an infinitesimal material surface element dAto its image dain the current configuration. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 3.18Configurations associated with motions χandχ+differing by a superposed rigid motion ¯χ+.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 4.1A surface Abounded by the curve C.. . . . . . . . . . . . . . . . . . . . . . 63 4.2A region Pwith boundary ∂Pand its image P0with boundary ∂P0in the reference configuration. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 4.3A limiting process used to define the mass density ρat a point xin the current configuration. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 4.4Setting for a derivation of Cauchy’s lemma. . . . . . . . . . . . . . . . . . . 73 4.5The Cauchy tetrahedron. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 4.6Interpretation of the Cauchy stress components on an orthog onal parallelepiped aligned with the {ei}-axes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 4.7Projection of the traction to its normal and tangential comp onents. . . . . . 80 6.1Traction acting on a surface of an inviscid fluid. . . . . . . . . . . . . . . . . 111 6.2A ball of ideal fluid in equilibrium under uniform pressure. . . . . . . . . . . 115 6.3Orthogonal transformation of the reference configuration. . . . . . . . . . . . 122 6.4Static continuation Eδ t(s)ofEt(s)byδ.. . . . . . . . . . . . . . . . . . . . . 131 6.5An interpretation of relaxation . . . . . . . . . . . . . . . . . . . . . . . . . . 132 7.1Flow down an inclined plane . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 7.2Couette flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 7.3Semi-infinite domain for Stokes’ Second Problem . . . . . . . . . . . . . . . . 141 7.4Function f(λ)in Rivlin’s cube . . . . . . . . . . . . . . . . . . . . . . . . . . 148 viii DRAFTList of Tables ix DRAFTChapter 1 Introduction 1.1 Solids and fluids as continuous media All matter is inherently discontinuous, as it is comprised o f distinct building blocks, the molecules. Each molecule consists of a finite number of atoms , which in turn consist of finite numbers of nuclei and electrons. Many important physical phenomena involve matter in large l ength and time scales. This is generally the case when matter is considered at lengt h scales much larger than the characteristic length of the atomic spacings and at time scales much larger than the characteristic times of atomic bond vibrations. The preced ing characteristic lengths and times can vary considerably depending on the state of the mat ter (e.g., temperature, precise composition, deformation). However, one may broadly estim ate such characteristic lengths and times to be of the order of up to a few Angstroms (1 ˚A= 10−10m) and a few femtoseconds (1 femtosecond = 10−15sec), respectively. As long as the physical problems of inte rest occur at length and time scales of several orders of magnitude high er that those noted previously, it is possible to consider matter as a continuous medium, nam ely to effectively ignore its discrete nature without introducing any even remotely sign ificant errors. A continuous medium may be conceptually defined as a finite amo unt of matter whose physical properties are independent of its actual size or th e time over which they are mea- sured. As a thought experiment, one may choose to perpetuall y dissect a continuous medium into smaller pieces. No matter how small it gets, its physica l properties remain unaltered. 1 DRAFT2 History of continuum mechanics Mathematical theories developed for continuous media (or “ continua”) are frequently referred to as “phenomenological”, in the sense that they capture the observed physical response without directly accounting for the discrete structure of m atter. Solids and fluids (including both liquids and gases) can be ac curately viewed as continuous media in many occasions. Continuum mechanics is concerned w ith the response of solids and fluids under external loading precisely when they can be v iewed as continuous media. 1.2 History of continuum mechanics Continuum mechanics is a modern discipline that unifies soli d and fluid mechanics, two of the oldest and most widely examined disciplines in applied s cience. It draws on classical scientific developments that go as far back as the Hellenisti c-era work of Archimedes (287- 212 A.C.) on the law of the lever and on hydrostatics. It is mot ivated by the imagination and creativity of da Vinci (1452-1519) and the rigid-body ex periments of Galileo (1564- 1642). It is founded on the laws of motion put forward by Newto n (1643-1727), later set on firm theoretical ground by Euler (1707-1783) and further dev eloped and refined by Cauchy (1789-1857). Continuum mechanics as taught and practiced today emerged i n the latter half of the 20th century. This “renaissance” period can be attributed t o several factors, such as the flourishing of relevant mathematics disciplines (particul arly linear algebra, partial differential equations and differential geometry), the advances in mater ials and mechanical systems tech- nologies, and the increasing availability (especially sin ce the late 1960s) of high-performance computers. A wave of gifted modern-day mechanicians, such a s Rivlin, Truesdell, Ericksen, Naghdi and many others, contributed to the rebirth and conso lidation of classical mechan- ics into this new discipline of continuum mechanics, which e mphasizes generality, rigor and abstraction, yet derives its essential features from the ph ysics of material behavior. ME185 DRAFTChapter 2 Mathematical preliminaries 2.1 Elements of set theory This section summarizes a few elementary notions and definit ions from set theory. A setX is a collection of objects referred to as elements . A set can be defined either by the properties of its elements or by merely identifying all elements. For ex ample, X={1,2,3,4,5} (2.1) or X={all integers greater than 0 and less than 6 }. (2.2) Two sets of particular interest in the remainder of the cours e are: N={all positive integers } (2.3) and R={all real numbers } (2.4) Ifxis an element of the set X, one writes x∈X. If not, one writes x /∈X. LetX,Ybe two sets. The set Xis asubset of the set Y(denoted as X⊆YorY⊇X) if every element of Xis also an element of Y. The set Xis aproper subset of the set (denoted asX⊂YorY⊃X) if every element of Xis also an element of Y, but there exists at least one element of Ythat does not belong to X. 3 DRAFT4 Vector spaces Theunion of sets XandY(denoted by X∪Y) is the set which is comprised of all elements of both sets. The intersection of sets XandY(denoted by X∩Y) is a set which includes only the elements common to the two sets. The empty set (denoted by ∅) is a set that contains no elements and is contained in every set, ther efore, X∪ ∅=X. TheCartesian product X×Yof sets XandYis a set defined as X×Y={(x, y) such that x∈X, y∈Y}. (2.5) Note that the pair ( x, y) in the preceding equation is ordered, i.e., the element ( x, y) is, in general, not the same as the element ( y, x). The notation X2,X3,. . ., is used to respectively denote the Cartesian products X×X,X×X×X,. . .. 2.2 Vector spaces Consider a set Vwhose members (typically called “points”) can be scalars, v ectors or func- tions, visualized in Figure 2.1. Assume that Vis endowed with an addition operation (+) and a scalar multiplication operation ( ·), which do not necessarily coincide with the classical addition and multiplication for real numbers. A "point" that belongs to V V Figure 2.1: Schematic depiction of a set Alinear (orvector )space {V,+;R,·}is defined by the following properties for any u,v,w∈ Vandα, β∈R: (i)α·u+β·v∈ V(closure), (ii) (u+v) +w=u+ (v+w) (associativity with respect to + ), ME185 DRAFTMathematical preliminaries 5 (iii)∃0∈ V | u+0=u(existence of null element), (iv)∃ −u∈ V | u+ (−u) = 0 (existence of negative element), (v)u+v=v+u(commutativity), (vi) (αβ)·u=α·(β·u) (associativity with respect to ·), (vii) ( α+β)·u=α·u+β·u(distributivity with respect to R), (viii) α·(u+v) =α·u+α·v(distributivity with respect to V), (ix) 1 ·u=u(existence of identity). Examples: (1)V=P2:={all second degree polynomials ax2+bx+c}with the standard polynomial addition and scalar multiplication. It can be trivially verified that {P2,+;R,·}is a linear function space. P2is also “equivalent” to an ordered triad ( a, b, c)∈R3. (2) Define V={(x, y)∈R2|x2+y2= 1}with the standard addition and scalar multi- plication for vectors. Notice that given u= (x1, y1) andv= (x2, y2) as in Figure 2.2, xy 1 uvu+v Figure 2.2: Example of a set that does not form a linear space property (i) is violated, i.e., since, in general, for α=β= 1, u+v= (x1+x2, y1+y2), ME185 DRAFT6 Vector spaces and (x1+x2)2+ (y1+y2)2/ne}ationslash= 1. Thus, {V,+;R,·}is not a linear space. /square Consider a linear space {V,+;R,·}and a subset UofV. Then Uforms a linear sub-space ofVwith respect to the same operations (+) and ( ·), if, for any u,v∈ Uandα, β,∈R α·u+β·v∈ U, i.e., closure is maintained within U. Example: (a) Define the set Pnof all algebraic polynomials of degree smaller or equal to n >2 and consider the linear space {Pn,+;R,·}with the usual polynomial addition and scalar multiplication. Then, P2is a linear subspace of {Pn,+;R,·}. /square In order to simplify the notation, in the remainder of these n otes the symbol ·used in scalar multiplication will be omitted. Letv1·v2, ...,vpbe elements of the vector space {V,+;R,·}and assume that α1v1+α2v2+. . .+αpvp= 0⇔α1=α2=...=αp= 0. (2.6) Then, {v1,v2, . . .,vp}is alinearly independent set in V. The vector space {V,+;R,·}is infinite-dimensional if, given any n∈N, it contains at least one linearly independent set withn+ 1 elements. If the above statement is not true, then there is ann∈N, such that all linearly independent sets contain at most nelements. In this case, {V,+;R,·}is afinite dimensional vector space (specifically, n-dimensional). Abasis of an n-dimensional vector space {V,+;R,·}is defined as any set of nlinearly independent vectors. If {g1,g2, ...,gn}form a basis in {V,+;R,·}, then given any non-zero v∈ V, α1g1+α2g2+. . .+αngn+βv= 0⇔not all α1, . . ., α n, βequal zero . (2.7) More specifically, β/ne}ationslash= 0 because otherwise there would be at least one non-zero αi,i= 1, . . ., n , which would have implied that {g1,g2, ...,gn}are not linearly independent. Thus, the non-zero vector vcan be expressed as v=−α1 βg1−α2 βg2−. . .−αn βgn. (2.8) ME185 DRAFTMathematical preliminaries 7 The above representation of vin terms of the basis {g1,g2, ...,gn}is unique. Indeed, if, alternatively, v=γ1g1+γ2g2+. . .+γngn, (2.9) then, upon subtracting the preceding two equations from one another, it follows that 0=/parenleftbigg γ1+α1 β/parenrightbigg g1+/parenleftbigg γ2+α2 β/parenrightbigg g2+. . .+/parenleftbigg γn+αn β/parenrightbigg gn, (2.10) which implies that γi=−αi β,i= 1,2, . . ., n , since {g1,g2, ...,gn}are assumed to be a linearly independent. Of all the vector spaces, attention will be focused here on th e particular class of Euclidean vector spaces in which a vector multiplication operation ( ·) is defined, such that for any u,v,w∈ Vandα∈R, (x)u·v=v·u(commutativity with respect to ·), (xi)u·(v+w) =u·v+u·w(distributivity), (xii) ( αu)·v=u·(αv) =α(u·v) (associativity with respect to ·) (xiii)u·u≥0 andu·u= 0⇔u= 0. This vector operation is referred to as the dot-product . An n-dimensional vector space obeying the above additional rules is referred to as a Euclidean vector space and is denoted byEn. Example: The standard dot-product between vectors in Rnsatisfies the above properties. /square The dot-product provide a natural means for defining the magnitude of a vector as /ba∇dblu/ba∇dbl= (u·u)1/2. (2.11) Two vectors u,v∈Enareorthogonal , ifu·v= 0. A set of vectors {u1,u2, ...}is called orthonormal , if ui·uj=/braceleftBigg 0 ifi/ne}ationslash=j 1 ifi=j=δij, (2.12) where δijis called the Kronecker delta symbol. ME185 DRAFT8 Points, vectors and tensors in the Euclidean 3-space Every orthonormal set {e1,e2...ek},k≤ninEnis linearly independent. This is because, if α1e1+α2e2+. . .+αkek=0, (2.13) then, upon taking the dot-product of the above equation with ei,i= 1,2, . . ., k , and invoking the orthonormality of {e1,e2...ek}, α1(e1·ei) +α2(e2·ei) +. . .+αk(ek·ei) =αi= 0. (2.14) Of particular importance to the forthcoming developments i s the observation that any vector v∈Encan be resolved to an orthonormal basis {e1,e2, ...,en}as v=v1e1+v2e2+. . .+vnen=n/summationdisplay i=1viei, (2.15) where vi=v·ei. In this case, videnotes the i-thcomponent ofvrelative to the orthonormal basis{e1,e2, ...,en}. 2.3 Points, vectors and tensors in the Euclidean 3- space Consider the Cartesian space E3with an orthonormal basis {e1,e2,e3}. As argued in the previous section, a typical vector vǫE3can be written as v=3/summationdisplay i=1viei;vi=v·ei. (2.16) Next, consider points x, yin the Euclidean point space E3, which is the set of all points in the ambient three-dimensional space, when taken to be devoid of the mathematical structure of vector spaces. Also, consider an arbitrary origin (or reference point) Oin the same space. It is now possible to define vectors x,y∈E3, which originate at Oand end at points xand y, respectively. In this way, one makes a unique association ( to within the specification of O) between points in E3and vectors in E3. Further, it is possible to define a measure of distance between xandy, by way of the magnitude of the vector v=y−x, namely d(x, y) =|x−y|= [(x−y)·(x−y)]1/2. (2.17) ME185 DRAFTMathematical preliminaries 9 InE3, one may define the vector product of two vectors as an operation with the following properties: for any vectors u,vandw, and any scalar α, (a)u×v=−v×u, (b) (u×v)·w= (v×w)·u= (w×u)·v, or, equivalently [ uvw] = [vwu] = [wuv], where [ uvw] = (u×v)·wis the triple product of vectors u,v, andw, (c)|u×v|=|u||v|sinθ , cosθ=u·v (u·u)1/2(v·v)1/2,0≤θ≤π. Appealing to property (a), it is readily concluded that u×u=0. Likewise, properties (a) and (b) can be used to deduce that ( u×v)·u= (u×v)·v= 0, namely that the vector u×vis orthogonal to both uandv. By definition, for a right-hand orthonormal coordinate basis {e1,e2,e3}, the following relations hold true: e1×e2=e3,e2×e3=e1,e3×e1=e2. (2.18) These relations, together with the implications of propert y (a) e1×e1=e2×e2=e3×e3=0 (2.19) and e2×e1=−e3,e3×e2=−e1,e1×e3=−e2 (2.20) can be expressed compactly as ei×ej=3/summationdisplay k=1ǫijkek, (2.21) where ǫijkis the permutation symbol defined as ǫijk=  1 if ( i, j, k) = (1,2,3), (2,3,1), or (3,1,2) −1 if ( i, j, k) = (2,1,3), (3,2,1), or (1,3,2) 0 otherwise. (2.22) It follows that u×v= (3/summationdisplay i=1uiei)×(3/summationdisplay j=1vjej) =3/summationdisplay i=13/summationdisplay j=1uivjei×ej=3/summationdisplay i=13/summationdisplay j=13/summationdisplay k=1uivjeijkek.(2.23) ME185 DRAFT10 Points, vectors and tensors in the Euclidean 3-space LetU,Vbe two sets and define a mapping ffromUtoVas a rule that assigns to each point u∈ Ua unique point v=f(u)∈ V, see Figure 2.3. The usual notation for a mapping is:f:U → V , u→v=f(u)∈ V. With reference to the above setting, Uis called the domain off, whereas Vis termed the range off. uvf U V Figure 2.3: Mapping between two sets A mapping T:E3→E3is called linear if it satisfies the property T(αu+βv) =αT(u) +βT(v), (2.24) for all u,v∈E3andα, β∈N. A linear mapping is also referred to as a tensor . Examples: (1)T:E3→E3,T(v) =vfor all v∈E3. This is called the identity tensor, and is typically denoted by T=I. (2)T:E3→E3,T(v) =0for all v∈E3. This is called the zerotensor, and is typically denoted by T=0. /square Thetensor product between two vectors vandwinE3is denoted by v⊗wand defined according to the relation (v⊗w)u= (w·u)v, (2.25) for any vector u∈E3. This implies that, under the action of the tensor product v⊗w, the vector uis mapped to the vector ( w·u)v. It can be easily verified that v⊗wis a tensor according to the previously furnished definition. Using the Cartesian components of vectors, ME185 DRAFTMathematical preliminaries 11 one may express the tensor product of vandwas v⊗w= (3/summationdisplay i=1viei)⊗(3/summationdisplay j=1wjej) =3/summationdisplay i=13/summationdisplay j=1viwjei⊗ej. (2.26) It will be shown shortly that the set of nine tensor products {ei⊗ej},i, j= 1,2,3, form a basis for the space L(E3, E3) of all tensors on E3. Before proceeding further with the discussion of tensors, i t is expedient to introduce a summation convention, which will greatly simplify the comp onent representation of both vectorial and tensorial quantities and their algebra. This originates with A. Einstein, who employed it first in his relativity work. The summation conve ntion has three rules, which, when adapted to the special case of E3, are as follows: Rule 1. If an index appears twice in a single component term or in a pro duct term, the summation sign is omitted and summation is automatically as sumed from value 1 to 3. Such an index is referred to as dummy . Rule 2. An index which appears once in a single component or in a produ ct expression is not summed and is assumed to attain a single value (1, 2, or 3). Such an index is referred to free. Rule 3. No index can appear more than twice in a single component or in a product term. Examples: 1. The vector representation u=3/summationdisplay i=1uieiis replaced by u=uieiand it involves the summation of three terms. 2. The tensor product u⊗v=3/summationdisplay i=13/summationdisplay j=1uivjei⊗ejis equivalently written as u⊗v= uivjei⊗ejand it involves the summation of nine terms. 3. The term uivjis a single term with two free indices iandj. 4. It is easy to see that δijui=δ1ju1+δ2ju2+δ3ju3=uj. This index substitution property is frequently used in component manipulations. ME185 DRAFT12 Points, vectors and tensors in the Euclidean 3-space 5. A similar index substitution property applies in the case of a two-index quantity, namely δijaik=δ1ja1k+δ2ja2k+δ3ja3k=ajk. /square With the summation convention in place, take a tensor T∈ L(E3, E3) and define its components Tij, such that Tej=Tijei. It follows that (T−Tijei⊗ej)v= (T−Tijei⊗ej)vkek =Tekvk−Tijvk(ei⊗ej)ek =Tikeivk−Tijvk(ej·ek)ei =Tikeivk−Tijvkδjkei =Tikeivk−Tikvkei =0, (2.27) hence, T=Tijei⊗ej. (2.28) This derivation demonstates that any tensor Tcan be written as a linear combination of the nine tensor product terms {ei⊗ej}. The components of the tensor Trelative to {ei⊗ej} can be put in matrix form as [Tij] = T11T12T13 T21T22T23 T31T32T33 . (2.29) Thetranspose TTof a tensor Tis defined by the property u·Tv=v·TTu, (2.30) for any vectors uandvinE3. Using components, this implies that uiTijvj=viAijuj=vjAjiui, (2.31) where Aijare the components of TT. It follows that ui(Tij−Aji)vj= 0. (2.32) ME185 DRAFTMathematical preliminaries 13 Since uiandvjare arbitrary, this implies that Aij=Tji, hence the transpose of Tcan be written as TT=Tijej⊗ei. (2.33) A tensor Tissymmetric ifTT=Tor, when both TandTTare resolved relative to the same basis, Tji=Tij. Likewise, a tensor Tisskew-symmetric ifTT=−Tor, again, upon resolving both on the same basis, Tji=−Tij. Note that, in this case, T11=T22=T33= 0. Given tensors T,S∈ L(E3, E3), the multiplication TSis defined according to (TS)v=T(Sv), (2.34) for any v∈E3. In component form, this implies that (TS)v=T(Sv) =T[(Sijei⊗ej)(vkek)] =T(Sijvkδjkei) =T(Sijvjei) =TkiSijvjei = (TkiSijek⊗ej)(vlel), (2.35) which leads to (TS) =TkiSijek⊗ej. (2.36) ThetracetrT:L(E3, E3)/ma√sto→Rof the tensor product of two vectors u⊗vis defined as tru⊗v=u·v, (2.37) hence, the trace of a tensor Tis deduced from equation (2.37) as trT=tr(Tijei⊗ej) =Tijei·ej=Tijδij=Tii. (2.38) Thecontraction (orinner product )T·S:L(E3, E3)× L(E3, E3)/ma√sto→Rof two tensors Tand Sis defined as T·S= tr(TST). (2.39) Using components, tr(TST) = tr( TkiSjiek⊗ej) =TkiSjiek·ej=TkiSjiδkj=TkiSki. (2.40) ME185 DRAFT14 Vector and tensor calculus A tensor Tisinvertible if, for any w∈E3, the equation Tv=w (2.41) can be uniquely solved for v. Then, one writes v=T−1w, (2.42) andT−1is the inverse ofT. Clearly, if T−1exists, then T−1w−v=0 =T−1(Tv)−v = (T−1T)v−v = (T−1T−I)v, (2.43) henceT−1T=Iand, similarly, TT−1=I. A tensor Tisorthogonal if TT=T−1, (2.44) which implies that TTT=TTT=I. (2.45) It can be shown that for any tensors T,S∈ L(E3, E3), (S+T)T=ST+TT,(ST)T=TTST. (2.46) If, further, the tensors TandSare invertible, then (ST)−1=T−1S−1. (2.47) 2.4 Vector and tensor calculus Define scalar, vector and tensor functions of a vector variab lexand a real variable t. The scalar functions are of the form φ1:R→R, t→φ=φ1(t) φ2:E3→R,x→φ=φ2(x) (2.48) φ3:E3×R→R,(x, t)→φ=φ3(x, t), ME185 DRAFTMathematical preliminaries 15 while the vector and tensor functions are of the form v1:R→E3, t→v=v1(t) v2:E3→E3,x→v=v2(x) (2.49) v3:E3×R→E3,(x, t)→v=v3(x, t) and T1:R→ L(E3, E3), t→T=T1(t) T2:E3→ L(E3, E3),x→T=T2(x) (2.50) T3:E3×R→ L(E3, E3),(x, t)→T=T3(x, t), respectively. Thegradient gradφ(x) (otherwise denoted as ∇φ(x) or∂φ(x) ∂x) of a scalar function φ=φ(x) is a vector defined by (gradφ(x))·v=/bracketleftbiggd dwφ(x+wv)/bracketrightbigg w=0, (2.51) for any v∈E3. Using the chain rule, the right-hand side of equation (2.51 ) becomes /bracketleftbiggd dwφ(x+wv)/bracketrightbigg w=0=/bracketleftbigg∂φ(x+wv) ∂(xi+wvi)d(xi+wvi) dw/bracketrightbigg w=0=∂φ(x) ∂xivi. (2.52) Hence, in component form one may write gradφ(x) =∂φ(x) ∂xiei. (2.53) In operator form, this leads to the expression grad = ∇=∂ ∂xiei. (2.54) Example: Consider the scalar function φ(x) =|x|2=x·x. Its gradient is gradφ=∂ ∂x(x·x) =∂(xjxj) ∂xiei=/parenleftbigg∂xj ∂xixj+xj∂xj ∂xi/parenrightbigg ei = (δijxj+xjδij)ei= 2xiei= 2x.(2.55) ME185 DRAFT16 Vector and tensor calculus Alternatively, using directly the definition (gradφ)·v=/bracketleftbiggd dw{(x+wv)·(x+uv)}/bracketrightbigg w=0 =/bracketleftbiggd dw{x·x+ 2wx·v+w2v·v}/bracketrightbigg w=0 = [2x·v+ 2wv·v]w=0 = 2x·v (2.56) /square Thegradient gradv(x) (otherwise denoted as ∇v(x) or∂v(x) ∂x) of a vector function v=v(x) is a tensor defined by the relation (gradv(x))w=/bracketleftbiggd dwv(x+ww)/bracketrightbigg w=0, (2.57) for any w∈E3. Again, using chain rule, the right-hand side of equation (2 .57) becomes /bracketleftbiggd dwv(x+ww)/bracketrightbigg w=0=/bracketleftbigg∂vi(x+ww) ∂(xj+wwj)d(xj+wwj) dw/bracketrightbigg w=0ei=∂vi(x) ∂xjwjei,(2.58) hence, using components, grad(v(x) =∂vi(x) ∂xjei⊗ej. (2.59) In operator form, this leads to the expression grad = ∇=∂ ∂xi⊗ej. (2.60) Example: Consider the function v(x) =αx. Its gradient is gradv=∂(αx) ∂x=∂(αxi) ∂xjei⊗ej=αδijei⊗ej=αei⊗ei=αI, (2.61) since (ei⊗ei)v= (v·ei)ei=viei=v. Alternatively, using directly the definition, (gradv)w=/bracketleftbiggd dw(v+ww)/bracketrightbigg w=0=αw, (2.62) hence grad v=αI. Thedivergence divv(x) (otherwise denoted as ∇ ·v(x)) of a vector function v=v(x) is a scalar defined as divv(x) = tr(grad v(x)), (2.63) ME185 DRAFTMathematical preliminaries 17 on, using components, divv(x) = tr/parenleftbigg∂vi ∂xjei⊗ej/parenrightbigg =∂vi ∂xjei·ej=∂vi ∂xjδij=∂vi ∂xi=vi,i. (2.64) In operator form, one writes div = ∇·=∂ ∂xi·ei. (2.65) Example: Consider again the function v(x) =αx. Its divergence is divv(x) =∂(αxi) ∂xi=α∂xi ∂xi=αδii= 3α . (2.66) /square Thedivergence divT(x) (otherwise denoted as ∇ ·T(x)) of a tensor function T=T(x) is a vector defined by the property that (divT(x))·c= div/parenleftbig (TT(x))c/parenrightbig , (2.67) for any constant vector c∈E3. Using components, divT) = div( TTc) = div[( Tijej⊗ei)(ckek)] = div[ Tijckδikej] = div[ Tijciej] = tr/bracketleftbigg∂Tijci ∂xkej⊗ek/bracketrightbigg =∂(Tijci) ∂xkδjk =∂(Tijci) ∂xj =∂Tij ∂xjci, (2.68) hence, divT=∂Tij ∂xjei. (2.69) ME185 DRAFT18 Vector and tensor calculus In the case of the divergence of a tensor, the operator form be comes div = ∇·=∂ ∂xiei. (2.70) Finally, the curlcurlv(x) of a vector function v(x) is a vector defined as curlv(x) =∇ ×v(x), (2.71) which translates using components to curlv(x) =/parenleftbigg∂ ∂xiei/parenrightbigg ×(vjej) =∂vj ∂xiei×ej=∂vj ∂xieijkek=eijk∂vk ∂xjei. (2.72) In operator form, the curl is expressed as curl = ∇×=∂ ∂xiei×. (2.73) ME185 DRAFTChapter 3 Kinematics of deformation 3.1 Bodies, configurations and motions Let a continuum body Bbe defined as a collection of material particles, which, when con- sidered together, endow the body with local (pointwise) phy sical properties which are inde- pendent of its actual size or the time over which they are meas ured. Also, let a typical such particle be denoted by P, while an arbitrary subset of Bbe denoted by S, see Figure 3.1. S BP Figure 3.1: A body Band its subset S. Letxbe the point in E3occupied by a particle Pof the body Bat time t, and let xbe its associated position vector relative to the origin Oof an orthonormal basis in the vector space E3. Then, define by ¯χ: (P, t)∈ B × R/ma√sto→E3themotion ofB, which is a differentiable mapping, such that x=¯χ(P, t) = ¯χt(P). (3.1) In the above, ¯χt:B /ma√sto→ E3is called the configuration mapping ofBat time t. Given ¯χ, 19 DRAFT20 Bodies, configurations and motions the body Bmay be mapped to its configuration R=¯χt(B, t) with boundary ∂Rat time t. Likewise, any part S ⊂ B can be mapped to its configuration P=¯χt(S, t) with boundary ∂Pat time t, see Figure 3.2. Clearly, RandPare point sets in E3. S BP P R∂P∂R x¯χ Figure 3.2: Mapping of a body Bto its configuration at time t. The configuration mapping ¯χtis assumed to be invertible, which means that, given any pointx∈ P, P=¯χ−1 t(x). (3.2) The motion ¯χof the body is assumed to be twice-differentiable in time. The n, one may define the velocity and acceleration of any particle Pat time taccording to v=∂¯χ(P, t) ∂t,a=∂2¯χ(P, t) ∂t2. (3.3) The mapping ¯χrepresents the material description of the body motion. This is because the domain of ¯χconsists of the totality of material particles in the body, a s well as time. This description, although mathematically proper, is of limite d practical use, because there is no direct quantitative way of tracking particles of the body. F or this reason, two alternative descriptions of the body motion are introduced below. Of all configurations in time, select one, say R0=¯χ(B, t0) at a time t=t0, and refer to it as the reference configuration . The choice of reference configuration may be arbitrary, although in many practical problems it is guided by the need f or mathematical simplicity. Now, denote the point which Poccupies at time t0asXand let this point be associated with position vector X, namely X=¯χ(P, t0) = ¯χt0(P). (3.4) ME185 DRAFTKinematics of deformation 21 Thus, one may write x=¯χ(P, t) = ¯χ(χ−1 t0(X), t) =χ(X, t). (3.5) The mapping χ:E3×R/ma√sto→E3, where x=χ(X, t) =χt(X) (3.6) represents the referential orLagrangian description of the body motion. In such a descrip- tion, it is implicit that a reference configuration is provid ed. The mapping χtis theplacement of the body relative to its reference configuration, see Figu re 3.3. BP R R0xX¯χt0¯χt χt Figure 3.3: Mapping of a body Bto its reference configuration at time t0and its current configuration at time t. Assume now that the motion of the body Bis described with reference to the configuration R0defined at time t=t0and let the configuration of Bat time tbe termed the current configuration . Also, let {E1,E2,E3}and{e1,e2,e3}be fixed right-hand orthonormal bases associated with the reference and current configuration, re spectively. With reference to the preceding bases, one may write the position vectors Xandxcorresponding to the points occupied by the particle Pat times t0andtas X=XAEA ,x=xiei, (3.7) respectively. Hence, resolving all relevant vectors to the ir respective bases, the motion χ may be expressed using components as xiei=χi(XAEA, t)ei, (3.8) ME185 DRAFT22 Bodies, configurations and motions or, in pure component form, xi=χi(XA, t). (3.9) The velocity and acceleration vectors, expressed in the ref erential description, take the form v=∂χ(X, t) ∂t,a=∂2χ(X, t) ∂t2, (3.10) respectively. Likewise, using the orthonormal bases, v=vi(XA, t)ei,a=ai(XA, t)ei. (3.11) Scalar, vector and tensor functions can be alternatively ex pressed using the spatial or Eulerian description , where the independent variables are the current position v ectorxand timet. Indeed, starting, by way of an example, with a scalar functi onf=ˇf(P, t), one may write f=ˇf(P, t) = ˇf(χ−1 t(x), t) = ˜f(x, t). (3.12) In analogous fashion, one may write f=˜f(x, t) = ˜f(χt(X), t) = ˆf(X, t). (3.13) The above two equations can be combined to write f=ˇf(P, t) = ˜f(x, t) = ˆf(X, t). (3.14) (X, t) (x, t) ˆv˜v vχ Figure 3.4: Schematic depiction of referential and spatial mappings fo r the velocity v. Any function (not necessarily scalar) of space and time can b e written equivalently in material, referential or spatial form. Focusing specifical ly on the referential and spatial ME185 DRAFTKinematics of deformation 23 descriptions, it is easily seen that the velocity and accele ration vectors can be equivalently expressed as v=ˆv(X, t) =˜v(x, t),a=ˆa(X, t) =˜a(x, t), (3.15) respectively, see Figure 3.4. In component form, one may wri te v= ˆvi(XA, t)ei= ˜vi(xj, t)ei,a= ˆai(XA, t)ei= ˜ai(xj, t)ei. (3.16) Given a function f=ˆf(X, t), define the material time derivative offas ˙f=∂ˆf(X, t) ∂t. (3.17) From the above definition it is clear that the material time de rivative of a function is the rate of change of the function when keeping the referential p osition X(therefore also the particle Passociated with this position) fixed. If, alternatively, fis expressed in spatial form, i.e., f=˜f(x, t), then ˙f=∂˜f(x, t) ∂t+∂˜f(x, t) ∂x·∂χ(X, t) ∂t =∂˜f(x, t) ∂t+∂˜f(x, t) ∂x·v =∂˜f(x, t) ∂t+ grad ˜f·v. (3.18) The first term on the right-hand side of (3.18) is the spatial time derivative offand corre- sponds to the rate of change of ffor afixedpointxin space. The second term is called the convective rate of change offand is due to the spatial variation of fand its effect on the material time derivative as the material particle which occ upies the point xat time ttends to travel away from xwith velocity v. A similar expression for the material time derivative applies for vector functions. Indeed, take, for example, th e velocity v=˜v(x, t) and write ˙v=∂˜v(x, t) ∂t+∂˜v(x, t) ∂x∂χ(X, t) ∂t =∂˜v(x, t) ∂t+∂˜v(x, t) ∂xv =∂˜v(x, t) ∂t+ (grad ˜v)v. (3.19) ME185 DRAFT24 Bodies, configurations and motions A volume/surface/curve which consists of the same material points in all configurations is termed material . By way of example, consider a surface in three dimensions, w hich can be expressed in the form F(X) = 0. This is clearly a material surface, because it contains the same material particles at all times, since its referent ial representation is independent of time. On the other hand, a surface described by the equatio nF(X, t) = 0 is generally non material, because the locus of its points contains differ ent material particles at different times. This distinction becomes less apparent when a surfac e is defined in spatial form, i.e., by an equation f(x, t) = 0. In this case, one may employ Lagrange’s criterion of materiality , which states the following: Lagrange’s criterion of materiality (1783) A surface described by the equation f(x, t) = 0 is material if, and only if, ˙f= 0. A sketch of the proof is as follows: if the surface is assumed m aterial, then f(x, t) =F(X) = 0 , (3.20) hence ˙f(χ, t) = ˙F(X) = 0 . (3.21) Conversely, if the criterion holds, then ˙f(χ, t) = ˙f(χ(X, t), t) =∂F ∂t(X, t) = 0 , (3.22) which implies that F=F(X), hence the surface is material. A similar analysis applies to curves in Euclidean 3-space. S pecifically, a material curve can be be viewed as the intersection of two material surfaces , sayF(X) = 0 and G(X) = 0. Switching to the spatial description and expressing these s urfaces as F(X) =F(χ−1 t(x)) = f(x, t) = 0 (3.23) and G(X) =G(χ−1 t(x)) = g(x, t) = 0 , (3.24) it follows from Lagrange’s criterion that a curve is materia l if˙f= ˙g= 0. It is easy to show that this is a sufficient, but not a necessary condition for the materiality of a curve. ME185 DRAFTKinematics of deformation 25 Some important definitions regarding the nature of the motio nχfollow. First, a motion χ issteady at a point x, if the velocity at that point is independent of time. If this is the case at all points, then v=v(x) and the motion is called steady . If a motion is not steady, then it is called unsteady . A point xin space where v(x, t) =0at all times is called a stagnation point. Letχbe the motion of body Band fix a particle P, which occupies a point Xin the reference configuration. Subsequently, trace its successi ve placements as a function of time, i.e., fix Xand consider the one-parameter family of placements x=χ(X, t),(Xfixed) . (3.25) The resulting parametric equations (with parameter t) represent in algebraic form the particle pathof the given particle, see Figure 3.5. Alternatively, one ma y express the same particle path in differential form as dx=ˆv(X, τ)dτ , x(t0) =X,(Xfixed) , (3.26) or, equivalently, dy=˜v(y, τ)dτ , y(t) =x,(xfixed) . (3.27) X xv R0 Figure 3.5: Particle path of a particle which occupies Xin the reference configuration. Now, let v=˜v(x, t) be the velocity field at a given time t. Define a stream line through x at time tas the space curve that passes through xand is tangent to the velocity field vat all of its points, i.e., it is defined according to dy=˜v(y, t)dτ , y(τ0) =x,(tfixed) , (3.28) ME185 DRAFT26 Bodies, configurations and motions where τis a scalar parameter and τ0some arbitrarily chosen value of τ, see Figure 3.6. Using components, the preceding definition becomes dy1 ˜v1(yj, t)=dy2 ˜v2(yj, t)=dy3 ˜v3(yj, t)=dτ , y i(τ0) =xi,(tfixed) . (3.29) xy˜v(y, t)dy Figure 3.6: Stream line through point xat time t. Thestreak line through a point xat time tis defined by the equation y=χ(χ−1 τ(x), t), (3.30) where τis a scalar parameter. In differential form, this line can be e xpressed as dy=˜v(y, t)dt , y(τ) =x. (3.31) It is easy to show that the streak line through xat time tis the locus of placements at time t of all particles that have passed or will pass through x. Equation (3.31) can be derived from (3.30) by noting that ˜v(y, t) is the velocity at time τof a particle which at time τoccupies the point x, while at time tit occupies the point y. Note that given a point xat time t, then the path of the particle occupying xattand the stream line through xatthave a common tangent. Indeed, this is equivalent to stating that the velocity at time tof the material point associated with Xhas the same direction with the velocity of the point that occupies x=χ(X, t). In the case of steady motion, the particle path for any partic le occupying xcoincides with the stream line and streak line through xat time t. To argue this property, take a stream line (which is now a fixed curve, since the motion is steady). C onsider a material point P which is on the curve at time t. Notice that the velocity of Pis always tangent to the stream line, this its path line coincides with the stream line throu ghx. A similar argument can be made for streak lines. ME185 DRAFTKinematics of deformation 27 In general, path lines can intersect, since intersection po ints merely mean that different particles can occupy the same position at different times. Ho wever, stream lines do not inter- sect, except at points where the velocity vanishes, otherwi se the velocity at an intersection point would have two different directions. Example: Consider a motion χ, such that χ1=χ1(XA, t) =X1et χ2=χ2(XA, t) =X2+tX3 (3.32) χ3=χ3(XA, t) =−tX2+X3, with reference to fixed orthonormal system {ei}. Note that x=Xat time t= 0, i.e., the body occupies the reference configuration at time t= 0. The inverse mapping χ−1 tis easily obtained as X1=χ−1 t1(xj) =x1e−t X2=χ−1 t2(xj) =x2−tx3 1 +t2(3.33) X3=χ−1 t3(xj) =tx2+x3 1 +t2. The velocity field, written in the referential description h as components ˆ vi(XA, t) = ∂χi(XA, t) ∂t, i.e., ˆv1(XA, t) =X1et ˆv2(XA, t) =X3 (3.34) ˆv3(XA, t) =−X2, while in the spatial description has components ˜ vi(χj, t) given by ˜v1(χj, t) = (x1e−t)et=x1 ˜v2(χj, t) =tx2+x3 1 +t2(3.35) ˜v3(χj, t) =−x2−tx3 1 +t2. Note that x=0is a stagnation point and, also, that the motion is steady on t hex1-axis. ME185 DRAFT28 Deformation gradient and other measures of deformation The acceleration in the referential description has compon ents ˆai(XA, t) =∂2χi(XA, t) ∂t2, hence, ˆa1(XA, t) =X1et ˆa2(XA, t) = 0 (3.36) ˆa3(XA, t) = 0 , while in the spatial description the components ˜ ai(χj, t) are given by ˜a1(xj, t) =x1 ˜a2(xj, t) = 0 (3.37) ˜a3(xj, t) = 0 . /square 3.2 The deformation gradient and other measures of deformation Consider a body Bwhich occupies its reference configuration R0at time t0and the current configuration Rat time t. Also, let {EA}and{ei}be two fixed right-hand orthonormal bases associated with the reference and current configurati on, respectively. Recall that the motion χis defined so that x=χ(X, t) and consider the deformation of an infinitesimal material line element dXlocated at the point Xof the reference configuration. This material element is mapped into another infinitesimal l ine element dxat point xin the current configuration at time t, see Figure 3.7. It follows from chain rule that dx=/bracketleftbigg∂χ ∂X/bracketrightbigg (X, t)dX=FdX, (3.38) where Fis the deformation gradient tensor, defined as F=∂χ(X, t) ∂X. (3.39) ME185 DRAFTKinematics of deformation 29 xdx XdX R R0 Figure 3.7: Mapping of an infinitesimal material line elements dXfrom the reference to the current configuration. Using components, the deformation gradient tensor can be ex pressed as ∂χi(XB, t) ∂XAei⊗EA=FiAei⊗EA. (3.40) It is clear from the above that the deformation gradient is a two-point tensor which has one “leg” in the reference configuration and the other in the curr ent configuration. It follows from (3.38) that the deformation gradient Fprovides the rule by which infinitesimal line element are mapped from the reference to the current configur ation. Using (3.40), one may rewrite (3.38) in component form as dxi=FiAdXA=χi,AdXA. (3.41) Recall now that the motion χis assumed invertible for fixed t. Also, recall the inverse function theorem of real analysis, which, in the case of the m apping χcan be stated as follows: For a fixed time t, letχt:R0→ R be continuously differentiable (i.e.,∂χt ∂Xexists and is continuous) and consider X∈ R 0, such that det∂χt ∂X(X)/ne}ationslash= 0. Then, there is an open neighborhood P0ofXinR0and an open neighborhood PofR, such that χt(P0) =P andχthas a continuously differentiable inverse χ−1 t, so that χ−1 t(P) =P0, as in Figure 3.8. Moreover, for any x∈ P,X=χ−1 t(x) and∂χ−1 t(x) ∂x= (F(X, t))−11. 1This means that the derivative of the inverse motion with res pect to xis identical to the inverse of the gradient of the motion with respect to X. ME185 DRAFT30 Deformation gradient and other measures of deformation x X R R0P P0χ Figure 3.8: Application of the inverse function theorem to the motion χat a fixed time t. The inverse function theorem states that the mapping χis invertible at a point Xfor a fixed time t, if the Jacobian determinant J= det∂χ(X, t) ∂X= detFsatisfies the condition J/ne}ationslash= 0. The condition det J/ne}ationslash= 0 will be later amended to be det J >0. Generally, the infinitesimal material line element dXstretches and rotates to dxunder the action of F. To explore this, write dX=MdSanddx=mds, where Mandmare unit vectors (i.e., M·M=m·m= 1) in the direction of dXanddx, respectively and dS > 0 andds >0 are the infinitesimal lengths of dXanddx, respectively. Next, define the stretch λof the infinitesimal material line element dXas λ=ds dS, (3.42) and note that dx=FdX=FMdS =mds , (3.43) hence λm=FM. (3.44) Since det F/ne}ationslash= 0, it follows that λ/ne}ationslash= 0 and, in particular, that λ >0, given that mis chosen to reflect the sense of x. In summary, the preceding arguments imply that λ∈(0,∞). To determine the value of λ, take the dot-product of each side of (3.44) with itself, whi ch ME185 DRAFTKinematics of deformation 31 leads to (λm)·(λm) =λ2(m·m) =λ2= (FM)·(FM) =M·FT(FM) =M·(FTF)M =M·CM, (3.45) where Cis the right Cauchy-Green deformation tensor , defined as C=FTF (3.46) or, upon using components, CAB=FiAFiB. (3.47) Note that, in general C=C(X, t), since F=F(X, t). Also, it is important to observe thatCis defined with respect to the basis in the reference configura tion. Further, Cis a symmetric and positive-definite tensor. The latter means th at,M·CM>0 for any M/ne}ationslash=0 andM·CM= 0, if, and only if M=0, both of which follow clear from (3.45). To summarize the meaning of C, it can be said that, given a direction Min the reference configuration, C allows the determination of the stretch λof an infinitesimal material line element dXalong Mwhen mapped to the line element dxin the current configuration. Recall that the inverse function theorem implies that, sinc e detF/ne}ationslash= 0, dX=∂χ−1 t(x) ∂xdx=F−1dx. (3.48) Then, one may write, in analogy with the preceding derivatio n ofC, that dX=F−1dx=F−1mds =MdS , (3.49) hence 1 λM=F−1m. (3.50) ME185 DRAFT32 Deformation gradient and other measures of deformation Again, taking the dot-products of each side with itself lead s to (1 λM)·(1 λM) =1 λ2(M·M) =1 λ2= (F−1m)·(F−1m) =m·F−T(F−1m) =M·(F−TF−1)m =m·B−1m, (3.51) where F−T= (F−1)T= (FT)−1andBis the left Cauchy-Green tensor , defined as B=FFT(3.52) or, using components, Bij=FiAFjA. (3.53) In contrast to C, the tensor Bis defined with respect to the basis in the current configu- ration. Like C, it is easy to establish that the tensor Bis symmetric and positive-definite. To summarize the meaning of B, it can be said that, given a direction min the current configuration, Ballows the determination of the stretch λof an infinitesimal element dx alongmwhich is mapped from an infinitesimal material line element dXin the reference configuration. Consider now the difference in the squares of the lengths of th e line elements dXanddx, namely, ds2−dS2and write this difference as ds2−dS2= (dx·dx)−(dX·dX) = (FdX)·(FdX)−(dX·dX) =dX·FT(FdX)−(dX·dX) =dX·(CdX)−(dX·dX) =dX·(C−I)dX =dX·2EdX, (3.54) where E=1 2(C−I) =1 2(FTF−I) (3.55) ME185 DRAFTKinematics of deformation 33 is the (relative) Lagrangian strain tensor . Using components, the preceding equation can be written as EAB=1 2(FiAFiB−δAB), (3.56) which shows that the Lagrangian strain tensor Eis defined with respect to the basis in the reference configuration. In addition, Eis clearly symmetric and satisfies the property that E=0when the body undergoes no deformation between the referenc e and the current configuration. The difference ds2−dS2may be also written as ds2−dS2= (dx·dx)−(dX·dX) = (dx·dx)−(F−1dx)·(F−1dx) = (dx·dx)−dx·F−1(F−1dx) =dx·dx−(dx·B−1dx) =dx·(i−B−1)dx =dX·2edX, (3.57) where e=1 2(i−B−1) (3.58) is the (relative) Eulerian strain tensor orAlmansi tensor , andiis the identity tensor in the current configuration. Using components, one may write eij=1 2(δij−B−1 ij). (3.59) LikeE, the tensor eis symmetric and vanishes when there is no deformation betwe en the current configuration remains undeformed relative to the re ference configuration. However, unlike E, the tensor eis naturally resolved into components on the basis in the cur rent configuration. While, in general, the infinitesimal material line element dXis both stretched and rotated due to F, neither C(orB) norE(ore) yield any information regarding the rotation of dX. To extract rotation-related information from F, recall the polar decomposition theorem , which states that any invertible tensor Fcan be uniquely decomposed into F=RU=VR, (3.60) ME185 DRAFT34 Deformation gradient and other measures of deformation where Ris an orthogonal tensor and U,Vare symmetric positive-definite tensors. In com- ponent form, the polar decomposition is expressed as FiA=RiBUBA=VijRjA. (3.61) The tensors ( R,U) or (R,V) are called the polar factors ofF. Given the polar decomposition theorem, it follows that C=FTF= (RU)T(RU) =UTRTRU=UU=U2(3.62) and, likewise, B=FFT= (VR)(VR)T=VRRTV=VV=V2. (3.63) The tensors UandVare called the right stretch tensor and the left stretch tensor , respec- tively. Given their relations to CandB, it is clear that these tensors may be used to deter- mine the stretch of the infinitesimal material line element dX, which explains their name. Also, it follows that the component representation of the tw o tensors is U=UABEA⊗EB andV=Vijei⊗ej, i.e., they are resolved naturally on the bases of the refere nce and current configuration, respectively. Also note that the relations U=C1/2andV=B1/2hold true, as it is possible to define the square-root of a symmetric posi tive-definite tensor. Of interest now is R, which is a two-point tensor, i.e., R=RiAei⊗EA. Recalling equation (3.38), attempt to provide a physical interpretat ion of the decomposition theorem, starting from the right polar decomposition F=RU. Using this decomposition, and taking into account (3.38), write dx=FdX= (RU)dX=R(UdX). (3.64) This suggests that the deformation of dXtakes place in two steps. In the first one, dXis deformed into dX′=UdX, while in the second one, dX′is further deformed into RdX′=dx. ME185 DRAFTKinematics of deformation 35 Note that, letting dX′=M′dS′, where M′·M′= 1 and dS′is the magnitude of dX′, dX′·dX′= (M′dS′)·(M′dS′) =dS′2 = (UdX)·(UdX) =dX·(CdX) = (MdS)·(CMdS) =dS2M·CM =dS2λ2, (3.65) which, since λ=ds dS, implies that dS′=ds. Thus, dX′is stretched to the same differential length as dxdue to the action of U. Subsequently, write dx·dx= (RdX′)·(RdX′) =dX′·(RTRdX′) =dX′·dX′, (3.66) which confirms that Rinduces a length-preserving transformation on X′. In conclusion, F=RUimplies that dXis first subjected to a stretch U(possibly accompanied by rotation) to its final length ds, then is reoriented to its final state dxbyR, see Figure 3.9. dx dX dX′U R Figure 3.9: Interpretation of the right polar decomposition. Turning attention to the left polar decomposition F=VR, note that dx=FdX= (VR)dX=V(RdX). (3.67) This, again, implies that the deformation of dXtakes place in two steps. Indeed, in the first one, dXis deformed into dx′=RdX, while in the second one, dx′is mapped into Vdx′=dx. For the first step, note that dx′·dx′= (RdX)·(RdX) =dX·(RTRdX) =dX·dX, (3.68) ME185 DRAFT36 Deformation gradient and other measures of deformation which means that the mapping from dXtodx′is length-preserving. For the second step, write dx′·dx′=dX·dX=dS2 = (V−1dx)·(V−1dx) =dx·(B−1dx) = (mds)·(B−1mds) =ds2m·B−1m =1 λ2ds2, (3.69) which implies that Vinduces the full stretch λduring the mapping of dx′todx. Thus, the left polar decomposition F=VRmeans that the infinitesimal material line element dXis first subjected to a length-preserving transformation, f ollowed by stretching (with further rotation) to its final state dx, see Figure 3.10. dx dX dx′V R Figure 3.10: Interpretation of the left polar decomposition. It is conceptually desirable to decompose the deformation g radient into a pure rotation and a pure stretch (or vice versa). To investigate such an opt ion, consider first the right polar decomposition of equation (3.66). In this case, for the stre tchUto be pure, the infinitesimal line elements dXanddX′need to be parallel, namely dX′=UdX=λdX, (3.70) or, upon recalling that dX=MdS, UM =λM. (3.71) ME185 DRAFTKinematics of deformation 37 Equation (3.71) represents a linear eigenvalue problem. Th e eigenvalues λA>0 of (3.71) are called the principal stretches and the associated eigenvectors MAare called the principal directions . When λAare distinct, one may write UM A=λ(A)M(A) UM B=λ(B)M(B), (3.72) where the parentheses around the subscripts signify that th e summation convention is not in use. Upon premultiplying the preceding two equations wit hMBandMA, respectively, one gets MB·(UM A) =λ(A)MB·M(A) (3.73) MA·(UM B) =λ(B)MA·M(B). (3.74) Subtracting the two equations from each other leads to (λ(A)−λ(B))M(A)·M(B)= 0, (3.75) therefore, since, by assumption, λA/ne}ationslash=λB, it follows that MA·MB=δAB, (3.76) i.e., that the principal directions are mutually orthogona l and that {M1,M2,M3}form an orthonormal basis on E3. It turns out that regardless of whether Uhas distinct or repeated eigenvalues, the classical spectral representation theorem guarantees that there exists an orthonormal basis MAofE3 consisting entirely of eigenvectors of Uand that, if λAare the associated eigenvalues, U=3/summationdisplay A=1λ(A)M(A)⊗M(A). (3.77) ME185 DRAFT38 Deformation gradient and other measures of deformation Thus, C=U2= (3/summationdisplay A=1λ(A)M(A)⊗M(A))(3/summationdisplay B=1λ(B)M(B)⊗M(B)) =3/summationdisplay A=13/summationdisplay B=1λ(A)λ(B)(M(A)⊗M(A))(M(B)⊗M(B)) =3/summationdisplay A=13/summationdisplay B=1λ(A)λ(B)(M(A)·M(B))(M(A)⊗M(B)) =3/summationdisplay A=1λ2 (A)M(A)⊗M(A) (3.78) and, more generally, Um=3/summationdisplay A=1λm (A)M(A)⊗M(A), (3.79) for any integer m. Now, attempt a reinterpretation of the right polar decompos ition, in light of the discussion of principal stretches and directions. Indeed, when Uapplies on infinitesimal material line elements which are aligned with the principal directions MA, then it subjects them to a pure stretch. Subsequently, the stretched elements are reorien ted to their final direction by the action of R, see Figure 3.11. xXM1M2M3 λ1M1λ2M2λ3M3 λ1RM 1λ2RM 2λ3RM 3 U R Figure 3.11: Interpretation of the right polar decomposition relative t o the principal directions MAand associated principal stretches λA. Following an analogous procedure for the left polar decompo sition, note for the stretch Vto be pure it is necessary that dx=Vdx′=λdx′(3.80) ME185 DRAFTKinematics of deformation 39 or, recalling that dx′=RMdS, VRM =λRM. (3.81) This is, again, a linear eigenvalue problem that can be solve d for the principal stretches λi and the (rotated) principal directions mi=RM i. Appealing to the spectral representation theorem, one may write V=3/summationdisplay i=1λ(i)(RM (i))⊗(RM (i)) =3/summationdisplay i=1λ(i)m(i)⊗m(i). (3.82) A reinterpretation of the left polar decomposition can be aff orded along the lines of the earlier corresponding interpretation of the right deco mposition. Specifically, here the infinitesimal material line elements that are aligned with t he principal stretches MAare first reoriented by Rand subsequently subjected to a pure stretch to their final le ngth by the action of V, see Figure 3.12. xXM1M2M3 RM 1RM 2RM 3 λ1RM 1λ2RM 2λ3RM 3 VR Figure 3.12: Interpretation of the left polar decomposition relative to the principal directions RM iand associated principal stretches λi. Returning to the discussion of the polar factor R, recall that it is an orthogonal tensor. This, in turn, implies that (det R)2= 1, or det R=±1. An orthogonal tensor Ris termed proper (resp. improper ) if det R= 1 (resp. det R=−1). Assuming, for now, that Ris proper orthogonal, and invoking elementary properties o f ME185 DRAFT40 Deformation gradient and other measures of deformation determinant, it is seen that RTR=I⇒RTR−RT=I−RT ⇒RT(R−I) =−(R−I)T ⇒detRTdet(R−I) =−det(R−I)T ⇒det(R−I) =−det(R−I) ⇒det(R−I) = 0 , (3.83) so that Rhas at least one unit eigenvalue. Denote by pa unit eigenvector associated with the above eigenvalue (there exist two such unit vectors), an d consider two unit vectors q andr=p×qthat lie on a plane normal to p. It follows that {p,q,r}form a right-hand orthonormal basis of E3and, thus, Rcan be expressed with reference to this basis as R=Rppp⊗p+Rpqp⊗q+Rprp⊗r+Rqpq⊗p+Rqqq⊗q+Rqrq⊗r +Rrpr⊗p+Rrqr⊗q+Rrrr⊗r.(3.84) Note that, since pis an eigenvector of R, Rp=p⇒Rppp+Rqpq+Rrpr=p, (3.85) which implies that Rpp= 1 , R qp=Rrp= 0. (3.86) Moreover, given that Ris orthogonal, RTp=R−1p=p⇒Rppp+Rpqq+Rprr=p, (3.87) therefore Rpq=Rpr= 0. (3.88) Taking into account (3.86) and (3.88), the orthogonality of Rcan be expressed as (p⊗p+Rqqq⊗q+Rqrr⊗q+Rrqq⊗r+Rrrr⊗r) (p⊗p+Rqqq⊗q+Rqrq⊗r+Rrqr⊗q+Rrrr⊗r) =p⊗p+q⊗q+r⊗r.(3.89) ME185 DRAFTKinematics of deformation 41 and, after reducing the terms on the left-hand side, p⊗p+ (R2 qq+R2 rq)q⊗q+ (R2 rr+R2 qr)r⊗r + (RqqRqr+RrqRrr)q⊗r+ (RrrRrq+RqrRqq)r⊗q =p⊗p+q⊗q+r⊗r.(3.90) The above equation implies that R2 qq+R2 rq= 1, (3.91) R2 rr+R2 qr= 1, (3.92) RqqRqr+RrqRrr= 0, (3.93) RrrRrq+RqrRqq= 0. (3.94) Equations (3.91) and (3.92) imply that there exist angles θandφ, such that Rqq= cos θ , R rq= sin θ , (3.95) and Rrr= cos φ , R qr= sin φ . (3.96) It follows from (3.93) (or (3.94)) that cosθsinφ+ sin θcosφ= sin ( φ+θ) = 0 , (3.97) thus φ=−θ+ 2kπ or φ=π−θ+ 2kπ , (3.98) where kis an arbitrary integer. It can be easily seen that the latter choice yields an improper orthogonal tensor R, thus φ=−θ+ 2kπ, and, consequently, Rqq= cos θ , (3.99) Rrq= sin θ , (3.100) Rrr= cos θ , (3.101) Rqr=−sinθ . (3.102) ME185 DRAFT42 Deformation gradient and other measures of deformation From (3.86), (3.88) and (3.99-3.102), it follows that Rcan be expressed as R=p⊗p+ cos θ(q⊗q+r⊗r)−sinθ(q⊗r−r⊗q). (3.103) The angle θthat appears in (3.103) can be geometrically interpreted as follows: let an arbitrary vector xbe written in terms of {p,q,r}as x=pp+qq+rr, (3.104) where p=p·x, q =q·x, r =r·x, (3.105) and note that Rx=pp+ (qcosθ−rsinθ)q+ (qsinθ+rcosθ)r. (3.106) Equation (3.106) indicates that, under the action of R, the vector xremains unstretched and it rotates by an angle θaround the p-axis, where θis assumed positive when directed fromqtorin the sense of the right-hand rule. qrθ Rxp x Figure 3.13: Geometric interpretation of the rotation tensor Rby its action on a vector x. The representation (3.103) of a proper orthogonal tensor Ris often referred to as the Rodrigues formula . IfRis improper orthogonal, the alternative solution in (3.98 2in connec- tion with the negative unit eigenvalue pimplies that, Rxrotates by an angle θaround the p-axis and is also reflected relative to the origin of the ortho normal basis {p,q,r}. ME185 DRAFTKinematics of deformation 43 A simple counting check can be now employed to assess the pola r decomposition (3.60). Indeed, Fhas nine independent components and U(orV) has six independent components. At the same time, Rhas three independent components, for instance two of the th ree components of the unit eigenvector pand the angle θ. Example: Consider a motion χdefined in component form as χ1=χ1(XA, t) = (√acosϑX1)−(√asinϑ)X2 χ2=χ2(XA, t) = (√asinϑ)X1+ (√acosϑ)X2 χ3=χ3(XA, t) =X3, where a=a(t)>0 and ϑ=ϑ(t). This is clearly a planar motion, specifically independent of X3. The components FiA=χi,Aof the deformation gradient can be easily determined as [FiA] = √acosϑ−√asinϑ0 √asinϑ√acosϑ0 0 0 1 . This is clearly a spatially homogeneous motion , i.e., the deformation gradient is independent ofX. Further, note that det( FiA) =a >0, hence the motion is always invertible. The components CABofCand the components UABofUcan be directly determined as [CAB] = a0 0 0a0 0 0 1  and [UAB] = √a0 0 0√a0 0 0 1 . Also, recall that CM=λ2M, which implies that λ1=λ2=√aandλ3= 1. ME185 DRAFT44 Deformation gradient and other measures of deformation Given that Uis known, one may apply the right polar decomposition to dete rmine the rotation tensor R. Indeed, in this case, [RiA] = √acosϑ−√asinϑ0 √asinϑ√acosϑ0 0 0 1  1√a0 0 01√a0 0 0 1 = cosϑ−sinϑ0 sinϑcosϑ0 0 0 1 . Note that this motion yield pure stretch for ϑ= 2kπ, where k= 0,1,2, . . .. /square Example: Sphere under homogeneous deformation Consider the part of a deformable body which occupies a spher eP0of radius σcentered at the fixed origin O of E3. The equation of the surface ∂P0of the sphere can be written as Π·Π=σ2, (3.107) where Π=σM, (3.108) andM·M= 1. Assume that the body undergoes a spatially homogeneous deformation with deformation gradient F(t), so that π=FΠ, (3.109) where π(t) is the image of Πin the current configuration. Recall that λm=FM, where λ(t) is the stretch of a material line element that lies along Min the reference configuration, O OMm P P0Ππ Figure 3.14: Spatially homogeneous deformation of a sphere. andm(t) is the unit vector in the direction this material line eleme nt at time t, as in the figure above. Then, it follows from (3.107) and (3.109) that π=σλm. (3.110) ME185 DRAFTKinematics of deformation 45 Also, since λ2m·B−1m= 1, where B(t) is the left Cauchy-Green deformation tensor, equation (3.110) leads to π·B−1π=σ2. (3.111) Recalling the spectral representation theorem, the left Ca uchy-Green deformation tensor can be resolved as B=3/summationdisplay i=1λ2 (i)m(i)⊗m(i), (3.112) where ( λ2 i,mi),i= 1,2,3, are respectively the eigenvalues (squares of the princip al stretches) and the associated eigenvectors (principal directions) of B. In addition, note that mi,i= 1,2,3, are orthonormal and form a basis of E3, therefore the position vector πcan be uniquely resolved in this basis as π=πimi. (3.113) Starting from equation (3.112), one may write B−1=3/summationdisplay i=1λ−2 (i)m(i)⊗m(i), (3.114) and using (3.113) and (3.114), deduce that π·B−1π=λ−2 iπ2 i. (3.115) It is readily seen from (3.115) that π2 1 λ2 1+π2 2 λ2 2+π2 3 λ2 3=σ2, which demonstrates that, under a spatially homogeneous def ormation, the spherical region P0 is deformed into an ellipsoid with principal semi-axes of le ngthσλialong the principal directions of B, see Figure 3.15. /square Consider now the transformation of an infinitesimal materia l volume element dVof the reference configuration to its image dvin the current configuration due to the motion χ. The referential volume element is defined as an infinitesimal parallelepiped with sides dX1, dX2, anddX3. Likewise, its spatial counterpart is the infinitesimal par allelepiped with sides dx1,dx2, and dx3, where each dxiis the image of dXidue to χ, see Figure 3.15. ME185 DRAFT46 Deformation gradient and other measures of deformation m1m2m3 Figure 3.15: Image of a sphere under homogeneous deformation. x X R0 RdX1dX2dX3 dx1dx2dx3 Figure 3.16: Mapping of an infinitesimal material volume element dVto its image dvin the current configuration. First, note that dV=dX1·(dX2×dX3) =dX2·(dX3×dX1) =dX3·(dX1×dX2), (3.116) where each of the representation of dVin (3.116) corresponds to the triple product [ dX1dX2dX3] of the vectors dX1, dX2anddX3. Recalling that the triple product satisfies the property [dX1dX2dX3] = det [[ dX1 A],[dX2 A],[dX3 A]], it follows that in the current configuration dv=dx1·(dx2×dx3) = (FdX1)·/parenleftbig (FdX2)×FdX3)/parenrightbig = det/bracketleftbig [FiAdX1 A],[FiAdX2 A],[FiAdX3 A]/bracketrightbig = det/bracketleftbig FiA[dX1 A, dX2 A, dX3 A]/bracketrightbig = (det F) det/bracketleftbig [dX1 A],[dX2 A],[dX3 A]/bracketrightbig =JdV . (3.117) ME185 DRAFTKinematics of deformation 47 A motion for which dv=dV(i.e.J= 1) for all infinitesimal material volumes dVat all times is called isochoric (orvolume preserving ). Here, one may argue that if, by convention, dV > 0 (which is true as long as dX1, dX2 anddX3observe the right-hand rule), then the relative orientatio n of line elements should be preserved during the motion, i.e., J >0 everywhere and at all time. Indeed, since the motion is assumed smooth, any changes in the sign of Jwould necessarily imply that there exists a time tat which J= 0 at some material point(s), which would violate the assump tion the assumption of invertibility. Based on the preceding obs ervation, the Jacobian Jwill be taken to be positive at all time. Consider next the transform ation of an infinitesimal material surface element dAin the reference configuration to its image dain the current configuration. The referential surface element is defined as the parallelog ram formed by the infinitesimal material line elements dX1anddX2, such that dA=dX1×dX2=NdA , (3.118) whereNis the unit normal to the surface element consistently with t he right-hand rule, see Figure 3.16. Similarly, in the current configuration, one ma y write da=dx1×dx2=nda , (3.119) where nis the corresponding unit normal to the surface element defin ed by dx1anddx2. x X R0 RdX1dX2 dx1dx2 Figure 3.17: Mapping of an infinitesimal material surface element dAto its image dain the current configuration. LetdXbe any infinitesimal material line element, such that N·dX>0 and consider the infinitesimal volumes dVanddvformed by {dX1, dX2, dX}and{dx1, dx2, dx=FdX}, ME185 DRAFT48 Velocity gradient and other measures of deformation rate respectively. It follows from (3.117) that dv=dx·(dx1×dx2) =dx·nda= (FdX)·nda =JdV =JdX·(dX1×dX2) =JdX·NdA , (3.120) which implies that (FdX)·nda=JdX·NdA , (3.121) for any infinitesimal material line element dX. This, in turn, leads to nda=JF−TNdA . (3.122) Equation (3.122) is known as Nanson’s formula . Dotting each side in (3.122) with itself and taking square roots yields |da|=J/radicalbig N·(C−1N)|dA|. (3.123) As argued in the case of the infinitesimal volume transformat ions, if an infinitesimal material line element satisfies dA > 0, then da >0 everywhere and at all time, since J >0 andC−1 is positive-definite. 3.3 Velocity gradient and other measures of deforma- tion rate In this section, interest is focused to measures of the rate a t which deformation occurs in the continuum. To this end, define the spatial velocity gradient tensor Las L= grad v=∂v ∂x. (3.124) This tensor is naturally defined relative to the basis {ei}in the current configuration. Next, recall any such tensor can be uniquely decomposed into a symm etric and a skew-symmetric part, so that Lcan be written as L=D+W, (3.125) where D=1 2(L+LT) (3.126) ME185 DRAFTKinematics of deformation 49 is the rate-of-deformation tensor , which is symmetric, and W=1 2(L−LT) (3.127) is the vorticity (orspin) tensor, which is skew-symmetric. Consider now the rate of change of the deformation gradient f or a fixed particle associated with point Xin the reference configuration. To this end, write the materi al time derivative ofFas ˙F=˙/parenleftbigg∂χ(X, t) ∂X/parenrightbigg =∂ ∂X˙χ(X, t) =∂ˆv(X, t) ∂X=∂˜v(x, t) ∂x∂χ(X, t) ∂X=LF,(3.128) where use is made of the chain rule. Also, in the above derivat ion the change in the order of differentiation between the derivatives with respect to Xandtis allowed under the assumption that the mixed second derivative∂2χ ∂X∂tis continuous. Given (3.128), one may express with the aid of the product rul e the rate of change of the right Cauchy-Green deformation tensor Cfor a fixed particle Xas ˙C=˙FTF=˙FTF+FT˙F= (LF)TF+FT(LF) =FT(LT+L)F= 2FTDF.(3.129) Likewise, for the left Cauchy-Green deformation tensor, on e may write ˙B=˙FFT=˙FFT+F˙FT= (LF)FT+F(LF)T=LB+BLT. (3.130) Similar results may be readily obtained for the rates of the L agrangian and Eulerian strain measures. Specifically, it can be shown that ˙E=FTDF (3.131) and ˙e=1 2(B−1L+LTB−1). (3.132) Proceed now to discuss physical interpretations for the rat e tensors DandW. Starting from (3.44), take the material time derivatives of both sides to o btain the relation ˙λm+λ˙m=˙FM+F˙M =LFM =L(λm) =λLm. (3.133) ME185 DRAFT50 Velocity gradient and other measures of deformation rate Note that ˙M=0, since Mis a fixed vector in the fixed reference configuration, hence do es not vary with time. Upon taking the dot-product of each side w ithm, it follows that ˙λm·m+λ˙m·m=λ(Lm)·m. (3.134) However, since mis a unit vector, it follows that ˙m·m= 0, so that the preceding equation simplifies to ˙λ=λm·Lm. (3.135) Further, since the skew-symmetric part WofLsatisfies m·Wm =m·(−WT)m=−m·Wm, (3.136) hencem·Wm= 0, one may rewrite (3.135) as ˙λ=λm·Dm (3.137) or, alternatively, as ˙lnλ=m·Dm=D·(m⊗m). (3.138) Thus, the tensor Dfully determines the material time derivative of the logari thmic stretch lnλfor a material line element along a direction min the current configuration. Before proceeding with the geometric interpretation of W, some preliminary development on skew-symmetric tensors is required. First, note that any skew-symmetric tensor in E3, such as W, has only only three independent components. This suggests that there exists a one-to-one correspondence between skew-symmetric tensor s and vectors in E3. To establish this correspondence, observe that W=1 2(W−WT), (3.139) so that, when operating on any vector z∈E3, Wz=1 2Wij(ei⊗ej−ej⊗ei)z =1 2Wij[(z·ej)ei−(z·ei)ej]. (3.140) ME185 DRAFTKinematics of deformation 51 Recalling the standard identity u×(v×w) = (u·w)v−(v·u)w, the preceding equation can be rewritten as Wz=1 2Wij[z×(ei×ej)] =−1 2Wij[(ei×ej)×z] =1 2Wji[(ei×ej)×z] =/bracketleftbigg/parenleftbigg1 2Wjiei/parenrightbigg ×ej]/bracketrightbigg ×z =w×z, (3.141) where the vector wis defined as w=1 2Wjiei×ej (3.142) and is called the axial vector of the skew-symmetric tensor W. Using components, one may represent Win terms of wand vice-versa. Specifically, starting from (3.142), w=wkek=1 2Wjiei×ej=1 2Wjieijkek, (3.143) hence, wk=1 2eijkWji. (3.144) Conversely, starting from (3.141), Wijzjei=eijkwjzkei=eikjwkzjei, (3.145) so that Wij=eikjwk=ejikwk. (3.146) The above relations apply to any skew-symmetric tensor. If Wis specifically the vorticity ME185 DRAFT52 Velocity gradient and other measures of deformation rate tensor, then the associated axial vector satisfies the relat ion w=1 4(vj,i−vi,j)ei×ej =1 4(vj,i−vi,j)eijkek =1 4(eijkvj,i−eijkvi,j)ek =1 4(eijkvj,i−ejikvi,j)ek =1 2eijkvj,iek =1 2curlv. (3.147) In this case, the axial vector wis called the vorticity vector . Also, a motion is termed irrotational ifW=0(or, equivalently, w=0). Letw=˜w(x, t) be the vorticity vector field at a given time t. The vortex line through x at time tis the space curve that passes through xand is tangent to the vorticity vector field ˜wat all of its points, hence is defined as dy=˜w(y, t)dτ , y(τ0) =x,(tfixed) . (3.148) For an irrotational motion, any line passing through xat time tis a vortex line. Returning to the geometric interpretation of W, take ¯mto be a unit vector that lies along a principal direction of Din the current configuration, namely (D−γi)¯m=0. (3.149) It follows from (3.149) that (D¯m)·¯m=γ¯m·¯m=γ=˙¯λ ¯λ=˙ln¯λ, (3.150) i.e., the eigenvalues of Dare equal to the material time derivatives of the logarithmi c stretches ln¯λof line elements along the eigendirections ¯min the current configuration. Recalling the identity (3.133), write ˙m=Lm−˙λ λm=/parenleftBigg L−˙λ λi/parenrightBigg m =/parenleftBigg D−˙λ λi/parenrightBigg m+Wm, (3.151) ME185 DRAFT3.4. SUPERPOSED RIGID-BODY MOTIONS 53 which holds for any direction min the current configuration. Setting in the above equation m=¯m, it follows that ˙¯m=W¯m=wׯm. (3.152) Therefore, the material time derivative of a unit vector ¯malong a principal direction of D is determined by (3.152). Recalling from rigid-body dynami cs the formula relating linear to angular velocities, one may conclude that wplays the role of the angular velocity of a line element which, in the current configuration, lies along the p rincipal direction mofD. 3.4 Superposed rigid-body motions Arigid motion is one in which the distance between any two material points r emains constant at all times. Denoting by XandYthe position vectors of two material points on the fixed reference configuration, define the distance function as d(X,Y) =|X−Y|. (3.153) According to the preceding definition, a motion is rigid if, f or any points XandY, d(X,Y) =d(χ(X, t),χ(Y, t)) = d(x,y), (3.154) for all times t. Now, introduce the notion of a superposed rigid-body motion . To this end, take a motion χof body Bsuch that, as usual, x=χ(X, t). Then, let another motion χ+be defined for B, such that x+=χ+(X, t), (3.155) so that χandχ+differ by a rigid-body motion. Then, with reference to Figure 3.18, one may write x+=χ+(X, t) =χ+(χ−1 t(x), t) = ¯χ+(x, t). (3.156) Said differently, x+=χ+ t(X) = ¯χ+ t(x) = ¯χ+ t(χt(X)) (3.157) or, equivalently, χ+ t=¯χ+ t◦χt. (3.158) ME185 DRAFT54 Superposed rigid-body motions xx+ X yy+ Y R0 RR+ χχ+ ¯χ+ Figure 3.18: Configurations associated with motions χandχ+differing by a superposed rigid motion ¯χ+. Clearly, the superposed motion ¯χ+(x, t) is invertible for fixed t, since χ+is assumed invert- ible for fixed t. Similarly, take a second point Yin the reference configuration, so that y=χ(Y, t) and write y+=χ+(Y, t) =χ+(χ−1 t(y), t) = ¯χ+(y, t). (3.159) Recalling that RandR+differ only by a rigid transformation, one may conclude that (x−y)·(x−y) = (x+−y+)·(x+−y+) =/bracketleftbig ¯χ+(x, t)−¯χ+(y, t)/bracketrightbig/bracketleftbig ¯χ+(x, t)−¯χ+(y, t)/bracketrightbig , (3.160) for all x,yin the region Rat time t. Since xandyare chosen independently, one may differentiate equation (3.160) first with respect to xto get x−y=/bracketleftbigg∂¯χ+(x, t) ∂x/bracketrightbiggT (¯χ+(x, t)−¯χ+(y, t)). (3.161) Then, equation (3.161) may be differentiated with respect to y, which leads to i=/bracketleftbigg∂¯χ+(x, t) ∂x/bracketrightbiggT/bracketleftbigg∂¯χ+(y, t) ∂y/bracketrightbigg . (3.162) Since the motion ¯χ+is invertible, equation (3.162) can be equivalently writte n as /bracketleftbigg∂¯χ+(x, t) ∂x/bracketrightbiggT =/bracketleftbigg∂¯χ+(y, t) ∂y/bracketrightbigg−1 . (3.163) ME185 DRAFTKinematics of deformation 55 Then, the left- and right-hand side should be necessarily fu nctions of time only, hence /bracketleftbigg∂χ+(x, t) ∂x/bracketrightbiggT =/bracketleftbigg∂χ+(y, t) ∂y/bracketrightbigg−1 =QT(t). (3.164) Since equation (3.164) implies that∂χ+(x, t) ∂x=∂χ+(y, t) ∂y=Q(t), it follows immediately thatQT(t)Q(t) =I, i.e.,Q(t) is an orthogonal tensor. Further, note that using the chain rule the deformation gradient F+of the motion χ+is written as F+=∂χ+ ∂X=∂χ+ ∂x∂χ ∂X=QF. (3.165) Since, by assumption, both motions χandχ+lead to deformation gradients with positive Jacobians, equation (3.165) implies that det Q>0, hence det Q= 1 and Qis proper orthogonal. SinceQis a function of time only, equation (3.164) 1can be directly integrated with respect to x, leading to x+=¯χ+(x, t) =Q(t)x+c(t), (3.166) wherec(t) is a vector function of time. Equation (3.166) is the genera l form of the superposed rigid motion ¯χon the original motion χ. Given (3.165) 3and recalling the right polar decomposition of F, write F+=R+U+ =QF=QRU ,(3.167) where R,R+are proper orthogonal tensors and U,U+are symmetric positive-definite tensors. Since, clearly, (QR)T(QR) = (RTQT)(QR) =RT(QTQ)R=RTR=I (3.168) and also det ( QR) = (det Q)(detR) = 1, therefore QRis proper orthogonal, the uniqueness of the polar decomposition, in conjunction with (3.167), ne cessitates that R+=QR (3.169) and U+=U. (3.170) ME185 DRAFT56 Superposed rigid-body motions Similarly, the left decomposition of Fyields F+=V+R+=V+(QR) =QF=Q(VR),(3.171) which implies that V+(QR) =Q(VR), (3.172) hence, V+=QVQT. (3.173) It follows readily from (3.165) 3that C+=F+TF+= (QF)T(QF) = (FTQT)(QF) =FTF=C (3.174) and B+=F+F+T= (QF)(QF)T= (QF)(FTQT) =QBQT. (3.175) Likewise, E+=1 2(C+−I) =1 2(C−I) =E (3.176) and e+=1 2(I−B+−1) =1 2[I−(QBQT)−1] =1 2(I−Q−TB−1Q−1) =1 2(I−QB−1QT) =1 2Q(I−B−1)QT =QeQT. (3.177) A vector or tensor is called objective if it transforms under superposed rigid motions in the same manner as its natural basis. Physically, this mea ns that the components of an objective tensor relative to its basis are unchanged if bo th the tensor and its natural basis are transformed due to a rigid motion superposed to the original motion. Specifically, a referential tensor is objective when it does not change und er superposed rigid motions, ME185 DRAFTKinematics of deformation 57 since its natural basis {EA⊗EB}remains untransformed under superposed rigid motions, as the latter do not affect the reference configuration. Hence , referential tensors such as C,UandEare objective. Correspondingly a spatial tensor is objecti ve if it transforms according to ( ·)+=Q(·)QT. This is because its tensor basis {ei⊗ej}transforms as {(Qei)⊗(Qej)}=Q{ei⊗ej}QT. Hence, spatial tensors such as B,Vandeare objective. It is easy to deduce that two-point tensors are objective if t hey transform as ( ·)+=Q(·) or (·)+= (·)QTdepending on whether the first or second leg of the tensor lies in the current configuration, respectively. Analogous conclusions may be drawn for vectors, which, when objective, remain untransformed if they are referential or transform as ( ·)+=Q(·) if they are spatial. Examine next the transformation of the velocity vector vunder a superposed rigid-body motion. In this case v+=˙χ+(X, t) =˙χ+(x, t) =˙[Q(t)x+c(t)] = ˙Q(t)x+Q(t)v+˙c(t). (3.178) SinceQQT=I, ˙QQT=˙QQT+Q˙QT=0. (3.179) Setting Ω(t) = ˙Q(t)QT(t), (3.180) it follows from (3.179) that Ωis skew-symmetric, hence there exists an axial vector ωsuch thatΩz=ω×z, for any z∈E3. Returning to (3.178), write, with the aid of (3.166) and (3.180), v+=ΩQ(t)x+Q(t)v+˙c(t) =Ω(x+−c(t)) +Q(t)v+˙c(t). (3.181) Invoking the definition of the axial vector ω, one may rewrite (3.181) as v+=ω×(x+−c) +Qv+˙c. (3.182) Equation (3.181) reveals that the velocity is not an objecti ve vector. ME185 DRAFT58 Superposed rigid-body motions Next, examine how the various tensorial measures of deforma tion rate transform under superposed rigid motions. Starting with the spatial veloci ty gradient, write L+=∂v+ ∂x+=∂ ∂x+[Ω(x+−c) +Qv+˙c] =Ω+∂(Q˜v) ∂x+ =Ω+∂(Q˜v) ∂x∂χ ∂x+ =Ω+Q∂˜v ∂x∂ ∂x+[QT(x+−c)] =Ω+QLQT, (3.183) where use is made of (3.181), the definition of Land the chain rule. It follows from (3.183) thatLis not objective. Also, the rate-of-deformation tensor Dtransforms according to D+=1 2(L++L+T) =1 2(Ω+QLQT) +1 2(Ω+QLQT)T =1 2(Ω+ΩT) +Q1 2(L+LT)QT =QDQT, (3.184) which implies that Dis objective. Turning to the vorticity tensor W, one may write W+=1 2(L+−L+T) =1 2(Ω+QLQT)−1 2(Ω+QLQT)T =1 2(Ω−ΩT) +Q1 2(L−LT)QT =Ω+QWQT, (3.185) from where it is seen that Wis not objective. Objectivity conclusions can be drawn for other kinematic qu antities of interest by appeal- ing to results already in place. For instance, infinitesimal material line elements transform as dx+=F+dX= (QF)dX=Q(FdX) =Qdx, (3.186) ME185 DRAFTKinematics of deformation 59 hence dxis objective. Similarly, infinitesimal material volume ele ments transform as dv+=J+dV= det(QF)dV= (det Q)(detF)dV= (det F)dV=JdV =dv ,(3.187) hence dvis an objective scalar quantity. For infinitesimal material area elements, write da+=n+da+=J+F+−TNdA =J(QF)−TNdA=J(Q−TF−T)NdA=JQF−TNdA=Qnda=Qda, (3.188) hence the vector dais objective. Now, taking the dot-product of each side of (3. 188) with itself yields (n+da+)·(n+da+) = (Qnda)·(Qnda), (3.189) therefore ( da+)2=da2, hence also da+=da(as long as dais taken positive from the outset) andn+=Qn. In closing, note that the superposed rigid motion operation (·)+commutes with the transposition ( ·)T, inversion ( ·)−1and material time derivative˙(·) operations. ME185 DRAFT60 Superposed rigid-body motions ME185 DRAFTChapter 4 Basic physical principles 4.1 The divergence and Stokes’ theorems By way of background, first review the divergence theorem for scalar, vector and tensor functions. To this end, let Pbe a bounded closed region with smooth boundary ∂Pin the Euclidean point space E3. Recall that Pisbounded if it can be enclosed by a sphere of finite radius, closed if it contains its boundary and smooth if the normal nto its boundary is uniquely defined at all boundary points. Next, define a scalar function φ:P → R, a vector function v:P →E3, and a tensor function T:P → L (E3, E3). All three functions are assumed continuously differentia ble. Then, the gradients of φandvsatisfy /integraldisplay Pgradφ dv=/integraldisplay ∂Pφnda , (4.1) and/integraldisplay Pgradvdv=/integraldisplay ∂Pv⊗nda . (4.2) In addition, the divergences of vandTsatisfy /integraldisplay Pdivvdv=/integraldisplay ∂Pv·nda , (4.3) and/integraldisplay PdivTdv=/integraldisplay ∂PTnda . (4.4) 61 DRAFT62 The divergence and Stokes’ theorems Equation (4.3) expresses the classical divergence theorem . The other three identities can be derived from this theorem. Indeed, the identity (4.4) can be derived by dotting the left-hand side by a constant vector cand using (4.3) and the definition of the divergence of a tenso r. This leads to /integraldisplay PdivTdv·c=/integraldisplay PdivT·cdv=/integraldisplay Pdiv(TTc)dv=/integraldisplay ∂P(TTc)·nda =/integraldisplay ∂P(Tn)·cda=/integraldisplay ∂PTnda·c.(4.5) Sincecis chosen arbitrarily, equation (4.4) follows immediately . Equation (4.1) can be deduced from (4.4) by setting T=φI, in which case, for any constant vector c, /integraldisplay Pdiv(φI)dv·c=/integraldisplay Pdiv(φc)dv=/integraldisplay Ptrgrad( φc)dv=/integraldisplay Ptr(c⊗gradφ)dv =/integraldisplay Pgradφ·cdv=/integraldisplay Pgradφ dv·c=/integraldisplay Pgradφ dv·c =/integraldisplay ∂Pφnda·c,(4.6) which, again due to the arbitrariness of c, implies (4.1). Lastly, (4.2) is obtained from (4.1) by taking any constant cand writing /bracketleftbigg/integraldisplay Pgradvdv/bracketrightbiggT c=/integraldisplay P(gradv)Tcdv=/integraldisplay Pgrad(v·c)dv =/integraldisplay ∂P(v·c)nda=/integraldisplay ∂P(n⊗v)cda=/bracketleftbigg/integraldisplay ∂P(n⊗v)da/bracketrightbigg c,(4.7) which proves the identity. Consider a closed non-intersecting curve Cwhich is parametrized by a scalar τ, 0≤τ≤1, so that the position vector of a typical point on Cisc(τ). Also, let Abe an open surface bounded by C, see Figure 4.1. Clearly, any point on Apossesses two equal and opposite unit vectors, each pointing outward to one of the two sides of the surface. To eliminate the ambiguity, choose one of the sides of the surface and denote i ts outward unit normal by n. This side is chosen so that c(¯τ)×c(¯τ+dτ) points toward it, for any ¯ τ∈[0,1). If now vis a continuously differentiable vector field, then Stokes’ the orem states that /integraldisplay Acurlv·ndA=/integraldisplay Cv·dx. (4.8) ME185 DRAFT4.2. THE REYNOLDS’ TRANSPORT THEOREM 63 This integral on the right-hand side of (4.8) is called the circulation of the vector field v around C. The circulation is the (infinite) sum of the tangential comp onents of valongC. If vis identified as the spatial velocity field, then Stokes’ theo rem states that the circulation of the velocity around Cequals to twice the integral of the normal component of the vo rticity vector on any open surface that is bounded by C. CA Figure 4.1: A surface Abounded by the curve C. 4.2 The Reynolds’ transport theorem LetPbe a closed and bounded region in E3with smooth boundary ∂Pand assume that the particles which occupy this region at time toccupy a closed and bounded region P0with smooth boundary ∂P0at another time t0, see Figure 4.2. χ P∂PP0∂P0 R∂RR0 ∂R0 Figure 4.2: A region Pwith boundary ∂Pand its image P0with boundary ∂P0in the reference configuration. Further, let a scalar field φbe defined by a referential function ˆφor a spatial function ˜φ, such that φ=ˆφ(X, t) = ˜φ(x, t). (4.9) ME185 DRAFT64 The Reynolds’ transport theorem Both ˆφand˜φare assumed continuously differentiable in both of their var iables. In the forthcoming discussion of balance laws, it is important to b e able to manipulate integrals of the form d dt/integraldisplay P˜φdv , (4.10) namely, material time derivatives of volume integrals defin ed in some subset of the current configuration. Example: Consider the integral in (4.10) for φ= 1. Here,d dt/integraldisplay Pdv=d dtvol{(P)}, which is the rate of change at time tof the total volume of the region occupied by the material particles that at time toccupy P. /square In evaluating (4.10), one first notes that the differentiatio n and integration operations cannot be directly interchanged, because the region Pover which the integral is evaluated is itself a function of time. Therefore, one needs to first map the integral to the (fixed) reference configuration, interchange the differentiation a nd integration operations, evaluate the derivative of the integrand, and then map the integral ba ck to the current configuration. Taking this approach leads to d dt/integraldisplay P˜φ dv=d dt/integraldisplay P0ˆφJ dV (pull-back to P0) =/integraldisplay P0d dt[ˆφJ]dV (interchanged dtand/integraldisplay ) =/integraldisplay P0/bracketleftBigg ∂ˆφ ∂tJ+ˆφ∂J ∂t/bracketrightBigg dV =/integraldisplay P0(˙φJ+ˆφJdivv)dV (˙J=Jdivv) =/integraldisplay P0(˙φ+ˆφdivv)J dV =/integraldisplay P(˙φ+˜φdivv)dv (push-forward to P). (4.11) This result is known as the Reynolds’ transport theorem . Note that the theorem applies also to vector and tensor functions without any modifications. To interpret the Reynolds’ transport theorem, note that the left-hand side of (4.11) is the rate of change of the integral of φover the region P, when following the set of particles that happen to occupy Pat time t. The right-hand side of (4.11) consists of the sum of two ME185 DRAFTBasic physical principles 65 terms. The first one is the integral of the rate of change of φfor all particles that happen to occupy Pat time t. The second one is due to the rate of change of the volume occup ied by the same particles assuming that ˜φremains constant in time for those particles. The Reynolds’ transport theorem can be restated in a number o f equivalent forms. To this end, write with the aid of the divergence theorem (4.3), d dt/integraldisplay P˜φdv=/integraldisplay P(˙φ+φdivv)dv =/integraldisplay P/bracketleftBigg ∂˜φ ∂t+∂˜φ ∂x·v+φdivv/bracketrightBigg dv =/integraldisplay P/bracketleftBigg ∂˜φ ∂t+ div( ˜φv)/bracketrightBigg dv =/integraldisplay P∂˜φ ∂tdv+/integraldisplay ∂P˜φv·nda . (4.12) An alternative interpretation of the theorem is now in order . Here, the right-hand side of (4.12) consists of the sum of two terms. The first term is the in tegral of the rate of change ofφat time tfor all points that form the fixed region P. The second term is the rate at which the volume of Pweighted by φchanges as particles exit Pacross ∂P. Example: Consider again the special case ˜φ(x, t) = 1, which corresponds to the transport of volume. Here, d dt/integraldisplay Pdv=/integraldisplay Pdivvdv=/integraldisplay ∂Pv·nda . (4.13) This means that the rate of change of the volume occupied by th e same material particles equals the boundary integral of the normal component of the v elocity v·nof∂P, i.e., the rate at which the volume of Pchanges as the particles exit across the boundary ∂¯Pof the fixed region ¯Pwhich equals to Pat time t. /square Starting from (4.12), note that/integraldisplay P∂˜φ ∂tdv=∂ ∂t/integraldisplay P˜φdv, since∂˜φ ∂tis the rate of change of φ for fixed position in space (hence, Pis also fixed in space). Therefore, one may alternatively express (4.12) as d dt/integraldisplay P˜φ dv=∂ ∂t/integraldisplay P˜φ dv+/integraldisplay ∂P˜φv·nda . (4.14) ME185 DRAFT66 The localization theorem 4.3 The localization theorem Another important result with implications in the study of b alance laws is presented here by way of background. Let ˜φ:R ×R→Rbe a function such that φ=˜φ(x, t), where R ⊂E3. Also, let ˜φbe continuous function in the spatial argument. Then, assum e that /integraldisplay P˜φ dv= 0, (4.15) for all P ⊂ R at a given time t. The localization theorem states that this is true if, and only if,˜φ= 0 everywhere in Rat time t. To prove this result, first note that the “if” portion of the th eorem is straightforward, since, if ˜φ= 0 in R, then (4.15) holds trivially true for any subset PofR. To prove the converse, note that continuity of ˜φin the spatial argument xat a point x0∈ Rmeans that for any time tand every ε >0, there is a δ=δ(ε)>0, such that |˜φ(x, t)−˜φ(x0, t)|< ε , (4.16) provided that |x−x0)|< δ(ε). (4.17) Now proceed by contradiction: assume that there exists a poi ntx0∈ R, such that, at any fixed time t,˜φ(x0, t) =φ0>0. Then, invoking continuity of ˜φ, there exists a δ=δ(φ0 2), such that |˜φ(x, t)−˜φ(x0, t)|=|˜φ(x, t)−φ0|<φ0 2, (4.18) whenever |x−x0)|< δ(φ0 2). (4.19) Now, define the region Pδthat consists of all points of Rfor which |x−x0|< δ(φ0 2). This is a sphere of radius δinE3with volume vol( Pδ) =/integraldisplay Pδdv >0. It follows from (4.18) that ˜φ(x, t)>φ0 2everywhere in Pδ. This, in turn, implies that /integraldisplay Pδ˜φ dv >/integraldisplay Pδφ0 2dv=φ0 2vol(Pδ)>0, (4.20) which constitutes a contradiction. Therefore, the localiz ation theorem holds. The localization theorem can be also proved with ease for vec tor and tensor functions. ME185 DRAFT4.4. MASS AND MASS DENSITY 67 4.4 Mass and mass density Consider a body Band take any arbitrary part S ⊆ B . Define a set function m:S /ma√sto→ R with the following properties: (i)m(S)≥0, for all S ⊆ B (i.e.,mnon-negative) (ii)m(∅) = 0. (iii)m(∪∞ i=1Si) =∞/summationdisplay i=1m(Si), where Si⊂ B,i= 1,2, . . .,andSi∩ Sj=∅, ifi/ne}ationslash=j(i.e.,m countably additive). A function mwith the preceding properties is called a measure onB. Assume here that there exists such a measure mand refer to m(B) as the mass of body Bandm(S) as the mass of the part SofB. In other words, consider the body as a set of particles with p ositive mass. Recall now that at time tthe body Boccupies a region R ⊂ E3and the part Soccupies a region P. Assuming that mis an absolutely continuous measure, it can be established t hat there exists a unique function ρ=ρ(x, t), such that, for any function f=ˇf(P, t) =˜f(x, t), one may write/integraldisplay Bˇf dm =/integraldisplay R˜fρ dv (4.21) and/integraldisplay Sˇf dm =/integraldisplay P˜fρ dv . (4.22) The function ρ >0 is termed the mass density . Its existence is a direct consequence of a classical result in measure theory, known as the Radon-Niko dym theorem. As a special case, one may consider the function f= 1, so that (4.21) and (4.22) reduce to/integraldisplay Bdm=/integraldisplay Rρ dv=m(B) (4.23) and/integraldisplay Sdm=/integraldisplay Pρ dv=m(S). (4.24) ME185 DRAFT68 Mass and mass density The mass density of a particle Poccupying point xin the current configuration may be defined by a limiting process as ρ= lim δ→0m(Sδ) vol(Pδ), (4.25) where Pδ⊂ E3denotes a sphere of radius δ >0 centered at xandSδthe part of the body that occupies Pδat time t, see Figure 4.3. Pδ RSδ Bxδ Figure 4.3: A limiting process used to define the mass density ρat a point xin the current configuration. An analogous definition of mass density can be furnished in th e reference configuration, where, for any function f=ˇf(P, t) =ˆf(X, t), /integraldisplay Bˇf dm =/integraldisplay R0ˆfρ0dV (4.26) and/integraldisplay Sˇf dm =/integraldisplay P0ˆfρ dV . (4.27) Here, the mass density ρ0=ρ0(X, t) in the reference configuration may be again defined by a limiting process, such that at a given point X, ρ0= lim δ→0m(Sδ) vol(P0,δ), (4.28) where P0,δ⊂ E3denotes a sphere of radius δ >0 centered at XandSδthe part of the body that occupies P0,δat time t0. Also, as in the spatial case, one may write /integraldisplay Bdm=/integraldisplay R0ρ0dV=m(B) (4.29) and/integraldisplay Sdm=/integraldisplay P0ρ0dV=m(S). (4.30) The continuum hypothesis is crucial in establishing the exi stence of ρandρ0. ME185 DRAFT4.5. THE PRINCIPLE OF MASS CONSERVATION 69 4.5 The principle of mass conservation The principle of mass conservation states that the mass of an y material part of the body remains constant at all times, namely that d dtm(S) = 0 (4.31) or, upon recalling (4.24), d dt/integraldisplay Pρ dv= 0. (4.32) The preceding is an integral form of the principle of mass con servation in the spatial descrip- tion. Using the Reynolds’ transport theorem, the above equa tion may be written as /integraldisplay P[ ˙ρ+ρdivv]dv= 0. (4.33) Assuming that the integrand in (4.33) is continuous and reca lling that S(hence, also P) is arbitrary, it follows from the localization theorem that ˙ρ+ρdivv= 0. (4.34) Equation (4.34) constitutes the local form of the principle of mass conservation in the spatial description.1It can be readily rewritten as ∂ρ ∂t+∂ρ ∂x·v+ρdivv= 0 (4.35) hence, also as ∂ρ ∂t+ div ( ρv) = 0 . (4.36) Example: In a volume-preserving flow of a material with uniform densit y, conservation of mass reduces to∂ρ ∂t= 0. Hence, recalling (4.14), one may write d dt/integraldisplay Pρ dv=/integraldisplay P∂ρ ∂tdv+/integraldisplay ∂Pρv·nda=/integraldisplay ∂Pρv·nda . (4.37) /square 1This is also sometimes referred to as the “continuity” equat ion, with reference to the continuity of fluid flow in a control volume P. ME185 DRAFT70 The principles of linear and angular momentum balance An alternative form of the mass conservation principle can b e obtained by recalling equations (4.24) and (4.30), from which it follows that m(S) =/integraldisplay Pρ dv=/integraldisplay P0ρ0dV . (4.38) Recalling also (3.117), one concludes that /integraldisplay P0ρJ dV =/integraldisplay P0ρ0dV . (4.39) This is an integral form of the principle of mass conservatio n in referential description. From it, one finds that/integraldisplay P0(ρJ−ρ0)dV= 0. (4.40) Taking into account the arbitrariness of P0, the localization theorem may be invoked to yield a local form of mass conservation in referential descriptio n as ρ0=ρJ . (4.41) 4.6 The principles of linear and angular momentum balance Once mass conservation is established, the principles of li near and angular momentum are postulated to describe the motion of continua. These two pri nciples originate from the work of Newton and Euler. By way of background, review Newton’s three laws of motion, a s postulate for particles in 1687. The first law says that a particle stays at rest or contin ues at constant velocity unless an external force acts on it. The second law says that the tota l external force on a particle is proportional to the rate of change of the momentum of the part icle. The third law says that every action has an equal and opposite reaction. As Euler rec ognized, Newton’s three laws of motion, while sufficient for the analysis of particles, are not strictly appropriate for the study of rigid and deformable continua. Rather, he postulat e a linear momentum balance law (akin to Newton’s second law) and an angular momentum bal ance law. The latter can be easily motivated from the analysis of systems of particle s. ME185 DRAFTBasic physical principles 71 To formulate Euler’s two balance laws, first define the linear momentum of the part of the body that occupies the infinitesimal volume element dvat time tasdmv, where dmis the mass of dv. Also, define the angular momentum of the same part relative to the origin of the fixed basis {ei}asx×(dmv), where xis the position vector associated with the infinitesimal volume element. Similarly, define the linear a nd angular momenta of the part Swhich occupies a region Pat time tas/integraltext Svdmand/integraltext Sx×vdm, respectively. Next, admit the existence of two types of external forces act ing on Bat any time t. There are: (a) body forces per unit mass (e.g., gravitational, magnetic) b=b(x, t) which act on the particles which comprise the continuum, and (b) contact forces per unit area t=t(x, t;n) =t(n)(x, t), which act on the particles which lie on boundary surfaces a nd depend on the orientation of the surface on which they act thr ough the outward unit normal nto the surface. The force t(n)is alternatively referred to as the stress vector or the traction vector . The principle of linear momentum balance states that the rate of change of linear mo- mentum for any region Poccupied at time tby a part Sof the body equals the total external forces acting on this part. In mathematical terms, this mean s that d dt/integraldisplay Svdm=/integraldisplay Sbdm+/integraldisplay ∂Pt(n)da (4.42) or, equivalently, d dt/integraldisplay Pρvdv=/integraldisplay Pρbdv+/integraldisplay ∂Pt(n)da . (4.43) Using conservation of mass, the left-hand side of the equati on can be written as d dt/integraldisplay Pρvdv=/integraldisplay Pd dt(ρv)dv+/integraldisplay P(ρv) divvdv =/integraldisplay P( ˙ρv+ρ˙v)dv+/integraldisplay P(ρv) divvdv =/integraldisplay P[( ˙ρ+ρdivv)v+ρ˙v]dv =/integraldisplay Pρadv , (4.44) hence, the principle of linear momentum balance can be also e xpressed as /integraldisplay Pρadv=/integraldisplay Pρbdv+/integraldisplay ∂Pt(n)da . (4.45) ME185 DRAFT72 The principles of linear and angular momentum balance The principle of angular momentum balance states that the rate of change of angular momentum for any region Poccupied at tby a part Sof the body equals the moment of all external forces acting on this part. Again, this principle c an be expressed mathematically as d dt/integraldisplay Sx×vdm=/integraldisplay Sx×bdm+/integraldisplay ∂Px×t(n)da (4.46) or, equivalently, d dt/integraldisplay Px×ρvdv=/integraldisplay Px×ρbdv+/integraldisplay ∂Px×t(n)da . (4.47) Again, appealing to conservation of mass, one may easily rew rite the term on the left-hand side of (4.47) as d dt/integraldisplay Px×ρvdv=/integraldisplay P/braceleftbiggd dt(x×ρv) + (x×ρv) divv/bracerightbigg dv =/integraldisplay P{[˙x×ρv+x×˙ρv+x×ρ˙v] + (x×ρvdivv)}dv =/integraldisplay P{x×( ˙ρ+ρdivv)v+x×ρa}dv =/integraldisplay Px×ρadv . (4.48) As a result, the principle of angular momentum balance may be also written as /integraldisplay Px×ρadv=/integraldisplay Px×ρbdv+/integraldisplay ∂Px×t(n)da . (4.49) The preceding two balance laws are also referred to as Euler’s laws . They are termed “balance” laws because they postulate that there exists a ba lance between external forces (and their moments) and the rate of change of linear (and angu lar) momentum. Euler’s laws are independent axioms in continuum mechanics. In the special case where b=0inPandt(n)=0on∂P, then equations (4.43) and (4.47) readily imply that the linear and the angular momentu m are conserved quantities in P. Another commonly encountered special case is when the veloc ityvvanishes identically. In this case, equations (4.43) and (4.47) imply that the sum of a ll external forces and the some of all external moments vanish, which gives rise to the class icalequilibrium equations of statics. ME185 DRAFT4.7. STRESS VECTOR AND STRESS TENSOR 73 4.7 Stress vector and stress tensor As in the case of mass balance, it is desirable to obtain local forms of linear and angular momentum balance. Recalling the corresponding integral st atements (4.43) and (4.47), it is clear that the Reynolds’ transport theorem may be employe d in the rate of change of momentum terms to yield volume integrals, while the body for ce terms are already in the form of volume integral. Therefore, in order to apply the loc alization theorem, it is essential that the contact form terms (presently written as surface in tegrals) be transformed into equivalent volume integral terms. P1 P1 P2 P2 ∂P ∂P′∂P′′σσσn1=n n 2 Figure 4.4: Setting for a derivation of Cauchy’s lemma. By way of background, consider some properties of the tracti on vector t(n). To this end, take an arbitrary region P ⊆ R and divide it into two mutually disjoint subregions P1and P2by means of an arbitrary smooth surface σ, namely P=P1∪ P2andP1∩ P2=∅, see Figure 4.4. Also, note that the boundaries ∂P1and∂P2ofP1andP2, respectively, can be expressed as ∂P1=∂P′∪σand∂P2=∂P′′∪σ. Now, enforce linear momentum balance separately in P1andP2to find that d dt/integraldisplay P1ρvdv=/integraldisplay P1ρbdv+/integraldisplay ∂P1t(n)da (4.50) and d dt/integraldisplay P2ρvdv=/integraldisplay P2ρbdv+/integraldisplay ∂P2t(n)da (4.51) Subsequently, add the two equations together to find that d dt/integraldisplay P1∪P2ρvdv=/integraldisplay P1∪P2ρbdv+/integraldisplay ∂P1∪∂P2t(n)da (4.52) ME185 DRAFT74 Stress vector and stress tensor or d dt/integraldisplay Pρvdv=/integraldisplay Pρbdv+/integraldisplay ∂P1∪∂P2t(n)da . (4.53) Further, enforce linear momentum balance on the union of P1andP2to find that d dt/integraldisplay Pρvdv=/integraldisplay Pρbdv+/integraldisplay ∂Pt(n)da . (4.54) Subtracting (4.54) from (4.53) leads to /integraldisplay ∂P1∪∂P2t(n)da=/integraldisplay ∂Pt(n)da . (4.55) Recalling the decomposition of ∂P1and∂P2, the preceding equation may be also expressed as /integraldisplay ∂P′∪σt(n)da+/integraldisplay ∂P′′∪σt(n)da=/integraldisplay ∂Pt(n)da (4.56) or/integraldisplay ∂P′∪∂P′′t(n)da+/integraldisplay σt(n1)da+/integraldisplay σt(n2)da=/integraldisplay ∂Pt(n)da , (4.57) so, finally, /integraldisplay σt(n1)da+/integraldisplay σt(n2)da=0, (4.58) which can be also written as /integraldisplay σt(n)da+/integraldisplay σt(−n)da=0. (4.59) Since σis an arbitrary surface, assuming that tdepends continuously on nandxalong σ, the localization theorem yields the condition t(n)+t(−n)=0. (4.60) This result is called Cauchy’s lemma ont(n). It states that the stress vectors acting at xon opposite sides of the same smooth surface are equal and oppos ite, namely that t(x, t;n) =−t(x, t;−n). (4.61) It is important to recognize here that in continuum mechanic s Cauchy’s lemma is not a principle. Rather, it is derivable from linear momentum bal ance, as above. This is in stark contrast with particle mechanics, where action-reaction i s a principle, also known as Newton’s Third Law. ME185 DRAFTBasic physical principles 75 At this stage, consider the following problem, originally c onceived by Cauchy: take a tetrahedral region P ⊂ R , such that three edges are parallel to the axes of {ei}and meet at a point x, as in Figure 4.5. Let σibe the face with unit outward normal −ei, andσ0the (inclined) face with outward unit normal n. Denote by Athe area of σ0, so that the area vector ndAcan be resolved as nA= (niei)A=Aniei=Aiei, (4.62) where Ai=Aniis the area of the face σi. In addition, the volume Vof the tetrahedron can be written as V=1 3Ah, where his the distance of xfrom face σ0. 321 Re1 e2e3 x Figure 4.5: The Cauchy tetrahedron. Now apply balance of linear momentum to the tetrahedral regi onPin the form of equa- tion (4.45) and concentrate on the surface integral term, wh ich, in this case, becomes /integraldisplay ∂Pt(n)da=/integraldisplay σ1t(−e1)da+/integraldisplay σ2t(−e2)da+/integraldisplay σ3t(−e3)da+/integraldisplay σ0t(n)da . (4.63) Upon using Cauchy’s lemma, this becomes /integraldisplay ∂Pt(n)da=−/integraldisplay σ1t(e1)da−/integraldisplay σ2t(e2)da−/integraldisplay σ3t(e3)da+/integraldisplay σ0t(n)da . (4.64) Returning now to the balance of linear momentum statement (4 .45), it can be written as /integraldisplay Pρ(a−b)dv=/integraldisplay σ0t(n)da−/integraldisplay σ1t(e1)da−/integraldisplay σ2t(e2)da−/integraldisplay σ3t(e3)da . (4.65) Assuming that ρ,a, andbare bounded, one can obtain an upper-bound estimate for the domain integral on the left-hand side as /vextendsingle/vextendsingle/vextendsingle/vextendsingle/integraldisplay Pρ(a−b)dv/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤/integraldisplay P|ρ(a−b)dv|=/integraldisplay PK(x, t)dv=K∗V=K∗1 3Ah , (4.66) ME185 DRAFT76 Stress vector and stress tensor where K(x, t) =|ρ(a−b)|andK∗=K(x∗, t), with x∗being some interior point of P. The preceding derivation made use of the mean-value theorem for integrals.2 Assuming that t(ei) are continuous in x, apply the mean value theorem for integrals component-wise to get /integraldisplay σit(ei)da=t∗ iAi, (4.67) so that summing up all three like equations 3/summationdisplay i=1/integraldisplay σit(ei)da=t∗ iAi=t∗ iAni. (4.68) Also, for the inclined face /integraldisplay σ0t(n)da=t∗ (n)A . (4.69) In the preceding two equations, t∗ iandt∗ (n)are the traction vectors at some interior points ofσiandσ0. Recalling from (4.65) and (4.66) that /vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/integraldisplay σ0t(n)da−3/summationdisplay i=1/integraldisplay σit(ei)da/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤1 3K∗Ah , (4.70) write /vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/integraldisplay σ0t(n)da−3/summationdisplay i=1/integraldisplay σit(ei)da/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=|t∗A−t∗ iAni|=A|t∗−t∗ ini| ≤1 3K∗Ah , (4.71) which simplifies to |t∗ (n)−t∗ ini| ≤1 3K∗h . (4.72) Now, upon applying the preceding analysis to a sequence of ge ometrically similar tetra- hedra with heights h1> h2> . . ., where lim i→∞hi= 0, one finds that |t∗ (n)−t∗ ini| ≤0, (4.73) where, obviously, all stress vectors are evaluated exactly atx. It follows from (4.73) that at pointx t(n)=tini, (4.74) 2Mean-value Theorem for Integrals: If Phas positive volume (vol( P)>0) and is compact and connected, andfis continuous in ε3, then there exists a point x∗∈ Pfor which/integraldisplay Pf(x)dv=f(x∗)vol(P). ME185 DRAFTBasic physical principles 77 where the superscript has been dropped, since, now, x∗=x. With the preceding development in place, define the tensor Tas T=ti⊗ei, (4.75) so that, when operating on n, Tn= (ti⊗ei)n=ti(ei·n) =tini=t(n), (4.76) as seen from (4.74). The tensor Tis called the Cauchy stress tensor. It can be readily seen from (4.75) that T=Tkiek⊗ei=ti⊗ei, (4.77) hence ti=Tkiek. (4.78) Also, since ti·ej=Tkiek·ej=Tji, (4.79) it is immediately seen that Tij=ei·tj. (4.80) Note that, from its definition in (4.75), it is clear that the C auchy stress tensor does not depend on the normal n. Return now to the integral statement of linear momentum bala nce and, taking into account (4.76), apply the divergence theorem to the boundar y integral term. This leads to /integraldisplay Pρadv=/integraldisplay Pρbdv+/integraldisplay ∂Pt(n)da =/integraldisplay Pρbdv+/integraldisplay ∂PTnda =/integraldisplay Pρbdv+/integraldisplay PdivTdv . (4.81) It follows from the preceding equation that the condition /integraldisplay P(ρa−ρb−divT)dv=0, (4.82) ME185 DRAFT78 Stress vector and stress tensor holds for an arbitrary area P, which, with the aid of the localization theorem leads to a lo cal form of linear momentum balance in the form divT+ρb=ρa. (4.83) An alternative statement of linear momentum balance can be o btained by noting from (4.75) that /integraldisplay Pρadv=/integraldisplay Pρbdv+/integraldisplay ∂Pt(n)da =/integraldisplay Pρbdv+/integraldisplay ∂Ptinida =/integraldisplay Pρbdv+/integraldisplay Pti,idv . (4.84) Again, appealing to the localization theorem, this leads to ti,i+ρb=ρa. (4.85) Turn attention next to the balance of angular momentum and ex amine the boundary integral term in (4.49). This can be written as /integraldisplay ∂Px×t(n)da=/integraldisplay ∂Px×tinida=/integraldisplay P(x×ti),idv=/integraldisplay P(x,i×ti+x×ti,i)dv .(4.86) Substituting the preceding equation into (4.49) yields /integraldisplay P(x×ρa)dv=/integraldisplay P(x×ρb)dv+/integraldisplay P(x,i×ti+x×ti,i)dv (4.87) or, upon rearranging the terms, /integraldisplay P[x×(ρa−ρb−ti,i) +x,i×ti]dv=0. (4.88) Recalling the local form of linear momentum balance equatio n (4.85), the above equation reduces to/integraldisplay Px,i×tidv=0. (4.89) The localization theorem can be invoked again to conclude th at x,i×ti=0 (4.90) ME185 DRAFTBasic physical principles 79 or ei×ti=0. (4.91) In component form, this condition can be expressed as ei×(Tjiej) =Tjiei×ej=Tjieijkek=0, (4.92) which means that Tij=Tji, i.e., the Cauchy stress tensor is symmetric (or, said differ ently, the skew-symmetric tensor T−TTvanishes identically). Hence, angular momentum balance restricts the Cauchy stress tensor to be symmetric. e1e2e3 T11T21T31 T12T13 T22T23 T32T33 Figure 4.6: Interpretation of the Cauchy stress components on an orthog onal parallelepiped aligned with the {ei}-axes. An interpretation of the components of Ton an orthogonal parallelepiped is shown in Figure 4.6. Indeed, recalling (4.78), it follows that t1=T11e1+T21e2+T31e3, (4.93) which means that Ti1is the i-th component of the traction that acts on the plane with outward unit normal e1. More generally, Tijis the i-th component of the traction that acts on the plane with outward unit normal ej. The components Tijof the Cauchy stress tensor can be put in matrix form as [Tij] = T11T12T13 T21T22T23 T31T32T33 , (4.94) ME185 DRAFT80 Stress vector and stress tensor where [ Tij] is symmetric. The linear eigenvalue problem (T−Ti)n=0 (4.95) yields three real eigenvalues T1≥T2≥T3, which are solutions of the characteristic polyno- mial equation T3−ITT2+IITT−IIIT= 0, (4.96) where IT,IITandIIITare the three principal invariants of the tensor T, defined as IT= tr(T) IIT=1 2[(tr(T))2−trT2] (4.97) IIIT=1 6[(tr(T))3−3 trT(trT2) + 2 tr T3] = det T. (4.98) As is well-known, the associated unit eigenvectors n(1),n(2)andn(3)ofTare mutually orthogonal provided the eigenvalues are distinct. Also, wh ether the eigenvalues are distinct or not, there exists a set of mutually orthogonal eigenvecto rs forT. The traction vector t(n)can be generally decomposed into normal and shearing compo- nents on the plane of its action. To this end, the normal traction (i.e., the projection of t(n) alongn) is given by (t(n)·n)n= (n⊗n)t(n), (4.99) as in Figure 4.7. Then, the shearing traction is equal to n t(n)(t(n)·n)n Figure 4.7: Projection of the traction to its normal and tangential comp onents. ME185 DRAFTBasic physical principles 81 t(n)−(t(n)·n)n=t(n)−(n⊗n)t(n)= (i−n⊗n)t(n). (4.100) Ifnis a principal direction of T, equations (4.95) and (4.100) imply that (i−n⊗n)t(n)= (i−n⊗n)Tn= (i−n⊗n)Tn=0, (4.101) namely that the shearing traction vanishes on the plane with unit normal n. Next, consider three special forms of the stress tensor that lead to equilibrium states in the absence of body forces. (a) Hydrostatic pressure In this state, the stress vector is always pointing in the dir ection normal to any plane that it is acting on, i.e., t(n)=−pn, (4.102) where pis called the pressure . It follows from (4.76) that T=−pi. (4.103) (b) Pure tension along the e-axis Without loss of generality, let e=e1. In this case, the traction vectors tiare of the form t1=Te1,t2=t3=0. (4.104) Then, it follows from (4.78) that T=T(e1⊗e1) =T(e⊗e). (4.105) (c) Pure shear on the ( e,k)-plane Here, let eandkbe two orthogonal vectors of unit magnitude and, without los s of generality, set e1=eande2=k. The tractions tiare now given by t1=Te2,t2=Te1,t3=0. (4.106) Appealing, again, to (4.78), it is easily seen that T=T(e1⊗e2+e2⊗e1) =T(e⊗k+k⊗e). (4.107) ME185 DRAFT82 Stress vector and stress tensor It is possible to resolve the stress vector acting on a surfac e of the current configuration using the geometry of the reference configuration. This is co nceivable when, for example, one wishes to measure the internal forces developed in the cu rrent configuration per unit area of the reference configuration. To this end, start by let tingdfbe the total force acting on the differential area dawith outward unit normal non the surface ∂Pin the current configuration, i.e., df=t(n)da . (4.108) Also, let dAbe the image of dain the reference configuration under χ−1 tand assume that its outward unit is N. Then, define p(N)to be the traction vector resulting from resolving the force df, which acts on ∂P, on the surface ∂P0, namely, df=p(N)dA . (4.109) Clearly, tandpare parallel, since they are both parallel to df. Returning to the integral statement of linear momentum bala nce in (4.45), note that this can be now readily “pulled-back” to the reference configurat ion, hence taking the form /integraldisplay P0ρ0adV=/integraldisplay P0ρ0bdV+/integraldisplay ∂P0p(N)dA . (4.110) Upon applying the Cauchy tetrahedron argument to P0, it is immediately concluded that p(N)=pANA, (4.111) where pAare the tractions developed in the current configuration, bu t resolved on the geometry of the reference configuration on surfaces with out ward unit normals EA. Further, let the tensor Pbe defined as P=pA⊗EA, (4.112) so that, when operating on N, PN= (pA⊗EA)N=pANA, (4.113) thus, from (4.111), it is seen that p=PN. (4.114) ME185 DRAFTBasic physical principles 83 The tensor Pis called the first Piola-Kirchhoff stress tensor and it is naturally unsymmetric, since it has a mixed basis, i.e., P=PiAei⊗EA. (4.115) It follows from (4.112) that P=pA⊗EA=PiAei⊗EA, (4.116) which implies that pA=PiAei. (4.117) Further, since pA·ej=PiAei·ej=PjA, (4.118) it is clear that PiA=ei·pA. (4.119) Turning attention to the integral statement (4.110), it is c oncluded with the aid of (4.114) and the divergence theorem that /integraldisplay P0ρ0adV=/integraldisplay P0ρ0bdV+/integraldisplay ∂P0p(N)dA =/integraldisplay P0ρ0bdV+/integraldisplay ∂P0PNdA =/integraldisplay P0ρ0bdV+/integraldisplay P0DivPdA (4.120) which, upon using the localization theorem, results in ρ0a=ρ0b+ DivP, (4.121) This is the local form of linear momentum balance in the refer ential description. Alternatively, equation (4.113) and the divergence theore m can be invoked to show that /integraldisplay P0ρ0adV=/integraldisplay P0ρ0bdV+/integraldisplay ∂P0p(N)dA =/integraldisplay P0ρ0bdV+/integraldisplay ∂P0pANAdA =/integraldisplay P0ρ0bdV+/integraldisplay P0pA,AdA (4.122) ME185 DRAFT84 Stress vector and stress tensor from which another version of the referential statement of l inear momentum balance can be derived in the form ρ0a=ρ0b+pA,A, (4.123) Starting from the integral form of angular momentum balance in (4.49) and pulling it back to the reference configuration, one finds that /integraldisplay P0x×ρ0adV=/integraldisplay P0x×ρ0bdV+/integraldisplay ∂P0x×p(N)dA . (4.124) Using (4.113) and the divergence theorem on the boundary ter m gives rise to /integraldisplay P0x×ρ0adV=/integraldisplay P0x×ρ0bdV+/integraldisplay ∂P0x×pANAdA =/integraldisplay P0x×ρ0bdV+/integraldisplay P0(x×pA),AdV . (4.125) Expanding and appropriately rearranging the terms of the ab ove equation leads to /integraldisplay P0[x×(ρ0a−ρ0b−PA,A) +x,A×pA]dV=0. (4.126) Appealing to the local form of linear momentum balance in (4. 123) and, subsequently, the localization theorem, one concludes that x,A×pA=0. (4.127) With the aid of (4.117) and the chain rule, the preceding equa tion can be rewritten as x,A×pA=FiA×PjAei×ej=FiAPjAeijkek=0, (4.128) which implies that FPT=PFT. This is a local form of angular momentum balance in the referential description. Recalling (4.108) and (4.108), one may conclude with the aid of (4.76), (4.114) and Nanson’s formula (3.122) that PNdA=Tnda=TJF−TNdA , (4.129) so that T=1 JPFT. (4.130) ME185 DRAFTBasic physical principles 85 Clearly, the above relation is consistent with the referent ial and spatial statements of an- gular momentum balance, namely (4.130) can be used to derive the local form of angular momentum balance in spatial form from the referential state ment and vice-versa. Likewise, it is possible to derive the local linear momentum balance st atement in the referential (resp. spatial) form from its corresponding spatial (resp. refere ntial) counterpart. Note that there is no approximation or any other source of err or associated with the use the balance laws in the referential vs. spatial descript ion. Indeed, the invertibility of the motion at any fixed time implies that both forms of the bala nce laws are completely equivalent. Other stress tensors beyond the Cauchy and first Piola-Kirch hoff tensors are frequently used. Among them, the most important are the nominal stress tensor Πdefined as the transpose of the first Piola-Kirchhoff stress, i.e., Π=PT=JF−1T, (4.131) theKirchhoff stress tensor τ, defined as τ=JT=PFT, (4.132) and the second Piola-Kirchhoff stress S, defined as S=F−1P=JF−1TF−T. (4.133) It is clear from (4.133) that Shas both “legs” in the reference configuration and is symmetr ic. All the stress measures defined here are obtainable from one a nother assuming that the deformation is known. 4.8 The transformation of mechanical fields under su- perposed rigid-body motions In this section, the transformation under superposed rigid motions is considered for mechan- ical fields, such as the stress vectors and tensors. To this en d, start with the stress vector t=t(x, t;n), and recalling the general form of the superposed rigid mot ion in (3.166), write ME185 DRAFT86 Mechanical fields under superposed rigid motions t+=t+(x+, t;n+). To argue how tandt+are related, first recall that n+=Qnand that tis linear in n, as established in (4.76). Since the two motions give rise to the same defor- mation, it is then reasonable to assume that, under a superposed rigid motion, t+will not change in magnitude relative to tand will have the same orientation relative to n+asthas relative to n. Therefore, it is postulated that t+=Qt. (4.134) The above transformation indeed implies that |t+|=|t|andt+·n+=t·n. Unlike the transformation of kinematic terms, which is gove rned purely by geometry, in the case of kinetic terms (both mechanical and thermal) the t ransformation is governed by the principle of invariance under superposed rigid motion . This, effectively, states that the balance laws are invariant under superposed rigid motions, in the sense that their mathemat- ical representation remains unchanged under such motions. To demonstrate this principle, consider the transformation of the Cauchy stress tensor. Ta king taking into account (4.76), invariance under superposed rigid motions implies that t+=T+n+. This means that if two motions differ by a mere rigid motion, then the relation betwe en the stress vector ton a plane with an outward unit normal nand the Cauchy stress tensor Tis the same with the relation of the stress vector t+on a plane with outward unit normal n+with the Cauchy stress tensor T+. Invariance of (4.76) under superposed rigid motions in con junction with (4.134) implies that t+=Qt=QTn =T+n+=T+Qn,(4.135) from where it is concluded that (QT−T+Q)n=0, (4.136) hence T+=QTQT. (4.137) Equation (4.137) implies that Tis an objective spatial tensor. Recall next the relation between the Cauchy and the first Piol a-Kirchhoff stress tensor in (4.130). Given that this relation is itself invariant under superposed rigid motions it follows ME185 DRAFTBasic physical principles 87 that P+=J+T+(F−T)+=J(QTQT)(QF−T) =Q(JTF−T) =QP, (4.138) where the kinematic transformations (3.165) and (3.187) ar e employed.3Equation (4.138) implies that Pis an objective two-point tensor. Similarly, turning to the second Piola- Kirchhoff stress tensor, it follows from (4.133) that S+=J+(F−1)+T+(F−T)+=J(F−1QT)(QTQT)(QF−T) =JF−1TF−T=S,(4.139) which implies that Sis an objective referential tensor. Since (4.139) holds true, it follows also that ˙S+=˙S, (4.140) namely that the rate of Sis objective. However, from (4.137) and the relation (3.180 ) it can be seen that ˙T+=˙QTQT+Q˙TQT+QT˙QT = (ΩQ)TQT+Q˙TQT+QT(ΩQ)T =Ω(QTQT) +Q˙TQT+ (QTQ )TΩT =ΩT++Q˙TQT−T+Ω, (4.141) which shows that, unlike T, the spatial tensor ˙Tis not objective. A similar conclusion may be drawn for the rate ˙Pof the first Piola-Kirchhoff stress tensor. Invariance under superposed rigid motions applies to the pr inciple of mass conservation. Indeed, using the local referential form (4.41) of this prin ciple gives rise to ρ0=ρ+J+=ρ+J =ρJ ,(4.142) which imply that ρ+=ρ . (4.143) 3Also, note that, by its definition, ( F−1)+= (F+)−1. ME185 DRAFT88 Mechanical energy balance theorem Finally, admitting invariance under superposed rigid moti ons of the local spatial form (4.83) of linear momentum balance leads to div+T++ρ+b+=ρ+a+. (4.144) Resorting to components, note that ∂T+ ij ∂x+ j=∂(QikTklQjl) ∂xm∂χm ∂x+ j =Qik∂Tkl ∂xmQjlQjm =Qik∂Tkl ∂xmδlm =Qik∂Tkl ∂xl, (4.145) where it is recognized from (3.166) that∂χ ∂x+=QT, therefore, in components,∂xm ∂x+ j=Qjm. Equation (4.145) can be written using direct notation as div+T+=QdivT. (4.146) Using (4.83), (4.144) and (4.146), one concludes that div+T+=ρ+(a+−b+) =ρ(a+−b+) =QdivT=Qρ(a−b), from where it follows that a+−b+=Q(a−b). (4.147) This means that, under superposed rigid motions, the body fo rces transform as b+=Qb+a+−Qa. (4.148) 4.9 The Theorem of Mechanical Energy Balance Consider again the body Bin the current configuration Rand take an arbitrary material region Pwith smooth boundary ∂P. ME185 DRAFTBasic physical principles 89 Recalling the integral statement of linear momentum balanc e (4.43), define the rate at which the external body forces band surface tractions tdo work in Pand on ∂P, respectively, as Rb(P) =/integraldisplay Pρb·vdv (4.149) and Rc(P) =/integraldisplay ∂Pt·vda , (4.150) respectively. Also, define the rate of work done by all extern al forces as R(P) =Rb(P) +Rc(P). (4.151) In addition, define the total kinetic energy of the material p oints contained in Pas K(P) =/integraldisplay P1 2v·vρ dv . (4.152) Starting from the local spatial form of linear momentum bala nce (4.83), one may dot both sides with the velocity vto obtain ρ˙v·v=ρb·v+ divT·v. (4.153) Now, note that (divT)·v= div( TTv)−T·gradv = div( Tv)−T·(D+W) = div( Tv)−T·D, (4.154) where angular momentum balance is invoked in the form of the s ymmetry of T. Equation (4.154) can be used to rewrite (4.153) as 1 2ρd dt(v·v) =ρb·v+ div(Tv)−T·D. (4.155) Integrating (4.155) over Pleads to /integraldisplay P1 2ρd dt(v·v)dv=/integraldisplay Pρb·vdv+/integraldisplay Pdiv(Tv)dv−/integraldisplay PT·Ddv . (4.156) or, upon using conservation of mass and the divergence theor em, d dt/integraldisplay P1 2ρ(v·v)dv+/integraldisplay PT·Ddv=/integraldisplay Pρb·vdv+/integraldisplay ∂P(Tv)·nda . (4.157) ME185 DRAFT90 Mechanical energy balance theorem Recalling (4.76), the preceding equation can be rewritten a s d dt/integraldisplay P1 2ρ(v·v)dv+/integraldisplay PT·Ddv=/integraldisplay Pρb·vdv+/integraldisplay ∂Pt·vda . (4.158) The second term on the right-hand side of (4.158), S(P) =/integraldisplay PT·Ddv , (4.159) is called the stress power and it represents the rate at which the stresses do work in P. Taking into account (4.149), (4.150), (4.152), (4.159) and (4.158), it is seen that d dtK(P) +S(P) =Rb(P) +Rc(P) =R(P). (4.160) Equation (4.160) (or, equivalently, equation (4.158)) sta tes that, for any region P, the rate of change of the kinetic energy and the stress power are balan ced by the rate of work done by the external forces. This is a statement of the mechanical energy balance theorem . It is important to emphasize that mechanical energy balance is derivable from the three basic principles of the mechanical theory, namely conservation o f mass and balance of linear and angular momentum. Returning to the stress power term S(P), note that /integraldisplay PT·Ddv=/integraldisplay PT·Ldv =/integraldisplay P/parenleftbigg1 JPFT/parenrightbigg ·Ldv =/integraldisplay P0(PFT)·LdV =/integraldisplay P0tr[(PFT)LT]dV =/integraldisplay P0tr[P(LF)T]dV =/integraldisplay P0tr[P˙FT]dV =/integraldisplay P0P·˙FdV , (4.161) ME185 DRAFTBasic physical principles 91 where use is made of (4.130) and (3.128). Further, /integraldisplay PT·Ddv=/integraldisplay P0P·˙FdV =/integraldisplay P0(FS)·˙FdV =/integraldisplay P0S·(FT˙F)dV =/integraldisplay P0S·1 2(˙FTF+FT˙F)dV =/integraldisplay P0S·˙EdV , (4.162) where, this time use is made of (4.133) and the definition of th e Lagrangian strain. Equations (4.161) and (4.162) reveal that Pis the work-conjugate kinetic measure to F inP0and, likewise, Sis work-conjugate to E. These equations appear to leave open the question of work-conjugacy for T. 4.10 The principle of energy balance The physical principles postulated up to this point are inca pable of modeling the intercon- vertibility of mechanical work and heat. In order to account for this class of (generally coupled) thermomechanical phenomena, one needs to introdu ce an additional primitive law known as balance of energy . Preliminary to stating the balance of energy, define a scalar fieldr=r(x, t) called the heat supply per unit mass (or specific heat supply ), which quantifies the rate at which heat is supplied (or absorbed) by the body. Also, define a scalar fie ldh=h(x, t;n) =h(n)(x, t) called the heat flux per unit area across a surface ∂Pwith outward unit normal n. Now, given any region P ⊆ R , define the total rate of heating H(P) as H(P) =/integraldisplay Pρr dv−/integraldisplay ∂Ph da , (4.163) where the negative sign in front of the boundary integral sig nifies the fact that the heat flux is assumed positive when it exits the region P. Next, assume the existence of a scalar function ε=ε(x, t) per unit mass, called the (specific) internal energy . This function quantifies all forms of energy stored in the bo dy ME185 DRAFT92 The principle of energy balance other than the kinetic energy. Examples of stored energy inc lude strain energy (i.e., energy due to deformation), chemical energy, etc. The internal ene rgy stored in Pis denoted by U(P) =/integraldisplay Pρε dv . (4.164) The principle of balance of energy is postulated in the form d dt/integraldisplay P/bracketleftbigg1 2ρv·v+ρε/bracketrightbigg dv=/integraldisplay Pρb·vdv+/integraldisplay ∂Pt·vda+/integraldisplay Pρr dv−/integraldisplay ∂Ph da . (4.165) This is also sometimes referred to as a statement of the “Firs t Law of Thermodynamics”. Equivalently, equation (4.165) can be written as d dt[K(P) +U(P)] = R(P) +H(P). (4.166) Subtracting (4.158) from (4.165) leads to a statement of thermal energy balance in the form d dt/integraldisplay Pρε dv =/integraldisplay PT·Ddv+/integraldisplay Pρr dv−/integraldisplay ∂Ph da . (4.167) Returning to the heat flux h=h(x, t;n), and note that one can apply a standard ar- gument to find the exact dependence of honn, as already done with the stress vector t=t(x, t;n). Indeed, one may apply thermal energy balance to a region Pwith boundary ∂Pand to each of two regions P1andP2with boundaries ∂P1and∂P2, where P1∪ P2=P andP1∪ P2=∅. Also, the boundaries ∂P1=∂P′∪σ,∂P2=∂P′′∪σhave a common surface σand∂P′∪∂P′′=∂P. It follows that d dt/integraldisplay Pρε dv =/integraldisplay PT·Ddv+/integraldisplay Pρr dv−/integraldisplay ∂Ph da . (4.168) and, also, d dt/integraldisplay P1ρε dv =/integraldisplay P1T·Ddv+/integraldisplay P1ρr dv−/integraldisplay ∂P1h da (4.169) and d dt/integraldisplay P2ρε dv =/integraldisplay P2T·Ddv+/integraldisplay P2ρr dv−/integraldisplay ∂P2h da (4.170) Adding the last two equations leads to d dt/integraldisplay P1∪P2ρε dv =/integraldisplay P1∪P2T·Ddv+/integraldisplay P1∪P2ρr dv−/integraldisplay ∂P1∪∂P2h da (4.171) ME185 DRAFTBasic physical principles 93 or, equivalently, d dt/integraldisplay Pρε dv =/integraldisplay PT·Ddv+/integraldisplay Pρr dv−/integraldisplay ∂P1∪∂P2h da . (4.172) Subtracting (4.168) from (4.172) results in /integraldisplay ∂P1∪∂P2h da−/integraldisplay ∂Ph da= 0, (4.173) or, equivalently,/integraldisplay ∂P′∪σh da+/integraldisplay ∂P′′∪σh da=/integraldisplay ∂Ph da . (4.174) As in the case of the stress vector, the preceding equation ma y be expanded to /integraldisplay ∂P′∪∂P′′h da+/integraldisplay σh(n1)da+/integraldisplay σh(n2)da=/integraldisplay ∂Ph da (4.175) or/integraldisplay σ(h(n)−h(−n))da= 0, (4.176) where n1=nandn2=−n. Since σis an arbitrary surface and his assumed to depend continuously on nandxalong σ, the localization theorem yields the condition h(n) =−h(−n). (4.177) or, more explicitly, h(x, t;n) =−h(x, t;−n). (4.178) This states that the flux of heat exiting a body across a surfac e with outward unit normal nat a point xis equal to the flux of heat entering a neighboring body at the s ame point across the same surface. Using the tetrahedron argument in connection with the therm al energy balance equation (4.167) and the flux continuity equation (4.178), gives rise to h=hini (4.179) where hiare the fluxes across the faces of the tetrahedron with outwar d unit normals ei. Thus, one may write h=q·n, (4.180) ME185 DRAFT94 The principle of energy balance where qis the heat flux vector with components qi=hi. Now, returning to the integral statement of energy balance i n (4.165), use mass conser- vation to rewrite it as /integraldisplay P(ρv·˙v+ρ˙ε)dv=/integraldisplay Pρb·vdv+/integraldisplay ∂Pt·vda+/integraldisplay Pρr dv−/integraldisplay ∂Ph da . (4.181) Using (4.76) and (4.180), the above equation may be put in the form /integraldisplay P(ρv·˙v+ρ˙ε)dv=/integraldisplay Pρb·vdv+/integraldisplay ∂P(Tn)·vda+/integraldisplay Pρr dv−/integraldisplay ∂Pq·nda . (4.182) Upon invoking the divergence theorem, it is easily seen that /integraldisplay ∂P(Tn)·vda=/integraldisplay P[(divT)·v+T·D]dv (4.183) and/integraldisplay ∂Pq·nda=/integraldisplay Pdivqdv . (4.184) If the last two equations are substituted in (4.182), one find s that /integraldisplay P/bracketleftbig {ρ˙v−ρb−divT} ·v+ρ˙ε−T·D−ρr+ divq/bracketrightbig dv=0. (4.185) Upon observing linear momentum balance and invoking the loc alization theorem, the pre- ceding equation gives rise to the local form of energy balanc e as ρ˙ε=T·D+ρr−divq. (4.186) This equation could be also derived along the same lines from the integral statement of thermal energy balance (4.167).4 Regarding invariance under superposed rigid-body motions , it is postulated that ε+=ε (4.187) and r+=r , q+=Qq. (4.188) 4The energy equation is frequently quoted in elementary ther modynamics texts as “ dU=δQ+δW”, where dUcorresponds to ρ˙ε,δQtoρr−divq, and δWtoT·D. ME185 DRAFTBasic physical principles 95 Equation (4.188) 2and invariance of the thermal energy balance imply that h+=q+·n+= (Qq)·(Qn) =q·n=h . (4.189) Example: Consider a rigid heat conductor, where rigidity implies tha tF=R(i.e.,U=I). This means that ˙F=˙R=ΩR (Ω=˙RRT) =LF=LR,(4.190) which implies that L=Ω, so that D=0. Further, assume that Fourier’s law holds, namely that q=−kgradT , (4.191) where Tis the empirical temperature and k >0 is the (isotropic) conductivity. These conditions imply that the balance of energy (4.186) reduces to ρ˙ε−div(kgradT)−ρr= 0. (4.192) Further, assume that the internal energy depends exclusive ly onTand that this dependence is linear, hencedε dt=c, where cis termed the heat capacity . It follows from (4.192) that ρc˙T= div( kgradT) +ρr , (4.193) which is the classical equation of transient heat conductio n. /square 4.11 The Green-Naghdi-Rivlin theorem Assume that the principle of energy balance remains invaria nt under superposed rigid mo- tions. With reference to (4.165), this means that d dt/integraldisplay P+/bracketleftbig ρ+ε++1 2ρ+v+·v+/bracketrightbig dv+ =/integraldisplay P+ρ+b+·v+dv++/integraldisplay ∂P+t+·v+da++/integraldisplay P+ρ+r+dv+−/integraldisplay ∂P+h+da+.(4.194) Now, choose a special superposed rigid motion, which is a pur erigid translation , such that Q=I, ˙Q=0 ,c=c0t , (4.195) ME185 DRAFT96 The Green-Naghdi-Rivlin theorem where c0is a constant vector in E3. It follows immediately from (3.178) and (4.195) that v+=v+c0, ˙v+=˙v. (4.196) Moreover, it is readily concluded from (4.148), (4.134), (4 .195) and (4.196) that under this superposed rigid translation b+=b ,t+=t. (4.197) It follows from (4.143), (4.196) and (4.197) that (4.194) ta kes the form d dt/integraldisplay P/bracketleftbig ρε+1 2ρ(v+c0)·(v+c0)/bracketrightbig dv =/integraldisplay Pρb·(v+c0)dv+/integraldisplay ∂Pt·(v+c0)da+/integraldisplay Pρr dv−/integraldisplay ∂Ph da . (4.198) Upon subtracting (4.165) from (4.198), it is concluded that c0·/bracketleftBigd dt/integraldisplay Pρvdv−/integraldisplay Pρbdv−/integraldisplay ∂Ptda/bracketrightbig +1 2(c0·c0)/bracketleftBigd dt/integraldisplay Pρ dv/bracketrightBig = 0. (4.199) Sincec0is an arbitrary constant vector, one may set c0=γ¯c0, where γis an arbitrary non-vanishing scalar constant. Substituting γwith−γin (4.199), it is proved that d dt/integraldisplay Pρ dv= 0 (4.200) hence, also, d dt/integraldisplay Pρvdv=/integraldisplay Pρbdv+/integraldisplay ∂Ptda . (4.201) This, in turn, means that conservation of mass and balance of linear momentum hold, re- spectively. Next, a second special superposed rigid-body motion is chos en, such that Q=I, ˙Q=Ω0,c=0, (4.202) where Ω0is a constant skew-symmetric tensor. Given (4.202), it can b e easily seen from (3.166) and (3.178) that v+=v+Ω0x=v+ω0×x (4.203) and ˙v+=˙v+ 2Ω0v+Ω2 0x=˙v+ω0×(2v+ω0×x), (4.204) ME185 DRAFTBasic physical principles 97 where ω0is the (constant) axial vector of Ω0. Equations (4.203) and (4.204) imply that the superposed motion is a rigid rotation with constant angular velocity ω0on the original current configuration of the continuum. Taking into account (4.134) (4.148), (4.202) and (4.204), it is established that in this case b+=b+ 2Ω0v+Ω2 0x=b+ω0×(2v+ω0×x) (4.205) and t+=t. (4.206) In addition, equations (4.203) 1and (4.205) 1lead to v+·v+=v·v+ 2(Ω0x)·v+ (Ω0x)·(Ω0x) (4.207) and b+·v+=b·v+b·(Ωx) + 2(Ω0v)·(Ωx) + 2(Ω0v)·v + (Ω2 0x)·(Ω0x) + (Ω2 0x)·v =b·v+b·(Ωx) + (Ω0v)·(Ωx) + (Ω2 0x)·(Ω0x), (4.208) where the readily verifiable identities (Ω0v)·v= 0 ,(Ω0v)·(Ω0x) + (Ω2 0x)·v= 0 (4.209) are employed. Similarly, using (4.203) 1and (4.205) 1, it is seen that 1 2d dt(v+·v+) = ˙v·v+˙v·(Ω0x) + 2(Ω0v)·(Ω0x) + 2(Ω0v)·v + (Ω2 0x)·(Ω0x) + (Ω2 0x)·v =˙v·v+˙v·(Ω0x) + (Ω0v)·(Ω0x) + (Ω2 0x)·(Ω0x). (4.210) Invoking now invariance of the energy equation under the sup erposed rigid rotation, it can be concluded from (4.194), as well as (4.207) (4.208) and (4.210), that d dt/integraldisplay Pρε dv+/integraldisplay Pρ/bracketleftbig ˙v·v+˙v·(Ω0x) + (Ω0v)·(Ω0x) + (Ω2 0x)·(Ω0x)/bracketrightbig dv =/integraldisplay Pρ/bracketleftbig b·v+b·(Ω0x) + (Ω0v)·(Ω0x) + (Ω2 0x)·(Ω0x)/bracketrightbig dv +/integraldisplay ∂Pt·(v+Ω0x)da+/integraldisplay Pρr dv−/integraldisplay ∂Ph da . (4.211) ME185 DRAFT98 The Green-Naghdi-Rivlin theorem After subtracting (4.165) from (4.211) and simplifying the resulting equation, it follows that /integraldisplay Pρ˙v·(Ω0x)dv=/integraldisplay Pρb·(Ω0x)dv+/integraldisplay ∂Pt·(Ω0x)da . (4.212) Observing that for any given vector zinE3 z·(Ω0x) =z·(ω0×x) =ω0·(x×z), (4.213) and recalling that ω0is a constant vector, equation (4.212) takes the form ω0·/bracketleftBig/integraldisplay Pρx×˙vdv−/integraldisplay Pρx×bdv−/integraldisplay ∂Px×tda/bracketrightBig . (4.214) This, in turn, implies that d dt/integraldisplay Pρx×vdv=/integraldisplay Pρx×bdv+/integraldisplay ∂Px×tda , (4.215) namely that balance of angular momentum holds. The preceding analysis shows that the integral forms of cons ervation of mass, and balance of linear and angular momentum are directly deduced from the integral form of energy bal- ance and the postulate of invariance under superposed rigid -body motions. This remarkable result is referred to as the Green-Naghdi-Rivlin theorem. The Green-Naghdi-Rivlin theorem can be viewed as an implica tion of the general covari- ance principle proposed by Einstein. According to this prin ciple, all physical laws should be invariant under any smooth time-dependent coordinate tr ansformation (including, as a special case, rigid time-dependent transformations). Thi s far-reaching principle stems from Einstein’s conviction that physical laws are oblivious to s pecific coordinate systems, hence should be expressed in a covariant manner, i.e., without bei ng restricted by specific choices of coordinate systems. In covariant theories, the energy eq uation plays a central role, as demonstrated by the Green-Naghdi-Rivlin theorem. ME185 DRAFTChapter 5 Infinitesimal deformations The development of kinematics and kinetics presented up to t his point does not require any assumptions on the magnitude of the various measures of defo rmation. In many realistic circumstances, solids and fluids may undergo “small” (or “in finitesimal”) deformations. In these cases, the mathematical representation of kinematic quantities and the associated kinetic quantities, as well as the balance laws, may be simpl ified substantially. In this chapter, the special case of infinitesimal deformati ons is discussed in detail. Pre- liminary to this discussion, it is instructive to formally d efine the meaning of “small” or “infinitesimal” changes of a function. To this end, consider a scalar-valued real function f=f(x), which is assumed to possess continuous derivatives up to a ny desirable order. To analyze this function in the neighborhood of x=x0, one may use a Taylor series expansion atx0in the form f(x0+v) =f(x0) +vf′(x0) +v2 2!f′′(x0) +v3 3!f′′′(x0) +. . . =f(x0) +vf′(x0) +v2 2!f′′(¯x),(5.1) where vis a change to the value of x0and ¯x∈(x0, x0+v). Denoting by εthe magnitude of the difference between x0+vandx0, i.e., ε=|v|, it follows that as ε→0 (i.e., as v→0), the scalar f(x0+v) is satisfactorily approximated by the linear part of the Ta ylor series expansions in (5.1), namely f(x0+v).=f(x0) +vf′(x0). (5.2) 99 DRAFT100 The Gˆ ateaux differential Recalling the expansion (5.1) 2, one says that ε=|v|is “small”, when the termv2 2!f′′(¯x) can be neglected in this expansion without appreciable erro r, i.e. when /vextendsingle/vextendsingle/vextendsingle/vextendsinglev2 2!f′′(¯x)/vextendsingle/vextendsingle/vextendsingle/vextendsingle≪ |f(x0+v)|, (5.3) assuming that f(x0+v)/ne}ationslash= 0. 5.1 The Gˆ ateaux differential Linear expansions of the form (5.2) can be readily obtained f or vector or tensor functions using the Gˆ ateaux differential . In particular, given F=F(X), where Fis a scalar, vector or tensor function of a scalar, vector or tensor variable X, the Gˆ ateaux differential DF(X0,V) ofFatX=X0in the direction Vis defined as DF(X0,V) =/bracketleftbiggd dωF(X0+ωV)/bracketrightbigg w=0, (5.4) where ωis a scalar. Then, F(X0+V) =F(X0) +DF(X0,V) +o(|V|2), (5.5) such that lim |V|→0o(|V|2) |V|= 0. (5.6) The linear part L[F;V]X0ofFatX0in the direction Vis then defined as L[F;V]X0=F(X0) +DF(X0,V). (5.7) Examples: (1) Let F(X) =f(x) =x2. Using the definition in (5.4), Df(x0, v) =/bracketleftbiggd dωf(x0+ωv)/bracketrightbigg ω=0 =/bracketleftbiggd dω(x0+ωv)2/bracketrightbigg ω=0 =/bracketleftbiggd dω(x2 0+ 2x0ωv+ω2v2)/bracketrightbigg ω=0 =/bracketleftbig 2x0v+ 2ωv2/bracketrightbig ω=0 = 2x0v . ME185 DRAFTInfinitesimal deformations 101 Hence, L[f;v]x0=x2 0+ 2x0v . /square (2) Let F(X) =T(x) =x⊗x. Using, again, the definition in (5.4), DT(x0,v) =/bracketleftbiggd dωT(x0+ωv)/bracketrightbigg ω=0 =/bracketleftbiggd dω{(x0+ωv)⊗(x0+ωv)}/bracketrightbigg ω=0 =/bracketleftbiggd dω{x0⊗x0+ω(x0⊗v+v⊗x0) +ω2v⊗v}/bracketrightbigg ω=0 = [(x0⊗v+v⊗x0) + 2ωv⊗v]ω=0 =x0⊗v+v⊗x0. It follows that L[T;v]x0=x0⊗x0+x0⊗v+v⊗x0. /square 5.2 Consistent linearization of kinematic and kinetic variables Start by noting that the position vector xof a material point Pin the current configuration can be written as the sum of the position vector Xof the same point in the reference config- uration plus the displacement uof the point from the reference to the current configuration, namely x=X+u. (5.8) As usual, the displacement vector field can be expressed equi valently in referential or spatial form as u=ˆu(X, t) =˜u(x, t). (5.9) ME185 DRAFT102 Consistent linearization of kinematic and kinetic variabl es It follows that the deformation gradient can be written as F=∂χ ∂X=∂(X+ˆu) ∂X=I+∂ˆu ∂X =I+H, (5.10) where His the relative displacement gradient tensor defined by H=∂ˆu ∂X. (5.11) Clearly, Hquantifies the deviation of Ffrom the identity tensor. Recalling the discussion in Section 5.1, a linearized count erpart of a given kinematic measure is obtained by first expressing the kinematic measur e as a function of H, i.e., as F(H) and, then, by expanding F(H) around the reference configuration, where H=0. This leads to F(H) =F(0) +DF(0,H) +o(/ba∇dblH/ba∇dbl2), (5.12) where /ba∇dblH/ba∇dblis the two-norm ofHdefined at a point Xand time tas/ba∇dblH/ba∇dbl= (H·H)1/2. Taking into account (5.7) and (5.12), the linearized counte rpart of F(H) is the linear part L(F;H)0ofFaround the reference configuration in the direction of H, given by L(F;H)0=F(0) +DF(0,H). (5.13) Here, a suitable scalar measure of magnitude for the deviati on ofFfrom identity can be defined as ε=ε(t) = sup X∈R0||H(X, t)||, (5.14) where “sup” denotes the least upper bound over all points Xin the reference configuration. Now one may say that the deformations are “small” (of “infinit esimal”) at a given time tif εis small enough so that the term o(||H||2) can be neglected when compared with F(H). Now, proceed to obtain infinitesimal counterparts of some st andard kinematic fields, starting with the deformation gradient F. To this end, first recall that F=¯F(H) =I+H. Then, note that the Gˆ ateaux differential DF(0,H) =/bracketleftbiggd dω¯F(0+ωH)/bracketrightbigg ω=0 =/bracketleftbiggd dω(I+ωH)/bracketrightbigg ω=0 =H. (5.15) ME185 DRAFTInfinitesimal deformations 103 Hence, L[F;H]0=¯F(0) +DF(0,H) =I+H. (5.16) Effectively, (5.16) shows that the linear part of FisFitself, which should be obvious from (5.10). Next, starting from FF−1=I, take the linear part of both sides in the direction of H. This leads to L[FF−1;H]0=¯F(0)¯F−1(0) +D(FF−1)(0,H) =L[I;H]0=I, (5.17) where D(FF−1)(0,H) =DF(0,H)¯F−1(0) +¯F(0)DF−1(0,H) =H+DF−1(0,H) =0.(5.18) The preceding equation implies that DF−1(0,H) =−H. (5.19) Hence, the linear part of F−1atH=0in the direction His L[F−1;H]0=I−H. (5.20) Next, consider the linear part of the spatial displacement g radient grad u. First, observe that, using the chain rule, one finds that gradu= (Grad u)F−1= (F−I)F−1=I−F−1, (5.21) therefore gradu=gradu(H) =I−(I+H)−1. (5.22) This means that, taking into account (5.19), D(gradu)(0,H) =−DF−1(0,H) =H. (5.23) As a result, L[gradu;H]0=gradu(0) +D(gradu)(0,H) =0+H=H. (5.24) ME185 DRAFT104 Consistent linearization of kinematic and kinetic variabl es The last result shows that the linear part of the spatial disp lacement gradient grad ucoincides with the referential displacement gradient Grad u(=H). This, in turn, implies that, within the context of infinitesimal deformations, there is no differ ence between the partial derivatives of the displacement uwith respect to Xorx. This further implies that the distinction between spatial and referential description of kinematic q uantities becomes immaterial in the case of infinitesimal deformations. To determine the linear part of the right Cauchy-Green defor mation tensor C, write C=¯C(H) = (I+H)T(I+H) =I+H+HT+HTH. (5.25) Then, DC(0,H) =/bracketleftbiggd dw¯C(0+wH)/bracketrightbigg w=0 =/bracketleftbiggd dw/braceleftbig I+w(H+HT) +w2HTH/bracerightbig/bracketrightbigg w=0 =/bracketleftbig H+HT+ 2wHTH/bracketrightbig w=0 =H+HT. (5.26) Consequently, the linear part of CatH=0in the direction His L[C;H]0=¯C(0) +DC(0,H) =I+ (H+HT). (5.27) Using (5.27), it can be immediately concluded that the linea r part of the Lagrangian strain tensor Eis L[E;H]0=1 2(H+HT). (5.28) At the same time, the Eulerian strain tensor ecan be written as e=¯e(H) =1 2/parenleftbig i−¯F−T(H)¯F−1(H)/parenrightbig , (5.29) hence its Gˆ ateaux differential is given, with the aid of (5.1 9), by De(0,H) =−1 2/parenleftbig DF−T(0,H) +DF−1(0,H)/parenrightbig =1 2(H+HT). (5.30) This means that the linear part of eis equal to L[e;H]0=¯e(0) +De(0,H) =1 2(H+HT). (5.31) ME185 DRAFTInfinitesimal deformations 105 It is clear from (5.28) and (5.31) that the linear parts of the Lagrangian and Eulerian strain tensors coincide. Hence, under the assumption of infinitesi mal deformations, the distinction between the two strains ceases to exist and one writes that L[E;H]0=L[e;H]0=ε, (5.32) where εis the classical infinitesimal strain tensor , with components εij=1 2(ui,j+uj,i). Proceed next with the linearization of the right stretch ten sorU. To this end, write U2=C=I+H+HT+HTH, (5.33) so that, with the aid of (5.26), DU2(0,H) =DU(0,H)¯U(0) +¯U(0)DU(0,H) = 2 DU(0,H) =H+HT,(5.34) hence, L[U;H]0=¯U(0) +DU(0,H) =I+1 2(H+HT). (5.35) Repeating the procedure used earlier in this section to dete rmine the Gˆ ateaux differential ofF−1, one easily finds that DU−1(0,H) =−1 2(H+HT) (5.36) and L/bracketleftbig U−1;H/bracketrightbig 0=I−1 2(H+HT). (5.37) It is now possible to determine the linear part of the rotatio n tensor R, written as R=¯R(H) = ¯F(H)¯U−1(H), (5.38) by first obtaining the Gˆ ateaux differential as DR(0,H) =DF(0,H)¯U−1(0) +¯F(0)DU−1(0,H) =H−1 2(H+HT) =1 2(H−HT), (5.39) and then writing L[R;H]0=¯R(0) +DR(0,H) =I+1 2(H−HT). (5.40) ME185 DRAFT106 Consistent linearization of kinematic and kinetic variabl es When His small, the tensor ω=1 2(H−HT) (5.41) is called the infinitesimal rotation tensor and has components ωij=1 2(ui,j−uj,i). Next, derive the linear part of the Jacobian Jof the deformation gradient. To this end, observe that D(detF)(0,H) =/bracketleftbiggd dωdet¯F(ωH)/bracketrightbigg ω=0 =/bracketleftbiggd dωdet(I+ωH)/bracketrightbigg ω=0 =/bracketleftbiggd dωdet{ω(H−(−1 ω)I)}/bracketrightbigg ω=0 =/bracketleftbiggd dω/bracketleftbig ω3/braceleftbigg −(−1 ω)3+IH(−1 ω)2−IIH(−1 ω) +IIIH/bracerightbigg/bracketrightbig/bracketrightbigg ω=0 =/bracketleftbiggd dω/braceleftbig 1 +ωIH+ω2IIH+ω3IIIH/bracerightbig/bracketrightbigg ω=0 =IH= trH, (5.42) where IH,IIH, and IIIHare the three principal invariants of H. This, in turn, leads to L[detF;H]0= det ¯F(0) +D(detF)(0,H) = 1 + tr H= 1 + tr ε. (5.43) The balance laws are also subject to linearization. For inst ance, the conservation of mass statement (4.41) can be linearized to yield L[ρ0;H]0=L[ρJ;H]0. (5.44) This means that ρ0= ¯ρ(0)¯J(0) +Dρ(0,H)¯J(0) + ¯ρ(0)DJ(0,H). (5.45) Since conservation of mass is assumed to hold in all configura tions therefore also including the reference configuration, it follows that ρ0= ¯ρ(0)¯J(0) = ¯ρ(0), (5.46) thus equation (5.45), with the aid of (5.42) results in Dρ(0,H) + ¯ρ(0) trε= 0, (5.47) ME185 DRAFTInfinitesimal deformations 107 or, equivalently, Dρ(0,H) =−̺0trε. (5.48) The linear part of the mass density relative to the reference configuration now takes the form L[ρ;H]0= ¯ρ(0) +Dρ(0,H) =ρ0(1−trε). (5.49) Equation (5.49) reveals that the linearized mass density do es not coincide with the mass density of the reference configuration. ME185 DRAFT108 Consistent linearization of kinematic and kinetic variabl es ME185 DRAFTChapter 6 Mechanical constitutive theories 6.1 General requirements The balance laws furnish a total of seven equations (one from mass balance, three from linear momentum balance and three from angular momentum bal ance) to determine thirteen unknowns, namely the mass density ρ, the position x(or velocity v) and the stress tensor (e.g., the Cauchy stress T). Clearly, without additional equations this system lacks closure. The latter is established by constitutive equations, which relate the stresses to the mass density and the kinematic variables. Before accounting for any possible restrictions or reducti ons, a general constitutive equa- tion for the Cauchy stress may be written as T=ˆT(x,v, . . .,F,˙F, . . .,GradF, . . ., ρ, ˙ρ, . . .) (6.1) or, in rate form, as ˙T=ˆ˙T(x,v, . . .,F,˙F, . . .,GradF, . . ., ρ, ˙ρ, . . .). (6.2) Analogous general functional representations may be writt en for other stress measures. A number of restrictions may be placed to the preceding equat ions on mathematical or physical grounds. Some of these restrictions appear to be un iversally agreed upon, while others tend to be less uniformly accepted. Three of these res trictions are reviewed below. First, all constitutive laws need to be dimensionally consi stent. This simply means that the units of the left- and right-hand side of (6.1) or (6.2) sh ould be the same. For example, 109 DRAFT110 Inviscid fluids when applied to a constitutive law of the form ˙T=αB, this would necessitate that the parameter αhave units of stress (since Bis unitless). Second, the same laws need to be consistent in their tensorial representation. This means t hat the right-hand side of (6.1) and (6.2) should have a tensorial representation in terms of the Eulerian basis. This is to maintain consistency with the left-hand sides, which are na turally resolved on this basis. This restriction would disallow, for example, a constituti ve law of the form T=βF, where βis a constant. The third source of restrictions is the postulate of invaria nce under superposed rigid motions (also referred to as objectivity), which is assumed to apply to constitutive laws just as it applies to balance laws. This postulate requires the fu nctions ˆTandˆ˙Tin (6.1) and (6.2) to remain unaltered under superposed rigid motions. B y way of example, applying the postulate of invariance under superposed rigid motions to t he constitutive law T=ˆT(F) (6.3) necessitates that T+=ˆT(F+). (6.4) Taking into account (3.165) and (4.137), equations (6.3) an d (6.4) lead to QˆT(F)QT=ˆT(QF), (6.5) for all proper orthogonal tensors Q. Equation (6.5) places a restriction on the choice of ˆT. An identical restriction is placed on the functionˆ˙T, where, for example, ◦ T=ˆT(F), (6.6) and◦ Tis the Jaumann rate of the Cauchy stress tensor. The same conclusion is reac hed when using any other objective rate of the Cauchy stress tens or in (6.6). 6.2 Inviscid fluids An inviscid fluid is defined by the property that the stress vec tortacting on any surface is always opposite to the outward normal nto the surface, regardless of whether the fluid ME185 DRAFTMechanical constitutive theories 111 is stationary or flowing. Said differently, an inviscid fluid c annot sustain shearing tractions under any circumstances. This means that t=Tn=−pn, (6.7) hence T=−pi, (6.8) see Figure 6.1. nt Figure 6.1: Traction acting on a surface of an inviscid fluid. On physical grounds, one may assume that the pressure pdepends on the density ρ, i.e., T=−p(ρ)i. (6.9) This constitutive relation defines a special class of invisc id fluids referred to as elastic fluids . It is instructive here to take an alternative path for the der ivation of (6.9). In particular, suppose that one starts from the more general constitutive a ssumption T=ˆT(ρ). (6.10) Upon invoking invariance under superposed rigid motions, i t follows that T+=ˆT(ρ+), (6.11) which, with the aid of (4.137) and (4.143) leads to QˆT(ρ)QT=ˆT(ρ), (6.12) for all proper orthogonal Q. Furthermore, substituting −QforQin (6.12), it is clear that (6.12) holds for all improper orthogonal tensors Qas well, hence it holds for all orthogonal tensors. A tensor function ˆT(φ) of a scalar variable is termed isotropic when QˆT(φ)QT=ˆT(φ), (6.13) ME185 DRAFT112 Inviscid fluids for all orthogonal Q. This condition may be interpreted as meaning that the compo nents of the tensor function remain unaltered when resolved on any tw o orthonormal bases. Clearly, the constitutive function ˆTin (6.10) is isotropic. Therepresentation theorem for isotropic tensor functions of a scalar variable states that a tensor function of a scalar variable is isotropic if, and on ly if, it is a scalar multiple of the identity tensor. In the case of ˆTin (6.10), this immediately leads to the constitutive equation (6.9). To prove the preceding representation theorem, first note th at the sufficiency argument is trivial. The necessity argument can be made by setting Q=Q1=e1⊗e1−e2⊗e3+e3⊗e2, (6.14) which, recalling the Rodrigues formula (3.103), correspon ds top=e1,q=e2,r=e3, and θ=π/2, i.e., to a rigid rotation of π/2 with respect to the axis of e1. It is easy to verify that, in this case, equation (6.13) yields  T11−T13T12 −T31T33−T32 T21−T23T22 = T11T12T13 T21T22T23 T31T32T33 . (6.15) This, in turn, means that T22=T33, T 12=T21=T13=T31= 0 , T 23=−T32. (6.16) Next, set Q=Q2=e2⊗e2−e3⊗e1+e1⊗e3, (6.17) which corresponds to p=e2,q=e3,r=e1, and θ=π/2. This is a rigid rotation of π/2 with respect to the axis of e2. Again, upon using this rotation in (6.13), it follows that  T33T32−T31 T23T22−T21 −T13−T12T11 = T11T12T13 T21T22T23 T31T32T33 , (6.18) which leads to T11=T33 , T 32=T12=−T12 , T 23=T21. (6.19) ME185 DRAFTMechanical constitutive theories 113 One may combine the results in (6.16) and (6.19) to deduce tha t T=TI, (6.20) where T=T11=T22=T33, which completes the proof. Invariance under superposed rigid motions may be also used t o exclude certain functional dependencies in the constitutive assumption for T. Indeed, assume that T=ˆT(x, ρ), (6.21) namely that the Cauchy stress tensor depends explicitly on t he position. Invariance of ˆT under superposed rigid motions implies that T+=ˆT(x+, ρ+) (6.22) hence QˆT(x, ρ)QT=ˆT(Qx+c, ρ), (6.23) for all proper orthogonal tensors Qand vectors c. Now, choose a superposed rigid motion, such that Q=Iandc=c0, where c0is constant. It follows from (6.23) that ˆT(x, ρ) = ˆT(x+c0, ρ). (6.24) However, given that c0is arbitrary, the condition (6.24) can be met only if ˆTis altogether independent of x. A similar derivation can be followed to argue that a constitu tive law of the form T=ˆT(v, ρ) (6.25) violates invariance under superposed rigid motions. Indee d, in this case, invariance implies that T+=ˆT(v+, ρ+), (6.26) which readily translates to QˆT(v, ρ)QT=ˆT(ΩQx +Qv+˙c, ρ). (6.27) ME185 DRAFT114 Inviscid fluids Now, choose Q=I,Ω=0andc=c0t, where c0is, again, a constant. It follows that for this particular choice of a superposed motion ˆT(v, ρ) = ˆT(v+c0, ρ), (6.28) which leads to the elimination of vas an argument in ˆT. Returning to the balance laws for the elastic fluid, note that angular momentum balance is satisfied automatically by the constitutive equation (6. 8) and the non-trivial equations that govern its motion are written in Eulerian form (which is suitable for fluids) as ˙ρ+ρdivv= 0 −gradp(ρ) +ρb=ρa(6.29) or, upon expressing the acceleration in Eulerian form in ter ms of the velocities ˙ρ+ρdivv= 0 −gradp(ρ) +ρb=ρ(∂v ∂t+Lv).(6.30) Equations (6.30) 2are referred to as the compressible Euler equations . Equations (6.30) form a set of four coupled non-linear partial differential equati ons in xandt, which, subject to the specification of suitable initial and boundary conditio ns and a pressure law p=p(ρ), can be solved for ρ(x, t) and ˜v(x, t). Recall the definition of an isochoric (or volume-preserving ) motion, and note that for such a motion conservation of mass leads to ρ0(X) =ρ(x, t) for all time. Then, upon appealing to the local statement of mass conservation (4.34), it is see n that div v= 0 for all isochoric motions. A material is called incompressible if it can only undergo isochoric motions. If the invis- cid fluid is assumed incompressible, then the constitutive e quation (6.9) loses its meaning, because the function p(ρ) does not make sense as the density ρis not a variable quantity. Instead, the constitutive equation T=−piholds with pbeing the unknown. In summary, the governing equations for an incompressible inviscid flui d (ofter also referred to as an ideal fluid) are divv= 0 −gradp+ρ0b=ρ0(∂v ∂t+Lv),(6.31) ME185 DRAFT6.3. VISCOUS FLUIDS 115 where now the unknowns are pandv. Notice that if a set ( p,v) satisfies the above equations, then so does another set of th e form (p+c,v), where cis a constant. This suggests that the pressure field in an inco mpressible elastic fluid is not uniquely determined by the equations of m otion. The indeterminacy is broken by specifying the value of the pressure on some part of the boundary of the domain. This point is illustrated by way of an example: consider a bal l composed of an ideal fluid, which is in equilibrium under uniform time-independent pre ssurep. The same “motion” of the ball can be also sustained by any pressure field p+c, where cis a constant, see Figure 6.2. p Figure 6.2: A ball of ideal fluid in equilibrium under uniform pressure. 6.3 Viscous fluids All real fluids exhibit some viscosity, i.e., some ability to sustain shearing forces. It is easy to conclude on physical grounds that the resistance to shear ing is related to the gradient of the velocity. Therefore, it is sensible to postulate a gener al constitute law for viscous fluids in the form T=ˆT(ρ,L) (6.32) or, recalling the unique additive decomposition of Lin (3.125), more generally as T=ˆT(ρ,D,W). (6.33) The explicit dependence of the Cauchy stress tensor on Wcan be excluded by invoking invariance under superposed rigid-body motions. Indeed, i nvariance requires that T+=ˆT(ρ+,D+,W+). (6.34) ME185 DRAFT116 Viscous fluids Now, consider a special superposed rigid motion for which Q(t) =I,˙Q(t) =Ω0(Ω0being constant skew-symmetric tensor), c(t) =0and˙c(t) =0. This is a superposed rigid rotation with constant angular velocity defined by the skew-symmetri c tensor Ω0(or, equivalently, its axial vector ω0). Recalling (3.184), (3.185) and (4.143), equation (6.34) takes the form QˆT(ρ,D,W)QT=ˆT(ρ,QDQT,QWQT+Ω), (6.35) for all proper orthogonal Q. Given the special form of the chosen superposed rigid motio n, equation (6.35) leads to ˆT(ρ,D,W) = ˆT(ρ,DQT,W+Ω0), (6.36) which needs to hold for any skew-symmetric tensor Ω0. This implies that the constitutive function ˆTcannot depend on W, thus it reduces to T=ˆT(ρ,D). (6.37) Invariance under superposed rigid motions for the constitu tive function in (6.37) gives rise to the condition T+=ˆT(ρ+,D+), (6.38) which, in turn, necessitates that QˆT(ρ,D)QT=ˆT(ρ,QDQT), (6.39) for all proper orthogonal tensors Q. In fact, since both sides of (6.39) are even functions of Q, it is clear that (6.39) must hold for all orthogonal tensors Q. Neglecting, for a moment, the dependence of ˆTonρin equation (6.39), note that a tensor function ˆTof a tensor variable Sis called isotropic if QˆT(S)QT=ˆT(QSQT), (6.40) for all orthogonal tensors Q. It can be proved following the process used earlier for isot ropic tensor functions of a scalar variable that a tensor function ˆTof a tensor variable Sis isotropic if, and only if, it can be written in the form ˆT(S) =a0I+a1S+a2S2, (6.41) ME185 DRAFTMechanical constitutive theories 117 where a0,a1, and a2are scalar functions of the three principal invariants IS,IISandIIIS ofS, i.e., a0= ˆa0(IS, IIS, III S) a1= ˆa1(IS, IIS, III S). (6.42) a2= ˆa2(IS, IIS, III S) The above result is known as the first representation theorem for isotropic tensors . Using this theorem, it is readily concluded that the Cauchy stress for a fluid that obeys the general constitutive law (6.37) is of the form T=a0i+a1D+a2D2, (6.43) where a0,a1anda2are functions of ID,IID,IIIDandρ. The preceding equation character- izes what is known as the Reiner-Rivlin fluid . Materials that obey (6.43) are also generally referred to as non-Newtonian fluids. At this stage, introduce a physically plausible assumption by way of which the Cauchy stressTreduces to hydrostatic pressure −p(ρ)iwhenD= 0. Then, one may rewrite the constitutive function (6.43) as T= (−p(ρ) +a∗ 0)i+a1D+a2D2, (6.44) where, in general, a∗ 0= ˆa∗ 0(ρ, ID, IID, III D). Clearly, when a∗ 0=a1=a2= 0, the viscous fluid degenerates to an inviscid one. From the above general class of viscous fluids, consider the s ub-class of those which are linear in D. To preserve linearity in D, the constitutive function in (6.44) is reduced to T= (−p(ρ) +a∗ 0)I+a1D, (6.45) where, a∗ 0=λIDanda1= 2µ, where λandµdepend, in general, on ρ. This means that the Cauchy stress tensors takes the form T=−p(ρ)i+λtrDi+ 2µD. (6.46) Viscous fluids that obey (6.46) are referred to as Newtonian viscous fluids orlinear viscous fluids. The functions λandµare called the viscosity coefficients . ME185 DRAFT118 Viscous fluids With the constitutive equation (6.46) in place, consider th e balance laws for the Newto- nian viscous fluid. Clearly, angular momentum balance is sat isfied at the outset, since Tin (6.46) is symmetric. Conservation of mass and linear moment um balance can be expressed as ˙ρ+ρdivv= 0 div(−p(ρ)i+λtrDi+ 2µD) +ρb=ρ˙v.(6.47) Assuming that either λandµare independent of ρ(which is very common) or that ρis spatially homogeneous, one may write balance take the form div(−p(ρ)i+λtrDi+ 2µD) =−gradp+λgraddiv v+µ(div grad v+ grad div v) =−gradp+ (λ+µ) graddiv v+µdiv grad v.(6.48) This implies that for this special case the governing equati ons of motions (6.47) can be recast as ˙ρ+ρdivv= 0 −gradp+ (λ+µ) graddiv v+µdiv grad v+ρb=ρ˙v.(6.49) Equations (6.49) 2are known as the Navier-Stokes equations for the compressible Newtonian viscous fluid. As in the case of the compressible inviscid flui d, there are four coupled non- linear partial differential equations in (6.49) and four unk nowns, namely the velocity vand the mass density ρ. If the Newtonian viscous fluid is incompressible, the Cauchy stress is given by T=−pi+ 2µD, (6.50) hence, the governing equations of (6.49) become divv= 0 −gradp+µdiv grad v+ρb=ρ˙v.(6.51) The former equation is a local statement of the constraint of incompressibility, while the latter is the reduced statement of linear momentum balance t hat reflects incompressibility. As in the inviscid case, the four unknowns now are the velocit yvand the pressure p. ME185 DRAFT6.4. NON-LINEARLY ELASTIC SOLID 119 The Navier-Stokes equations (compressible or incompressi ble) are non-linear in vdue to the acceleration term, which is expanded in the form ˙v=∂˜v ∂t+∂˜v ∂xv. In the special case of very slow and nearly steady flow, referred to as creeping flow orStokes flow , the acceleration term may be ignored, giving rise to a system of linear partial differential equations. 6.4 Non-linearly elastic solid Recalling the definition of stress power in the mechanical en ergy balance theorem of equation (4.158), define the non-linearly elastic solid by admitting the existence of a strain energy function Ψ =ˆΨ(F) per unit mass, such that T·D=ρ˙Ψ. (6.52) It follows that the stress power in the region Ptakes the form /integraldisplay PT·Ddv=/integraldisplay Pρ˙Ψdv=d dt/integraldisplay PρΨdv=d dtW(P), (6.53) where W(P) =/integraldisplay PρΨdvis the total strain energy of the material occupying the region P. As a result, the mechanical energy balance theorem for this c lass of materials takes the form d dt[K(P) +W(P)] = Rb(P) +Rc(P) =R(P). (6.54) In words, equation (6.54) states that the rate of change of th e kinetic and strain energy (which together comprise the total internal energy of the non-line arly elastic material) equals the rate of work done by the external forces. Recall that the strain energy function at a given time depend s exclusively on the defor- mation gradient. With the aid of the chain rule, this leads to ˙Ψ =∂ˆΨ ∂F·˙F, (6.55) so that, upon recalling (6.52), T·D=ρ˙Ψ = ρ∂ˆΨ ∂F·(LF), (6.56) ME185 DRAFT120 Non-linearly elastic solid which, in turn, leads to T·L=ρ∂ˆΨ ∂F·(LF) =ρ∂ˆΨ ∂FFT·L. (6.57) The preceding equation can be also written as (T−ρ∂ˆΨ ∂FFT)·L= 0. (6.58) Observing that this equation holds for any L, it is immediately concluded that T=ρ∂ˆΨ ∂FFT. (6.59) Upon enforcing angular momentum balance, equation (6.59) l eads to ∂ˆΨ ∂FFT=F/parenleftBigg ∂ˆΨ ∂F/parenrightBiggT . (6.60) This places a restriction on the form of the strain energy fun ction ˆΨ. Instead of explicitly enforcing this restriction, one may simply write the Cauchy stress as T=1 2ρ ∂ˆΨ ∂FFT+F/parenleftBigg ∂ˆΨ ∂F/parenrightBiggT . (6.61) Alternative expressions for the strain energy of the non-li nearly elastic solid can be ob- tained by invoking invariance under superposed rigid motio ns. Specifically, invariance of the function ˆΨ implies that Ψ+=ˆΨ(F+) = ˆΨ(QF) = Ψ = ˆΨ(F),(6.62) for all proper orthogonal tensors Q. Selecting Q=RT, where Ris the rotation stemming from the polar decomposition of F, it follows from (6.62) that ˆΨ(F) = ˆΨ(QF) = ˆΨ(RTRU) = ˆΨ(U). (6.63) Therefore, one may write Ψ = ˆΨ(F) = ˆΨ(U) = ¯Ψ(C) = ˇΨ(E), (6.64) ME185 DRAFTMechanical constitutive theories 121 by merely exploiting the one-to-one relations between tens orsU,CandE. Then, the material time derivative of Ψ can be expressed as ˙Ψ =∂¯Ψ ∂C·˙C=∂¯Ψ ∂C·(2FTDF), (6.65) where (3.129) is invoked. It follows from (6.52) that T·D=ρ∂¯Ψ ∂C·(2FTDF) = 2 ρF∂¯Ψ ∂CFT·D, (6.66) which readily leads to/parenleftbigg T−2ρF∂¯Ψ ∂CFT/parenrightbigg ·D= 0. (6.67) Given the arbitrariness of D, it follows that T= 2ρF∂¯Ψ ∂CFT. (6.68) Using an analogous procedure, one may also derive a constitu tive equation for the Cauchy stress in terms of the strain energy function ˇΨ as T=ρF∂ˇΨ ∂EFT. (6.69) Now, consider a body made of non-linearly elastic material t hat undergoes a special motion χ, for which there exist times t1andt2(> t1), such that x=χ(X, t1) =χ(X, t2) v=˙χ(X, t1) = ˙χ(X, t2).(6.70) for allX. This motion is referred to as a closed cycle . Recall next the theorem of mechanical energy balance in the form of (6.54) and integrate this equat ion in time between t1andt2 to find that [K(P) +W(P)]t2 t1=/integraldisplayt2 t1[Rb(P) +Rc(P)]dt . (6.71) However, since the motion is a closed cycle, it is immediatel y concluded from (6.70) that [K(P) +W(P)]t2 t1=/bracketleftbigg/integraldisplay P1 2ρv·vdv+/integraldisplay PρˆΨ(F)dv/bracketrightbiggt2 t1= 0, (6.72) thus, also /integraldisplayt2 t1[Rb(P) +Rc(P)]dt=/integraldisplayt2 t1/bracketleftbigg/integraldisplay Pρb·vdv+/integraldisplay ∂Pt·vda/bracketrightbigg dt= 0. (6.73) ME185 DRAFT122 Non-linearly elastic solid This proves that the work done on a non-linearly elastic soli d by the external forces during a closed cycle is equal to zero. Equation (6.54) further implies that [K(P) +W(P)]t t1=/integraldisplayt t1/bracketleftbigg/integraldisplay Pρb·vdv+/integraldisplay ∂Pt·vda/bracketrightbigg dt . (6.74) This means that the work done by the external forces taking th e body from its configuration at time t1to time t(> t1) depends only on the end states at tandt1and not on the path connecting these two states. This is the sense in which the no n-linearly elastic material is characterized as path-independent . The preceding class of non-linearly elastic materials for w hich there exists a strain energy function ˆΨ is sometimes referred to as Green-elastic orhyperelastic materials. A more general class of non-linearly elastic materials is defined by the con stitutive relation T=ˆT(F). (6.75) Such materials are called Cauchy-elastic and, in general, do not satisfy the condition of worklessness in a closed cycle. Recalling the constitutive equation (6.61), it is clear that any Green-elastic material is also Cauchy-elastic. PP P0 P′ 0 Figure 6.3: Orthogonal transformation of the reference configuration. The concept of material symmetry is now introduced for the cl ass of Cauchy-elastic ma- terials. To this end, let Pbe a material particle that occupies the point Xin the reference configuration. Also, take an infinitesimal volume element P0which contains Xin the ref- erence configuration. Since the material is assumed to be Cau chy-elastic, it follows that the Cauchy stress tensor for Pat time tis given by (6.75). Now, subject the reference configuration to an orthogonal transformation characteriz ed by the orthogonal tensor Q, see ME185 DRAFTMechanical constitutive theories 123 Figure 6.3. This transforms the region P0toP′ 0. Note, however, that the stress at a point is independent of the specific choice of reference configurat ion. Hence, when expressed in terms of the deformation relative to the orthogonally trans formed reference configuration, the Cauchy stress at point Pis, in general, given by T=ˆT′(FQ). (6.76) The preceding analysis shows that the constitutive law depe nds, in general, on the choice of reference configuration. For this reason, one may choose, at the expense of added notational burden, to formally write (6.75) and (6.76) as T=ˆTP0(F). (6.77) and T=ˆTP′ 0(FQ), (6.78) respectively. By way of background, recall here that a group Gis a set together with an operation ∗, such that the following properties hold for all elements a,b,cof the set: (i)a∗bbelongs to the set (closure), (ii) (a∗b)∗c=a∗(b∗c) (associativity), (iii) There exists an element i, such that i∗a=a∗i=a(existence of identity), (iv) For every a, there exists an element −a, such that a∗(−a) = (−a)∗a=i(existence of inverse). It is easy to confirm that the set of all orthogonal transforma tionsQof the original refer- ence configuration forms a group under the usual tensor multi plication, called the orthogonal group orO(3). In this group, the identity element is the identity tens orIand the inverse element is the inverse (or transpose) Q−1(orQT) of any given element Q. The subgroup1 1A subset of the group set together with the group operation is called a subgroup if it satisfies the closure property within the subset. ME185 DRAFT124 Non-linearly elastic solid GP0⊆O(3) is called a symmetry group for the Cauchy-elastic material with respect to the reference configuration P0if ˆTP0(F) = ˆTP0(FQ), (6.79) for all Q∈ GP0. Physically, equation (6.79) identifies orthogonal transf ormation Qwhich produce the same stress at Punder two different loading cases. The first one subjects the reference configuration to a deformation gradient F. The second one subjects the reference configuration to an orthogonal transformation Qand then to the deformation gradient F. If the stress in both cases is the same, then the orthogonal tran sformation characterizes the ma- terial symmetry of the body in the neighborhood of Prelative to the reference configuration P0. If equation (6.79) holds for all Q∈O(3), then the Cauchy-elastic material is termed isotropic relative to the configuration P0. An isotropic material is insensitive to any orthog- onal transformation of its reference configuration. Choosi ngQ=RTand recalling the left polar decomposition of the deformation gradient, equation (6.79) implies that T=ˆTP0(F) = ˆTP0(FRT) = ˆTP0(VRRT) = ˆTP0(V). (6.80) In addition, invariance of ˆTP0under superposed rigid motions leads to the condition QˆTP0(V)QT=ˆTP0(QVQT), (6.81) for all proper orthogonal tensors Q(hence, given that (6.80) is quadratic in Q, all orthog- onalQ). This means that ˆTP0is an isotropic tensor function of V. Invoking the first representation theorem for isotropic tensor functions, it follows that T=ˆTP0(V) =a0i+a1V+a2V2, (6.82) where a0,a1, and a2are functions of the three principal invariants of V. An alternative representation of the Cauchy stress of a Cauchy-elastic mat erial is T=¯TP0(B) =b0i+b1B+b2B2, (6.83) where, now, b0,b1, and b2are functions of the three principal invariants of V. This can be trivially derived by noting that ¯TP0(B) =ˆTP0(V). ME185 DRAFT6.5. LINEARLY ELASTIC SOLID 125 6.5 Linearly elastic solid In this section, a formal procedure is followed to obtain the equations of motion and the constitutive equations for a linearly elastic solid. To thi s end, start by writing the linearized version of linear momentum balance as L[divT;H]0+L[ρb;H]0=L[ρa;H]0. (6.84) Now, proceed by making three assumptions: First, let the reference configuration be stress-free, i.e. , assume that if T=ˆT(F) =¯T(H), then ˆT(I) =¯T(0) =0. (6.85) It follows that L[T;H]0=¯T(0) +D¯T(0,H) =D¯T(0,H), (6.86) where D¯T(0,H) =/bracketleftbiggd dω¯T(0+ωH)/bracketrightbigg ω=0=C CH. (6.87) The quantity C Cis called the elasticity tensor . This is a fourth-order tensor that can be resolved in components as C C=Cijklei⊗ej⊗ek⊗el. (6.88) The product C CHin (6.87) can be written explicitly as C CH= (Cijklei⊗ej⊗ek⊗el)(Hmnem⊗en) =CijklHmnei⊗ej[(ek⊗el)·(em⊗en)] =CijklHmnei⊗ej[δkmδln] =CijklHklei⊗ej. (6.89) At this state, note that invariance under superposed rigid m otions implies that QˆT(F)QT=ˆT(QF), (6.90) for all proper orthogonal tensors Q. Taking into account (6.85), it is concluded from the above equation that ˆT(Q(t0)) =0, (6.91) ME185 DRAFT126 Linearly elastic solid namely that any rigid rotation results in no stress. Hence, o ne may choose a special such rotation for which Q(t0) =I(hence, also H(t0) =0) and ˙Q(t0) =Ω0(hence, also ˙H=Ω0), where Ω0is a constant skew-symmetric tensor. In this case, one may wr ite that /bracketleftbiggd dω¯T(0+ω˙H)/bracketrightbigg ω=0=D¯T(0,˙H) = C C˙H=C CΩ0=0. (6.92) SinceΩ0is an arbitrarily chosen skew-tensor, this means that C CΩ=0for any skew- symmetric tensor Ω. Recalling (6.86), (6.87) and that H=ε+ω, it follows that L[T;H]0=C C(ε+ω) = C Cε=σ. (6.93) Here, σis the stress tensor of the theory of linear elasticity. Since the distinction between partial derivatives with res pect to Xandxdisappears in the infinitesimal case, it is clear that so does the distincti on between the “Div” and “div” operators. Therefore, L[divT;H]0=L[DivT;H]0= Div L[T;H]0= Div σ. (6.94) By way of a second assumption, write L[ρa;H]0= ¯ρ(0)¯a(0) + [Dρ(0,H)]¯a(0) + ¯ρ(0)[Da(0,H)]. (6.95) With reference to (6.95), assume that in the linearized theo ry the acceleration asatis- fies¯a(0) =0(i.e., there is no acceleration in the reference configurati on) and also that [Da(0,H)] =a(i.e., the acceleration is linear in H). It follows from (6.95) and the preced- ing assumptions that L[ρa;H]0=ρ0a. (6.96) Finally, noting first that L[ρb;H]0= ¯ρ(0)¯b(0) + [Dρ(0,H)]¯b(0) +ρ(0)[Db(0,H)] (6.97) assume that in the linearized theory ¯b(0) =0(which, in view of the earlier two assumptions, is tantamount to admitting that equilibrium holds in the ref erence configuration) and also that [Db(0,H)] =b(i.e., the body force is linear in H). With (6.97) and the preceding assumptions on bin place, it is easily seen that L[ρb;H]0=ρ0b. (6.98) ME185 DRAFTMechanical constitutive theories 127 Taking into account (6.94), (6.96) and (6.98), equation (6. 84) reduces to Divσ+ρ0b=ρ0a. (6.99) Of course, neither anorbareexplicit functions of the deformation. The acceleration a depends on the deformation implicitly in the sense that the l atter is obtained from the motion χwhose second time derivative equals a. On the other hand, bdepends on the motion (and deformation) implicitly through the balance of linear mome ntum. In the context of linear elasticity, all measures of stress c oincide, namely the distinction between the Cauchy stress Tand other stress tensors such as P,S, etc, disappears. To see this point, recall the relation between TandPin (4.130) and take the linear part of both sides to conclude that L[T;H]0=L/bracketleftbigg1 JPFT;H/bracketrightbigg 0, (6.100) which, in light of (6.93), implies that σ=1 ¯J(0)¯P(0)¯FT(0)+/bracketleftbigg D1 J(0,H)/bracketrightbigg ¯P(0)¯FT(0)+1 ¯J(0)[DP(0,H)]¯FT(0)+1 ¯J(0)¯P(0)[DFT(0,H)]. (6.101) Recalling that the reference configuration is assumed stres s-free, it follows from the above equation that σ=DP(0,H), (6.102) which implies further that L[P;H]0=¯P(0) +DP(0,H) =σ, (6.103) hence, L[P;H]0=L[T;H]0. (6.104) Similar derivations apply to deduce equivalency of Tto other stress tensors in the infinites- imal theory. Return now to the constitutive law (6.93) and write it in comp onent form as σij=Cijklεkl. (6.105) ME185 DRAFT128 Linearly elastic solid In general, the fourth-order elasticity tensor possesses 3 ×3×3×3 = 81 material constants Cijkl. However, since balance of angular momentum implies that σij=σjiand also, by the definition of εin (5.32), εij=εji, it follows that Cijkl=Cjikl=Cijlk=Cjilk, (6.106) which readily implies that only 6 ×6 = 36 of these components are independent.2 Next, note that in the infinitesimal theory, equation (6.93) may be derived from a strain energy function W=W(ε) per unit volume as σ=∂W ∂ε, (6.107) where W=1 2ε·C Cε. (6.108) It follows from (6.108) and (6.107) that ∂σij ∂εkl=∂2W ∂εij∂εkl=Cijkl, (6.109) which, in turn, implies that Cijkl=Cklij. The preceding identity reduces the number of independent material parameters from 36 to 36 −36 2+ 3 = 21.3 The number of independent material parameters can be furthe r reduced by material symmetry. To see this point, recall the constitutive equati ons (6.82) or (6.83) for the isotropic non-linearly elastic solid and note that a corresponding eq uation can be easily deduced in terms of the Almansi strain e, namely T=˜TP0(e) =c0i+c1e+c2e2, (6.110) where c0,c1, and c2are function of the three principal invariants Ie,IIeandIIIeofe. Recalling (5.32) 2and noting that, by assumption, the reference configuration is stress-free, it is easy to see that the linearization of (6.110) leads to σ=c∗ 0IεI+c1ε, (6.111) 2To see this, take each pair ( i, j) or (k, l) and use (6.106) to conclude that only 6 combinations of each pair are independent. 3To see this, write the 36 parameters as a 6 ×6 matrix and argue that only the terms on and above (or below) the major diagonal are independent. ME185 DRAFT6.6. VISCOELASTIC SOLID 129 where c∗ 0andc1are constants. Setting c∗ 0=λandc1= 2µ, one may rewrite the preceding equation as σ=λtrεI+ 2µε. (6.112) The parameters λandµare known as the Lam´ e constants . These are related to the Young’s modulus Eand the Poisson’s ratio νby λ=Eν (1 +ν)(1−2ν), µ =E 2(1 + ν)(6.113) and, inversely, E=µ(3λ+ 2µ) λ+µ, ν =λ 2(λ+µ). (6.114) 6.6 Viscoelastic solid Most materials exhibit some memory effects, i.e., their curr ent state of stress depends not only on the current state of deformation, but also on the defo rmation history. Consider first a broad class of materials with memory, whose C auchy stress is given by T(X, t) = ˆT(H τ≤t[F(X, τ)];X). (6.115) This means that the Cauchy stress at time tof some material particle Pwhich occupies pointXin the reference configuration depends on the history of the d eformation gradient of that point up to (and including) time t. Materials that satisfy the constitutive law (6.115) are called simple . Invoking invariance under superposed rigid motions and sup pressing, in the interest of brevity, the explicit dependence of functions on X, it is concluded that Q(t)ˆT(H τ≤t[F(τ)])QT(t) = ˆT(H τ≤t[Q(τ)F(τ)]), (6.116) for all proper orthogonal Q(τ), where −∞< τ≤t. Choosing Q(τ) =RT(τ), for all τ∈(−∞, t], it follows that RT(t)ˆT(H τ≤t[F(τ)])R(t) = ˆT(H τ≤t[U(τ)]), (6.117) where, as usual, F(τ) =R(τ)U(τ). Equation (6.117) can be readily rewritten as T(t) =R(t)ˆT(H τ≤t[U(τ)])RT(t). (6.118) ME185 DRAFT130 Viscoelastic solid or, equivalently, T(t) =F(t)U−1(t)ˆT(H τ≤t[U(τ)])U−1(t)FT(t). (6.119) Upon recalling (4.133) 2, this means that S(t) =J(t)U−1(t)ˆT(H τ≤t[U(τ)])U−1(t) =ˆS(H τ≤t[U(τ)]). (6.120) As usual, one may alternatively write S(t) =¯S(H τ≤t[C(τ)]) = ˇS(H τ≤t[E(τ)]). (6.121) Now, proceed by distinguishing between the past ( τ < t) and the present ( τ=t) in the preceding constitutive laws. To this end, define Et(s) =E(t−s)−E(t), (6.122) where, obviously, Et(0) = 0. Clearly, for any given time t, the variable sis probing the history of the Lagrangian strain looking further in the past assincreases. Next, rewrite (6.121) 2as S(t) =ˇS(H τ≤t[E(τ)]) = ˇS(¯H s≥0[Et(s),E(t)]). (6.123) Then, define the elastic response function Seas Se(E(t)) = ˇS(¯H s≥0[0,E(t)]). (6.124) and the memory response function Smas Sm(¯H s≥0[Et(s),E(t)]) = ˇS(¯H s≥0[Et(s),E(t)])−ˇS(¯H s≥0[0,E(t)]). (6.125) In summary, the stress response is given by S(t) =Se(E(t)) +Sm(¯H s≥0[Et(s),E(t)]). (6.126) The first term on the right-hand side of (6.126) is the stress w hich depends exclusively on the present state of the Lagrangian strain. The second term o n the right-hand side of (6.126) contains all the dependency of the stress response on past La grangian strain states. ME185 DRAFTMechanical constitutive theories 131 Note that, by definition, the stress during a state deformati on, i.e., which E(t) =E0, where E0is a constant, is equal to S(t) =Se(E0), or, equivalently, Sm(¯H s≥0[0,E(t)]) =0. All viscoelastic materials can be described by the constitu tive equation (6.126). For such materials, Smis rate-dependent (i.e., it depends on the rate ˙Eof Lagrangian strain) and also exhibits fading memory . The latter means that the impact on the stress at time tof the deformation at time t−s(s >0) diminishes as s→ ∞. This condition can be expressed mathematically as lim δ→∞Sm(¯H s≥0[Eδ t(s),E(t)]) = 0, (6.127) where Eδ t(s) =/braceleftBigg 0 if 0 ≤s < δ Et(s−δ) ifδ≤s <∞(6.128) is thestatic continuation ofEt(s) byδ. With reference to Figure 6.4, it is seen that the static sδ Et(s) Eδ t(s) Figure 6.4: Static continuation Eδ t(s)ofEt(s)byδ. continuation is a rigid translation of the argument Et(s) of the memory response function Smbyδ. Therefore, the fading memory condition (6.127) implies th at, as time elapses, the effect of earlier Lagrangian strain states on Smcontinuously diminishes and, ultimately, disappears altogether. The condition (6.127) is often refe rred to as the relaxation property . This is because it implies that any time-dependent Lagrangi an strain process which reaches a steady-state results in memory response which ultimately relaxes to zero memory stress (plus, possibly, elastic stress), see Figure 6.5. Under special regularity conditions, the memory response f unction bSmcan be reduced to the linear functional form Sm(¯H s≥0[Et(s),E(t)]) =/integraldisplay∞ 0L L(E(t), s)Et(s)dS , (6.129) ME185 DRAFT132 Viscoelastic solid s t t1t2Et1(s) Et2(s) Figure 6.5: An interpretation of relaxation where L L(E(t), s) is a fourth-order tensor function of E(t) and s. Of course, L L(E(t), s) needs to be chosen so that Smsatisfy the relaxation property (6.127). Upon Taylor expansion of Et(s) around t−s, one finds that Et(s) =E(t−s)−E(t) =−s˙E(t−s) +o(s2). (6.130) Ignoring the second-order term in (6.130), which is tantamo unt to neglecting long-term memory effects, one may substitute Et(s) in (6.129) to find that Sm(¯H s≥0[Et(s),E(t)]) =/integraldisplay∞ 0L L(E(t), s){−s˙E(t−s)}ds=/integraldisplay∞ 0¯L L(E(t), s)˙E(t−s)ds . (6.131) Alternatively, upon Taylor expansion of Et(s) around t, one finds that Et(s) =E(t−s)−E(t) =−s˙E(t) +o(s2), (6.132) which leads to Sm(¯H s≥0[Et(s),E(t)]) =/integraldisplay∞ 0L L(E(t), s){−s˙E(t)}ds=/bracketleftbigg −/integraldisplay∞ 0L L(E(t), s)s ds/bracketrightbigg ˙E(t) =MM(E(t))˙E(t), (6.133) where MM(E(t)) = −/integraldisplay∞ 0L L(E(t), s)s ds . (6.134) Now, attempt to reconcile the general constitutive framewo rk developed here with the clas- sical one-dimensional viscoelasticity models of Kelvin-V oigt and Maxwell. Start with the Kelvin-Voigt model, which is comprised of a spring and a dash pot in parallel, and let the ME185 DRAFTMechanical constitutive theories 133 spring constant be Eand the dashpot constant be η. It follows that the uniaxial stress σis related to the uniaxial strain εby σ=Eε+η˙ε . (6.135) Clearly, this law is a simple reduction of (6.126), where the memory response is obtained from a one-dimensional counterpart of (6.133). Next, consider the Maxwell model of a spring and dashpot in se ries with material prop- erties as in the Kelvin-Vogt model. In this case, the constit utive law becomes ˙ε=˙σ E+σ η, (6.136) and take the accompanying initial condition to be σ(0) = 0. The general solution of (6.136) is σ(t) =c(t)e−E ηt, (6.137) which, upon substituting into (6.136) leads to ˙c(t) =EeE ηt˙ε(t). (6.138) This, in turn, may be integrated to yield c(t) =c(0) +/integraldisplayt 0EeE ητ˙ε(τ)dτ . (6.139) Noting that the initial condition results in c(0) = 0, one may write that σ(t) =/bracketleftbigg/integraldisplayt 0EeE ητ˙ε(τ)dτ/bracketrightbigg e−E ηt=/integraldisplayt 0EeE η(τ−t)˙ε(τ)dτ =/integraldisplayt 0Ee−E ηs˙ε(t−s)ds . (6.140) Clearly, the stress response of the Maxwell model falls with in the constitutive framework of (6.126), where the elastic response function vanishes iden tically and the memory response can be deduced from (6.131). ME185 DRAFT134 Viscoelastic solid ME185 DRAFTChapter 7 Boundary- and Initial/boundary-value Problems 7.1 Incompressible Newtonian viscous fluid Parts of this chapter include material taken from the contin uum mechanics notes of the late Professor P.M. Naghdi. 7.1.1 Gravity-driven flow down an inclined plane Consider an incompressible Newtonian viscous fluid under st eady flow down an inclined plan due to gravity, see Figure 7.1. Let the pressure of the free su rface be p0and assume that the fluid region has constant depth h. g h θe1e2e3 Figure 7.1: Flow down an inclined plane 135 DRAFT136 Flow down an inclined plane Assume that the velocity and pressure fields are of the form v= ˜v(x2, x3)e2 (7.1) and p= ˜p(x1, x2, x3), (7.2) respectively. Incompressibility implies that divv=∂˜v ∂x2= 0, (7.3) which means that the velocity field is independent of x2, namely that v= ¯v(x3)e2. (7.4) Given the reduced velocity field in (7.4), the rate-of-defor mation tensor has components [Dij] = 0 0 0 0 01 2∂¯v ∂x3 01 2∂¯v ∂x30 . (7.5) Recalling the constitutive equation (6.50), the Cauchy str ess is given by [Tij] = −p0 0 0−p µ∂¯v ∂x3 0µ∂¯v ∂x3−p . (7.6) Note that gravity induces a body force per unit mass equal to b=g(sinθe2−cosθe3), (7.7) where gis the gravitational constant. Given (7.2), (7.4), (7.6) an d (7.7), the equations of linear momentum balance assume the form −∂˜p ∂x1= 0 −∂˜p ∂x2+µd2¯v dx2 3+ρgsinθ=ρ/bracketleftbigg∂¯v ∂t+∂¯v ∂xkvk/bracketrightbigg . (7.8) −∂˜p ∂x3−ρgcosθ= 0 ME185 DRAFTBoundary- and initial-value problems 137 It follows from (7.8) 1,3that p= ¯p(x2, x3) =−ρgcosθx3+f(x2). (7.9) Imposing the boundary condition on the free surface leads to ¯p(x2, h) =−ρgcosθh+f(x2) =p0, (7.10) which implies that the function f(x2) is constant and equal to f(x2) =p0+ρgcosθh . (7.11) Substituting this equation to (7.9) results in an expressio n for the pressure as p=p0+ρgcosθ(h−x3). (7.12) Substituting the pressure from (7.12) into the remaining mo mentum balance equation (7.8) 2 yields µd2¯v dx2 3+ρgsinθ= 0, (7.13) which may be integrated to ¯v(x3) =−ρgsinθ 2µx2 3+c1x+c2. (7.14) Enforcing the boundary conditions ¯ v(0) = 0 (no slip condition on the solid-fluid interface) andT23(h) = 0 (no shear stress on the free surface), gives c2= 0 and c1=ρghsinθ µ, which, when substituted into (7.14) yield ¯v(x3) =ρgsinθ µx3(h−x3 2). (7.15) It is seen from (7.15) that the velocity distribution is para bolic along x3and attains a maximum value of vmax=ρgsinθ 2µh2on the free surface. 7.1.2 Couette flow This is the steady flow between two concentric cylinder of rad iiRo(outer cylinder) and Ri (inner cylinder) rotating with constant angular velocitie sωo(outer cylinder) and ωi(inner ME185 DRAFT138 Couette flow r RoRiωo ωi ereθ Figure 7.2: Couette flow cylinder), see Figure 7.2. The fluid is assumed Newtonian and incompressible. Also, the effect of body forces is neglected. The problem lends itself naturally to analysis using cylind rical polar coordinates with basis vectors {er,eθ,ez}. The velocity and pressure fields are assumed axisymmetric a nd, using the cylindrical polar coordinate representation, ca n be expressed as v=¯v(r)eθ (7.16) and p= ¯p(r). (7.17) Taking into account (A.9), the spatial velocity gradient ca n be written as L=d¯v(r) dreθ⊗er−¯v rer⊗eθ, (7.18) so that D=r 2d dr/parenleftbigg¯v(r) r/parenrightbigg (er⊗eθ+eθ⊗er). (7.19) It follows from (7.19) that T=−pi+µrd dr/parenleftbigg¯v(r) r/parenrightbigg (er⊗eθ+eθ⊗er). (7.20) Given the form of the velocity in (7.16), it is clear that div v= 0, hence the incompress- ibility condition is satisfied at the outset. Also, the accel eration becomes a=/parenleftbiggd¯v(r) dreθ⊗er−¯v1 rer⊗eθ/parenrightbigg ¯veθ=−¯v2 rer. (7.21) ME185 DRAFTBoundary- and initial-value problems 139 Taking into account (7.20) and (7.21) the linear momentum ba lance equations in the r- andθ-directions become −dp dr=−ρ¯v2(r) dr µd dr/bracketleftbigg rd dr/parenleftBigv r/parenrightBig/bracketrightbigg + 2µd dr/parenleftBigv r/parenrightBig = 0,(7.22) respectively. The second of the above equations may be integrated twice to g ive ¯v(r) =c1r+c2 r. (7.23) The integration constants c1andc2can be determined by imposing the boundary conditions ¯v(Ri) =ωiRiand ¯v(Ro) =ωoRo. Upon determining these constants, the velocity of the flow takes the form ¯v(r) =ωoRoRo Ri/parenleftbiggr Ri−Ri r/parenrightbigg +ωi ωo/parenleftbiggRo r−r Ro/parenrightbigg /parenleftbiggRo Ri/parenrightbigg2 −1. (7.24) Finally, integrating equation (7.22) 1and using a pressure boundary condition such as, e.g., ¯p(R0) =p0, leads to an expression for the pressure ¯ p(r). 7.1.3 Poiseuille flow This is the flow of an incompressible Newtonian viscous fluid t hrough a straight cylindrical pipe of radius R. Assuming that the pipe is aligned with the e3-axis, assume that the velocity of the fluid of the general form v=v(r)ez, (7.25) while the pressure is p= ¯p(r, z). (7.26) Clearly, the assumed velocity field satisfies the incompress ibility condition at the outset. Taking again into account (A.9), the velocity gradient for t his flow is L=∂v ∂rez⊗er, (7.27) ME185 DRAFT140 Stokes’ First Problem hence the rate-of-deformation tensor is D=1 2∂v ∂r(er⊗ez+ez⊗er). (7.28) Then, the Cauchy becomes T=−pi+µ∂v ∂r(er⊗ez+ez⊗er). (7.29) Noting that the acceleration vanishes identically, the equ ations of linear momentum bal- ance in the absence of body force take the form −∂p ∂r= 0 −1 r∂p ∂θ= 0, (7.30) −∂p ∂z+µd2v dr2+µ rdv dr= 0 where (A.11) is employed. Equation (7.30) 2is satisfied identically due to the assumption (7.26), while equation (7.30) 1implies that p=p(z). However, given that vdepends only on r, equation (7.30) 3requires that dp dz=c , (7.31) where cis a constant. Upon integrating (7.30) 3inr, one finds that v=cr2 4µ+c1lnr+c2, (7.32) where c1andc2are also constants. Admitting that the solution should rema in finite at r= 0 and imposing the no-slip condition v(R) = 0, it follows that v(r) =c 4µ(r2−R2 0). (7.33) Two additional boundary conditions are necessary (either a velocity boundary condition on one end and a pressure boundary condition on another or press ure boundary conditions on both ends of the pipe) in order to fully determine the velocit y and pressure field. 7.2 Compressible Newtonian viscous fluids 7.2.1 Stokes’ First Problem See Problem 5 in Homework 11. ME185 DRAFTBoundary- and initial/boundary-value problems 141 7.2.2 Stokes’ Second Problem Consider the semi-infinite domain R={(x1, x2, x3)|x3>0}, which contains a Newtonian viscous fluid, see Figure 7.3. The fluid is subjected to a perio dic motion of its boundary x3= 0 in the form vp=Ucosωte1, (7.34) where Uandωare given positive constants. In addition, the body force is zero everywhere in the domain. vpe1e2e3 Figure 7.3: Semi-infinite domain for Stokes’ Second Problem Adopting a semi-inverse approach, assume a general form of t he solution as v=f(x3) cos (ωt−αx3)e1, (7.35) where the function fand the constant αare to be determined. The solution (7.35) assumes that the velocity field varies along x3and is also phase-shifted by αx3relative to the boundary velocity. Further, note that the assumed solution renders t he motion isochoric. In view of (7.35), the acceleration field is a=−ωfsin (ωt−αx3)e1. (7.36) Further, mass conservation implies that ˙ ρ= 0, i.e., if the fluid is initially homogeneous (which it is assumed to be), then it remains homogeneous for a ll time. The only non-vanishing components of the rate-of-deformat ion tensor are D23=D32=1 2/bracketleftbiggdf dxcos (ωt−αx3) +αfsin (ωt−αx3)/bracketrightbigg . (7.37) ME185 DRAFT142 Stokes’ Second Problem Recalling (6.46), with tr D= divv= 0, and noting that pandµare necessarily constant, since the mass density is homogeneous, it follows that [Tij] = −p0T13 0−p0 T310−p , (7.38) where T13=T31=µ/bracketleftbiggdf dxcos (ωt−αx3) +αfsin (ωt−αx3)/bracketrightbigg . (7.39) Taking into account (7.36) and (7.38) it is easy to see that th e linear momentum balance equations in the e2ande3directions hold identically. In the e1direction, the momentum balance equation reduces to ∂T13 ∂x3=ρa1 (7.40) or µ/bracketleftbiggd2f dx2 3cos (ωt−αx3) + 2αdf dx3sin (ωt−αx3)−α2fcos (ωt−αx3)/bracketrightbigg =ρωfsin (ωt−αx3).(7.41) The preceding equation can be also written as µ/bracketleftbiggd2f dx2 3−α2f/bracketrightbigg cos (ωt−αx3) +/bracketleftbigg 2µαdf dx3+ρωf/bracketrightbigg sin (ωt−αx3) = 0 . (7.42) For this equation to be satisfied identically for all x3andt, it is necessary and sufficient that d2f dx2 3−α2f= 0 (7.43) and 2µαdf dx3+ρωf= 0. (7.44) These two equations can be directly integrated to give f(x3) =c1eαx3+c2e−αx3(7.45) and f(x3) =c3e−ρω 2µαx3, (7.46) ME185 DRAFT7.3. LINEAR ELASTIC SOLIDS 143 respectively. To reconcile the two solutions, one needs to t ake f(x3) =ce−ρω 2µαx3, (7.47) where α=/radicalbiggρω 2µ. With this expression in place, the velocity field of equatio n (7.35) takes the form v=ce−ρω 2µαx3cos (ωt−αx3)e1. (7.48) Applying the boundary condition v(0) = Ucosωte1leads to c=U, so that, finally, v=Ue−ρω 2µαx3cos (ωt−αx3)e1. (7.49) In this problem, the pressure pis constitutively specified, yet is here constant throughou t the semi-infinite domain owing to the homogeneity of the mass density. 7.3 Linear elastic solids Consider a homogeneous and isotropic linearly elastic soli d, and recall that its stress-strain relation is given by (6.112). Taking the trace of both sides o n (6.112) and assuming that λ+2 3µis non-zero, it is easily seen that trε=1 3λ+ 2µtrσ, (7.50) hence ε=1 2µ/bracketleftbigg σ−λ 3λ+ 2µtrσI/bracketrightbigg (7.51) or, using the elastic materials parameters Eandν, ε=1 E[(1 +ν)σ−νtrσI]. (7.52) where (6.113) is used. 7.3.1 Simple tension and simple shear In the case of simple tension along the e3-axis, σ33/ne}ationslash= 0, while all other components σij= 0. It follows from (7.52) that ε33=σ33 E, ε 11=ε22=−νσ33 E, (7.53) ME185 DRAFT144 Saint-Venant torsion of a circular cylinder while all shearing components of strain vanish. A simple ten sion experiment can be used to determine the material constants Eandνas E=σ33 ε33, ν =−ε11 ε33. (7.54) In the case of simple shear on the plane of e1ande2, the only non-zero components of stress is σ23=σ32. Recalling (7.52), it follows that ε12=σ12 2µ, (7.55) while all other strain components vanish. The elastic const antµcan be experimentally measured by arguing that 2 εis the change in the angle between the axes e1ande2. 7.3.2 Uniform hydrostatic pressure Suppose that a homogeneous and isotropic linearly elastic s olid is subjected to a uniform hydrostatic pressure σ=−pI. Taking into account (7.51), it follows that ε=−p1 3λ+ 2µI=−p1 3KI, (7.56) where K=3λ+ 2µ 3=E 3(1−2ν)is the bulk modulus of elasticity. Equation (7.56) can be used in an experiment to determine the bulk modulus by noti ng that tr ε=−p Kis the infinitesimal change of volume due to the hydrostatic pressu rep. 7.3.3 Saint-Venant torsion of a circular cylinder Consider an isotropic homogeneous linearly elastic cylind er under quasistatic conditions. The cylinder has length Land radius Rand is fixed on the one end ( x3= 0), while at the opposite end ( x3=L) it is subjected to a resultant moment Me3relative to the point with coordinates (0 ,0, L). Also, the lateral sides of the cylinder are assumed tracti on-free. Due to symmetry, it is assumed that the cross-section remain s circular and that plane sections of constant x3remain plane after the deformation. With these assumptions in place, assume that the displacement of the cylinder may be written a s u=αx3reθ, (7.57) ME185 DRAFTBoundary- and initial-value problems 145 where αis the angle of twist per unit x3-length. Recalling that eθ=−e1x2 r+e2x1 r, one may rewrite the displacement as u=α(−x2x3e1+x1x3e2). (7.58) It follows that the infinitesimal strain tensor has componen ts [εij] = 0 0 −1 2αx2 0 01 2αx1 −1 2αx21 2αx10 . (7.59) Hence, the stress tensor has components [σij] =µα 0 0 −x2 0 0 x1 −x2x10 . (7.60) It can be readily demonstrated that, in the absence of body fo rces, all the equilibrium equations are satisfied identically. Further, for the later al surfaces, the tractions vanish, since [ti] = [ σij][nj] =µα 0 0 −x2 0 0 x1 −x2x10 1 r x1 x2 0 = 0 0 0 . (7.61) On the other hand, the traction at x=Lis [ti] = [ σij][nj] =µα 0 0 −x2 0 0 x1 −x2x10  0 0 1 =µα −x2 x1 0 , (7.62) so that the resultant force is given by /integraldisplay x3=L[ti]dA=µα/integraldisplay2π 0/integraldisplayR 0r −sinθ cosθ 0 r drdθ =µαR3 3/integraldisplay2π 0 −sinθ cosθ 0 dθ=µαR3 3 cosθ sinθ 0 2π 0= 0 0 0 .(7.63) ME185 DRAFT146 Rivlin’s cube Moreover, the resultant moment with respect to (0 ,0, L) equals Me3=/integraldisplay x3=L(x1e1+x2e2)×µα(−x2e1+x1e2)dA =µα/integraldisplay x3=L(x2 1+x2 2)dA=µα/integraldisplay2π 0/integraldisplayR 0r2rdrdθ =µαI ,(7.64) where I=πR4 2is the polar moment of inertia of the circular cross-section. 7.4 Non-linearly elastic solids 7.4.1 Rivlin’s cube Consider a unit cube made of a homogeneous, isotropic and inc ompressible non-linearly elastic material. First, observe that the pressure pis work-conjugate to the volume change in that T·D= (T′+1 3trTi)·(D′+1 3trDi) =T′·D′+ (−p) divv, (7.65) where T′andD′are the deviatoric parts of TandD, respectively. Next, recall the general form of the constitutive equations for isotropic non-linea rly elastic materials in (6.83) and, assuming, as a special case, b2= 0, write T=−pi+b1B, (7.66) where b1(>0) is a constant, and pis a Lagrange multiplier to be determined upon enforcing the incompressibility constraint. Returning to the unit cube, assume that its edges are aligned with the common orthonor- mal basis {e1,e2,e3}of the reference and current configuration, and is loaded by t hree pairs of equal and opposite tensile forces, all of equal magnitude , and distributed uniformly on each face. Taking into account (4.130) and (7.66), one may write P=J(−pi+b1B)F−T=−pF−T+b1F, (7.67) where J= 1 due to the assumption of incompressibility. The traction s, when resolved on the geometry of the reference configuration, satisfy Pei=cei, (7.68) ME185 DRAFTBoundary- and initial/boundary-value problems 147 where c >0 is the magnitude of the tractions per unit area in the refere nce configuration. Note that cis the same for all faces of the cube, since, by assumption, th e force on each face is constant and uniform. Therefore, one may also write that P=c(e1⊗e1+e2⊗e2+e3⊗e3). (7.69) Solutions of the form F=λ1e1⊗e1+λ2e2⊗e2+λ3e3⊗e3 (7.70) are sought for this boundary-value problem, subject to the c ondition λ1λ2λ3= 1. It is clear from (7.68) that the equilibrium equations are satisfied ide ntically in the absence of body forces. Returning to the constitutive equations, combine ( 7.67) and (7.69) to conclude that c=−p λi+b1λi, (7.71) or b1λ2 i=cλi+p , (7.72) where i= 1,2,3. Eliminating the pressure in the preceding equation leads to b1(λ2 i−λ2 j) =c(λi−λj), (7.73) where i/ne}ationslash=j. This, in turn, means that λ1=λ2orb1(λ1+λ2) =c λ2=λ3orb1(λ2+λ3) =c , (7.74) λ3=λ1orb1(λ3+λ1) =c subject to λ1λ2λ3= 1, where not all conditions can be chosen from the same colum n in (7.74). One solution of (7.74) is obviously λ1=λ2=λ3= 1. (7.75) This corresponds to the cube remaining rigid under the influe nce of the tensile load. Another possible set of solutions is λ1=λ2/ne}ationslash=λ3,λ2=λ3/ne}ationslash=λ1andλ3=λ1/ne}ationslash=λ2. Explore one of these solutions, say λ1=λ2/ne}ationslash=λ3by setting λ3=λand noting from (7.74) that λ2+λ3=λ3+λ1=c b1=η , (7.76) ME185 DRAFT148 Rivlin’s cube so that λ1λ2λ3= (η−λ)2λ= 1. (7.77) The preceding equation may be rewritten as f(λ) =λ3−2ηλ2+η2λ−1 = 0 . (7.78) To examine the roots of f(λ) = 0, note that f′(λ) = 3 λ2−4ηλ+η2, f′′(λ) = 6 λ−4η , (7.79) hence the extrema of foccur at λ1,2=  η 3where f′′(η 3) =−2η <0 (maximum) ηwhere f′′(η) =η >0 (minimum)(7.80) and are equal to f(η 3) =4 27η3−1, f(η) =−1. (7.81) It is also obvious from the definition of f(λ) that f(0) =−1,f(−∞) =−∞, andf(∞) =∞. The plot in Figure 7.4 depicts the essential features of f(λ). 0 −1η/3 η λf Case ICase II Case III Figure 7.4: Function f(λ)in Rivlin’s cube In summary, λ1=λ2=λ3= 1 is always a solution. Further, 1. If4 27η3<1, then there are no additional solutions (Case I). 2. If4 27η3= 1, then there is one set of three additional solutions corre sponding to λ=η 3 (Case II). ME185 DRAFT7.5. MULTISCALE PROBLEMS 149 3. If4 27η3>1, then there are two sets of three additional solutions corr esponding to the two roots of f(λ) which are smaller than η(Case III). Note that the root λ=λ3> ηis inadmissible because it leads to λ1=λ2=η−λ <0. 7.5 Multiscale problems It is sometimes desirable to relate the theory of continuous media to theories of particle mechanics. This is, for example, the case, when one wishes to analyze metals and semi- conductors at very small length and time scales, at which the continuum assumption is not unequivocally satisfied. In such cases, multiscale analyse s offer a means for relating kinematic and kinetic information between the continuum and the discr ete system. 7.5.1 The virial theorem The virial theorem is a central result in the study of continu a whose constitutive behavior is derived from an underlying microscale particle system. Pre liminary to the derivation of the theorem, recall that the mean Cauchy stress Tin a material region Pcan be written as (volP)¯T=/integraldisplay ∂Pt⊗xda−/integraldisplay PdivT⊗xdv . (7.82) Taking into account (4.83), the preceding equation may be re written as (volP)¯T=/integraldisplay ∂Pt⊗xda+/integraldisplay Pρ(b−a)⊗xdv (7.83) =/integraldisplay ∂Pt⊗xda+/integraldisplay Pρb⊗xdv−d dt/integraldisplay Pρ˙x⊗xdv+/integraldisplay Pρ˙x⊗˙xdv . (7.84) Define next the (long) time-average /an}b∇acketle{tφ/an}b∇acket∇i}htof a time-dependent quantity φ=φ(t) as /an}b∇acketle{tφ/an}b∇acket∇i}ht= lim T→∞1 T/integraldisplayt0+T t0φ(t)dt . (7.85) and note that /angbracketleftBigd dt/integraldisplay Pρ˙x⊗xdv/angbracketrightBig = lim T→∞1 T/bracketleftbigg/integraldisplay Pρ˙x⊗xdv|t=t0+T−/integraldisplay Pρ˙x⊗xdv|t=t0/bracketrightbigg =0.(7.86) ME185 DRAFT150 The virial theorem The preceding time-average vanishes due to the assumed boun dedness of the integral/integraltext Pρ˙x⊗ xdvthroughout time. Using (7.86), the time-averaged counterp art of the mean-stress for- mula (7.83) is /an}b∇acketle{t(volP)¯T/an}b∇acket∇i}ht=/angbracketleftBig/integraldisplay ∂Pt⊗xda/angbracketrightBig +/angbracketleftBig/integraldisplay Pρb⊗xdv/angbracketrightBig +/angbracketleftBig/integraldisplay Pρ˙x⊗˙xdv/angbracketrightBig . (7.87) Turn attention now to a system of Nparticles whose motion is governed by Newton’s Second Law, namely mα¨xα=fα, (7.88) where mαandxαare the mass and the current position of particle α, while fαis the total force acting on particle α. It is easy to show that the above equation leads to mαd dt(˙xα⊗xα)−mα˙xα⊗˙xα=fα⊗xα. (7.89) Taking time averages of (7.89) for the totality of the partic les and assuming boundedness of the term/summationtextN α=1mα˙xα⊗xα, it is seen that −/angbracketleftBigN/summationdisplay α=1mα˙xα⊗˙xα/angbracketrightBig =/angbracketleftBigN/summationdisplay α=1fα⊗xα/angbracketrightBig . (7.90) Recognizing that, in each particle, the total force fαis comprised of an internal part fext,α (due to inter-particle potentials) and an external part fint,α, the preceding equation may be rewritten as −/angbracketleftBigN/summationdisplay α=1mα˙xα⊗˙xα/angbracketrightBig =/angbracketleftBigN/summationdisplay α=1fint,α⊗xα/angbracketrightBig +/angbracketleftBigN/summationdisplay α=1fext,α⊗xα/angbracketrightBig . (7.91) Upon ignoring the body forces in the continuum problem and co mparing (7.87) to (7.91), it can be argued that there is a one-to-one correspondence be tween the three terms in each statement. Specifically, for the mean stress, one may argue t hat /an}b∇acketle{t(volP)¯T/an}b∇acket∇i}ht.=−/angbracketleftBigN/summationdisplay α=1fint,α⊗xα/angbracketrightBig , (7.92) which leads to an estimate of the mean Cauchy stress in terms o f the underlying particle system dynamics as /an}b∇acketle{t(volP)¯T/an}b∇acket∇i}ht.=/angbracketleftBigN/summationdisplay α=1mα˙xα⊗˙xα/angbracketrightBig +/angbracketleftBigN/summationdisplay α=1fext,α⊗xα/angbracketrightBig . (7.93) Equation (7.93) is a statement of the virial theorem. ME185 DRAFT7.6. EXPANSION OF THE UNIVERSE 151 7.6 Expansion of the universe ME185 DRAFTAppendix A A.1 APPENDIX A A.1 Cylindrical polar coordinate system The basis vectors of the cylindrical polar coordinate syste m are {er,eθ,ez}, where er=e1cosθ+e2sinθ eθ=−e1sinθ+e2cosθ , (A.1) ez=e3 (A.2) Here, θis the angle formed between the e1andervectors. Conversely, e1=ercosθ−eθsinθ e2=ersinθ+eθcosθ . (A.3) ez=e3 (A.4) Further, one may easily conclude that r=/radicalBig x2 1+x2 2, θ = arctanx2 x1, z =x3 (A.5) and, conversely, x1=rcosθ , x 2=rsinθ , x 3=z . (A.6) Using chain rule, one may express the gradient of a scalar fun ctionfin polar coordinates as gradf=∂f ∂rer+1 r∂f ∂θeθ+∂f ∂zez=/parenleftbigg∂ ∂rer+1 r∂ ∂θeθ+∂ ∂zez/parenrightbigg f . (A.7) When the gradient operator grad is applied on a vector functi onv=vrer+vθeθ+vzez, then gradTv=/parenleftbigg∂ ∂rer+1 r∂ ∂θeθ+∂ ∂zez/parenrightbigg ⊗v, (A.8) or, upon expanding, gradv=∂vr ∂rer⊗er+∂vθ ∂reθ⊗er+∂vz ∂rez⊗er+ 1 r/parenleftbigg∂vr ∂θ−vθ/parenrightbigg er⊗eθ+1 r/parenleftbigg∂vθ ∂θ+vr/parenrightbigg eθ⊗eθ+1 r∂vz ∂θez⊗eθ+ ∂vr ∂zer⊗ez+∂vθ ∂zeθ⊗ez+∂vz ∂zez⊗ez.(A.9) ME185 DRAFTA.2 Cylindrical polar coordinate system Also, given a symmetric tensor function T=Trrer⊗er+Trθ(er⊗eθ+eθ⊗er) +Trz(er⊗ez+ez⊗er)+ Tθθeθ⊗eθ+Tθz(eθ⊗ez+ez⊗eθ) +Tzzez⊗ez,(A.10) its divergence can be expanded to divT=/bracketleftbigg∂Trr ∂r+Trr−Tθθ r+1 r∂Trθ ∂θ+∂Trz ∂z/bracketrightbigg er+ /bracketleftbigg∂Trθ ∂r+2Trθ r+1 r∂Tθθ ∂θ+∂Tθz ∂z/bracketrightbigg eθ+ /bracketleftbigg∂Trz ∂r+Trz r+1 r∂Tθz ∂θ+∂Tzz ∂z/bracketrightbigg ez.(A.11) ME185