piaras Part III

11 Vectors & Tensors This chapter is divided into three parts. The first part covers vectors (§1.1-1.7). The second part is concerned with second, and higher-order, tensors (§1.8-1.15). The second part covers much of the same ground as done in the first part, mainly generalizing the vector concepts and expressions to tensors. The final part (§1.16-1.19) is concerned with generalizing the earlier work to curvilinear coordinate systems. The first part comprises basic vector alge bra, such as the do t product and the cross product; the mathematics of how the component s of a vector transform between different coordinate systems; the symbolic, index and matrix notations for vectors; the differentiation of vectors, including the gradient, the divergence and the curl; the integration of vectors, including line, double, surface and volume integrals, and the integral theorems. The second part comprises the definition of the tensor (and a re-defini tion of the vector); dyads and dyadics; the manipulation of tensors; properties of tensors, such as the trace, transpose, norm, determinant and principal valu es; special tensors, su ch as the spherical, identity and orthogonal tensors; the tran sformation of tensor components between different coordinate systems; th e calculus of tensors, including the gradient of vectors and higher order tensors and the divergence of hi gher order tensors and special fourth order tensors. In the first two parts, attention is restricted to rectangular Cartesian coordinates. In the third part, curvilinear coordinates are intr oduced, including covari ant and contravariant vectors and tensors, the metric coefficien ts, the physical com ponents of vectors and tensors, the metric, coordinate transforma tion rules, tensor calculus, including the Christoffel symbols and covariant differentia tion, and curvilinear co ordinates for curved surfaces. 2 Section 1.1 Solid Mechanics Part III Kelly 31.1 Vector Algebra 1.1.1 Scalars A physical quantity which is completely described by a single real number is called a scalar . Physically, it is something which has a magnitude, and is completely described by this magnitude. Examples are temperature, density and mass . In the following, lowercase (usually Greek) letters, e.g. γβα ,, , will be used to represent scalars. 1.1.2 Vectors The concept of the vector is used to describe physical quantities which have both a magnitude and a direction associated with them. Examples are force , velocity , displacement and acceleration . Geometrically, a vector is represented by an arrow; the arrow defines the direction of the vector and the magnitude of the vector is represented by the length of the arrow, Fig. 1.1.1a. Analytically, vectors will be represented by lowercase bold-face Latin letters, e.g. a, r, q. The magnitude (or length ) of a vector is denoted by a or a. It is a scalar and must be non-negative. Any vector whos e length is 1 is called a unit vector ; unit vectors will usually be denoted by e. Figure 1.1.1: (a) a vector; (b) addition of vectors 1.1.3 Vector Algebra The operations of addition, subtraction and multiplication familiar in the algebra of numbers (or scalars) can be extended to an algebra of vectors. ab c (a) (b) Section 1.1 Solid Mechanics Part III Kelly 4The following definitions and properties fundamentally define the vector: 1. Sum of Vectors: The addition of vectors a and b is a vector c formed by placing the initial point of b on the terminal point of a and then joining the initial point of a to the terminal point of b. The sum is written bac+= . This definition is called the parallelogram law for vector addition be cause, in a geometrical interpretation of vector addition, c is the diagonal of a parallelogram formed by the two vectors a and b, Fig. 1.1.1b. The following properties hold for vector addition: a+b=b+a … commutative law a+(b+c)= (a+b)+c … associative law 2. The Negative Vector: For each vector a there exists a negative vector . This vector has direction opposite to that of vector a but has the same magnitude; it is denoted by a−. A geometrical interpretation of the negative vector is shown in Fig. 1.1.2a. 3. Subtraction of Vectors and the Zero Vector: The subtraction of two vectors a and b is defined by )(b aba −+=− , Fig. 1.1.2b. If ba= then ba− is defined as the zero vector (or null vector ) and is represented by the symbol o. It has zero magnitude and unspecified direction. A proper vector is any vector other than the null vector. Thus the following properties hold: () oa aaoa =−+=+ 4. Scalar Multiplication: The product of a vector a by a scalar α is a vector aα with magnitude α times the magnitude of a and with direction the same as or opposite to that of a, according as α is positive or negative. If 0=α , aα is the null vector. The following properties hold for scalar multiplication: () a a aβαβα +=+ … distributive law, over addition of scalars () b a ba αα α +=+ … distributive law, over addition of vectors a a )()(αββα= … associative law for scalar multiplication Figure 1.1.2: (a) negative of a vector; (b) subtraction of vectors (a) (b) a a− ab− ba−b a Section 1.1 Solid Mechanics Part III Kelly 5Note that when two vectors a and b are equal, they have the same direction and magnitude, regardless of the position of their initial points. Thus a=b in Fig. 1.1.3. A particular position in space is not assigned here to a vector – it just has a magnitude and a direction. Such vectors are called free, to distinguish them from certain special vectors to which a particular position in space is actually assigned. Figure 1.1.3: equal vectors The vector as something with “magnitude and direction” and defined by the above rules is an element of one case of the mathematical structure, the vector space . The vector space will be discussed in the next section. 1.1.4 The Dot Product The dot product of two vectors a and b (also called the scalar product ) is denoted by ba⋅. It is a scalar defined by θcosbaba=⋅ . (1.1.1) θ here is the angle between the vectors when their initial points coincide and is restricted to the range πθ≤≤0 , Fig. 1.1.4. Figure 1.1.4: the dot product An important property of the dot product is that if for two (proper) vectors a and b, the relation 0=⋅ba , then a and b are perpendicular. The two vectors are said to be orthogonal . Also, )0cos(aaaa=⋅ , so that the length of a vector is aa a⋅= . Another important property is that the projection of a vector u along the direction of a unit vector e is given by eu⋅. This can be interpreted geometrically as in Fig. 1.1.5. a ba bθa b Section 1.1 Solid Mechanics Part III Kelly 6 Figure 1.1.5: the projection of a vector along the direction of a unit vector It follows that any vector u can be decomposed into a component parallel to a (unit) vector e and another component perpendicular to e, according to ()()[]eeuueeu u ⋅−+⋅= (1.1.2) The dot product possesses the following propert ies (which can be proved using the above definition) { ▲Problem 6}: (1) abba⋅=⋅ (commutative) (2) () cabacba ⋅+⋅=+⋅ (distributive) (3) () ( ) ba ba α α ⋅=⋅ (4) 0≥⋅aa ; and 0=⋅aa if and only if oa= 1.1.5 The Cross Product The cross product of two vectors a and b (also called the vector product ) is denoted by ba×. It is a vector with magnitude θsinbaba=× (1.1.3) with θ defined as for the dot product. It can be seen from the figure that the magnitude of ba× is equivalent to the area of the parallelogram determined by the two vectors a and b. Figure 1.1.6: the magnitude of the cross product The direction of this new vector is perpendicular to both a and b. Whether ba× points “up” or “down” is determined from the fact that the three vectors a, b and ba× form a right handed system . This means that if the thumb of the right hand is pointed in the a bθba×u eu θ θcosueu=⋅ Section 1.1 Solid Mechanics Part III Kelly 7direction of ba×, and the open hand is directed in the direction of a, then the curling of the fingers of the right hand so that it closes should move the fingers through the angle θ, πθ≤≤0 , bringing them to b. Some examples are shown in Fig. 1.1.7. Figure 1.1.7: examples of the cross product The cross product possesses the following properties (which can be proved using the above definition): (1) ab ba ×−=× ( not commutative) (2) () cabacba ×+×=+× (distributive) (3) () ( ) b aba α α ×=× (4) oba=× if and only if a and b ()o≠ are parallel (“linearly dependent”) The Triple Scalar Product The triple scalar product , or box product , of three vectors wvu,, is defined by () () ()vuw uwv wvu ⋅×=⋅×=⋅× Triple Scalar Product (1.1.4) Its importance lies in the fact that, if the thre e vectors form a right-handed triad, then the volume V of a parallelepiped spanned by the three vectors is equal to the box product. To see this, let e be a unit vector in the direction of vu×, Fig. 1.1.8. Then the projection of w on vu× is ew⋅=h , and ()() Vh =×=×⋅=×⋅ vuevuw vuw (1.1.5) abba× θa b ba×θ Section 1.1 Solid Mechanics Part III Kelly 8 Figure 1.1.8: the triple scalar product Note : • if the three vectors do not form a right handed triad, then the triple scalar product yields the negative of the volume. For example, using the vectors above, () V−=⋅× uvw 1.1.6 Vectors and Points Vectors are objects which have magnitude and direction, but they do not have any specific location in space. On the other hand, a point has a certain position in space, and the only characteristic that distinguishes one point from another is its position. Points cannot be “added” together like vectors. On the other hand, a vector v can be added to a point p to give a new point q, pvq+= , Fig. 1.1.9. Similarly, the “difference” between two points gives a vector, vpq=− . Note that the notion of point as defined here is slightly different to the familiar point in space with axes and origin – the concept of origin is not necessary for these points a nd their simple operations with vectors. Figure 1.1.9: adding vectors to points 1.1.7 Problems 1. Which of the following are scalars and which are vectors? (i) weight (ii) specific heat (iii) momentum (iv) energy (v) volume 2. Find the magnitude of the sum of three unit vectors drawn from a common vertex of a cube along three of its sides. 3. Consider two non-collinear (not parallel) vectors a and b. Show that a vector r lying in the same plane as these vectors can be written in the form b a r q p+= , w uv e h •• pq v Section 1.1 Solid Mechanics Part III Kelly 9where p and q are scalars. [Note: one says that all the vectors r in the plane are specified by the base vectors a and b.] 4. Show that the dot product of two vectors u and v can be interpreted as the magnitude of u times the component of v in the direction of u. 5. The work done by a force, represented by a vector F, in moving an object a given distance is the product of the component of force in the given direction times the distance moved. If the vector s represents the direction and magnitude (distance) the object is moved, show that the work done is equivalent to sF⋅. 6. Prove that the dot product is commutative, abba⋅=⋅ . [Note: this is equivalent to saying, for example, that the work done in problem 5 is also equal to the component of s in the direction of the force, times the magnitude of the force.] 7. Sketch ab× if a and b are as shown below. 8. Show that 2 2 2 2ba ba ba =⋅+× . 9. Suppose that a rigid body rotates about an axis O with angular speed w, as shown below. Consider a point p in the body with position vector r. Show that the velocity v of p is given by rωv×= , where ω is the vector with magnitude ω and whose direction is that in which a righ t-handed screw would advance under the rotation. [Note: let s be the arc-length traced out by the particle as it rotates through an angle θ on a circle of radius r, then ωr v==v (since )/( /, dtdr dtdsrs θ θ= = ).] 10. Show, geometrically, that the dot and cross in the triple scalar product can be interchanged: () ()cbacba ×⋅=⋅× . 11. Show that the triple vector product ()cba×× lies in the plane spanned by the vectors a and b. ω v rω Oprab Section 1.2 Solid Mechanics Part III Kelly 101.2 Vector Spaces The notion of the vector presented in the previ ous section is here re-cast in a more formal and abstract way. This might seem at first to be unnecessarily complicating matters, but this approach turns out to be helpful in unifying and bringing clarity to much of the theory which follows. Some background theory which complements this material is given in Appendix A to this Chapter, §1.A. 1.2.1 The Vector Space The vectors introduced in the previous section obey certain rules, those listed in §1.1.3. It turns out that many other mathematical objects obey the same list of rules. For that reason, the mathematical structure defined by these rules is given a special name, the linear space or vector space . First, a set is any well-defined list, collection, or cl ass of objects, which could be finite or infinite. An example of a set might be {}3 |≤= xx B (1.2.1) which reads “ B is the set of objects x such that x satisfies the property 3≤x ”. Members of a set are referred to as elements . Consider now the field1 of real numbers R. The elements of R are referred to as scalars . Let V be a non-empty set of elements K,,,cba with rules of addition and scalar multiplication , that is there is a sum V∈+ba for any V∈ba, and a product V∈aα for any V∈a , R∈α . Then V is called a (real )2 vector space over R if the following eight axioms hold: 1. associative law for addition : for any V∈cba,, , one has ) ( ) ( cbacba ++=++ 2. zero element : there exists an element V∈o , called the zero element, such that aaooa =+=+ for every V∈a 3. negative (or inverse ): for each V∈a there exists an element V∈−a , called the negative of a, such that 0 )()( =+−=−+ aa a a 4. commutative law for addition : for any V∈ba, , one has abba+=+ 5. distributive law, over addition of elements of V : for any V∈ba, and scalar R∈α , b a ba αα α +=+) ( 6. distributive law, over addition of scalars : for any V∈a and scalars R∈βα, , a a aβαβα +=+) ( 1 A field is another mathematical structure (see Appendix A to this Chapter, §1.A). For example, the set of complex numbers is a field. In what follows, the only field which will be used is the familiar set of real numbers with the usual operations of addition and multiplication. 2 “real”, since the associated field is the reals. The word real will usually be omitted in what follows for brevity. Section 1.2 Solid Mechanics Part III Kelly 117. associative law for multiplication : for any V∈a and scalars R∈βα,, a a )()(αββα= 8. unit multiplication : for the unit scalar R∈1 , aa=1 for any V∈a . The set of vectors as objects with “magnitude and direction” discussed in the previous section satisfy these rules and therefore form a vector space over R. However, despite the name “vector” space, other objects, which are not the familiar geometric vectors, can also form a vector space over R, as will be seen in a later section. 1.2.2 Inner Product Space Just as the vector of the previous section is an element of a vector space, next is introduced the notion that the vector dot product is one example of the more general inner product . First, a function (or mapping ) is an assignment which assigns to each element of a set A a unique element of a set B, and is denoted by B Af→: (1.2.2) An ordered pair ()ba, consists of two elements a and b in which one of them is designated the first element and the other is designated the second element The product set (or Cartesian product ) BA× consists of all ordered pairs ()ba, where Aa∈ and Bb∈: (){ }BbAaba BA ∈∈ =× , |, (1.2.3) Now let V be a real vector space. An inner product (or scalar product ) on V is a mapping that associates to each ordered pair of elements x, y, a scalar, denoted by yx,, R VV→×⋅⋅:, (1.2.4) that satisfies the following properties, for V∈zyx,,, R∈α : 1. additivity : zy zx zyx , , , +=+ 2. homogeneity : yx yx , ,αα= 3. symmetry : xy yx , ,= 4. positive definiteness : 0 ,>xx when ox≠ From these properties, it follows that, if 0 ,=yx for all V∈y , then 0=x A vector space with an associated inner product is called an inner product space . Two elements of an inner product space are said to be orthogonal if Section 1.2 Solid Mechanics Part III Kelly 12 0 ,=yx (1.2.5) and a set of elements of V, {}K,,,zyx , are said to form an orthogonal set if every element in the set is orthogonal to every other element: ,0 , ,0 , ,0 , = = = zy zx yx etc. (1.2.6) The above properties are those listed in §1.1.4, and so the set of vectors with the associated dot product forms an inner product space. Inner products other than the dot product will be introduced later. Euclidean Vector Space The set of real triplets ()3 2 1,,xxx under the usual rules of addition and multiplication forms a vector space 3R. With the inner product defined by 33 22 11 , yx yxyx ++=yx one has the inner product space known as (three dimensional) Euclidean vector space , and denoted by E. This inner product allows one to take distances (and angles) between elements of E through the norm (length) and metric (distance) concepts discussed next. 1.2.3 Normed Space Let V be a real vector space. A norm on V is a real-valued function, R V→: (1.2.7) that satisfies the following properties, for V∈yx, , R∈α : 1. positivity : 0≥x 2. triangle inequality : y x yx +≤+ 3. homogeneity : x xαα= 4. positive definiteness : 0=x if and only if ox= A vector space with an associated norm is called a normed vector space . Many different norms can be defined on a given vector space, each one giving a different normed linear space. A natural norm for the inner product space is xx x ,≡ (1.2.8) It can be seen that this norm indeed satisfies the defining properties. When the inner Section 1.2 Solid Mechanics Part III Kelly 13product is the vector dot product, the norm defined by 1.2.8 is the familiar vector “length”. One important consequence of the defini tions of inner product and norm is the Schwarz inequality , which states that yx yx≤, (1.2.9) One can now define the angle between two elements of V to be ()⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛≡ →×− yxyxyx,cos , , :1θ θ R VV (1.2.10) The quantity inside the curved brackets here is necessarily between 1 − and 1+, by the Schwarz inequality, and hence the angle θ is indeed a real number. 1.2.4 Metric Spaces Metric spaces are built on the concept of “distance” between objects. This is a generalization of the familiar distance between two points on the real line. Consider a set X. A metric is a real valued function, () R XX d →×⋅⋅:, (1.2.11) that satisfies the following properties, for X∈yx,: 1. positive: 0),(≥yxd and 0),(=xxd , for all X∈yx, 2. strictly positive: if 0),(=yxd then yx=, for all X∈yx, 3. symmetry: ) ,( ),( xy yx d d= , for all X∈yx, 4. triangle inequality: ) ,( ),( ),( yz zx yx d d d +≤ , for all X∈zyx,, A set X with an associated metric is called a metric space . The set X can have more than one metric defined on it, with different metrics producing different metric spaces. Consider now a normed vector space. This space naturally has a metric defined on it: () yx yx−=,d (1.2.12) and thus the normed vector space is a metric space. For the set of vectors with the dot product, this gives the “distance” between two vectors yx,. Section 1.2 Solid Mechanics Part III Kelly 141.2.5 The Affine Space Consider a set P, the elements of which are called points . Consider also an associated vector space V. An affine space consists of the set P, the set V, and two operations which connect P and V: (i) given two points P∈qp, , one can define a difference , pq− which is a unique element v of V, i.e. V∈−= pqv (ii) given a point P∈p and V∈v , one can define the sum pv+ which is a unique point q of P, i.e. P∈+= pvq and for which the following property holds, for P∈rqp,, : ()()( ) pq pr rq −=−+− . From the above, one has for the affine space that opp=− and ()qp pq −−=− , for all P∈qp, . Note that one can take the sum of vectors, according to the structure of the vector space, but one cannot take the sum of points, only the difference between two points. Further, there is no notion of origin in the affine space. One can choose some fixed P∈o to be an origin. In that case, opv−= is called the position vector of p relative to o. Suppose now that the associated vector space is a Euclidean vector space, i.e. an inner product space. Define the distance between two points through the inner product associated with V, () pqpq pq qp −−=−= , ,d (1.2.13) It can be shown that this mapping R PPd→×: is a metric, i.e. it satisfies the metric properties, and thus P is a metric space (although it is not a vector space). In this case, P is referred to as Euclidean point space , Euclidean affine space or, simply, Euclidean space . Whereas in Euclidean vector space there is a zero element, the origin ) 0,0,0(, i n Euclidean point space there is none – apart fro m that, the two spaces are the same and, apart from certain special cases, one does not need to distinguish between them. Section 1.3 Solid Mechanics Part III Kelly 151.3 Cartesian Vectors So far the discussion has been in symbolic notation1, that is, no reference to ‘axes’ or ‘components’ or ‘coordinates’ is made, implied or required. The vectors exist independently of any coordinate system. It turns out that much of vector (tensor) mathematics is more concise and easier to ma nipulate in such notation than in terms of corresponding component notations. However, there are many circumstances in which use of the component forms of vectors (and tensors) is more helpful – or essential. In this section, vectors are discussed in terms of components – component form . 1.3.1 The Cartesian Basis Consider three dimensional (Euc lidean) space. In this space, consider the three unit vectors 3 2 1,, eee having the properties 01 3 3 2 2 1 =⋅=⋅=⋅ ee ee ee , (1.3.1) so that they are mutually perpendicular (mutually orthogonal ), and 13 3 2 2 1 1 =⋅=⋅=⋅ ee ee ee , (1.3.2) so that they are unit vectors. Such a set of orthogonal unit vectors is called an orthonormal set, Fig. 1.3.1. Note further that this orthonormal system {}3 2 1,,eee is right-handed , by which is meant 3 2 1 e ee=× (or 1 3 2 e ee=× or 2 1 3 e ee=× ). This set of vectors {}3 2 1,,eee forms a basis, by which is meant that any other vector can be written as a linear combination of these vectors, i.e. in the form 33 22 11 e e e a a a a ++= (1.3.3) Figure 1.3.1: an orthonormal set of base vectors and Cartesian components 1 or absolute or invariant or direct or vector notation 1e2e3e3 3 ea⋅≡aa 2 2 ea⋅≡a 1 1 ea⋅≡a Section 1.3 Solid Mechanics Part III Kelly 16By repeated application of Eqn. 1.1.2 to a vector a, and using 1.3.2, the scalars in 1.3.3 can be expressed as (see Fig. 1.3.1) 3 2 2 2 1 1 , , ea ea ea ⋅=⋅=⋅= a a a (1.3.4) The scalars 2 1,aa and 3a are called the Cartesian components of a in the given basis {}3 2 1,,eee . The unit vectors are called base vectors when used for this purpose. Note that it is not necessary to have three mutually orthogonal vectors, or vectors of unit size, or a right-handed system, to form a ba sis – only that the thr ee vectors are not co- planar. The right-handed orthonormal set is ofte n the easiest basis to use in practice, but this is not always the case – for example, wh en one wants to describe a body with curved boundaries (see later). The dot product of two vectors u and v, referred to the above basis, can be written as () ( ) () () () () () () () () () 33 22 113 3 33 2 3 23 1 3 133 2 32 2 2 22 1 2 123 1 31 2 1 21 1 1 1133 22 11 33 22 11 vuvuvuvu vu vuvu vu vuvu vu vuv v v u u u ++=⋅+⋅+⋅+⋅+⋅+⋅+⋅+⋅+⋅=++⋅++=⋅ ee ee eeee ee eeee ee eee e e e e e vu (1.3.5) Similarly, the cross product is () ( ) ()()() ()()() ()()() () () ()3 12 21 2 13 31 1 23 323 3 33 2 3 23 1 3 133 2 32 2 2 22 1 2 123 1 31 2 1 21 1 1 1133 22 11 33 22 11 e e eee ee eeee ee eeee ee eee e e e e e vu vuvu vuvu vuvuvu vu vuvu vu vuvu vu vuv v v u u u −+−−−=×+×+×+×+×+×+×+×+×=++×++=× (1.3.6) This is often written in the form 3 2 13 2 13 2 1 v vvu uue ee vu=× , (1.3.7) that is, the cross product is e qual to the determinant of the 33× matrix ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ 3 2 13 2 13 2 1 v vvu uue ee Section 1.3 Solid Mechanics Part III Kelly 171.3.2 The Index Notation The expression for the cross product in term s of components, Eqn. 1.3.6, is quite lengthy – for more complicated quanti ties things get unmanageably long. Thus a short-hand notation is used for these component equations, and this index notation2 is described here. In the index notation, the expression for the vector a in terms of the components 3 2 1,,aaa and the corresponding basis vectors 3 2 1,,eee is written as ∑ ==++=3 133 22 11 iiia a a a e e e e a (1.3.8) This can be simplified further by using Einstein’s summation convention , whereby the summation sign is dropped and it is unde rstood that for a repeated index ( i in this case) a summation over the range of the index (3 in this case3) is implied. Thus one writes iiae a= . This can be further shortened to, simply, ia. The dot product of two vectors wri tten in the index notation reads iivu=⋅vu Dot Product (1.3.9) The repeated index i is called a dummy index , because it can be replaced with any other letter and the sum is the same; for exampl e, this could equally well be written as jjvu=⋅vu or kkvu . For the purpose of writing the vector cross product in index notation, the permutation symbol (or alternating symbol ) ijkε can be introduced defined by ⎪⎩⎪⎨⎧ −+ = equal are indices moreor twoif 0)3,2,1( ofn permutatio oddan is ),,( if1)3,2,1( ofn permutatio even an is ),,( if1 kjikji ijkε (1.3.10) For example (see Fig. 1.3.2), 011 122132123 =−=+= εεε 2 or indicial or subscript or suffix notation 3 2 in the case of a two-dimensional space/analysis Section 1.3 Solid Mechanics Part III Kelly 18 Figure 1.3.2: schematic for the permut ation symbol (clockwise gives +1) Note that ikj kji jik kij jki ijk εεεεεε −=−=−=== (1.3.11) and that, in terms of the base vectors { ▲Problem 7}, k ijk j i e eeε=× (1.3.12) and {▲Problem 7} ()k j i ijk eee⋅×=ε . (1.3.13) The cross product can now be written concisely as { ▲Problem 8} kji ijkvue vuε=× Cross Product (1.3.14) Introduce next the Kronecker delta symbol ijδ, defined by ⎩⎨⎧ =≠=jiji ij,1,0δ (1.3.15) Note that 111=δ but, using the index notation, 3 =iiδ . The Kronecker delta allows one to write the expressions defi ning the orthonormal basis vect ors (1.3.1, 1.3.2) in the compact form ij j iδ=⋅ee Orthonormal Basis Rule (1.3.16) The triple scalar product (1.1.4) can now be written as ()() 3 2 13 2 13 2 1 w w wv v vu u uwvuwvuw vu kji ijkkmmji ijkmm kji ijk ===⋅ =⋅× εδεε e e wvu (1.3.17) 1 2 3 Section 1.3 Solid Mechanics Part III Kelly 19 Note that, since the determinant of a matrix is equal to the determinant of the transpose of a matrix, this is equivalent to () 3 3 32 2 21 1 1 wvuwvuwvu =⋅× wvu (1.3.18) Here follow some useful formulae involving th e permutation and Krone cker delta symbol {▲Problem 13}: pk ijp ijkjp iq jq ip kpq ijk δεεδδδδεε 2=−= (1.3.19) Finally, here are some other important identit ies involving vectors; th e third of these is called Lagrange’s identity : () () () () ( ) () () () () () () [] ()[] ()[] ()[] ()[] ()[] cdbabcdaacbddcbadcbacdba dc badbdacbcadcbacbabca cbaba ba baba ×⋅+×⋅+×⋅=×⋅×⋅−×⋅=×××⋅⋅⋅⋅=×⋅×⋅−⋅=××⋅−=×⋅×2 2 2 (1.3.20) 1.3.3 Matrix Notation for Vectors The symbolic notation v and index notation iive (or simply iv) can be used to denote a vector. Another notation is the matrix notation : the vector v can be represented by a 13× matrix (a column vector ): ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ 321 vvv Matrices will be denoted by square brack ets, so a shorthand notation for this matrix/vector would be []v. The elements of the matrix []v can be written in the element form iv. The element form for a matrix is esse ntially the same as the index notation for the vector it represents. Formally, a vector can be represented by the ordered triplet of real numbers, ()3 2 1,,vvv . The set of all vectors can be represented by 3R, the set of all ordered triplets of real numbers: Section 1.3 Solid Mechanics Part III Kelly 20(){ }R vvvvvv R ∈ =3 2 1 3 2 13,,|,, (1.3.21) It is important to note the distinction betw een a vector and a matrix : the former is a mathematical object independent of any basis, th e latter is a representation of the vector with respect to a particular basis – use a diffe rent set of basis vector s and the elements of the matrix will change, but the matrix is stil l describing the same vector. Said another way, there is a difference between an element (vector) v of Euclidean vector space and an ordered triplet 3Rvi∈ . This notion will be discussed more fully in the next section. As an example, the dot product can be written in the matrix notation as Here, the notation []Tu denotes the 31× matrix (the row vector ). The result is a 11× matrix, i.e. a scalar, in element form iivu. 1.3.4 Cartesian Coordinates Thus far, the notion of an origin has not been used. Choose a point o in Euclidean (point) space, to be called the origin . An origin together with a right-handed orthonormal basis {}ie constitutes a ( rectangular ) Cartesian coordinate system , Fig. 1.3.3. Figure 1.3.3: a Cartesian coordinate system A second point v then defines a position vector ov−, Fig. 1.3.3. The components of the vector ov− are called the ( rectangular ) Cartesian coordinates of the point v 4. For brevity, the vector ov− is simply labelled v, that is, one uses th e same symbol for both the position vector and associated point. 4 That is, “components” are used for vectors and “coordinates” are used for points ov(point) ovv−= (vector) 1e2e3e (point) “short” matrix notation “full” matrix notation[][] [ ] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = 321 3 2 1T vvv u uu vu Section 1.3 Solid Mechanics Part III Kelly 21 1.3.5 Problems 1. Evaluate vu⋅ where 3 2 1 2 3 e e eu −+= , 3 2 1 4 2 4 e e e v +−= . 2. Prove that for any vector u, 3 3 2 2 1 1 ) ( ) ( ) ( eeu eeu eeu u ⋅+⋅+⋅= . [Hint: write u in component form.] 3. Find the projection of the vector 3 2 12 e e eu +−= on the vector 3 2 1 7 4 4 e e e v +−= . 4. Find the angle between 3 2 1 6 2 3 e e e u −+= and 3 2 13 4 e e e v +−= . 5. Write down an expression for a unit vector parallel to the resultant of two vectors u and v (in symbolic notation). Find this vector when 3 2 1 5 4 2 e e e u −+= , 3 2 1 3 2 e e ev ++= (in component form). Check that your final vector is indeed a unit vector. 6. Evaluate vu×, where 3 2 1 2 2 e e e u +−−= , 3 2 12 2 e e e v +−= . 7. Verify that m ijm j i e eeε=× . Hence, by dotting each side with ke, show that ()k j i ijk eee⋅×=ε . 8. Show that kji ijkvue vuε=× . 9. The triple scalar product is given by ()kji ijk wvuε=⋅× wvu . Expand this equation and simplify, so as to express the triple scalar product in full (non-index) component form. 10. Write the following in index notation: v, 1ev⋅, kev⋅. 11. Show that jiijbaδ is equivalent to ba⋅. 12. Verify that 6 =ijk ijkεε . 13. Verify that jp iq jq ip kpq ijk δδδδεε −= and hence show that pk ijp ijkδεε 2= . 14. Evaluate or simplify th e following expressions: (a) kkδ (b) ijijδδ (c) jk ijδδ (d) kj jkv3 1δε 15. Prove Lagrange’s identity 1.3.20b. 16. If e is a unit vector and a an arbitrary vector, show that ()()eaeeeaa ××+⋅= which is another representation of Eqn. 1.1.2, where a can be resolved into components parallel and perpendicular to e. Section 1.4 Solid Mechanics Part III Kelly 221.4 Matrices and Element Form 1.4.1 Matrix – Matrix Multiplication In the next section, §1.5, rega rding vector transf ormation equations, it will be necessary to multiply various matrices with each other (of sizes 13×, 31× and 33×). It will be helpful to write these matrix multipli cations in a short-hand element form. First, it has been seen th at the dot product of two v ectors can be represented by [][]vuT, or iivu. Similarly, the matrix multiplication [][]Tvu gives a 33× matrix with element form jivu or, in full, ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ 33 23 1332 22 1231 21 11 vu vuvuvu vuvuvu vu vu This type of matrix represents the tensor product of two vectors, written in symbolic notation as vu⊗ (or simply uv). Tensor products will be di scussed in detail in a later section. Next, the matrix multiplication [] [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ≡ 321 33 32 3123 22 2113 12 11 uuu Q Q QQ Q QQ Q Q uQ is a 13× matrix with elements [][]()j ij i uQ≡uQ . The elements of [][ ]uQ are the same as those of [][]T TQu , which in element form reads [][]()ijj i Qu≡T TQu . The expression [][]Qu is meaningless, but [][]QuT {▲Problem 1} is a 31× matrix with elements [][]()ji j i Qu≡ QuT. This leads to the following rule: 1. if a vector pre-multiplies a matrix []Q → it is the transpose []Tu 2. if a matrix []Q pre-multiplies the vector → it is []u 3. if summed indices are “b eside each other”, as the j in ji jQu or j ijuQ → the matrix is []Q 4. if summed indices are not beside each other, as the j in ijjQu → the matrix is the transpose, []TQ Section 1.4 Solid Mechanics Part III Kelly 23 Finally, consider the multiplication of 33× matrices. Again, this follows the “beside each other” rule for the summed index. For example, [][]BA gives the 33× matrix {▲Problem 5} [] []()kj ik ij BA= BA , and the multiplication [][]BAT is written as [][]()kj ki ij BA= BAT. There is also the important identity [][]()[][]T T TAB BA= (1.4.1) Note also the following (which applies to both the index notation and element form): (i) if there is no free index, as in iivu, there is one element (representing a scalar) (ii) if there is one free index, as in ji jQu , it is a 13× (or 31×) matrix (representing a vector) (iii) if there are two free indices, as in kj kiBA , it is a 33× matrix (representing, as will be seen later, a second-order tensor) 1.4.2 The Trace of a Matrix Another important notation i nvolving matrices is the trace of a matrix, defined to be the sum of the diagonal terms, and denoted by []iiA A A A ≡++=33 22 11 trA The Trace (1.4.2) 1.4.3 Proble ms 1. Show that [][]QuT is a 31× matrix with elements jijQu (write the matrices out in full) 2. Show that [] []()[] []T T TQu uQ= 3. Are the three elements of [][]uQ the same as those of [][]QuT? 4. What is the element form for the matrix representation of ()cba⋅? 5. Write out the 33× matrices A and B in full, i.e. in terms of ,,12 11AA etc. and verify that []kj ik ij BA= AB for 1 ,2== j i . 6. What is the element form for (i) [][]TBA (ii) [][][]vAvT (there is no ambiguity here, since [][]()[][][] []()vAv vAvT T= ) (iii) [][][]BABT 7. Show that []Atr=ij ijAδ . 8. Show that 3 2 1 3 2 1 ] det[k ji ijk k j i ijk AAA AAA ε ε = =A . Section 1.5 Solid Mechanics Part III Kelly 241.5 Coordinate Transformation of Vector Components Very often in practical problems, the compone nts of a vector are known in one coordinate system but it is necessary to find them in some other coordinate system. For example, one might know that the force f acting “in the 1x direction” has a certain value, Fig. 1.5.1 – this is equivalent to knowing the 1x component of the force, in an 2 1xx− coordinate system. One might then want to know what force is “acting” in some other direction – for example in the 1x′ direction shown – this is equivalent to asking what the 1x′ component of the force is in a new 2 1xx′−′ coordinate system. Figure 1.5.1: a vector represented usin g two different coordinate systems The relationship between the components in one coordinate system and the components in a second coordinate system are called the transformation equations . These transformation equations are derived and discussed in what follows. 1.5.1 Rotations and Translations Any change of Cartesian coordinate systems can be split up into a translation of the base vectors and a rotation of the base vectors. A transl ation of the base vectors does not change the components of a vector. Mathema tically, this can be e xpressed by saying that the components of a vector a are ae ⋅i, and these three quantities do not change under a translation of base vectors. 1.5.2 Components of a Vector in Different Systems Vectors are mathematical objects which exist independently of any coordinate system. Introducing a coordinate system for the pur pose of analysis, one could choose, for example, a certain Cartesian coordi nate system with base vectors ie and origin o, Fig. 1.5.2. In that case the vector can be written as 33 22 11 e e e u u u u ++= , and 3 2 1,, uuu are its components. 1x component of force 1x2x f 1x′2x′1x′ component of force Section 1.5 Solid Mechanics Part III Kelly 25 Now a second coordinate system can be introduced (with the same origin), this time with base vectors ie′. In that case, the vector can be written as 33 22 11 e e e u ′′+′′+′′= u u u , where 3 2 1,, uuu′′′ are its components in this second coordi nate system, as shown in the figure. Thus the same vector can be written in more than one way: 33 22 11 33 22 11 e e e e e e u ′′+′′+′′=++= u u u u u u The first coordinate system is often referred to as “the 321xxox system” and the second as “the 321xxxo′′′ system”. Figure 1.5.2: a vector represented usin g two different coordinate systems Note that the new coordinate system is obtained from the first one by a rotation of the base vectors. The figure shows a rotation θ about the 3x axis (the sign convention for rotations is positive counterclockwise). Two Dimensions Concentrating for the moment on the two dimensions 2 1xx−, from trigonometry (refer to Fig. 1.5.3), [] [] [] []22 1 12 12 122 11 cos sin sin cos e ee ee e u u u u uCP BD AB OBu u ′+′+′−′=++−=+= θθ θθ and so 2x′2x 1x1x′ 1u2u′1u′ 2u θθ o1e′ 2e′ vector components in second coordinate system vector components in first coordinate system 2 1 22 1 1 cos sinsin cos u u uu u u ′+′=′−′= θθθθ Section 1.5 Solid Mechanics Part III Kelly 26In matrix form, these transforma tion equations can be written as ⎥⎦⎤ ⎢⎣⎡ ′′ ⎥⎦⎤ ⎢⎣⎡−=⎥⎦⎤ ⎢⎣⎡ 21 21 cos sinsin cos uu uu θθθθ Figure 1.5.3: geometry of the 2D coordinate transformation The 22× matrix is called the transformation or rotation matrix []Q. By pre- multiplying both sides of these equations by the inverse of []Q, []1−Q , one obtains the transformation equations transforming from []T 2 1uu to []T 2 1uu′′ : ⎥⎦⎤ ⎢⎣⎡ ⎥⎦⎤ ⎢⎣⎡ −=⎥⎦⎤ ⎢⎣⎡ ′′ 21 21 cos sinsin cos uu uu θθθθ An important property of the tran sformation matrix is that it is orthogonal , by which is meant that [][]T 1Q Q=− Orthogonality of Transformation/Rotation Matrix (1.5.1) Three Dimensions It is straight forward to show that, in the full three dimensions, Fig. 1.5.4, the components in the two coordinate sy stems are related through ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ′′′ ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ′ ′ ′′ ′ ′′ ′ ′ = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ 321 3 3 2 3 1 33 2 2 2 1 23 1 2 1 1 1 321 ), cos(), cos(), cos(), cos(), cos(), cos(), cos(), cos(), cos( uuu xx xx xxxx xx xxxx xx xx uuu where ), cos(j ixx′ is the cosine of the angle between the ix and jx′ axes. These nine quantities are called the direction cosines of the coordinate transformation. Again denoting these by the letter Q, ), cos( ),, cos(2 1 12 1 1 11 xx Qxx Q ′ =′ = , etc., so that ), cos(j i ij xx Q ′ = , (1.5.2) 2x′2x 1x1x′ 1u2u′1u′ 2u θθθ A BP D oC Section 1.5 Solid Mechanics Part III Kelly 27 one has the matrix equations ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ′′′ ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡321 33 32 3123 22 2113 12 11 321 uuu Q Q QQ Q QQ Q Q uuu or, in element form and short-hand matrix notation, [][][]uQ u′= ′= Kjij i uQ u (1.5.3) Figure 1.5.4: two different coordinate systems in a 3D space Note : • some authors define the matrix of direction cosines to consist of the components ), cos(j i ij xx Q ′= , so that the subscript i refers to the new coordinate system and the j to the old coordinate system, rather than the other way around as used here Transformation of Cartesian Base Vectors The direction cosines introduced above also re late the base vectors in any two Cartesian coordinate systems. It can be seen that ij j i Q=′⋅ee (1.5.4) This relationship is illu strated in Fig. 1.5.5 for 1=i . 1x2x 1x′2x′ 3x3x′u Section 1.5 Solid Mechanics Part III Kelly 28 Figure 1.5.5: direction cosines Formal Derivation of the Transformation Equations In the above, the transformation equations j ij i uQ u′= were derived geometrically. They can also be derived algebrai cally using the index notation as follows: start with the relations jj kk u u e e u ′′== and post-multiply both sides by ie to get (the corresponding matrix representation is to the right (also, see Problem 2 in §1.4.3)): [][] [] [] [] [] uQ uQu uee ee ′= ′=→′= ′=→′=→⋅′′=⋅ KK jij iijj iijj kiki jj i kk uQ uQu uQu uu u T T Tδ The inverse equations are { ▲Problem 3} [][][]uQ uT=′ =′ Kjji i uQ u (1.5.5) Orthogonality of the Transformation Matrix []Q As in the two dimensional case, the transformation matrix is orthogonal, [][]1 T −=Q Q . This follows from 1.5.3, 1.5.5. Example Consider a Cartesian coordinate system with base vectors ie. A coordinate transformation is carried out with the new basis given by 3)3( 3 2)3( 2 1)3( 1 33)2( 3 2)2( 2 1)2( 1 23)1( 3 2)1( 2 1)1( 1 1 e e e ee e e ee e e e n n nn n nn n n ++=′++=′++=′ What is the transformation matrix? 1e′2e′ 1e 3e′2 1 2 1), cos( ee′⋅=′xx 3 1 3 1), cos( ee′⋅=′xx1 1 1 1), cos( ee′⋅=′xx Section 1.5 Solid Mechanics Part III Kelly 29 Solution The transformation matrix consists of the direction cosines j i j i ij xx Q ee′⋅=′ = ), cos( , so ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ′′′ ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ 321 )3( 3)2( 3)1( 3)3( 2)2( 2)1( 2)3( 1)2( 1)1( 1 321 uuu n n nn n nn n n uuu ■ 1.5.3 Problems 1. The angles between the axes in two coor dinate systems are gi ven in the tables below. 1x 2x 3x 1x′ o135 o60 o120 2x′ o90 o45 o45 3x′ o45 o60 o120 Construct the correspondi ng transformation matrix []Q and verify that it is orthogonal. 2. The 321xxxo′′′ coordinate system is obtained from the 321xxox coordinate system by a positive (counterclockwise) rotation of θ about the 3x axis. Find the (full three dimensional) transformation matrix []Q. A further positive rotation β about the 2x axis is then made to give the 321xxxo′′′′′′ coordinate system. Find the corresponding transformation matrix []P. Then construct the transformation matrix []R for the complete transformation from the 321xxox to the 321xxxo′′′′′′ coordinate system. 3. Beginning with the expression i kk i jj u u ee ee ′⋅′′=′⋅ , formally derive the relation jji i uQ u=′ ([][][]uQ uT=′ ). Section 1.6 Solid Mechanics Part III Kelly 301.6 Vector Calculus 1 - Differentiation Calculus involving vectors is discussed in this section, rather intuitively at first and more formally toward the end of this section. 1.6.1 The Ordinary Calculus Consider a scalar-valued function of a scalar , for example the time-dependent density of a material )(tρρ= . The calculus of scalar valued functions of scalars is just the ordinary calculus. Some of the important concepts of the ordinary calculus are reviewed in Appendix B to this Chapter, §1.B.2. 1.6.2 Vector-valued Functions of a scalar Consider a vector-valued function of a scalar , for example the time-dependent displacement of a particle )(tuu= . In this case, the derivative is defined in the usual way, tt t t dtd tΔ−Δ+=→Δ)() (lim0u u u, which turns out to be simply the derivative of the coefficients1, ii dtdu dtdu dtdu dtdu dtde e e eu≡++=33 22 11 Partial derivatives can also be defined in the usual way. For example, if u is a function of the coordinates, ),,(3 2 1 xxxu , then 13 2 1 3 2 1 1 0 1),,(),, (lim 1xxxx xxx x xxΔ− Δ+=∂∂ →Δu u u Differentials of vectors are also defined in the usual way, so that when 3 2 1,, uuu undergo increments 3 3 2 2 1 1 , , u duu duu du Δ=Δ=Δ= , the differential of u is 33 22 11 e e e u du du du d ++= and the differential and actual increment uΔ approach one another as 0 , ,3 2 1 →ΔΔΔ u u u . 1 assuming that the base vectors do not depend on t Section 1.6 Solid Mechanics Part III Kelly 31Space Curves The derivative of a vector can be interpreted geometrically as shown in Fig. 1.6.1: uΔ is the increment in u consequent upon an increment tΔ in t. As t changes, the end-point of the vector )(tu traces out the dotted curve Γ shown – it is clear that as 0→Δt , uΔ approaches the tangent to Γ, so that dtd/u is tangential to Γ. The unit vector tangent to the curve is denoted by τ: dtddtd // uuτ= (1.6.1) Figure 1.6.1: a space curve; (a) the tangen t vector, (b) increment in arc length Let s be a measure of the length of the curve Γ, measured from some fixed point on Γ. Let sΔ be the increment in arc-length corresponding to increments in the coordinates, []T 3 2 1 , , u u uΔΔΔ=Δu , Fig. 1.6.1b. Then, from the ordinary calculus (see Appendix 1.A.2), ()()()()2 32 22 12du du du ds ++= so that 2 32 22 1⎟ ⎠⎞⎜ ⎝⎛+⎟ ⎠⎞⎜ ⎝⎛+⎟ ⎠⎞⎜ ⎝⎛=dtdu dtdu dtdu dtds But 33 22 11e e eu dtdu dtdu dtdu dtd++= so that dtds dtd=u (1.6.2) )(tu ) ( t tΔ+uuΔτ •sΓ 1x2x •• 1du2du dssΔ (a) (b) Section 1.6 Solid Mechanics Part III Kelly 32 Thus the unit vector tangent to the curve can be written as dsd dtdsdtd u uτ ==// (1.6.3) If u is interpreted as the position vector of a particle and t is interpreted as time, then dtd/u v= is the velocity vector of the particle as it moves with speed dtds/ along Γ. Example (of particle motion) A particle moves along a curve whose parametric equations are 2 12t x= , t t x 42 2−= , 533−=t x where t is time. Find the component of the velocity at time 1=t in the direction 3 2 1 2 3 e e ea +−= . Solution The velocity is () () {} 1 at 3 2 453 4 2 3 2 13 22 12 = +−=−+−+ == tt t t tdtd dtd e e ee e erv The component in the given direction is avˆ⋅, where aˆ is a unit vector in the direction of a, giving 7/148 . ■ Curvature The curvature )(sκ of a space curve is defined to be the length of the rate of change of the unit tangent vector: 22 )(dsd dsdsuτ==κ Note that τΔ is in a direction perpendicular to τ, Fig. 1.6.2. In fact, this can be proved as follows: since τ is a unit vector, ττ⋅ is a constant ( 1=), and so () 0 /=⋅ds dττ , but also, ()dsd dsd ττττ⋅=⋅ 2 and so τ and dsd/τ are perpendicular. The unit vector defined in this way is called the principal normal vector : Section 1.6 Solid Mechanics Part III Kelly 33dsdτνκ1= Figure 1.6.2: the curvature This can be seen geometrically in Fig. 1.6.2: from the small triangle, τΔ is a vector of magnitude sΔκ in the direction of the vector normal to τ. The radius of curvature R is defined as the reciprocal of the curvature; it is the radius of the circle which just touches the curve at s, Fig. 1.6.2. Finally, the unit vector perpendicular to both the tangent vector and the principal normal vector is called the unit binormal vector : ντb×= The planes defined by these vectors are shown in Fig. 1.6.3; they are called the rectifying plane , the normal plane and the osculating plane . Figure 1.6.3: the unit tangent, principal normal and binormal vectors and associated planes )(sτ•)(1 sRκ= •) (dss+ττΔ)(sν ) (dss+νsΔκ sΔκ τ•ν bNormal plane Osculating plane Rectifying plane Section 1.6 Solid Mechanics Part III Kelly 34Rules of Differentiation The derivative of a vector is also a vector and the usual rules of differentiation apply, () ()dtd dtdtdtddtd dtd dtd ααα vvvv uvu +=+=+ )( (1.6.4) Also, it is straight forward to show that { ▲Problem 2} () () av av av av av av ×+×=× ⋅+⋅=⋅dtd dtd dtd dtd dtd dtd (1.6.5) (The order of the terms in the cross-product expression is important here.) 1.6.3 Fields In many applications of vector calculus, a sc alar or vector can be associated with each point in space x. In this case they are called scalar or vector fields . For example )(xθ temperature a scalar field (a scalar-valued function of position) )(xv velocity a vector field (a vect or valued function of position) These quantities will in general depend also on time, so that one writes ),(txθ or ),(txv . Partial differentiation of scalar and vector fields with respect to the variable t is symbolised by t∂∂/. On the other hand, partial differentiation with respect to the coordinates is symbolised by ix∂∂/. The notation can be made more compact by introducing the subscript comma to denote partial differentiation with respect to the coordinate variables, in which case i i x∂∂= /,φφ , k j i jki xx u u ∂∂∂= /2 , , and so on. 1.6.4 The Gradient of a Scalar Field Let )(xφ be a scalar field. The gradient of φ is a vector field defined by (see Fig. 1.6.4) xee e e ∂∂≡∂∂=∂∂+∂∂+∂∂=∇ φφφφφφ i ixx x x3 32 21 1 Gradient of a Scalar Field (1.6.6) The gradient φ∇ is of considerable importance because if one takes the dot product of φ∇ with xd, it gives the increment in φ : Section 1.6 Solid Mechanics Part III Kelly 35 )() ( x x xe e x d dddxxdxxd i ijj i i φ φφφφφ −+==∂∂=⋅∂∂=⋅∇ (1.6.7) Figure 1.6.4: the gradient of a vector If one writes xd as e ex dx d= , where e is a unit vector in the direction of dx, then ne e dd dxd φ φφ ≡⎟ ⎠⎞⎜ ⎝⎛=⋅∇ direction in (1.6.8) This quantity is called the directional derivative of φ, in the direction of e, and will be discussed further in §1.6.11. The gradient of a scalar field is also called the scalar gradient , to distinguish it from the vector gradient (see later)2, and is also denoted by φφ∇≡ grad (1.6.9) Example (of the Gradient of a Scalar Field) Consider a two-dimensio nal temperature field 2 22 1x x+=θ . Then 22 11 2 2 e e x x+=∇θ For example, at )0,1( , 1=θ , 12e=∇θ and at )1,1(, 2=θ , 2 12 2 e e+=∇θ , Fig. 1.6.5. Note the following: (i) θ∇ points in the direction normal to the curve const.=θ (ii) the direction of maximum rate of change of θ is in the direction of θ∇ 2 in this context, a gradient is a derivative with respect to a position vector, but the term gradient is used more generally than this, e.g. see §1.12 •• xxdφ∇ Section 1.6 Solid Mechanics Part III Kelly 36(iii) the direction of zero θd is in the direction perpendicular to θ∇ Figure 1.6.5: gradient of a temperature field The curves () const. ,2 1=xxθ are called isotherms (curves of constant temperature). In general, they are called iso-curves (or iso-surfaces in three dimensions). ■ Many physical laws are given in terms of the gradient of a scalar field. For example, Fourier’s law of heat conduction relates the heat flux q (the rate at which heat flows through a surface of unit area3) to the temperature gradient through θ∇−=k q (1.6.10) where k is the thermal conductivity of the material, so that heat flows along the direction normal to the isotherms. The Normal to a Surface In the above example, it was seen that θ∇ points in the direction normal to the curve const.=θ Here it will be seen generally how and why the gradient can be used to obtain a normal vector to a surface. Consider a surface represented by the scalar function c xxxf =),,(3 2 1 , c a constant4, and also a space curve C lying on the surface, defined by the position vector 3 3 2 2 1 1 )( )( )( e e e r tx tx tx ++= . The components of r must satisfy the equation of the surface, so c txtxtxf =))(),(),((3 2 1 . Differentiation gives 03 32 21 1=∂∂+∂∂+∂∂=dtdx xf dtdx xf dtdx xf dtdf 3 the flux is the rate of flow of fluid, particles or energy through a given surface; the flux density is the flux per unit area but, as here, this is more commonly referred to simply as the flux 4 a surface can be represented by the equation c xxxf =),,(3 2 1 ; for example, the expression 42 32 22 1 =++ x x x is the equation for a sphere of radius 2 (with centre at the origin). Alternatively, the surface can be written in the form ),(2 1 3 xxg x= , for example 2 22 1 3 4 x x x −−= 1=θ 2=θ)0,1(θ∇)1,1(θ∇ Section 1.6 Solid Mechanics Part III Kelly 37which is equivalent to the equation ()0 / grad =⋅ dtdfr and, as seen in §1.6.2, dtd/r is a vector tangential to the surface. Thus f grad is normal to the tangent vector; f grad must be normal to all the tangents to all the curves through p, so it must be normal to the plane tangent to the surface. Taylor’s Series Writing φ as a function of three variables (omitting time t), so that ),,(3 2 1 xxxφφ= , then φ can be expanded in a three-dimensional Taylor’s series: () ⎭⎬⎫ ⎩⎨⎧+ ∂∂+⎭⎬⎫ ⎩⎨⎧ ∂∂+∂∂+∂∂+ =+++ L2 1 2 123 32 21 13 2 1 3 3 2 2 1 1 21),,( ) , , ( dx xdxxdxxdxxxxx dxxdxxdxx φφφφφ φ Neglecting the higher order terms, this can be written as xxx x x d d ⋅∂∂+=+φφ φ )() ( which is equivalent to 1.6.6, 1.6.7. 1.6.5 The Nabla Operator The symbolic vector operator ∇ is called the Nabla operator5. One can write this in component form as iix x x x ∂∂=∂∂+∂∂+∂∂=∇ e e e e 33 22 11 (1.6.11) One can generalise the idea of the gradient of a scalar field by defining the dot product and the cross product of the vector operator ∇ with a vector field ()•, according to the rules () () () ()•×∂∂=•×∇•⋅∂∂=•⋅∇ ii iix xe e , (1.6.12) The following terminology is used: u uu u ×∇=⋅∇=∇= curldivgrad φφ (1.6.13) 5 or del or the Gradient operator Section 1.6 Solid Mechanics Part III Kelly 38 These latter two are discussed in the following sections. 1.6.6 The Divergence of a Vector Field From the definition (1.6.12), the divergence of a vector field )(xa is the scalar field () 33 22 11div xa xa xaxaaxii jj ii ∂∂+∂∂+∂∂=∂∂=⋅⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂=⋅∇= e e a a Divergence of a Vector Field (1.6.14) Differential Elements & Physical interpretations of the Divergence Consider a flowing compressible6 material with velocity field ),,(3 2 1 xxxv . Consider now a differential element of this material, with dimensions 3 2 1 , , x xxΔΔΔ , with bottom left-hand corner at ) ,,(3 2 1 xxx , fixed in space and through which the material flows7, Fig. 1.6.6. The component of the velocity in the1x direction, 1v, will vary over a face of the element but, if the element is small , the velocities will vary linearly as shown; only the components at the four corners of the face are shown for clarity. Since [distance = time × velocity], the volume of material flowing through the right-hand face in time tΔ is tΔ times the “volume” bounded by the four corner velocities (between the right-hand face and the plane surface denoted by the dotted lines); it is straightforward to show that this volume is equal to the volume shown to the right, Fig. 1.6.6b, with constant velocity equal to the average velocity avev, which occurs at the centre of the face. Thus the volume of material flowing out is8 t vxxaveΔΔΔ3 2 and the volume flux , i.e. the rate of volume flow, is avevxx3 2ΔΔ . Now ) , , (3 21 3 2 21 2 1 1 1 x xx xx xv vave Δ+Δ+Δ+= Using a Taylor’s series expansion, and neglecting higher order terms, 31 3 21 21 2 21 11 1 3 2 1 1 ),,(xvxxvxxvx xxxv vave∂∂Δ+∂∂Δ+∂∂Δ+ ≈ 6 that is, it can be compressed or expanded 7 this type of fixed volume in space, used in analysis, is called a control volume 8 the velocity will change by a small amount during the time interval tΔ. One could use the average velocity in the calculation, i.e. () ) ,( ),(1 1 21t t vt v Δ++ x x , but in the limit as 0→Δt , this will reduce to ),(1t vx Section 1.6 Solid Mechanics Part III Kelly 39with the partial derivatives evaluated at ),,(3 2 1 xxx , so the volume flux out is ⎭⎬⎫ ⎩⎨⎧ ∂∂Δ+∂∂Δ+∂∂Δ+ ΔΔ 31 3 21 21 2 21 11 1 3 2 1 1 3 2 ),,(xvxxvxxvx xxxvxx Figure 1.6.6: a differential element; (a) fl ow through a face, (b) volume of material flowing through the face The net volume flux out (rate of volume flow out through the right-hand face minus the rate of volume flow in through the left-hand face) is then ()1 1 3 2 1 /xvxxx ∂∂ΔΔΔ and the net volume flux per unit volume is 1 1/x v∂∂ . Carrying out a similar calculation for the other two coordinate directions leads to net unit volume flux out of an elemental volume : vdiv 33 22 11≡∂∂+∂∂+∂∂ xv xv xv (1.6.15) which is the physical meaning of the divergence of the velocity field. If 0div>v , there is a net flow out and the density of material is decreasing. On the other hand, if 0 div=v , the inflow equals the outflow and the density remains constant – such a material is called incompressible9. A flow which is divergence free is said to be isochoric . A vector v for which 0 div=v is said to be solenoidal . Notes : • The above result holds only in the limit when the element shrinks to zero size – so that the extra terms in the Taylor series tend to zero and the velocity field varies in a linear fashion over a face • consider the velocity at a fixed point in space, ),(txv . The velocity at a later time, ) ,( t tΔ+xv , actually gives the velocity of a different material pa rticle. This is shown in Fig. 1.6.7 below: the material particles 3,2,1 are moving through space and whereas ),(txv represents the velocity of particle 2, ) ,( t tΔ+xv now represents the velocity of particle 1, which has moved into position x. This point is important in the considerat ion of the kinematics of materials, to be discussed in Chapter 2 9 a liquid , such as water, is a material which is very incompressible ),, (3 2 1 1 1 xxx xvΔ+ ),,(3 2 1 xxx 1xΔ2xΔ ) ,, (3 3 2 1 1 1 x xxx xv Δ+ Δ+) , , (3 3 2 2 1 1 1 x xx xx xv Δ+Δ+Δ+ ), , (3 2 2 1 1 1 xx xx xv Δ+Δ+3xΔ avev (a) (b) Section 1.6 Solid Mechanics Part III Kelly 40 Figure 1.6.7: moving material particles Another example would be the divergence of the heat flux vector q. This time suppose also that there is some generator of heat inside the element (a source ), generating at a rate of r per unit volume, r being a scalar field. Again, assuming the element to be small, one takes r to be acting at the mid-point of the element, and one considers ), (1 21 1Lx xrΔ+ . Assume a steady-state heat flow, so that the (heat) energy within the elemental volume remains constant with time - the law of balance of (heat) energy then requires that the net flow of heat out must equal the heat generated within, so ⎭⎬⎫ ⎩⎨⎧ ∂∂Δ+∂∂Δ+∂∂Δ+ ΔΔΔ=∂∂ΔΔΔ+∂∂ΔΔΔ+∂∂ΔΔΔ 33 21 22 21 11 21 3 2 1 3 2 133 3 2 1 22 3 2 1 11 3 2 1 ),,(xrxxrxxrx xxxrxxxxqxxxxqxxxxqxxx Dividing through by 3 2 1 xxxΔΔΔ and taking the limit as 0 , ,3 2 1 →ΔΔΔ x xx , one obtains r=qdiv (1.6.16) Here, the divergence of the heat flux vector fi eld can be interpreted as the heat generated (or absorbed) per unit volume per unit time in a temperature field. If the divergence is zero, there is no heat being generated (or absorbed) and the heat leaving the element is equal to the heat entering it. 1.6.7 The Laplacian Combining Fourier’s law of heat conduction (1.6.10), θ∇−=k q , with the energy balance equation (1.6.16), r=qdiv , and assuming the conductivity is constant, leads to r k=∇⋅∇−θ . Now 2 32 2 22 2 1222 x x xx x x x xiij j ij j ii ∂∂+ ∂∂+ ∂∂=∂∂=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂ ∂∂=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂⋅∂∂=∇⋅∇ θθθθδθ θθ e e (1.6.17) This expression is called the Laplacian of θ. By introducing the Laplacian operator ∇⋅∇≡∇2, one has 12 3 x xΔ− x x xΔ+12 3t time ttΔ+ time),(txv ) ,( ttΔ+xv Section 1.6 Solid Mechanics Part III Kelly 41 kr−=∇θ2 (1.6.18) This equation governs the steady state heat flow for constant conductivity. In general, the equation a=∇φ2 is called Poisson’s equation . When there are no heat sources (or sinks), one has Laplace’s equation , 02=∇θ . Laplace’s and Poisson’s equation arise in many other mathematical models in fluid mechanics, electromagnetism, etc. 1.6.8 The Curl of a Vector Field From the definition 1.6.12 and 1.6.11, the curl of a vector field )(xa is the vector field () k ij ijk j i ijjj ii xa xaax e eee ea a ∂∂=×∂∂=×∂∂=×∇= εcurl Curl of a Vector Field (1.6.19) It can also be expressed in the form j ki ijk i jk ijk k ij ijkxa xa xaa a ax x x e e ee e e a a ∂∂=∂∂=∂∂=∂∂ ∂∂ ∂∂=×∇= ε ε ε3 2 13 2 13 2 1 curl (1.6.20) Note : • the divergence and curl of a vector field are inde pendent of any coordinate system (for example, the divergence of a vector and the length and direction of acurl are independent of a coordinate system) – these will be re-defined without refere nce to any particular coordinate system when discussing tensors (see §1.14) Physical interpretation of the Curl Consider a particle with position vector r and moving with velocity rωv×= , that is, with an angular velocity ω about an axis in the direction of ω. Then {▲Problem 7} ()ω rω v 2 curl =××∇= (1.6.21) Thus the curl of a vector field is associated with rotational properties. In fact, if v is the velocity of a moving fluid, then a small paddle wheel placed in the fluid would tend to rotate in regions where 0 curl≠v , in which case the velocity field v is called a vortex field . The paddle wheel would remain stationary in regions where 0 curl=v , in which case the velocity field v is called irrotational . Section 1.6 Solid Mechanics Part III Kelly 42 1.6.9 Identities Here are some important identities of vector calculus { ▲Problem 8}: () ()() v u vuv u vu curl curl curldiv div divgrad grad grad +=++=++=+ ψφψφ (1.6.22) () () () () () () φλφλφλφφ φφφ φφφψψφφψ grad grad grad div0 curldivgrad curlcurl curl divgrad curl curlgrad div divgrad grad )(grad 2⋅+∇===⋅−⋅=××+=⋅+=+ = uov uu vvuu u uu u u (1.6.23) 1.6.10 Cylindrical and Spherical Coordinates Cartesian coordinates have been used exclusively up to this point. In many practical problems, it is easier to carry out an anal ysis in terms of cylindrical or spherical coordinates. Differentiation in these coordi nate systems is discussed in what follows10. Cylindrical Coordinates Cartesian and cylindrical coordinates are related through (see Fig. 1.6.8) zzryrx === θθ sincos , () zzxyy x r ==+= −/ tan12 2 θ (1.6.24) Then the Cartesian partial derivatives become θθθθθθθθθθ ∂∂+∂∂=∂∂ ∂∂+∂∂ ∂∂=∂∂∂∂−∂∂=∂∂ ∂∂+∂∂ ∂∂=∂∂ r r y ryr yr r x rxr x cossinsincos (1.6.25) 10 this section also serves as an introduction to the more general topic of Curvilinear Coordinates covered in §1.14-§1.17 Section 1.6 Solid Mechanics Part III Kelly 43 Figure 1.6.8: cylindrical coordinates The base vectors are related through z zr yr x e ee e ee e e =+=−= θθθθ θθ cos sinsin cos , z zy xy x r e ee e ee e e =+−=+= θθθθ θ cos sinsin cos (1.6.26) so that from Eqn. 1.6.11, after some algebra, the Nabla operator in cylindrical coordinates reads as { ▲Problem 9} z r rz r∂∂+∂∂+∂∂=∇ e e eθθ1 (1.6.27) which allows one to take the gradient of a scalar field in cylindrical coordinates: z rz r re e e∂∂+∂∂+∂∂=∇φ θφφφθ1 (1.6.28) Cartesian base vectors are independent of position. However, the cylindrical base vectors, although they are always of unit magnit ude, change direction with position. In particular, the directions of the base vectors θee,r depend on θ, and so these base vectors have derivatives with respect to θ: from Eqn. 1.6.26, rr e ee e −=∂∂=∂∂ θθ θθ (1.6.29) with all other derivatives of the base vectors with respect to zr,,θ equal to zero. The divergence can now be evaluated: xx≡1y x≡2zx≡3 •()()zr zyx ,, ,,θ≡ θ rxe• ze yeze reθe Section 1.6 Solid Mechanics Part III Kelly 44() zv v rrv rvv v vz r r z r rzz rr z r ∂∂+∂∂++∂∂=++⋅⎟ ⎠⎞⎜ ⎝⎛ ∂∂+∂∂+∂∂=⋅∇ θθ θθθ θ 11e e e e e e v (1.6.30) Similarly the curl of a vector and the Laplacian of a scalar are { ▲Problem 10} () 22 22 2 22 2 1 11 1 z rrr rvrvrr rv zv zv v rzr z r rz ∂∂+ ∂∂+∂∂+ ∂∂=∇⎥⎦⎤ ⎢⎣⎡⎟ ⎠⎞⎜ ⎝⎛ ∂∂−∂∂+⎟ ⎠⎞⎜ ⎝⎛ ∂∂−∂∂+⎟ ⎠⎞⎜ ⎝⎛ ∂∂−∂∂=×∇ φ θφφφφθ θθ θθe e e v (1.6.31) Spherical Coordinates Cartesian and spherical coordinates ar e related through (see Fig. 1.6.9) θφθφθ cossin sincos sin rzryrx === , () ()xyzy xz y x r / tan/ tan 12 2 12 2 2 −− =+ =++= φθ (1.6.32) and the base vectors are related through φφθφθφθθφθφθθθφφθφθφφθφθ φθθφ θφ θ cos sinsin sin cos cos coscos sin sin cos sinsin coscos sin cos sin sinsin cos cos cos sin y xz y xz y x rr zr yr x e e ee e e ee e e ee e ee e e ee e e e +−=− + =+ + =−=+ + =− + = (1.6.33) Figure 1.6.9: spherical coordinates In this case the non-zero derivatives of the base vectors are xe• ze yeφere θe xyz •()()φθ,, ,, r zyx≡ φrθ Section 1.6 Solid Mechanics Part III Kelly 45 rr e ee e −=∂∂=∂∂ θθ θθ, θ φφ θφ θθφθφθφ e e ee ee e cos sincossin −−=∂∂=∂∂=∂∂ rr (1.6.34) and it can then be shown that { ▲Problem 11} () () 22 2 2 2 22 2 22 22 2 sin1 cot 1 2sin1sinsin1 1sin1 1 φϕ θ θϕθ θϕϕϕϕφθθθθφϕ θ θϕϕϕ φ θφ θ ∂∂+∂∂+∂∂+∂∂+∂∂=∇∂∂+∂∂+∂∂=⋅∇∂∂+∂∂+∂∂=∇ r r rrr rv rvrvrrrr r r rr ve e e (1.6.35) 1.6.11 The Dire ctional Derivative Consider a function ()xφ . The difference between its values at position x and at position wx+, where w is some vector, Fig. 1.6.10, is ()()x wxφ φφ −+=d (1.6.36) Figure 1.6.10: the directional derivative φ x w ε1=ε0=ε)(xφ) (wx+φ )(D xwφ Section 1.6 Solid Mechanics Part III Kelly 46An approximation to φd can be obtained by introducing a parameter ε and by considering the function () wxεφ+ ; one has ()()x wx φεφε=+=0 and ()( ) wx wx +=+=φεφε1 . If one treats φ as a function of ε, a Taylor’s series about 0=ε gives () ()L+ + += = = 022 2 02)0( )( ε ε εεφε εεφεφεφ dd dd or, writing it as a function of wxε+ , () L++ +=+ =wx x wx εφεεφεφ ε0)( ) (dd By setting 1=ε , the derivative here can be seen to be a linear approximation to the increment φd, Eqn. 1.6.36. This is defined as the directional derivative of the function )(xφ at the point x in the direction of w, and is denoted by () wx wx εφεφ ε+ =∂ =0][dd The Directional Derivative (1.6.37) The directional derivative is also written as ()xwφD. The power of the directional derivative as defined by Eqn. 1.6.37 is its generality, as seen in the following example. Example (the Directional Derivative of the Determinant) Consider the directional derivative of the determinant of the 22× matrix A, in the direction of a second matrix T (the word “direction” is obviously used loosely in this context). One has () () () () () ()[] 1221 21 12 1122 22 1121 21 12 12 22 22 11 11 00det ][ det TA TA TA TAT A T A T AT Adddd −−+=+ +−+ + =+ = ∂ == ε ε ε εεεε εεT A TAA ■ The Directional Derivative and The Gradient Consider a scalar-valued function φ of a vector z. Let z be a function of a parameter ε, () () ()() εεεφφ3 2 1 , , z z z≡ . Then Section 1.6 Solid Mechanics Part III Kelly 47εφ εφ εφ dd ddz z ddi iz z⋅∂∂=∂∂= Thus, with wxzε+= , ()() wxz zz wx ⋅∂∂=⎟ ⎠⎞⎜ ⎝⎛⋅∂∂= =∂ = =φ εφεφεφ ε ε 0 0][dd dd (1.6.38) which can be compared with Eqn. 1.6.8. Note that for Eqns. 1.6.8 and 1.6.38 to be consistent definitions of the directional derivative, w here should be a unit vector. 1.6.12 Formal Treat ment of Vect or Calculus Consider a vector h, an element of the Euclidean vector space E, E∈h . In order to be able to speak of limits as elements become “s mall” or “close” to each other in this space, one requires a norm. Here, take the standard Euclidean norm on E, Eqn. 1.2.8, hh hh h ⋅=≡ , (1.6.39) Consider next a scalar function R Ef→: . If there is a constant 0>M such that () h h M f≤ as o h→ , then one writes ()()h h O f= as o h→ (1.6.40) This is called the Big Oh (or Landau ) notation. Eqn. 1.6.40 states that ()hf goes to zero at least as fast as h. An expression such as ()()()h h h O g f += (1.6.41) then means that () () h hg f− is smaller than h for h sufficiently close to o. Similarly, if ()0→hhf as o h→ (1.6.42) then one writes ()()h h o f= as o h→ . This implies that ()hf goes to zero faster than h. A field is a function which is defined in a Euclidean (point) space 3E. A scalar field is then a function R Ef→3: . A scalar field is differentiable at a point 3E∈x if there exists a vector () E Df∈x such that Section 1.6 Solid Mechanics Part III Kelly 48 ()()()()h hx x hx o Df f f +⋅+=+ for all E∈h (1.6.43) In that case, the vector ()xDf is called the derivative (or gradient ) of f at x (and is given the symbol ()xf∇ ). Now setting w hε= in 1.6.43, where E∈w is a unit vector, dividing through by ε and taking the limit as 0→ε , one has the equivalent statement () () wx wx εεε+ =⋅∇ =fddf 0 for all E∈w (1.6.44) which is 1.6.38. In other words, for the derivative to exist, the scalar field must have a directional derivative in all directions at x. Using the chain rule as in §1.6.11, Eqn. 1.6.44 can be expressed in terms of the Cartesian basis {}ie, ()jj i ii iwxfwxff e e wx ⋅∂∂=∂∂=⋅∇ (1.6.45) This must be true for all w and so, in a Cartesian basis, ()i ixff e x∂∂=∇ (1.6.46) which is Eqn. 1.6.6. 1.6.13 Problems 1. A particle moves along a curve in space defined by ()()()33 2 22 133 8 4 4 e e e r t t t t t t −+++−= Here, t is time. Find (i) a unit tangent vector at 2=t (ii) the magnitudes of the tangential and nor mal components of acceleration at 2=t 2. Use the index notation (1.3.12) to show that () av av av ×+×=×dtd dtd dtd. Verify this result for 2 12 32 1 , 3 e e ae e v t t t t +=−= . [Note: the permutation symbol and the unit vectors are independent of t; the components of the vectors are scalar functions of t which can be differentiated in the usual way, for example by using the product rule of differentiation.] 3. The density distribution throughout a material is given by xx⋅+=1ρ . (i) what sort of function is this? (ii) the density is given in symbolic notation - write it in index notation (iii) evaluate the gradient of ρ Section 1.6 Solid Mechanics Part III Kelly 49(iv) give a unit vector in the direction in which the density is increasing the most (v) give a unit vector in any direction in which the density is not increasing (vi) take any unit vector other than the base vectors and the other vectors you used above and calculate dxd/ρ in the direction of this unit vector (vii) evaluate and sketch all these quantities for the point (2,1). In parts (iii-iv), give your answer in (a) symbolic, (b) index, and (c) full notation. 4. Consider the scalar field defined by z yx x 2 32++=φ . (i) find the unit normal to the surface of constant φ at the origin (0,0,0) (ii) what is the maximum value of the directional derivative of φ at the origin? (iii) evaluate dxd/φ at the origin if ) (3 1ee x+=ds d . 5. If 31 221 1321 e e e u x xx xxx ++ = , determine udiv and ucurl . 6. Determine the constant a so that the vector () ()()3 3 1 2 3 2 1 2 1 2 3 e e e v axx x x x x ++−++= is solenoidal. 7. Show that ω v2 curl= . 8. Verify the identities (1.6.23). 9. Use (1.6.11) to derive the Nabla operator in cylindrical coordinates (1.6.27). 10. Derive Eqn. (1.6.31), the curl of a vector and the Laplacian of a scalar in the cylindrical coordinates. 11. Derive (1.6.35), the gradient, divergence and Laplacian in spherical coordinates. 12. Show that the directional derivative ) (D uvφ of the scalar-valued function of a vector uu u⋅=)(φ , in the direction v, is vu⋅2 . 13. Show that the directional derivative of the functional () ∫ ∫−⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛=l l dxxvxp dx dxvdEI xvU 0 02 22 )()(21)( in the direction of ) (xω is given by ∫ ∫−l l dxx xp dx dxx d dxxvdEI 0 022 22 )()()( )(ωω. Section 1.7 Solid Mechanics Part III Kelly 501.7 Vector Calculus 2 - Integration 1.7.1 Ordinary Integrals of a Vector A vector can be integrated in the ordinary way to produce another vector, for example (){}3 2 12 13 22 123215 653 2 e e e e e e −+−=−+−∫dt t tt 1.7.2 Line Integrals Discussed here is the notion of a definite integral involving a vector function that generates a scalar. Let 33 22 11 e e e x x x x ++= be a position vector tracing out the curve C between the points 1p and 2p. Let f be a vector field. Then {}∫∫∫++=⋅=⋅ C Cp pdxf dxf dxf d d3 3 2 2 1 12 1xf xf is an example of a line integral. Example (of a Line Integral) A particle moves along a path C from the point )0,0,0( to )1,1,1( , where C is the straight line joining the points, Fig. 1.7.1. The particle moves in a force field given by ()32 31 232 1 22 1 20 14 6 3 e e e f xx xx x x + −+= What is the work done on the particle? Figure 1.7.1: a particle moving in a force field •• Cf xd Section 1.7 Solid Mechanics Part III Kelly 51Solution The work done is (){}∫∫+ −+=⋅= C Cdxxx dxxx dxx x d W32 31 2 32 1 22 1 20 14 6 3 xf The straight line can be written in the parametric form t xt xtx ===3 2 1 , ,, s o t h a t ()3136 11 201 02 3=+−=∫dtt t t W or ()313 3 2 1 =++⋅=⋅=∫∫ C Cdt dtdtdW eeefxf ■ If C is a closed curve, i.e. a loop, the line integral is often denoted ∫⋅ Cdxv . Note : • in fluid mechanics and aerodynamics, when v is the velocity field, this integral ∫⋅C dxv is called the circulation of v about C 1.7.3 Conservative Fields If for a vector f one can find a scalar φ such that φ∇=f ( 1 . 7 . 1 ) then (1) ∫⋅2 1p pdxf is independent of the path C joining 1p and 2p (2) 0=⋅∫ Cdxf around any closed curve C In such a case, f is called a conservative vector field and φ is its scalar potential1. For example, the work done by a conservative force field f is )()(1 22 12 12 12 1p p d dxxd dp pp pi ip pp pφφφφφ −==∂∂=⋅∇=⋅ ∫∫∫∫x xf which clearly depends only on the values at the end-points 1p and 2p, and not on the path taken between them. It can be shown that a vector f is conservative if and only if of= curl {▲Problem 3}. 1 in general, of course, there does not exist a scalar field φ such that φ∇=f ; this is not surprising since a vector field has three scalar components whereas φ∇ is determined from just one Section 1.7 Solid Mechanics Part III Kelly 52 Example (of a Conservative Force Field) The gravitational force field 3e f mg−= is an example of a cons ervative vector field. Clearly, of= curl , and the gravitational scalar potential is 3mgx−=φ . Also, [] () ()1 2 1 3 2 3 3 3 )( )(2 12 1p p px pxmg dx mg d mg Wp pp pφφ−=− −= −=⋅−= ∫ ∫x e ■ Example (of a Conservative Force Field) Consider the force field 32 31 22 1 13 3 21 3 ) 2( e e e f xx x x xx +++= Show that it is a conservative force field, fi nd its scalar potential and find the work done in moving a particle in this field from )1,2,1(− to ) 4,1,3(. Solution One has oe e e f = +∂∂∂∂∂∂= 2 312 13 3 213 2 13 2 1 3 2/ / / curl xx x x xxx x x so the field is conservative. To determine the scalar potential, let 3 32 21 133 22 11 e e e e e ex x xf f f∂∂+∂∂+∂∂=++φφφ. Equating coefficients and integrating leads to ),(),(),( 1 13 313 1 22 13 23 31 22 1 xxr xxxxq xxxxp xx xx + =+ =++= φφφ which agree if one chooses 22 13 31, ,0 xxrxxq p === , so that 3 31 22 1 xx xx+=φ , to which may be added a constant. The work done is simply Section 1.7 Solid Mechanics Part III Kelly 53 202)1,2,1()4,1,3( =−−= φφ W ■ Helmholtz Theory As mentioned, a conservative vector field which is irrotational, i.e. φ∇=f , implies of=×∇ , and vice versa . Similarly, it can be shown that if one can find a vector a such that a f×∇= , where a is called the vector potential , then f is solenoidal, i.e. 0=⋅∇f {▲Problem 4}. Helmholtz showed that a vector can always be represented in terms of a scalar potential φ and a vector potential a:2 Type of Vector Condition Representation General a f ×∇+∇=φ Irrotational (conservative) of=×∇ φ∇=f Solenoidal 0=⋅∇f a f×∇= 1.7.4 Double Integrals The most elementary type of two-dimensional integral is that over a plane region. For example, consider the integral over a region R in the 2 1xx− plane, Fig. 1.7.2. The integral ∫∫ Rdxdx2 1 then gives the area of R and, just as the one dimensiona l integral of a function gives the area under the curve, the integral ∫∫ Rdxdxxxf2 1 2 1),( gives the volume under the (in general, curved) surface ) ,(2 1 3 xxf x= . These integrals are called double integrals . 2 this decomposition can be made unique by requiring that 0 f→ as ∞→x ; in general, if one is given f, then φ and a can be obtained by solving a number of differential equations Section 1.7 Solid Mechanics Part III Kelly 54 Figure 1.7.2: integration over a region Change of variables in Double Integrals To evaluate integrals of the type ∫∫Rdxdxxxf2 1 2 1),( , it is often convenient to make a change of variable. To do this, one must find an elemental surface area in terms of the new variables, 21,tt say, equivalent to that in the 2 1,xx coordinate system, 2 1dxdx dS= . The region R over which the integration ta kes place is the plane surface 0),(2 1=xxg . Just as a curve can be represented by a position vector of one single parameter t (cf. §1.6.2), this surface can be represented by a position vector with two parameters3, 1t and 2t: 2 212 1 211 ),( ),( e e x ttx ttx + = Parameterising the plane surface in this way, one can calculate the element of surface dS in terms of 21,tt by considering curves of constant 21,tt, as shown in Fig. 1.7.3. The vectors bounding the element are 2 2const )2( 1 1const )1( 1 2, dttd d dttd dt t∂∂= =∂∂= =xx xxx x (1.7.2) so the area of the element is given by dtdtJ dtdtt td d dS1 2 1 2 1)2( )1(=∂∂×∂∂=×=x xx x (1.7.3) where J is the Jacobian of the transformation, 3 for example, the unit circle 012 22 1=−+x x can be represented by 22 1 12 1sin cos e e x t t t t + = , 1 01≤<t, π2 02≤<t (21,tt being in this case the polar coordinates r, θ, respectively) 1x2x3x R),(2 1 3 xxf x= Section 1.7 Solid Mechanics Part III Kelly 5522 1221 11 22 2112 11 or tx txtx tx J tx txtx tx J ∂∂ ∂∂∂∂ ∂∂ = ∂∂ ∂∂∂∂ ∂∂ = (1.7.4) The Jacobian is also often written using the notation () ()212 1 2 1 2 1,,,ttxxJ dtJdt dxdx∂∂= = The integral can now be written as ∫∫ RdtJdtttf2 1 21),( Figure 1.7.3: a surface element Example Consider a region R, the quarter unit-circle in the first quadrant, 2 1 2 1 0 x x−≤≤ , 1 01≤≤x . The moment of inertia about the 1x – axis is defined by ∫∫≡ Rx dxdxx I2 12 21 Transform the integral into the new coordinate system 21,tt by making the substitutions4 2 1 2 2 1 1 sin , cos t t xt tx = = . Then 1 2 1 22 1 2 22 1221 11 cos sinsin costt t tt t t tx txtx tx J =−= ∂∂ ∂∂∂∂ ∂∂ = 4 these are the polar coordinates, 21,tt equal to r, θ, respectively )1(xd )2(xd 1t1 1 t tΔ+ 2t2 2 t tΔ+ 1x2x dS Section 1.7 Solid Mechanics Part III Kelly 56 so 16sin2 12/ 0221 03 11ππ = =∫∫dtdtt t Ix ■ 1.7.5 Surface Integrals Up to now, double integrals over a plane region have been considered. In what follows, consideration is given to integrals over more complex, curved, surfaces in space, such as the surface of a sphere. Surfaces Again, a curved surface can be parameterized by 21,tt, now by the position vector 3 213 2 212 1 211 ),( ),( ),( e e e x ttx ttx ttx + + = One can generate a curve C on the surface S by taking )(1 1 stt= , )(2 2 stt= so that C has position vector, Fig. 1.7.4, ()())(),(2 1 stst sx x= A vector tangent to C at a point p on S is, from Eqn. 1.6.3, dsdt t dsdt t dsd2 21 1∂∂+∂∂=x x x Figure 1.7.4: a curved surface Many different curves C pass through p, and hence there are many different tangents, with different corr esponding values of ds dtdsdt / ,/2 1 . Thus the partial derivatives 2 1 /,/ t t∂∂∂∂ x x must also both be tangential to C and so a normal to the surface at p is given by their cross-product, and a unit normal is 1x2x3x ),(21ttx)(sxC Ss Section 1.7 Solid Mechanics Part III Kelly 57 2 1 2 1/t t t t ∂∂×∂∂ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂×∂∂=x x x xn (1.7.5) In some cases, it is possible to use a non-pa rametric form for the surface, for example c xxxg =),,(3 2 1 , in which case the normal can be obtained simply from g ggrad/ grad=n . Example (Parametric Representati on and the Normal to a Sphere) The surface of a sphere of radius a can be parameterised as5 {}31 22 1 12 1 cos sin sin cos sin e e e x t t t t t a + + = , π π 2 0, 02 1 ≤≤≤≤ t t Here, lines of const1=t are parallel to the 2 1xx− plane (“parallels”), whereas lines of const2=t are “meridian” lines, Fig. 1.7.5. If one takes the simple expressions s tst −== 2/ ,2 1π , over 2/ 0π≤≤s , one obtains a curve 1C joining ) 1,0,0( and ) 0,0,1(, and passing through )2/1,2/1,2/1( , as shown. Figure 1.7.5: a sphere The partial derivatives with respect to the parameters are {} {}22 1 12 1 231 22 1 12 1 1 cos sin sin sinsin sin cos cos cos e exe e ex t t t t att t t t t at + −=∂∂− + =∂∂ so that {}31 1 22 12 12 12 2 2 1cos sin sin sin cos sin e e ex xt t t t t t at t+ + =∂∂×∂∂ 5 these are the spherical coordinates (see §1.6.10); φθ==2 1,t t 1x2x3x 1C2/1π=tn Section 1.7 Solid Mechanics Part III Kelly 58 and a unit normal to the spherical surface is 31 22 1 12 1 cos sin sin cos sin e e e n t t t t t + + = For example, at 4/2 1π==tt (this is on the curve 1C), one has ()321 221 1214/,4/ e e e n ++=ππ and, as expected, it is in the same direction as r. ■ Surface Integrals Consider now the integral dS S∫∫f where f is a vector function and S is some curved surface. As for the integral over the plane region, 2 1 2 1const const 1 2dtdtt td d dSt t∂∂×∂∂= × =x xx x , only now dS is not “flat” and x is three dimensional. The integral can be evaluated if one parameterises the surface with 21,tt and then writes 2 1 2 1dtdtt tS∂∂×∂∂∫∫x xf One way to evaluate this cross product is to use the relation ( Lagrange’s identity , Problem 15, §1.3) () () ()()()()cbda dbca dcba ⋅⋅−⋅⋅=×⋅× (1.7.6) so that 2 2 1 2 2 1 1 2 1 2 12 2 1⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂⋅∂∂−⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂⋅∂∂ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂⋅∂∂=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂×∂∂⋅⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂×∂∂=∂∂×∂∂ tt t t tt t t t t t tx x x x xx x x x x x x (1.7.7) Example (Surface Area of a Sphere) Using the parametric form for a s phere given above, one obtains 12 42 2 1sint at t=∂∂×∂∂ x x so that Section 1.7 Solid Mechanics Part III Kelly 592 2 12 00124 sin area a dtdtt a dS Sπππ = ==∫∫∫∫ ■ Flux Integrals Surface integrals often involve the normal to the surface, as in the following example. Example If 332 22 2 1314 e e e f xx x xx +−= , evaluate dS S∫∫⋅nf , where S is the surface of the cube bounded by 1 ,0 ;1,0 ;1,03 2 1 === x x x , and n is the unit outward normal, Fig. 1.7.6. Figure 1.7.6: the unit cube Solution The integral needs to be evaluated ov er the six faces. For the face with 1e n+= , 11=x and () 2 4 43 21 01 03 3 21 01 01 332 22 2 13 = = ⋅+− =⋅ ∫∫ ∫∫∫∫dxdxx dxdx xx x x dS See e e nf Similarly for the other five sides, whence 23=⋅∫∫dS Snf . ■ Integrals of the form dS S∫∫⋅nf are known as flux integrals and arise quite often in applications. For example, consider a material flowing with velocity v, in particular the flow through a small surface element dS with outward unit normal n, Fig. 1.7.7. The volume of material flowing through the surface in time dt is equal to the volume of the slanted cylinder shown, which is the base dS times the height. The slanted height is (= 1x2x3x 2en=3en= 1en=2e n−= Section 1.7 Solid Mechanics Part III Kelly 60velocity × time) is dtv, and the vertical height is then dtnv⋅ . Thus the rate of flow is the volume flux (volume per unit time) through the surface element: dSnv⋅ . Figure 1.7.7: flow through a surface element The total (volume) flux out of a surface S is then6 volume flux : dS S∫∫⋅nv (1.7.8) Similarly, the mass flux is given by mass flux : dS S∫∫⋅nvρ (1.7.9) For more complex surfaces, one can write using Eqn. 1.7.3, 1.7.5, 2 1 2 1dtdtt tdS SS ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂×∂∂⋅=⋅∫∫∫∫x xf nf Example (of a Flux Integral) Compute the flux integral dS S∫∫⋅nf , where S is the parabolic cylinder represented by 3 0,2 0,3 12 1 2 ≤≤≤≤= x x x x and 331 2 12 2 e e e f xx x ++= , Fig. 1.7.8. Solution Making the substitutions 2 3 1 1 , t xtx == , so that 2 1 2t x=, the surface can be represented by the position vector 6 if v acts in the same direction as n, i.e. pointing outward, the dot produc t is positive and this integral is positive; if, on the other hand, material is flowing in through th e surface, v and n are in opposite directions and the dot product is negative, so the integral is negative nv dtnv⋅dtv Section 1.7 Solid Mechanics Part III Kelly 6132 22 1 11 e e e x t t t ++= , 3 0,2 02 1 ≤≤≤≤ t t Then 3 2 21 1 1 / , 2 / e x e e x =∂∂+=∂∂ t t t and 2 11 2 12 eex x−=∂∂×∂∂tt t so the integral becomes () () 12 2 23 02 02 1 2 11 321 2 12 1 = −⋅++∫∫dtdt t tt t ee e e e Figure 1.7.8: flux through a parabolic cylinder ■ Note : • in the above example, the value of the integral depends on the choice of n. If one chooses n− instead of n, one would obtain 12−. The normal in the opposite direction (on the “other side” of the surface) can be obtained by simply switching 1t and 2t, since 1 2 2 1 / / / / t t t t ∂∂×∂−∂=∂∂×∂∂ x x x x . Surface flux integrals can also be evaluated by first converting them into double integrals over a plane region. For example, if a surface S has a projection R on the 2 1xx− plane, then an element of surface dS is related to the projected element 2 1dxdx through (see Fig. 1.7.9) ()2 1 3 cos dxdx dS dS =⋅= enθ and so ∫∫∫∫⋅⋅=⋅ R Sdxdx dS2 1 31 ennf nf 1x2x3x nf• Section 1.7 Solid Mechanics Part III Kelly 62 Figure 1.7.9: projection of a su rface element onto a plane region The Normal and Surface Area Elements It is sometimes convenient to associate a special vector Sd with a differential element of surface area dS, where dS d nS= so that Sd is the vector with magnitude dS and direction of the unit normal to the surface. Flux integrals can then be written as ∫∫∫∫⋅=⋅ S Sd dS Sf nf 1.7.6 Volume Integrals The volume integral, or trip le integral, is a generalis ation of the double integral. Change of Variable in Volume Integrals For a volume integral, it is often convenie nt to make the change of variables ),,(),,( 321 3 2 1 ttt xxx→ . The volume of an element dV is given by the triple scalar product (Eqns. 1.1.5, 1.3.17) 3 2 1 3 2 1 3 2 1dtdtJdt dtdtdtt t tdV =∂∂⋅⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂×∂∂=x x x (1.7.10) where the Jacobian is now n 3e 2x 1xθ Section 1.7 Solid Mechanics Part III Kelly 6333 23 1332 22 1231 21 11 33 32 3123 22 2113 12 11 or tx tx txtx tx txtx tx tx J tx tx txtx tx txtx tx tx J ∂∂ ∂∂ ∂∂∂∂ ∂∂ ∂∂∂∂ ∂∂ ∂∂ = ∂∂ ∂∂ ∂∂∂∂ ∂∂ ∂∂∂∂ ∂∂ ∂∂ = (1.7.11) so that () ()()() ( ) ∫∫∫ ∫∫∫= V VdtdtdtJtttxtttxtttx dxdydzxxx3 2 1 3213 3212 3211 3 2 1 ,, ,,, ,,, ,, f f 1.7.7 Integral Theorems A number of integral theorems and relations are presented he re (without proof), the most important of which is the divergence theorem. These theorems can be used to simplify the evaluation of line, double, surface and trip le integrals. They can also be used in various proofs of other important results. The Divergence Theorem Consider an arbitrary di fferentiable vector field ),(txv defined in some finite region of physical space. Let V be a volume in this sp ace with a closed surface S bounding the volume, and let the outward normal to this bounding surface be n. The divergence theorem of Gauss states that (in symbo lic and index notation) ∫∫∫∫∂∂= =⋅ V ii Sii V SdVxvdSnv dV dS v nv div Divergence Theorem (1.7.12) and one has the following useful identities { ▲Problem 10} ∫∫∫∫∫∫ =×==⋅ V SV SV S dV dSdV dSdV dS u unnu nu curlgrad)(div φ φφ φ ( 1 . 7 . 1 3 ) By applying the divergence theorem to a very small volume, one finds that VdS S V∫⋅ = →nv v 0lim div that is, the divergence is equal to the out ward flux per unit volume, the result 1.6.15. Section 1.7 Solid Mechanics Part III Kelly 64Stoke’s Theorem Stoke’s theorem transforms line integrals into surface integrals and vice versa . It states that ()∫ ∫∫⋅=⋅ C Sds dSτf nfcurl (1.7.14) Here C is the boundary of the surface S, n is the unit outward normal and dsd/rτ= is the unit tangent vector. As has been seen, Eqn. 1.6.21, the curl of th e velocity field is a measure of how much a fluid is rotating. The direction of this vect or is along the direction of the local axis of rotation and its magnitude measures the local angular velocity of the fluid. Stoke’s theorem then states that the amount of rotati on of a fluid can be measured by integrating the tangential velocity around a curve (the line integral), or by integrating the amount of vorticity “moving through” a su rface bounded by the same curve. Green’s Theorem and Related Identities Green’s theorem relates a line integral to a doub le integral, and states that {} ∫∫ ∫ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−∂∂=+ R Cdxdxx xdx dx2 1 21 12 2 2 1 1ψψψψ , (1.7.15) where R is a region in the 2 1xx− plane bounded by the curve C. In vector form, Green’s theorem reads as ∫∫∫⋅=⋅ R Cdxdx d2 1 3 curl ef xf where 22 11 e e f ψψ+= (1.7.16) from which it can be seen that Green’s theorem is a special case of Stoke’s theorem, for the case of a plane surface (region) in the 2 1xx− plane. It can also be shown that (this is Green’s first identity ) () { }dV dS V S∫∫∫ ∫∫⋅+∇= ⋅ φψφψ φψ grad grad grad2n (1.7.17) Note that the term φgrad⋅n is the directional derivative of φ in the direction of the outward unit normal. This is often denoted as n∂∂/φ . Green’s first identity can be regarded as a multi-dimensional “integ ration by parts” – compare the rule ∫∫−= vdu uv udv with the identity re-written as () ()()()dV dS dV V S V∫∫∫ ∫∫ ∫∫∫∇⋅∇−⋅∇=∇⋅∇ φψ φψφψ n (1.7.18) Section 1.7 Solid Mechanics Part III Kelly 65or () ()()dV dS dV V S V∫∫∫∫∫ ∫∫∫⋅∇−⋅=⋅∇ u nu u ψ ψ ψ (1.7.18) One also has the relation (this is Green’s second identity ) () () {} { }dV dS V S∫∫∫ ∫∫∇−∇= ⋅−⋅ ψφφψ ψφφψ2 2grad grad n n (1.7.19) 1.7.8 Problems 1. Find the work done in moving a part icle in a force field given by 11 23 121 10 5 3 e e e f x x xx +−= along the curve 12 1+=tx , 2 22t x= , 3 3t x=, from 1=t to 2=t . (Plot the curve.) 2. Show that the following vectors are conser vative and find their scalar potentials: (i) 33 22 11 e e e x x x x ++= (ii) ()21 1221e e v x x exx+ = (iii)33 22 2 1 1 2 )/( )/1( e e e u x xx x + − = 3. Show that if φ∇=f then of= curl . 4. Show that if a f×∇= then 0=⋅∇f . 5. Find the volume beneath the surface 032 22 1 =−+ x x x and above the square with vertices ) 0,0(, ) 0,1(, ) 1,1( and ) 1,0( in the 2 1xx−plane. 6. Find the Jacobian (and sketch lines of constant 2 1,tt ) for the rotation θθθθ cos sinsin cos 2 1 22 1 1 t t xt tx +=−= 7. Find a unit normal to the circular cylinder with parametric representation 1 0,2 0, sin cos ),(1 1 32 21 11 21 ≤≤≤≤ + + = t t t t a t a tt π e e e x 8. Evaluate dS S∫ψ where 3 2 1 x xx++=ψ and S is the plane surface 2 1 3 xx x+= , 1 2 0 x x≤≤ , 1 01≤≤x . 9. Evaluate the flux integral dS S∫⋅nf where 3 2 1 2 2 e e ef ++= and S is the cone () a xx xa x ≤+=32 22 1 3 , [Hint: first paramete rise the surface with 2 1,tt .] 10. Prove the relations in (1.7.13). [Hint: firs t write the expression s in index notation.] 11. Use the divergence theorem to show that V dS S3=⋅∫nx , where V is the volume enclosed by S (and x is the position vector). 12. Verify the divergence theorem for 33 3 23 2 13 1 e e e v x x x ++= where S is the surface of the sphere 2 2 32 22 1 a x x x =++ . 13. Interpret the divergence theore m (1.7.12) for the case when v is the velocity field. See (1.6.15, 1.7.8). Interpret also the case of 0 div=v . Section 1.7 Solid Mechanics Part III Kelly 6614. Verify Stoke’s theorem for 31 23 12 e e e f x x x ++= where S is 0 12 22 1 3 ≥−−= x x x (so that C is the circle of radius 1 in the 2 1xx− plane). 15. Verify Green’s theorem for the case of 2 1 2 22 1 1 ,2 xx x x +=−= ψ ψ , with C the unit circle 12 22 1=+x x . The following relations might be useful: 0 cos sin cos sin , cos sin2 022 02 022 02= = = = ∫ ∫ ∫∫π π π πθθθ θθθ πθθ θθ d d d d 16. Evaluate ∫⋅ Cdxf using Green’s theorem, where 23 1 13 2 e e f x x+−= and C is the circle 42 22 1=+x x . 17. Use Green’s theorem to show that th e double integral of the Laplacian of p over a region R is equivalent to the integral of n⋅=∂∂ p np grad / around the curve C bounding the region: dsnpdxdxp C R∫ ∫∫∂∂= ∇2 12 [Hint: Let 1 2 2 1 / ,/ xp xp ∂+∂=∂−∂= ψ ψ . Also, show that 21 12e e ndsdx dsdx−= is a unit normal to C, Fig. 1.7.10] Figure 1.7.10: projection of a su rface element onto a plane region ds 1dx2dx C Section 1.8 Solid Mechanics Part III Kelly 671.8 Tensors Here the concept of the tensor is introduced. Tensors can be of different orders – zeroth- order tensors, first-order tens ors, second-order tensors, and so on. Apart from the zeroth and first order tensors (see below), the s econd-order tensors are the most important tensors from a practical point of view, being important quan tities in, amongst other topics, continuum mechanics, relativity, electromagnetism and quantum theory. 1.8.1 Zeroth and First Order Tensors A tensor of order zero is simply another name for a scalar α. A first-order tensor is simply another name for a vector u. 1.8.2 Second Order Tensors Notation Vectors: lowercase bold-face Latin letters, e.g. a, r, q 2nd order Tensors: uppercase bold-face Latin letters, e.g. F, T, S Tensors as Linear Operators A second -order tensor T may be defined as an operator that acts on a vector u generating another vector v, so that v uT=)(, o r1 v Tu vuT = =⋅ or Second-order Tensor (1.8.1) The second-order tensor T is a linear operator (or linear transformation )2, which means that () Tb Ta baT +=+ … distributive ()()Ta aTαα= … associative This linearity can be viewed geom etrically as in Fig. 1.8.1. Note: • the vector may also be defined in this way, as a mapping u that acts on a vector v, this time generating a scalar α, α=⋅vu . This transformation (the dot product) is linear (see properties (2,3) in §1.1.4). Thus a first-order tensor (vect or) maps a first-order tensor into a zeroth-order tensor (scalar), whereas a second-order tensor maps a first-order tensor into a first-order tensor. It will be seen that a third-order tensor maps a first-order tensor into a second-order tensor, and so on 1 both these notations for the tensor operation are us ed; here, the convention of omitting the “dot” will be used 2 An operator or transformation is a special function wh ich maps elements of one type into elements of a similar type; here, vectors into vectors Section 1.8 Solid Mechanics Part III Kelly 68 Figure 1.8.1: Linearity of the second order tensor Further, two tensors T and S are said to be equal if and only if Tv Sv= for all vectors v. Example (of a Tensor) Suppose that F is an operator which transforms every vector into its mirror-image with respect to a given pl ane, Fig. 1.8.2. F transforms a vector into another vector and the transformation is linear, as can be seen geometrically from the figure. Thus F is a second-order tensor. Figure 1.8.2: Mirror-imaging of vect ors as a second order tensor mapping ■ Example (of a Tensor) The combination ×u linearly transforms a vector into another vector and is thus a second-order tensor3. For example, consider a force f applied to a spanner at a distance r from the centre of the nut, Fig. 1.8.3. Then it can be said that the tensor ()×r maps the force f into the (moment/torque) vector fr×. 3 Some authors use the notation u~ to denote ×u a bba+ TaTb()baT+ u vuα vu+ vF⋅Fu()uFα ()vuF+ Section 1.8 Solid Mechanics Part III Kelly 69 Figure 1.8.3: the force on a spanner ■ 1.8.3 The Dyad (the tensor product) The vector dot product and vector cross product have been considered in previous sections. A third vector product, the tensor product (or dyadic product ), is important in the analysis of tensors of order 2 or mo re. The tensor product of two vectors u and v is written as4 vu⊗ Tensor Product (1.8.2) This tensor product is itself a tens or of order two, and is called dyad : vu⋅ is a scalar (a zeroth order tensor) vu× is a vector (a first order tensor) vu⊗ is a dyad (a second order tensor) It is best to define this dyad by what it does: it transforms a vector w into another vector with the direction of u according to the rule5 ) ( ) ( wvuwvu ⋅=⊗ The Dyad Transformation (1.8.3) This relation defines the symbol “ ⊗”. The length of the new vector is u times wv⋅, and the new vector has the same direction as u, Fig. 1.8.4. It can be seen that the dya d is a second order tensor, because it operates linearly on a vector to give another vector { ▲Problem 2}. Note that the dyad is not commutative, uvvu⊗≠⊗ . Indeed it can be seen clearly from the figure that () () wuv wvu ⊗≠⊗ . 4 many authors omit the ⊗ and write simply uv 5 note that it is the two vectors that are beside each other (separated by a bracket) that get “dotted” together f r Section 1.8 Solid Mechanics Part III Kelly 70 Figure 1.8.4: the dyad transformation The following important relations follow from the above definition { ▲Problem 4}, ()()()() () ( ) wvu wvuxuwv x wvu ⋅=⊗⊗⋅=⊗⊗ (1.8.4) It can be seen from these that the operati on of the dyad on a vector is not commutative: ()()uwv wvu ⊗≠⊗ (1.8.5) Example (The Projection Tensor) Consider the dyad ee⊗. From the definition 1.8.3, ()()eueuee ⋅=⊗ . But ue⋅ is the projection of u onto a line through the unit vector e. Thus ()eue⋅ is the vector projection of u on e. For this reason ee⊗ is called the projection tensor . It is usually denoted by P. Figure 1.8.5: the projection tensor ■ u v e Pu Pvwvu ) (⊗uwv Section 1.8 Solid Mechanics Part III Kelly 711.8.4 Dyadics A dyadic is a linear combination of these dyads (w ith scalar coefficients). An example might be ()()()fe dc ba ⊗−⊗+⊗ 2 3 5 This is clearly a second-order tensor . It will be seen in §1.9 that every second-order tensor can be represented by a dyadic, that is ()()()L+⊗+⊗+⊗= fe dc ba T γ β α (1.8.6) Note : • second-order tensors cannot, in ge neral, be written as a dyad, baT⊗= – when they can, they are called simple tensors Example (Angular Momentum and th e Moment of Inertia Tensor) Suppose a rigid body is rotating so that every particle in the body is instantaneously moving in a circle abou t some axis fixed in space, Fig. 1.8.6. Figure 1.8.6: a particle in motion about an axis The body’s angular velocity ω is defined as the vector whose magnitude is the angular speed ω and whose direction is along the axis of rotation. Then a particle’s linear velocity is rωv×= where wdv= is the linear speed, d is the distance between the axis and the particle, and r is the position vector of the particle from a fixed point O on the axis. The particle’s angular momentum (or moment of momentum) h about the point O is defined to be vr h×=m where m is the mass of the particle. The a ngular momentum can be written as d rω v θ Section 1.8 Solid Mechanics Part III Kelly 72ωIhˆ= (1.8.8) where Iˆ, a second-order tensor, is the moment of inertia of the particle about the point O, given by ()rrIr I ⊗−=2ˆm (1.8.9) where I is the identity tensor, i.e. a Ia= for all vectors a. To show this, it must be shown that ()ωrrIr vr ⊗−=×2. First examine vr×. It is evidently a vector perpendicular to both r and v and in the plane of r and ω; its magnitude is θsin2ωrvrvr ==× Now (see Fig. 1.8.7) () () ()re eωrωrrωrωrrIr θωcos22 2 − =⋅−=⊗− where ωe and re are unit vectors in the directions of ω and r respectively. From the diagram, this is equal to heωrθsin2. Thus both expressions are equivalent, and one can indeed write ωIhˆ= with Iˆ defined by Eqn. 1.8.9: the second-order tensor Iˆ maps the angular velocity vector ω into the angular momentum vector h of the particle. Figure 1.8.7: geometry of unit vector s for angular momentum calculation ■ 1.8.5 The Vector Space of Second Order Tensors The vector space of vectors and associated space s were discussed in §1.2. Here, spaces of second order tensors are discussed. As mentioned above, the second order te nsor is a mapping on the vector space V, ωe reθ reθcos−θcos he θsin Section 1.8 Solid Mechanics Part III Kelly 73 V V→:T (1.8.10) and follows the rules () () () Ta aTTb Ta baT αα=+=+ (1.8.11) for all V∈ba, and R∈α . Denote the set of all second order tensors by 2V. Define then the sum of two tensors 2, V∈TS through the relation () Tv SvvTS +=+ (1.8.12) and the product of a scalar R∈α and a tensor 2V∈T through () Tv vTαα= (1.8.13) Define an identity tensor 2V∈I through v Iv=, for all V∈v (1.8.14) and a zero tensor 2V∈O through o Ov=, for all V∈v (1.8.15) It follows from the definition 1.8.11 that 2V has the structure of a real vector space, that is, the sum 2V∈+TS , the product 2V∈Tα , and the following 8 axioms hold: 1. for any 2,, V∈CBA , one has ) ( ) ( CB AC BA ++=++ 2. there exists an element 2V∈O such that TTOOT =+=+ for every 2V∈T 3. for each 2V∈T there exists an element 2V∈−T , called the negative of T, such that 0 )()( =+−=−+ T T T T 4. for any 2, V∈TS , one has STTS+=+ 5. for any 2, V∈TS and scalar R∈α , T S TS αα α +=+) ( 6. for any 2V∈T and scalars R∈βα,, T T Tβαβα +=+) ( 7. for any 2V∈T and scalars R∈βα,, T T )()(αββα= 8. for the unit scalar R∈1 , TT=1 for any 2V∈T . Section 1.8 Solid Mechanics Part III Kelly 741.8.6 Problems 1. Consider the function f which transforms a vector v into β+⋅va . Is f a tensor (of order one)? [Hint: test to see whether the transformation is linear, by examining () vuf+α .] 2. Show that the dyad is a linear operator, in other words, show that () xvu wvu x wvu ) ( ) ( ) ( ⊗+⊗=+⊗ β αβα 3. When is abba⊗=⊗ ? 4. Prove that (i) () () ( ) () xuwv x wvu ⊗⋅=⊗⊗ [Hint: post-“multiply” both sides of the definition (1.8.3) by x⊗; then show that ()()()()x wvux wvu ⊗⊗=⊗⊗ .] (ii) ()()wvu wvu ⋅=⊗ [hint: pre “multiply” both sides by ⊗x and use the result of (i)] 5. Consider the dyadic (tensor) bbaa⊗+⊗ . Show that this tensor orthogonally projects every vector v onto the plane formed by a and b (sketch a diagram). 6. Draw a sketch to show the meaning of ()Pvu⋅ , where P is the projection tensor. What is the order of the resulting tensor? 7. Prove that ()××=⊗−⊗ ababba . Section 1.9 Solid Mechanics Part III Kelly 751.9 Cartesian Tensors As with the vector, a (highe r order) tensor is a mathema tical object which represents many physical phenomena and which exists indepe ndently of any coordinate system. In what follows, a Cartesian coordinate sy stem is used to describe tensors. 1.9.1 Cartesian Tensors A second order tensor and the vector it operates on can be described in terms of Cartesian components. For example, cba ) (⊗ , with 3 2 12 eee a −+= , 3 2 12 ee eb ++= and 3 2 1 e ee c ++−= , is 3 2 1 2 2 4)( ) ( e e e cbacba −+=⋅=⊗ Example (The Unit Dyadic or Identity Tensor) The identity tensor , or unit tensor , I, which maps every vector onto itself, has been introduced in the previous section. The Cartesian representation of I is i ie e e ee ee e ⊗≡⊗+⊗+⊗3 3 2 2 1 1 (1.9.1) This follows from () ()()() ()()() ue e eueeueeueeue eue eue e ue e e ee e =++=⋅+⋅+⋅=⊗+⊗+⊗=⊗+⊗+⊗ 33 22 113 3 2 2 1 13 3 2 2 1 1 3 3 2 2 1 1 u u u Note also that the identity tensor can be written as ()j i ij e e I⊗=δ , in other words the Kronecker delta gives the components of the identity tensor in a Cartesian coordinate system. ■ Second Order Tensor as a Dyadic In what follows, it will be shown that a se cond order tensor can al ways be written as a dyadic involving the Cartesian base vectors ei 1. Consider an arbitrar y second-order tensor T which operates on a to produce b, b aT=)( , or b eT=) (iia . From the linearity of T, 1 this can be generalised to the case of non-Cartesian base vectors, which might not be orthogonal nor of unit magnitude (see §1.14) Section 1.9 Solid Mechanics Part III Kelly 76b eT eT eT = + + )( )( )(3 3 2 2 1 1 a a a Just as T transforms a into b, it transforms the base vectors ei into some other vectors; suppose that w eTv eTu eT = = = )(, )(, )(3 2 1 , then () () () () () ( ) [] aew eveuaewaevaeuweaveaueaw v u b 3 2 13 2 13 2 13 2 1 ⊗+⊗+⊗=⊗+⊗+⊗=⋅+⋅+⋅=++= a a a and so 3 2 1 ew eveuT ⊗+⊗+⊗= (1.9.2) which is indeed a dyadic. Cartesian components of a Second Order Tensor The second order tensor T can be written in terms of components and base vectors as follows: write the vectors u, v and w in (1.9.2) in component form, so that () ()() LL L +⊗+⊗+⊗=⊗+⊗+⊗++= 1 33 1 22 1 113 2 1 33 22 11 e e e e e ee e e e e e T u u uu u u Introduce nine scalars ijT by letting 3 2 1 , ,i i i i i i T wTvT u === , so that 3 3 33 2 3 32 1 3313 2 23 2 2 22 1 2 213 113 2 1 12 1 111 e e e e e ee e e e e ee e e e e e T ⊗+⊗+⊗+⊗+⊗+⊗+⊗+⊗+⊗= T T TT T TT T T Second-order Cartesian Tensor (1.9.3) These nine scalars ijT are the components of the second order tensor T in the Cartesian coordinate system. In index notation, ()j i ijT e e T⊗= Thus whereas a vector has three co mponents, a second order tensor has nine components. Similarly, whereas the three vectors {}ie form a basis for the space of vectors, the nine dyads {}j ie e⊗ from a basis for the space of tensors, i.e. all second order tensors can be expressed as a linear combination of these basis tensors. It can be shown that the components of a s econd-order tensor can be obtained directly from {▲Problem 1} j i ijT Tee= Components of a Tensor (1.9.4) Section 1.9 Solid Mechanics Part III Kelly 77 which is the tensor expression an alogous to the vector expression ue⋅=i iu . Note that, in Eqn. 1.9.4, the components can be written simply as j iTee , since j i j i eTe Tee ⋅=⋅ . Example (The Stress Tensor) Define the traction vector t acting on a surface element within a material to be the force acting on that element2 divided by the area of the element, Fig. 1.9.1. Let n be a vector normal to the surface. The stress σ is defined to be that seco nd order tensor which maps n onto t, according to σn t= The Stress Tensor (1.9.5) Figure 1.9.1: stress acting on a plane If one now considers a coordina te system with base vectors ie, then j iij e eσ⊗=σ and, for example, 331 2 21 111 1 e e eσe σσσ ++= Thus the components 11σ, 21σ and 31σ of the stress tensor ar e the three components of the traction vector which act s on the plane with normal 1e. ■ Higher Order Tensors The above can be generalised to tensors of order three and higher. The following notation will be used: α, β, γ … 0th-order tensors (“scalars”) a, b, c … 1st-order tensors (“vectors”) A, B, C … 2nd-order tensors (“dyadics”) A, B, C … 3rd-order tensors (“triadics”) 2 this force would be due, for exampl e, to intermolecular forces within the material: the particles on one side of the surface element exert a force on the particles on the other side 1x 3x2x 1et11σ 21σ31σ n t Section 1.9 Solid Mechanics Part III Kelly 78A, B, C … 4th-order tensors (“tetradics”) An important third-or der tensor is the permutation tensor , defined by k j i ijk e e e⊗⊗=εE (1.9.6) whose components are those of the perm utation symbol, Eqns. 1.3.10-1.3.13. A fourth-order tensor can be written as l k j i ijklA e e e e ⊗⊗⊗=A (1.9.7) It can be seen that a zerot h-order tensor (scalar) has 1 30= component, a first-order tensor has 3 31= components, a second-order tensor has 9 32= components, so A has 27 33= components and A has 81 components. 1.9.2 Simple Contraction Tensor/vector operations can be written in component form, for example, () ()[] ij ijijkkijk j i kijkk j i ij aTaTaTa T eeee ee e e Ta ==⊗ =⊗= δ (1.9.8) This operation is called simple contraction , because the order of the tensors is contracted – to begin there was a tensor of order 2 and a tensor of or der 1, and to end there is a tensor of order 1 (it’s called “simple” to distinguish it from “d ouble” contraction – see below). This is always the case – when a tens or operates on another in this way, the order of the result will be two less than the sum of the original orders. An example of simple contraction of two s econd order tensors has already been seen in Eqn. 1.8.4a; the tensors there were simple te nsors (dyads). Here is another example: ()() ()()[] () ()l i jl ijl i jk klijl k j i klijl k kl j i ij STSTSTS T e ee ee ee ee e e e TS ⊗ =⊗ =⊗⊗ =⊗ ⊗= δ (1.9.9) From the above, the simple contraction of two second order tensors results in another second order tensor. If one writes TSA= , then the components of the new tensor are related to those of the original tensors through kj ik ij ST A= . Note that, in general, Section 1.9 Solid Mechanics Part III Kelly 79 BA AB≠ ()()BCACAB= … associative () AC AB CBA +=+ … distributive The associative and distributive properties follow from the fact that a tensor is by definition a linear operator, §1. 8.2; they apply to tensors of any order, for example, ()()BvAvAB= To deal with tensors of any or der, all one has to remember is how simple tensors operate on each other – the two vectors which are be side each other are the ones which are “dotted” together: ()() () () ( ) () () ( ) ( ) ( ) () () ( ) ( ) febadc fedcbaedacb edcbadacb dcbaacbcba ⊗⊗⊗⋅=⊗⊗⊗⊗⊗⊗⋅=⊗⊗⊗⊗⋅=⊗⊗⋅=⊗ (1.9.10) An example involving a hi gher order tensor is ( )() ()n k j i nl ijkln m l k j i mn ijkl EAEA e e e ee ee e e e E ⊗⊗⊗ =⊗⊗⊗⊗ =⋅A and C A=====⋅ BCbv AuC ABvu Aα Note the relation ()()()CD AB DCBA ⊗=⊗ (1.9.11) Powers of Tensors Integral powers of tensors are defined inductively by I T=0, TT T1−=n n, so, for example, TT T=2 The Square of a Tensor (1.9.12) TTT T=3, etc. Section 1.9 Solid Mechanics Part III Kelly 80 1.9.3 Double Contraction Double contraction, as the name implies, c ontracts the tensors twi ce as much a simple contraction. Thus, where the sum of the orde rs of two tensors is reduced by two in the simple contraction, the sum of the orders is reduced by four in doubl e contraction. The double contraction is denoted by a colon (:), e.g. ST:. First, define the double contraction of simple tensors (dyads) through ()()()()dbca dcba ⋅⋅=⊗⊗ : ( 1 . 9 . 1 3 ) So in double contraction, one takes the scalar product of four vectors which are adjacent to each other, according to the following rule: For example, ()() ()()[] ijijl j k i klijl k kl j i ij STSTS T =⋅⋅ =⊗ ⊗= eeeee e e e ST : : (1.9.14) which is, as expected, a scalar. Here is another example, the contraction of the two second order tensors I (see Eqn. 1.9.1) and vu⊗, ()() () () vuveuevu e evuI ⋅==⋅⋅=⊗⊗=⊗ iii ii i vu: : ( 1 . 9 . 1 5 ) so that the scalar product of two vectors can be writte n in the form of a double contraction. An example of double contraction involvi ng the permutation tensor 1.9.6 is { ▲Problem 10} () uv vu ×=⊗:E (1.9.16) It can be shown that the components of a four th order tensor are given by (compare with Eqn. 1.9.4) () ()()()()faecdb fedcba ⊗⋅⋅=⊗⊗⊗⊗ : Section 1.9 Solid Mechanics Part III Kelly 81()()l k j i ijklA e e e e ⊗ ⊗= ::A ( 1 . 9 . 1 7 ) In summary then, β=BA: γ=b:A cB=:A CB=:A Note the following identities: () ()() ( ) () () () () ( ) () () ( ) CBD A D ACB DC BACBA BAC CBAACB CBACBA : : :: : :: : : ⊗=⊗ =⊗⊗= =⊗= =⊗ (1.9.18) Note : • There are many operations that can be define d and performed with tensors. The two most important operations, the ones which arise most in practice, are the simple and double contractions defined above. Other possibilities are: (a) double contraction with two “horizontal” dots, ST⋅⋅, b⋅⋅A , etc., which is based on the definition of the following operation as applied to simple tensors: () ()()()()fadceb fedcba ⊗⋅⋅≡⊗⊗⋅⋅⊗⊗ (b) operations involving one cross ()×: ()()()()cb da dc ba ×⊗⊗≡⊗×⊗ (c) “double” operations involving the cross ()× and dot: () () ( ) () () () ( ) ( )() () ( ) () dbca dc badbca dc badb ca dc ba ×⋅≡⊗⋅ ×⊗⋅×≡⊗× ⋅⊗×⊗×≡⊗× ×⊗ 1.9.4 Index Notation The index notation for single and double contrac tion of tensors of any order can easily be remembered. From the above, a single contract ion of two tensors imp lies that the indices “beside each other” are the same 3, and a double contrac tion implies that a pair of indices are repeated. Thus, for example, in both symbolic and index notation: i jk ijkijk mk ijm c BAC BA = == = cBB :ACA (1.9.19) 3 compare with the “beside each other rule” for matrix multiplication given in §1.4.1 Section 1.9 Solid Mechanics Part III Kelly 821.9.5 Matrix Notation Here the matrix notation of §1.4 is extended to include second-order tensors4. The Cartesian components of a second-order te nsor can conveniently be written as a 33× matrix, [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = 33 32 3113 22 2113 12 11 T T TT T TT T T T The operations involving vector s and second-order tensors can now be written in terms of matrices, for example, The tensor product can be written as (see §1.4.1) [][] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ==⊗ 33 23 1332 22 1231 21 11 T vuvuvuvuvuvuvu vuvu vuvu (1.9.20) which is consistent with the definition of the dyadic transformation, Eqn. 1.8.3. 1.9.6 Problems 1. Use Eqn. 1.9.3 to show that the component 11T of a tensor T can be evaluated from 1 1Tee , and that 2 1 12 Tee=T (and so on, so that j i ijT Tee= ). 2. Evaluate aT using the index notation (f or a Cartesian basis). What is this operation called? Is your result equal to Ta, in other words is this operation commutative? Now carry out this operati on for two vectors, i.e. ba⋅. Is it commutative in this case? 3. Evaluate the simple contractions bA and BA, with respect to a Cartesian coordinate system (use index notation). 4. Evaluate the double contraction B:A (use index notation). 4 the matrix notation cannot be used for higher-order tensors [] [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ++++++ = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ==3 33 2 32 1 313 23 2 22 1 213 13 2 12 1 11 321 33 32 3113 22 2113 12 11 T T TT T TT T T uT uT uTuT uT uTuT uT uT uuu uT Tu symbolic notation “short” matrix notation “full” matrix notation Section 1.9 Solid Mechanics Part III Kelly 835. Show that, using a Cartesian coordinate system and the i ndex notation, that the double contraction b:A is a scalar. Write this scalar out in full in terms of the components of A and b. 6. Consider the second-order tensors 3 3 2 3 2 2 3 13 3 3 2 2 2 1 1 3 6 45 2 3 e ee e e e e e Fe e e ee e e e D ⊗+⊗−⊗+⊗=⊗+⊗−⊗+⊗= Compute DF and DF:. 7. Consider the second-order tensor 3 3 2 2 1 2 2 1 1 1 2 4 3 e ee ee e e e e e D ⊗+⊗+⊗+⊗−⊗= . Determine the image of the vector 3 2 1 5 2 4 e e e r ++= when D operates on it. 8. Write the following out in full – are these the components of s calars, vectors or second order tensors?: (a) iiB (b) kkjC (c) mnB (d) ijjiAba 9. Write ()()dcba⊗⊗ : in terms of the components of the four vectors. What is the order of the resulting tensor? 10. Show that () uv vu ×=⊗:E – see (1.9.6, 1.9.16). [Hin t: use the definition of the cross product in terms of the permutati on symbol, (1.3.14), and the fact that kji ijkεε−= .] Section 1.10 Solid Mechanics Part III Kelly 841.10 Special Second Order Tensors & Properties of Second Order Tensors In this section will be examined a number of special second order tensors, and special properties of second order tensors, which play important roles in tensor analysis. The following will be discussed: • The Identity tensor • Transpose of a tensor • Trace of a tensor • Norm of a tensor • Determinant of a tensor • Inverse of a tensor • Orthogonal tensors • Rotation Tensors • Change of Basis Tensors • Symmetric and Skew-symmetric tensors • Axial vectors • Spherical and Deviatoric tensors • Positive Definite tensors 1.10.1 The Ident ity Tensor The linear transformation which transforms every tensor into itself is called the identity tensor . This special tensor is denoted by I so that, for example, a Ia= for any vector a In particular, 3 3 2 2 1 1 , , e Iee Iee Ie === , from which it follows that, for a Cartesian coordinate system, ij ijIδ= . In matrix form, [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = 100010001 I (1.10.1) 1.10.2 The Transpose of a Tensor The transpose of a second order tensor A with components ijA is the tensor TA with components jiA; so the transpose swaps the indices, j iji j iij A A e e A e e A ⊗= ⊗=T, Transpose of a Second-Order Tensor (1.10.2) In matrix notation, Section 1.10 Solid Mechanics Part III Kelly 85 [] [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ =33 23 1332 22 1231 21 11 T 33 32 3123 22 2113 12 11 , A A AA A AA A A A A AA A AA A A A A Some useful properties and relations involving the transpose are { ▲Problem 2}: () () () () ()() () B AC CAB BCAvA uAvuB ABAAB ABuT uT uT Tuuv vuB A B AA A : : :) ( ) (: :, T TTT TT T TT TTT T TTT = =⊗=⊗=== =⊗=⊗+=+= βαβα (1.10.3) A formal definition of the transpose which does not rely on any particular coordinate system is as follows: the transpose of a sec ond-order tensor is that tensor which satisfies the identity1 uAv AvuT⋅=⋅ (1.10.4) for all vectors u and v. To see that Eqn. 1.10.4 implies 1.10.2, first note that, for the present purposes, a convenient way of writing the components ijA of the second-order tensor A is ()ijA. From Eqn. 1.9.4, ()j i ij Aee A⋅= and the components of the transpose can be written as ()j i ij eAe AT T⋅= . Then, from 1.10.4, () ()ji ji i j j i ij A==⋅=⋅= A Aee eAe AT T. 1.10.3 The Trace of a Tensor The trace of a second order tensor A, denoted by Atr, is a scalar equal to the sum of the diagonal elements of its matrix representation. Thus (see Eqn. 1.4.2) iiA=Atr Trace (1.10.5) A more formal definition, again not relyi ng on any particular coordinate system, is AIA : tr= Trace (1.10.6) 1 note that, from the linearity of tensors, Avuv uA ⋅=⋅ ; for this reason, this expression is usually written simply as uAv Section 1.10 Solid Mechanics Part III Kelly 86and Eqn. 1.10.5 follows from 1.10.6 { ▲Problem 4}. For the dyad vu⊗ {▲Problem 5}, () vuvu ⋅=⊗tr (1.10.7) Another example is () () qi iqr p qr pq j i ij EEEE =⊗ ⊗== e e e eEI E :: ) tr(2 2 δ (1.10.8) This and other important traces, and functions of the trace are listed here { ▲Problem 6}: () ()kk jj iijj iiki jk ijji ijii AAAAAAAAAAA ===== 3232 trtrtrtrtr AAAAA (1.10.9) Some useful properties and relations involving the trace are { ▲Problem 7}: () () () () ()() () ()T T T TT tr tr tr tr :tr trtr tr trtr trtr tr BA AB AB BA BAA AB A BABA ABA A = = = ==+=+== αα (1.10.10) The double contraction of two tensors was ear lier defined with re spect to Cartesian coordinates, Eqn. 1.9.14. This last expression allows one to re-define the double contraction in terms of the trace, i ndependent of any coordinate system. Consider again the real vector space of second order tensors 2V introduced in §1.8.5. The double contraction of two tensors as de fined by 1.10.10e clearly satisfies the requirements of an inner product listed in §1.2.2. Thus this scalar quantity serves as an inner product for the space 2V: ()BA BA BATtr : , =≡ (1.10.11) and generates an inner product space. Just as the base vectors {}ie form an orthonormal set in the inner product (vector dot product) of the space of vectors V, so the base dyads {}j ie e⊗ form an orthonormal set in the inner product 1.10.11 of the space of second order tensors 2V. For example, Section 1.10 Solid Mechanics Part III Kelly 87()()1 : ,1 1 1 1 1 1 1 1 =⊗⊗=⊗⊗ e e e e e ee e (1.10.12) Similarly, just as the dot product is zero for orthogonal vectors, when the double contraction of two tensors A and B is zero, one says that the tensors are orthogonal , ()0 tr :T= = BA BA , BA, orthogonal (1.10.13) 1.10.4 The Norm of a Tensor Using 1.2.8 and 1.9.10, the norm of a second order tensor A, denoted by A (or A), is defined by AA A := (1.10.14) This is analogous to the norm a of a vector a, aa⋅. 1.10.5 The Determinant of a Tensor The determinant of a second order tensor A is defined to be the determinant of the matrix []A of components of the tensor: k j i ijkk j i ijk AAAAAAA A AA A AA A A 3 2 13 2 133 32 3123 22 2113 12 11 det det εε ==⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ =A (1.10.15) Some useful properties of the determinant are { ▲Problem 8} () () () ( ) ( ) [] cbaT Tc Tb TaAvuA AA AB A AB ⋅× =⋅×==⊗=== detdet0 detdet ) det(det detdet det) det( 3T kr jq ip ijk pqr AAAε εαα (1.10.16) Note that Adet , like Atr, is independent of the choice of coordinate system / basis. Section 1.10 Solid Mechanics Part III Kelly 881.10.6 The Inver se of a Tensor The inverse of a second order tensor A, denoted by 1−A, is defined by AAI AA1 1 − −== (1.10.17) The inverse of a tensor exists only if it is non-singular (a singular tensor is one for which 0 det=A ), in which case it is said to be invertible . Some useful properties and rela tions involving the inverse are: 1 11 1 11 11 1 ) (det) det()()/1( )() ( − −−− −− −−− ==== A AAB ABA AA A α α (1.10.18) Since the inverse of the transpose is equivalent to the transpose of the inverse, the following notation is used: 1 T T1 T)( ) (− − −=≡ A A A (1.10.19) 1.10.7 Orthogonal Tensors An orthogonal tensor Q is a linear vector transformation satisfying the condition vu QvQu ⋅=⋅ (1.10.20) for all vectors u and v. Thus u is transformed to Qu, v is transformed to Qv and the dot product vu⋅ is invariant under the transformation. Thus the magnitude of the vectors and the angle between the vectors is preserved, Fig. 1.10.1. Figure 1.10.1: An orthogonal tensor Since ()vQQu Qv uQ QvQu ⋅⋅⋅=⋅=⋅T T (1.10.21) θ vu θQ QvQu Section 1.10 Solid Mechanics Part III Kelly 89it follows that for vu⋅ to be preserved under the transformation, IQQ=T, which is also used as the definition of an orthogonal tensor. Some useful properties of orthogonal tensors are{ ▲Problem 10}: 1 det, T 1T T ±==== == − QQ QQQI QQkj ki ij jk ik QQ QQδ (1.10.22) 1.10.8 Rotation Tensors If for an orthogonal tensor, 1 det+=Q , Q is said to be a proper orthogonal tensor, corresponding to a rotation . If 1det−=Q , Q is said to be an improper orthogonal tensor, corresponding to a reflection . Proper orthogonal tensors are also called rotation tensors . 1.10.9 Change of Basis Tensors Consider a rotation tensor Q which rotates the base vectors 3 2 1,,eee into a second set, 3 2 1,,eee′′′ , Fig. 1.10.2. 3,2,1= =′ ii iQee (1.10.23) Such a tensor can be termed a change of basis tensor from {}ie to {}ie′. The transpose QT rotates the base vectors ie′ back to ie and is thus change of basis tensor from {}ie′ to {}ie. The components of Q in the ie coordinate system are, from 1.10.4, j i ijQ Qee= and so j i ij j iij Q Q ee e e Q ′⋅= ⊗= , , (1.10.24) which are the direction cosines between the axes (see Fig. 1.5.5). Figure 1.10.2: Rotation of a set of base vectors The change of basis tensor can also be written in the explicit form 2e Q3e 1e 1e′2e′3e′ Section 1.10 Solid Mechanics Part III Kelly 90 i ie eQ⊗′= (1.10.25) from which the above relations can easily be derived, for example i iQee=′ , I QQ=T, etc. Consider now the operation of the change of basis tensor on a vector: ()ii i i v v e Qe Qv ′== (1.10.26) Thus Q transforms v into a second vector v′, but this new vector has the same components with respect to the basis ie′, as v has with respect to the basis ie, i ivv=′ . Example Consider the two-dimensional rotation tensor ()i i j i e e e e Q ⊗′≡⊗⎥⎦⎤ ⎢⎣⎡ +−=011 0 which corresponds to a rotation of the base vectors through 2/π . The vector []T11=v then transforms into (see Fig. 1.10.3) i i e e Qv ′⎥⎦⎤ ⎢⎣⎡ ++=⎥⎦⎤ ⎢⎣⎡ +−=11 11 Figure 1.10.3: a rotated vector ■ Similarly, for a second order tensor A, the operation ()( )()j iij j i ij j i ij j iij A A A A e e Qe Qe Qe Qe Qe e Q QAQ ′⊗′=⊗=⊗=⊗=T T T (1.10.27) results in a new tensor which has the same components with respect to the ie′, as A has with respect to the ie, ij ijA A=′ . v Qv 1e2e1e′ 2e′ Section 1.10 Solid Mechanics Part III Kelly 911.10.10 Symmetric and Skew Tensors A tensor T is said to be symmetric if it is identical to the transposed tensor, TTT= , and skew (antisymmetric ) if TT T−= . Any tensor A can be (uniquely) decomposed into a symmetric tensor S and a skew tensor W, where () ()TT 21skew21sym AA WAAA SA −=≡+=≡ (1.10.28) and T T, W W SS −= = (1.10.29) In matrix notation one has [] [ ] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −−−= ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = 000 ,23 1323 1213 12 33 23 1323 22 1213 12 11 W WW WW W W S SSS SSS S S S (1.10.30) Some useful properties of symmetric and skew tensors are { ▲Problem 13}: () () () () inverse no has0 det00 tr0 :: : :: : : T 21 TT 21 T ==⋅==−=−=+== WWvvSWWSB W BW BWBB S BSBS (1.10.31) where v and B denote any arbitrary vector and second-order tensor respectively. Note that symmetry and skew-symmetry are te nsor properties, independent of coordinate system. 1.10.11 Axial Vectors A skew tensor W has only three independent coefficients, so it behaves “like a vector” with three components. Indeed, a skew tensor can always be written in the form uω Wu×= (1.10.32) Section 1.10 Solid Mechanics Part III Kelly 92where u is any vector and ω characterises the axial (or dual ) vector of the skew tensor W. The components of W can be obtained from the components of ω through ()() () k ijkk kji p kjpk ij kk i j i j i ijW ωεωεεωω −== ⋅=×⋅=×⋅=⋅= e eee e eωe Wee (1.10.33) If one knows the components of W, one can find the components of ω by inverting this equation, whence { ▲Problem 14} 3 12 2 13 123 e e eω W W W −+−= (1.10.34) Example (of an Axial Vector) Decompose the tensor [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ == 111124321 ijT T into its symmetric and skew parts. Also find the axial vector for the skew part. Verify that aω Wa×= for 2 1eea+= . Solution One has [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = ⎪⎭⎪⎬⎫ ⎪⎩⎪⎨⎧ ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ + ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ =+= 112123231 113122141 111124321 21 21TTT S [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −− = ⎪⎭⎪⎬⎫ ⎪⎩⎪⎨⎧ ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ + ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ =−= 00100 111 0 113122141 111124321 21 21TTT W The axial vector is 3 2 3 12 2 13 123 ee e e eω +=−+−= W W W and it can be seen that 3 2 13 33 31 2 23 21 1 13 113 1 3 1 3 1 ) ( ) ( ) () ( ) ( ) )( ( eeee e ee e eee e Wa −+=+++++=+=+=+⊗= W W W W W WW W W Wi i i i j j ij j i ij δδ Section 1.10 Solid Mechanics Part III Kelly 93and 3 2 13 2 1 101110 eeeeee aω −+= =× ■ The Spin Tensor The velocity of a particle rotating in a rigid body motion is given by xωv×= , where ω is the angular velocity vector and x is the position vector relative to the origin on the axis of rotation (see Problem 9, §1.1). If the velo city can be written in terms of a skew- symmetric second order tensor w, such that v wx=, then it follows from xω wx×= that the angular velocity vector ω is the axial vector of w. In this context, w is called the spin tensor . 1.10.12 Spherical and Deviatoric Tensors Every tensor A can be decomposed into its so-called spherical part and its deviatoric part, i.e. A A A dev sph+= (1.10.35) where () () () () () () () ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ++−++−++− =−=⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ++++++ == 33 22 11 31 33 32 3123 33 22 11 31 22 2113 12 33 22 11 31 1133 22 11 3133 22 11 3133 22 11 3131 sph dev0 00 00 0tr sph A A A A A AA A A A A AA A A A A AA A AA A AA A A A AAIA A (1.10.36) Any tensor of the form Iα is known as a spherical tensor , while Adev is known as a deviator of A, or a deviatoric tensor . Some important properties of the spherical and deviatoric tensors are 0 sph: dev0) dev(sph0) dev(tr === B AAA (1.10.37) Section 1.10 Solid Mechanics Part III Kelly 94 1.10.13 Positive De finite Tensors A positive definite tensor A is one which satisfies the relation 0> vAv , ov≠∀ (1.10.38) The tensor is called positive semi-definite if 0≥ vAv . In component form, L+++++=2 2 22 1221 3113 21122 111 vAvvA vvA vvA vA vAvjiji (1.10.39) and so the diagonal elements of the matrix representation of a positive definite tensor must always be positive. It can be shown that the following conditions are necessary for a tensor A to be positive definite (although they are not sufficient): (i) the diagonal elements of []A are positive (ii) the largest element of []A lies along the diagonal (iii) 0 det>A (iv) ij jj ii A A A 2>+ for ji≠ (no sum over ji,) These conditions are seen to hold for the following matrix representation of a positive definite tensor: [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −= 100041022 A A necessary and sufficient condition for a tensor to be positive definite is given in the next section, during the discus sion of the eigenvalue problem. One of the key properties of a positiv e definite tensor is that, since 0 det>A , positive definite tensors are always invertible. An alternative definition of positive definiteness is the equivalent expression 0 :>⊗vvA ) 40.10.1( 1.10.14 Problems 1. Show that the components of the (second-order) identity tensor are given by ij ijIδ= . 2. Show that Section 1.10 Solid Mechanics Part III Kelly 95(a) ) ( ) (TvA uAvu ⊗=⊗ (b) ()()()B AC CAB BCA : : :T T= = 3. Use (1.10.4) to show that I I=T. 4. Show that (1.10.6) implies (1.10.5) for the trace of a tensor. 5. Show that () vuvu ⋅=⊗tr . 6. Formally derive the index notation for the functions 3 2 3 2)tr(,)tr(,tr,tr A A A A 7. Show that ) (tr :TBA BA= . 8. Prove (1.10.16f), () ()()[]cbaT Tc Tb Ta ⋅× =⋅× det . 9. Show that 3 :) (T1=−A A . [Hint: one way of doing this is using the result from Problem 7.] 10. Use 1.10.16b and 1.10.18d to prove 1.10.22c, 1 det±=Q . 11. Use the explicit dyadic representation of the rotation tensor, i ie eQ⊗′= , to show that the components of Q in the “second”, 321xxxo′′′ , coordinate system are the same as those in the first system [hint: use the rule j i ijQ eQe′⋅′=′ ] 12. Consider the tensor D with components (in a certain coordinate system) ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −− 2/12/1 2/12/12/1 02/1 2/12/1 Show that D is a rotation tensor (just show that D is proper orthogonal). 13. Show that () 0 tr= SW . 14. Multiply across (1.10.32), k ijk ijWωε−= , by ijpε to show that kij ijkWe ωε21−= . [Hint: use the relation 1.3.19b,pk ijk ijpδεε 2= .] 15. Show that ( )abba⊗−⊗21 is a skew tensor W. Show that its axial vector is ()abω×=21. [Hint: first prove that ()()()( ) uab baubuaaub ××=××=⋅−⋅ .] 16. Find the spherical and deviatoric parts of the tensor A for which 1=ijA . Section 1.11 Solid Mechanics Part III Kelly 961.11 The Eigenvalue Problem and Polar Decomposition 1.11.1 Eigenvalues, Eigenvectors and Invariants of a Tensor Consider a second-order tensor A. Suppose that one can find a scalar λ and a (non-zero) normalised, i.e. unit, vector nˆ such that n nA ˆ ˆλ= (1.11.1) In other words, A transforms the vector nˆ into a vector parallel to itself, Fig. 1.11.1. If this transformation is possibl e, the scalars are called the eigenvalues (or principal values ) of the tensor, and th e vectors are called the eigenvectors (or principal directions or principal axes ) of the tensor. It will be seen that there are three vectors nˆ (to each of which corresponds some scalar λ) for which the above holds. Figure 1.11.1: the action of a tensor A on a unit vector Equation 1.11.1 can be solved for the eigenva lues and eigenvectors by rewriting it as ()0ˆ=− nI Aλ (1.11.2) or, in terms of a Cartesian coordinate system, ()() () 0 ˆ ˆ0 ˆ ˆ0 ˆ ˆ =−→=− →= ⊗ − ⊗ ii jijrr ijijrr q p pq kk j i ij n nAn nAn n A ee ee e e e e e λλλδ In full, [] [] [] 0 ˆ) (ˆ ˆ0 ˆ ˆ) (ˆ0 ˆ ˆ ˆ) ( 3 3 33 2 32 1312 3 23 2 22 12113 13 2 12 1 11 =−++= +−+= ++− eee n A nAnAnA n A nAnA nAn A λλλ (1.11.3) Dividing out the base vectors, this is a set of three hom ogeneous equations in three unknowns (if one treats λ as known). From basic linear algebra, this system has a solution (apart from 0 ˆ=in ) if and only if the determinant of the coefficient matrix is zero, i.e. if A nˆ nˆλ Section 1.11 Solid Mechanics Part III Kelly 970 det) det( 33 32 3123 22 2113 12 11 = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −−− =− λλλ λ A A AA A AA A A I A (1.11.4) Evaluating the determinant, one has the following cubic characteristic equation of A, 0 III II I2 3=−+−A A Aλλλ Tensor Characteristic Equation (1.11.5) where () [] AA AA AAA detIII) tr()(trIItrI 3 2 12 2 2121 ==− =− === k j i ijkijji jj iiii AAAAA AAA ε (1.11.6) It can be seen that there are three roots 3 2 1,,λλλ , to the characteristic equation. Solving for λ, one finds that 32113 32 213 2 1 IIIIII λλλλλλλλλλλλ =++=++= AAA (1.1.7) The eigenvalues (principal values) iλ must be independent of a ny coordinate system and, from Eqn. 1.11.5, it follows that the functions A A A III,II,I are also independent of any coordinate system. They are called the principal scalar invariants (or simply invariants ) of the tensor. Once the eigenvalues are found, the eigenvect ors (principal direct ions) can be found by solving 0ˆ) (ˆ ˆ0ˆ ˆ) (ˆ0ˆ ˆ ˆ) ( 3 33 2 32 1313 23 2 22 1213 13 2 12 1 11 =−++=+−+=++− n A nAnAnA n A nAnA nAn A λλλ (1.11.8) for the three components of the principal direction vector 3 2 1 ˆ,ˆ,ˆ nnn , in addition to the condition that 1 ˆˆˆˆ ==⋅iinnnn . There will be three vectors iine nˆˆ= , one corresponding to each of the three principal values. Note : • a unit eigenvector nˆ has been used in the above discussion, but any vector parallel to nˆ, for example nˆα, is also an eigenvector (with the same eigenvalue λ): Section 1.11 Solid Mechanics Part III Kelly 98()()()()n n nA nA ˆ ˆ ˆ ˆ αλλααα === Example (of Eigenvalues and Eigenvectors of a Tensor) A second order tensor T is given with respect to the axes 321xxOx by the values [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −−−== 1 12012 6 00 05 ijT T . Determine (a) the principal values, (b) the principal directions (and sketch them). Solution: (a) The principal values are the solu tion to the characteristic equation 0) 15)( 5)( 10( 1 12 012 6 00 0 5 =+−+−= −−−−−− λλλ λλλ which yields the three principal values 15 ,5 ,103 2 1 −=== λλλ . (b) The eigenvectors are now obtained from () 0=−j ij ij n Tλδ . First, for 101=λ , 0 9 12 00 12 16 00 0 0 5 3 2 13 2 13 2 1 =−−=−−=++− n n nn n nn n n and using also the equation 1 =iinn leads to 3 2 1 )5/4( )5/3( ˆ e e n +−= . Similarly, for 52=λ and 153−=λ , one has, respectively, 0 4 12 00 12 11 00 0 0 0 3 2 13 2 13 2 1 =−−=−−=++ n n nn n nn n n and 0 16 12 00 12 9 00 0 0 20 3 2 13 2 13 2 1 =+−=−+=++ n n nn n nn n n which yield 1 2ˆ e n= and 3 2 3 )5/3( )5/4( ˆ e e n + = . The principal directions are sketched in Fig. 1.11.2. Note : • the three components of a principal direction, 3 2 1 ,, nnn , are the direction cosines between that direction and the three coordinate axes respectively. For example, for 1λ with 5/4 ,5/3 ,013 2 =−== n n n , the angles made with the coordinate axes 3 2 1 ,, xxx , are 0, 127o and 37o Section 1.11 Solid Mechanics Part III Kelly 99 Figure 1.11.2: eigenvecto rs of the tensor T ■ 1.11.2 Real Symmetric Tensors Suppose now that A is a real symmetric tensor (real meaning that its components are real). In that case it ca n be proved (see below) that1 (i) the eigenvalues are real (ii) the three eigenvectors form an orthonormal basis {}inˆ. In that case, the components of A can be written relative to the basis of principal directions as (see Fig. 1.11.3) ()j i ijA n n A ˆ ˆ⊗= (1.11.9) Figure 1.11.3: eigenvectors forming an orthonormal set The components of A in this new basis can be obtained from Eqn. 1.9.4, () ⎩⎨⎧ ≠==⋅=⋅= jijiA ijj ij i ij ,0,ˆ ˆˆ ˆ λλn nnAn (no summation over j) (1.11.10) 1 this was the case in the previous example – the tens or is real symmetric and the principal directions are orthogonal 1ˆn 2ˆn3ˆn 1e2e3e3x 1x2x3ˆn1ˆn 2ˆn Section 1.11 Solid Mechanics Part III Kelly 100where iλ is the eigenvalue correspond ing to the basis vector inˆ. Thus2 ∑ =⊗=3 1ˆ ˆ ii ii n n Aλ Spectral Decomposition (1.11.11) This is called the spectral decomposition (or spectral representation ) of A. In matrix form, [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = 321 000 000 λλλ A (1.11.12) For example, the tensor used in the previous example can be written in terms of the basis vectors in the principal directions as ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −= 15 000 500 0 10 T , basis: j in n ˆ ˆ⊗ To prove that real symmetric tensors have real eigenvalues and or thonormal eigenvectors, take 3 2 1ˆ,ˆ,ˆ nnn to be the eigenvectors of an arbitrary tensor A, with components i i i nnn3 2 1 ˆ,ˆ,ˆ , which are solutions of (the 9 equations – see Eqn. 1.11.2) () ()() 0 ˆ0 ˆ0 ˆ 3 32 21 1 =−=−=− nI AnI AnI A λλλ ( 1 . 9 . 1 3 ) Dotting the first of these by 1ˆn and the second by 1ˆn, leads to () () 0 ˆˆ0 ˆˆ 2 12 1 22 11 2 1 =⋅−⋅=⋅−⋅ nn n Annn n An λλ Using the fact that TAA= , subtracting these equations leads to () 0 ˆˆ2 1 1 2 =⋅− nnλλ (1.11.14) Assume now that the eigenvalu es are not all real. Sin ce the coefficients of the characteristic equation are all real, this imp lies that the eigenvalues come in a complex conjugate pair, say 1λ and 2λ, and one real eigenvalue 3λ. It follows from Eqn. 1.11.13 that the components of 1ˆn and 2ˆn are conjugates of each other, say iba n+=1ˆ , iba n−=2ˆ , and so 2 it is necessary to introduce the summation sign here, because the summation convention is only used when two indices are the same – it cannot be used wh en there are more than two indices the same Section 1.11 Solid Mechanics Part III Kelly 101 ()() 0 ˆˆ2 2 2 1 >+=−⋅+=⋅ b a ba ba nn i i It follows from 1.11.14 that 01 2=−λλ which is a contradiction, since this cannot be true for conjugate pairs. Thus the original assu mption regarding complex roots must be false and the eigenvalues are all real. With thr ee distinct eigenvalues, Eqn. 1.11.14 (and similar) show that the eigenvectors form an orthonormal set. When the eigenvalues are not distinct, more than one set of eigenvector s may be taken to form an orthonormal set (see the next subsection). Equal Eigenvalues There are some special tensors for which two or three of the principal directions are equal. When all three are equal, λλλλ ===3 2 1 , one has I Aλ= , and the tensor is spherical: every direction is a principal direction, since n nI nA ˆ ˆ ˆλλ== for all nˆ. When two of the eigenvalues are equa l, one of the eigenvectors wi ll be unique but the other two directions will be arbitrary – one can choose any two principal directions in the plane perpendicular to the uniquely determined direct ion, so that the three form an orthonormal set. Eigenvalues and Positi ve Definite Tensors Since n nA ˆ ˆλ= , then λλ=⋅=⋅ nnnAn ˆˆˆˆ . Thus if A is positive definite, Eqn. 1.10.38, the eigenvalues are all positive . In fact, it can be shown that a tensor is posi tive definite if and only if its symmetric part has all positive eigenvalues. Note : • if there exists a non-zero eigenvector correspondi ng to a zero eigenvalue, then the tensor is singular. This is the case for the skew tensor W, which is singular. Since ω oωω Wω 0==×= (see , §1.10.11), the axial vector ω is an eigenvector corresponding to a zero eigenvalue of W 1.11.3 Maximum and Minimum Values The diagonal components of a tensor A, 33 22 11, AAA , have different values in different coordinate systems. However, the three ei genvalues include the extreme (maximum and minimum) possible values that any of these three components can take, in any coordinate system. To prove this, consider an arbitrary set of unit base vectors 3 2 1,,eee , other than the eigenvectors. From Eqn. 1.9.4, the components of A in a new coordinate system with these base vectors are j i ijA Aee=′ Express 1e using the eigenvectors as a basis, 3 2 1 1ˆ ˆ ˆ n n n e γβα ++= Then Section 1.11 Solid Mechanics Part III Kelly 102[]32 22 12 321 11 0 00 00 0 λγλβλα γβα λλλ γβα ++= ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ =′A Without loss of generality, let 3 2 1λλλ≥≥ . Then, with 12 2 2=++γβα , one has ( ) ()11 32 22 12 2 2 2 3 311 32 22 12 2 2 2 1 1 AA ′=++≤++=′=++≥++= λγλβλαγβαλλλγλβλαγβαλλ which proves that the eigenvalues include th e largest and smallest possible diagonal element of A. 1.11.4 The Cayley-Hamilton Theorem The Cayley-Hamilton theorem states that a tensor A (not necessarily symmetric) satisfies its own characteristic equation 1.11.5: 0I A A AA A A =−+− III II I2 3 (1.11.15) This can be proved as follows: one has n nA ˆ ˆλ= , where λ is an eigenvalue of A and nˆ is the corresponding eigenvector . A repeated application of A to this equation leads to n nA ˆ ˆn nλ= . Multiplying 1.11.1 by nˆ then leads to 1.11.15. The third invariant in Eqn. 1.11.6 can now be written in terms of traces by a double contraction of the Cayley-Hamilton equation with I, and by using the definition of the trace, Eqn.1.10.6: [] []3 21 2 23 3 312 2 21 2 32 32 3 )tr( trtr tr III0 III3 tr tr)tr( trtr tr0 III3 trII trI tr0 : III: II: I: A AA AA A A AA AA A AII IA IA IA AAA A AA A A + −=→=− − + −→=−+−→=−+ − (1.11.16) The three invariants of a tensor can now be listed as [] []3 21 2 23 3 312 2 21 )tr( trtr tr III) tr()(tr IItr I A AA AA AA AAA + −=− == Invariants of a Tensor (1.11.17) The Deviatoric Tensor Denote the eigenvalues of the deviatoric tensor dev A, Eqn. 1.10.36, 3 2 1,,sss and the principal scalar invariants by 3 2 1,,JJJ . The characteristic e quation analogous to Eqn. 1.11.5 is then Section 1.11 Solid Mechanics Part III Kelly 103 03 22 13=−−− JsJ sJs (1.11.18) and the deviatoric invariants are3 ()() ( ) () [] () ()321 313 32 212 2 21 23 2 1 1 devdetdevtr devtr) dev(tr sss Jssssss Jsss J = =++−= − −=++= = AA AA (1.11.19) From Eqn. 1.10.37, 01=J (1.11.20) The second invariant can also be expressed in the useful forms { ▲Problem 4} ()2 32 22 1 21 2 s s s J ++= , (1.11.21) and, in terms of the eigenvalues of A, {▲Problem 5} () () ()[ ]2 1 32 3 22 2 1 261λλλλλλ −+−+−=J . (1.11.22) Further, the deviatoric invariants are re lated to the tensor invariants through {▲Problem 6} ()( )A A A A A A III27 III9 I2 ,II3 I3 271 32 31 2 +−= −= J J (1.11.23) 1.11.5 Coaxial Tensors Two tensors are coaxial if they have the same eigenvectors. It can be shown that a necessary and sufficient c ondition that two tensors A and B be coaxial is that their simple contraction is commutative, BA AB= . Since for a tensor T, TT TT1 1− −= , a tensor and its inverse ar e coaxial and have the same eigenvectors. 3 there is a convention (adhered to by most authors) to write the characteristic e quation for a general tensor with a λAII+ term and that for a deviatoric tensor with a sJ2− term (which ensures that 02>J - see 1.11.18 below) ; this means that the formulae for J2 in Eqn. 1.11.19 are the negative of those for AII in Eqn. 1.11.6 Section 1.11 Solid Mechanics Part III Kelly 1041.11.6 Fractional Powers of Tensors Integer powers of tensors were defined in §1. 9.2. Fractional power s of tensors can be defined provided the tensor is real, symmetric and positive definite (so that the eigenvalues are all positive). Contracting both sides of n nT ˆ ˆλ= with T gives n nT ˆ ˆ2 2λ= . It follows that, if T has eigenvectors inˆ and corresponding eigenvalues iλ, then nT is coaxial, having the same eigenvectors, but corresponding eigenvalues n iλ. Because of this, fractional powers of tensors are defined as follows: mT, where m is any real number, is that tensor which has the same eigenvectors as T but which has corresponding eigenvalues m iλ. For example, the square root of the positive definite tensor ∑ =⊗=3 1ˆ ˆ ii ii n n Tλ is ∑ =⊗ =3 12/1ˆ ˆ ii ii n n T λ (1.11.24) and the inverse is ∑ =−⊗ =3 11ˆ ˆ)/1( ii i i n n T λ (1.11.25) These new tensors are also positive definite. 1.11.7 Polar Decomposition of Tensors Any (non-singular second-order) tensor F can be split up multiplicatively into an arbitrary proper orthogonal tensor R ( IRR=T, 1 det=R ) and a tensor U as follows: RUF= Polar Decomposition (1.11.26) The consequence of this is that any transformation of a vector a according to Fa can be decomposed into two transformations, one involving a transformation U, followed by a rotation R. The decomposition is not, in general, unique; one can often find more than one orthogonal tensor R which will satisfy the above relation. In practice, R is chosen such that U is symmetric. To this end, consider FFT. Since 02 T>=⋅=⋅ Fv FvFv FvFv , FFT is positive definite. Further, jk jiFF≡FFT is clearly symmetric, i.e. the same result is obtained upon an interchange of i and k. Thus the square-root of FFT can be taken: let U in 1.11.26 be given by Section 1.11 Solid Mechanics Part III Kelly 105 ()2/1TFF U= (1.11.27) and U is also symmetric positive definite. Then, with 1.9.3e, 1.9.19, ()() IUUUUFUFUFU FU RR ==== − −− −− − 1 T1 T T1T1 T (1.11.28) Thus if U is symmetric, R is orthogonal. Further, from (1.9.16a,b) and (1.9.18d), F Udet det= and 1 det/ det det = = U F R so that R is proper orthogonal. It can also be proved that this decomposition is unique. An alternative decomposition is given by VRF= (1.11.29) Again, this decomposition is unique and R is proper orthogona l, this time with ()2/1TFF V= (1.11.30) 1.11.8 Problems 1. Find the eigenvalues, (normalised) eigenve ctors and principal invariants of 1 2 2 1 e e e eIT ⊗+⊗+= 2. Derive the spectral decomposition 1.11.11 by writing the identity tensor as i in nI ˆ ˆ⊗= , and writing AIA= . [Hint: inˆ is an eigenvector.] 3. Derive the characteristic equation and Ca yley-Hamilton equation for a 2-D space. Let A be a second order tensor with square root A S= . By using the Cayley-Hamilton equation for S, and relating SStr, det to AAtr, det through the corresponding eigenvalues, show that A AIA AA det2 trdet ++= . 4. The second invariant of a deviatoric tensor is given by Eqn. 1.11.19b, ( )13 32 21 2 ssssss J ++−= By squaring the relation 03 2 1 1 =++= sss J , derive Eqn. 1.11.21, ()2 32 22 1 21 2 s s s J ++= 5. Use Eqns. 1.11.21 (and your work from Problem 4) and the fact that 2 1 2 1 ss−=−λλ , etc. to derive Eqn. 1.11.22. 6. Use the fact that 03 2 1 =++ sss to show that Section 1.11 Solid Mechanics Part III Kelly 1063 13 32 21 3212 13 32 21 ) ( III3) ( II3 I m mmm ssssss sssssssss λ σλλ +++ +=+++== AAA where ii m A31=λ . Hence derive Eqns. 1.11.23. 7. Consider the tensor ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡− = 10001102 2 F (a) Verify that the polar decomposition for F is RUF= where ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡− = 1 0 002/12/102/1 2/1 R , ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −− = 1 0 002/32/102/1 2/3 U (verify that R is proper orthogonal). (b) Evaluate FbFa,, where T]0,1,1[=a , T]0,1,0[=b by evaluating the individual transformations UbUa, followed by ()()UbR UaR , . Sketch the vectors and their images. Note how R rotates the vectors into their final positions. Why does U only stretch a but stretches and rotates b? (c) Evaluate the eigenvalues iλ and eigenvectors inˆ of the tensor FFT. Hence determine the spectral decomposition (d iagonal matrix representation) of FFT. Hence evaluate FF UT= with respect to the basis {}inˆ – again, this will be a diagonal matrix. Section 1.12 Solid Mechanics Part III Kelly 1071.12 Higher Order Tensors In this section are discussed some importan t higher (third and fourth) order tensors. 1.12.1 Fourth Order Tensors After second-order tensors, the most commonl y encountered tensors are the fourth order tensors A, which have 81 components. Some pr operties and relations involving these tensors are listed here. Transpose The transpose of a fourth-order tensor A, denoted by TA, by analogy with the definition for the transpose of a second or der tensor 1.10.4, is defined by B CC B :: ::TA A= (1.12.1) for all second-order tensors B and C. It has the property () A A=TT and its components are klij ijkl )( )(TA A= . It also follows that () AB BA ⊗=⊗T (1.12.2) Identity Tensors There are two fourth-order identity tensors . They are defined as follows: T:: AAAA == II (1.12.3) and, from 1.9.7, they have components i j j i l k j i jk ilj i j i l k j ijl ik e e e ee e e ee e e ee e e e ⊗⊗⊗=⊗⊗⊗≡⊗⊗⊗=⊗⊗⊗≡ δδδδ II (1.12.4) For a symmetric second order tensor S, SS S==: : I I . Another important fourth-order tensor is II⊗, j j i i l k j i kl ij e e e e e e e e II ⊗⊗⊗=⊗⊗⊗=⊗δδ (1.12.5) Functions of the trace can be writ ten in terms of these tensors { ▲Problem 1}: Section 1.12 Solid Mechanics Part III Kelly 108() () 2T2 tr ::tr ::tr ::)tr( : A AAAA AAA AAIIIA AII === ⊗=⊗ II (1.12.6) Projection Tensors The symmetric and skew-symmetric parts of a second order tensor A can be written in terms of the identity tensors: () () A AA A :21skew:21sym IIII −=+= (1.12.7) The deviator of A, 1.9.30, can be written as A A II IAI AIA AA :ˆ:) (31):(31)tr(31dev P I ≡⎟ ⎠⎞⎜ ⎝⎛⊗−= −= −= (1.12.8) which defines Pˆ, the so-called fourth-order projection tensor . From Eqns. 1.10.6, 1.10.37a, it has the property that 0 ::ˆ=IAP . Note also that it has the property PP PP Pˆˆ::ˆ:ˆ ˆ= =Ln. For example, P III I PP P ˆ) (:) (91 31 31:31:31 ˆ:ˆ ˆ2 =⊗⊗+⊗−⊗−=⎟ ⎠⎞⎜ ⎝⎛⊗−⎟ ⎠⎞⎜ ⎝⎛⊗−== IIII II IIII II (1.12.9) The tensors ()()2/ /2, II II−+ in Eqn. 1.12.7 are also proj ection tensors, projecting the tensor A onto its symmetric and skew-symmetric parts. 1.12.2 Higher-Order Tensors and Symmetry A higher order tensor possesses complete symmet ry if the interchange of any indices is immaterial, for example if L=⊗⊗=⊗⊗=⊗⊗= ) ( ) ( ) (k j i jik k j i ikj k j i ijk A A A e e e e e e e e e A It is symmetric in two of its indices if the in terchange of these indices is immaterial. For example the above tensor A is symmetric in j and k if ) ( ) (k j i ikj k j i ijk A A e e e e e e ⊗⊗=⊗⊗=A Section 1.12 Solid Mechanics Part III Kelly 109 This applies also to antisymmetry. For example, the permutation tensor ( )k j i ijk e e e⊗⊗=εE is completely antisymmetric, since L==−=kij ikj ijk εεε . A fourth-order tensor C possesses the minor symmetries if ijlk ijkl jikl ijkl C C C C = = , (1.12.10) in which case it has only 36 independent components. The first equality here is for left minor symmetry, the second is for right minor symmetry. It possesses the major symmetries if it also satisfies klij ijkl C C= (1.12.11) in which case it has only 21 inde pendent components. From 1.12.1, this can also be expressed as A BB A :: :: C C= (1.12.12) for arbitrary second-order tensors A, B. Note that II⊗,,II posses the major symmetries {▲Problem 2}. 1.12.3 Problems 1. Derive the relations 1.12.6. 2. Use 1.12.12 to show that II⊗,,II possess the major symmetries. Section 1.13 Solid Mechanics Part III Kelly 1101.13 Coordinate Transformation of Tensor Components It has been seen in §1.5.2 that the transformation equations fo r the components of a vector are jij i uQ u′= , where []Q is the transformation matrix. Note that these ijQ’s are not the components of a tensor – these sQij' are mapping the components of a vector onto the components of the same vector in a second coordinate syst em – a (second-order) tensor, in general, maps one vector onto a different vector. The equation jij i uQ u′= is in matrix element form, and is not to be confused with the index notation for vectors and tensors. 1.13.1 Relationship between Base Vectors Consider two coordinate systems with base vectors ie and ie′. It has been seen in the context of vectors that, Eqn. 1.5.4, ), cos(j i ij j i xx Q ′ ≡=′⋅ee . (1.13.1) Recal that the i’s and j’s here are not referring to the three different components of a vector, but to different vectors (nine differe nt vectors in all). It is interesting that the relationshi p 1.13.1 can also be derived as follows: jijj i ji j j i i Qeeeeee e Ie e ′=′⋅′=′⊗′== ) () ( (1.13.2) Dotting each side here with ke′ then gives 1.13.1. Eqn. 1.13.2, together with the corresponding inverse relations, read jij iQe e′= , jji iQe e=′ (1.13.3) Note that the components of the transformation matrix []Q are the components of the change of basis tensor 1.10.24-25. 1.13.2 Tensor Transformation Rule As with vectors, the components of a (second- order) tensor will change under a change of coordinate system. In this case, using 1.13.3, n m pq nq mpn nq m mp pqq p pq j iij TQQQ QTT T e ee ee e e e ⊗′ =⊗′=′⊗′′≡⊗ (1.13.4) Section 1.13 Solid Mechanics Part III Kelly 111so that (and the inverse relationship) pq qj pi ij pq jq ip ij TQQ T TQQ T =′′ = , Tensor Transformation Formulae (1.13.5) or, in matrix form, [] [][][][][][][]QTQ T QTQ TT T, =′ ′= (1.13.6) Note : • as with vectors, second-order tensors are often defined as mathematical entities whose components transform according to the rule 1.13.5 • the transformation rule for higher order tensors can be established in the same way, for example, pqr rk qj pi ijk TQQQ T=′ , and so on Example (Mohr Transformation) Consider a two-dimensional space with base vectors 2 1,ee . The second order tensor S can be written in component form as 2 2 22 1 2 21 2 1 12 1 111 e e e e e e e e S ⊗+⊗+⊗+⊗= S S S S Consider now a second coordinate system, with base vectors 2 1,ee′′, obtained from the first by a rotation θ. The components of the transformation matrix are ⎥⎦⎤ ⎢⎣⎡−=⎥⎦⎤ ⎢⎣⎡ −+=⎥⎦⎤ ⎢⎣⎡ ′⋅′⋅′⋅′⋅=′⋅=θθθθ θ θθ θ cos sinsin cos cos ) 90cos() 90cos( cos 2 2 1 22 1 1 1 eeeeee eeeej i ijQ and the components of S in the second coordinate system are [][][] []QSQ ST=′ , so ⎥⎦⎤ ⎢⎣⎡− ⎥⎦⎤ ⎢⎣⎡ ⎥⎦⎤ ⎢⎣⎡ −=⎥⎦⎤ ⎢⎣⎡ ′′′′ θθθθ θθθθ cos sinsin cos cos sinsin cos 22 2112 11 22 2112 11 S SS S S SS S For S symmetric, 21 12S S= , and this simplifies to θ θθθ θ θθ θ θ 2cos cos sin) (2sin cos sin2sin sin cos 12 11 22 12122 222 11 22122 222 11 11 S S S SS S S SS S S S + −=′− + =′+ + =′ The Mohr Transformation (1.13.7) ■ 1.13.3 Isotropic Tensors An isotropic tensor is one whose components are the sa me under arbitrary rotation of the basis vectors, i.e. in any coordinate system. Section 1.13 Solid Mechanics Part III Kelly 112All scalars are isotropic. There is no isotropic vector (first-order tensor), i.e. there is no vector u such that jij i uQ u= for all orthogonal []Q (except for the zero vector o). To see this, consider the particular orthogonal transformation matrix [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −= 100001010 Q , (1.13.8) which corresponds to a rotation of 2/π about 3e. This implies that [] [ ]T 3 1 2T 3 2 1 uu u u uu −= or 02 1==u u . The matrix correspo nding to a rotation of 2/π about 1e is [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −= 01 0100001 Q , (1.13.9) which implies that 03=u . The only isotropic second-order tensor is ijαδα≡I , where α is a constant, that is, the spherical tensor, §1.10.12. To see this, first note that, by substituting Iα into 1.13.6, it can be seen that it is indeed isotropic. To see that it is the only isotropic second order tensor, first use 1.13.8 in 1.13.6 to get [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −−−− =′ 33 32 3123 22 2113 12 11 33 31 3213 11 1223 21 22 T T TT T TT T T T T TT T TT T T T (1.13.10) which implies that 0 , ,32 31 23 13 21 12 22 11 ====−== T T T TT TT T . Repeating this for 1.13.9 implies that 0 ,12 33 11 == TT T , so [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = 111111 0 00 00 0 TTT T or I T11T= . Multiplying by a scalar does not affect 1.13.6, so one has Iα. The only third-order isotropic tensors are sc alar multiples of the permutation tensor, ( )k j i ijk e e e⊗⊗=εE . Using the third order transformation rule, pqr rk qj pi ijk TQQQ T=′ , Section 1.13 Solid Mechanics Part III Kelly 113one has pqr rk qj pi ijk QQQε ε=′ . From 1.10.16e this reads ()ijk ijk ε ε Qdet=′ , where Q is the change of basis tensor, with components ijQ. When Q is proper orthogonal, i.e. a rotation tensor, one has indeed, ijk ijkεε=′ . That it is the only isotropic tensor can be established by carrying out specific rotations as done above for the first and second order tensors. Note that orthogonal tensors in gene ral, i.e. having the possibility of being reflection tensors, with 1 det−=Q are not used in the definition of isotropy , otherwise one would have the less desirable ijk ijkεε−=′ . Note also that this issue does not ari se with the second order tensor (or the fourth order tensor –see below), since the above result, that Iα is the only isotropic second order tensor, holds regardless of whether Q is proper orthogonal or not. There are three independent fourth-order isotropic tensors – these are the tensors encountered in §1.12.1, Eqns. 1.12.4-5, II⊗,,I I For example, () ()()( )ijkl klij lr kr jp ip rs pq ls kr jq ip pqrs ls kr jq ip QQ QQ QQQQ QQQQ II II ⊗== = =⊗ δδ δδ The most general isotropic fourth order tens or is then a linear combination of these tensors: I I C γμλ ++⊗= II Most General Isotropic Fourth-Order Tensor (1.13.11) 1.13.4 Invariance of Tensor Components The components of (non-isotropic) tensors wi ll change upon a rotation of base vectors. However, certain combinations of these components are the same in every coordinate system. Such quantities are called invariants . For example, the following are examples of scalar invariants {▲Problem 2} iijiijii AaaTaa ==⋅=⋅ AaTaaa tr (1.13.12) The first of these is the only independent sc alar invariant of a v ector. A second-order tensor has three independent scalar invariants, the first, second and third principal scalar invariants, defined by Eqn. 1.11.17 (or linear combinations of these). Example (of Invariance) Consider a tensor A with eigenvector nˆ and corresponding eigenvalue λ. Then n nA ˆ ˆλ= , or i jij n nA ˆ ˆλ= . The components of A and nˆ in a second coordinate system are pq qj pi ij TQQ A=′ and m mj j nQ n ˆ ˆ=′ . Thus Section 1.13 Solid Mechanics Part III Kelly 114 i p pi m pm pi m mj pq qj pi jij n n Q nAQ nQAQQ nA ′== = =′′ ˆ ˆ ˆ ˆ ˆ λλ Thus λ is an eigenvalue and nQn ˆ ˆ=′ is an eigenvector, showi ng that the eigenvalue is invariant, but a new eigenvector is obta ined, the original ei genvector rotated by Q. ■ 1.13.5 Problems 1. Consider a coordinate system 321xxox with base vectors ie. Let a second coordinate system be represented by the set {}ie′ with the transformation law 3 32 1 2 cos sin e ee e e =′+−=′ θθ (a) find 1e′ in terms of the old set {}ie of basis vectors (b) find the orthogonal matrix []Q and express the old coordi nates in terms of the new ones (c) express the vector 3 2 13 6 ee e u +−−= in terms of the new set {}ie′ of basis vectors. 2. Show that (a) the trace of a tensor A, iiA=Atr , is an invariant. (b) jiijaaT=⋅aTa is an invariant. 3. Consider Problem 7 in §1.11. Take the tensor FF UT= with respect to the basis {}inˆ and carry out a coordinate transformation of its tensor components so that it is given with respect to the original {}ie basis – in which case the matrix representation for U given in Problem 7, §1.11, should be obtained. Section 1.14 Solid Mechanics Part III Kelly 1151.14 Tensor Calculus I: Tensor Fields In this section, the concepts from the calculus of vectors are generalise d to the calculus of higher-order tensors. 1.14.1 Tensor-valued Functions Tensor-valued functions of a scalar The most basic type of calculus is that of te nsor-valued functions of a scalar, for example the time-dependent stress at a point, )(tSS= . If a tensor T depends on a scalar t, then the derivative is defined in the usual way, tt t t dtd tΔ−Δ+=→Δ)( ) (lim0T T T, which turns out to be j iij dtdT dtde eT⊗= (1.14.1) The derivative is also a tensor and th e usual rules of differentiation apply, () () () () ()T T)( ⎟ ⎠⎞⎜ ⎝⎛=+=+=+=+=+ dtd dtddtd dtd dtddtd dtd dtddtd dtdtdtddtd dtd dtd TTBT BT TBaT aT TaTTTB TBT ααα For example, consider the time derivative of TQQ , where Q is orthogonal. By the product rule, using I QQ=T, () 0QQ QQ QQ QQQQ =⎟ ⎠⎞⎜ ⎝⎛+=+=T TT T T dtd dtd dtd dtd dtd Thus, using Eqn. 1.10.3e ()TT T TQQ QQ QQ && & −=−= (1.14.2) Section 1.14 Solid Mechanics Part III Kelly 116which shows that TQQ& is a skew-symmetric tensor. 1.14.2 Vector Fields The gradient of a scalar field and the divergence and curl of vector fields have been seen in §1.6. Other important quantities are the gr adient of vectors and higher order tensors and the divergence of higher order tensors. First, the gradient of a vector field is introduced. The Gradient of a Vector Field The gradient of a vector field is de fined to be the second-order tensor j i ji j j xa xe e eaa ⊗∂∂=⊗∂∂≡ grad Gradient of a Vector Field (1.14.3) In matrix notation, ⎥⎥⎥⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢⎢⎢⎢ ⎣⎡ ∂∂ ∂∂ ∂∂∂∂ ∂∂ ∂∂∂∂ ∂∂ ∂∂ =33 23 1332 22 1231 21 11 grad xa xa xaxa xa xaxa xa xa a (1.14.4) One then has () )() (grad xax xaaee e e xa d dddxxadxxad ij jikk j i ji −+==∂∂=⊗∂∂= (1.14.5) which is analogous to Eqn 1.6.7 for the gradient of a scalar field. As with the gradient of a scalar field, if one writes xd as exd, where e is a unit vector, then direction in grad eaea⎟ ⎠⎞⎜ ⎝⎛=dxd (1.14.6) Thus the gradient of a vector field a is a second-order tensor which transforms a unit vector into a vector de scribing the gradient of a in that direction. For a space curve parameterised by s, one has Section 1.14 Solid Mechanics Part III Kelly 117 () τaτeaeτa a agrad=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛⊗∂∂=⋅∂∂=∂∂=i ii ii i x x dsdx x dsd where τ is a tangent vector to C (see §1.6.2). Although for a scalar field φgrad is equivalent to φ∇, note that the gradient defined in 1.14.3 is not the same as a⊗∇ . In fact, () a a gradT=⊗∇ (1.14.7) since j i ij jj iixaaxe e e ea ⊗∂∂=⊗∂∂=⊗∇ (1.14.8) These two different definitions of the gradient of a vector, j ij ix a e e⊗∂∂/ and j ii jx a e e⊗∂∂/ , are both commonly used. In what follows, they will be distinguished by labeling the former as agrad (which will be called the gradient of a) and the latter as a⊗∇ . Note : • in much of the literature, a⊗∇ is written in the contracted form a∇, but the more explicit version is used here • some authors define the operation of ⊗∇ on a vector or tensor ()• not as in 1.14.8, but through () ()()i ix e⊗∂•∂≡•⊗∇ / so that ()j ij ix a e e a a ⊗∂∂==⊗∇ / grad Example (The Displacement Gradient) Consider a particle 0p of a deforming body at position X (a vector) and a neighbouring point 0q at position Xd relative to 0p, Fig. 1.14.1. As the material deforms, these two particles undergo displacements of, respectively, )(Xu and ) ( X Xu d+ . The final positions of the particles are fp and fq. Then Xu XXu XXu X XuX x d dd dd d d grad)()() ( +=+=−++= Section 1.14 Solid Mechanics Part III Kelly 118 Figure 1.14.1: displacement of material particles Thus the gradient of the displacement field u encompasses the mapping of (infinitesimal) line elements in the undeformed body into li ne elements in the deformed body. For example, suppose that 0 ,3 22 2 1 == = u u kX u . Then 2 122 2 0 0 00 0 00 20 grad e e u ⊗= ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ =∂∂= kXkX Xu ji A line element iidX d e X= at iiXe X= maps onto () ( ) 12 233 22 11 2 12 22 e Xe e e e e X x dXkX ddX dX dX kX d d +=++ ⊗ += The deformation of a box is as shown in Fig. 1.14.2. For example, the vector 2e Xαd d= (defining the left-hand side of the box) maps onto 2 1 2 e e x ααα+ = dk d . Figure 1.14.2: deformation of a box Note that the map x X d d→ does not specify where in spa ce the line element moves to. It translates too according to uXx+= . ■ The Divergence and Curl of a Vector Field The divergence and curl of vectors have been defined in §1.6.6, §1.6.8. Now that the gradient of a vector has been introduc ed, one can re-defin e the divergence of a independent of any coordinate system: it is the scalar field given by the trace of the gradient of a {▲Problem 4}, 1X2X finalXXd xd) ( X Xu d+ )(Xu final initial 0p0q fpfq Section 1.14 Solid Mechanics Part III Kelly 119 a Ia a a ⋅∇= = = : grad) grad(tr div Divergence of a Vector Field (1.14.9) Similarly, the curl of a can be defined to be the vector field given by twice the axial vector of the antisymmetric part of agrad . 1.14.3 Tensor Fields A tensor-valued function of the positi on vector is called a tensor field, )(xkijTL . The Gradient of a Tensor Field The gradient of a second order tensor field T is defined in a manner analogous to that of the gradient of a vector, Eqn. 1.14. 2. It is the third-order tensor k j i kij k k xT xe e e eTT ⊗⊗∂∂=⊗∂∂= grad Gradient of a Tensor Field (1.14.10) This differs from the quantity ()k j i ijk k j jk iixTTxe e e e e eT ⊗⊗∂∂=⊗⊗∂∂=⊗∇ (1.14.11) The Divergence of a Tensor Field Analogous to the definition 1.14.9, the divergence of a second order tensor T is defined to be the vector i jiji ik j jk i i xTxT x eee eeTIT T ∂∂=∂⊗∂=∂∂= =) (: grad div Divergence of a Tensor (1.14.12) The divergence of a tensor can also be equiva lently defined as that vector field which satisfies the relation ()()vT vTTdiv div=⋅ for all constant vectors v. One also has Section 1.14 Solid Mechanics Part III Kelly 120i jji k j jk iixTTxe e e eT∂∂=⊗⋅∂∂=⋅∇ ) ( (1.14.13) so that Tdiv T T⋅∇= (1.14.14) As with the gradient of a vector, both ()i j ijx T e∂∂/ and ()i j jix T e∂∂/ are commonly used as definitions of the divergence of a tensor,. They are distinguished here by labelling the former as Tdiv (called here th e divergence of T) and the latter as T⋅∇ . Note that the operations Tdiv and T⋅∇ are equivalent for the case of T symmetric. Note : • some authors define the operation of ⋅∇ on a vector or tensor ()• not as in (1.14.13), but through () ()()i ixe⋅∂•∂≡•⋅∇ / so that ()ij ijx T e T T ∂∂==⋅∇ / div . • using the convention that the “dot” is omitted in the contraction of tensors, one should write T∇ for T⋅∇ , but the “dot” is retained here because of the familiarity of this latter notation from vector calculus. • another operator is the Hessian , ()j ij ixx e e⊗∂∂∂=∇⊗∇ /2. Identities Here are some important identities involving the grad ient, divergence and curl {▲Problem 5}: () () ( ) ( ) () () () ( ) ( ) uv vu uvvuvuuv vu vuuv vu vuvv v grad grad div div curl) div( grad divgrad grad gradgrad grad grad T T − +−=×+ =⊗+ =⋅⊗+= φ φφ (1.14.15) () () ( ) () ()()( ) ( ) () φ φφφ φφφφ φ grad grad gradgrad div div: grad div divgradtr div divdiv grad div T ⊗+=+ =+=+⋅=+= AA ABA AB BABA B A ABv A A v AvA A A (1.11.16) Note also the following identities, which involve the Laplacian of both vectors and scalars: () () u u uv uv u vu vu 22 2 2 div grad curlcurlgrad: grad2 ∇− =∇⋅+ +⋅∇=⋅∇ (1.14.17) Section 1.14 Solid Mechanics Part III Kelly 1211.14.4 Cylindrical and Spherical Coordinates Cylindrical and spherical coordinates were introduced in §1.6.10 and the gradient and Laplacian of a scalar field and the divergence and curl of vector fi elds were derived in terms of these coordinates. The calculus of higher order tensors can also be cast in terms of these coordinates. For example, from 1.6.27, the gradient of a vector in cylindrical coordinates is ()Tgrad u u⊗∇= with () z zz zz r zzzr rz rr rr r rrzz rr z r zu u r ruzu ru u r ruzu ru u r ruu u uz r r e e e e e ee e e e e ee e e e e ee e e e e e u ⊗∂∂+⊗∂∂+⊗∂∂+⊗∂∂+⊗⎟ ⎠⎞⎜ ⎝⎛+∂∂+⊗∂∂+⊗∂∂+⊗⎟ ⎠⎞⎜ ⎝⎛−∂∂+⊗∂∂=⎥⎦⎤ ⎢⎣⎡++⊗⎟ ⎠⎞⎜ ⎝⎛ ∂∂+∂∂+∂∂= θθθ θθθ θθθθθθ θ θθθθ 1111gradT (1.14.18) and from 1.6.27, 1.14.12, the divergence of a tensor in cylindrical coordinates is {▲Problem 6} zzz z zr zrr r z rrrr rz r rr zA A r rA rArA A zA A r rArA A zA A r rA eee A A ⎟ ⎠⎞⎜ ⎝⎛ ∂∂+∂∂++∂∂+⎟ ⎠⎞⎜ ⎝⎛ ++∂∂+∂∂+∂∂+⎟ ⎠⎞⎜ ⎝⎛ −+∂∂+∂∂+∂∂=⋅∇= θθθ θθθ θ θ θθ θθθ θ 111divT (1.14.19) 1.14.5 The Divergence Theorem The divergence theorem 1.6.12 can be extended to the case of higher-order tensors. Consider an arbitrary di fferentiable tensor field ),(t TkijxL defined in some finite region of physical space. Let S be a closed surface bounding a volume V in this space, and let the outward normal to S be n. The divergence theorem of Gauss then states that ∫∫∂∂= V kkij Skkij dVxTdSn TL L (1.14.20) For a second order tensor, ∫∫ ∫∫∂∂ = = V jij Sj ij V SdVxT dSnT dV dS , divT Tn (1.14.21) Section 1.14 Solid Mechanics Part III Kelly 122 One then has the important identities { ▲Problem 7} () ∫∫∫∫∫∫ =⋅=⊗= V SV SV S dV dSdV dSdV dS ) (divgrad)(div TuT Tnuu nuT nT φ φ (1.14.22) 1.14.6 Formal Treatment of Tensor Calculus As in §1.6.12, here a more formal treatment of the tensor calculus of fields is briefly presented. Vector Gradient What follows is completely analogous to Eqns. 1.6.43-46. A vector field V E→3:v is differentiable at a point 3E∈x if there exists a scond order tensor () E D∈xv such that () ( ) ()()h hxv xv hxv o D++=+ for all E∈h (1.14.23) In that case, the tensor ()xvD is called the derivative (or gradient ) of v at x (and is given the symbol ()xv∇ ). Setting w hε= in 1.14.23, where E∈w is a unit vector, dividing through by ε and taking the limit as 0→ε , one has the equivalent statement () () wxv wxv εεε+ =∇ =0 dd for all E∈w (1.14.24) Using the chain rule as in §1.6.11, Eqn. 1.14.24 can be expressed in terms of the Cartesian basis {}ie, () ()kk j i ji ik kiwxvwxve e e e wxv ⊗∂∂=∂∂=∇ (1.14.25) This must be true for all w and so, in a Cartesian basis, ()j i ji xve e xv ⊗∂∂=∇ (1.14.26) which is Eqn. 1.14.3. Section 1.14 Solid Mechanics Part III Kelly 123 1.14.7 Problems 1. Consider the vector field 32 2 22 3 12 1 e e e v x x x ++= . (a) find the matrix representation of the gradient of v, (b) find the vector ()vvgrad . 2. If 31 221 1321 e e e u x xx xxx ++ = , determine u2∇ . 3. Suppose that the displacement field is given by 1 3 2 1 ,1 ,0 X u u u === . By using ugrad , sketch a few (undeformed) line elements of material and their positions in the deformed configuration. 4. Use the matrix form of ugrad and u⊗∇ to show that the definitions (i) ) grad(tr div a a= (ii) ω a2 curl= , where ω is the axial vector of the skew part of agrad agree with the definitions 1.6.14, 1. 6.19 given for Cartesian coordinates. 5. Prove the following: (i) () φ φφ grad grad grad ⊗+= vv v (ii) ()( ) ()uv vu vuT Tgrad grad grad + =⋅ (iii) ()( ) uv vu vu ) div( grad div + =⊗ (iv) () ()()uv vu uvvuvu grad grad div div curl − +−=× (v) () A A A div grad div φφ φ + = (vi) () ()v A A v Av gradtr div divT+⋅= (vii) () BA B A AB : grad div div += (viii) ()()()()φ φφ grad div div BA AB BA + = (ix) () φ φφ grad grad grad ⊗+= AA A 6. Derive Eqn. 1.14.19, the divergence of a tensor in cylindrical coordinates. 7. Deduce the Divergence Theorem identities in 1.14.22 [Hint: write them in index notation.] Section 1.15 Solid Mechanics Part III Kelly 1241.15 Tensor Calculus 2: Tensor Functions 1.15.1 Vector-valued functions of a vector Consider a vector-valued function of a vector )( ),(j i i ba a= =baa This is a function of three independent variables 3 2 1,,bbb , and there are nine partial derivatives j ib a∂∂/ . The partial derivative of the vector a with respect to b is defined to be a second-order tensor with these partial derivatives as its components: j i ji bae ebba⊗∂∂≡∂∂ )( (1.15.1) It follows from this that 1− ⎟ ⎠⎞⎜ ⎝⎛ ∂∂=∂∂ ab ba or ij jm mi ab baδ=∂∂ ∂∂=∂∂ ∂∂,Iab ba (1.15.2) To show this, with ) ( ),(j i i j i i abbba a = = , note that the differential can be written as ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂ ∂∂+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂ ∂∂+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂ ∂∂=∂∂ ∂∂=∂∂= 31 3 21 2 11 11 1 1ab badaab badaab bada daab badbbadaj jj jj ji ij jj j Since 3 2 1 ,, dadada are independent, one may set 03 2==da da , so that 1 11=∂∂ ∂∂ ab ba j j Similarly, the terms inside the other brackets are zero and, in this way, one finds Eqn. 1.15.2. 1.15.2 Scalar-valued functions of a tensor Consider a scalar valued func tion of a (second-order) tensor j iijT e e T T ⊗= = ),(φφ. This is a function of nine independent variables, ) (ijTφφ= , so there are nine different partial derivatives: Section 1.15 Solid Mechanics Part III Kelly 12533 32 31 23 22 21 13 12 11, , , , , , , ,T T T T T T T T T ∂∂ ∂∂ ∂∂ ∂∂ ∂∂ ∂∂ ∂∂ ∂∂ ∂∂ φφφφφφφφφ The partial derivative of φ with respect to T is defined to be a second-order tensor with these partial derivatives as its components: j i ijTe eT⊗∂∂≡∂∂φφ Partial Derivative with respect to a Tensor (1.15.3) The quantity T T∂∂ /)(φ is also called the gradient of φ with respect to T. Thus differentiation with respec t to a second-order tensor ra ises the order by 2. This agrees with the idea of the gradient of a scal ar field where differentia tion with respect to a vector raises the order by 1. Derivatives of the Trace and Invariants Consider now the trace: the derivative of Atr, with respect to A can be evaluated as follows: Ie ee ee ee e e e e eA A AAA =⊗+⊗+⊗=⊗∂∂+⊗∂∂+⊗∂∂=∂∂+∂∂+∂∂=∂∂ 3 3 2 2 1 133 22 1133 22 11tr j i ijj i ijj i ij AA AA AAA A A (1.15.4) Similarly, one finds that { ▲Problem 1} () () ()() () ()IAAAIAAAAAAAAAIAA 23 2T23 T2 )tr(3)tr()tr(2)tr(3tr2tr tr =∂∂=∂∂=∂∂=∂∂=∂∂ (1.15.5) Derivatives of Trace Functions From these and 1.10.17, one can evaluate the derivatives of the invariants { ▲Problem 2}: ()T T2TT III II IIIIIIII −=+−=∂∂−=∂∂=∂∂ A I A AAAIAIA A A AAAAA Derivatives of the Invariants (1.15.6) Section 1.15 Solid Mechanics Part III Kelly 126Derivative of the Determinant An important relation is () ()Tdet det−=∂∂AA AA (1.15.7) which follows directly from 1.15.6c. Other Relations The total differential can be written as TTddTTdTTdTTd :13 1312 1211 11 ∂∂≡+∂∂+∂∂+∂∂= φφ φ φφ L (1.15.8) This total differential gives a good appr oximation to the total increment in φ when the increments of the independent variables L,11T are small. The second partial derivativ e is defined similarly: q p j i pq ijTTe e e eTT⊗⊗⊗∂∂∂≡∂∂∂ φφ2 (1.15.9) the result being in this case a fourth-order tensor. Consider a scalar-valued function of a tensor, )(Aφ , but now suppose that the components of A depend upon some scalar parameter t: ))((tAφφ= . By means of the chain rule of differentiation, dtdA Aij ij∂∂=φφ& (1.15.10) which in symbolic notation reads (see Eqn. 1.10.10e) ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ⎟ ⎠⎞⎜ ⎝⎛ ∂∂=∂∂=dtd dtd dtd A AA AT tr :φ φφ (1.15.11) Identities for Scalar-valued functions of Symmetric Tensor Functions Let C be a symmetric tensor, TCC= . Then the partial derivative of () )(TCφφ= with respect to T can be written as { ▲Problem 3} Section 1.15 Solid Mechanics Part III Kelly 127(1) CTT∂∂=∂∂ φφ2 f o r TTCT= (2) CT T∂∂=∂∂φφ2 for TTTC= (1.15.12) (3) TC CTTC CTT ∂∂+∂∂=∂∂=∂∂=∂∂ φφφφφ2 2 f o r TTC= and symmetric T Scalar-valued functions of a Symmetric Tensor Consider the expression () () ijij ijAAB∂∂=∂∂=φ φ AAB (1.15.13) If A is a symmetric tensor, there are a number of ways to consider this expression: two possibilities are that φ can be considered to be (i) a symmetric function of the 9 variables ijA (ii) a function of 6 independent variables: ( )33 23 22 13 12 11 ,,,,, AAAAAAφφ= where () () ()32 23 32 23 2331 13 31 13 1321 12 21 12 12 212121 A A A A AA A A A AA A A A A ==+===+===+= Looking at (i) and writing ()() ( )L L , ,, ,12 21 12 12 11 AA AAAφφ= , one has, for example, 12 21 12 1221 21 1212 12 122A A A AA A AA A A ∂∂=∂∂+∂∂=∂∂ ∂∂+∂∂ ∂∂=∂∂ φφφ φ φφ, the last equality following from the fact that φ is a symmetrical function of the ijA. Thus, depending on how the scalar func tion is presented, one could write (i) etc., , , 1313 1212 1111ABABAB∂∂=∂∂=∂∂=φ φ φ (ii) etc.,21,21, 1313 1212 1111ABABAB∂∂=∂∂=∂∂=φ φ φ Section 1.15 Solid Mechanics Part III Kelly 1281.15.3 Tensor-valued functions of a tensor The derivative of a (second-order) tensor A with respect to another tensor B is defined as q p j i pqij BAe e e eBA⊗⊗⊗∂∂≡∂∂ (1.15.14) and forms therefore a fourth-order tensor. The total differential Ad can in this case be written as BBAA d d :∂∂= (1.15.15) Consider now l k j i klij AAe e e eAA⊗⊗⊗∂∂=∂∂ The components of the tensor are independent, so .etc ,0 ,1 1211 1111L=∂∂=∂∂ AA AA nq mp pqmn AAδδ=∂∂ (1.15.16) and so I=⊗⊗⊗=∂∂ j i j i e e e eAA, (1.15.17) the fourth-order identity tensor of Eqn. 1.12.4. Example Consider the scalar-valued function φ of the tensor A and vector v (the “dot” can be omitted from the following and similar expression), () Avv vA⋅=,φ The gradient of φ with respect to v is ()vAA vA AvvvAv Avvv vT+=+=∂∂⋅+⋅∂∂=∂∂φ On the other hand, the gradient of φ with respect to A is vvvvvAAvA⊗=⋅=∂∂⋅=∂∂Iφ Section 1.15 Solid Mechanics Part III Kelly 129■ Consider now the derivative of the inverse, A A∂∂−/1. One can differentiate 0AA=−1 using the product rule to arrive at AAA AAA ∂∂−=∂∂−− 11 One needs to be careful with derivatives becau se of the position of the indices in 1.15.14); it looks like a post-operati on of both sides with the inverse leads to ()l k j i jl ikAA e e e e AAA A A A ⊗⊗⊗ −=∂∂−=∂∂−− − − − 1 1 1 1 1/ / . However, this is not correct (unless A is symmetric). Using the index notation (there is no clear symbolic notation), one has () ()l k j i jl ik klijjn jl mk im mn klimjn klmj im jn mj kliml k j i klmj im mj klim AAAAA AAAAAAA AAAAAAA AAA e e e ee e e e ⊗⊗⊗ −=∂∂→−=∂∂→∂∂−=∂∂→⊗⊗⊗∂∂−=∂∂ −−−− −−− − −−−− 1 111 111 1 1111 δδ δ (1.15.18) ■ 1.15.4 The Directional Derivative The directional derivative was introduced in §1.6.11. The ideas introduced there can be extended to tensors. For example, the dire ctional derivative of the trace of a tensor A, in the direction of a tensor T, is () () () T T A T A TAA tr tr tr tr ][tr 0 0=+ =+ = ∂ = =εεεεε ε dd dd (1.15.19) As a further example, consider the scalar function Avu A⋅=)(φ , where u and v are constant vectors. Then () ()[] TvuvT Au TvuAA ⋅=+⋅ = ∂ =εεφ ε0][,,dd (1.15.20) Also, the gradient of φ with respect to A is () vu AvuA A⊗=⋅∂∂=∂∂φ (1.1.5.21) Section 1.15 Solid Mechanics Part III Kelly 130 and it can be seen that this is an example of the more general relation TATA : ][∂∂=∂φφ (1.15.22) which is analogous to 1.6.38. Indeed, wuvwvTATwxw uAx ∂∂=∂∂∂=∂⋅∂∂=∂ ][: ][][ φφφφ (1.15.23) Example (the Directional De rivative of the Determinant) It was shown in §1.6.11 that the directi onal derivative of the determinant of the 22× matrix A, in the direction of a second matrix T, is ()[]1221 21 12 1122 22 11 det TA TA TA TA −−+= ∂ TAA This can be seen to be equal to ()T AA : detT−, which will now be proved more generally for tensors A and T: () () ()[] () TAI ATAIAT A TAA 1 01 00 det detdetdet ][ det − =− == + =+ =+ = ∂ εεεεεε εεε dddddd The last line here follows from (1.9.16a). Now the characteristic equation for a tensor B is given by (1.11.4, 1.11.5), () () ()()I Bλ λλλλλλ −==−−− det03 2 1 where iλ are the three eigenvalues of B. Thus, setting 1−=λ and TA B1−=ε , Section 1.15 Solid Mechanics Part III Kelly 131() ()()() ()()() () ()TA AAAA TA TA TA TATA TA TATA TA TA A 13 2 13 2 1 03 2 1 0 tr detdet1 1 1 det1 1 1 det][ det 1 1 11 1 11 1 1 −== =++ =+ + + =+ + + = ∂ − − −− − −− − − λλλλε λε λεελ λ λε εε ε ε ε dddd and, from (1.10.10e), () ()T AA TAA : det][ detT−= ∂ (1.15.24) ■ Example (the Directional Deri vative of a vector function) Consider the n homogeneous algebraic equations ()oxf=: ( ) () () 0 ,,,0 ,,,0 ,,, 2 12 1 22 1 1 === n nnn x xxfx xxfx xxf LLLL The directional derivative of f in the direction of some vector u is ()()( ) () Kuz zzfuxz zf uxfx =⎟ ⎠⎞⎜ ⎝⎛ ∂∂=+= =∂ == 00])[( εε εε εε dddd (1.15.25) where K, called the tangent matrix of the system, is ⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢ ⎣⎡ ∂∂ ∂∂∂∂ ∂∂∂∂∂∂ ∂∂∂∂ =∂∂= n n nnn x f x fxf xf xfxf xf xf / // / // / / 12 2 2 1 21 2 1 1 1 LM ML xfK , () uf ufx grad][=∂ which can be compared to (1.15.23c). ■ Properties of the Di rectional Derivative The directional derivative is a linear operator and so one can apply the usual product rule. For example, consider the directional derivative of 1−A in the direction of T: Section 1.15 Solid Mechanics Part III Kelly 132 () ()1 01][− =−+ = ∂ T A T AA εεεdd To evaluate this, note that ()() 0 TI TAAA A =∂= ∂−][ ][1, since I is independent of A. The product rule then gives () ()][ ][1 1TA A AT AA A ∂−= ∂− −, so that ()1 1 1 1 1][ ][−− − − −−= ∂−= ∂ TAA ATA A T AA A (1.15.26) Another important property of th e directional derivative is the chain rule , which can be applied when the function is of the form ()()xBf xf ˆ)(= . To derive this rule, consider (see §1.6.11) ][ )() ( uf xf uxfx∂+≈+ , (1.15.27) where terms of order )(uo have been neglected, i.e. 0)(lim0=→uu uo. The left-hand side of the previous expression can also be written as ()() ( ) () ( ) ]][ [ˆ )(ˆ][ )(ˆ ˆ uB Bf xBfuB xBf uxBf x Bx ∂∂+≈∂+≈+ Comparing these expressions, one arrives at the chain rule, () ]][ [ˆ ][ uB Bf ufx B x ∂∂=∂ Chain Rule (1.15.28) As an application of this rule, consider the directional derivative of 1det−A in the direction T; here, f is 1det−A and ())(ˆˆ ABff= . Let 1−=AB and B fdetˆ= . Then, from Eqns. 1.15.24, 1.15.25, 1.10.3h, f, ()() ()()() ()() () T AATAA AATAA BBTA B T AA B A : det: det: det]][ [ det ][ det T 11 1 T 11 1 T1 1 −−−− −−− −− − −=−=− =∂∂= ∂ (1.15.29) 1.15.5 Formal Treatment of Tensor Calculus As in §1.6.12, derivatives can be defined formally as follows: Section 1.15 Solid Mechanics Part III Kelly 133A scalar function R Vf→2: i s differentiable at 2V∈A if there exists a second order tensor ()2V Df∈A such that () ( ) ()()H HA A HA o Df f f + +=+ : for all 2V∈H (1.15.30) In that case, the tensor ()ADf is called the derivative of f at A. It follows from this that ()ADf is that tensor for which []() () B A BA BA εεε+ = =∂ =fddDf f 0: for all 2V∈B (1.15.31) For example, from 1.15.24, () ()()T AA T AA TAA : det : det][ detT T − −= = ∂ (1.15.32) from which it follows, from 1.15.31, that Tdet det−=∂∂AA AA (1.15.33) which is 1.1.5.7. Similarly, a tensor-valued function 2 2: V V→ T is differentiable at 2V∈A if there exists a fourth order tensor ()4V D∈AT such that ()( ) ()()H HAT AT HAT o D+ +=+ for all 2V∈H (1.15.34) In that case, the tensor ()ATD is called the derivative of T at A. It follows from this that ()ATD is that tensor for which []() () B AT BAT BTA εεε+ = =∂ =0:ddD for all 2V∈B (1.15.35) 1.15.6 Problems 1. Evaluate the derivatives (use the chai n rule for the last two of these) ()()()() AA AA AA AA ∂∂ ∂∂ ∂∂ ∂∂2 2 3 2)tr(,)tr(,tr,tr 2. Derive the derivatives of the invariants, Eqn. 1.15.5. [Hint: use the Cayley-Hamilton theorem, Eqn. 1.11.15, to express the deriva tive of the third invariant in terms of the third invariant.] 3. (a) Consider the scalar valued function ()()FCφφ= , where FFCT= . Use the chain rule Section 1.15 Solid Mechanics Part III Kelly 134j i ijmn mn FC Ce eF⊗∂∂ ∂∂=∂∂φφ to show that kjik ij CFF ∂∂=∂∂ ∂∂=∂∂ φ φ φφ2 , 2CFF (b) Show also that UC CUU ∂∂=∂∂=∂∂ φφφ2 2 for UUC= with U symmetric. [Hint: for (a), use the index notation: first evaluate ij mn F C∂∂ / using the product rule, then evaluate ijF∂∂/φ using the fact that C is symmetric.] 4. Show that (a) 1 11 :−−− −=∂∂BAA BAA, (b) 1 1 11 :− − −− ⊗−=⊗∂∂A A A AAA 5. Show that TT : BBAA=∂∂ 6. By writing the norm of a tensor A, 1.10.13, where A is symmetric, in terms of the trace (see 1.10.10), show that AA AA=∂∂ 7. Evaluate (i) ()][2TAA∂ (ii) ()][ tr2TAA∂ (see 1.10.10e) 8. Derive 1.15.29 by using the de finition of the directional derivative and the relation 1.15.7, () ()Tdet / det−=∂∂ AA A A . Section 1.16 Solid Mechanics Part III Kelly 1351.16 Curvilinear Coordinates Up until now, a rectangular Cartesian coordi nate system has been used, and a set of orthogonal unit base vectors ie has been employed as the basis for representation of vectors and tensors. This basis is independent of position and provides a simple formulation. Two exceptions were in §1.6.10 and §1.14.4, where cylindrical and spherical coordinate systems were used. These differ from the Cartesian system in that the cylindrical and spherical base vectors do depend on position. However, although the directions of these base vectors may change with position, they are always orthogonal to each other. In this section, arbitrary bases, with base vectors not necessarily orthogonal nor of unit length, are considered. It will be seen how these systems reduce to the special cases of orthogonal (e.g. cylindrical and s pherical systems) and Cartesian systems. 1.16.1 Curvilinear Coordinates A Cartesian coordinate system is defined by the fixed base vectors 3 2 1,,eee and the coordinates ) ,,(3 2 1xxx , and any point p in space is then determined by the position vector iixe x= (see Fig. 1.16.11). This can be expressed in terms of curvilinear coordinates ),,(3 2 1ΘΘΘ by the transformation (and inverse transformation) () ),,(,, 3 2 13 2 1 ΘΘΘ=Θ=Θ i ii i x xxxx ( 1 . 1 6 . 1 ) In order to be able to solve for the iΘ given the ix, and to solve for the ix given the iΘ, it is necessary and sufficient that the following determinants are non-zero – see Appendix 1.A.2 (the first here is termed the Jacobian J of the transformation): J x xx xJji ji ji ji1det , det = ∂Θ∂=⎥⎦⎤ ⎢⎣⎡ ∂Θ∂ Θ∂∂=⎥⎦⎤ ⎢⎣⎡ Θ∂∂≡ , (1.16.2) the last equality following from (1.15.2, 1.10.18d). If 1Θ is varied while holding 2Θ and 3Θ constant, a space curve is generated called a 1Θ coordinate curve . Similarly, 2Θ and 3Θ coordinate curves may be generated. Three coordinate surfaces intersect in pairs along the coordinate curves. On each surface, one of the curvilinear coordinates is constant. Note : • This Jacobian is the same as that used in ch anging the variable of integration in a volume integral, §1.7; from Cartesian coordinates to curvilinear coordinates, one has ∫∫ΘΘΘ→ V Vdd Jd dxdxdx3 2 1 3 2 1 1 superscripts are used here and in much of what follows for notational consistency (see later) Section 1.16 Solid Mechanics Part III Kelly 136 Figure 1.16.1: curvilinear coordinate system and coordinate curves 1.16.2 Base Vectors in the Moving Frame Covariant Base Vectors From §1.6.2, writing ()iΘ=xx , tangent vectors to the coordinate curves at x are given by2 mim i ixexgΘ∂∂=Θ∂∂= Covariant Base Vectors (1.16.3) with inverse ()mi m i xg e ∂Θ∂= /. T h e ig emanate from the point p and are directed towards the site of increasing coordinate iΘ. They are called covariant base vectors . Increments in the two coordinate systems are related through i ii id ddd Θ=ΘΘ∂= gxx Note that the triple scalar product ()3 2 1 ggg×⋅ , Eqns. 1.2.15-16, is equivalent to the determinant in 1.16.2, 2 in the Cartesian system, with the coordinate curves parallel to the coordinate axes, these equations reduce trivially to ()m mi mi m i x x e e e δ=∂∂= / x 1x2x3xconst3=Θ 1g2g 3g 1e2e3ecurve1−Θcurve2−Θ curve3−Θp 1g Section 1.16 Solid Mechanics Part III Kelly 137()()()() ()() () () () ()⎥⎦⎤ ⎢⎣⎡ Θ∂∂== =×⋅jixJdet 33 23 1332 22 1231 21 11 3 2 1 g g gg g gg g g ggg (1.16.4) so that the condition that the determinant does not vanish is equivalent to the condition that the vectors ig are linearly independent, and so the ig can form a basis. Contravariant Base Vectors Unlike in Cartesian coordinates, where ij j iδ=⋅ee , the covariant base vectors do not necessarily form an orthonormal basis, and ij j iδ≠⋅gg . In order to deal with this complication, a second set of base vectors are introduced, which are defined as follows: introduce three contravariant base vectors ig such that each vector is normal to one of the three coordinate surfaces through the point p. From §1.6.4, the normal to the coordinate surface const1=Θ is given by the gradient vector 1gradΘ, with Cartesian representation m mxe∂Θ∂=Θ1 1grad and, in general, one may define the contravariant base vectors through m mi i xe g∂Θ∂= Contravariant Base Vectors (1.16.5) The contravariant base vector 1g is shown in Fig. 1.16.1. As with the covariant base vectors, the triple scalar product ()3 2 1ggg×⋅ is equivalent to the determinant in 1.16.2, ()()()() ()() () () () ()⎥⎦⎤ ⎢⎣⎡ ∂Θ∂== =×⋅ij x Jdet1 33 23 1332 22 1231 21 11 3 2 1 g g gg g gg g g ggg (1.16.6) and again the condition that the determinant does not vanish is equivalent to the condition that the vectors ig are linearly independent, and so the contravariant vectors also form a basis. 1.16.3 Metric Coefficients It follows from the definitions of the covariant and contravariant vectors that { ▲Problem 1} Section 1.16 Solid Mechanics Part III Kelly 138i j jiδ=⋅gg (1.16.7) which is the defining relationship between reciprocal pairs of general bases. Of course the ig were chosen precisely because they satisfy this relation. Here, j iδ is again the Kronecker delta3, with a value of 1 when ji= and zero otherwise. One needs to be careful to distinguish betwee n subscripts and superscripts when dealing with arbitrary bases, but the rules to follow are straightforward. For example, each free index which is not summed over, such as i or j in 1.16.7, must be either a subscript or superscript on both sides of an equation. Hence the new notation for the Kronecker delta symbol. The relation i j jiδ=⋅gg implies that each base vector ig is orthogonal to two of the reciprocal base vectors ig. For example, 1g is orthogonal to both 2g and 3g. Unlike the orthogonal base vectors, the dot product of a covariant/contravariant base vector with another base vector is not necessarily one or zero. Because of their importance in curvilinear coordinate systems, the dot products are given a special symbol: define the metric coefficients to be j i ijj i ij gg gggg ⋅=⋅= Metric Coefficients (1.16.8) The following important and useful relati ons may be derived by manipulating the equations already introduced: { ▲Problem 2} jij ij ij i gg g gg g == (1.16.9) and {▲Problem 3} i ki k kjijg gg ≡=δ (1.16.10) Note here another rule about indices in eq uations involving general bases: summation can only take place over a dummy index if one is a subscript and the other is a superscript – they are paired off as with the j’s in these equations. The metric coefficients can be written explicitly in terms of the curvilinear components: kj ki j i ij jk ik j i ijx xgx xg∂Θ∂ ∂Θ∂=⋅=Θ∂∂ Θ∂∂=⋅= gg gg , (1.16.11) Note here also a rule regarding derivatives with general bases: the index i on the right hand side of 1.16.11a is a superscript of Θ but it is in the denominator of a quotient and 3 although in this context it is called the mixed Kronecker delta Section 1.16 Solid Mechanics Part III Kelly 139so is regarded as a subscript to the entire symbol, matching the subscript i on the g on the left hand side4. One can also write 1.16.11 in the matrix form [] []T T , ⎥⎦⎤ ⎢⎣⎡ ∂Θ∂ ⎥⎦⎤ ⎢⎣⎡ ∂Θ∂= ⎥⎦⎤ ⎢⎣⎡ Θ∂∂ ⎥⎦⎤ ⎢⎣⎡ Θ∂∂=kj ki ij jk ik ijx xgx xg and, from 1.9.13a,b, [] []22 22 1det det , det det J xg Jxgji ij ji ij =⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ⎥⎦⎤ ⎢⎣⎡ ∂Θ∂= =⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ⎥⎦⎤ ⎢⎣⎡ Θ∂∂= (1.16.12) These determinants play an important role, and are denoted by g: [][]ij ijgg g det1det== (1.16.13) Note : • The matrix []i kxΘ∂∂/ is called the Jacobian matrix J, so []ijg=JJT Scale Factors The covariant and contravariant base vectors ar e not unit vectors: in particular, consider the covariant base vectors and introduce the unit triad igˆ, with i i i gg g⋅= : iii ii igg ggg==ˆ (no sum) (1.16.14) The lengths of the covariant base vectors are denoted by h and are called the scale factors : ii i i g h==g (no sum) (1.16.15) 1.16.4 The Covariant and Cont ravariant Comp onents of a Vector A vector can now be represented in terms of either basis: ()()ii i i u u g g u3 2 1 3 2 1,, ,, ΘΘΘ=ΘΘΘ= (1.16.16) 4 the rule for pairing off indices has been broken in (1.12.11) for clarity; more precisely, these equations should be written as ()()mnj n i m ijx x g δΘ∂∂Θ∂∂= / / and ()()mn n j m i ijx x g δ∂Θ∂∂Θ∂= / / Section 1.16 Solid Mechanics Part III Kelly 140The iu are the covariant components of u and iu are the contravariant components of u. Thus the covariant components are the coefficients of the contravariant base vectors and vice versa – subscripts denote covariance whil e superscripts denote contravariance. When u is written with covariant components, i iug u= , it is called a covariant vector . When u is written with contravariant components, iiug u= , it is called a contravariant vector . Analogous to the orthonormal case, where i iu=⋅eu {▲Problem 4}: i i i i u u =⋅=⋅ gu gu , (1.16.17) Note the following useful formula involving the metric coefficients, for raising or lowering the index on a vector component, relating the covariant and contravariant components, { ▲Problem 5} j ij i jij iug u ug u = = , (1.16.18) Physical Components of a Vector The contravariant and covariant components of a vector do not have the same physical significance in a curvilinear coordinate syst em as they do in a rectangular Cartesian system; in fact, they often have different dimensions. For example, the differential xd of the position vector has in cylindrical coordinates the contravariant components ),,( dzddrθ , that is, 33 22 11g g g x Θ+Θ+Θ= d d d d with r=Θ1, θ=Θ2, z=Θ3 (this will be discussed in detail below). Here, θd does not have the same dimensions as the others. The physical components in this example are ), ,( dz rddrθ . The physical components iu of a vector u are defined to be the components along the covariant base vectors (and hence are obtained from the contravariant components), referred to unit vectors. Thus, ii iiiiii u huu g gg u ˆ ˆ3 1≡ == ∑ = (1.16.19) and iii igu u= (no sum) Physical Components of a Vector (1.16.20) The Dot Product The dot product of two vectors can be written in one of two ways: { ▲Problem 6} ii i i vu vu==⋅vu Dot Product of Two Vectors (1.16.21) Section 1.16 Solid Mechanics Part III Kelly 141 1.16.5 The Vector Cross Product The triple scalar product is an important qu antity in analysis with general bases, particularly when evaluating cross products. From Eqns. 1.16.4, 1.16.6 and 1.16.12-13, [][] [] []ijij gg g det1 1det 23 2 12 3 2 1 = ×⋅==×⋅= gggggg (1.16.22) Introducing permutation symbols ijk ijkee,, one can in general write5 ge g eijk k j i ijk ijk k j i ijk1, ε ε =×⋅≡ =×⋅≡ ggg ggg where ijk ijkεε= is the Cartesian permutation symbol (Eqn. 1.3.8). The cross product of the base vectors can now be written in terms of the reciprocal base vectors as (note the similarity to the Cartesian relation 1.3.11) { ▲Problem 7} kijk j ik ijk j i ee g ggg gg =×=× Cross Products of Base Vectors (1.16.23) Further, from 1.3.17, i qj pj qi p pqkijk pqrijk pqrijkee ee δδδδ εε −= =, (1.16.24) The Cross Product The cross product of vectors can be written as { ▲Problem 8} 3 2 13 2 13 2 13 2 13 2 13 2 1 1 v vvu uu gvuev vvu uug vue kjiijkkji ijk g gg gg gg g vu = == =× Cross Product of Two Vectors (1.16.25) 5 assuming the base vectors form a right handed set, otherwise a negative sign needs to be included Section 1.16 Solid Mechanics Part III Kelly 1421.16.6 The Covariant, Contravari ant and Mixed Co mponents of a Tensor Tensors can be represented in any of four ways, depending on which combination of base vectors is being utilised: jij ij ii jj i ij j iijA A A A g g g g g g g g A ⊗=⊗=⊗=⊗=⋅ ⋅ (1.16.26) Here, ijA are the contravariant components , ijA are the covariant components , i jA⋅ and j iA⋅ are the mixed components of the tensor A. On the mixed components, the subscript is a covariant index, whereas the superscript is called a contravariant index. Note that the “first” index always refers to the first base vector in the tensor product. An “index switching” rule for tensors is ik k jij ikj kij A A A A = = δ δ , (1.16.27) and the rule for obtaining the components of a tensor A is (compare with 1.9.4), {▲Problem 9} () () ()() j ij ij iji i ji jj i ij ijj i ij ij AAAA Agg AAgg AAgg AAgg A ⋅=≡⋅=≡⋅=≡⋅=≡ ⋅⋅⋅⋅ (1.16.28) As with the vectors, the metric coefficients can be used to lower and raise the indices on tensors: kj ikj ikljl ik ij Tg TTgg T == ⋅ (1.16.29) In matrix form, these expressions can be conveniently used to evaluate tensor components, e.g. (note that the matrix of metric coefficients is symmetric) [][][][]lj klik ijgTg T= . An example of a higher order te nsor is the permutation tensor E, whose components are the permutation symbols introduced earlier: k j iijk k j i ijk e e g g g g g g E ⊗⊗=⊗⊗= . Section 1.16 Solid Mechanics Part III Kelly 143Physical Components of a Tensor Physical components of tensors can also be defined. For example, if two vectors a and b have physical components as defined earlier, then the physical components of a tensor T are obtained through6 j ij ibT a= . (1.16.30) As mentioned, physical components are defined with respect to the covariant base vectors, and so the mixed componen ts of a tensor are used, since ()ii iji j kk j ii j a bT b T g g g g g Tb ≡= ⊗=⋅ ⋅ as required. It follows from 1.16.22 that iii jjj i jga gbT=⋅ (no sum on the g) and so from 1.16.30, i j jjii ijT ggT⋅= (no sum) Physical Components of a Tensor (1.16.31) The Identity Tensor The components of the identity tensor I in a general basis can be obtained as follows: Iuug gggugg u ≡⊗=⋅=== ) () ( j iiji jijijijii ggugu Thus the contravariant components of the identity tensor are the metric coefficients ijg and, similarly, the covariant components are ijg. For this reason the identity tensor is also called the metric tensor . On the other hand, the mixed components are the Kronecker delta, i jδ (also denoted by i jg). In summary7, 6 these are called right physical components; left physical components are defined through bTa= 7 there is no distinction between i jj iδδ,; they are often written as i jj igg, and there is no need to specify which index comes first, for example by j ig⋅ Section 1.16 Solid Mechanics Part III Kelly 144() () () () () () () ()ii ji j ij ij ii ij ii ji ji jj iij ij ijj i ij ij ij g gg g g g g g I Ig g g g I Ig g I Ig g I I ⊗=⊗= =⊗=⊗= =⊗= =⊗= = ⋅⋅ δ δδ δ (1.16.32) Symmetric Tensors A tensor S is symmetric if S S=T, i.e. if vSu uSv= . If S is symmetric, then k mim jki ji j ji ijji ijSgg S S S S S S⋅⋅ ⋅== = = , , In terms of matrices, [][][][][][]T, ,i ji jT ij ijTij ijS S S S S S⋅ ⋅≠ = = 1.16.7 Genera lising Cartesian Relations to the Case of General Bases The tensor relations and defin itions already derived for Carte sian vectors and tensors in previous sections, for example in §1.10, are valid also in curvilinear coordinates, for example I AA=−1, AIA : tr= and so on. Formulae involving the index notation may be generalised to arbitrary components by: (1) raising or lowering the indices appropriately (2) replacing the (ordinary) Kronecker delta ijδ with the metric coefficients ijg (3) replacing the Cartesian permutation symbol ijkε with ijke in vector cross products Some examples of this are given in Table 1.16.1 below. Note that there is only one way of repr esenting a scalar, there are two ways of representing a vector (in terms of its covari ant or contravariant components), and there are four ways of representing a (second-order) tensor (in terms of its covariant, contravariant and both types of mixed components). Cartesian General Bases ba⋅ iiba ii i i ba ba= aB ijiBa j ii ij ijiji i j i j Ba BaBa Ba ⋅⋅ ==== )()( aBaB Ab jijbA j i j jij ijj ij ij i bA bAbA bA ⋅⋅ ==== )()( AbAb Section 1.16 Solid Mechanics Part III Kelly 145 AB kj ikBA () () () ()k ji k kjik i jkj ikj kk ij ikj i kj kik ijkjk ik j ik ij BA BABA BABA BABA BA ⋅⋅ ⋅⋅⋅⋅⋅⋅⋅ ⋅ ======== ABABABAB ba× ji ijkbaε () ()jiijk kji ijk k baebae =×=× baba ba⊗ jiba () () () ()ji i jj ij iji ijji ij babababa =⊗=⊗=⊗=⊗ ⋅⋅ babababa BA: ijijBA i jj ij ii j ijij ij ij BA BA BA BA⋅⋅⋅ ⋅=== AIA : tr≡ iiA i ii iA A⋅⋅= Adet 3 2 1 k j i ijk AAAε kj i ijk AAA3 2 1⋅⋅⋅ε TA ()ji ijTA= A ()() () ()i jj ij ij ii ji jjiij ji ij A A A AA A ⋅ ⋅⋅ ⋅⋅ ⋅ ≠= ≠== = T TT T ,, A AA A Table 1.16.1: Tensor relations in Cartesian and general curvilinear coordinates 1.16.8 Line, Surface and Volume Elements In order to carry out integration along curves, over regions, or throughout volumes, it is necessary to have expressions for the length of a line element sΔ, the area of a surface element SΔ and the volume of a volume element VΔ, in terms of the increments in the curvilinear coordinates 3 2 1, ,ΔΘΔΘΔΘ . The Metric Consider a differential line element, Fig. 1.16.2, ii iid dx d g e x Θ== (1.16.33) The square of the length of this line element, denoted by ()2sΔ and called the metric of the space, is then () ()()j i ij jj iiddg d d dd s ΘΘ=Θ⋅Θ=⋅=Δ g g xx2 (1.16.34) This relation ()j i ij ddg s ΘΘ=Δ2 is called the fundamental differential quadratic form . The sgij' can be regarded as a set of scale factors for converting increments in iΘ to changes in length. Section 1.16 Solid Mechanics Part III Kelly 146 Figure 1.16.2: a line element in space Surface Area and Volume Elements The surface area 1SΔ of a face of the elemental parallelepiped on which 1Θ is constant (to which 1g is normal) is, using 1.7.6, () () () () () () 3 2 113 2 2 23 33 223 2 3 2 3 2 3 3 2 23 2 3 2 3 23 2 3 233 22 1 )() () ( ΔΘΔΘ=ΔΘΔΘ− =ΔΘΔΘ⋅⋅−⋅⋅=ΔΘΔΘ×⋅×=ΔΘΔΘ×=ΔΘ×ΔΘ=Δ ggg ggS gggg gggggg ggggg g (1.16.35) and similarly for th e other surfaces. The volume VΔ of the parallelepiped is 3 2 1 3 2 1 3 2 1 ΔΘΔΘΔΘ=ΔΘΔΘΔΘ×⋅=Δ g V ggg (1.16.36) 1.16.9 Orthogonal Cu rvilinear Coordinates Orthogonal curvilinear coordinates are considered in this section). In this case, [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = = =⋅= 2 32 22 1 0 00 00 0 , hhh g hh gij jiij j i ij j i ij δ δ gg gg (1.16.37) x 1x2x3x 11gΘd33gΘd curve1−Θcurve2−Θcurve3−Θ xd22gΘd Section 1.16 Solid Mechanics Part III Kelly 147The contravariant base vectors are collinear with the covariant, but the vectors are of different magnitudes: i ii ii ihh g g g g ˆ1,ˆ= = (1.16.38) It follows that () 3 2 1 3212 1 21 31 3 13 23 2 32 12 32 32 22 22 12 12 ΔΘΔΘΔΘ=ΔΔΘΔΘ=ΔΔΘΔΘ=ΔΔΘΔΘ=ΔΘ+Θ+Θ=Δ hhhVhh Shh Shh Sdh dh dh s (1.16.39) Examples 1. Cylindrical Coordinates Consider the cylindrical coordinates, ()()3 2 1,, ,, ΘΘΘ=zrθ , cf. §1.6.10, Fig. 1.16.3: 3 32 1 22 1 1 sincos Θ=ΘΘ=ΘΘ= xxx , ()() () 3 31 2 1 22221 1 / tan xxxx x =Θ=Θ+=Θ − with ∞<Θ<∞−<Θ≤≥Θ3 2 1,2 0,0 π which give 1det Θ=⎥⎦⎤ ⎢⎣⎡ Θ∂∂=jixJ so that there is a one-to-one correspondence between the Cartesian and cylindrical coordinates at all point except for 01=Θ (which corresponds to the axis of the cylinder). These points are called singular points of the transformation. Section 1.16 Solid Mechanics Part III Kelly 148 Figure 1.16.3: Cylindrical Coordinates The unit vectors and scale factors are { ▲Problem 11} () () () () () ( )zr hr hh e g g gegg ge g g g == ==== Θ= =Θ==== === 3 3 3 312 21 2 21 1 1 1 ˆ 1 1ˆˆ 1 1 θ The physical components of a vector v are 3 21 1, , vv vΘ and one has Metric: ()()()() ()( )2 2 22322 221 2dz rd dr d d d s ++=Θ+ΘΘ+Θ=Δ θ Surface Element: 2 1 1 31 3 23 2 1 1 ΔΘΔΘΘ=ΔΔΘΔΘ=ΔΔΘΔΘΘ=Δ SSS Volume Element: ( )z rr V ΔΔΔ=ΔΘΔΘΔΘΘ=Δ θ3 2 1 1 2. Spherical Coordinates Consider the spherical coordinates, ()()3 2 1,, ,, ΘΘΘ=φθr , cf. §1.6.10, Fig. 1.16.4: 2 1 33 2 1 23 2 1 1 cossin sincos sin ΘΘ=ΘΘΘ=ΘΘΘ= xxx , ()()() () () () ()()()2122 1 3232221 1 2232221 1 / tan/ tan x xx x xx x x −− =Θ⎟⎠⎞⎜⎝⎛+ =Θ++=Θ with π π 2 0, 0,03 2 1 <Θ≤≤Θ≤≥Θ which give ()221sin det ΘΘ=⎥⎦⎤ ⎢⎣⎡ Θ∂∂=jixJ 1x2x3x • 1Θ1e• 3e 2e3g 1g2g 2Θ3Θcurve1−Θcurve2−Θcurve3−Θ Section 1.16 Solid Mechanics Part III Kelly 149so that there is a one-to-one correspondence between the Cartesian and spherical coordinates at all point except for the singular points along the 3x axis. Figure 1.16.4: Spherical Coordinates The unit vectors and scale factors are { ▲Problem 11} () () () () () ()φθ θ egg gegg ge g g g =ΘΘ= =ΘΘ=== Θ= = Θ=== = = == 2 13 32 1 3 312 21 2 21 1 1 1 sinˆ sin sinˆˆ 1 1 r hr hhr The physical components of a vector v are 32 1 21 1sin, , v vΘΘΘΘ , and one has Metric: ()()()( ) ()( )()2 2 223 2 122 121 2 sinsin φθ θ d r rd drd d d s ++=ΘΘΘ+ΘΘ+Θ=Δ Surface Element: () 2 1 1 31 3 2 1 23 2 221 1 sinsin ΔΘΔΘΘ =ΔΔΘΔΘΘΘ=ΔΔΘΔΘΘΘ=Δ SSS Volume Element: () ( )φθθΔΔΔ=ΔΘΔΘΔΘΘΘ=Δ r r V sin sin2 3 2 1 221 1.16.10 Rectangular Ca rtesian (Orthonormal) Coordinate System In an orthonormal Cartesian coordinate system, ii i e g g== , ij ijgδ= , 1=g , 1=ih and ) (ijk ijk ijke εε== . 1.16.11 Problems 1. Derive the fundamental relation i j jiδ=⋅gg . 2. Show that j ij igg g= [Hint: assume that one can write k ik iag g= and then dot both sides with jg.] •1g 1x2x3x • 3Θ1Θ2Θcurve1−Θ curve2−Θcurve3−Θ 1e3e 2e3g 2g Section 1.16 Solid Mechanics Part III Kelly 1503. Use the relations 1.16.9 to show that i k kjijggδ=. Write these equations in matrix form. 4. Show that i iu=⋅gu . 5. Show that j ij i ug u= . 6. Show that ii i i vu vu==⋅vu 7. Use the relation g eijk k j i ijk ε=×⋅≡ ggg to derive the cross product relation k ijk j i eg gg=× . [Hint: show that ()k k j i j i gggg gg ⋅×=× .] 8. Derive equation 1.16.25 for the cross product of vectors 9. Show that ()j i ij Agg A⋅= . 10. Given 3 1 3 2 2 1 1 , , ee ge ge g +=== , 3 2 1 e eev ++= . Find j i ijk ijivveg ,,,,g (write the metric coefficients in matrix form). 11. Derive the scale factors for the (a) cylindri cal and (b) spherical coordinate systems. 12. Parabolic Cylindrical (orthogonal) coordinates are given by ()()()3 3 2 1 22221 21 1, , Θ=ΘΘ= Θ−Θ= x x x with ∞<Θ<∞−≥Θ∞<Θ<∞−3 2 1,0 , Evaluate: (i) the scale factors (ii) the Jacobian – are there any singular points? (iii) the metric, surface elements, and volume element Verify that the base vectors ig are mutually orthogonal. [These are intersecting parabolas in the2 1xx− plane, all with the same axis] 13. Repeat Problem 7 for the Elliptical Cylindrical (orthogonal) coordinates : 3 3 2 1 2 2 1 1, sin sinh , cos cosh Θ=ΘΘ=ΘΘ= x a x a x with ∞<Θ<∞−<Θ≤≥Θ3 2 1,2 0,0 π [These are intersecting ellipses and hyperbolas in the 2 1xx− plane with foci at a x±=1.] 14. Consider the non-orthogonal curvilinear system illustrated in Fig. 1.16.5, with transformation equations 3 32 22 1 1 3231 xxx x =Θ=Θ−=Θ Derive the inverse transformation equations, i.e. ) ,,(3 2 1ΘΘΘ=i ix x , the Jacobian matrices ⎥⎦⎤ ⎢⎣⎡ ∂Θ∂=⎥⎦⎤ ⎢⎣⎡ Θ∂∂=− ji ji xx1,J J , the covariant and contravariant base vectors, the matrix representation of the metric coefficients [][]ij ijg g, from 1.16.8, verify that [][]ij ij g g = =−− T 1 T,JJ JJ and evaluate g. Section 1.16 Solid Mechanics Part III Kelly 151 Figure 1.16.5: non-orthogoanl curvilinear coordinate system 15. Consider a (two dimensional) curvilinear coordinate system with covariant base vectors 2 1 2 1 1 , ee g e g +== (a) Evaluate the contravariant base vectors and the metric coefficients ij ijgg, (b) Consider the vectors 2 1 2 1 2 ,3 g g v g gu +−= += Evaluate the corresponding covariant vectors. Evaluate vu⋅ (this can be done in a number of different ways – by using the relations ii i i vuvu, , or by directly dotting the vectors in terms of the base vectors i igg, and using the metric coefficients ) (c) Evaluate the contravariant vector Auw= , given that the mixed components i jA⋅ are ⎥⎦⎤ ⎢⎣⎡ − 1101 Evaluate the contravariant components ijA using the index lowering/raising rule 1.16.28. Re-evaluate the contravariant vector w using these components. 16. Consider iji jA g g A ⊗=⋅. Verify that any of the four versions of I in 1.16.32 results in I IA=. 17. Use the definitions 1.16.3-5 to convert j iijA g g⊗ , j i ijA g g⊗ and j ii jA g g⊗⋅ to the Cartesian bases. Hence show that Adet is given by the determinant of the matrix of mixed components, []i jA⋅ det , and not by []ijAdet or []ijAdet . 1x2x• 1g2g 1e2ecurve1−Θcurve2−Θ O60 01=Θ Section 1.17 Solid Mechanics Part III Kelly 1521.17 Curvilinear Coordinates: Transformation Laws 1.17.1 Coordinate Transformation Rules Suppose that one has a second se t of curvilinea r coordinates ),,(3 2 1ΘΘΘ , with ),,( ),,,(3 2 1 3 2 1ΘΘΘΘ=ΘΘΘΘΘ=Θi i i i (1.17.1) By the chain rule, the covariant base vectors in the second coordinate system are given by j ij j ij i i gx xg Θ∂Θ∂= Θ∂∂ Θ∂Θ∂= Θ∂∂= A similar calculation can be carried out for the inverse relation and for the contravariant base vectors, giving j ji i j ji ij ij i j ij i g g g gg g g g Θ∂Θ∂= Θ∂Θ∂=Θ∂Θ∂= Θ∂Θ∂= ,, (1.17.2) The coordinate transformation formulae for vectors u can be obtained from ii iiu u g g u== and i ii i u u g g u== : j ij i j ij ij ji i j ji i u u u uu u u u Θ∂Θ∂= Θ∂Θ∂=Θ∂Θ∂=Θ∂Θ∂= ,, Vector Transformation Rule (1.17.3) These transformation laws have a simple structure and pattern – the subscripts/superscripts on th e transformed coordinates Θ quantities match those on the transformed quantities, g,u, and similarly for the first coordinate system. Note: • Covariant and contravariant vectors (and other quantities) are often defined in terms of the transformation rules which they obey. For example, a covariant vector can be defined as one whose components transform according to the ru les in the second line of the box Eqn. 1.17.3 The transformation laws can be extended to higher-order tensors, Section 1.17 Solid Mechanics Part III Kelly 153n m im nj j in m im nj j im n jn mi i jm n jn mi i jmn nj mi ij mn nj mi ijmn jn im ij mn jn im ij A A A AA A A AA A A AA A A A ⋅ ⋅ ⋅ ⋅⋅ ⋅ ⋅ ⋅ Θ∂Θ∂ Θ∂Θ∂=Θ∂Θ∂ Θ∂Θ∂=Θ∂Θ∂ Θ∂Θ∂=Θ∂Θ∂ Θ∂Θ∂=Θ∂Θ∂ Θ∂Θ∂=Θ∂Θ∂ Θ∂Θ∂=Θ∂Θ∂ Θ∂Θ∂=Θ∂Θ∂ Θ∂Θ∂= ,,,, Tensor Transformation Rule (1.17.4) From these transformation expressions, the following important theorem can be deduced: If the tensor components are zero in any one coordinate system, they also vanish in any other coordinate system Reduction to Cartesian Coordinates For the Cartesian system, let i i ii i i g g eg ge ==′== , and ji ji ijxxQ′∂∂=Θ∂Θ∂= (1.17.5) It follows from 1.17.2 that 1−=→Θ∂Θ∂=Θ∂Θ∂ ij ji ji ij Q Q (1.17.6) so the transformation is ort hogonal, as expected. Also, as in Eqns. 1.5.3 and 1.5.5. jji j ij ij ji ijij ij ji i uQ uQ u u uuQ u u u ==′→ Θ∂Θ∂=′=→Θ∂Θ∂= −1 (1.17.7) Transformation Matrix Transforming coordinates from i ig g→ , one can write ()jj i jj i iM ggg g g ⋅==⋅ (1.17.8) The transformation for a vector can then be expressed, in index notation and matrix notation, as [][][]jj i i jj i i v M v vMv⋅ ⋅= = , (1.17.9) and the transformation matrix is Section 1.17 Solid Mechanics Part III Kelly 154 [] []j i ij j iM gg⋅=⎥⎦⎤ ⎢⎣⎡ Θ∂Θ∂=⋅ Transformation Matrix (1.17.10) The rule for contravariant components is then, from 1.17.4, [][][][]j nmn i mij mnj ni mijM A M A AMM A⋅ ⋅ ⋅⋅= =T, (1.17.11) The Identity Tensor The identity tensor transforms as i ik jj kk j ki ij i ij ii j g g g g g g g g g g I ⊗=⊗=⊗Θ∂Θ∂ Θ∂Θ∂=⊗=⊗= δ δ (1.17.12) Note that mn jn im n m jn im j i ijmn jn im n m jn im j i ij g gg g Θ∂Θ∂ Θ∂Θ∂=⋅ Θ∂Θ∂ Θ∂Θ∂=⋅=Θ∂Θ∂ Θ∂Θ∂=⋅Θ∂Θ∂ Θ∂Θ∂=⋅= gg gggg gg (1.17.13) so that, for example, n m mnn nj m mi mn jn im j i ij g g g g g g g g g I ⊗=⊗Θ∂Θ∂ Θ∂Θ∂ Θ∂Θ∂ Θ∂Θ∂=⊗= (1.17.14) 1.17.2 The Metric of the Space In a second coordinate system, th e metric 1.16.34 transforms to () ()22)( sgg s q p pqq p m km qk pq qj p pi m jm k ikj i j ij i ij Δ=ΔΘΔΘ=ΔΘΔΘ⋅ =ΔΘΘ∂Θ∂ΔΘΘ∂Θ∂ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ Θ∂Θ∂⋅Θ∂Θ∂=ΔΘΔΘ⋅=ΔΘΔΘ=Δ ggg ggg δδ (1.17.15) confirming that the metric is a scalar invariant. Section 1.17 Solid Mechanics Part III Kelly 155 1.17.3 Problems 1 Show that nm mnvug is an invariant. 2 How does g transform between different coordinate systems (in terms of the Jacobian of the transformation, []p mJ Θ∂Θ∂= / det )? [Note that g, although a scalar, is not invariant; it is thus called a pseudoscalar .] 3 The components ijA of a tensor A are ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −− 21 00 101 21 in cylindrical coordinates, at the point 3 ,4/ ,1 === z rπθ . Find the contravariant components of A at this point in spherical coordinates. [Hint: use matrix multiplication.] Section 1.18 Solid Mechanics Part III Kelly 1561.18 Curvilinear Coordinates: Tensor Calculus 1.18.1 Differentiation of the Base Vectors Differentiation in curvilinear coordinates is more involved than that in Cartesian coordinates because the base vectors are no long er constant and their derivatives need to be taken into account, for example the partial derivative of a vector with respect to the Cartesian coordinates is i ji j xv xev ∂∂=∂∂ but1 ji i i ji jvv Θ∂∂+ Θ∂∂= Θ∂∂ ggv The Christoffel Symbol s of the Second Kind First, from Eqn. 1.16.3 – and using the inverse relation, k mk j im m im j ji xx xg eg ∂Θ∂ ΘΘ∂∂=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ Θ∂∂ Θ∂∂=Θ∂∂2 (1.18.1) this can be written as kk ij jiggΓ=Θ∂∂ Partial Derivatives of Covariant Base Vectors (1.18.2) where mk j im k ijxx ∂Θ∂ ΘΘ∂∂=Γ2 , (1.18.3) and k ijΓ is called the Christoffel symbol of the second kind ; it can be seen to be equivalent to the kth contravariant com ponent of the vector j iΘ∂∂/g . One then has {▲Problem 1} k ij k ji k jik ij gggg⋅Θ∂∂=⋅Θ∂∂=Γ=Γ Christoffel Symbols of the 2nd kind (1.18.4) and the symmetry in the indices i and j is evident2. Looking now at the derivatives of the contravariant base vectors ig: differentiating the relation k ik iδ=⋅gg leads to k ijk mm ijk ji i jk Γ=⋅Γ=⋅Θ∂∂=⋅Θ∂∂− gg gggg 1 of course, one could express the ig in terms of the ie, and use only the first of these expressions 2 note that, in non-Euclidean space, this symmetry in the indi ces is not necessarily valid Section 1.18 Solid Mechanics Part III Kelly 157 and so k i jk ji ggΓ−=Θ∂∂ Partial Derivatives of Co ntravariant Base Vectors (1.18.5) Transformation formulae for the Christoffel Symbols The Christoffel symbols are not the components of a (third order) tensor. This follows from the fact that these components do not transform according to the tensor transformation rules given in §1.17. In fact, sk j is r pq rk jq ip k ijΘ∂Θ∂ ΘΘ∂Θ∂+ΓΘ∂Θ∂ Θ∂Θ∂ Θ∂Θ∂=Γ2 The Christoffel Symbol s of the First Kind The Christoffel symbols of the second kind rela te derivatives of cova riant (contravariant) base vectors to the covariant (c ontravariant) base vectors. A second set of symbols can be introduced relating the base vectors to the de rivatives of the reciprocal base vectors, called the Christoffel symbols of the first kind : k ij k ji jik ijk gggg⋅Θ∂∂=⋅Θ∂∂=Γ=Γ Christoffel Symbols of the 1st kind (1.18.6) so that the partial derivati ves of the covariant base v ectors can be written in the alternative form k ijk jiggΓ= Θ∂∂, (1.18.7) and it also follows from Eqn. 1.18.2 that mk ijmk ij mkm ij ijk g g Γ=ΓΓ=Γ , (1.18.8) showing that the index k here can be raised or lowered using the metric coefficients as for a third order tensor (but the first two indexes, i and j, cannot and, as stat ed, the Christoffel symbols are not the components of a third order tensor). Example: Newton’s Second Law The position vector can be expressed in terms of curvilin ear coordinates, ()iΘ=xx . The velocity is then ii i idtd dtd dtdgx xvΘ=Θ Θ∂∂== Section 1.18 Solid Mechanics Part III Kelly 158and the acceleration is ik j i jki k kjj ii dtd dtd dtd dtd dtd dtd dtdgggva⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ΘΘΓ+Θ=Θ Θ∂∂Θ+Θ==22 22 Equating the contravariant com ponents of Newton’s second law a fm= then gives the general curvilinear expression ( )kj i jki im f ΘΘΓ+Θ= &&&& ■ Partial Differentiation of the Metric Coefficients The metric coefficients can be differentiated with the aid of the Christoffel symbols of the first kind { ▲Problem 3}: jki ikj kijgΓ+Γ=Θ∂∂ (1.18.9) Using the symmetry of the metric coefficients and the Christoffel symbols, this equation can be written in a number of different ways: jki kij kijg Γ+Γ=, , kij ijk ijkg Γ+Γ=, , ijk jki jkig Γ+Γ=, Subtracting the first of these fr om the sum of the second and th ird then leads to the useful relations (using also 1.18.8) () ()mij jmi ijmmk k ijkij jki ijk ijk g g ggg g g , , ,, , , 2121 −+ =Γ−+=Γ (1.18.10) which show that the Christoffel symbols depe nd on the metric coefficients only. Alternatively, one can write the derivatives of th e metric coefficients in the form (the first of these is 1.18.9) i kmjm j kmim kijjki ikj kij g g gg Γ−Γ−=Γ+Γ= ,, (1.18.11) Also, directly from 1.15.7, one has the relations ij ijij ijgg gggggg= ∂∂=∂∂, (1.18.12) Section 1.18 Solid Mechanics Part III Kelly 159and from these follow other useful relations, for example { ▲Problem 4} () j j ji ijJJg gg Θ∂∂=Θ∂∂=Θ∂∂=Γ−1 1 log (1.18.13) and ()n mnijk n mnijk mijk mijkn mn ijkn mn ijk m ijk mijk e gg ee gg e Γ−=Γ−=Θ∂∂=Θ∂∂Γ=Γ=Θ∂∂=Θ∂∂ 1 /1ε εε ε (1.18.14) 1.18.2 Partial Differentiation of Tensors The Partial Derivative of a Vector The derivative of a vector in curvi linear coordinates can be written as ijikk iji i jiji i i ji j vvvvv gg gggv ≡Γ+Θ∂∂=Θ∂∂+Θ∂∂=Θ∂∂ or i jik i jkii jiji ii ji j vvvvv gg gggv ≡Γ− Θ∂∂=Θ∂∂+ Θ∂∂= Θ∂∂ (1.18.15) where kk ij ji jiki kjji ji v v vv v v Γ−=Γ+= ,, || Covariant Derivative of Vector Components (1.18.16) The first term here is the ordinary partial derivative of the vector components. The second term enters the expressi on due to the fact that the curvilinear base vectors are changing. The complete quantity is defined to be the covariant derivative of the vector components. The covariant derivative reduces to the ordinary partial derivative in the case of rectangular Cartesian coordinates. The jiv| is the ith component of the j – derivative of v. The jiv| are also the components of a second order covariant tensor, transforming under a change of coordinate system according to the tensor tran sformation rule 1.17.4 (see the gradient of a vector below). Section 1.18 Solid Mechanics Part III Kelly 160The covariant derivative of vector components is given by 1.18.10. In the same way, the covariant derivative of a vector is defined to be the complete expression in 1.18.9, j,v, with i ji jv g v |,= . The Partial Derivative of a Tensor The rules for covariant differentiation of vector s can be extended to higher order tensors. The various partial derivatives of a second-order tensor j ii j jij ij i ij j iijA A A A g g g g g g g g A ⊗=⊗=⊗=⊗=⋅⋅ are indicated using the following notation: j i ki j ji kj ij i kij j i kij kA A A A g g g g g g g gA⊗=⊗=⊗=⊗=Θ∂∂ ⋅⋅| | | | (1.18.17) Thus, for example, []j i imm jk mjm ik kijm j kmi ijj m i mkijj i kijkj i ijj ki ijj i kij k A A AA A AA A A g gg g g g g gg g g g g g A ⊗Γ−Γ−=Γ⊗−⊗Γ−⊗=⊗+⊗+⊗= ,,, ,, , and, in summary, i mm jkm ji mki kj ki jj mm ikm ij mk kj i kj iim j mkmj i mkkij kijimm jk mjm ik kij kij A A A AA A A AA A A AA A A A ⋅ ⋅ ⋅ ⋅⋅ ⋅ ⋅ ⋅ Γ−Γ+=Γ−Γ+=Γ+Γ+=Γ−Γ−= ,,,, |||| (1.18.18) Covariant Derivative of Tensor Components The covariant derivative formulas can be re membered as follows: the formula contains the usual partial derivative plus • for each contravariant index a term containing a Christoffel symbol in which that index has been inserted on the upper level, multiplied by the tensor component with that index replaced by a dummy summation index which al so appears in the Christoffel symbol • for each covariant index a term prefixed by a minus sign and containing a Christoffel symbol in which that index has been inserted on the lower level, multiplied by the tensor with that index re placed by a dummy which also appears in the Christoffel symbol. • the remaining symbol in all of the Christoffel symbols is the index of the variable with respect to which the covariant derivative is taken. For example, i jmm kli mkm jlm jki mli ljk li jk A A A A A⋅ ⋅ ⋅ ⋅ ⋅ Γ−Γ−Γ+=, | Section 1.18 Solid Mechanics Part III Kelly 161 Note that the covariant derivative of a pr oduct obeys the same rules as the ordinary differentiation, e.g. ()mjk ijk mi mjk i Au A u Au | | ||+ = Covariantly Constant Coefficients It can be shown that the metric coefficients are covariantly constant3 {▲Problem 5}, 0| |==kij kijg g , This implies that the metric (identity) tensor I is constant, 0,=kI (see Eqn. 1.16.32) – although its components ijg are not constant. Simila rly, the components of the permutation tensor, are covariantly constant 0| |==mijk mijk e e . In fact, specialising the identity tensor I and the permutation tensor E to Cartesian coordinates, one has ijij ijg g δ→= , ijkijk ijke e ε→= , which are clearly constant. Specialising the derivatives, kij kijg, |δ→ , mijk m ijke, |ε→ , and these are clearly zero. From §1.17, since if the components of a tens or vanish in one coordinate system, they vanish in all coordinate systems, the curvilinea r coordinate versions vanish also, as stated above. The above implies that any time any of these factors appears in a covariant derivative, they may be extracted, as in ()()ki ij ki ij ug ug | |= . The Riemann-Christoffel Curvature Tensor Higher-order covariant derivatives are defined by repeated applicati on of the first-order derivative. This is straight-f orward but can lead to algebr aically lengthy expressions. For example, to evaluate mniv| , first write the first covariant derivative in the form of a second order covariant tensor B, im kk im mi mi B v v v ≡Γ−=, | so that () () ( )ll ik kik mn ll km mkk in nkk im miikk mn kmk in nimn im mni v v v v v vB B BB v Γ−Γ−Γ−Γ−Γ−=Γ−Γ−== , , , ,,| | (1.18.19) 3 Section 1.18 Solid Mechanics Part III Kelly 162The covariant derivative nmiv| is obtained by interchaning m and n in this expression. Now investigate the difference ()()()() () ()ll ik kik nm ll ik kik mnll kn nkk im ll km mkk in mkk in ni nkk im mi nmi mni v v v vv v v v v v v v v v Γ−Γ+Γ−Γ−Γ−Γ+Γ−Γ−Γ−−Γ−=− , ,, , , , , , | | The last two terms cancel here because of the symmetry of the Christoffel symbol, leaving () ()ll kn nkk im ll km mkk inmkk in kk min nmi nkk im kk nim mni nmi mni v v v vv v v v v v v v Γ−Γ+Γ−Γ−Γ+Γ+−Γ−Γ−=− , ,, , , , , , | | The order on the ordinary partial differenti ation is interchangeab le and so the second order partial derivative terms cancel, () ()ll kn nkk im ll km mkk inmkk in kk min nmi nkk im kk nim mni nmi mni v v v vv v v v v v v v Γ−Γ+Γ−Γ−Γ+Γ+−Γ−Γ−=− , ,, , , , , , | | After further cancellation one arrives at jj imn nmi mni vR v v⋅=−| | (1.18.20) where R is the fourth-order Riemann-Christoffel curvature tensor , with (mixed) components j knk imj kmk inj nimj minj imnR ΓΓ−ΓΓ+Γ−Γ=⋅ , , (1.18.21) Since the Christoffel symbols vanish in a Cartesian coordinate system, then so does j imnR⋅. Again, any tensor that vanishes in one coordi nate system must be zero in all coordinate systems, and so 0=⋅j imnR , implying that the order of covariant differentiation is immaterial, nmi mni v v | |= . From 1.18.10, it follows that 0=++=−=−= iljk iklj ijklklij ijlk jikl ijkl R R RR R R R The latter of these known as the Bianchi identities . In fact, only six components of the Riemann-Christoffel tensor ar e independent; the expression 0=⋅j imnR then represents 6 equations in the 6 independent components ijg. This analysis is for a Euclidean space – the usual three-dimensional space in which quantities can be expressed in terms of a Ca rtesian reference system – such a space is Section 1.18 Solid Mechanics Part III Kelly 163called a flat space . These ideas can be extended to other, curved spaces, so-called Riemannian spaces (Riemannian manifolds ), for which the Riemann-Christoffel tensor is non-zero (see §1.19). 1.18.3 Differential Operators and Tensors In this section, the concepts of the gradie nt, divergence and curl from §1.6 and §1.14 are generalized to the case of curvilinear components. Space Curves and the Gradient Consider first a scalar function ()xf, where iixe x= is the position vector, with ()j i ix xΘ= . Let the curvilinear coordinates depend on some parameter s, ()sj jΘ=Θ , so that )(sx traces out a space curve C. For example, the cylindrical coordinates ()sj jΘ=Θ , with ar= , cs/=θ , csbz /= , c sπ2 0≤≤ , generate a helix. From §1.6.2, a tangent to C is iii idsd dsdgx xτ τ=Θ Θ∂∂== so that ds di/Θ are the contravariant components of τ. Thus ()jj i ii if f dsdfg gτ τ ⋅⎟ ⎠⎞⎜ ⎝⎛ Θ∂∂=Θ∂∂= . For Cartesian coordinates, τ⋅∇=f dsdf/ (see the discussion on normals to surfaces in §1.6.4). For curvilinear coordinates, theref ore, the Nabla operator of 1.6.11 now reads ii Θ∂∂=∇g (1.18.22) so that again the directional derivative is τ⋅∇=fdsdf The Gradient of a Scalar In general then, the gradient of a scalar valued function Φ is defined to be i igΘ∂Φ∂=Φ≡Φ∇ grad Gradient of a Scalar (1.18.23) Section 1.18 Solid Mechanics Part III Kelly 164and, with ii iid dx d g e x Θ== , one has xd d di i⋅Φ∇=ΘΘ∂Φ∂≡Φ (1.18.24) The Gradient of a Vector Analogous to Eqn. 1.14.3, the gradient of a v ector is defined to be the tensor product of the derivative jΘ∂/u with the contravariant base vector jg: j i jij i jij j uu g gg g guu ⊗=⊗=⊗Θ∂∂= || grad Gradient of a Vector (1.18.25) Note that ji ij j i ij ii iiu u g g g gugu gu ⊗=⊗=Θ∂∂⊗=⊗Θ∂∂=⊗∇ | | so that again one arri ves at Eqn. 1.14.7, () u u gradT=⊗∇ . Again, one has for a space curve parameterised by s, () τuτug gτu u u⋅=⋅⎟ ⎠⎞⎜ ⎝⎛ Θ∂∂⊗=⋅Θ∂∂=Θ∂∂= gradT ii i ii idsdτ Similarly, from 1.18.18, the gradie nt of a second-order tensor is k j i ki jk ji kj ik j i kijk j i kij k k AAAA g g gg g gg g gg g g gAA ⊗⊗ =⊗⊗ =⊗⊗ =⊗⊗ =⊗Θ∂∂= ⋅⋅ |||| grad Gradient of a Tensor (1.18.26) The Divergence From 1.14.9, the divergence of a vector is { ▲Problem 6} ⎟ ⎠⎞⎜ ⎝⎛⋅Θ∂∂= = =j j iiu guIu u | : grad div Divergence of a Vector (1.18.27) This is equivalent to the divergen ce operation involving th e Nabla operator, . div u u⋅∇= An alternative expression can be obtained from 1.18.13 { ▲Problem 7}, Section 1.18 Solid Mechanics Part III Kelly 165()() ii ii ii JuJug guΘ∂∂=Θ∂∂==−1 1| divu Similarly, using 1.14.12, the diverg ence of a second-order tensor is i jj ij j i jij AA ggAg IA A || : grad div ⋅=⎟ ⎠⎞⎜ ⎝⎛ Θ∂∂= = = Divergence of a Tensor (1.18.28) Here, one has the alte rnative definition, L= =Θ∂∂⋅=⋅Θ∂∂=⋅∇i jji ii iiA gAgA gA | so that again one arri ves at Eqn. 1.14.14, Tdiv A A⋅∇= . The Curl The curl of a vector is defined by { ▲Problem 8} k ij ijk k ijijk kkue ue g gugu uΘ∂∂= =Θ∂∂×=×∇= | curl Curl of a Vector (1.18.29) the last equality following from the fact that all the Christoffel symbols cancel out. Covariant derivatives as Tensor Components Equation 1.18.25 shows clearly that the cova riant derivatives of vector components are themselves the components of second order tensors. It follows that they can be manipulated as other tensors, for example, ji j mimu ug | |= and it is also helpful to in troduce the following notation: mj mi ji mj mij i g u u g u u | | , | | = = . The divergence and curl can then be written as { ▲Problem 10} k ij ijk k ijijki i ii ue ueu u g g uu | | curl| | div = === . Section 1.18 Solid Mechanics Part III Kelly 166Generalising Tensor Calculus from Cartesian to Curvilinear Coordinates It was seen in §1.16.7 how formulae could be generalised from the Cartesian system to the corresponding formulae in curvilinear coor dinates. In addition, formulae for the gradient, divergence and curl of tensor fields may be generalised to curvilinear components simply by replacing the partial deri vatives with the covariant derivatives. Thus: Cartesian Curvilinear Of a scalar field ix∂∂=∇ / , grad φφφ i i i Θ∂∂≡= / |, φφφ of a vector field j ix u∂∂= / gradu jiu| Gradient of a tensor field k ijx T∂∂= / gradT kijT| of a vector field i ix u∂∂=⋅∇ / ,div u u iiu| Divergence of a tensor field j ijx T∂∂= / divT jijT| Curl of a vector field i j ijk x u∂∂=×∇ / , curl εu u ijijkue | Table 1.18.1: generalising formulae fr om Cartesian to General Curvilinear Coordinates All the tensor identities derived for Cartes ian bases (§1.6.9, §1.14.3) hold also for curvilinear coordinates, for example { ▲Problem 11} () () v AA v vAvv v grad: div divgrad grad grad +⋅=⊗+= α αα 1.18.4 Partial Derivatives with respect to a Tensor The notion of differentiation of one tensor with respect to another can be generalised from the Cartesian differentiation discussed in §1.15. For example: n mn mn mj in jm i n mj i mnijn mj i mnijj i j ij i ij AAABA A g g g gg g g g g g g gAAg g g gABg g g gA ⊗⊗⊗=⊗⊗⊗=⊗⊗⊗∂∂=∂∂=⊗⊗⊗∂∂=∂∂=⊗∂Φ∂=⊗∂Φ∂=∂Φ∂ ⋅ δδLL 1.18.5 Orthogonal Curvilinear Coordinates This section is based on the groundwork carried out in §1.16.9. In orthogonal curvilinear systems, it is best to write all equations in te rms of the covariant base vectors, or in terms of the corresponding physical components, using the identities (see Eqn. 1.16.38) Section 1.18 Solid Mechanics Part III Kelly 167 i ii ii h hg g g ˆ1 1 2== (no sum) (1.18.30) The Gradient of a Scalar Field From the definition 1.18.23 for the gradient of a scalar field, and Eqn. 1.18.30, one has for an orthogonal curvilinear coordinate system, 3 3 32 2 21 1 13 3 2 32 2 2 21 1 2 1 ˆ1ˆ1ˆ11 1 1 g g gg g g Θ∂Φ∂+Θ∂Φ∂+Θ∂Φ∂=Θ∂Φ∂+Θ∂Φ∂+Θ∂Φ∂=Φ∇ h h hh h h (1.18.31) The Christoffel Symbols The Christoffel symbols simplify considerably in orthogonal coordinate systems. First, from the definition 1.18.4, k ji kk ijhgg⋅Θ∂∂=Γ21 (1.18.32) Note that the introduction of the scale factors h into this and the following equations disrupts the summation and index notation c onvention used hitherto. To remain consistent, one should use the metric coeffici ents and leave this equation in the form mkm ji k ij ggg⋅ Θ∂∂=Γ Now ()i iji i ji i i jhΓ=⎟ ⎠⎞⎜ ⎝⎛⋅Θ∂∂=⋅Θ∂∂22 2 gggg and 2 i i i h=⋅gg so, in terms of the derivatives of the scale factors, ji iikk iji ijh hΘ∂∂=Γ=Γ =1 (no sum) (1.18.33) Similarly, it can be shown that { ▲Problem 14} i jk ij kiji jk ik ijk h h h h Γ=Γ−=Γ−=Γ2 2 2 2 when kji≠≠ (1.18.34) so that the Christoffel symbols are zero when the indices are distinct, so that there are only 21 non-zero symbols of the 27. Further, { ▲Problem 15} Section 1.18 Solid Mechanics Part III Kelly 168ki ki jik ijk iih hh Θ∂∂−=Γ=Γ =2, ki≠ (no sum) (1.18.35) From the symmetry condition (see Eqn. 1.18. 4), only 15 of the 21 non-zero symbols are distinct: 3 333 323 233 223 313 133 112 332 322 232 222 212 122 111 331 221 311 131 211 121 11 , ,, ,, ,, ,,, , , ΓΓ=ΓΓΓ=ΓΓΓΓ=ΓΓΓ=ΓΓΓΓΓ=ΓΓ=ΓΓ Note also that these are related to each other through the rela tion between (1.18.33, 1.18.35), i.e. i ik ki k iihhΓ−=Γ22 , ki≠ (no sum) so that 2 33 2 32 2 3 323 231 33 2 32 1 3 313 133 22 2 22 3 2 322 231 22 2 22 1 2 212 123 11 2 12 3 1 311 132 11 2 12 2 1 211 123 332 221 11 , ,, ,, , Γ−=Γ=ΓΓ−=Γ=ΓΓ−=Γ=ΓΓ−=Γ=ΓΓ−=Γ=ΓΓ−=Γ=ΓΓΓΓ hh hh hhhh hh hh (1.18.36) The Gradient of a Vector From the definition 1.18.25, the gradient of a vector is j iji jj ijivhv g g g g v ⊗ =⊗=21grad (no sum over jh) (1.18.37) In terms of physical components, j ik i kj ki ii ij ji jj ii kjk ji j vhhvv hvv h g gg g v ˆ ˆ11grad2 ⊗⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛Γ+Γ− Θ∂∂=⊗⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛Γ+ Θ∂∂= (1.18.38) The Divergence of a Vector From the definition 1.18.27, th e divergence of a vector is iiv=vdiv or {▲Problem 16} Section 1.18 Solid Mechanics Part III Kelly 169()()() ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ Θ∂∂+Θ∂∂+Θ∂∂=Γ+Θ∂∂=3213 2312 1321 3211divhhv hhv hhv hhhvvi kik iiv (1.18.39) The Curl of a Vector From §1.16.5 and 1.16.37, the permutation symb ol in orthogonal curv ilinear coordinates reducec to ijk ijk hhhe ε 3211= (1.18.40) where ijkijkεε= is the Cartesian permutation symbol . From the definition 1.18.29, the curl of a vector is then ()() ()() ⎪⎭⎪⎬⎫ ⎪⎩⎪⎨⎧ + ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ Θ∂∂− Θ∂∂=⎪⎭⎪⎬⎫ ⎪⎩⎪⎨⎧+⎥ ⎦⎤ ⎢ ⎣⎡ Θ∂∂− Θ∂∂=⎭⎬⎫ ⎩⎨⎧+⎥⎦⎤ ⎢⎣⎡ Θ∂∂−Θ∂∂=Θ∂∂= LLL 33 211 122 3213 22 11 12 22 3213 21 12 321 321 ˆ111 1curl ggg g v hhv hv hhhhv hv hhhv v hhhv hhhkij ijkε (1.18.41) or 3 32 21 13 2 133 22 11 321ˆ ˆ ˆ 1curl vh vh vhh h h hhh Θ∂∂ Θ∂∂ Θ∂∂=g g g v (1.18.42) The Laplacian From the above results, the Laplacian is given by ⎥ ⎦⎤ ⎢ ⎣⎡ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ Θ∂Φ∂ Θ∂∂+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ Θ∂Φ∂ Θ∂∂+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ Θ∂Φ∂ Θ∂∂=Φ∇⋅∇=Φ∇3 321 3 2 231 2 1 132 1 3212 1 hhh hhh hhh hhh Divergence of a Tensor From the definition 1.18.28, and using 1.16.31, 1.16.29 { ▲Problem 17} Section 1.18 Solid Mechanics Part III Kelly 170imj m ij jim im j mj mij ji j ii j mm ijm ij mj jj i i jj i AhhhAhAhh hA AAA gg g A ˆ1 1| div ⎪⎭⎪⎬⎫ ⎪⎩⎪⎨⎧ Γ−Γ+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ Θ∂∂=⎭⎬⎫ ⎩⎨⎧Γ−Γ+Θ∂∂= =⋅ ⋅⋅ ⋅ (1.18.43) Examples 1. Cylindrical Coordinates Gradient of a Scalar Field (see 1.6.28): 3 3 2 22 1 1ˆ ˆ1ˆ g g g Θ∂Φ∂+Θ∂Φ∂ Θ+ Θ∂Φ∂=Φ∇ Christoffel symbols: With 1 , ,131 2 1 =Θ== h h h , there are two distinct non-zero symbols: 12 212 121 1 22 1 Θ=Γ=ΓΘ−=Γ Derivatives of the base vectors: The non-zero derivatives are 11 22 2 1 12 21,1gggg gΘ−= Θ∂∂ Θ= Θ∂∂= Θ∂∂ and in terms of physical components, the non-zero derivatives are 1 22 2 21ˆˆ,ˆˆgggg−=Θ∂∂=Θ∂∂ which agree with 1.5.29. The Divergence (see 1.6.30), Curl (see 1.6.31) and Gradient { ▲Problem 18} (see 1.14.18) of a vector: ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ Θ∂∂ +Θ∂∂ Θ+Θ+Θ∂∂ =33 22 1 11 11 1divv v v v v 3 21 13 2 13 21 1 1ˆ ˆ ˆ 1curl v v vΘΘ∂∂ Θ∂∂ Θ∂∂Θ Θ=g g g v Section 1.18 Solid Mechanics Part III Kelly 1713 3 33 3 2 32 3 1 31 2 3 23 12 2 1 22 1 2 1 2 21 11 3 13 1 2 12 1 1 11 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ1ˆ ˆ1ˆ ˆ1ˆ ˆ ˆ ˆ ˆ ˆ grad g g g g g g g gg g g gg g g g g g v ⊗ Θ∂∂ +⊗ Θ∂∂ +⊗ Θ∂∂ +⊗⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ Θ∂∂ Θ+⊗⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ + Θ∂∂ Θ+⊗⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ − Θ∂∂ Θ+⊗ Θ∂∂ +⊗ Θ∂∂ +⊗ Θ∂∂ = v v v vvv vvv v v The Divergence of a tensor { ▲Problem 19} (see 1.15.16): 3 333 232 1 131 1312 112 21 323 222 1 1211 122 11 313 212 1 111 ˆ1ˆ1ˆ1div ggg A ⎪⎭⎪⎬⎫ ⎪⎩⎪⎨⎧ Θ∂∂+ Θ∂∂ Θ+ Θ+ Θ∂∂+⎪⎭⎪⎬⎫ ⎪⎩⎪⎨⎧ Θ++Θ∂∂+Θ∂∂ Θ+Θ∂∂+⎪⎭⎪⎬⎫ ⎪⎩⎪⎨⎧ Θ−+ Θ∂∂+ Θ∂∂ Θ+ Θ∂∂= A A A AA A A A AA A A A A 2. Spherical Coordinates Gradient of a Scalar Field (see 1.6.35): 3 3 2 1 2 2 1 1 1ˆ sin1ˆ1ˆ g g gΘ∂Φ∂ ΘΘ+Θ∂Φ∂ Θ+Θ∂Φ∂=Φ∇ Christoffel symbols: With 2 1 31 2 1 sin , ,1 ΘΘ=Θ== h h h , there are six distin ct non-zero symbols: 2 3 323 23 13 313 132 2 2 33 12 212 122 2 1 1 331 1 22 cot ,1cos sin ,1sin , Θ=Γ=ΓΘ=Γ=ΓΘΘ−=ΓΘ=Γ=ΓΘΘ−=ΓΘ−=Γ Derivatives of the base vectors: The non-zero derivatives are 22 2 12 2 1 33 32 23 3211 22 3 1 13 31 2 1 12 21 cos sin sin , cot,1,1 g gggg ggggg ggg g ΘΘ−ΘΘ−=Θ∂∂Θ=Θ∂∂=Θ∂∂Θ−= Θ∂∂ Θ= Θ∂∂= Θ∂∂ Θ= Θ∂∂= Θ∂∂ and in terms of physical components, the non-zero derivatives are 22 12 33 32 321 22 32 31 2 21 ˆ cosˆ sinˆ,ˆ cosˆˆˆ,ˆ sinˆ,ˆˆ g gggggggggg Θ−Θ−=Θ∂∂Θ=Θ∂∂−=Θ∂∂Θ=Θ∂∂=Θ∂∂ which agree with 1.6.34. Section 1.18 Solid Mechanics Part III Kelly 172 The Divergence (see 1.6.35), Curl and Gradient of a Vector: ()()() () 33 2 1 222 2 1 1121 21 sin1 sin sin1 1div Θ∂∂ ΘΘ+ Θ∂Θ∂ ΘΘ+ Θ∂Θ∂ Θ=v v v v ()3 2 1 21 13 2 132 1 21 1 221 sinˆ sin ˆ ˆ sin1curl v v v ΘΘΘΘ∂∂ Θ∂∂ Θ∂∂ΘΘΘ ΘΘ=g g g v 3 3 12 2 11 33 2 12 3 23 1 1 3 133 2 13 2 32 2 1 2 2 11 22 11 2 12 3 1 13 31 2 12 1 12 21 1 1 1 11 ˆ ˆ cot sin1ˆ ˆ1ˆ ˆˆ ˆ cot sin1ˆ ˆ1ˆ ˆ ˆ ˆ sin1ˆ ˆ1ˆ ˆ grad g gg g g gg g g gg g g gg g g g v ⊗⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ΘΘ+ Θ+ Θ∂∂ ΘΘ+⊗ Θ∂∂ Θ+⊗ Θ∂∂ +⊗⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ΘΘ− Θ∂∂ ΘΘ+⊗⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ Θ+ Θ∂∂ Θ+⊗ Θ∂∂ +⊗⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ Θ− Θ∂∂ ΘΘ+⊗⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ Θ− Θ∂∂ Θ+⊗ Θ∂∂ = v v vv vv v v vv v vv v v The Divergence of a tensor { ▲Problem 20} () () 3 132 23 2 31 13 333 2 1 232 2 1 1312 133 22 2 21 12 323 2 1 222 1 1211 133 22 12 2 11 313 2 1 212 1 111 ˆcot 2 sin1 sin1ˆcot 2 sin1 1ˆcot 2 sin1 1div ggg A ⎪⎭⎪⎬⎫ ⎪⎩⎪⎨⎧ Θ+Θ+++ Θ∂∂ ΘΘ+ Θ∂∂ ΘΘ+ Θ∂∂+⎪⎭⎪⎬⎫ ⎪⎩⎪⎨⎧ Θ−Θ+++ Θ∂∂ ΘΘ+ Θ∂∂ Θ+ Θ∂∂+⎪⎭⎪⎬⎫ ⎪⎩⎪⎨⎧ Θ−−Θ++ Θ∂∂ ΘΘ+ Θ∂∂ Θ+ Θ∂∂= A A A A A A AA A A A A A AA A A A A A A 1.18.6 Problems 1 Show that the Christoffel symbol of the second kind is symmetric, i.e. k jik ijΓ=Γ , and that it is explicitly given by k ji k ij gg⋅ Θ∂∂=Γ . 2 Consider the scalar-valued function ()ji ijvuA=⋅=Φ v Au . By taking the gradient of this function, and using the relation for the covariant derivative of A, i.e. imm jk mjm ik kij kij A A A A Γ−Γ−=, | , show that ()()kji ij kji ijvuAvuA| =Θ∂∂, Section 1.18 Solid Mechanics Part III Kelly 173i.e. the partial derivative and covariant de rivative are equivalent for a scalar-valued function. 3 Prove 1.18.9: (i) jki ikj kijgΓ+Γ=Θ∂∂, (ii) i kmjm j kmim kij g ggΓ−Γ−=Θ∂∂ [Hint: for (ii), first differentiate Eqn. 1.16.10, i k kjijggδ=.] 4 Derive 1.18.13, relating the Christoffel symbols to the partial derivatives of g and ()g log . [Hint: begin by using the chain rule jmn mnjg gg g Θ∂∂ ∂∂= Θ∂∂.] 5 Use the definition of the covariant derivativ e of second order tensor components, Eqn. 1.18.18, to show that (i) 0|=kijg and (ii) 0 |=kijg . 6 Use the definition of the gradient of a vector, 1.18.25, to show that iiu| : grad div = = Iu u . 7 Derive the expression ()()i iug g Θ∂∂= / /1 divu 8 Use 1.16.23 to show that ()k ijijk k kue g u g | /=Θ∂∂× . 9 Use the relation j nk mk nj m imnijkδδδδεε −= (see Eqn. 1.3.17) to show that () () () ()12 21 13 31 23 323 3 2 2 1 13 2 1 curl vu vu vuvu vuvuk kk kk k − −−−Γ+Θ∂∂Γ+Θ∂∂Γ+Θ∂∂=×g g g vu . 10 Show that (i) i i iiu u | |= , (ii) k ij ijk k ijijkue ue g g | |= 11 Show that (i) () α αα grad grad grad ⊗+= vv v , (ii) () v AA v vA grad: div div +⋅= [Hint: you might want to use the relation ()baTbaT ⊗=⋅ : for the second of these.] 12 Derive the relation () IA A=∂∂ /tr in curvilinear coordinates. 13 Consider a (two dimensional) curvilinear coordinate system with covariant base vectors 11 2 2 12 1 ,2 e g e e g Θ= −Θ= . (a) Evaluate the transformation equations ) (j i ix xΘ= and the Jacobian J. (b) Evaluate the inverse tr ansformation equations ()j i ixΘ=Θ and the contravariant base vectors ig. (c) Evaluate the metric coefficients ij ijgg, and the function g: (d) Evaluate the Christoffel sy mbols (only 2 are non-zero (e) Consider the scalar field 2 1Θ+Θ=Φ . Evaluate Φgrad . (f) Consider the vector fields ()2 121 22 1 2 , g g v g gu +Θ−= Θ+= : (i) Evaluate the covariant components of the vectors u and v (ii) Evaluate v udiv,div (iii)Evaluate v ucurl, curl (iv) Evaluate v ugrad, grad (g) Verify the vector identities Section 1.18 Solid Mechanics Part III Kelly 174() () () () () 0 curldivgrad curlcurl curl divgrad curl curlgrad div div ==Φ⋅−⋅=××Φ+Φ=Φ⋅Φ+Φ=Φ uov uu vvuu u uu u u (h) Verify the identities () () ( ) ( ) () () () ( ) ( ) uv vu uvvuvuuv vu vuuv vu vuvv v grad grad div div curl) div( grad divgrad grad gradgrad grad grad T T − +−=×+ =⊗+ =⋅Φ⊗+Φ=Φ (i) Consider the tensor field ()j ig g A ⊗⎥⎦⎤ ⎢⎣⎡ Θ−=220 1 Evaluate all contravariant and mixed components of the tensor A 14 Use the fact that 0 =⋅i kgg , ik≠ to show that k iji jk ik ijk h h Γ−=Γ2 2. Then permutate the indices to show that i jk ij kiji jk ik ijk h h h h Γ=Γ−=Γ−=Γ2 2 2 2 when kji≠≠ . 15 Use the relation () jij i i≠=⋅Θ∂∂,0 gg to derive i ij ji j iihhΓ−=Γ22 . 16 Derive the expression 1.18.39 for th e divergence of a vector field v. 17 Derive 1.18.43 for the divergence of a te nsor in orthogonal coordinate systems. 18 Use the expression 1.18.38 to derive the expr ession for the gradient of a vector field in cylindrical coordinates. 19 Use the expression 1.18.43 to derive the e xpression for the divergence of a tensor field in cylindrical coordinates. 20 Use the expression 1.18.43 to derive the e xpression for the divergence of a tensor field in spherical coordinates. Section 1.19 Solid Mechanics Part III Kelly 1751.19 Curvilinear Coordinates: Curved Geometries In this section is examined the special case of a two-dimensional curved surface. 1.19.1 Monoclinic Coordinate Systems Base Vectors A curved surface can be defined using two covariant base vectors 2 1,aa , with the third base vector, 3a, everywhere of unit size and normal to the other two, Fig. 1.19.1 These base vectors form a monoclinic reference frame, that is, onl y one of the angles between the base vectors is not n ecessarily a right angle. Figure 1.19.1: Geometry of the Curved Surface In what follows, in the index notation, Greek letters such as βα, take values 1 and 2; as before, Latin letters take values from 1..3. Since 33a a= and 03 3 =⋅= aaα αa , 03 3=⋅= aaα αa (1.19.1) the determinant of metric coefficients is 10 000 22 2112 11 2g gg g J= , 1 0 000 122 2112 11 2g gg g J= (1.19.2) The Cross Product Particularising the results of §1.16.5, define the surface permutation symbol to be the triple scalar product 1a2a3a 1Θ2Θ Section 1.19 Solid Mechanics Part III Kelly 176ge g e1,3 3αβ βα αβ αβ βα αβ ε ε =×⋅≡ =×⋅≡ a aa a aa (1.19.3) where αβ αβεε= is the Cartesian permutation symbol, 112+=ε , 121−=ε , and zero otherwise, with ημβαα ηβ μβ ηα μ μηαβ μηαβ μηαβδδδδ εε ee ee ee =−= = , (1.19.4) From 1.19.3, 33 a a aa a a αββααββα ee =×=× (1.19.5) and so g2 1 3aaa×= (1.19.6) The cross product of surface vect ors, that is, vectors with component in the normal (3g) direction zero, can be written as 3 2 12 1 33 2 12 1 3 1a aa a vu vvuu gvuevvuug vue = == =× βααββα αβ (1.19.7) The Metric and Surface elements Considering a line element lying within the surface, so that 03=Θ , the metric for the surface is () ()()βα αβ ββ ααΘΘ=Θ⋅Θ=⋅=Δ ddg d d dd s a a ss2 (1.19.8) which is in this context known as the first fundamental form of the surface . Similarly, from 1.16.35, a surface element is given by 2 1ΔΘΔΘ=Δ g S (1.5.9) Christoffel Symbols The Christoffel symbols can be simplifie d as follows. A differentiation of 13 3=⋅aa leads to Section 1.19 Solid Mechanics Part III Kelly 1773 ,3 3 ,3 a a a a ⋅−=⋅α α (1.19.10) so that, from Eqn 1.18.6, 0 33 33 =Γ=Γα α (1.19.11) Further, since 0 /3 3=Θ∂∂a , 0 ,0333 33 =Γ=Γα (1.19.12) These last two equations imply that the ijkΓ vanish whenever two or more of the subscripts are 3. Next, differentiate 1.19.1 to get αβ βα a a a a ⋅−=⋅,3 3 , , α β βαa a a a ⋅−=⋅ ,3 3 , (1.19.13) and Eqns. 1.18.6 now lead to αβ βα βα αβ 3 3 3 3 Γ−=Γ−=Γ=Γ (1.19.14) From 1.18.8, using 1.19.11, 0 3333 333 33 3333 33 3 =Γ=Γ+Γ=ΓΓ=Γ+Γ=Γ α αβ αβ ααβ αβγ αβγ αβ g gg g (1.19.15) and, similarly { ▲Problem 1} 03 33 333 3 =Γ=Γ=Γα α (1.19.16) 1.19.2 The Curvature Tensor In this section is introduced a tensor which, with the metr ic coefficients, completely describes the surface. First, although the base vector 3a maintains unit length, its direction changes as a function of the coordinates 2 1,ΘΘ , and its derivative is, from 1.18.2 or 1.18.5 (and using 1.19.15) ββ α α αa aa 3 33Γ=Γ= Θ∂∂ kk, β αβ α αa aa3 33 Γ−=Γ−=Θ∂∂k k (1.19.17) Define now the curvature tensor K to have the c ovariant components αβK, through Section 1.19 Solid Mechanics Part III Kelly 178β αβ αaaK−=Θ∂∂3 (1.19.18) and it follows from 1.19.13, 1.19.15a and 1.19.14, βα αβ αβ αβ 3 33Γ−=Γ=Γ= K (1.19.19) and, since these Christoffel symbols are symmetric in the βα,, the curvature tensor is symmetric . The mixed and contravariant components of th e curvature tensor follows from 1.16.28-9: γγ α γγβ αββ αβ αγλβλαγ αβ γβ αγ αγγββ α a a aaK gK KKgg K Kg Kg K −= −=−≡Θ∂∂= == 3, (1.19.20) and the “dot” is not necessary in the mixed notation because of the symmetry property. From these and 1.18.8, it follows that α βα β βγγα γβγαα β 3 3 3 Γ−=Γ−=Γ−== g Kg K (1.19.21) Also, ()() () ( ) βα αβββ γα αγββ α α ΘΘ−=Θ⋅Θ−=Θ⋅Θ=⋅ ddKd dKd d d d a aa as a,3 3 (1.19.22) which is known as the second fundamental fo rm of the surface . From 1.19.19 and the definitions of the Christ offel symbols, 1.18.4, 1.18.6, the curvature can be expressed as αβ βα αβ aaaa⋅ Θ∂∂−=⋅ Θ∂∂=3 3 K (1.19.23) showing that the curvature is a measure of the change of the base vector αa along the βΘ curve, in the direction of the normal vector; alternatively, the rate of change of the normal vector along βΘ, in the direction αa−. Looking at this in more detail, consider now the change in the normal vector 3a in the 1Θ direction, Fig. 1.19.2. Then γγa a a1 11 1,3 3 Θ−=Θ= dK d d (1.19.24) Section 1.19 Solid Mechanics Part III Kelly 179 Figure 1.19.2: Curvature of the Surface Taking the case of 0 ,02 11 1=≠K K , one has 11 1 1 3 a a Θ−= dK d . From Fig. 1.19.2, and since the normal vector is of unit length, the magnitude 3ad equals φd, the small angle through which the normal vector ro tates as one travels along the 1Θ coordinate curve. The curvature of the surface is defined to be the rate of change of the angle φ:1 1 1 1111 1 1K ddK dsd= ΘΘ− = aa φ (1.19.25) and so the mixed component 1 1K is the curvature in the 1Θ direction. Similarly, 2 2K is the curvature in the 2Θ direction. Assume now that 0 ,02 11 1≠=K K . Eqn. 1.19.24 now reads 21 2 1 3 a a Θ−= dK d and, referring Fig. 1.19.3, the twist of the surface with respec t to the coordinates is 12 2 1 1121 2 1 aa aa K ddK dsd= ΘΘ− =ϕ (1.5.26) Figure 1.19.3: Twisting over the Surface When 2 1a a= , 2 1K is the twist; when they are not equal, 2 1K is closely related to the twist. 1 this is essentially the same definition as for the space curve of §1 .5.2; there, the angle sΔ=κφ 11aΘd3a 1Θ3 3a a d+3ad ϕd2Θ11aΘd3a 1Θ3 3a a d+3ad φd1a Section 1.19 Solid Mechanics Part III Kelly 180Two important quantities are often used to desc ribe the curvature of a surface. These are the first and the third principal scalar invariants: βα αβε2 12 11 22 21 1 2 22 11 21 12 21 1 det IIII KK KK KK K KK KKK K K i ji i =−= ==+== ⋅⋅ KK (1.19.27) The first invariant is twice the mean curvature MK whilst the third invariant is called the Gaussian curvature (or Total curvature ) GK of the surface. Example (Curvature of a Sphere) The surface of a sphere of radius a can be described by the coordinates ()2 1,ΘΘ , Fig. 1.19.4, where 1 3 2 1 2 2 1 1cos , sin sin , cos sin Θ=ΘΘ=ΘΘ= a x a x a x Figure 1.19.4: a spherical surface Then, from the definitions 1.16.3, 1.16.8-9, 1.16.13, { ▲Problem 2} 2 1 2 221 2122 1 12 1 231 22 1 12 1 1 sin11cos sin sin sinsin sin cos cos cos a aa ae e ae e e a Θ==ΘΘ+ΘΘ−=Θ−ΘΘ+ΘΘ+= aaa aa a a (1.19.28) 1 2 4 1 2 22 sin , sin 00Θ= Θ= ag aagαβ From 1.19.6, 31 22 1 12 1 3 cos sin sin cos sin e e e a Θ+ΘΘ+ΘΘ= (1.19.29) 2a 1x2x3x • 2Θa1Θ1Θ 2Θ2a 1a Section 1.19 Solid Mechanics Part III Kelly 181and this is clearly an orthogonal coordinate system with scale factors (see 1.16.15) 1 , sin ,31 2 1 =Θ== h a ha h (1.19.30) The surface Christoffel symbols are, from 1.18.33, 1.18.36, 1 1 1 22 1 21 2 212 122 222 111 211 121 11 cos sin ,sincos,0 ΘΘ−=ΓΘΘ=Γ=Γ=Γ=Γ=Γ=Γ=Γ (1.19.31) Using the definitions 1.18.4, { ▲Problem 3} 1 2 3 223 213 123 112 322 232 312 131 321 231 311 13 sin ,0 ,1,00 ,1 Θ−=Γ=Γ=Γ−=Γ=Γ=Γ=Γ=Γ=Γ=Γ=Γ=Γ a aaa (1.19.32) with the remaining symbols 03 33 333 33 3 =Γ=Γ=Γ=Γα α α . The components of the curvature tensor are then, from 1.19.21, 1.19.19, [] []⎥⎦⎤ ⎢⎣⎡ Θ−−= ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −− =1 2sin 00,1001 aaK aaKαβα β (1.19.33) The mean and Gaussian curvature of a sphere are then 212 aKaK GM =−= (1.19.34) The principal curvatures are evidently 1 1K and 2 2K. As expected, they are simply the reciprocal of the radius of curvature a. ■ 1.19.3 Covariant Derivatives Vectors Consider a vector v, which is not necessarily a surface vector, that is, it might have a normal component 3 3v v=. The covariant derivative is Section 1.19 Solid Mechanics Part III Kelly 182γ γαααγ γαααγα γααγα γα αα βγα γββα βα v vv v v vv vv v v vv v v v 3 ,33 3 33 ,3 333,3 33 33,33 3, ||| Γ+=Γ+Γ+=Γ+=Γ+Γ+=Γ+Γ+= , γγ αααγγ αααγγ α ααγγ α α ααβγγ αββαβα v vv v v vv vv v v vv v v v 3 ,333 3 3 ,3 33 3,33 3 3 3, 333 , ||| Γ−=Γ−Γ−=Γ−=Γ−Γ−=Γ−Γ−= (1.19.35) Define now a two-dimensional analogue of the three-dimensional covariant derivative through γγ αββαβαγα γββα βα v v vv v v Γ−=Γ+= ,, |||| (1.19.36) so that, using 1.19.19, 1.19.21, the cova riant derivative can be expressed as 33 || ||| | vK v vvK v v αββαβαα ββα βα −=−= (1.19.37) In the special case when the vector is a plane vector, then 03 3==v v , and there is no difference between the three-dimensional and two-dimensional covari ant derivatives. In the general case, the covariant derivatives can now be expressed as () ()3 3 3,33 3, | |||| ||| a aa va aa v βα αββαβ ββ αα ββαβ β v vK vvv vK vv i iii + −==+ −== (1.19.38) From 1.18.25, the gradient of a surface vector is (using 1.19.21) ()3 3 || grad a a a a v ⊗ +⊗ −=α γγ αβα αββα vK vK v (1.19.39) Tensors The covariant derivatives of second order te nsor components are given by 1.18.18. For example, 3 33 3,, | i j i j j i j i ijim j mmj i mij ij A A A A AA A A A γλ λγ γλ λγγγ γγγ Γ+Γ+Γ+Γ+=Γ+Γ+= (1.19.40) Here, only surface tensors will be examined, th at is, all components with an index 3 are zero. The two dimensional (p lane) covariant derivative is Section 1.19 Solid Mechanics Part III Kelly 183αλβ λγλβα λγγαβ γαβA A A A Γ+Γ+≡ , || (1.19.41) Although 03 3==α αA A for plane tensors, one still has non-zero λβ λγλβ λγλβ λγλβ λγγβ γβαλ λγαλ λγαλ λγλα λγγα γα AKAA A A AAKAA A A A =Γ=Γ+Γ+==Γ=Γ+Γ+= 33 3 ,3 333 3 ,3 3 || (1.19.42) with 0|33=γA . From 1.18.28, the divergence of a surface tensor is 3 || div a a Aβγ βγαβαβAK A + = (1.19.43) 1.19.4 The Gauss-Codazzi Equations Some useful equations can be derived by cons idering the second deri vatives of the base vectors. First, from 1.18.2, 333 , a aa a a αβλλ αβαβλλ αββα K+Γ=Γ+Γ= (1.19.44) A second derivative is γαβ γαβγλλ αβλλ γαββγα ,3 3 , , , , a a a a a K K++Γ+Γ= (1.19.45) Eliminating the base vectors derivati ves using 1.19.44 and 1.19.20b leads to { ▲Problem 4} ( )( )3 , , , a a aγαβ λγλ αβλλ γαβλ ηγη αβλ γαβ βγα K K KK +Γ+ −ΓΓ+Γ= (1.19.46) This equals the partial derivative γβα,a . Comparison of the coefficient of 3a for these alternative expressions for the se cond partial derivative leads to λγλ αββαγ λβλ αγγαβ K K K K Γ−=Γ−, , (1.19.47) From Eqn. 1.18.18, Section 1.19 Solid Mechanics Part III Kelly 184αλλ βγλβλ αγγαβγαβ K K K K Γ−Γ−=, || (1.19.48) and so βαγγαβ || ||K K= (1.19.49) These are the Codazzi equations , in which there are only two independent non-trivial relations: 2 12 1 22 1 12 2 11 || || ,|| || K K K K = = (1.19.50) Raising indices using the metric coefficients leads to the similar equations βα γγα β || ||K K= (1.19.51) The Riemann-Christoffel Curvature Tensor Comparing the coefficients of λa in 1.19.46 and the similar expression for the second partial derivative shows that λ γαβλ βαγλ ηγη αβλ ηβη αγλ γαβλ βαγ KK KK−=ΓΓ−ΓΓ+Γ−Γ, , (1.19.52) The terms on the left are the two-dimensional Riemann-Christoffel, Eqn. 1.18.21, and so λ γαβλ βαγλ αβγ KK KK R −=⋅ (1.19.53) Further, γλαβ βλαγη γληαβη βληαγη αβγλη λαβγ KK KK KgK KgK Rg R − = − = =⋅ (1.19.54) These are the Gauss equations . From 1.18.21 et seq. , only 4 of the Riemann-Christoffel symbols are non-zero, and th ey are related through 2121 1221 2112 1212 R R R R =−=−= (1.19.55) so that there is in fact only one indepe ndent non-trivial Gauss relation. Further, () ()γνβρν ηρ μρ ην μη λμ αγηβμ βηγμη λμ αγλαβ βλαγ λαβγ δδδδ gg KKgg ggKKKK KK R − =− =− = (1.19.56) Using 1.19.4b, 1.19.3, Section 1.19 Solid Mechanics Part III Kelly 185η λμ αημβγημβγη λμ αγνβρημρνη λμ α λαβγ εε KK geeKKggeeKK R === (1.19.57) and so the Gauss relation can be expressed succinctly as gRKG1212= (1.19.58) where GK is the Gaussian curvature, 1.19.27b. Thus the Riemann-Christoffel tensor is zero if and only if the Gaussian curvature is zero, and in this case only can the order of the two covariant differen tiations be interchanged. The Gauss-Codazzi equations, 1.19.50 and 1.19.58, are equivalent to a set of two first order and one second order differential equati ons that must be satisfied by the three independent metric coefficients αβg and the three independent curvature tensor coefficients αβK. Intrinsic Surface Properties An intrinsic property of a surface is any quantit y that remains unchanged when the surface is bent into another shape without st retching or shrinking. Some examples of intrinsic properties are the length of a curv e on the surface, surface area, the components of the surface metric tensor αβg (and hence the components of the Riemann-Christoffel tensor) and the Guassian curvature (which follows from the Gauss equation 1.19.58). A developable surface is one which can be obtained by bending a plane, for example a piece of paper. Examples of developable su rfaces are the cylindrical surface and the surface of a cone. Since the Riemann-Chri stoffel tensor and hence the Gaussian curvature vanish for the plane, they vanish for all developable surfaces. 1.19.5 Geodesics The Geodesic Curvature and Normal Curvature Consider a curve C lying on the surface, with arc length s measured from some fixed point. As for the space curve, §1.6.2, one can define the unit tangent vector τ, principal normal ν and binormal vector b (Eqn. 1.5.3 et seq. ): αα axτdsd dsdΘ== , dsdτνκ1= , ντb×= (1.19.59) so that the curve passes along the intersec tion of the osculating plane containing τand ν (see Fig. 1.6.3), and the surface These vect ors form an orthonormal set but, although ν is normal to the tangent, it is not necessarily nor mal to the surface, as illustrated in Fig. Section 1.19 Solid Mechanics Part III Kelly 1861.19.5. For this reason, form the new orthonormal triad ()3 2,,aττ , so that the unit vector 2τ lies in the plane tangent to the surface. From 1.19.59, 1.19.3, βα αβαα a aaτaτdsdedsd Θ=×Θ=×=3 3 2 (1.19.60) Figure 1.19.5: a curve lying on a surface Next, the vector dsd/τ will be decomposed into components along 2τ and the normal 3a. First, differentiate 1.19.59a and use 1.19.44b to get { ▲Problem 5} 3 22 a aτ dsd dsdKdsd dsd dsd dsdβα αβγβα γ αβγΘΘ+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ΘΘΓ+Θ= (1.19.61) Then 3 2 aττ n gdsdκκ+= (1.19.62) where dsd dsdKdsd dsd dsd dsde ng βα αββα γ αβγ λ λγ κκ ΘΘ=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ΘΘΓ+ΘΘ=22 (1.19.63) These are formulae for the geodesic curvature gκ and the normal curvature nκ. Many different curves with representations ) (sαΘ can pass through a certain point with a given tangent vector τ. Form 1.19.59, these will all have the same value of ds d /αΘ and so, from 1.19.63, these curves will have the same normal curvature but, in general, different geodesic curvatures. ν C3a 1x2x3x τ x1Θ2Θ Section 1.19 Solid Mechanics Part III Kelly 187A curve passing through a normal section , that is, along the in tersection of a plane containing τand 3a, and the surface, will have zero geodesic curvature. The normal curvature can be expressed as τKτ=nκ (1.19.64) If the tangent is al ong an eigenvector of K, then nκ is an eigenvalue, and hence a maximum or minimum normal curvature. Surface curves with the property that an eigenvector of the curvature tensor is ta ngent to it at every point is called a line of curvature . A convenient coordinate system for a surface is one in which the coordinate curves are lines of curvature. Such a system, with 1Θ containing the maximum values of nκ, has at every point a curvature tensor of the form []() ()⎥⎦⎤ ⎢⎣⎡=⎥ ⎦⎤ ⎢ ⎣⎡= minmax 2 21 1 00 00 nn j iKKKκκ (1.19.65) This was the case with the spherical surface example discussed in §1.19.2. The Geodesic A geodesic is defined to be a curve wh ich has zero geodesic curvature at every point along the curve. Form 1.19.63, parametric e quations for the geodesics over a surface are 022 =ΘΘΓ+Θ dsd dsd dsdβα γ αβγ (1.19.64) It can be proved that the geodesic is the cu rve of shortest distan ce joining two points on the surface. Thus the geodesic curvature is a measure of the deviance of the curve from the shortest-path curve. The Geodesic Coordinate System If the Gaussian curvature of a surface is not ze ro, then it is not po ssible to find a surface coordinate system for which the metric tensor components αβg equal the Kronecker delta αβδ everywhere. Such a geometry is called Riemannian . However, it is always possible to construct a coordinate system in which αβ αβδ=g , and the derivatives of the metric coefficients are zero, at a particular point on the surface. This is the geodesic coordinate system . 1.19.6 Problems 1 Derive Eqns. 1.19.16, 03 33 333 3 =Γ=Γ=Γα α . 2 Derive the Cartesian components of the cu rvilinear base vectors for the spherical surface, Eqn. 1.19.28. Section 1.19 Solid Mechanics Part III Kelly 1883 Derive the Christoffel symbols for the spherical surface, Eqn. 1.19.32. 4 Use Eqns. 1.19.44-5 and 1.19.20b to derive 1.19.46. 5 Use Eqns. 1.19.59a and 1.19.44b to derive 1.19.61. Section 1.A Solid Mechanics Part III Kelly 1891.A Appendix to Chapter 1 1.A.1 The Algebraic Structures of Groups, Fields and Rings Definition: The nonempty set G with a binary operation, that is, to each pair of elements Gba∈, there is assigned an element G ab∈, is called a group if the following axioms hold: 1. associative law : )( )( bcacab= for any Gcba∈,, 2. identity element : there exists an element Ge∈, called the identity element, such that a ea ae== 3. inverse : for each Ga∈, there exists an element G a∈−1, called the inverse of a, such that eaa aa ==−− 1 1 Examples : (a) An example of a group is the set of integers under addition. In this case the binary operation is denoted by +, as in ba+; one has (1) addition is associative, cba++) ( equals ) (cba++ , (2) the identity element is denoted by 0, aa a =+=+ 00 , (3) the inverse of a is denoted by -a, called the negative of a, and 0 )()( =+−=−+ aa a a Definition: An abelian group is one for which the commutative law holds, that is, if ba ab= for every Gba∈, . Examples : (a) The above group, the set of integers under addition, is commutative, abba+=+ , and so is an abelian group. Definition: A mapping f of a group G to another group G′, G Gf′→: , is called a homomorphism if ) ()( )( bfaf abf= for every Gba∈,; i f f is bijective (one-one and onto), then it is called an isomorphism and G and G′ re said to be isomorphic Definition: If G Gf′→: is a homomorphism, then the kernel of f is the set of elements of G which map into the identity element of G′, { }e afGa k ′= ∈= )(| Examples (a) Let G be the group of non-zero complex numbers under multiplication, and let G′ be the non-zero real numbers unde r multiplication. The mapping G Gf′→: defined by z zf=)( is a homomorphism, because )()( ) (2 1 2 1 21 21 zfzf zz zz zzf === The kernel of f is the set of elements which map into 1, that is, the complex numbers on the unit circle Section 1.A Solid Mechanics Part III Kelly 190 Definition: The non-empty set A with the two binary operations of addition (denoted by +) and multiplication (denoted by juxtaposition) is called a ring if the following are satisfied: 1. associative law for addition : for any Acba∈,,, ) ( ) ( cbacba ++=++ 2. zero element (additive identity): there exists an element A∈0 , called the zero element, such that aa a =+=+ 00 for every Aa∈ 3. negative (additive inverse): for each Aa∈ there exists an element Aa∈− , called the negative of a, such that 0 )()( =+−=−+ aa a a 4. commutative law for addition : for any Aba∈,, abba+=+ 5. associative law for multiplication : for any Acba∈,,, ) ( )( bcacab= 6. distributive law of multiplication over addition (both left and right distributive): for any Acba∈,, , (i) ac abcba +=+) (, ( i i ) ca baacb +=+) ( Remarks : (i) the axioms 1-4 may be summarized by saying that A is an abelian group under addition (ii) the operation of subtraction in a ring is defined through ) (b aba −+≡− (iii) using these axioms, it can be shown that 0 00==a a , ab ba ba −=−=− )()(, ab ba=−− ))(( for all Aba∈, Definition: A commutative ring is a ring with the additional property: 7. commutative law for multiplication : for any Aba∈,, ba ab= Definition: A ring with a unit element is a ring with the additional property: 8. unit element (multiplicative identity): there exists a nonzero element A∈1 such that aa a==11 for every Aa∈ Definition: A commutative ring with a unit element is an integral domain if it has no zero divisors, that is, if 0=ab , then 0=a or 0=b Examples : (a) the set of integers Z is an integral domain Definition: A commutative ring with a unit element is a field if it has the additional property: 9. multiplicative inverse : there exists an element A a∈−1 such that 11 1==−−aa aa Remarks : (i) note that the number 0 has no multiplicative inverse. When constructing the real numbers R, 0 is a special element which is not allowed have a multiplicative inverse. For this reason, division by 0 in R is indeterminate Section 1.A Solid Mechanics Part III Kelly 191Examples : (a) The set of real numbers R with the usual operations of addition and multiplication forms a field (b) The set of ordered pairs of real numbers with addition and multiplication defined by ) , (),)(,() , (),(),( bc adbdac dcbadbca dc ba +−=++=+ is also a field - this is just the set of complex numbers C 1.A.2 The Linear (Vector) Space Definition: Let F be a given field whose elements are called scalars . Let V be a non-empty set with rules of addition and scalar multiplication, that is there is a sum ba+ for any Vba∈, and a product aα for any Va∈, F∈α . Then V is called a linear space over F if the following eight axioms hold: 1. associative law for addition : for any Vcba∈,, , one has ) ( ) ( cbacba ++=++ 2. zero element : there exists an element V∈0 , called the zero element, or origin, such that aa a =+=+ 00 for every Va∈ 3. negative : for each Va∈ there exists an element Va∈− , called the negative of a, such that 0 )()( =+−=−+ aa a a 4. commutative law for addition : for any Vba∈, , we have abba+=+ 5. distributive law, over addition of elements of V : for any Vba∈, and scalar F∈α , b a ba αα α +=+) ( 6. distributive law, over addition of scalars : for any Va∈ and scalars F∈βα, , a a aβαβα +=+) ( 7. associative law for multiplication : for any Va∈ and scalars F∈βα,, a a )()(αββα= 8. unit multiplication : for the unit scalar F∈1, aa=1 for any Va∈. Section 1.B Solid Mechanics Part III Kelly 1921.B Appendix to Chapter 1 1.B.1 The Ordinary Calculus Here are listed some important conc epts from the ordinary calculus. The Derivative Consider u, a function f of one independent variable x. The derivative of u at x is defined by ()( ) xxf x xf xuxfdxdu x xΔ−Δ+=ΔΔ=′≡→Δ →Δ 0 0 lim lim)( (1.B.1) where uΔ is the increment in u due to an increment xΔ in x. The Differential The differential of u is defined by xxf du Δ′= )( (1.B.2) By considering the special case of x xfu== )( , one has x dx du Δ== , so the differential of the independent variable is equivalent to the increment. x dxΔ= . Thus, in general, the differential can be written as dxxf du )(′= . The differential of u and increment in u are only approximately equal, u duΔ≈ , and approach one another as 0→Δx . This is illustrated in Fig. 1.B.1. Figure 1.B.1: the differential If x is itself a function of another variable, t say, ()()txu , then the chain rule of differentiation gives differentialslope x dxΔ+x dxΔ= x)( ) ( xf x xfu −Δ+=Δ)(xf′duu increment Section 1.B Solid Mechanics Part III Kelly 193dtdxxfdtdu)(′= (1.B.3) Arc Length The length of an arc, measured from a fixed point a on the arc, to x, is, from the definition of the integral, () dxdxdy dx ds sx ax ax a∫∫∫+= ==2/ 1 secψ (1.B.4) where ψ is the angle the tangent to the arc makes with the x axis, Fig 1.B.2, with ψtan)/(=dxdy and () () ()2 2 2dy dx ds += (ds is the length of the dotted line in Fig. 1.B.2b). Also, it can be seen that () () 1limlim lim 2 22 2 02 2 0 arcchord 0 =⎟ ⎠⎞⎜ ⎝⎛+⎟ ⎠⎞⎜ ⎝⎛=⎟ ⎠⎞⎜ ⎝⎛ ΔΔ+⎟ ⎠⎞⎜ ⎝⎛ ΔΔ=ΔΔ+Δ= →Δ→Δ →Δ dsdy dsdxsy sxsy x pqpq ss s (1.B.5) so that, if the increment sΔ is small, ()()()2 2 2y x s Δ+Δ≈Δ . Figure 1.B.2: arc length The Calculus of Two or More Variables Consider now two independent variables, ) ,(yxfu= . We can define partial derivatives so that, for example, xψ as (a) (b) sΔ ds x dxΔ=dyyΔ pq Section 1.B Solid Mechanics Part III Kelly 194xyxf yx xf xu xu x yxΔ−Δ+=ΔΔ=∂∂ →Δ →Δ),( ), (lim lim0 constant 0 (1.B.6) The total differential du due to increments in both x and y can in this case be shown to be yyuxxudu Δ∂∂+Δ∂∂= (1.B.7) which is written as () ()dyyu dxxu du ∂∂+∂∂= / / , by setting y dyx dx Δ=Δ=, . Again, the differential du is only an approximation to the actual increment uΔ (the increment and differential are shown in Fig. 1.B.3 for the case 0 =Δ=y dy ). It can be shown that this expression for the differential du holds whether x and y are independent, or whether they are functions themselves of an independent variable t, () )(),( tytxuu≡ , in which case one has the total derivative of u with respect to t, dtdy yu dtdx xu dtdu ∂∂+∂∂= (1.B.8) Figure 1.B.3: the partial derivative The Chain rule for Two or More Variables Consider the case where u is a function of the two variables yx,, ),(yxfu= , but also that x and y are functions of the tw o independent variables s and t, ()()() tsytsxfu ,,,= . Then xyu • x dxΔ=differential uΔxu ∂∂du incrementslope Section 1.B Solid Mechanics Part III Kelly 195dtty yu tx xudssy yu sx xudttydssy yudttxdssx xudyyudxxudu ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂ ∂∂+∂∂ ∂∂+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂ ∂∂+∂∂ ∂∂=⎟ ⎠⎞⎜ ⎝⎛ ∂∂+∂∂ ∂∂+⎟ ⎠⎞⎜ ⎝⎛ ∂∂+∂∂ ∂∂=∂∂+∂∂= (1.B.9) But, also, dttudssudu∂∂+∂∂= (1.B.10) Comparing the two, and since dtds, are independent and arbitr ary, one obtains the chain rule ty yu tx xu tusy yu sx xu su ∂∂ ∂∂+∂∂ ∂∂=∂∂∂∂ ∂∂+∂∂ ∂∂=∂∂ (1.B.11) In the special case when x and y are functions of only one variable, t say, so that [] )(),( tytxfu= , the above reduces to the to tal derivative given earlier. One can further specialise: In the case when u is a function of x and t, with ) (txx= , [] ttxfu ),(= , one has tu dtdx xu dtdu ∂∂+∂∂= (1.B.12) When u is a function of one variable only, x say, so that [])(txfu= , the above reduces to the chain rule for ordinary differentiation. Taylor’s Theorem Suppose the value of a function ) ,(yxf is known at ) ,(0 0yx . Its value at a neighbouring point ) , (0 0 y yx x Δ+Δ+ is then given by () () () () ()() ()L+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂Δ+∂∂∂ΔΔ+∂∂Δ+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂Δ+∂∂Δ+ =Δ+Δ+ 0 0 0 0 0 00 0 0 0 ,22 2 ,2 ,22 2, ,0 0 0 0 21),( ) , ( yx yx yxyx yx yfxyxfyxxfxyfyxfx yxf y yx xf (1.B.13) Section 1.B Solid Mechanics Part III Kelly 196The Mean Value Theorem If )(xf is continuous over an interval bxa<< , then abaf bff−−=′)( )()(ξ (1.B.14) Geometrically, this is equivalent to saying that there exists at least one point in the interval for which the tangent line is parallel to the line joining ) (af and ) (bf. This result is known as the mean value theorem . Figure 1.B.4: the mean value theorem The law of the mean can also be written in term s of an integral: there is at least one point ξ in the interval []ba, such that dxxflfb a∫= )(1)(ξ (1.B.15) where l is the length of the interval, abl−= . The right hand side here can be interpreted as the average value of f over the interval. The theorem th erefore states that the average value of the function lies somewhere in the interval. The equiva lent expression for a double integral is that ther e is at least one point ()2 1,ξξ in a region R such that 2 1 2 1 2 1 ),(1),( dxdxxxfAf R∫∫=ξξ (1.B.16) where A is the area of the region of integration R, and similarly for a triple/volume integral. a bξ)(af)(bf )(ξf Section 1.B Solid Mechanics Part III Kelly 1971.B.2 Transformation of Coordinate System Let the coordinates of a point in space be ()3 2 1,,xxx . Introduce a second set of coordinates ()3 2 1 ,,ΘΘΘ , related to the firs t set through the transformation equations ()3 2 1,,xxxfi i=Θ ( 1 . B . 1 7 ) with the inverse equations ()3 2 1 ,,ΘΘΘ=i ig x ( 1 . B . 1 8 ) A transformation is termed an admissible transformation if the inverse transformation exists and is in one-to-one corresponden ce in a certain region of the variables ()3 2 1,,xxx , that is, each set of numbers ()3 2 1 ,,ΘΘΘ defines a unique set ()3 2 1,,xxx in the region, and vice versa . Now suppose that one has a point with coordinates 0 ix, 0 iΘ which satisfy 1.B.17. Eqn. 1.B.17 will be in general non-linear, but differentiating leads to j ji i dxxfd∂∂=Θ , (1.B.19) which is a system of three linear equations. From basic linear algebra, this system can be solved for the jdx if and only if the determinant of the coefficients does not vanish, i.e 0 det≠ ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ∂∂= ji xfJ , (1.B.20) with the partial derivatives evaluated at 0 ix (the one dimensional situation is shown in Fig. 1.B.5). If 0≠J , one can solve for the idx: j ij i dA dxΘ= , (1.B.21) say. This is a linear approximation of the inverse equations 1.B.18 and so the inverse exists in a small region near ()0 30 20 1 ,,xxx . This argument can be extended to other neighbouring points and the region in which 0≠J will be the region for which the transformation will be admissible. f the Jacobian is positive everywhere, then a right handed set will be transformed into another right handed set, and the transformation is said to be proper . Section 1.B Solid Mechanics Part III Kelly 198 Figure 1.B.5: linear approximation x 0xdx0Θlinear approximation 1992 Kinematics Kinematics is concerned with characterisi ng movement. The goal is to express in mathematical form the deformation and motion of materials. In what follows, a number of important quantities, mainly vectors and second-or der tensors, are introd uced. Each of these quantities, for example the velocity, deformatio n gradient or rate of deformation tensor, allows one to describe a particular aspect of a deforming material. No consideration is given to what is causing the deformation and move ment – the cause is the action of forces on the material, and these will be discussed in the next chapter. The first section introduces the material and sp atial coordinates and descriptions. The second and third sections discuss the strain tensors. The fourth, fifth and sixt h sections deal with rates of deformation and rates of change of kinematic quantities. The theory is specialised to small strain deformations in section 7. The no tion of objectivity is discussed in section 8 and the final sections, 9-13, deal with kinematics using the convected coordinate system, and include the important notions of push-forw ard, pull-back and the Lie time derivative. 200 Section 2.1 Solid Mechanics Part III Kelly 2012.1 Motion 2.1.1 The Material Body and Motion Physical materials in the real world are mode led using an abstract mathematical entity called a body . This body consists of an infinite number of material particles 1. Shown in Fig. 2.1.1a is a body B with material particle X. One distinguishes between this body and the space in which it resides and through which it travels. Shown in Fig. 2.1.1b is a certain place x in familiar Euclidean space E. Figure 2.1.1: (a) a material partic le in a body, (b) a place in space, (c) a configuration of the body By fixing the material particles of the body to places in space, one has a configuration of the body χ, Fig. 2.1.1c. A configuration can be ex pressed as a mapping of the particles X to the places x, ()Xχx= (2.1.1) A motion of the body is a family of configurations parameterised by time t, ()t,Xχx= (2.1.2) At any time t, Eqn. 2.1.2 gives the location in space x of the material particle X, Fig. 2.1.2. There are a number of different ways in wh ich the motion can be described. Eqn. 2.1.2 is the material description , in which the motion is described in terms of the material particles which make up the body. 1 these particles are not the discrete mass particles of Newtonian mechanics, rather they are very small portions of continuous matter; the meaning of partic le is made precise in the definitions which follow BX• • x (a) (b)E (c)χ • Section 2.1 Solid Mechanics Part III Kelly 202 Figure 2.1.2: a motion of material The Reference and Current Configurations Choose now some reference configuration , Fig. 2.1.3. The motion can then be measured relative to this configuration. The reference configuration might be the configuration occupied by the material at time 0=t , in which case it is often called the initial configuration . For a solid, it might be natural to choose a configuration for which the material is stress-free, in which case it is often called the undeformed configuration . However, the choice of reference conf iguration is completely arbitrary. Introduce a Cartesian coordinate system with base vectors iE for the reference configuration. A materi al particle (point) X in the reference confi guration can then be assigned a unique position vector i iXE X= relative to the origin of the axes. The coordinates ()3 2 1 ,, XXX of the particle are called material coordinates (or Lagrangian coordinates or referential coordinates ). Some time later, say at time t, the material occupies a diffe rent configuration, which will be called the current configuration (or deformed configuration ). Introduce a second Cartesian coordinate syst em with base vectors ie for the current configuration, Fig. 2.1.3. In the current configuration, the same particle X now occupies the location (point) x, which can now also be assigned a position vector iixe x= . The coordinates ()3 2 1,, xxx are called spatial coordinates (or Eulerian coordinates ). Each particle thus has two sets of coordinates associated with it. The particle’s material coordinates stay with it thr oughout its motion. The particle’s spatial coordinates change as it moves. 1t2t X• X• Section 2.1 Solid Mechanics Part III Kelly 203 Figure 2.1.3: reference and current configurations In practice, the material and spatial axes are us ually taken to be coincident so that the base vectors iE and ie are the same, as in Fig. 2.1.4. Nevert heless, the use of different base vectors E and e for the reference and current confi gurations is useful even when the material and spatial axes are coincident, since it helps distingu ish between quantities associated with the reference configurati on and those associated with the spatial configuration (see later). Figure 2.1.4: reference and current config urations with coincident axes In terms of the position vectors, the mo tion 2.1.2 can be expressed as a relationship between the material and spatial coordinates, () tXXX x ti i ,,, ),,(3 2 1χ= =Xχx Material description (2.1.3) or the inverse relation ()txxx X ti i ,,, ),,(3 2 11 1 − −= = χ xχX Spatial description (2.1.4) If one knows the material coordinates of a particle then its position in the current configuration can be determined from 2.1.3. Alternatively, if one focuses on some location in space, in the current configuration, then the material particle occupying that position can be determined from 2.1.4. This is illustrated in the following example. 1 1,xX2 2,xX 3 3,xXX 2 2,eEx 1 1,eEcurrent configuration 1X2X 3XX 1E2Ereference configuration 1e2e2x 3xxt •• X X Section 2.1 Solid Mechanics Part III Kelly 204Example (Extension of a Bar) Consider the motion 3 3 2 2 1 1 1 , , 3 X x X xt XtX x = =++= (2.1.5) These equations are of the form 2.1.3 and say that “the particle that was originally at position X is now, at time t, at position x”. They represent a simple translation and uniaxial extension of material as sh own in Fig. 2.1.5. Note that xX= at 0=t . Figure 2.1.5: translation and extension of material Relations of the form 2.1.4 can be obtained by inverting 2.1.5: 3 3 2 21 1 , ,31x X x Xtt xX = =+−= These equations say that “the pa rticle that is now, at time t, at position x was originally at position X”. ■ Convected Coordinates The material and spatial coordinate systems us ed here are fixed Cartesian systems. An alternative method of de scribing a motion is to attach the material coordinate system to the material and let it deform with the materi al. The motion is then described by defining how this coordinate system changes. This is the convected coordinate system . In general, the axes of a convected system will not remain mutually orthogonal and a curvilinear system is required. Convected coordinates will be examined in §2.10. 2.1.2 The Material and Spatial Descriptions Any physical property (such as density, temper ature, etc.) or kinematic property (such as displacement or veloci ty) of a body can be described in terms of either the material coordinates X or the spatial coordinates x, since they can be transformed into each other using 2.1.3-4. A material (or Lagrangian ) description of events is one where the 1x2x 1X2X configuration at 0=tconfigurations at 0>t •X • xχ Section 2.1 Solid Mechanics Part III Kelly 205material coordinates are the independent variables. A spatial (or Eulerian ) description of events is one where the spatial coordinates are used. Example (Temperature of a Body) Suppose the temperature θ of a body is, in material coordinates, 3 13),( X X t−= Xθ (2.1.6) but, in the spatial description, 311 ),( xtxt −−=xθ . (2.1.7) According to the material desc ription 2.1.6, the temperature is different for different particles, but the temperature of each particle remains cons tant over time. The spatial description 2.1.7 describes the time-dependent temperature at a specific location in space, x, Fig. 2.1.6. Different material particle s are flowing through this location over time. Figure 2.1.6: particles flowing through space ■ In the material description, then, attention is focused on specific material . The piece of matter under consideration may change shape, density, velocity, a nd so on, but it is always the same piece of material. On the othe r hand, in the spatial description, attention is focused on a fixed location in space . Material may pass thr ough this location during the motion, so different material is under consideration at different times. The spatial description is the one most ofte n used in Fluid Mechanics since there is no natural reference configuration of the material as it is cont inuously moving. However, both the material and spatial descriptions are used in Solid Mechanics, where the reference configuration is usually the stress-free configuration. 2.1.3 Small Perturbations A large number of important problems invol ve materials which deform only by a relatively small amount. An example would be the steel structural columns in a building under modest loading. In this type of problem there is virtually no distinction to be made xmotion of individual material particles Section 2.1 Solid Mechanics Part III Kelly 206between the two viewpoints take n above and the analysis is simplified greatly (see later, on Small Strain Theory, §2.7). 2.1.4 Problems 1. The density of a material is given by 2 13 X X+=ρ and the motion is given by the equations t x Xt x Xx X −=−==3 3 2 2 1 1 , ,. (a) what kind of description is this for the density, and what kind of description is this for the motion? (b) re-write the density in terms of x – what is the name give n to this description of the density? (c) is the density of any given mate rial particle changing with time? (d) invert the motion equations so that X is the independent variable – what is the name given to this description of the motion? (e) draw the line element joining the origin to )0,1,1( and sketch the position of this element of material at times 1=t and 2=t . Section 2.2 Solid Mechanics Part III Kelly 2072.2 Deformation and Strain A number of useful ways of describing the de formation of a material are discussed in this section. Attention is restricted to th e reference and current configurations. No consideration is given to the particular sequence by which the current configuration is reached from the reference configuration and so the deformation can be considered to be independent of time. In what follows, particles in the reference configuration will often be termed “undeformed” and those in the current configuration “deformed”. In a change from Chapter 1, lower case letters will now be reserved for both vector- and tensor- functions of the spatial coordinates x, whereas upper-case letters will be reserved for functions of material coordinates X. There will be exceptions to this, but it should be clear from the context what is implied. 2.2.1 The Deformation Gradient The deformation gradient F is the fundamental measure of deformation in continuum mechanics. It is the second order tensor which maps line elements in the reference configuration into line elements (consisting of the same material particles) in the current configuration. Consider a line element Xd emanating from position X in the reference configuration which becomes xd in the current configuration, Fig. 2.2.1. Then, using 2.1.3, ()() () XχXχ X Xχx dd d Grad=−+= (2.2.1) A capital G is used on “Grad” to emphasise that this is a gradient with respect to the material coordinates1, the material gradient , Xχ∂∂/. Figure 2.2.1: the Deformation Gradient acting on a line element 1 one can have material gradients and spatial gradients of material or spatial fields – see later X xF Xdxd Section 2.2 Solid Mechanics Part III Kelly 208It is customary to denote the motion vector-function χ in 2.1.3 simply by x, i.e. ()t,Xxx= , so that Ji iJXxF∂∂= =∂∂= , Grad xXxF Deformation Gradient (2.2.2) with J iJ i dXF dx d d = = ,XFx action of F (2.2.3) Lower case indices are used in the index notat ion to denote quantities associated with the spatial basis {}ie whereas upper case indices are used for quantities associated with the material basis {}IE. Note that XXxx d d∂∂= is a differential quantity and this expression has some error associated with it; the error (due to terms of order 2)(Xd and higher, neglected from a Taylor series) tends to zero as the differential 0→Xd . The deformation gradient (whose components are finite) thus characterises the deformation in the neighbourhood of a point X, mapping infinitesimal line elements Xd emanating from X in the reference configuration to the infinitesimal line elements xd emanating from x in the current configuration, Fig. 2.2.2. Figure 2.2.2: deformation of a material particle Example Consider the cube of material with sides of unit length illustrated by dotted lines in Fig. 2.2.3. It is deformed into the rectangular prism illustrated (this could be achieved, for example, by a continuous rotation and stretching motion). The material and spatial coordinate axes are coincident. The material description of the deformation is 33 21 1231 216 )( e e e Xfx X X X ++−== and the spatial description is before after Section 2.2 Solid Mechanics Part III Kelly 209 33 21 1213612)( E E E xfX x x x +−==− Figure 2.2.3: a deforming cube Then ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡− =∂∂= 3/10 00 02/106 0 Ji XxF Once F is known, the position of any element can be determined with ease. For example, taking a line element T]0,0,[da d=X , T]0,2/,0[da d d == XFx . ■ Homogeneous Deformations A homogeneous deformation is one where the deformation gradient is uniform, i.e. independent of the coordinates, a nd the associated motion is termed affine . Every part of the material deforms as the whole does, and straight parallel lines in the reference configuration map to straight parallel lines in the current configuration, as in the above example. Most examples to be considered in what follows will be of homogeneous deformations; this keeps the algebra to a minimum, but homogeneous deformation analysis is very useful in itself since most of the basic experimental testing of materials, e.g. the uniaxial tensile test, involve homogeneous deformations. Rigid Body Rotations and Translations One can add a constant vector c to the motion, cxx+= , without changing the deformation, () x cx Grad Grad =+ . Thus the deformation gradient does not take into account rigid-body translations of bodies in space. If a body only translates as a rigid body in space, then IF=, and cXx+= . If there is no motion, then not only is IF=, but Xx=. 1 1,xX2 2,xX 3 3,xXD AB CE E′D′ B′ C′ Section 2.2 Solid Mechanics Part III Kelly 210If the body rotates as a rigid body (with no translation), then RF=, a rotation tensor (§1.10.8). The Inverse of the Deformation Gradient The inverse deformation gradient 1−F carries the spatial line element d x to the material line element d X. It is defined as jI jIxXF∂∂= =∂∂=− − 1 1, grad XxXF Inverse Deformation Gradient (2.2.4) so that j Ij I dxF dX d d1 1,− −= = x FX action of 1−F (2.2.5) with (see Eqn. 1.15.2) I FFFF ==− − 1 1 ij jM iMFFδ=−1 (2.2.6) Cartesian Base Vectors Explicitly, in terms of the material and spatial base vectors (see 1.14.3), j I jI j jJ i Ji J J xX xXx X e E eXFE e ExF ⊗∂∂=⊗∂∂=⊗∂∂=⊗∂∂= −1 (2.2.7) so that () ()() x e E E e XF d dXXx dX Xx diJ J i M M J i J i = ∂∂= ⊗∂∂= / / and X x F d d=−1. Because F and 1−F act on vectors in one configuration to produce vectors in the other configuration, they are termed two-point tensors . They are defined in both configurations. This is highlighted by their having both reference and current base vectors E and e in their Cartesian representation 2.2.7. Here follow some important relations whic h relate scalar-, vector- and second-order tensor-valued functions in the mate rial and spatial descriptions { ▲Problem 1}. T11 : Grad divGrad gradGrad grad −−− === FA aFV vFφφ (2.2.8) Here, φ is a scalar; V and v are the same vector, the former being a function of the material coordinates, the material descript ion, the latter a function of the spatial Section 2.2 Solid Mechanics Part III Kelly 211coordinates, the spatial description. Similarly, A is a second order tensor in the material form and a is the equivalent spatial form. Example Consider the deformation () ()( ) () ( ) ( )3 2 1 2 2 1 3 2 13 3 2 1 2 2 1 3 2 2 53 2 E E E Xe e e x x x x x x xX X X X X X −−+−+++=+++−+−= so that ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −−−= ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −− =− 02 101 015 1 , 1 3101 01 20 1F F Consider the vector () ()()3 3 1 2 32 2 1 2 1 3 2)( e e e xv xx x x xx +++−+−= which, in the material description, is () ( )()3 2 1 22 2 3 2 1 1 3 2 5 3 3 2 5)( E E E XV X X X X X X X X ++ −+++−= The material and spatial gradients are ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −− = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −− = 1 0 11 6 00 1 2 grad , 0 5 11 6312 5 0 Grad2 2 x X v V and it can be seen that v FV grad 1 0 11 6 00 1 2 1 0 11 6001 2 Grad2 21= ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −− = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡− =−x X ■ 2.2.2 The Cauchy-Green Strain Tensors The deformation gradient describes how a line element in the reference configuration maps into a line element in the current c onfiguration. Other useful measures of deformation are the Left Cauchy-Green Strain and Right Cauchy-Green Strain tensors. They give a measure of how the lengths of line elements and angles between line elements (through the vector dot pro duct) change between configurations. Section 2.2 Solid Mechanics Part III Kelly 212The Right Cauchy-Green Strain Consider two line elements in the reference configuration )2( )1(,X X d d which are mapped into the line elements )2( )1(,x xd d in the current configuration. Then, using 1.10.3d, ()() () )2( )1()2( T )1()2( )1( )2( )1( XCXXFFXXF XF x x d dd dd d d d ==⋅ =⋅ action of C (2.2.9) where, by definition, C is the right Cauchy-Green Strain2 Jk Ik JkIk IJXx XxFF C∂∂ ∂∂== = ,TFFC Right Cauchy-Green Strain (2.2.10) It is a symmetric, positive definite ( cf. §1.11.2), tensor (see Eqn. 2.2.17 below), which implies that it has real positive eigenvalue s, and this has important consequences (see later). Explicitly in terms of the base vectors, J I Jk Ik J m Jm k I ik Xx Xx Xx XxE E E e e E C ⊗∂∂ ∂∂=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛⊗∂∂ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛⊗∂∂= . (2.2.11) Just as the line element Xd is a vector defined in and associated with the reference configuration, C is defined in and associated with the reference configuration, acting on vectors in the reference configuration, and so is called a material tensor . The inverse of C, C-1, is called the Piola deformation tensor . The Left Cauchy-Green Strain Consider now the following, using Eqn. 1.10.18c: ()() () )2( 1 )1()2( 1 T )1()2( 1 )1( 1 )2( )1( xbxx FFxxF xF X X d dd dd d d d −−−− − ==⋅ =⋅ action of 1b− (2.2.12) where, by definition, b is the left Cauchy-Green Strain, also known as the Finger tensor : Kj Ki Kj Ki ijXx XxFF b∂∂ ∂∂== = ,TFFb Left Cauchy-Green Strain (2.2.13) Again, this is a symmetric, positive definite tensor, only here, b is defined in the current configuration and so is called a spatial tensor . The inverse of b, b-1, is called the Cauchy deformation tensor . 2 “right” because F is on the right of the formula Section 2.2 Solid Mechanics Part III Kelly 213 It can be seen that the right and left Cauchy-Green tensors are related through -1 -1, FCFb bFFC = = (2.2.14) Note that tensors can be material (e.g. C), two-point (e.g. F) or spatial (e.g. b). Whatever type they are, they can always be described using material or spatial coordinates through the motion mapping 2.1.3, that is, using the material or spatial descriptions. Thus one distinguishes between, for example, a spatial tensor, which is an intrinsic property of a tensor, and the spatial description of a tensor. The Principal Scalar Invariants of the Cauchy-Green Tensors Using 1.10.10b, ()() b FF FF C tr tr tr trT T=== (2.2.15) This holds also for arbitrary powers of these tensors, n nb C tr tr= , and therefore, from Eqn. 1.11.17, the invariants of C and b are equal. 2.2.3 The Stretch The stretch (or the stretch ratio ) λ is defined as the ratio of the length of a deformed line element to the length of the corresponding undeformed line element: Xx dd=λ The Stretch (2.2.16) From the relations involving the Cauchy-Green Strains, letting X X X d d d ≡=)2( )1(, x x x d d d ≡=)2( )1(, and dividing across by the square of the length of Xd or xd, xbxxXXCXXxˆ ˆ ,ˆ ˆ12 22 2d dddd ddd− −=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛= =⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛= λ λ (2.2.17) Here, the quantities XX X dd d / ˆ= and xx x dd d / ˆ= are unit vectors in the directions of Xd and xd. Thus, through these relations, C and b determine how much a line element stretches (and, from 2.2.17, C and b can be seen to be indeed positive definite). One says that a line element is extended , unstretched or compressed according to 1>λ , 1=λ or 1<λ . Section 2.2 Solid Mechanics Part III Kelly 214 Stretching along the Coordinate Axes Consider now three line elements lying along the three coordinate axes3. Suppose that the material deforms in a special way, such that these line elements undergo a pure stretch , that is, they change length with no change in the right angles between them. If the stretches in these directions are 1λ, 2λ and 3λ, then 3 3 3 2 2 2 1 1 1 , , X x X x X x λ λ λ = = = (2.2.18) and the deformation gradient has only diagonal elements in its matrix form: Jii JiFδλ λλλ = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = , 0 00 00 0 321 F (no sum) (2.2.19) The axes are in this case called the principal axes of the material and the stretches are called the principal stretches . Line elements not on these three axes will in general stretch/contract and rotate relative to each other. A spherical element of material will deform into an ellipsoid, with the axes of the ellipsoid coincident with the base vectors. For example, a line element T]0,,[αα=Xd stretches by () 2/ ˆ ˆ2 22 1Tλλ λ += = X FFX d d with T 2 1 ]0, ,[αλαλ=xd , and rotates if 2 1λλ≠. The Case of F Real and Symmetric Consider now another special deformation, where F is a real symmetric tensor, in which case the eigenvalues are real and the eigenvectors form an orthonormal basis ( cf. §1.11.2). In any given coordinate system, F will in general result in the stretching of line elements and the changing of the angles between line elements. However, if one chooses a coordinate set to be the eigenvectors of F, then from Eqn. 1.10.11-12 one can write [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = ⊗=∑ = 3213 10 00 00 0 ,ˆ ˆ λλλ λ F N n F ii ii (2.2.20) where 3 2 1,,λλλ are the eigenvalues of F – the eigenvectors4 are the principal axes and the eigenvalues are the principal stretches. This indicates that as long as F is real and symmetric, one can always find a coordinate system along whose axes the material undergoes a pure stretch, with no rotation. Again a spherical element of material will deform into an ellipsoid, but now the axes of the ellipsoid coincide with the principal axes of F. 3 with the material and spatial basis vectors coincident 4 inˆ are the eigenvectors of ie, INˆ of iEˆ, with inˆ, INˆ coincident; when the bases are not coincident, the notion of rotating line elements becomes ambiguous – this topic will be examined later in the context of objectivity Section 2.2 Solid Mechanics Part III Kelly 215 2.2.4 The Green-Lagrange and Eu ler-Almansi Strain Tensors Whereas the left and right Cauchy-Green tensors give information about the change in angle between line elements and the stretch of line elements, the Green-Lagrange strain and the Euler-Almansi strain tensors directly give information about the change in the squared length of elements. Specifically, when the Green-Lagrange strain E operates on a line element d X, it gives (half) the change in the squares of the undeformed and deformed lengths: {} (){} XXEXICXXX X XCX x d dd ddd d dd d ≡− =⋅− =− 2121 22 2 action of E (2.2.21) where ()() ()JI JI JI C E δ−= −=−=21,21 21TIFF IC E Green-Lagrange Strain (2.2.22) It is a symmetric positive definite material tensor. Similarly, the (symmetric spatial) Euler-Almansi strain tensor is defined through xexX xddd d=− 22 2 action of e (2.2.23) and ()()1 T 1 21 21−− −−=−= FFI bI e Euler-Almansi Strain (2.2.24) Physical Meaning of the Components of E Take a line element in the 1-direction, []T 1 )1( 0,0,dX d=X , so that []T )1( 0,0,1 ˆ=Xd . The square of the stretch of this element is ()()121121ˆ ˆ2 )1( 11 11 11 )1( )1(2 )1( −=−=→= = λ λ C E C d d XCX The unit extension is () 1 /−= − λ X X x d d d . Denoting the unit extension of )1(Xd by )1(E, one has 2 )1( )1( 1121E E+=E (2.2.25) Section 2.2 Solid Mechanics Part III Kelly 216 and similarly for the other diagonal elements 33 22,E E . When the deformation is small, 2 )1(E is small in comparison to )1(E, so that )1( 11E=E . For small deformations then, the diagonal terms are equivalent to the unit extensions. Let 12θ denote the angle between the deformed elements which were initially parallel to the 1X and 2X axes. Then 1 21 22cos 22 1112)2()1(12 )2()2( )1()1( )2( )1()2( )1( )2()2( )1()1( 12 ++== ⎪⎭⎪⎬⎫ ⎪⎩⎪⎨⎧ ⋅ =⋅= E EEC dd dd d dd d dd dd λλθ XXC XX x xX X xx xx (2.2.26) and similarly for the other off-dia gonal elements. Note that if 2/12πθ= , so that there is no angle change, then 012=E . Again, if the deformation is small, then 22 11,EE are small, and 12 12 12 12 2 cos2sin2E≈=⎟ ⎠⎞⎜ ⎝⎛−≈− θ θπθπ (2.2.27) In words: for small defo rmations, the component 12E gives half the change in the original right angle. 2.2.5 Stretch and Rotation Tensors The deformation gradient can always be deco mposed into the product of two tensors, a stretch tensor and a rotation tensor (in one of two different ways, material or spatial versions). This is known as the polar decomposition , and is discussed in §1.11.7. One has RUF= Polar Decomposition (Material) (2.2.28) Here, R is a proper orthogonal tensor, i.e. IRR=T with 1 det=R , called the rotation tensor . It is a measure of the local rotation at X. U is a unique symmetric tensor, called the right stretch tensor . It is a measure of the local stretching (or contr action) of material at X. Consider a line element dX. Then X RUXFx ˆ ˆ ˆ d d d ==λ (2.2.29) and so {▲Problem 1} XUUX ˆ ˆ2d d⋅=λ (2.2.30) Section 2.2 Solid Mechanics Part III Kelly 217 Thus (this is a definition of U) ()UUC C U = = The Right Stretch Tensor (2.2.31) From 2.2.30, the right Cauchy-Green strain C (and by consequence the Euler-Lagrange strain E) only give information about the stretch of line elements; it does not give information about the rotation that is experienced by a particle during motion. The deformation gradient F, however, contains information about both the stretch and rotation. It can also be seen from 2.2.30-1 that U is a material tensor. Note that, since ()XURx d d= , the undeformed line element is first stretched by U and is then rotated by R into the deformed element xd (the element may also undergo a rigid body translation c), Fig. 2.2.4. R is a two-point tensor. Figure 2.2.4: the polar decomposition Evaluation of U In order to evaluate U, it is necessary to evaluate C. To evaluate the square-root, C must first be obtained in relation to its principal axes, so that it is diagonal, and then the square root can be taken of the diagonal elem ents (see §1.11.6). Then the tensor needs to be transformed back to the original coordinate system. Example Consider the motion 3 3 2 1 2 2 1 1 , ,2 2 X x X X x X X x = += −= The (homogeneous) deformation of a unit square in the 2 1xx− plane is as shown in Fig. 2.2.5. undeformed stretched final configuration principal material axes R Section 2.2 Solid Mechanics Part III Kelly 218 Figure 2.2.5: deformation of a square One has [] () [][] ()j i j i E E FF C E e F ⊗ ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −− == ⊗ ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡− = : basis 10 005303 5 , : basis 10001102 2 T Note that F is not symmetric, so that it might have only one real eigenvalue (in fact here it does have complex eigenvalues), and the eigenvectors may not be orthonormal. C, on the other hand, by its very definition, is symmetric; it is in fact positive definite and so has positive real eigenvalues forming an orthonormal set. To determine the principal axes of C, it is necessary to evaluate the eigenvalues/eigenvectors of the tensor. The eigenvalues are the roots of the characteristic equation 1.9.34, 0 III II I2 3=−+−C C Cααα and the first, second and third invariants of the tensor are given by 1.9.37 so that 0 16 26 112 3=−+− ααα , with roots 1,2,8=α . The three corresponding eigenvectors are found from 1.9.40, 0 ˆ) 1(0 ˆ) 5(ˆ30 ˆ3ˆ) 5( 0 ˆ) (ˆ ˆ0 ˆ ˆ) (ˆ0 ˆ ˆ ˆ) ( 32 12 1 3 33 2 32 1 313 23 2 22 1 213 13 2 12 1 11 =−=−+−=−− → =−++=+−+=++− NN NN N N C NC NCNC N C NCNC NC N C ααα ααα Thus (normalizing the eigenvectors so that th ey are unit vectors, and form a right-handed set, Fig. 2.2.6): (i) for 8=α , 0 ˆ7,0 ˆ3ˆ3,0 ˆ3ˆ33 2 1 2 1 =−=−−=−− N N N N N , 2 21 1 21 1ˆ E E N −= (ii) for 2=α , 0 ˆ,0 ˆ3ˆ3,0 ˆ3ˆ33 2 1 2 1 =−=+−=− N N N N N , 2 21 1 21 2ˆ E E N += (iii) for 1=α , 0 ˆ0,0 ˆ4ˆ3,0 ˆ3ˆ43 2 1 2 1 ==+−=− N N N N N , 3 3ˆ E N= 1 1,xX3 2,xX Section 2.2 Solid Mechanics Part III Kelly 219 Figure 2.2.6: deformation of a square Thus the right Cauchy-Green strain tensor C, with respect to coordinates with base vectors 1 1ˆN E=′ , 2 2ˆN E=′ and 3 3ˆN E=′ , that is, in terms of principal coordinates, is []j iN N C ˆ ˆ: basis 100020008 ⊗ ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ = This result can be checked using the tensor transformation formulae 1.10.3b, [][] [] [] QCQ CT=′ , where Q is the transformation matrix of direction cosines (see also the example at the end of §1.4.2), ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −= ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ′⋅′⋅′⋅′⋅′⋅′⋅′⋅′⋅′⋅ = 1 0 002/12/102/12/1 ˆ ˆ ˆ 3 2 1 3 3 2 3 1 33 2 2 2 1 23 1 2 1 1 1 MMMMMM N N N ee eeeeeeeeeeee eeee ijQ . The stretch tensor U, with respect to the principal directions is [][]j iN N C U ˆ ˆ: basis 0 00 00 0 10 002 000 22 321 ⊗ ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ≡ ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ == λλλ These eigenvalues of U (which are the square root of those of C) are the principal stretches and, as before, they are labeled 3 2 1,,λλλ . In the original coordinate system, using the inverse tensor transformation rule 1.10.3a, [][] [] []TQUQ U ′=, []j iE E U ⊗ ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −− = : basis 1 0 002/32/102/1 2/3 so that 1X2X 2ˆNprincipal material directions 1ˆN Section 2.2 Solid Mechanics Part III Kelly 220[][]j iE e FU R ⊗ ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡− ==−: basis 1 0 00 2/12/102/1 2/1 1 and it can be verified that R is a rotation tensor, i.e. is proper orthogonal. Returning to the deformation of the unit square, the stretch and rotation are as illustrated in Fig. 2.2.7 – the action of U is indicated by the arrows, deforming the unit square to the dotted parallelogram, whereas R rotates the parallelogram through o45 as a rigid body to its final position; note that the line elemen t along the diagonal (indicated by the heavy line) lies along a principal direction of U and therefore undergoes a pure stretch. Figure 2.2.7: stretch and rotation of a square ■ Spatial Description A polar decomposition can be made in the spatial description. In that case, vRF= Polar Decomposition (Spatial) (2.2.32) Here v is a symmetric, positive definite second order tensor called the left stretch tensor , and b vv=, where b is the left Cauchy-Green tensor. R is the same rotation tensor as appears in the material description. Thus an elemental sphere can be regarded as first stretching into an ellipsoid, whose axes are the principal material axes (the principal axes of U), and then rotating; or first rotating, and then stretching into an ellipsoid whose axes are the principal spatial axes (the principal axes of v). The end result is the same. The development in the spatial description is similar to that given above for the material description, and one finds by analogy with 2.2.30, xv vx ˆ ˆ 1 1 2d d−− −⋅=λ (2.2.33) In the above example, it turns out that v takes the simple diagonal form 1 1,xX2 2,xX Section 2.2 Solid Mechanics Part III Kelly 221[]j ie e v ⊗ ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = : basis 10 002 000 22 . so the unit square rotates first and then undergoes a pure stretch along the coordinate axes, which are the principal spatial axes, and the sequence is now as shown in Fig. 2.2.9. Figure 2.2.8: stretch and rotation of a square in spatial description Relationship between the Material and Spatial Decompositions Comparing the two decompositions, one sees that the material and spatial tensors involved are related through TRURv= , TRCRb= (2.2.34) Further, suppose that U has an eigenvalue λ and an eigenvector Nˆ. Then N NU ˆ ˆλ= , so that RN RUNλ= . But vR RU= , so ()()NR NRv ˆ ˆλ= . Thus v also has an eigenvalue λ, but an eigenvector NRn ˆ ˆ= . From this, it is seen that the rotation tensor R maps the principal material axes into the principal spatial axes. It also follows that R and F can be written explicitly in terms of the material and spatial principal axes (compare the first of these with 1.10.25)5: i iN nR ˆ ˆ⊗= , ∑ ∑ = =⊗=⊗ ==3 13 1ˆ ˆ ˆ ˆ ii ii ii ii N n N N R RUF λ λ (2.2.35) and the deformation gradient acts on the principal axes base vectors according to {▲Problem 4} ii i i ii i ii ii i N nF N nF n NF n NF ˆ ˆ ,ˆ1ˆ ,ˆ1ˆ ,ˆ ˆT 1 Tλλ λλ = = = =− − (2.2.36) 5 this is not a spectral decomposition of F (unless F happens to be symmetric, which it must be in order to have a spectral decomposition) 1 1,xX2 2,xX Section 2.2 Solid Mechanics Part III Kelly 222 The representation of F and R in terms of both material and spatial principal base vectors in 2.3.35 highlights their two-point character. Other Strain Measures Some other useful measures of strain are The Hencky strain measure: U Hln≡ (material) or v hln= (spatial) The Biot strain measure: IUB−= (material) or Ivb−= (spatial) The Hencky strain is evaluated by first evaluating U along the principal axes, so that the logarithm can be taken of the diagonal elements. The material tensors H, B, C, U and E are coaxial tensors, with the same eigenvectors iNˆ. Similarly, the spatial tensors h, b, b, v and e are coaxial with the same eigenvectors inˆ. From the definitions, the spectral de compositions of these tensors are () () () () () ()∑ ∑∑ ∑∑ ∑∑ ∑∑ ∑ = == == == == = ⊗−= ⊗−=⊗ = ⊗ =⊗−= ⊗−=⊗= ⊗ =⊗= ⊗ = 3 13 13 13 13 12 213 12 213 123 123 13 1 ˆ ˆ1 ˆ ˆ1ˆ ˆ ln ˆ ˆ lnˆ ˆ/11 ˆ ˆ1ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ii i i ii i iii i i ii i iii i i ii i iii ii ii i iii ii ii ii n n b N N Bn n h N N Hn n e N N En n b N N Cn n v N N U λ λλ λλ λλ λλ λ (2.2.37) 2.2.6 Some Sim ple Deformations In this section, some elementary deformations are considered. Pure Stretch This deformation has already been seen, but now it can be viewed as a special case of the polar decomposition. The motion is 3 3 3 2 2 2 1 1 1 , , X x X x X x λ λ λ = = = Pure Stretch (2.2.38) and the deformation gradient is Section 2.2 Solid Mechanics Part III Kelly 223⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ == ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ =321 321 0 00 00 0 100010001 0 00 00 0 λλλ λλλ RU F Here, IR= and there is no rotation. FU= and the principal material axes are coincident with the material coordinate axes. 3 2 1,,λλλ , the eigenvalues of U, are the principal stretches. Stretch with rotation Consider the motion 3 3 2 1 2 2 1 1 , , X x X kX x kX X x = += −= so that ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡− == ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡− =1 0 00 sec 00 0 sec 1 0 00 cos sin0 sin cos 100010 1 θθ θθθθ RU F kk where θtan=k . This decomposition shows that the deformation consists of material stretching by ) 1( sec2k+=θ , the principal stretches, along each of the axes, followed by a rigid body rotation through an angle θ about the 03=X axis, Fig. 2.2.9. The deformation is relatively simple because the principal material axes are aligned with the material coordinate axes (so that U is diagonal). The deformation of the unit square is as shown in Fig. 2.2.9. Figure 2.2.9: stretch with rotation Pure Shear Consider the motion 3 3 2 1 2 2 1 1 , , X x X kX x kX X x = += += Pure Shear (2.2.39) 1 1,xX2 2,xX 21k+θ Section 2.2 Solid Mechanics Part III Kelly 224so that ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ == ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = 100010 1 100010001 100010 1kk kk RU F where, since F is symmetric, there is no rotation, and UF=. Since the rotation is zero, one can work directly with U and not have to consider C. The eigenvalues of U, the principal stretches, are 1,1, 1 k k−+ , with corresponding principal directions 2 21 1 21 1ˆ E E N += , 2 21 1 21 2ˆ E E N +−= and 3 3ˆ E N= . The deformation of the unit square is as shown in Fig. 2.2.10. The diagonal indicated by the heavy line stretches by an amount k+1 whereas the other diagonal contracts by an amount k−1 . An element of material along the diagonal will undergo a pure stretch as indicated by the stretching of the dotted box. Figure 2.2.10: pure shear Simple Shear Consider the motion 3 3 2 2 2 1 1 , , X x X x kX X x = = += Simple Shear (2.2.40) so that ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ += ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = 1 0 00 10 1 , 1000100 12k kk k C F The invariants of C are 1 III, 3 II, 3 I2 2=+=+=C C C k k and the characteristic equation is 01) 1() 3(2 3=−−++ λλ λ k , so the principal values of C are 1, 4 12 21 2 21k k k +±+=λ . The principal values of U are the (positive) square-roots of these: 1, 421 2 21k k±+=λ . 1 1,xX2 2,xX kk 1ˆN Section 2.2 Solid Mechanics Part III Kelly 225These can be written as 1, tan sec θθλ±= by letting k21tan=θ . Three corresponding eigenvectors of C are 3 3 2 12 21 2 212 2 12 21 2 211ˆ, 4ˆ, 4ˆ E N E E N E E N = + +−= + ++= k k kk k k kk or, normalizing so that they are of unit size, and writing in terms of θ, 3 3 2 1 2 2 1 1ˆ,2sin1 2sin1ˆ,2sin1 2sin1ˆ E N E E N E E N =−++−=++−=θ θ θ θ The transformation matrix of direction cosines is then []()() () () ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ − ++− − = 1 0 002/ sin1 2/ sin102/ sin1 2/ sin1 θ θθ θ Q so that, using the inverse transformation formula, [][][][]TQUQ U ′= , one obtains U in terms of the original coordinates, and hence ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ + ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −== ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = 1 0 00 cos/) sin1( sin0 sin cos 1 0 00 cos sin0 sin cos 1000100 1 2θθ θθ θ θθθθ RU Fk The deformation of the unit square is shown in Fig. 2.2.11 (for o7.5 ,2.0==θ k ). The square first undergoes a pure stretch/contract ion along the principal axes, and is then brought to its final position by a negative (clockwise) rotation of θ. For this deformation, 1 det=F and, as will be shown below, this means that the simple shear deformation is volume-preserving. Figure 2.2.11: simple shear 1 1,xX2 2,xX 2ˆN1ˆN θ Section 2.2 Solid Mechanics Part III Kelly 226 2.2.7 Displacement & Displacement Gradients The displacement of a material particle 6 is the movement it undergoes in the transition from the reference configuration to the current configuration. Thus, Fig. 2.2.12,7 X Xx XU −= ),( ),( t t Displacement (Material Description) (2.2.41) ),( ),( t t xXx xu −= Displacement (Spatial Description) (2.2.42) Note that U and u have the same values, they just have different arguments. Figure 2.2.12: the displacement Displacement Gradients The displacement gradient in the material and spatial descriptions, X XU∂∂ /),(t and x xu∂∂ /),(t , are related to the deformation gradient and the inverse deformation gradient through 1 ) (grad) (Grad −−=∂−∂=∂∂=−=∂−∂=∂∂= FIxXx xuuIFXXx XUU ji ij jiij ji ji xX xuXx XU ∂∂−=∂∂−∂∂=∂∂ δδ (2.2.43) and it is clear that the displacement gradients are related through (see Eqn. 2.2.8) 1Grad grad−= FU u (2.2.44) The deformation can now be written in terms of either the material or spatial displacement gradients: 6 In solid mechanics, the motion and deformation are often described in terms of the displacement u. In fluid mechanics, however, the primary field quantity describing the kinematic properties is the velocity v (and the acceleration va&=) – see later. 7 The material displacement U here is not to be confused with the right stretch tensor discussed earlier. X xuU= Section 2.2 Solid Mechanics Part III Kelly 227xu X xu X xXU X XU X x d d d d dd d d d d grad )(Grad )( +=+=+=+= (2.2.45) Strains in terms of Displacement Gradients The strains can be written in terms of the displacement gradients. Using 1.10.3b, () () ()() () ()() ⎭⎬⎫ ⎩⎨⎧ ∂∂ ∂∂+∂∂+∂∂= + + =−+ + =−= JK IK IJ JI JIXU XU XU XUE21, Grad Grad Grad Grad21Grad Grad2121 T TTT U U U UIIU IUIFF E (2.2.46a) () () ()() () ()() ⎪⎭⎪⎬⎫ ⎪⎩⎪⎨⎧ ∂∂ ∂∂−∂∂+∂∂= − + =− −−=−=−− jk ik ij ji ijxu xu xu xue21, grad grad grad grad21grad grad2121 T TT1 T u u u uu Iu IIFFI e (2.2.46b) Small Strain If the displacement gradients are small, then the quadratic terms, their products, are small relative to the gradients themselves, and may be neglected. With this assumption, the Green-Lagrange strain E (and the Euler-Almansi strain) reduces to the small-strain tensor , ()()⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂= + = IJ JI JIXU XU 21, Grad Grad21 Tε U U ε (2.2.47) Since in this case the displacement gradients are small, it does not matter whether one refers the strains to the reference or current c onfigurations – the error is of the same order as the quadratic terms already neglected8, so the small strain tensor can equally well be written as ()()⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂= + = ij ji ijxu xu 21, grad grad21 Tε u u ε Small Strain Tensor (2.2.48) 8 although large rigid body rotations must not be allowed – see §2.7 Section 2.2 Solid Mechanics Part III Kelly 2282.2.8 The Deformation of Area and Volume Elements Line elements transform between the reference and current configurations through the deformation gradient. Here, the transformation of area and volume elements is examined. The Jacobian Determinant The Jacobian determinant of the deformation is defined as the determinant of the deformation gradient, F X det),(=t J 33 23 1332 22 1231 21 11 det Xx Xx XxXx Xx XxXx Xx Xx ∂∂ ∂∂ ∂∂∂∂ ∂∂ ∂∂∂∂ ∂∂ ∂∂ =F The Jacobian Determinant (2.2.49) Equivalently, it can be considered to be the Jacobian of the transformation from material to spatial coordinates (see Appendix 1.B.2). From Eqn. 1.3.17, the Jacobian can also be written in the form of the triple scalar product ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂×∂∂⋅∂∂= 3 2 1 X X XJx x x (2.2.50) Consider now a volume element in the reference configuration, a parallelepiped bounded by the three line-elements )1(Xd , )2(Xd and )3(Xd . The volume of the parallelepiped9 is given by the triple scalar product (Eqns. 1.1.4): ( ))3( )2( )1(X X X d d d dV ×⋅= (2.2.51) After deformation, the volume element is bounded by the three vectors )(idx, so that the volume of the deformed element is, using 1.9.16f, () () () dVd d dd d dd d d dv FX X XFXF XF XFx x x detdet)3( )2( )1()3( )2( )1()3( )2( )1( =×⋅ =×⋅=×⋅= (2.2.52) Thus the scalar J is a measure of how the volume of a material element has changed with the deformation and for this reason is often called the volume ratio . dVJ dv= Volume Ratio (2.2.53) 9 The vectors should form a right-handed set so that the volume is positive. Section 2.2 Solid Mechanics Part III Kelly 229Since volumes cannot be negative, one must insist on physical grounds that 0>J . Also, since F has an inverse, 0≠J . Thus one has the restriction 0>J (2.2.54) Note that a rigid body rotation does not alter the volume, so the volume change is completely characterised by the stretching tensor U. Three line elements lying along the principal directions of U form an element with volume dV, and then undergo pure stretch into new line elements defining an element of volume dV dv321λλλ= , where iλ are the principal stretches, Fig. 2.2.13. The unit change in volume is therefore also 1321−=−λλλdVdV dv (2.2.55) Figure 2.2.13: change in volume For example, the volume change for pure shear is 2k− (volume decreasing) and, for simple shear, is zero ( cf. Eqn. 2.2.39 et seq. , 01)1)( tan )(sec tan (sec =− − + θθθθ ). An incompressible material is one for which the volume change is zero, i.e. the deformation is isochoric. For such a material, 1=J , and the three principal stretches are not independent, but are constrained by 1321=λλλ Incompressibility Constraint (2.2.56) Nanson’s Formula Consider an area element in the reference configuration, with area dS, unit normal Nˆ, and bounded by the vectors )2( )1(,X X d d , Fig. 2.2.14. Then )2( )1( ˆ X X N d d dS ×= (2.2.57) The volume of the element bounded by the vectors )2( )1(,X X d d and some arbitrary line element Xd is X N ddS dV ⋅=ˆ . The area element is now deformed into an element of current configuration reference configuration principal material axes dVdV dv321λλλ= Section 2.2 Solid Mechanics Part III Kelly 230area ds with normal nˆ and bounded by the line elements )2( )1(,x xd d . The volume of the new element bounded by the area element and XFx d d= is then X N XF nx n ddSJ d ds dds dv ⋅≡⋅=⋅= ˆ ˆ ˆ (2.2.58) Figure 2.2.14: change of surface area Thus, since dX is arbitrary, and using 1.10.3d, dS J ds NF n ˆ ˆT−= Nanson’s Formula (2.2.59) Nanson’s formula shows how the vector element of area dsnˆ in the current configuration is related to the vector element of area dSNˆ in the reference configuration. 2.2.9 Inextensibility and Orientation Constraints A constraint on the principal stretches was introduced for an incompressible material, 2.2.56. Other constraints arise in practice. For example, consider a material which is inextensible in a certain direction, defined by a unit vector Aˆ in the reference configuration. It follows that 1ˆ=AF and the constraint can be expressed as 2.2.17, 1ˆˆ=ACA Inextensibility Constraint (2.2.60) If there are two such directions in a plane, defined by Aˆ and Bˆ, making angles θ and φ respectively with the principal material axes 2 1ˆ,ˆNN , then [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = 0sincos 0 00 00 0 0 sin cos1 2 32 22 1 θθ λλλ θθ and () ()φλλλθλλ2 2 22 12 22 2 22 1 cos 1 cos −=−= − . It follows that θφ=, πθφ+= , πφθ=+ or πφθ 2=+ (or 12 1==λλ , i.e. no deformation). )1(Xd)2(XdXd Nˆ )1(xd)2(xdxd nˆ Section 2.2 Solid Mechanics Part III Kelly 231Similarly, one can have orientation constraints. For example, suppose that the direction associated with the vector Aˆ maintains that direction. Then A AF ˆ ˆμ= Orientation Constraint (2.2.61) for some scalar 0 >μ . 2.2.10 Problems 1. In equations 2.2.8, one has from the chain rule 1Grad grad−=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛⊗∂∂ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂=∂∂ ∂∂=∂∂= F e E E e e φφ φφφi m im j ji im mi i xX X xX X x Derive the other two relations. 2. Use IRR=T, U U=T, and 1.10.3e to show that XXUUXX dd dd⋅=2λ 3. For the deformation 3 2 1 3 3 2 2 3 1 1 2 2 ,2 ,2 X X X x X X x X X x ++−= −= += (a) Determine the Deformation Gradient and the Right Cauchy-Green tensors (b) Consider the two line elements 2)2( 1)1(, e Xe X = = d d (emanating from (0,0,0)). Use the Right Cauchy Green tensor to determine whether these elements in the current configuration ()2( )1(,x xd d ) are perpendicular. (c) Use the right Cauchy Green tensor to evaluate the stretch of the line element 2 1eeX+=d , and hence determine whether the element contracts, stretches, or stays the same length after deformation. (d) Determine the Green-Lagrange and Eulerian strain tensors (e) Decompose the deformation into a stretching and rotation (check that U is symmetric and R is orthogonal). What are the principal stretches? 4. Derive Equations 2.2.36. 5. For the deformation 3 2 3 3 2 2 1 1 , , X aX x X X x X x += += = (a) Determine the displacement vector in both the material and spatial forms (b) Determine the displaced location of the particles in the undeformed state which originally comprise (i) the plane circular surface ) 1/(1 ,02 2 32 2 1 a X X X −=+= (ii) the infinitesimal cube with edges along the coordinate axes of length ε=idX Sketch the displaced configurations if 2/1=a 6. For the deformation 3 1 3 3 2 2 2 1 1 , , X aX x aX X x aX X x += += += (a) Determine the displacement vector in both the material and spatial forms (b) Calculate the full material strain tensor and the full spatial strain tensor (c) Calculate the infinitesimal strain tensor as derived from the material and spatial tensors, and compare them for the case of very small a. Section 2.2 Solid Mechanics Part III Kelly 2327. In the example given above on the polar decomposition, §2.2.5, check that the relations 3 ,2,1,== ii i n Cnλ are satisfied (with respect to the original axes). Check also that the relations 3 ,2,1,=′=′ ii i n nCλ are satisfied (here, the eigenvectors are the unit vectors in the second coordinate system, the principal directions of C, and C is with respect to these axes, i.e. it is diagonal). Section 2.3 Solid Mechanics Part III Kelly 2332.3 Deformation and Strain: Further Topics 2.3.1 Volumetric and Isochoric Deformations When analysing materials which are only slig htly incompressible, it is useful to decompose the deformation gradient multiplicatively, according to () F FI F3/1 3/1J J= = (2.3.1) From this definition { ▲Problem 1}, 1 det=F (2.3.2) and so F characterises a volume preserving ( distortional or isochoric ) deformation. The tensor I3/1J characterises the volume-changing ( dilational or volumetric ) component of the deformation, with () J J == F Idet det3/1. This concept can be carried on to other kinematic tensors. For example, with FFCT= , C FF C3/2 T 3/2J J≡ = . (2.3.3) F and C are called the modified deformation gradient and the modified right Cauchy-Green tensor , respectively. The square of the stretch is given by {}XCX XCX ˆ ˆ ˆ ˆ3/2 2d d J d d==λ ( 2 . 3 . 4 ) so that λλ3/1J= , where λ is the modified stretch , due to the action of C. Similarly, the modified principal stretches are i iJλλ3/1−= , 3 ,2,1=i (2.3.5) with 1 det321==λλλF (2.3.6) The case of simple shear discussed earlier is an example of an isochoric deformation, in which the deformation gradient and the m odified deformation gr adient coincide, II=3/1J . 2.3.2 Relative Deformation It is usual to use the configuration at )0,(=tX as the reference configuration, and define quantities such as the deformation gradient re lative to this reference configuration. As mentioned, any configuration can be taken to be the reference configuration, and a new Section 2.3 Solid Mechanics Part III Kelly 234deformation gradient can be constructed with re spect to this new refe rence configuration. Further, the reference configur ation does not have to be fixe d, but could be moving also. In many cases, it is useful to choose the current configuration ) ,(tx to be the reference configuration, for example when evaluating ra tes of change of kine matic quantities (see later). To this end, introduce a third configur ation: this is the configuration at some time τ=t and the position of a material particle X here is denoted by ) ,( ˆτXχx= , where χ is the motion function. The deformation at this time τ relative to the current configuration is called the relative deformation , and is denoted by ),( ˆ)(τxχxt= , as illustrated in Fig. 2.3.1. Figure 2.3.1: the relative deformation The relative deformation gradient tF is defined through x xFx d dt ),( ˆτ= , xxF∂∂=ˆ t (2.3.7) Also, since X XFx dt d ),(= and X XFx d d ),( ˆτ= , one has the relation ),(),( ),( tt XFxF XF ττ= (2.3.8) Similarly, relative strain measures can be defined, for example the relative right Cauchy- Green strain tensor is ()()()τττt t t F F CT= ( 2 . 3 . 9 ) Example Consider the two-dimensional motion initial configuration current configuration configuration at τ=t t,X ),(tXχ),(τXχ ),()(τxχt),(),( ˆ )(ττ xχXχx t== ),(),( )( tt txχXχx ==tF Frelative deformation Section 2.3 Solid Mechanics Part III Kelly 235)1( ,2 2 1 1 += = tX x eX xt Inverting these gives the spatial description ) 1/( ,2 2 1 1 += =−t x X ex Xt, and the relative deformation is )1/()1( )1( ),(ˆ),(ˆ 2 2 21 1 1 ++=+===− t x X xex eX xt τ ττττ τ xx The deformation gradients are 2 2 1 12 2 1 1 )1/()1(ˆ),()1( ),( e e e e e e xFE e E e E e XF ⊗+++⊗=⊗∂∂=⊗++⊗=⊗∂∂= −t exxt eXxt t j i ji tt j i ji τ ττ ■ 2.3.3 Derivatives of the Stretch In this section, some useful formulae invol ving the derivatives of the stretches with respect to the Cauchy-Green st rain tensors are derived. Derivatives with respect to b First, take the stretches to be functions of the left Cauchy-Green strain b. Write b using the spatial principal directions inˆ as a basis, 2.2.37, so that the total differential can be expressed as [] ∑ =⊗+⊗+⊗ =3 12ˆ ˆˆ ˆ ˆ ˆ 2 ii i i i i i ii i d d d d n nn n n n b λ λλ (2.3.10) Since ij j iδ=⋅nnˆˆ , then []i i i i i i i i i i i d d d d d λλ λλλ 2 ˆˆ ˆ ˆ 2 ˆ ˆ2=⋅+⋅+= nn n n nbn (no sum over i) (2.3.11) This last follows since the change in a vector of constant length is always orthogonal to the vector itself (as in the curvature an alysis of §1.6.2). Using the property ) (: vuT uTv ⊗= , one has (summing over the k but not over the i; here ik i kd d δλλ= /) 1)ˆ ˆ(:212)ˆ ˆ(: )ˆ ˆ(: =⊗∂∂→ =⊗∂∂≡⊗i i i ii i i i k ki i d d d n nbn nbn nbλλλλ λλ (2.3.12) Then, since b b ∂∂∂∂ / : /i iλλ is also equal to 1, one has Section 2.3 Solid Mechanics Part III Kelly 236)ˆ ˆ(21: )ˆ ˆ(:21 i i ii i ii i i in nb bbn nb⊗=∂∂→∂∂ ∂∂=⊗∂∂ λλ λ λ λλ (2.3.13) The chain rule then gives the second derivative. The above analysis is for distin ct principal stretches. When λλλλ ≡==3 2 1 , then I b2λ= , I bλλd d 2= . Also, ()λλd d ∂∂= / 3b b , so () I b λλ 2 / 3=∂∂ , or bIbb ∂∂=∂∂ ∂∂ λλλ λ:2 : 3 (2.3.14) But 1 /:/ =∂∂∂∂ b bλλ and II: 3= , and so in this case, λ λ 2/ / Ib=∂∂ . A similar calculation can be carri ed out for two equal eigenvalues 3 2 1 λλλλ ≠== . In summary, () () 1 3 2 1 3 223 2 13 13 2 1 3 3 332 2 1 11 3 2 1 ) over sum no(ˆ ˆ ˆ ˆ 4121ˆ ˆ 21ˆ ˆ 21ˆ ˆ ˆ ˆ 21) over sum no( ˆ ˆ 21 λλλλλλλλλλλ λλλλλλ λλλλλλλλλλ ≠≠≠ ⊗⊗⊗−=∂∂=== =⊗ =∂∂≠== ⊗=∂∂⊗+⊗=∂∂≠≠≠ ⊗=∂∂ ∑= ii i i i i iii i ii i ii n n n nbI n nbn nbn n n nbn nb (2.3.15) Derivatives with respect to C The stretch can also be considered to be a function of the right Cauchy-Green strain C. The derivatives of the stretches with respect to C can be found in exactly the same way as for the left Cauchy-Green strain. The result s are the same as given in 2.3.15 except that, referring to 2.2.37, b is replaced by C and nˆ is replaced by Nˆ. 2.3.4 The Directional Derivative of Kinematic Quantities The directional derivative of vectors and tensors was introduced in §1.6.11 and §1.15.4. Taking directional derivatives of kinematic qua ntities is often very useful, for example in linearising equations in order to apply numerical solution algorithms The Deformation Gradient First, consider the deformation gradient as a function of th e current position x (or motion χ) and examine its value at ax+: Section 2.3 Solid Mechanics Part III Kelly 237 []()a aF xF axFx o+∂+=+ )( ) ( (2.3.16) The directional derivative []()axF aFx ∂∂=∂ / can be expressed as [] () () () FaaXaxaxF aFx gradGrad00 ==∂+∂=+ =∂ == ε εεε εε dddd (2.3.17) the last line resulting from 2.2.8b. It follo ws that the directional derivative of the deformation gradient in the dire ction of a displacement vector u from the current configuration is []()Fu uFx grad=∂ (2.3.18) On the other hand, consider the deformation gradient as a function of X and examine its value at AX+: []AF XF AXFX∂+=+ )( ) ( (2.3.19) and now [] () () () () aFAAFxXA XxXA XF AFX GradGrad000 ==+∂∂=+∂∂=+ =∂ === εεεεεε εεε dddddd (2.3.20) where FAa= . Other Kinematic Quantities The directional derivative of the Green-Lagr ange strain, the right and left Cauchy-Green tensors and the Jacobian in th e direction of a displacement u from the current configuration are { ▲Problem 2} Section 2.3 Solid Mechanics Part III Kelly 238() () u uu bbu ubεFF uCεFF uE xxxx div][grad grad][2][][ TTT J J=∂+ =∂=∂=∂ (2.3.21) where ε is the small-strain tensor, 2.2.48. The directional derivative is also useful for deriving various relations between the kinematic variables. For example, for an arbitrary vector a, using the chain rule 1.15.28, 2.3.20, 1.15.24, the trace relations 1.10.10e and 1.10.10b, and 2.2.8b, 1.14.9, () [] [][] ()[] () ()() ()() ()() ()FaFaFFaFa FFa FFaaFa a FX FX divgradtrGradtrGrad trGrad:GradGrad 11T JJJJJJJJ J =====∂=∂∂=∂=⋅ −−− (2.3.22) so that, from 1.14.16b with a constant, Tdiv Grad F J J= (2.3.23) 2.3.5 Problems 1. Use 1.10.16c to show that 1 det=F . 2. (a) use the relation ()IFF E −=T 21, Eqn. 2.3.18, ()Fu uFx grad][=∂ , and the product rule of differentiation to derive 2.3.21a, εFF uExT][=∂ , where ε is the small strain tensor. (b) evaluate []uCx∂ (in terms of F and ε, the small strain tensor) (c) evaluate []ubx∂ (in terms of ugrad and b) (d) evaluate []uxJ∂ (in terms of J and udiv ; use the chain rule [] [][] uF ux F x ∂∂=∂ J J ˆ , with F F det)(ˆ= J , [] u uFx Grad=∂ ) Section 2.4 Solid Mechanics Part III Kelly 2392.4 Material Time Derivatives The motion is now allowed to be a function of time, ()t,Xχx= , and attention is given to time derivatives, both the material time derivative and the local time derivative . 2.4.1 Velocity & Acceleration The velocity of a moving particle is the time rate of change of the position of the particle. From 2.1.3, by definition, dtt dt),(),(XχXV≡ (2.4.1) In the motion expression ()t,Xχx= , X and t are independent variables and so X is independent of time, denoting th e particle for which the velocity is being calculated. The velocity can thus be written as tt∂∂ /),(Xχ or, denoting the motion by ),(tXx , as dtt d /),(Xx or tt∂∂ /),(Xx . The spatial description of the velocity field may be obtained from the material description by simply replacing X with x, i.e. ()tt t ),,( ),(1xχV xv−= (2.4.2) As with displacements in both descriptions, there is only one velocity, ),( ),( t t xv XV= – they are just given in term s of different coordinates. The velocity is most often expresse d in the spatial description, as dtdtxx xv ==&),( velocity (2.4.3) To be precise, the right hand side here involves x which is a function of the material coordinates, but it is understood that the substitution back to spatial coordinates, as in 2.4.2, is made. Similarly, the acceleration is defined to be 22 22 22),( ),(),(tt dtd dtd dtt dt∂∂=== =Xχ V x XχXA (2.4.4) The Local Rate of Change Note that the derivative dtd/V in 2.4.4 (with X fixed) is not the same as the derivative t∂∂/v (with x fixed). The former is the accel eration of a material particle X. The latter is the time rate of change of the velocity of particles at a fixed location in space – this is Section 2.4 Solid Mechanics Part III Kelly 240called the local rate of change of v; in general, different material particles will occupy position x at different times. 2.4.2 The Material Derivative Suppose that the velocity in te rms of spatial coordinates, ),(txvv= is known; for example, one could have a measuring instrume nt which records the velocity at a specific location, but the motion χ itself is unknown. In that case, to evaluate the acceleration, the chain rule of differentia tion must be applied: ()dtd tttdtd x xv vxv v∂∂+∂∂= ≡ ),( & or () vvva grad+∂∂=t acceleration (spatial description) (2.4.5) The acceleration can now be determined, becau se the derivatives can be determined (measured) without knowing the motion. In the above, the material derivative , or total derivative , of the particle’s velocity was taken to obtain the acceleration. In general, one can take the time derivative of any physical or kinematic property ()• expressed in the spatial description: ()() () v•+•∂∂=• gradt dtd Material Time Derivative (2.4.6) For example, the rate of change of the density ),(txρρ= of a particle instantaneously at x is v⋅+∂∂=≡ ρρρρ gradt dtd& (2.4.7) The first term, t∂∂/ρ , gives the local rate of change of density at x whereas the second term ρ grad⋅v gives the change due to the particle’s motion, and is called the convective rate of change . The material derivative dtd/ can be applied to any sc alar, vector or tensor: () () vAA AAvaa aav gradgradgrad +∂∂=≡+∂∂=≡⋅+∂∂=≡ t dtdt dtdt dtd &&& αααα (2.4.8) Section 2.4 Solid Mechanics Part III Kelly 241 Another notation often used fo r the material derivative is DtD/: fixedX⎟ ⎠⎞⎜ ⎝⎛ ∂∂≡≡≡tffdtdf DtDf& (2.4.9) Steady and Uniform Flows In a steady flow , quantities are independent of time, so the local rate of change is zero and, for example, v⋅=ρρgrad& . In a uniform flow , quantities are independent of position so that, for example, t∂∂= /ρρ& Example Consider the motion 3 3 12 2 2 22 1 1 , , X x Xt X x Xt X x = += += The velocity and acceleration can be evaluated through 21 12 22 21 12 2 2 ),( , 2 2 ),( e exXA e exXV X Xdtdt tX tXdtdt +== +== One can write the motion in the spatial descri ption by inverting the ma terial description: 3 3 412 2 2 422 1 1 ,1,1x XtxtxXtxtxX =−−=−−= Substituting in these equations then gives th e spatial description of the velocity and acceleration: () ()2 422 1 1 412 2 12 422 1 1 412 2 1 1212 ),,( ),(12 12 ),,( ),( e e xfA xae e xfV xv txtx txtxtt ttxtxt txtxt tt t −−+−−= =−−+ −−= = −− Alternatively, the acceleration can be obtained directly from the spatial velocity field: Section 2.4 Solid Mechanics Part III Kelly 242() 2 422 1 1 412 2422 1412 2 43 44 43 2 422 1 1 412 2 121201212 0 0 0012 12012 12 1212grad ),( e ee evvvxa txtx txtxtxtxttxtxt tt tttt tt txtxttxtxtttt −−+−−=⎥⎥⎥⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢⎢⎢⎢ ⎣⎡ −−−− ⎥⎥⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢⎢⎢ ⎣⎡ −−−−−− +⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ −−+−− ∂∂=+∂∂= as before. ■ The Relationship between the Displacement and Velocity The velocity can be derived direc tly from the displacement 2.2.42: dtd dtd dtd u Xu xv =+==) (, (2.4.10) or () vuu uv grad+∂∂==t dtd (2.4.11) When the displacement field is given in material form one has dtdUV= (2.4.12) 2.4.3 Problems 1. The density of a material is given by xx⋅=−te2 ρ The velocity field is given by 2 1 3 1 3 2 3 2 1 2 ,2 ,2 x x v x x v x x v += −= += Determine the time derivative of the density (a) at a certain position x in space, and (b) of a material particle instantaneously occupying position x. Section 2.5 Solid Mechanics Part III Kelly 2432.5 Deformation Rates In this section, rates of change of the deformation tensors introduced earlier, F, C, E, etc., are evaluated, and special tensors used to measure deformation rates are discussed, for example the velocity gradient l, the rate of deformation d and the spin tensor w. 2.5.1 The Velocity Gradient The velocity gradient is used as a measure of the rate at which a material is deforming. Consider two fixed neighbouring points, x and x xd+ , Fig. 2.5.1. The velocities of the material particles at these points at any given time instant are )(xv and ) ( x xv d+ , and xxvxv x xv d d∂∂+=+ )( ) ( , The relative velocity between the points is xlxxvv d d d ≡∂∂= (2.5.1) with l defined to be the (spatial) velocity gradient, ji ijxvl∂∂= =∂∂= , grad vxvl Spatial Velocity Gradient (2.5.2) Figure 2.5.1: velocity gradient The spatial velocity gradient is commonly used in both solid and fluid mechanics. Less commonly used is the material velocity gradient, which is related to the rate of change of the deformation gradient: FXXx Xx X XXVV &=⎟ ⎠⎞⎜ ⎝⎛ ∂∂ ∂∂=⎟ ⎠⎞⎜ ⎝⎛ ∂∂ ∂∂=∂∂=),( ),( ),(Gradt t tt t (2.5.3) and use has been made of the fact that, since X and t are independent variables, material time derivatives and material gradients commute. •• xx xd+ ()xv()x xv d+vd xd Section 2.5 Solid Mechanics Part III Kelly 244 2.5.2 Material Derivatives of the Deformation Gradient The spatial velocity gradient may be written as xX Xx xXx X xX Xv xv ∂∂⎟ ⎠⎞⎜ ⎝⎛ ∂∂ ∂∂=∂∂⎟ ⎠⎞⎜ ⎝⎛ ∂∂ ∂∂=∂∂ ∂∂=∂∂ t t or 1−=FFl& so that the material derivative of F can be expressed as FlF=& Material Time Derivative of th e Deformation Gradient (2.5.4) Also, it can be shown that { ▲Problem 1} T T. T1. 1T. T − −− − −=−== Fl FlF FF F& (2.5.5) 2.5.3 The Rate of Defo rmation and Spin Tensors The velocity gradient can be decomposed into a symmetric tensor and a skew-symmetric tensor as follows (see §1.10.10): wdl+= (2.5.6) where d is the rate of deformation tensor (or rate of stretching tensor ) and w is the spin tensor (or rate of rotation , or vorticity tensor ), defined by () ()⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−∂∂= −=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂= += ij ji ijij ji ij xv xvwxv xvd 21,2121,21 TT ll wll d Rate of Deformation and Spin Tensors (2.5.7) The physical meaning of these tensors is next examined. The Rate of Deformation Consider first the rate of deformation tensor d and note that ()x v xl ddtdd d== (2.5.8) Section 2.5 Solid Mechanics Part III Kelly 245 The rate at which the square of the length of xd is changing is then () () ()() () xdx xlx x x xx xx x x d d dd ddtdd dddtdddtdddtdd ddtd 2 2 2, 2 22 = =⋅=⋅== (2.5.9) the last equality following from 2.5.6 and 1.10.31e. Dividing across by 22xd, then leads to ndnˆˆ=λλ& Rate of stretching per un it stretch in the direction nˆ (2.5.10) where Xxd d/=λ is the stretch and xx n dd/ ˆ= is a unit normal in the direction of xd. Thus the rate of deformation d gives the rate of stretching of line elements. The diagonal components of d, for example 1 1 11 dee=d , represent unit rates of extension in the coordinate directions. Note: • Eqn. 2.5.10 can also be derived as follows: let Nˆ be a unit normal in the direction of Xd, and nˆ be the corresponding unit normal in the direction of xd. Then XNFxn d d ˆ ˆ= , or NF n ˆ ˆ=λ . Differentiating gives NlFNF n n ˆ ˆ ˆ ˆ ==+&&&λλ or λλλ nl n n ˆ ˆ ˆ=+&&. Contracting both sides with nˆ leads to () nln nnnn ˆˆ /ˆˆˆˆ =⋅+⋅ λλ&&. But 0 )ˆˆ( 1ˆˆ =⋅→=⋅ dt d nn nn so, by the chain rule, 0ˆˆ=⋅nn& (confirming that a vector nˆ of constant length is orthogonal to a change in that vector nˆd), and the result follows Consider now the rate of change of the angle θ between two vectors )2( )1(,x x d d . Using 2.5.8 and 1.10.3d, () () () () )2( )1()2( )1( T)2( )1( )2( )1()2( )1( )2( )1( )2( )1( 2 xdxx xllxl x x xlx x x x x x d dd dd d d dddtdd d ddtdd ddtd =⋅+=⋅+⋅=⋅+⋅ =⋅ (2.5.11) which reduces to 2.5.9 when )2( )1(x x d d= . An alternative expression for this dot product is Section 2.5 Solid Mechanics Part III Kelly 246() () () () () )2( )1( )2()2( )1()1()2( )1( )1( )2( )2( )1( )2( )1( sin cos cossin cos cos cos x x xx xxx x x x x x x x d d dddtd dddtdd d d ddtdd ddtdd ddtd ⎟⎟⎟⎟ ⎠⎞ ⎜⎜⎜⎜ ⎝⎛ − + =− + = θθθ θθθθ θ θ && (2.5.12) Equating 2.5.11 and 2.5.12 leads to θθθλλ λλ&&& sin cos ˆˆ2 22 11 2 1 −⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+=ndn (2.5.13) where )( )(/i i i d d X x=λ is the stretch and )( )(/ ˆi i i d d x x n= is a unit normal in the direction of )(idx. It follows from 2.5.13 that the off-diagonal terms of the rate of deformation tensor represent shear rates : the rate of change of the right angle between line elements aligned with the coordinate directions. For example, taking the base vectors 1 1ˆn e=, 2 2ˆn e= , 2.5.13 reduces to 12 1221θ&−=d (2.5.14) where 12θ is the original right angle between the axes. The Spin Consider now the spin tensor w; since it is skew-symmetric, it can be written in terms of its axial vector ω (Eqn. 1.10.34), called the angular velocity vector : ve e eew ewewω curl2121 21 21 3 21 12 2 13 31 1 32 233 12 2 13 123 =⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−∂∂+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−∂∂+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−∂∂=−+−= xv xv xv xv xv xv (2.5.15) (The vector ω2 is called the vorticity (or spin ) vector .). Thus when d is zero, the motion consists of a rotation about some axis at angular velocity ω=ω (cf. the end of §1.10.11), with rωv×= , r measured from a point on the axis, and vrω wr =×= . On the other hand, when dl=, 0w=, one has oω=, and the motion is called irrotational . Section 2.5 Solid Mechanics Part III Kelly 247Example (Shear Flow) Consider a simple shear flow in which the velocity profile is “triangular” as shown in Fig. 2.5.2. This type of flow can be generated (at least approximately) in many fluids by confining the fluid between plates a distance h apart, and by sliding the upper plate over the lower one at constant velocity V . If the material particles adjacent to the upper plate have velocity 1eV, then the velocity field is 12e v xγ&= , where hV/=γ& . Then 2 1e el⊗=γ& and ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −= ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = 000000 0 21, 000000 0 21γγ γγ && && w d and, from 2.5.14, 12θγ&&−= , the rate of change of the angle shown in Fig. 2.5.2. Figure 2.4.2: shear flow The eigenvalues of d are 2 / ,0γλ&±= ( 0 det=d ) and the principal invariants, Eqn. 1.11.17, are 0 III, II,0 I2 41= −==d d d γ& . For 2 /γλ&+= , the eigenvector is []T0111=n and for 2/γλ&−= , it is []T0112−=n (for 0=λ it is 3e). (The eigenvalues and eigenvectors of w are complex.) Relative to the basis of eigenvectors, ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −= 0 0 002/ 00 0 2/γγ && d so at o45 there is an instantaneous pure rate of stretching/contraction of material. Choose the undeformed state to be the reference configuration, so 3 3 2 2 2 1 13 3 2 2 2 1 1 , ,, , x X x Xtx x XX x X xtX X x = = −== = += γγ && and 12 12 E e v X xγγ&&== . Then h1 2 1)(e v xv=V γ Section 2.5 Solid Mechanics Part III Kelly 248() ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ += ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = 0 0 00 10 1 , 0000100 12t tt t γγγ γ &&& & C F This is a simple shear with t kγ&= in Eqn. 2.2.40. ■ 2.5.4 Other Rates of Strain Tensors From 2.2.9, 2.2.22, () XEX XCX xx d d d d dddtd& &= =⋅21 21 (2.5.16) This can also be written in terms of spatial line elements: []x FEFx XEX d d d d1 T−−=& & (2.5.17) But from 2.5.9, these also equal xxddd , which leads to expressions for the material time derivatives of the right Cauchy-Green and Gree n-Lagrange strain tensors (also given here are expressions for the time derivatives of the left Cauchy-Green and Euler-Almansi tensors {▲Problem 3}) eleldebllbbdFFEdFF C −−=+=== TTTT2 &&&& (2.5.18) Note that ∫∫= E E d dt& so that the integral of the rate of Green-Lagrange strain is path independent and, in particular, the integral of E& around any closed loop (so that the final configuration is the same as the initial configuration) is zero. However, in general, the integral of the rate of deformation, dt∫d is not independent of the path – there is no universal function h such that dt d/h d= with ∫∫= h d d dt . Thus the integral dt∫d over a closed path may be non-zero, and hence the integral of the rate of deformation is not a good measure of the total strain. Section 2.5 Solid Mechanics Part III Kelly 249The Hencky Strain The Hencky strain is, Eqn. 2.2.37, ()∑=⊗ =3 1ˆ ˆ lni i i i n n h λ , where in are the principal spatial axes. Thus, if the principal spatial axes do not change with time, ()∑=⊗ =3 1ˆ ˆ/i i i i i n n h λλ& & . With the left stretch ∑=⊗ =3 1ˆ ˆ i i ii n n vλ , it follows that (and similarly for the corresponding material tensors), 1 1ln , ln−⋅ −⋅ =≡ =≡ vvv h UUU H & && & . For example, consider an extension in the coordinate directions, so ∑ ∑==⊗ =⊗ ===3 13 1ˆ ˆ ˆ ˆ ii i i i i ii N N n n vUF λ λ . The motion and velocity are ()sum no ,i ii i i i i i i x X x X xλλλ λ&&&== = so i i idλλ/&= (no sum), and hd&=. Further, dt∫=d h . Note that, as mentioned above, this expression does not hold in general, but does in this case of uniform extension. 2.5.5 Material Derivatives of Line, Area and Volume Elements The material derivative of a line element dt dd /)(x has been derived (defined) through 2.4.8. For area and volume elements, it is necessary first to evaluate the material derivative of the Jacobian determinant J. From the chain rule, one has (see Eqns 1.15.11, 1.15.7) () F F FFF & & & : : )(T−=∂∂= = JJJdtdJ (2.5.19) Hence {▲Problem 4} vvl div) grad(tr)(tr JJJ J ===& (2.5.20) Since wdl+= and 0 tr=w , it also follows that dtrJ J=& . As mentioned earlier, an isochoric motion is one for which the volume is constant – thus any of the following statements characterise the necessary and sufficient conditions for an isochoric motion: 0 : ,0 tr,0 div,0 ,1T= == ==−F F d v & &J J (2.5.21) Applying Nanson’s formula 2.2.59, the material derivative of an area vector element is {▲Problem 6} Section 2.5 Solid Mechanics Part III Kelly 250()()ds dsdtdnlv n ˆ div ˆT−= (2.5.22) Finally, from 2.2.53, the material time derivative of a volume element is ()() dv dVJ JdVdtddvdtdvdiv== = & (2.5.23) Example (Shear and Stretch) Consider a sample of material undergoing the following motion, Fig. 2.4.3. 3 32 22 1 1 X xX xXk X x ==+= λλ , 3 32 22 1 1 1 x Xx Xkx x X ==−= λ Figure 2.4.3: shear and stretch The deformation gradient and material strain tensors are () ()() ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −+ = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ += ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = 0 0 001 10 0 , 1 0 00 10 1 , 1000 00 12 2 21 2121 2 2k kk k kk k λλλ λ λλ λλ E C F , the Jacobian λ== FdetJ , and the spatial strain tensors are () ⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢ ⎣⎡ −−= ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡+ = 0 0 001 10 0 , 1 0 000 122 2 21 2121 2 22 22 λλλλλλ kkk kk k e b 1 1,xX2 2,xX δ k γλλk Section 2.5 Solid Mechanics Part III Kelly 251This deformation can also be expressed as a stretch followed by a simple shear: ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ =1000 0001 1000100 1 λk F The velocity is () ()() () ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡+ = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡+ == 0// , 022 22 xx kk XXk k dtdλλλλ λλλ &&& &&& vxV The velocity gradient is () ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡+ == 0 0 00 / 00 / 0 λλλλ &&&kk dd xvl and the rate of deformation and spin are ()[] ()[]()[] ()[] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ +−+ = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ++ = 0 0 00 0 /0 / 0 , 0 0 00 / /0 / 0 2121 2121 λλλλ λλλλλλ &&&& & &&&& kkkk kkkk w d Also ()() ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ++ ++ == 0 0 00 1 20 0 22λ λλλλλλ & & &&&& & k kk kkkk TdFF C As expected, from 2.5.20, ()λλλ&& & === / )(tr J J J d ■ 2.5.6 Problems 1. (a) Differentiate the relation 1−=FFI and use 2.5.4, FlF=& , to derive 2.5.5b, lF F1. 1 − −−= . Section 2.5 Solid Mechanics Part III Kelly 252(b) Differentiate the relation T T−= FFI and use 2.5.4, FlF=& , and 1.10.3e to derive 2.5.5c, T T. T − −−= Fl F . 2. For the velocity field 321 3 32 2 2 22 1 1 3 , 2 , xxx v xx v xx v = = = determine the rate of stretching per unit stretch at (2,0,1) in the direction of the unit vector ()5/3 42 1e e− And in the direction of 1e? 3. (a) Derive the relation 2.5.18a, dFF CT2=& directly from FFCT= (b) Use the definitions TFFb= and 2 /) (1−−= bIe to derive the relations 2.5.18c,d: elelde bllbb −−=+=T T,& & 4. Use 2.5.4, 2.5.19, 1.10.3h, 1.10.6, to derive 2.5.20. 5. For the motion 3 3 2 1 22 1 1 , , 3 tX xtX X xttX x =+=−= , verify that lFF=& . What is the ratio of the volume element currently occupying )1,1,1( to its volume in the undeformed configuration? And what is the rate of change of this volume element, per unit current volume? 6. Use Nanson’s formula 2.2.59, the product rule of differentiation, and 2.5.20, 2.5.5c, to derive the material time derivative of a vector area element, 2.5.22 (note that Nˆ, a unit normal in the undeformed configuration, is constant). Section 2.6 Solid Mechanics Part III Kelly 2532.6 Deformation Rates: Further Topics 2.6.1 Relationship between l, d, w and the rate of change of R and U Consider the polar decomposition RUF= . Since R is orthogonal, I RR=T, and a differentiation of this equation leads to T TRR RRΩR& &−=≡ (2.6.1) with RΩ skew-symmetric (see Eqn. 1.14.2). Us ing this relation, the expression 1−=FFl& , and the definitions of d and w, Eqn. 2.5.7, one finds that { ▲Problem 1} () [] () []T 1T 1 1T 1T 1 1T 1 sym21skew21 RUU RRUU UUR dΩ RUU RΩ RUU UUR wΩ RUURl RRR −− −−− −− =+ =+ =+ − =+ = && &&& && (2.6.2) Note that RΩ being skew-symmetric is consistent with w being skew-symmetric, and that both w and d involve R, and the rate of change of U. When the motion is a rigid body rotation, then 0U=& , and TRRΩwR&== (2.6.3) 2.6.2 Deformation Rate Tensors and the Principal Material and Spatial Bases The rate of change of the stretch tensor in te rms of the principal material base vectors is {}∑ =⊗+⊗+⊗ =3 1ˆ ˆ ˆ ˆ ˆ ˆ ii ii i ii i ii N N N N N N U&& & && λ λ λ (2.6.4) Consider the case when the principal material axes stay constant, as can happen in some simple deformations. In that case, U& and 1−U are coaxial (see §1.11.5): ∑ =⊗ =3 1ˆ ˆ ii ii N N Uλ&& and ∑ =−⊗ =3 11 ˆ ˆ1 ii i iN N Uλ (2.6.5) Section 2.6 Solid Mechanics Part III Kelly 254with UU UU & &1 1− −= and, as expected, from 2.5.25b, TRRΩwR&== , that is, any spin is due to rigid body rotation. Similarly, from 2.2.37, and differentiating I N N=⊗i iˆ ˆ , {}∑ =⊗+⊗+⊗ =3 12 21 2 21 ˆ ˆ ˆ ˆ ˆ ˆ ii i i i i i i iii N N N N N N E& & & & λ λ λλ . (2.6.6) Also, differentiating ij j iδ=⋅NNˆˆ leads to j i j i NN NN& & ˆˆ ˆˆ ⋅−=⋅ and so the expression m mim i WN N ˆ ˆ3 1∑ ==& (2.6.7) is valid provided ijW are the components of a skew-symmetric tensor, ji ij W W−= . This leads to an alternative expressi on for the Green-Lagrange tensor: () ∑∑ = ≠=⊗− +⊗ =3 12 23 1,21 ˆ ˆ ˆ ˆ in m n m nmnmmn i iii W N N N N E λλ λλ& & (2.6.8) Similarly, from 2.2.37, the left Cauchy-Green tensor can be expre ssed in terms of the principal spatial base vectors: { } ∑ ∑ = =⊗+⊗+⊗ =⊗=3 12 23 12ˆ ˆ ˆ ˆ ˆ ˆ 2 ,ˆ ˆ ii ii i ii i iii ii ii n n n n n n b n n b& & & & λ λ λλ λ (2.6.9) Then, from inspection of 2.5.18c, Tbllbb+=& , the velocity gradient can be expressed as {▲Problem 2} ∑ ∑ = = ⎭⎬⎫ ⎩⎨⎧⊗−⊗ = ⎭⎬⎫ ⎩⎨⎧⊗+⊗ =3 13 1ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ii i i i ii ii i i i iin nn n n n n n l&&&& λλ λλ (2.6.7) 2.6.3 Rates of Change and the Relative Deformation Just as the material time derivative of the deformation gradient is defined as ⎟ ⎠⎞⎜ ⎝⎛ ∂∂ ∂∂=∂∂=XxXF Fttt),(& one can define the material time derivative of the relative deformation gradient, cf. §2.3.2, the rate of change relative to the current configuration : t t tt=∂∂=τττ),( ),( xF xF& (2.6.8) Section 2.6 Solid Mechanics Part III Kelly 255 From 2.3.8, 1),(),( ),(−= tt XF XF xF ττ , so taking the derivative with respect to τ (t is now fixed) and setting t=τ gives 1),(),( ),(−= t t tt XFXF xF & & Then, from 2.5.4, ),(t txFl&= ( 2 . 6 . 9 ) as expected – the velocity gradient is the ra te of change of deformation relative to the current configuration. Further, using the polar decomposition, ),(),( ),( ττ τ xUxR xFt t t= Differentiating with respect to τ and setting t=τ then gives ),(),( ),(),( ),( t t t t tt t t t t xUxR xUxR xF & & & + = Relative to the current configuration, I xU xR == ),( ),( t tt t , so, from 2.4.34, ),( ),( t tt t xR xUl & &+= ( 2 . 6 . 1 0 ) With U symmetric and R skew-symmetric, ),( ),,( t tt t xR xU & & are, respectively, symmetric and skew-symmetric, and it follows that ),(),( tt tt xRwxUd && == (2.6.11) again, as expected – the rate of deformation is the instantaneous rate of stretching and the spin is the instantaneous rate of rotation. The Corotational Derivative The corotational derivative of a vector a is waaa−≡&o . Formally, it is defined through {} []{} []{} {} waaaw a aa w I aa R R aa R a a −=−−Δ+Δ=+Δ+−Δ+Δ=+Δ+−Δ+Δ=Δ+−Δ+Δ= →Δ→Δ→Δ→Δ &LL& )()( )() (1lim)() )( ) (1lim)() )( )( ) (1lim)() ( ) (1lim 0000o tt t t ttt tt t ttt tt t t tttt t t tt ttt tttt (2.6.12) Section 2.6 Solid Mechanics Part III Kelly 256 The definition shows that the corotati onal derivative involve s taking a vector a in the current configuration and rotati ng it with the rigid body rotati on part of the motion, Fig. 2.6.1. It is this new, rotated, vect or which is compared with the vector ) ( t tΔ+a , which has undergone rotation and stretch. Figure 2.6.1: rotation and stretch of a vector 2.6.4 Rivlin-Ericksen Tensors The n-th Rivlin-Ericksen tensor is defined as () L,2,1,0 , )( = = =nddt tt nn n τττC A (2.6.13) where ()τtC is the relative right Cauchy-Green strain. Since () I C==t tττ , I A=0 . To evaluate the next Rivlin-Ericksen tensor, one needs the derivatives of the relative deformation gradient; from 2.5.4, 2.3.8, ()[] () () () ττ ττ ττττt t t tdd ddFl FFl FF F = = =− − 1 1)()( )()( (2.6.14) Then, with 2.5.5a, ()()()()T T T/ ττττ l F Ft t d d = , and ()()()()() [ ] () ()() dl lF l l F A 2)( TT T 1 =+=+ == t ttt t t τττττ Thus the tensor 1A gives a measure of the rate of stre tching of material line elements (see Eqn. 2.5.10). Similarly, higher Rivlin-Erick sen tensors give a m easure of higher order stretch rates, λλ&&&&&, , and so on. )(ta)( ) ( t tttt t a F aΔ+==Δ+τ )(ttt t a RΔ+=τ Section 2.6 Solid Mechanics Part III Kelly 2572.6.5 The Directional Derivat ive and the Material Time Derivative The directional derivative of a function ) (tT in the direction of an increment in t is, by definition (see, for example, Eqn. 1.15.27), )( ) ( ][ t t t tt T T T −Δ+=Δ∂ (2.6.15) or tdtdtt Δ=Δ∂TT ][ (2.6.16) Setting 1=Δt , and using the chain rule 1.15.28, [][] []vTx TT T xx ∂=∂∂=∂= 1]1[ tt& (2.6.17) The material time derivative is thus equivalent to the directional derivative in the direction of the velocity vector. 2.6.6 Problems 1. Derive the relations 2.6.2. 2. Use 2.6.9 to verify 2.5.18, Tbllbb+=& . Section 2.7 Solid Mechanics Part III Kelly 2582.7 Small Strain Theory When the deformation is small, from 2.2.43-4, () u IFu IU IF gradgradGrad +≈+=+= (2.7.1) neglecting the product of ugrad with U Grad , since these are small quantities. Thus one can take u Ugrad Grad= and there is no dis tinction to be made between the undeformed and deformed configurations. The de formation gradient is of the form αIF+= , where α is small. 2.7.1 Decomposition of Strain Any second order tensor can be decomposed into its symmetric and antisymmetric part according to 1.10.28, so that ij ij ij ji ij ji ji xu xu xu xu xuΩ+=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−∂∂+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂=∂∂+=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ⎟ ⎠⎞⎜ ⎝⎛ ∂∂−∂∂+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ⎟ ⎠⎞⎜ ⎝⎛ ∂∂+∂∂=∂∂ ε21 2121 21T T Ωεxu xu xu xu xu (2.7.2) where ε is the small strain tensor 2.2.48 and Ω, the anti-symmetric part of the displacement gradient, is the small rotation tensor , so that F can be written as ΩεIF++= Small Strain Decomposition of the Deformation Gradient (2.7.3) It follows that (for the calculation of e, one can use the relation ()δIδI −≈+−1 for small δ) εeEεIbC ==+== 2 (2.7.4) Rotation Since Ω is antisymmetric, it can be wri tten in terms of an axial vector ω, cf. §1.10.11, so that for any vector a, 3 12 113 123 , e e e ω aωΩa Ω−Ω+Ω−= ×= (2.7.5) The relative displacement can now be written as Section 2.7 Solid Mechanics Part III Kelly 259() XωXεXu u d dd d ×+==grad (2.7.6) The component of relative displacement given by Xωd× is perpendicular to Xd, and so represents a pure rotation of the ma terial line element, Fig. 2.7.1. Figure 2.7.1: a pure rotation Principal Strains Since ε is symmetric, it must have three mutually orthogonal eigenvectors, the principal axes of strain , and three corresponding real eigenvalues, the principal strains, 3 2 1,, eee ), which can be positive or negative, cf. §1.11. The effect of ε is therefore to deform an elemental unit sphere into an elem ental ellipsoid, whose axes are the principal axes, and whose lengths are 3 2 1 1, 1, 1 e e e +++ . Material fibres in these principal directions are stretched only, in wh ich case the deformation is called a pure deformation ; fibres in other directions will be stretched and rotated. The term Xεd in 2.7.6 therefore corresponds to a pure stretch along the principal axes. The total deformation is the sum of a pure deformation, represented by ε, and a rigid body rotation, represented by Ω. This result is similar to that obtained for the exact finite strain theory, but here the decomposition is additive rather than multiplicative . Indeed, here the corresponding small strain stretch and rotation tensors are εIU+= and ΩIR+= , so that ΩεI RUF ++== ( 2 . 7 . 7 ) Example Consider the simple shear ( c.f. Eqn. 2.2.40) 3 3 2 2 2 1 1 , , X x X x kX X x = = += where k is small. The displacement vector is 12e ukx= so that ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = 0000000 0 gradk u xd Xdud Section 2.7 Solid Mechanics Part III Kelly 260 The deformation can be written as the additive decomposition XΩXεu d d d+= or XωXεu d d d ×+= with ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −= ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = 0 0 00 0 2/02/ 0 , 0 0 00 0 2/02/ 0 kk kk Ω ε and 3)2/( e ω k−= . For the rotation component, one can write ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −=+= 1 0 00 1 2/02/ 1 kk ΩIR which, since for small θ, θθθ ≈≈sin,1 cos , can be seen to be a rotation through an angle 2/k−=θ (a clockwise rotation). The principal values of ε are 0,2/k± with corresponding principal directions 2 1 1 )2/1( )2/1( e e n + = , 2 1 2 )2/1( )2/1( e e n + −= and 3 3e n=. Thus the simple shear with small displacemen ts consists of a ro tation through an angle 2/k superimposed upon a pure shear with angle 2/k, Fig. 2.6.2. Figure 2.6.2: simple shear ■ 2.7.2 Rotations and Small Strain Consider now a pure rotation about the 3X axis (within the exact finite strain theory), XRx d d= , with 2n1n θ+ = 2/k=θ Section 2.7 Solid Mechanics Part III Kelly 261 ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡− = 1 0 00 cos sin0 sin cos θθθθ R (2.7.8) This rotation does not change the length of line elements Xd. According to the small strain theory, however, ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −− = 0 0 001 cos 00 0 1 cos θθ ε , ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡− = 0 0 00 0 sin0 sin 0 θθ Ω which does predict line element length ch anges, but which can be neglected if θ is small. For example, if the rotation is of the order rad 102−, then 4 22 11 10−==εε . However, if the rotation is large, the e rrors will be appreciable; in that case, ri gid body rotation introduces geometrical non-linearities which must be dealt with using the finite deformation theory. Thus the small strain theory is restricted to not only the case of small displacement gradients, but also small rigid body rotations. 2.7.3 Volume Change An elemental cube with edges of unit length in the directions of the principal axes deforms into a cube with edges of lengths 3 2 1 1, 1, 1 e e e +++ , so the unit change in volume of the cube is () () () )2( 1 1 1 13 2 1 3 2 1 O e e e e e edVdV dv+++=−+++=− (2.7.9) Since second order quantities have already b een neglected in introducing the small strain tensor, they must be neglected here. Hence the increase in volume per unit volume, called the dilatation (or dilation ) is uεdiv tr3 2 1 ===++=iie e e eVVδ Dilatation (2.7.10) Since any elemental volume can be construc ted out of an infinite number of such elemental cubes, this result holds for a ny elemental volume irrespective of shape. 2.7.4 Rate of Deformation, St rain Rate and Spin Tensors Take now the expressions 2.4.7 for the rate of deformation and spin tensors. Replacing v in these expressions by u&, one has Section 2.7 Solid Mechanics Part III Kelly 262 () ()⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−∂∂= −=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂= += ij ji ijij ji ij xu xuwxu xud &&&& 21,2121,21 TT ll wll d (2.7.11) For small strains, one can take the time derivative outside (by considering the ix to be material coordinates independent of time): ⎪⎭⎪⎬⎫ ⎪⎩⎪⎨⎧ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−∂∂=⎪⎭⎪⎬⎫ ⎪⎩⎪⎨⎧ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂= ij ji ijij ji ij xu xu dtdwxu xu dtdd 2121 (2.7.12) The rate of deformation in this context is seen to be the rate of strain , εd&=, and the spin is seen to be the rate of rotation , Ωw&=. The instantaneous motion of a ma terial particle can hence be regarded as the sum of three effects: (i) a translation given by u& (so in the time interval tΔ the particle has been displaced by tΔu&) (ii) a pure deformation given by ε& (iii) a rigid body rotation given by Ω& 2.7.5 Compatibility Conditions Suppose that the strains ijε in a body are known. If the displacements are to be determined, then the strain-displacement partial differential equations ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂= ij ji ijxu xu 21ε (2.7.13) need to be integrated. However, there ar e six independent strain components but only three displacement components. This implies that the strains are not independent but are related in some way. The relatio ns between the strains are called compatibility conditions , and it can be shown that they are given by 0, , , , =−−+ikjm jmik ijkm kmij εεεε (2.7.14) These are 81 equations, but only six of them are distinct, and these six equations are necessary and sufficient to evaluate the displacement field. Section 2.8 Solid Mechanics Part III Kelly 2632.8 Objectivity and Objective Tensors 2.8.1 Dependence on Observer Consider a rectangular block of material resti ng on a circular table. A person stands and observes the material deform , Fig. 2.8.1a. The dashed lines indicate the undeformed material whereas the solid line indicates the current state. A second observer is standing just behind the first, but on a step ladder – th is observer sees the material as in 2.8.1b. A third observer is standing around the table, o45 from the first, and sees the material as in Fig. 2.8.1c. The deformation can be described by each observer using concepts like displacement, velocity, strain and so on.. However, it is clear that the three observers will in general record different values for these measures, since their perspectives differ. The goal in what follows is to determine which of the kinematical tensors are in fact independent of observer. Since the laws of physics describing the response of a deforming material must be independent of any observer, it is thes e particular tensors which will be more readily used in expr essions to describe material response. Figure 2.8.1: a deforming material as seen by different observers Note that Fig. 2.8.1 can be interpreted in another, equivalent, way. One can imagine one static observer, but this time with the material moved into three different positions. This viewpoint will be returned to in the next section. 2.8.2 Change of Reference Frame Consider two frames of reference , the first consisting of the origin o and the basis {}ie, the second consisting of the origin *o and the basis {}* ie, Fig. 2.8.2. A point x in space is then identified as having position vector iixe x= in the first frame and position vector ** * iixe x= in the second frame. When the origins o and *o coincide, *xx= and the vector components ix and * ix are related through Eqn. 1.5.3, * jij i xQx= , or ijij ii xQ x e e x*== , where []Q is the (a) (b) (c) Section 2.8 Solid Mechanics Part III Kelly 264transformation matrix 1.5.4, * j i ijQ ee⋅= . Alternatively, one has Eqn. 1.5.5, jji i xQ x=*, or * ** * ijji ii xQ x e e x== . Figure 2.8.2: two frames of reference With the shift in origin *ooa−= , one has ** * ** * ii ijji ii a xQ x e e e x +== (2.8.1) where ** iiae a= . Alternatively, ii ijij ii a xQ x e e e x −==* (2.8.2) where iiae a= , with jji i aQ a=*. Formulae 2.8.1-2 relate the coordinates of the position vector to a point in space as observed from one frame of reference to th e coordinates of the position vector to the same point as observed from a diffe rent frame of reference. Finally, consider the position vector x, which is defined relative to the frame ()ieo,. T o an observer in the frame ()* *,ieo , the same position vector would appear as ()*x, Fig. 2.8.3. Rotating this vector ()*x through TQ (the tensor which rotates the basis {}* ie into the basis {}ie) and adding the vector a then produces *x: ()a xQ x +=* T * (2.8.3) This relation will be discussed further below. x o*x 2e 1e ax * 1e•••* 2e *o• Section 2.8 Solid Mechanics Part III Kelly 265 Figure 2.8.3: Relation between vectors in Eqn. 2.8.3 2.8.3 Change of Observer The change of frame encompassed by Eqns . 2.8.1-2 is more precisely called a passive change of frame , and merely involves a transformati on between vector components. One would say that there is one observer but th at this observer is using two frames of reference. Here follows a different concept, an active change of frame , also called a change in observer , in which there are two observers , each with their own frame of reference. An observer is someone who can measure relative positions in space (with a ruler) and instants of time (with a clock). An event in the physical world (for example a material particle) is perceived by an observer as occurr ing at a particular point in space and at a particular time. One can regard an observer O to be a map of an event E in the physical world to a point x in point space ( cf. §1.2.5) and a real number t. A single event E is recorded as the pair ()t,x by an observer O and, in general, by a different pair ()*t,*x by a second observer *O, Fig. 2.8.4. Figure 2.8.4: recordings by two observers of the same event Let the two observers record th ree points corresponding to three events, Fig. 2.8.5. These points define vectors in space, as the difference between the points ( cf. §1.2.5). It is assumed that both observers “see” the same Eu clidean geometry, that is, if one observer sees an ellipse, then the other observer will see the same ellipse, but perhaps positioned differently in space. To ensure that this is so, observed vectors must be related through some orthogonal tensor Q, for example, ()0* 0*xxQ x x −=− (2.8.4) x *x O*O•E •t *to*x 2e 1e ax * 1e•••* 2e *o• ()*x TQa ()* TxQ Section 2.8 Solid Mechanics Part III Kelly 266 since this transformation will automatically pr eserve distances between points, and angles between vectors (see §1.10.7), for example, ()()()()()()0 0 1 0 0 1* 0* * 0* 1 xx xx xxQxxQ x x x x −⋅−=−⋅−=−⋅− (2.8.5) Figure 2.8.5: recordings of two observers of three separate events Although all orthogonal tensors Q do indeed preserve length and angles, it is taken that the Q in 2.8.4-5 is proper orthogona l, i.e. a rotation tensor ( cf. §1.10.8), so that orientation is also preserved. Further, it is assumed that ) (tQQ= , which expresses the fact that the observers can move relative to each other over time. Observers must also agree on time intervals between events. Let an observer O record a certain event at time t and a second observer *O record the same event as occurring at time *t. Then the times must be related through α+=tt* Observer Time Transformation (2.8.6) where α is a constant . If now the observers record a second event as occurring at 1t and * 1t say, one has tttt −=−1* * 1 as required. The observer transformation 2.8.4 involves the vectors 0xx−and * 0*x x− and as such does not require the notion of origin or coor dinate system; it is an abstract symbolic notation for an observer transfor mation. However, an origin o for O and *o for *O can be introduced and then the points * * 0 0 ,,, xxxx can be regarded as position vectors in space, Fig. 2.8.6. The transformation 2.8.4 can now be expressed in the oft-used format xQ c x )( )(*t t+= Observer (Spatial) Transformation (2.8.7) where 0* 0 )( )( xQ x c t t−= (2.8.8) The transformation 2.8.7 is called a Euclidean transformation , since it preserves the Euclidean geometry. 0xx 0xx− * 0*xx−*x* 0x O*O1x 0 1xx− * 0* 1xx− * 1x••• • •• Section 2.8 Solid Mechanics Part III Kelly 267 Figure 2.8.6: position vectors for two observers of the same events Coordinate Systems Each observer can introduce any Cartesian coordinate system, with basis vectors {}ie and {}* ie say. They can then resolve the position v ectors into vector com ponents. These basis vectors can be oriented with re spect to each other in any way, that is, they will be related through i iRe e=*, where R is any rotation tensor. Indeed, each observer can change their basis, effecting a coordinate transformation. No attempt to introduce specific coordinate systems will be made here since they are co mpletely unnecessary to the notion of observer transformation and would only greatly confuse the issue. Relationship to Passive Change of Frame Recall the passive change of frame encompassed in Eqns. 2.8.1-2. If one substitutes the actual x for ()*x in Eqn. 2.8.3, one has: axQ x+=T * (2.8.9) This is clearly an observer transformation, relating the position v ector as seen by one observer to the position vector as seen by a second observer, through an orthogonal tensor and a vector, as in Eqn. 2.8.7. In the passive change of frame, ijQ are the components of the orthogonal tensor i ie eQ⊗=*, Eqn. 1.10.25, which maps the bases onto each other: i iQe e=*. Thus the transformation 2.8.1-2 can be defined uniquely by the pair Q and a. In that sense, the passive change of frame doe s indeed define an active change of frame, i.e. a change of observer, through Eqn. 2.8.9. However, the concept of observer discussed above is the preferred way of de fining an observer transformation. 2.8.4 Objective Vectors and Tensors The observer transformation 2.8. 7 encapsulates the different vi ewpoints observers have of the physical world. They will see the same objects, but in general they will see these objects oriented differently and located at di fferent positions. The goal now is to see o0xx0xx− *o* 0*xx− *x* 0x O*O Section 2.8 Solid Mechanics Part III Kelly 268which of the kinematical tensors are independent of these different viewpoints. As a first step, next is introduced the concept of an objective tensor . Suppose that different observers are examining a deforming material. In order to describe the material, the observers take measurements. This will involve measurements of spatial objects associated with the current configuration, for example the velocity or spin. It will also involve material objects associated with the refere nce configuration, for example line elements in that configura tion. It will also involve two-point tensors such as the rotation or deformation gradient, which are associ ated with both the current and reference configurations. It is assumed that all observer s observe the reference configurat ion to be the same, that is, they record the same set of points for the material particles in the reference configuration 1. The observers then move relative to each other and their measurements of objects associated with the current configuration will in general differ. One would expect (want) different observers to make the same measurem ent of material objects despite this relative movement; thus one says that ma terial vectors and tensors are objective (material) vectors and objective (material) tensors if they remain unchanged under the observer transformation 2.8.6-7. A spatial vector u on the other hand is said to be an objective (spatial) vector if it satisfies the observer tr ansformation (see 2.8.4):2 Qu u=* Objectivity Requirement for a Spatial Vector (2.8.10) for all rotation tensors Q. An objective (spatial) tensor is defined to be one which transforms an objective vector into an object ive vector. Consider a tensor observed as Tand *T by two different observers. Take an objective vector which is observed as v and *v, and let Tvu= and ** *vT u= . Then, for u to be objective, *T *v QTQ QTv Qu u === (2.8.11) and so the tensor is objective provided T *QTQ T= Objectivity Requirement for a Spatial Tensor (2.8.12) Various identities can be derived; for example, for objective vectors a and b, and objective tensors A and B, {▲Problem 1} 1 this does not affect the generality of what follows; the notion of objective tensor is independent of the chosen reference configuration 2 the time transformation 2.8.6 is trivial and do es not affect the relations to be derived Section 2.8 Solid Mechanics Part III Kelly 269() ()()()() ()( ) () ()* * *** *1**1** *** ** * ** * ** * * : : BA BABA ABA ABA ABbA Abba bab a bab a ba =====⋅=⋅⊗=⊗+=+ −− (2.8.13) For a scalar, φφ=* Objectivity Requirement for a Scalar (2.8.14) In other words, an objective scalar is one which has the same value to all observers. Finally, consider a two-point tens or. Such a tensor is said to be objective if it maps an objective material vector into an objective sp atial vector. Consider then a two-point tensor observed as Tand *T. Take an objective material vector which is observed as v and *v, and let Tvu= and ** *vT u= . A material vector is objective if it is unaffected by an observer transformation, so * *QTv QTv Qu u === (2.8.15) and so the tensor is objective provided QT T=* Objectivity Requirement for a Two-point Tensor (2.8.16) Thus the objectivity requirement for a two-point tensor is the same as that for a spatial vector. 2.8.5 Objective Kinematics Next are examined the various kinematic v ectors and tensors introduced in the earlier sections, and their objectivit y status is determined. The motion is observed by one observer as ),(tXχx= and by a second observer as ),(* *tXχ x= . The observer transformation gives )(),()( ),(* *t t t t c XχQ Xχ + = , α+=tt* (2.8.17) and so the motion is not an objective vector, i.e. Qχχ≠*. Section 2.8 Solid Mechanics Part III Kelly 270The Velocity and Acceleration Differentiating 2.8.17 (and using the notation x& instead of ()t,Xχ& for brevity), the velocity under the obser ver transformation is cxQxQ x &&&& ++=* (2.8.18) which does not comply with the objectivity re quirement for spatial vectors, 2.8.10. In other words, different observers will measure different magnitudes for the velocity. The velocity expression can be put in a form sim ilar to that of elementary mechanics (the “non-objective” terms are on the right), ()ccxΩxQxQ& && +−=−* * (2.8.19) where TQQΩQ&= (2.8.20) is skew-symmetric (see Eqn. 1.14.2); this tens or represents the rigi d body angular velocity between the observers (see Eqn. 2.6.1). Note that the velocity is objective provided oc0Q==&& ,, f o r w h i c h 0 0*cxQ x+= , which is called a time-independent rigid transformation . Similarly, for the accelera tion, it can be shown that ()()()ccxΩ cxΩcxΩxQxQ Q Q&&&& &&&&& +−+−−−=− 2* 2 * * (2.8.21) The first three terms on the ri ght-hand side are called the Euler acceleration , the centrifugal acceleration and the Coriolis acceleration respectively. The acceleration is objective provided c& and Q are constant, for which ) (0*tcxQ x+= with oc=&& , which is called a Galilean transformation – where the two configura tions are related by a rigid rotation and a translational motion with constant velocity. The Deformation Gradient Consider the motion ),(tXχx= . As mentioned, observers observe the reference configuration to be the same: X X=*. The deformation is then observed as XFx d d= and XF x d d* *= , so that X QFX QFxQ x d d d d ===* (2.8.22) and QF F=* (2.8.23) and so, according to 2.8.16, the deformation gradient is objective. Section 2.8 Solid Mechanics Part III Kelly 271The Cauchy-Green Strain Tensors For the right and left Cauchy-Green tensors, T T T *T* *T T * T* * QbQ Q QFF FF bC QFQF FF C = === == (2.8.24) Thus the material tensor C and the spatial tensor b are objective3. The Jacobian Determinant For the Jacobian determinant, using 1.10.16a, () J J == = == F F Q QF F det det det det det* * (2.8.25) and4 so is objective according to 2.8.14. The Rotation and Stretch Tensors The polar decomposition is RUF= , where R is the orthogonal rotation tensor and U is the right stretch tensor. Then * * *UR QRU QF F ≡== . Since QR is orthogonal, the expression * *UR QRU= is valid provided U U QR R = =* *, ( 2 . 8 . 2 6 ) Thus the two-point tensor R and the material tensor U are objective. The Velocity Gradient Allowing Q to be a function of time, for the velo city gradient, using 2.5.4, 1.9.18c, QΩ QlQ QFFQFQ FFl += += =− −⋅ T T 1 1 * * *) ( )( && (2.8.27) where QΩ is the angular velocity tensor 2.8.20. On the other hand, with wdl+= , and separating out the symmetric and skew-symmetric parts, QΩ QwQ w QdQ d += =T * T *, (2.8.28) Thus the velocity gradient is not objective. Th is is not surprising given that the velocity is not objective. However, signifi cantly, the rate of deformation, a measure of the rate of stretching of material, is objective. 3 Some authors define a second order tensor to be objective only if 2.8.12 is satisfied, regardless of whether it is spatial, two-point or material; with this definition, F and C would be defined as non-objective 4 Note that Q must be a rotation tensor, not just an orthogonal tensor, here Section 2.8 Solid Mechanics Part III Kelly 272The Spatial Gradient Consider the spatial gradient of an objective vector t: xtt∂∂= grad , ()** *gradxtt∂∂= (2.8.29) Since Qtt=*, the chain rule gives () xtQxQt xx xt xt ∂∂=∂∂≡∂∂ ∂∂=∂∂* ** * (2.8.30) It follows that ()T *grad QxtQ t∂∂= (2.8.31) Thus the spatial gradient is objective. In gene ral, it can be shown that the spatial gradient of a tensor field of order n is objective, for example the gradient of a scalar φ, {▲Problem 2} φgrad . Further, for a vector v, {▲Problem 3} vdiv is objective. Objective Rates Consider an objective vector field u. The material derivative u& is not objective. However, the co-rotational derivative, Eqn. 2.6.12, wuuu−=&o is objective. To show this, contract 2.8.28b, T T *QQ QwQ w &+= , to the right with Q to get an expression for Q&: QwQwQ−=*& (2.8.32) and then o & && uQ Quw wuuQ QuwuQuQ u Qu u +=−+=+=→=⋅ * * * *) ( (2.8.33) Then o uQ uwu =−⋅ ** *, or o o uQ u=*)( , so that the co-rotational derivative of a vector is an objective vector. Rates of spatial tensors can also be modified in order to construct objective rates. For example, consider an objective spatial tensor T, so T *QTQ T= . Then T T T *QQT TQQ QTQ T & && ++=⋅ (2.8.34) which is clearly not objective. However, this can be re-arranged using 2.8.32 into Section 2.8 Solid Mechanics Part III Kelly 273()T * * ** *QTw wTTQ wT Tw T +−=+−⋅ & (2.8.35) and so the quantity Tw wTT+−& (2.8.36) is an objective rate, called the Jaumann rate . Other objective rates of tensors can be constructed in a similar fashion, for example the Cotter-Rivlin rate , defined by {▲Problem 4} TlTlT++T& (2.8.37) Summary of Objective Kinematic Objects Table 2.8.1 summarises the objectivity of some important kinematic objects: objective definition Type Transformation Jacobian determinant 9 Scalar J J=* Deformation gradient 9 2-point QF F=* Rotation 9 Fv FUR1 1−−== 2-point QR R=* Right Cauchy-Green strain 9 FFCT= Material C C=* Green-Lagrange strain 9 ()IC E−=21 Material E E=* Rate of Green- Lagrange strain 9 Material E E&=⋅ * Right Stretch 9 C U= Material U U=* Left Cauchy-Green strain 9 TFFb= Spatial T *QbQ b= Euler-Almansi strain 9 ()1 21 −−= bI e Spatial T *QeQ e= Left Stretch 9 b v= Spatial T *QvQ v= Spatial Velocity Gradient × v lgrad= Spatial T T *QQ QQll &+= Rate of Deformation 9 ()T 21ll d+= Spatial T *QdQ d= Spin × ()T 21ll w−= Spatial T T *QQ QQw w &+= Table 2.8.1: Objective kinematic objects 2.8.6 Objective Functions In a similar way, functions are de fined to be objective as follows: • A scalar-valued function φ of, for example, a tensor A, is objective if it transforms in the same way as an objective scalar, Section 2.8 Solid Mechanics Part III Kelly 274()()A Aφφ=* (2.8.38) • A (spatial) vector-valued function a of a tensor A is objective if it transforms in the same way as an objective vector )( )(*AQv Av= (2.8.39) • A (spatial) tensor-valued function f of a tensor A is objective if it transforms according to T *)( )( QAQf Af= (2.8.40) Objective functions of the Deformation Gradient Consider an objective scalar-valued function φ of the deformation gradient F, )(Fφ . The function is objective if )(*Fφφ= . But also, ()()QF Fφφφ ==* * (2.8.41) Using the polar deco mposition theorem, ()() QRU RUφφ= . Choosing the particular rigid-body rotation TRQ= then leads to ()()U RUφφ= (2.8.42) which leads to the reduced form ()()U Fφφ= (2.8.43) Thus for the scalar function φ to be objective, it must be independent of the rotational part of F, and depends only on the stretching part ; it cannot be a function of the nine independent components of the deformation gr adient, but only of the six independent components of the right stretch tensor. Consider next an objective (s patial) tensor-valued function f of the deformation gradient F, )(Ff . According to the definition of object ivity of a second order tensor, 2.8.12: ()T *QFQf f= (2.8.44) But also, ()()QFf Ff f ==* * (2.8.45) Again, using the polar decomposition theo rem and choosing the particular rigid-body rotation TRQ= leads to ()()RRUfR UfT= (2.8.46) Section 2.8 Solid Mechanics Part III Kelly 275 which leads to the reduced form ()()TRURf Ff= (2.8.47) Thus for f to be objective, its dependence on F must be through an arbitrary function of U together with a more explicit dependence on R, the rotation tensor Example Consider the tensor function ()2T)( FF Ffα= . Then [] []()T T2T2T) )(( )( QFQf Q FFQ QF QF QFf = = = α α and so the objectivity requirement is satisfi ed. According to the above, then, one can evaluate () ( ) ()2T TUU RRUfR Uf α= = , and the reduced form is ()T 4 T2TR RU R UURf α α = = Also, since 2UC= and () IC E−=21, alternative reduced forms are () ()T 3T 2 , RERff RCRff = = ■ Finally, consider a spatial tensor function f of a material tensor T. Then )()()( ,)( )(* * T *Tf Tf Tf QTQf Tf == = (2.8.48) It follows that TQfQf= (2.8.49) This is true only in the special case IQ= and so is not true in general. It follows that the function f is not objective. 2.8.7 Problems 1. Derive the relations 2.8.13 2. Show that the spatial gradient of a scalar φ is objective. 3. Show that the divergence of a spatial vector v is objective. [Hint: use the definition 1.11.9 and identity 1.9.10e] 4. Verify that the Rivlin-Cotter rate of a tensor T, TlTlT++T, is objective. Section 2.9 Solid Mechanics Part III Kelly 2762.9 Rigid Body Rotations of Configurations In this section are discussed rigid body rotations to the current and reference configurations. 2.9.1 A Rigid Body Rotation of the Current Configuration As mentioned in §2.8.1, the circumstance of tw o observers, moving relative to each other and examining a fixed configuration (the curre nt configuration) is equivalent to one observer taking measurements of two different configurations, moving relative to each other1. The objectivity requirements of the va rious kinematic object s discussed in the previous section can thus also be examined by consid ering rigid body rotations and translations of the current configuration. Any rigid body rotation and tran slation of the current confi guration can be expressed in the form ()() )( , )( ,*t t t t c XxQ Xx + = (2.9.1) where Q is a rotation tensor. This is illustrate d in Fig. 2.9.5. The current configuration is denoted by S and the rotated configuration by *S. Just as XFx d d= , the deformation gradient for the configuration *S relative to the reference configuration 0S is defined through XF x d d* *= . From 2.9.1, as in §2.8.5 (see Eqn. 2.8.23), and similarly for the right and left Cauchy-Green tensors, T T** ** T* ** QbQ FF bC FF CQF F ===== (2.9.2) Thus in the deformations S S→0:F and * 0*: S S→ F , the right Cauchy Green tensors, C and *C, are the same, but the left Cauchy Gr een tensors are different, and related through T *QbQ b= . All the other results obtained in the last sec tion in the context of observer transformations, for example for the Jacobian, stretch tensors, etc., hold also for the case of rotations to the current configuration. 1 Although equivalent, there is a difference: in one, th ere are two observers who record one event (a material particle say) as at two different points, in the other there is one observer who records two different events (the place where the one material particle is in two different configurations) Section 2.9 Solid Mechanics Part III Kelly 277 Figure 2.9.1: a rigid body rotation and translation of the current configuration 2.9.2 A Rigid Body Rotation of the Reference Configuration Consider now a rigid-body rotation to the reference configuration. Such rotations play an important role in the notion of ma terial symmetry (see Chapter 5). The reference configuration is denoted by 0S and the rotated/transl ated configuration by ◊S, Fig. 2.9.2. The deformation grad ient for the curre nt configuration S relative to ◊S is defined through XQF XFx d d d◊◊◊== . But XFx d d= and so (and similarly for the right and left Cauchy-Green tensors) b FF bQCQ FF CFQ F ===== ◊◊◊◊◊◊◊ TT TT (2.9.3) Thus the change to the right (left) Cauchy-Green strain tensor under a rotation to the reference configuration is the same as the chan ge to the left (right) Cauchy-Green strain tensor under a rotation of the current configuration. X reference configuration *S*x *FF x 0SSxd *xd cQXd Section 2.9 Solid Mechanics Part III Kelly 278 Figure 2.9.2: a rigid body rotation of the reference configuration XQ reference configuration S◊S ◊X◊F Fx0STQ S Section 2.10 Solid Mechanics Part III Kelly 2792.10 Convected Coordinates In this section, the deformation and strain tensors described in §2.2-3 are now described using convected coordinates (see §1.16). Note that all the tensor relations expressed in symbolic notation already discussed, such as C U= , ii i n NFλ=ˆ , lFF=& , are independent of coordinate system, and hold al so for convected coordinates. 2.10.1 Convected Coordinates Introduce the curvilinear coordinates iΘ. The material coordinates can then be written as ),,(3 2 1ΘΘΘ=XX (2.10.1) so iiXE X= and ii iid dX d G E X Θ== , (2.10.2) where iG are the covariant base vectors in the re ference configuration, with corresponding contravariant base vectors iG, Fig. 2.10.1, with i j jiδ=⋅GG (2.10.3) Figure 2.10.1: Curvilinear Coordinates The coordinate curves, curves of constant iΘ, form a net in the undeformed configuration. One says that the curvilinear coordinates are convected or embedded , that is, the coordinate curves are attached to material particles and deform with the body, so that each material 1 1,xX2 2,xX 3 3,xXX 1 1,eE2 2,eE1g2gcurrent configuration reference configuration 1G2G x Section 2.10 Solid Mechanics Part III Kelly 280particle has the same values of the coordinates iΘ in both the reference and current configurations. In the current configuration, the spatial coor dinates can be expressed in terms of a new, “current”, set of curvilinear coordinates ),,,(3 2 1tΘΘΘ=xx , (2.10.4) with corresponding covariant base vectors ig and contravariant base vectors ig, with ii iid dx d g e x Θ== , (2.10.5) Example Consider a motion whereby a cube of material, with sides of length 0L, is transformed into a cylinder of radius R and height H, Fig. 2.10.2. Figure 2.10.2: a cube deformed into a cylinder A plane view of one quarter of the cube and cylinder are shown in Fig. 2.10.3. Figure 2.10.3: a cube deformed into a cylinder 0LR 0LH 1X2X 1x2x 0L RXx• • P p Section 2.10 Solid Mechanics Part III Kelly 281 The motion and inverse motion are given by )(Xχx= , () ()() ()() 3 0322212 1 02222121 01 22 XLHxX XXX LRxX XX LRx =+=+= (basis: ie) and )(1xχX−= , () () () () 3 0 32221 12 0 22221 0 1 22 xHLXx xxx RLXx xRLX =+ =+ = (basis: iE) Introducing a set of convected coordinates, Fig. 2.10.4, the material and spatial coordinates are ),,( 3 2 1ΘΘΘ=XX , 3 0 32 1 0 21 0 1 tan22 Θ=ΘΘ⎟ ⎠⎞⎜ ⎝⎛=Θ⎟ ⎠⎞⎜ ⎝⎛= HLXRLXRLX and (these are simply cylindrical coordinates) ),,(3 2 1ΘΘΘ=xx , 3 32 1 22 1 1 sincos Θ=ΘΘ=ΘΘ= xxx A typical material particle (denoted by p) is shown in Fig. 2.10.4. Note that the position vectors for p have the same iΘ values, since they represent the same material particle. Section 2.10 Solid Mechanics Part III Kelly 282 Figure 2.10.4: curvilinear coordinate curves ■ 2.10.2 The Deformation Gradient With convected curvilinear coordinate s, the deformation gradient is i iG gF⊗= , (2.10.6) which is consistent with () XF GG g g x d d d dji ij jj=⊗Θ=Θ= (2.10.7) The deformation gradient F, the transpose TF and the inverses T 1,−−F F , map the base vectors in one configuration onto the base vectors in the other configuration: iiiii ii i g G FG g Fg G FG gF ⊗=⊗=⊗=⊗= −− TT1 Æ i ii ii ii i G gFg GFG gFg FG ==== −− TT1 Deformation Gradient (2.10.8) Thus the tensors F and 1−F map the covariant base vectors into each other, whereas the tensors T−F and TF map the contravariant base vectors into each other, as illustrated in Fig. 2.10.5. 1X2X 1x2x 1Θ2Θ 1Θ2Θ R=Θ142π=Θ • • p p Section 2.10 Solid Mechanics Part III Kelly 283 Figure 2.10.5: the deformation gradie nt, its transpose and the inverses Components of F F has different components with re spect to the different bases: j ii j jij i j iij j i ijj ii j ji j i j iij j i ij f f f fF F F F g g g g g g g gG G G G G G G G F ⊗=⊗=⊗=⊗=⊗=⊗=⊗=⊗= ⋅ ⋅⋅⋅ ⋅⋅ () () () () () () () ()j ii j jij i j iijj i ijj ii j jij i j iijj i ij f f f fF F F F g g g g g g g gG G G G G G G G F ⊗ =⊗ =⊗ =⊗ =⊗ =⊗ =⊗ =⊗ = ⋅ ⋅−⋅− − −⋅ ⋅−⋅− − − − 1 1 1 11 1 1 1 1 () () () () () () () ()j ii j jij i j iijj i ijj ii j jij i j iijj i ij f f f fF F F F g g g g g g g gG G G G G G G G F ⊗ =⊗ =⊗ =⊗ =⊗ =⊗ =⊗ =⊗ = ⋅ ⋅⋅⋅ ⋅⋅ T T T TT T T T T () () () () () () () ()j ii j jij i j iijj i ijj ii j jij i j iijj i ij f f f fF F F F g g g g g g g gG G G G G G G G F ⊗ =⊗ =⊗ =⊗ =⊗ =⊗ =⊗ =⊗ = ⋅ ⋅−⋅− − −⋅ ⋅−⋅− − − − T T T TT T T T T (2.10.9) The components of F with respect to the reference bases {}{}i iG G, are 1G2G 1G2G 1g2g 1g2g covariant basis contravariant basis T−F F 1−FTF Section 2.10 Solid Mechanics Part III Kelly 284jm mi ji ji i jk ijk j ij iki jk j i ijjm im j i j i ij x XFG FG Fx XF Θ∂∂ ∂Θ∂=⋅= =⋅= =⋅= =Θ∂∂ Θ∂∂=⋅= = ⋅⋅ gG FGGgG FGGgG FGGgG FGG (2.10.10) and similarly for the components with respect to the current bases. Components of the Base Vectors in different Bases Now ()() mm im mij imm jj im mjij mm j ij m mj i i F FF FF F G GG GGG G GG G FG g ⋅⋅⋅ = == =⊗=⊗== δ δ (2.10.11) showing that some of the components of the deformation gradient can be viewed also as components of the base vectors. Similarly, ()()mm im mi i i f f g g gF G⋅− − −= ==1 1 1 (2.10.12) For the contravariant base vectors, one has ()()()() () () () ()mi m mmii jmj mi j mmji jmj mi j mmji i F FF FF F G GG GGG G GG G GF g ⋅− −⋅− −⋅− − − = == =⊗ =⊗ == T TT TT T T δ δ (2.10.13) and ()()mi m mmii if f g g gF G⋅= ==T T T (2.10.14) 2.10.3 Reduction to Materi al and Spatial Coordinates Material Coordinates Suppose that the material coordinates iX with Cartesian basis are used (rather than the convected coordinates with curvilinear basis iG), Fig. 2.10.6. Then Section 2.10 Solid Mechanics Part III Kelly 285j ji j ji ij ij j ij i i j ji j ji ii j ij j ij ii i xX xXx x XX XXX X X e e ge e g E E E GE E E G ∂∂= ∂Θ∂=∂∂= Θ∂∂= = ∂∂= ∂Θ∂== ∂∂= Θ∂∂= →Θ , , (2.10.15) and X e E g E g G Fx E e E g G gF gradGrad 1=⊗∂∂=⊗=⊗==⊗∂∂=⊗=⊗= − j i ji i ii ii j ij i ii i xXXx (2.10.16) which are Eqns. 2.2.2, 2.2.4. Thus x Grad is the notation for F to be used when the material coordinates iX are used to describe the deformation. Figure 2.10.6: Material coordinates and deformed basis Spatial Coordinates Similarly, when the spatial coordinates ix are to be used as independent variables, then i j ji j ji ii j ij j ij i j ji j ji ij ij j ij ii i xx xxx x Xx XxX X x e e e ge e e g E E GE E G = ∂∂= ∂Θ∂== ∂∂= Θ∂∂= ∂∂= ∂Θ∂=∂∂= Θ∂∂= →Θ , , (2.10.17) and 1X2X 3XX1E2E 1g2gcurrent configuration reference configuration Section 2.10 Solid Mechanics Part III Kelly 286X e E e G g G Fx E e G e G gF gradGrad 1=⊗∂∂=⊗=⊗==⊗∂∂=⊗=⊗= − i j ij i ii ij iji i ii i xXXx (2.10.18) The descriptions are illustrated in Fig. 2.10.7. Note that the base vectors iG, ig are not the same in each of these cases (curvilinear, material and spatial). Figure 2.10.7: deformation described using different independent variables 1X2X 1x2x 1X2X 1x2x 1X2X 1x2x1G2G 1g2g 1E2E 1g2g 2G 1G1e2ei iG gF⊗= x E e F Grad=⊗∂∂=j iji Xxi ig G F⊗=−1 X e E F grad1=⊗∂∂=− j i ji xX Section 2.10 Solid Mechanics Part III Kelly 2872.10.4 Strain Tensors The Cauchy-Green tensors The right Cauchy-Green tensor C and the left Cauchy-Green tensor b are defined by Eqns. 2.2.10, 2.2.13, ()() () () ( ) () ( ) () () ( )j i ijj i ijj j iij iij j iij jj i ij iij j iij jj i ij i ijj i ijj j ii b Gb GC gC g g g g g g GG g FF bg g g g g GG g FFbG G G G G gg G FF CG G G G G gg G FFC ⊗ ≡⊗=⊗ ⊗==⊗≡⊗=⊗ ⊗==⊗ ≡⊗=⊗⊗==⊗≡⊗=⊗⊗== − −− −− −− − 1 1 T 1T1 T 1 1T (2.10.19) Thus the covariant components of the right Cauchy-Green tensor are the metric coefficients ijg, the covariant components of the identity tensor with respect to the convected bases in the current configuration, j i ijg g g gI ⊗=≡ . It is possible to evaluate other components of C, e.g. ijC, and also its components with respect to the current basis through 2.10.14, but only the components ijC with respect to the reference basis will be used in the analysis. Similarly, for 1−b, the components ()ijb1− with respect to the current configuration will be used. The Stretch Now, analogous to 2.2.9, 2.2.12, x xb XXX XC xx d d dd dSd d dd ds 1 22 −=⋅==⋅= (2.10.20) so that the stretches are, analogous to 2.2.17, ()j ijij iji xd bxd d ddd dd dsdSXdCXd d ddd dd dSds ˆ ˆ ˆ ˆ1ˆ ˆ ˆ ˆ 1 1 1 22 222 2 − − −→ = ==→ = == xbxxxbxxXCXXXCXX λλ (2.10.21) The Green-Lagrange and Euler-Almansi Tensors The Green-Lagrange strain tensor E and the Euler-Almansi strain tensor e are defined through 2.2.22, 2.2.24, Section 2.10 Solid Mechanics Part III Kelly 288() () xxe x bIxXXE XIC X dd d ddS dsdd d ddS ds ≡−=−≡− =− −12 22 2 21 221 2 (2.10.22) The components of E and e can be evaluate d through (writing I G≡, the identity tensor expressed in terms of the base vector s in the reference configuration, and Ig≡, the identity tensor expressed in terms of the base vectors in the current configuration) ()() () ()() ()j i ijj i ij ijj i ijj i ijj i ijj i ij ijj i ijj i ij e G g G gE G g G g g g g g g g g g bg eG G G G G G G G GC E ⊗≡⊗−=⊗−⊗=−=⊗≡⊗−=⊗−⊗ =−= − 21 21 2121 21 21 1 (2.10.23) Note that the components of E and e with respect to their bases are equal, ij ije E= (although this is not true regarding their other components, e.g. ij ije E≠). 2.10.5 Intermediate Configurations Stretch and Rotation Tensors The polar decompositions vR RUF== have been described in §2.2.5. The decompositions are illustrated in Fig. 2.10.8. In the material decomposition, the material is first stretched by U and then rotated by R. Let the base vectors in the associated intermediate configuration be {}igˆ. Similarly, in the spatial decomposition, the material is first rotated by R and then stretched by v. Let the base vectors in the associated intermediate configuration in this case be {}iG. Then, analogous to Eqn. 2.10.8, { ▲Problem 1} iiiii ii i g G UG g Ug G UG gU ˆˆˆˆ TT1 ⊗=⊗=⊗=⊗= −− Æ i ii ii ii i G gUg GUG gUg UG ==== −− ˆˆˆˆ TT1 (2.10.24) iiiii ii i g G vG g vg G vG gv ⊗=⊗=⊗=⊗= −− ˆˆˆˆ TT1 Æ i ii ii ii i G gvg GvG gvg Gv ˆˆˆˆ TT1 ==== −− (2.10.25) Section 2.10 Solid Mechanics Part III Kelly 289Note that U and v symmetric, TUU= , Tvv= , so ii i iii i i G g g G Ug G G gU ⊗=⊗=⊗=⊗= −ˆ ˆˆ ˆ 1 Æ i i i ii i i i g GU G gUG gU g UG ˆ , ˆˆ ,ˆ 1 1= == = − − (2.10.26) ii i iii i i G g g G vg G G gv ˆ ˆˆ ˆ 1⊗=⊗=⊗=⊗= − Æ i i i ii i i i g Gv G gvG vg g Gv = == = − − ˆ ,ˆˆ , ˆ 1 1 (2.10.27) Similarly, for the rotation tensor, with R orthogonal, T 1R R=−, ii i iii i i G G G G RG G G GR ˆ ˆˆ ˆ T⊗=⊗=⊗=⊗= Æ i i i ii i i i G GR G GRG RG G RG = == = ˆ , ˆˆ ,ˆ T T (2.10.28) ii i iii i i g g g g Rg g g gR ⊗=⊗=⊗=⊗= ˆ ˆˆ ˆ T Æ i i i ii i i i g gR g gRg gR g gR ˆ ,ˆˆ , ˆ T T= == = (2.10.29) The above relations can be ch ecked using Eqns. 2.10.8 and RUF= , vRF= , 1 1 − −=RF v , etc. Figure 2.10.8: the material and spatial polar decompositions Various relations between the base ve ctors can be derived, for example, ()() j ij iji jij i j ij i j i j i j i gG gGgG gGgG gGgG gRRG gR RG gG ˆ ˆˆ ˆˆ ˆˆ ˆ ˆ ˆ T ⋅= =⋅⋅= =⋅⋅= =⋅⋅= =⋅=⋅ LLL (2.10.30) {}iG {}igˆUR{}ig{}iGˆ R v Section 2.10 Solid Mechanics Part III Kelly 290Deformation Gradient Relationship between Bases The various base vectors are related above th rough the stretch and rotation tensors. The intermediate bases are related directly through the deformation gradient. For example, from 2.10.26a, 2.10.28b, i i i i GF G UR UG g ˆ ˆ ˆT T= == (2.10.31) In the same way, i ii ii ii i gF GgF GGF gGF g ˆ ˆˆ ˆˆ ˆˆ ˆ T1T ==== −− (2.10.32) Tensor Components The stretch and rotation tensors can be decomposed along any of the bases. For U the most natural bases would be {}iG and {}iG, for example, j ij ij i ji j iji ji i jj ii jmj im j i ij j iijj i j i ijj i ij U UU UG U UU U Gg UGG G G UgG UGG G G Ug G UGG G G UgG UGG G G U ⋅= = ⊗=⋅= = ⊗=⋅= = ⊗=⋅= = ⊗= ⋅ ⋅⋅ ⋅ ˆ ,ˆ ,ˆ ,ˆ , (2.10.33) with j ij ii ji jji ij ji ij U UU U U UU U⋅⋅⋅ ⋅ = = = = , , , . One also has j ij ij i ji j iji ji i jj ii jmj im j i ij j iijj i j i ijj i ij v vv vG v vv v Gg GvG G G vgG GvG G G vg G GvG G G vgG GvG G G v ˆ ˆˆ ,ˆ ˆˆ ˆˆ ,ˆ ˆˆˆ ˆˆ ,ˆ ˆˆ ˆˆ ,ˆ ˆ ⋅== ⊗=⋅== ⊗=⋅== ⊗=⋅== ⊗= ⋅ ⋅⋅ ⋅ (2.10.34) with similar symmetry. Also, Section 2.10 Solid Mechanics Part III Kelly 291() () () () () () () ()j ij ij i jij iji jii jj ii jj mim j iij j iijj i j i ijj i ij U UU Ug U UU U gG gUg g g UGg gUg g g Ug G gUg g g UgG gUg g g U ˆ ˆ ˆ ,ˆ ˆˆ ˆ ˆ ,ˆ ˆˆ ˆ ˆ ˆ ,ˆ ˆˆ ˆ ˆ ,ˆ ˆ 1 1 1 11 1 1 11 1 1 11 1 1 1 ⋅= = ⊗ =⋅= = ⊗ =⋅= = ⊗ =⋅= = ⊗ = −⋅−⋅− −− ⋅− ⋅− −− − − −− − − − (2.10.35) and () () () () () () () ()j ij ij i jij iji jii jj ii ji mmj j iij j iijj i j i ijj i ij v vv vg v vv v gG gvg g g vGg gvg g g vg G gvg g g vgG gvg g g v ⋅= = ⊗ =⋅= = ⊗ =⋅= = ⊗ =⋅= = ⊗ = −⋅−⋅− −− ⋅− ⋅− −− − − −− − − − ˆ ,ˆ ,ˆ ,ˆ , 1 1 1 11 1 1 11 1 1 11 1 1 1 (2.10.36) with similar symmetry. Note that, compar ing 2.10.33a, 2.10.34a, 2.10.35a, 2.10.36a and using 2.10.30, () ()() ( )ij ij ij ij j i ijj i ijj i ijj i ij v v U U vUvU 1 1 1 11 1ˆ ˆˆ ˆ − − − −− −=== ⊗ =⊗ =⊗=⊗= g g vg g UG G vG G U (2.10.37) Now note that rotations preserve vectors lengths and, in particular, preserve the metric, i.e., j i ij j i ijj i ij j i ij g gG G gg ggGG GG ˆˆ ˆˆˆ ˆ ⋅==⋅=⋅==⋅= (2.10.38) Thus, again using 2.10.30, a nd 2.10.33-2.10.36, the contrava riant components of the above tensors are also equal, () ()ijijijijv v U U1 1 − −=== . As mentioned, the tensors can be decomp osed along other bases, for example, j i j i ij j iijv v gG vgg g g v ⋅== ⊗= ˆ , (2.10.39) 2.10.6 Eigenvectors and Eigenvalues Analogous to §2.2.5, the eigenvalues of C are determined from the eigenvalue problem Section 2.10 Solid Mechanics Part III Kelly 292()0 det =−I CCλ (2.10.40) leading to the characteristic equation 1.11.5 0 III II I2 3=−+−C CC CC C λλλ (2.10.41) with principal scalar invariants 1.11.6-7 [] () 3 2 1 3 211 3 3 2 2 1 21 2 2 213 2 1 det III) tr()(tr IItr I C C C CC C C C C C CC C C C CC CC λλλ ελλλλλλλλλ = ==++=− =− =++=== kj i ijkj ii jj ji ii i CCCCC CCA (2.10.42) The eigenvectors are the principal material directions iNˆ, with () 0 NI C =−i iˆλ (2.10.43) The spectral decomposition is then ∑ =⊗ =3 12 ˆ ˆ ii i i N N Cλ (2.10.44) where 2 i iλλ=C and the iλ are the stretches. The remaining spectral decompositions in 2.2.37 hold also. Note also that the rotation tens or in terms of principal directions is (see 2.2.35) ii i i N n N nR ˆ ˆ ˆ ˆ ⊗=⊗= (2.10.45) where inˆ are the spatial principal directions. 2.10.7 Displacement and Displacement Gradients Consider the displacement u of a material particle. This can be written in terms of covariant components iU and iu: i ii i u U g G Xxu =≡−= . (2.10.46) The covariant derivative of u can be expressed as m imm im iu U g Gu= =Θ∂∂ (2.10.47) Section 2.10 Solid Mechanics Part III Kelly 293 The single line refers to covariant differentia tion with respect to the undeformed basis, i.e. the Christoffel symbols to use are functions of the ijG. The double line refers to covariant differentiation with respect to the deformed ba sis, i.e. the Christoffel symbols to use are functions of the ijg. Alternatively, the covariant de rivative can be expressed as i i i i iGgX x u−= Θ∂∂− Θ∂∂= Θ∂∂ (2.10.48) and so () () ()mm i mim m im im i imm i mim m im im i i f u uF U U g g g g GG G G G g ⋅−⋅ = −=−== += += 1δδ (2.10.49) The last equalities following from 2.10.11-12. The components of the Green-Lagrange and Euler-Almansi strain tensors 2.10.23 can be written in terms of displaceme nts using relations 2.10.49 { ▲Problem 2}: ()( ) ()( ) jn inij ji ij ij ijjn inij ji ij ij ij uu u u G g eUU U U G g E −+=−=++=−= 21 2121 21 (2.10.50) In terms of spatial coordinates, ()ji j i i ii iX x X e gE G ∂∂===Θ / , , , j i ji X U U ∂∂= / , the components of the Euler-Lagrange strain tensor are ()⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂ ∂∂+∂∂+∂∂=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛−∂∂ ∂∂=−=jk ik ij ji ij mn jn im ij ij ijXU XU XU XU Xx XxG g E21 21 21δδ (2.10.51) which is 2.2.46. 2.10.8 The Deformation of Area and Volume Elements Differential Volume Element Consider a differential volume element formed by the elements iidGΘ in the undeformed configuration, Eqn. 1.16.36: Section 2.10 Solid Mechanics Part III Kelly 294 3 2 1ΘΘΘ= dddG dV (2.10.52) where, Eqn. 1.16.13, []j i ij ij G G G GG⋅= = , det (2.10.53) The same volume element in the deformed configuration is determined by the elements iidgΘ : 3 2 1ΘΘΘ= dddg dv (2.10.54) where []j i ij ij g g g gg⋅= = , det (2.8.55) From 1.16.22 et seq. , 2.10.11, FG GGggg det3 2 13 2 13 2 1 GG FFFFFFg ijkk j ik j ik j i ==×⋅ =×⋅= ⋅⋅⋅⋅⋅⋅ ε (2.10.56) where ijkε is the Cartesian permutation symbol, and so the Jacobian determinant is (see 2.2.53) Fdet=== Gg dVdvJ (2.10.57) and Fdet is the determinant of the matrix with componentsi jF⋅. Differential Area Element Consider a differential surface (parallelogram ) element in the undeformed configuration, bounded by two vector elements )1(Xd and )2(Xd , and with unit normal Nˆ. Then the vector normal to the surface element and with magnitude equal to the area of the surface is, using 1.16.23, given by () k j i ijk jj iid de d d d d dS G G G X X NG )2( )1( )2( )1( )2( )1( ˆ ΘΘ=Θ×Θ=×= (2.10.58) Section 2.10 Solid Mechanics Part III Kelly 295where ()G ijke is the permutation symbol associated with the basis iG, i.e. ()G eijk k j i ijk ijk ε ε =×⋅= G GGG. (2.10.59) Using k kgF GT= , one has k j i ijk d dG dS gF NT )2( )1( ˆ ΘΘ=ε (2.10.60) Similarly, the surface vector in the deformed configuration with unit normal nˆ is () kj i ijk jj iid de d d d d ds g g g x x ng )2( )1( )2( )1( )2( )1(ˆ ΘΘ=Θ×Θ=×= (2.10.61) where ()g ijke is the permutation symbol associated with the basis ig, i.e. ()g eijk k j i ijk ijk ε ε =×⋅= g ggg. (2.10.62) Comparing the two expressions for the areas in the undeformed and deformed configurations, 2.10.60-61, one finds that () dS dSGgds NFF NF nT Tdet ˆ− −= = (2.10.63) which is Nanson’s relation, Eqn. 2.2.59. 2.10.9 Problems 1. Derive the relations 2.10.24. 2. Use relations 2.10.49, with j i ijg gg⋅= and j i ijG GG⋅= , to derive 2.10.50 ()( ) ()( ) jn inij ji ij ij ijjn inij ji ij ij ij uu u u G g eUU U U G g E −+=−=++=−= 21 2121 21 Section 2.11 Solid Mechanics Part III Kelly 296Convected Coordinates: Time Rates of Change In this section, the time derivatives of kinematic tensors described in §2.4-2.6 are now described using convected coordinates. 2.11.1 Deformation Rates Time Derivatives of the Base Vectors and the Deformation Gradient First, the material time derivatives of the deformed base vectors are, from 2.10.8, i i i ii i i i gFF gFF GFggFF gFF GFg T T T T T1 1 & &&& &&& − − −− − −= ==−=== (2.11.1) with, again from 2.10.8, iiiii ii i g G FGg Fg G FGgF & &&&& &&& ⊗=⊗=⊗=⊗= −− TT1 (2.11.2) The Velocity Gradient The velocity gradient is defined by 2.5.2, v lgrad= , so that, using 1.16.5, j ji ij ji ix xgvevev xvl ⊗ Θ∂∂=⊗ ∂Θ∂ Θ∂∂=⊗ ∂∂=∂∂= (2.11.3) Also, from 1.16.3, i i iΘ∂∂=Θ∂∂=v xg&& (2.11.4) so that, as an alternative to 2.11.3, i iggl⊗=& (2.11.5) This is consistent with Eqn. 2.5.4, lFF=& , which gives, with 1.11.2a and 2.10.8b, ()()i ij ji i gg g GGg FFl ⊗=⊗ ⊗==−& &&1 (2.11.6) Section 2.11 Solid Mechanics Part III Kelly 297 The components of the spa tial velocity gradient are j i j i ijj i m imj j ij iji ji i jj i j i ij lg lll gg lgggg gg lgggg lgggg lgg &&&&& ⋅==⋅=⋅==⋅==⋅== ⋅⋅ (2.11.7) Further, from 2.11.1, 2.11.2 and 2.11.5, lg lggl g lgg i ii i i i −= =−= = TT& & (2.11.8) Contracting the first of these with idΘ leads to i ii i d d Θ=Θ lg g& (2.11.9) which is equivalent to 2.5.1, xlv d d= . The Rate of Deformation and Spin Tensors From 2.5.6, wdl+= . The covariant components of the rate of deformation and spin are () () () () () ()j i j i j mm m m i j i ijj i j i j i j mm m m i j i ij wd gggg gg g g gg gllggg gggg gg g g gg gllg ⋅−⋅=⊗−⊗=−=⋅=⋅+⋅=⊗+⊗=+=⋅ && & &&& & & 21 21 2121 21 21 21 TT (2.11.10) Section 2.12 Solid Mechanics Part III Kelly 2982.12 Pull Back, Push Forward and Lie Time Derivatives 2.12.1 Push-Forward and Pull-Back The concepts of pull-back and push-forward have a number of uses, in particular they will be used to define the Lie derivative further below. Vectors Consider a vector V given in terms of the reference configuration base vectors: ii i i V V G G V== (2.12.1) The push-forward of V, ()V*χ , is defined to be the vector with the same components, but with respect to the current configuration base ve ctors. The push-forward of a vector depends on the type of components; the symbol b is used for covariant components iV and the symbol # for contravariant components iV. Thus, using 2.10.8, () () FV FG g VVF GF g V ==== ==− − ii iii ii ib V VV V # *T T * χχ. (2.12.2) A special case is the push forward of a line element in the reference configuration, Eqn. 2.10.7, () x g X d d dii=Θ=# *χ . (2.12.3) which is consistent with the fact that, wi th convected coordinates, the line element Xd has the same coordinates with respect to th e reference configuration basis as does xd with respect to the current configuration basis. Similarly, consider a vector v given in terms of the current configuration basis: ii i i v v g g v== (2.12.4) The pull-back of v, ()v1 *−χ , is defined to be the vector with components iv (or iv) with respect to the reference configuration base vectors iG (or iG). Thus, using 2.10.8, () () vF gF G vvF gF G v 1 1 # 1 *T T 1 * − −− ====== i-i iii ii ib v vv v χχ. (2.12.5) Section 2.12 Solid Mechanics Part III Kelly 299and, for a line element in the current configuration, () X xF G x d d dx dii===− − 1 # 1 *χ . (2.12.6) Note that a push-forward and pull-back appl ied successively to a vector with the same component type will result in the initial vector. From the above, for two material vectors U and V and two spatial vectors u and v, ()()()() () () () ()b bb b v u v u vuV U V U VU 1 *# 1 *# 1 *1 **# *# * * − − − −⋅=⋅=⋅⋅=⋅=⋅ χχ χχχχχχ (2.12.7) Tensors Consider a material tensor A: ji j ij ii j j iij j i ij A A A A G G G G G G G G A ⊗=⊗=⊗=⊗=⋅ ⋅ (2.12.8) As for the vector, the push-forward of A, ()A*χ , is defined to be the tensor with the same components, but with respect to the deform ed base vectors. Thus, using 2.10.8, () ( ) () () () () () ()T T T / *1 T \ *T # *1 T T T * AFF FG GF g g AFAF GF FG g g AFAF FG FG g g AAFF GF GF g g A − −⋅ ⋅− − ⋅ ⋅−− − − =⊗ =⊗== ⊗ =⊗== ⊗ =⊗== ⊗ =⊗= ji j i jij ij ii jj ii jj i ij j iijj i ijj i ijb A AA AA AA A χχχχ . (2.12.9) Similarly, consider a spatial tensor a: jij ij ii j j iij j i ij a a a a g g g g g g g g a ⊗=⊗=⊗=⊗=⋅ ⋅ (2.12.10) The pull-back is () ( ) () () () () () ()T T 1 T / 1 *1 T 1 \ 1 *T 1 1 1 # 1 *T T T 1 * − ⋅ ⋅ −− − ⋅ ⋅−−− − −− = ⊗ =⊗==⊗ =⊗==⊗ =⊗== ⊗ =⊗= aFF gF gF G G aaFF gF gF G G aaFF gF gF G G aaFF gF gF G G a j- i j i ji j ij ii jj ii jj- i ij j iijj i ijj i ijb a aa aa aa a χχχχ (2.12.11) Section 2.12 Solid Mechanics Part III Kelly 300Push-Forward and Pull-Back relations for Vectors and Tensors For two material tensors A and B and two spatial tensors a and b, the scalar product is i jj ij ii j ijij ij iji jj ij ii j ijij ij ij ba ba ba baBA BA BA BA ⋅⋅⋅ ⋅⋅⋅⋅ ⋅ ======== baBA :: (2.12.12) This scalar product then push-forwards and pull-backs as { ▲Problem 1} ()()()() () () () () () () () ()() () () () / 1 *\ 1 *\ 1 */ 1 *1 *# 1 *# 1 *1 */ *\ *\ */ **# *# * * : :: : :: :: : : b a b ab a b a baB A B AB A B A BA − − − −− − − − = == == == = χχχχχχ χχχχχχχχχχ b bb b (2.12.13) For material tensor A and material vectors VU,, and spatial tensor a and spatial vectors vu,, jj ii j i jij iji jij ijj ii j i j ij iji jij i vau vau vau vauVAU VAU VAU VAU ⋅ ⋅⋅ ⋅ ===== = = = uavUAV (2.12.14) Then ()()()()()() () () () () () () () () () () () ()() () () () () () b bb b bb bb b b v a u v a uv a u v a u uavV A U V A UV A U V A U UAV 1 */ 1 *# 1 *# 1 *\ 1 *1 *# 1 *1 *# 1 *1 *# 1 *1 **/ *# *# *\ * *# * *# * *# * * − − − − − −− − − − − − = == == == = χχχ χχχχχχ χχχχχχχχχχχχχχχ (2.12.15) For material tensor A and material vector V, and spatial tensor a and spatial vector v, the contractions AV and av are jij j i j jj ij ijjij j i j jj ij ij va va va vaVA VA VA VA ======== ⋅⋅⋅⋅ avAV (2.12.16) and so transform as Section 2.12 Solid Mechanics Part III Kelly 301() ()()()() ( ) () () () () ( ) () () () () ( ) () () () ()# 1 *\ 1 *1 *# 1 *# 1 *1 */ 1 *# 1 *1 *1 *# *\ * *# *# **/ *# * * * v a v a avv a v a avV A V A AVV A V A AV − − − − −− − − − − = == == == = χχ χχ χχχ χχ χχχχχ χχχχχ χ bb b bbb b b (2.12.17) Finally, for material tensors A, B and spatial tensors a, b, LL =⊗=⊗=⊗=⊗ ==⊗ =⊗ =⊗ =⊗ = ⋅⋅ ⋅⋅⋅⋅ ⋅⋅ j ik j ikj i kjk i jij kk i ji kj ikj i k j ikj i kjk i ji j kk i ji kj ik ba ba ba baBA BA BA BA g g g g g g g g abG G G G G G G G AB (2.12.18) and so ()()()()() () ( )( ) ( )( ) ( ) () () () () ( ) () () () ()/ 1 */ 1 *# 1 *1 *1 */ 1 */ 1 *# 1 *1 */ 1 *\ * * */ * */ */ *# * */ * b a b a abb a b a abB A B A ABB A B A AB − − − − −− − − − − = == == == = χχ χχ χχχ χχ χχχχχ χχχχχ χ b bbb b bb M (2.12.19) Push-Forward and Pull-Back operations for Strain Tensors The push-forward of the covariant right Cauchy-Gr een strain and its contravariant inverse are () ()()T 1#1 *1 T * FCF g g CCFF g g C =⊗ ==⊗= − −−− j iijj i ijb CC χχ . (2.12.20) From 2.10.19, ij ijg C= , the covariant components of the id entity tensor expressed in terms of the convected base vectors in the current c onfiguration, i.e. the spatial metric tensor, i i ijg g g g⊗= , and ()ijijg C=−1, the contravariant components of g, so the push-forward of covariant C is g and the pull-back of covariant g is C, and the push-forward of contravariant 1−C is g and the pull-back of contravariant g is 1−C: () () () ()1 # 1 *#1 *1 * * ,, − − −− = == = C g g CC g g C χ χχ χb b . (2.12.21) Section 2.12 Solid Mechanics Part III Kelly 302Similarly, the pull-back of covariant 1−b is G and the push-forward of covariant G is 1−b, and the pull-back of contravariant b is G and the push-forward of contravariant G is b. () () () () G b b GG b b G = == = −−− − # 1 *# *1 1 *1 * ,, χ χχ χb b . (2.12.22) For the covariant Green-Lagrange strain, the push-forward is ()1 T *−−=⊗= EFF g g Ej i ijbEχ . (2.12.23) From 2.10.23, ij ije E=, the covariant components of the Euler-Almansi strain tensor, and so the push-forward of covariant E is e and the pull-back of covariant e is E. () ()E e e E = =− b b 1 * * ,χ χ . (2.12.24) 2.12.2 Push-Forward and Pull-Ba ck with Polar Decomposition Intermediate Configurations Pull backs and push-forwards can be defined re lative to any two config urations. Consider the polar decomposition and the intermediate configurations discussed in §2.10. Pushing forward a material tensor A from the reference configuration {}iG to the configuration {}iGˆ leads naturally to (see Fig. 2.10.8) ()() ( ) ()() () ()() () ()() ()T T T T / *T 1 T \ *T # *T 1 T T T * ˆ ˆˆ ˆˆ ˆˆ ˆ RAR ARR RG GR G G ARAR RAR GR RG G G ARAR RG RG G G ARAR ARR GR GR G G A GRGRGRGR = =⊗ =⊗=== ⊗ =⊗== ⊗ =⊗== = ⊗ =⊗= − −⋅ ⋅− − ⋅ ⋅−− − − ji j i ji j ij ii jj ii jj iij j iijj i ijj i ijb A AA AA AA A χχχχ . (2.12.25) and the pull back of a tensor Aˆ from the intermediate configuration {}iGˆ to the reference configuration {}iG is ()() ()() ()() ()() RAR G G ARAR G G ARAR G G ARAR G G A GRGRGRGR ˆ ˆ ˆˆ ˆ ˆˆ ˆ ˆˆ ˆ ˆ Tˆ/1 *Tˆ\1 *Tˆ#1 *Tˆ1 * =⊗==⊗==⊗==⊗= ⋅ −⋅−−− ji j ij ii jj iijj i ijb AAAA χχχχ (2.12.26) Section 2.12 Solid Mechanics Part III Kelly 303Similarly, the push-forward of a tensor aˆ from {}igˆ to {}g and the corresponding pull-back of a spatial tensor a is ()() ()() ()() ()()Tˆ/ *Tˆ\ *Tˆ# *Tˆ* ˆ ˆ ˆˆ ˆ ˆˆ ˆ ˆˆ ˆ ˆ RaR g g aRaR g g aRaR g g aRaR g g a gRgRgRgR =⊗==⊗==⊗==⊗= ⋅⋅ jij ij ii jj iijj i ijb aaaa χχχχ , ()() ()() ()() ()() aRR g g aaRR g g aaRR g g aaRR g g a gRgRgRgR T / 1 *T \ 1 *T # 1 *T 1 * ˆ ˆˆ ˆˆ ˆˆ ˆ =⊗==⊗==⊗==⊗= ⋅ −⋅−−− jij ij ii jj iijj i ijb aaaa χχχχ (2.12.27) The push-forwards and pull-backs due to the stretch tensors are ()() ( ) ()() () ()() () ()() () AUU AUU UG GU g g AUAU GU UG g g AUAU UAU UG UG g g AAUU AUU GU GU g g A GUGUGUGU 1 T T T / *1 T \ *T # *1 1 1 T T T * ˆ ˆˆ ˆˆ ˆˆ ˆ − − −⋅ ⋅− − ⋅ ⋅−− −− − − = =⊗ =⊗== ⊗ =⊗=== ⊗ =⊗== = ⊗ =⊗= ji j i jij ij ii jj ii jj iij j iijj i ijj i ijb A AA AA AA A χχχχ . (2.12.28) ()() ()() ()() ()()1ˆ/ 1 *1ˆ\ 1 *1 1ˆ# 1 *ˆ1 * ˆ ˆ ˆˆ ˆ ˆˆ ˆ ˆˆ ˆ ˆ − ⋅ −− ⋅−−− −− =⊗==⊗==⊗==⊗= UaU G G aUaU G G aUaU G G aUaU G G a gUgUgUgU ji j ij ii jj iijj i ijb aaaa χχχχ (2.12.29) and ()() ()() ()() ()() vAv g g AvAv g g AvAv g g AvAv g g A GvGvGvGv ˆ ˆ ˆˆ ˆ ˆˆ ˆ ˆˆ ˆ ˆ 1ˆ/ *1ˆ\ *ˆ# *1 1ˆ* − ⋅− ⋅−− =⊗==⊗==⊗==⊗= jij ij ii jj iijj i ijb AAAA χχχχ , ()() ()() ()() ()()1 / 1 *1 \ 1 *1 1 # 1 *1 * ˆ ˆˆ ˆˆ ˆˆ ˆ − ⋅ −− ⋅−−− −− =⊗==⊗==⊗==⊗= vav G G aavv G G aavv G G avav G G a gvgvgvgv ji j ij ii jj iijj i ijb aaaa χχχχ (2.12.30) Push-forwards and pull-backs can also be defined using TF (in the place of F) and these move between the interm ediate configurations, g G ˆ ˆ⇔ . Recall Eqn. 2.10.37, which state that the covariant components of 1 1, ,,−−v UvU with respect to the bases i i i iggGG ,ˆ,ˆ, respectively, are equal. This can be explained also in terms of push-forwards and pull-backs. For example, with TRURv= and T 1 1R RU v− −= , one can write (in fact these relations are valid for all component types) Section 2.12 Solid Mechanics Part III Kelly 304()() ()()gR GR U v U vˆ1 *1 * ,− −= = χ χ (2.12.31) The first of these shows that the components of U with respect to G are the same as those of v with respect to Gˆ (for all component types). The second shows that the components of 1−U with respect to gˆ are the same as those of 1−v with respect to g. As another example, with 2UC= , (compare with Eqn. 2.10.19) ()() ()()gU gU g C g C ˆ# 1 *1ˆ1 *ˆ , ˆ− − −= = χ χb (2.12.32) 2.12.3 The Lie Time Derivative Vectors The material time derivative of a spatial vector u is ii iii ii i u uu u g gg g u &&&&& +=+= (2.12.33) The Lie (time) derivative avL is the material derivative holding the deformed basis constant , that is, the first terms on the right hand side of 2.12.33: ii vi ib v uu g ug u && == #LL (2.12.34) In terms of the pull-back and push-forward, ()[]⎟ ⎠⎞⎜ ⎝⎛=−u u1 * * L χχdtd v The Lie Time Derivative (2.12.35) This is illustrated in the Fig. 2.12.1. The spat ial vector is first pulled back to the reference configuration, there the differentiation is carr ied out, where the base vectors are constant, then the vector is pushed forward again to the spatial description. Section 2.12 Solid Mechanics Part III Kelly 305 Figure 2.12.1: The Lie Derivative For covariant components, one first pulls back the vector i iug to i iuG, the derivative is taken, i iuG&, and then it is pushed forward to i iug&, which is consistent with the definition 2.12.34a. The definition 2.12.35 allows one to calculate the Lie derivative in absolute notation: using 2.4.4-5, ()[] [] () () uluuFulFFuFuFFuF F u u TT TT TT T TT T 1 * * L +=+ =+ =⎟ ⎠⎞⎜ ⎝⎛=⎟ ⎠⎞⎜ ⎝⎛= −−− − &&&&dtd dtdb b v χχ (2.12.36) The Lie derivative for the contravariant components can be calculated in a similar way, and in summary: { ▲Problem 3} luu g uulu g u −==+== &&&& ii vi ib v uu #T LL Lie Derivatives of Vectors (2.12.37) Tensors The material time derivative of a spatial tensor a is jij i jij i jij ij ii jj ii jj ii jj iij j iij j iijj i ijj i ijj i ij a a aa a aa a aa a a g g g g g gg g g g g gg g g g g gg g g g g g a & & && & && & && & && ⊗+⊗+⊗=⊗+⊗+⊗=⊗+⊗+⊗=⊗+⊗+⊗= ⋅ ⋅ ⋅⋅ ⋅ ⋅ (2.12.38) The Lie (time) derivative avL is then u 1 *−χ*χ )(1 *u−χ)(1 *u−χdtd⎟ ⎠⎞⎜ ⎝⎛−)(1 * * uχχdtd Section 2.12 Solid Mechanics Part III Kelly 306 jij i vj ii j vj iij vj i ijb v aaaa g g ag g ag g ag g a ⊗=⊗=⊗=⊗= ⋅⋅ &&&& /\# LLLL (2.12.39) For covariant components, one first pulls back the tensor j i ija g g⊗ to j i ija G G⊗ , the derivative is taken, j i ija G G⊗& , and then it is pushed forward to j i ija g g⊗& . With 2.4.4-5, ()[] [] () () alaalFalFFFaF aFlFFFFaFFaF aFFFFaFF F a a ++=++ =++ =⎟ ⎠⎞⎜ ⎝⎛=⎟ ⎠⎞⎜ ⎝⎛= − −− −− − − &&& & & T1 T T TT T1 T T T T1 T T 1 * * Ldtd dtdb b v χχ (2.12.40) The Lie derivative for the other components ca n be calculated in a similar way, and in summary: { ▲Problem 4} T T /\T #T LLLL alala g g aallaa g g aallaa g g aalala g g a −+=⊗=+−=⊗=−−=⊗=++=⊗= ⋅⋅ & && && && & jij i vj ii j vj iij vj i ijb v aaaa Lie Derivatives of Tensors (2.12.41) Lie Derivatives of Strain Tensors From 2.5.18, 0 bllbbeleled =−−++= TT && (2.12.42) and so the Lie derivative of the covariant Euler-A lmansi strain is the rate of deformation and the Lie derivative of the contravariant left Ca uchy-Green tensor is zero. Further, from 2.12.21, 2.12.41, ()bb v C g& * Lχ= , dllglglgg 2 LT T=+=++=&b v (2.12.43) Section 2.12 Solid Mechanics Part III Kelly 307 Lie Derivatives and Objective Rates One of the most important uses of the Lie derivative is that Lie derivatives of objective spatial tensors are objective spatial tensors . Thus the rates given in 2.12.41 are all objective. Further, any linear combination of them is objective, for example, () ()[] ()() [ ] aw waa llaall a allaa alala +−=−+−−+=−−+++ & & & &T T T T 21 21 (2.12.44) is objective, provided a is. This is the Jaumann rate intr oduced in Eqn. 2.8.36. The Cotter- Rivlin rate of Eqn. 2.8.37 is equivalent to Tb vL. The Lie Derivative and the Directional Derivative Recall that the material time derivative of a te nsor can be written in terms of the directional derivative, §2.6.5. Hence the Lie derivative can also be expressed as ()()[] ( )vT Tf1 * * L−∂=χχv (2.12.45) and hence the subscript v on the L. Thus one can say that the Lie derivative is the push forward of the directional derivative of the material field ()T1 *−χ in the direction of the velocity vector. 2.12.4 Problems 1. Eqns. 2.12.13 follow immediately from 2.12. 12. However, use Eq ns. 2.12.9, 2.12.11, i.e. ()1 T *−−= AFF Abχ , etc., directly, and 1.10.3h, to verify relations 2.12.13. 2. Derive the Lie derivatives of a vector u, Eqns. 2.12.37. 3. Derive the Lie derivatives of a tensor a, Eqns. 2.12.41. Section 2.13 Solid Mechanics Part III Kelly 308Variation and Linearisation of Kinematic Tensors 2.13.1 The Variation of Kinematic Tensors The Variation In this section is reviewed the concept of the variation, introduced in Part I, §5.5. The variation is defined as follows: consider a function )(xu , with )(xu* a second function which is at most infinitesimally different from )(xu at every point x, Fig. 2.13.1 Figure 2.13.1: the variation Then define )()( xuxuu −=*δ The Variation (2.13.1) The operator δ is called the variation symbol and uδ is called the variation of ) (xu . The variation of ) (xu is understood to represent an in finitesimal change in the function at x. Note from the figure that a variation uδ of a function u is different to a differential ud. The ordinary differentiation gives a measure of the change of a function resulting from a specified change in the independent variable (in this case x). Also, note that the independent variable does not participate in the variation process; the variation operator imparts an infinitesimal change to the function u at some fixed x – formally, one can write this as 0=xδ . The Commutative Properties of the variation operator (1) xuux dd ddδδ= ( 2 . 1 3 . 2 ) )(xuδ )(xuxdud x)(*xu Section 2.13 Solid Mechanics Part III Kelly 309Proof : () )()*( * *xux xu u xu xu xu xu xuδ δdd dd dd dd dd dd dd=−=−=−⎟ ⎠⎞⎜ ⎝⎛= (2) ∫∫=2 12 1)( )(x xx xd d xxu xxu δ δ ( 2 . 1 3 . 3 ) Proof : [] ∫ ∫∫∫∫=− = − =2 12 12 12 12 1)( )()(* )( )(* )(x xx xx xx xx xxxu xxuxu xxu xxu xxu d d d d d δ δ Variation of a Function Consider A, a scalar-, vector-, or tensor-valued function of u. The value of A at u uδ+ , where uδ is a variation of u is, as in, for example, 1.15.27, ][ )( ) ( uA uA u uAuδ δ ∂+≈+ (2.13.4) The directional derivative in this context is also denoted by ()uuAδδ , and is called the variation of A: () () u uA uA uuAu εδεδ δδ ε+ =∂≡ =0][ ,dd (2.13.5) The variation of A is thus the directional derivative of A in the direction of the variation uδ. For example, consider the scalar function EP:=φ , where P and E are second order tensors. Then () () EP E EP E EEE δ εδεδφδδφ ε: : ][ , 0=+ =∂≡ =dd (2.13.6) The second variation is defined as () () u uA uA A Au εδδεδδδδδ ε+ =∂== =02][dd (2.13.7) For example, for a scalar function ()uφ of a vector u, Section 2.13 Solid Mechanics Part III Kelly 310uuuu u uuuuuuuuu δφδδδφδφδδδφφδδφδφ ∂∂∂=⋅⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂∂=⋅⎟ ⎠⎞⎜ ⎝⎛ ∂∂=⋅∂∂=⋅∂∂= 2 2 2 (2.13.8) Variation of Functions of the Displacement In what follows is discussed the change (variation) in functions )(uA when the displacement (or velocity) fields undergo a variation. These ideas are useful in formulating variational prionciples of mechanics (see, for example, §3.8). Shown in Fig. 2.13.2 is the current configuration frozen at some instant in time. The displacement field is then allowed to undergo a variation u δ. This change to the displacement field evidently changes kinematic tensors, and these changes are now investigated. Note that this variation to the displacement induces a variation to x, xδ, but X remains unchanged, 0=Xδ . Figure 2.13.2: a variation of the displacement To evaluate the variation of the deformation gradient F, ()uuFδδ ,, where u is the displacement field, note that Xxu−= and Eqn. 2.2.43, () Iu uF +=Grad . One has, from 2.13.5, () [] () () ( ) ()uu uFu uF uF uuFu δδεεεδεδ δδ εε GradGrad, 00 =+ =+ =∂= == dddd (2.13.9) Noting the first commutative property of the vari ation, 2.13.2, this can also be expressed as reference confi guration current confi guration uδ)(xu Xx Section 2.13 Solid Mechanics Part III Kelly 311() u uuF Grad ,δδδ= (2.13.10) Note that uδ is completely independent of the function u. Here are some other examples, involving the inverse deformation gradient, the Green-Lagrange strain, the inverse right Cauchy-Gre en strain and the spatial line element: {▲Problem 1-3} T 1 1T1 1 2grad −− −− − −==−= εFF CεF FEu F F δδδδ δ (2.13.11) where ε is the small strain tensor, Eqn. 2.2.48. One also has, using the chain rule for the direc tional derivative, Eqn. 1.15.28, the directional derivative for the determinant, Eqn. 1.15.32, the trace relation 1.10.10e, Eqn. 2.2.8b, () () [] [][] ()[] ()[] ()() ()() ()uuFuu FFu FuFFuFuuF uu Fu Fu δδδδδδδδδδδ divgradtrGradtrGrad: detGrad detdetdet, det , 1T JJJJ ====∂=∂∂=∂== −− (2.13.12) The Lie Variation The Lie-variation is defined for spatial vectors and tensors as a variation holding the deformed basis constant. For example, analogous to 2.12.33a, j i ija g g a⊗=δδb L (2.13.13) The object is first pulled-back, the variation is then taken and finally a push-forward is carried out. For example, analogous to 2.12.40, () ()()[] ( )ua uuauδχχδδ1 * * L,−∂≡ (2.13.14) For example, consider the Lie-vari ation of the Euler-Almansi strain e. First, from 2.12.24, () E e=−b * 1χ . Then 2.13.11b gives ()()[] εF FE u eu δδδχT * 1 == ∂−b. From 2.12.9a, Section 2.13 Solid Mechanics Part III Kelly 312 () ( ) ()[] ( )()εεF F u e uueu δδχδχχδδ = = ∂=−b bb T ** 1 * L, (2.13.15) 2.13.2 Linearisation of Kinematic Functions Linearisation of a Function As for the variation, consider A, a scalar-, vector-, or tensor-valued function of u. If u undergoes an increment uΔ, then, analogous to 2.13.4, ()() ][uA uA u uAuΔ∂+≈Δ+ (1.13.16) The directional derivative ][uAuΔ∂ in this context is also denoted by ()uuAΔΔ , . The linearization of A with respect to u is defined to be ()()()uuA uA uuA ΔΔ+=Δ , , L (1.13.17) Using exactly the same method of calculation as was used for the variations above, the linearization of F and E, for example, are ()()[] () ( ) [] εF FEuE uE uuEu FuF uF uuF uu Δ+=Δ∂+=ΔΔ+=Δ∂+=Δ T, LGrad , L (2.13.18) where () ()( )u u ε Δ+Δ=Δ grad gradT 21 is the linearised small strain tensor ε. Linearisation of Variations of a Function One can also linearise the variation of a function. For example, () ()()uuA uuA uuA ΔΔ+ =Δ , , , L δδδ δ (2.13.19) The second term here is the directional derivative () u uAuA uuAu Δ+ =Δ∂=ΔΔ =εδεδ δ ε0][ ],[ dd (2.13.20) This leads to an expression similar to A2δ . For example, for a scalar function ()uφ of a vector u, Section 2.13 Solid Mechanics Part III Kelly 313uuuu uuδφ δφδφ∂∂∂Δ=Δ⋅∂∂=Δ2 (2.13.21) Consider now the virtual Gree n-Lagrange stra in, 2.13.11b, εF FEδδT= . To carry out the linearization of Eδ, it is convenient to first write it in the form ()[] ()[] u FFuFu u FεF FE δ δδ δδδ Grad Gradgrad grad T T 21T T 21T + =+ == (2.13.22) Then [] ()[ ] { }[]u u FFu uE Eu u Δ + ∂=Δ∂=Δ δ δ δδ Grad GradT T 21 (2.13.23) Recall that the variation uδ is independent of u; this equation is being linearised with respect to u, and uδ is unaffected by the linearization (see Fi g. 2.13.3 below). However, the motion, and in particular F, are affected by the increment in u. Thus {▲Problem 4} ()( )u u E δ δ Grad Grad symTΔ =Δ (2.13.24) Figure 2.13.3: linearisation As with the variational operator, one can defi ne the linearization of a spatial tensor as involving a pull back, followed by the directional derivative, and finally the push forward operation. Thus () ()()[] ( )u a uuauΔ ∂≡ΔΔ−1 * * , χχ (2.13.25) reference configuration current configuration uuδ uδ uΔ Section 2.13 Solid Mechanics Part III Kelly 3142.13.3 Problems 1. Use Eqn. 2.2.22, ()IFF E −=T 21, Eqn. 2.13.9, ()()u uuF δ δδ Grad ,= , and Eqn. 2.2.8b, ()1Grad grad−= Fv v , to show that εF FEδδT= , where ε is the small strain tensor, Eqn. 2.2.48. 2. Use 2.13.9 to show that the variation of the inverse deformation gradient 1−F is u F F δ δ grad1 1 − −−= . [Hint: differente the relation IFF=−1 by the product rule and then use the relation ()1Grad grad−= Fv v for vector v.] 3. Use the definition FFCT= to show that T 1 12−− −−=εFF Cδ . 4. Use the relation ()A A A +=T 21sym to show that ()[ ] { }[]()( )u u u u FFu Eu δ δ δ δ Grad Grad sym Grad GradT T T 21Δ =Δ + ∂=Δ 5. Use ()[ ]u u ε e δ δ δδ grad gradT 21+ == and Eqn. 2.7.21 to show that the ()()[] () ()( ) ()[] u uu u ue eu δδ χ δχχδ grad grad symGrad Grad sym TT *1 * * ⋅Δ =Δ =Δ ∂=Δ− 3153 Stress and the Balance Principles Three basic laws of physics are discussed in this Chapter: (1) The Law of Conservation of Mass (2) The Balance of Linear Momentum (3) The Balance of Angular Momentum together with the conservation of mechanical energy and the princi ple of virtual work, which are different versions of (2). (2) and (3) involve the concep t of stress, which allows one to describe the action of forces in materials. 316 Section 3.1 Solid Mechanics Part III Kelly 3173.1 Conservation of Mass 3.1.1 Mass and Density Mass is a non-negative scalar measure of a body’s tendency to resist a change in motion. Consider a small volume element vΔ whose mass is mΔ. Define the average density of this volume element by the ratio vm ΔΔ=AVEρ (3.1.1) If p is some point within the volume element, then define the spatial mass density at p to be the limiting value of this ratio as the volume shrinks down to the point, vmtvΔΔ=→Δ 0 lim),(xρ Spatial Density (3.1.2) In a real material, the incremental volume element vΔ must not actually get too small since then the limit ρ would depend on the atomistic st ructure of the material; the volume is only allowed to decrease to some minimum value which contains a large number of molecules. The spatial mass dens ity is a representative average obtained by having vΔ large compared to the atomic scale, but small compared to a typical length scale of the problem under consideration. The density, as with displacement, velocity, and other quantities, is defined for specific particles of a continuum, and is a continuous function of coordinates and time, ),(txρρ= . However, the mass is not defined this way – one writes for the mass of an infinitesimal volume of material – a mass element , dvt dm ),(xρ= (3.1.3) or, for the mass of a volume v of material at time t, ()∫= vdvt m ,xρ (3.1.4) 3.1.2 Conservation of Mass The law of conservation of mass states that ma ss can neither be created nor destroyed. Consider a collection of matter located somewher e in space. This qua ntity of matter with well-defined boundaries is termed a system . The law of conservation of mass then implies that the mass of this given system remains constant, Section 3.1 Solid Mechanics Part III Kelly 3180=DtDm Conservation of Mass (3.1.5) The volume occupied by the matter may be cha nging and the density of the matter within the system may be changing, but the mass remains constant. Considering a differential mass element at position X in the reference configuration and at x in the current configuration, Eqn. 3.1.5 can be rewritten as ),( )( t dm dm x X= (3.1.6) The conservation of mass equation can be expressed in terms of densities. First, introduce 0ρ, the reference mass density (or simply the density ), defined through Vm VΔΔ=→Δ 0 0 lim)(Xρ Density (3.1.7) Note that the density 0ρ and the spatial mass density ρ are not the same quantities1. Thus the local (or differential ) form of the conservation of ma ss can be expressed as (see Fig. 3.1.1) const ),( )( 0 = = = dvt dV dm x Xρ ρ (3.1.8) Figure 3.1.1: Conservation of Ma ss for a deforming mass element Integration over a finite re gion of material gives the global (or integral ) form , const ),( )(0 = = = ∫∫ v Vdvt dV m x Xρ ρ (3.1.9) or 0 ),(= ==∫ vdvtdtd dtdmm xρ & (3.1.10) 1 they not only are functions of different variables, but also have different values; they are not different representations of the same thing, as were, for example, the velocities v and V. One could introduce a material mass density, )),,(( ),( ttXx tXρ=Ρ , but such a quantity is not useful in analysis reference configuration Xx 0,ρdVρ,dv•• current configuration Section 3.1 Solid Mechanics Part III Kelly 3193.1.3 Control Mass and Control Volume A control mass is a fixed mass of material whose volume and density may change, and which may move through space, Fig. 3.1.2. There is no mass transport through the moving surface of the control mass. Fo r such a system, Eqn. 3.1.10 holds. Figure 3.1.2: Control Mass By definition, the derivative in 3.1.10 is th e time derivative of a property (in this case mass) of a collection of materi al particles as they move through space, and when they instantaneously occupy the volume v, Fig. 3.1.3, or 0 ),( ) ,(1lim ) () (0 = ⎪⎭⎪⎬⎫ ⎪⎩⎪⎨⎧ −Δ+Δ= ∫∫ ∫ Δ+→Δ ttvt vt dvdvt dvt ttdvdtdx x ρ ρ ρ (3.1.11) Alternatively, one can take the material derivative inside the integral sign: [] 0 ),(= =∫ vdvtdtd dtdmxρ (3.1.12) This is now equivalent to the sum of the ra tes of change of mass of the mass elements occupying the volume v. Figure 3.1.3: Control Mass occupying different volumes at different times A control volume , on the other hand, is a fixed volume (region) of space through which material may flow, Fig. 3.1.4, and for which the mass may change . For such a system, one has )(),),((,1 1 1 tvtt m xρ )(),),((,2 2 2 tvtt m xρ time t time t tΔ+ Section 3.1 Solid Mechanics Part III Kelly 320 [] 0 ),( ),( ≠∂∂=∂∂=∂∂∫∫dvttdvtt tm v vx x ρ ρ (3.1.13) Figure 3.1.4: Control Volume 3.1.4 The Continuity Eq uation (Spatial Form) A consequence of the law of conservation of mass is the continuity equation , which (in the spatial form) relates the density and veloci ty of any material particle during motion. This equation can be derived in a number of ways: Derivation of the Continui ty Equation using a Contro l Volume (Global Form) The continuity equation can be derived directly by considering a control volume - this is the derivation appropriate to fluid mechanic s. Mass inside this fixed volume cannot be created or destroyed, so that the rate of increase of mass in the volume must equal the rate at which mass is flowing into the volume through its bounding surface. The rate of increase of mass inside the fixed volume v is ∫∫∂∂=∂∂=∂∂ v vdvtdvtt tm ρρ ),(x (3.1.14) The mass flux (rate of flow of mass) out through the surface is given by Eqn. 1.7.9, ∫ ∫⋅ sii sdsnv ds ρ ρ ,nv where n is the unit outward normal to the surface and v is the velocity. It follows that 0 ,0 = +∂∂=⋅+∂∂∫∫ ∫∫ sii v s vdsnv dvtds dvtρρρρnv (3.1.15) Use of the divergence theorem 1.7.12 leads to ()()0 ,0 div =⎥⎦⎤ ⎢⎣⎡ ∂∂+∂∂=⎥⎦⎤ ⎢⎣⎡+∂∂∫ ∫ v ii vdvxv tdvtρρρρv (3.1.16) vt tm ),,(),( xρ Section 3.1 Solid Mechanics Part III Kelly 321leading to the continuity equation, ()() 000 0 div grad0 div0 div =∂∂+∂∂+∂∂=∂∂+=∂∂+∂∂ =+⋅+∂∂=+=+∂∂ ii i iiiii xvvx txv dtdxv t tdtdt ρρρρρρρ ρρρρρρρ v vvv Continuity Equation (3.1.17) This is (these are) the continuity equation in spatial form. The second and third forms of the equation are obtained by re-writing the lo cal derivative in terms of the material derivative 2.4.7 (see also 1.6.23b). If the material is incompressible, so the de nsity remains constant in the neighbourhood of a particle as it moves, then th e continuity equation reduces to 0 ,0 div =∂∂= ii xvv Continuity Eqn. for Inco mpressible Material (3.1.18) Derivation of the Continuity Equation using a Control Mass Here follow two ways to derive the c ontinuity equation using a control mass. 1. Derivation using the Formal Definition From 3.1.11, adding and subtracting a term: ⎪⎭⎪⎬⎫ ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ −Δ+ +⎪⎩⎪⎨⎧ ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ Δ+−Δ+Δ= ∫∫∫∫ ∫ Δ+→Δ )() () () (0 ),( ) ,() ,( ) ,(1lim tvt vttvt vt dv dvt dvt tdvt t dvt ttdvdtd x xx x ρ ρρ ρ ρ (3.1.19) The terms in the second square brack et correspond to holding the volume v fixed and evidently equals the local rate of change: ∫ ∫∫ −Δ+→ΔΔ+Δ+∂∂= )() (0) ,(1lim tvttvt v dvdvt ttdvtdvdtdxρρρ (3.1.20) The region ) () ( tvt tv−Δ+ is swept out in time tΔ. Superimposing the volumes ) (tv and ) ( t tvΔ+ , Fig. 3.1.5, it can be seen that a small element vΔ of ) () ( tvttv−Δ+ is given by (see the example associated with Fig. 1.7.7) s t vΔ⋅Δ=Δ nv (3.1.21) Section 3.1 Solid Mechanics Part III Kelly 322where s is the surface. Thus ∫ ∫ ∫⋅ =⋅Δ+ΔΔ=Δ+Δ→Δ −Δ+→Δ s st tvttvtds t ds t t ttdvt ttnvx nv x x ),( ) ,(1lim ) ,(1lim 0 )() (0ρ ρ ρ (3.1.22) and 3.1.15 is again obtained, from which the continuity equation re sults from use of the divergence theorem. Figure 3.1.5: Evaluation of Eqn. 3.1.22 2. Derivation by Converting to Mass Elements This derivation requires the kinematic relation for the material time derivative of a volume element, 2.5.23: dv dt dvd vdiv /)(= . One has () () 0 div ),(. ≡ +=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+= = = ∫ ∫∫∫ v v v vdv dv dv dvdtddvtdtd dtdmρρρρρ ρ v x & & (3.1.23) The continuity equation then follows, since th is must hold for any arbitrary region of the volume v. Derivation of the Continui ty Equation using a Control Volume (Local Form) The continuity equation can also be derived us ing a differential control volume element. This calculation is similar to that given in §1.6.6, with the velocity v replaced by vρ. 3.1.5 The Continuity Equation (Material Form) From 3.1.9, and using 2.2.53, JdV dv= , [] 0 ),()),,(( )(0 = −∫ VdVt Jtt X Xχ Xρρ (3.1.24) ) ( t tvΔ+)(tv )(ts) ( t tsΔ+ nv s sΔ vΔ Section 3.1 Solid Mechanics Part III Kelly 323Since V is an arbitrary region, the integra nd must vanish everywhere, so that ),()),,(( )(0 t Jtt X Xχ Xρρ= Continuity Equation (Material Form) (3.1.25) This is known as the continuity (mass) equation in the material description. Since 00=ρ& , the rate form of this equation is simply 0)(=Jdtdρ (3.1.26) The material form of the continuity equation, Jρρ=0 , is an algebraic equation, unlike the partial differential equation in the spatial form. However, the two must be equivalent, and indeed the spatial form can be derived dire ctly from this material form: using 2.5.20, vdiv / J dtdJ= , () vdiv)( ρρρρρ +=+= &&& JJ J Jdtd (3.1.27) This is zero, and 0>J , and the spatial continuity equation follows. Example (of Conservation of Mass) Consider a bar of material of length 0l, with density in the undeformed configuration 0ρ and spatial mass density ),(txρ , undergoing the 1-D motion ) 1/( At+=xX , X Xx At+= . The volume ratio (taking unit cross-sectional area) is At J+=1 . The continuity equation in the material form 3.1.25 specifies that ) 1(0 At+=ρρ Suppose now that X X2 02)(lm o=ρ so that the total mass of the bar is ∫=olm d 00)( XXρ . It follows that the spatial mass density is 2 2 02 00 ) 1(2 12 ) 1( At lm At lm At +=+=+=x X ρρ Evaluating the total mass of the bar at time t leads to ∫ ∫+ + +=) 1( 02 2 0) 1( 0 ) 1(1 2),(At l At lo odAt lmdt xx x xρ Section 3.1 Solid Mechanics Part III Kelly 324 which is again m, as required. Figure 3.1.6: a stretching bar The density could have been derived from the equation of continuity in the spatial form: since the velocity is AtAtt t Adtt dt+= = = =− 1)),,(( ),( ,),(),(1 xxχV xv XXxXV one has 01 1=++∂∂ ++∂∂=∂∂+∂∂+∂∂ AtA AtA tv tρρ ρρρρ xx x xv Without attempting to solve this first order partial differentia l equation, it can be seen by substitution that the value for ρ obtained previously satisfies the equation. ■ 3.1.6 Material Derivatives of Integrals Reynold’s Transport Theorem In the above, the material derivative of the total mass carried by a control mass, ∫ vdvtdtd),(xρ , was considered. It is quite of ten that one needs to evaluate material time derivatives of similar volume (and line and surface) integral s, involving other properties, for example momentum or energy. Thus, suppose that ) ,(txA is the distributio n of some property (per unit volume) throughout a volume v (A is taken to be a sec ond order tensor, but what follows applies also to vectors and scalars). Then the rate of change of the total amount of the property carried by the mass system is end of bar (0l==Xx ) at 0=t end of bar (0 0 ), 1( l At l =+= X x ) at time t Section 3.1 Solid Mechanics Part III Kelly 325∫ vdvtdtd),(xA Again, this integral can be evaluated in a number of ways. For example, one could evaluate it using the formal definition of the material derivative, as done above for ρ=A . Alternatively, one can evaluate it using the relation 2.5.23, dv dt dvd vdiv /)(= , through [] []∫ ∫ ∫∫+=⎥⎦⎤ ⎢⎣⎡+= = v v v vdv dv dv dvtdtddvtdtdAv A A A xA xA div ),( ),(. & & (3.1.28) Thus one arrives at Reynold’s transport theorem ()() ()⎪⎪⎪⎪⎪ ⎩⎪⎪⎪⎪⎪ ⎨⎧ +∂∂⋅+∂∂⎥ ⎦⎤ ⎢ ⎣⎡ ∂∂+∂∂ ⎥⎦⎤ ⎢⎣⎡⊗+∂∂⎥ ⎦⎤ ⎢ ⎣⎡ ∂∂+∂∂+∂∂ ⎥⎦⎤ ⎢⎣⎡+⋅+∂∂⎥ ⎦⎤ ⎢ ⎣⎡ ∂∂+⎥⎦⎤ ⎢⎣⎡+ = ∫∫ ∫∫∫ ∫∫ ∫∫ ∫ ∫ skkij vij s vv kkij ij vvij kk k kij ij vvij kk ij v v dsnvA dvtAds dvtdvxvA tAdvtdvAxvvxA tAdvtdvAxv dtdAdvdtd dvtdtd nvAAv AAAv vAAAvA xA divdiv graddiv ),( Reynold’s Transport Theorem (3.1.29) The index notation is shown for the case when A is a second order tens or. In the last of these forms2 (obtained by application of the diverg ence theorem), the first term represents the amount (of A) created within the volume v whereas the second term (the flux term) represents the (volume) rate of flow of the property through the surface. In the last three versions, Reynold’s transport theorem gives th e material derivative of the moving control mass in terms of the derivative of the instanta neous fixed volume in sp ace (the first term). Of course when ρ=A , the continuity equation is recovered. Another way to derive this result is to first c onvert to the reference c onfiguration, so that integration and differentiation commute (since dV is independent of time): () () ( ) ()∫∫ ∫∫ ∫ ∫ + =+=+== = vV VV V v dvt tJdV dVJ JdVJtdtdJdVtdtddvtdtd ),( div),(div),( ),( ),( xAv xAAv A A AXA XA xA && && (3.1.30) 2 also known as the Leibniz formula Section 3.1 Solid Mechanics Part III Kelly 326 Reynold’s Transport Theorem for Specific Properties A property that is given per unit mass is called a specific property . For example, specific heat is the heat per unit mass. Consider then a property B, a scalar, vector or tensor, which is defined per unit mass through a volume. Then the rate of change of the total amount of the prope rty carried by the mass system is simply [] [] ∫∫∫∫∫= = = = v v v v vdvdtddmdtddmdtddvdtddvtdtd B BB B xB ρ ρ ρ ),( (3.1.31) Material Derivatives of Li ne and Surface Integrals Material derivatives of line and surface inte grals can also be eval uated. From 2.5.8, xl x d dt dd=/)( , []∫∫+= xAlA x xA d dtdtd& ),( (3.1.32) and, using 2.5.22, ()()ds dtdsd nlv n ˆ div /ˆT−= , () [ ] ∫∫−+= s sds dstdtdnlv AA nxA ˆ div ˆ),(T & (3.1.33) 3.1.7 Problems 1. A motion is given by the equations 3 3 22 1 2 2 1 1 ),1( , 3 X x tX tX xtX X x =++−= += (a) Calculate the spatial mass density ρ in terms of the density 0ρ (b) Derive a first order ordinary differential equation for the density ρ (in terms of x and t only) assuming that it is independent of position x Section 3.2 Solid Mechanics Part III Kelly 3273.2 The Momentum Principles In Parts I and II, the basic dynamics principl es used were Newton’s Laws, and these are equivalent to force equilibrium and moment e quilibrium. For example, they were used to derive the stress transformation equations in Part I, §3.4 and the Equations of Motion in Part II, §1.1. Newton’s laws there were ap plied to differential material elements. An alternative but completely equi valent set of dynamics laws are Euler’s Laws ; these are more appropriate for finite-sized collectio ns of moving particles, and can be used to express the force and moment equilibrium in terms of integrals. Euler’s Laws are also called the Momentum Principles : the principle of linear momentum (Euler’s first law) and the principle of angular momentum (Euler’s second law). 3.2.1 The Principle of Linear Momentum Momentum is a measure of the tendency of an object to keep movi ng once it is set in motion. Consider first the particle of rigid body dynamics: the (linear) momentum p is defined to be its mass times velocity, v pm= . The rate of change of momentum p& is av v pmdtdmdtmd dtd===)( (3.2.1) and use has been made of the fact that 0 /=dtdm . Thus Newton’s second law, a Fm= , can be rewritten as )(v F mdtd= ( 3 . 2 . 2 ) This equation, formulated by Euler, states that the rate of change of momentum is equal to the applied force . It is called the principle of linear momentum , or balance of linear momentum . If there are no forces applied to a sy stem, the total momentum of the system remains constant; the law in this case is known as the law of conservation of (linear) momentum . Eqn. 3.2.2 as applied to a particle can be ge neralized to the mechanics of a continuum in one of two ways. One could consider a differential element of material, of mass dm and velocity v. Alternatively, one can consider a finite portion of material, a control mass in the current configuration with spatial mass density ),(txρ and spatial velocity field ),(txv . The total linear momentum of this mass of material is ()∫= vdvt t t ,),( )( xvx Lρ Linear Momentum (3.2.3) The principle of linear momentum states that () )( ,),( )( t dvt tdtdt vF xvx L = =∫ρ & (3.2.4) Section 3.2 Solid Mechanics Part III Kelly 328 where ) (tF is the resultant of the forces acting on the portion of material. Note that the volume over which the integrati on in Eqn. 3.2.4 takes place is not fixed; the integral is taken over a fixed portion of material particles , and the space occupied by this matter may change over time. By virtue of the Transport theorem re lation 3.1.31, this can be written as () )( ,),( )( t dvt t t vF xvx L = =∫& &ρ (3.2.5) The resultant force acting on a body is due to the surface tractions t acting over surface elements and body forces b acting on volume elements, Fig. 3.2.1: dvb dst F dv ds t vi si i v s∫∫ ∫∫+= += , )( b t F Resultant Force (3.2.6) and so the principle of linear momentum can be expressed as ∫∫∫=+ v v sdv dv ds v b t &ρ Principle of Linear Momentum (3.2.7) Figure 3.2.1: surface and body forces acting on a finite volume of material The principle of linear momentum, Eqns. 3.2.7, will be used to prove Cauchy’s Lemma and Cauchy’s Law in the next section and, in §3.6, to derive the Equations of Motion. 3.2.2 The Principle of Angular Momentum Considering again the mechanic s of a single particle: the angular momentum is the moment of momentum about an axis, in other words, it is the product of the linear momentum of the particle and the perpendicu lar distance from the axis of its line of action. In the notation of Fig. 3.2.2, the angular momentum h is v rh m×= (3.2.8) which is the vector with magnitude vmd× and perpendicular to the plane shown. t n ds bdv Section 3.2 Solid Mechanics Part III Kelly 329 Figure 3.2.2: surface and body forces acting on a finite volume of material Consider now a collection of particles. The principle of angular momentum states that the resultant moment of the external fo rces acting on the system of particles, M, equals the rate of change of the total a ngular momentum of the particles: dtdhFr M =×= (3.2.9) Generalising to a continuum, the angular momentum is ∫×= vdvv r Hρ Angular Momentum (3.2.10) and the principle of angular momentum is dvvxdtddvbx dstxdvdtddv ds Vk j ijk vkj ijk skj ijkv v s ∫ ∫∫∫∫∫ = +×=×+× ρε ε ερ )()( nnv r br tr Principle of Angular Momentum (3.2.11) The principle of angular momentum, 3.2.11, will be used, in §3.6, to deduce the symmetry of the Cauchy stress. • vmo dr Section 3.3 Solid Mechanics Part III Kelly 3303.3 The Cauchy Stress Tensor 3.3.1 The Traction Vector The traction vector was introduced in Part I, §3.3. To recall, it is the limiting value of the ratio of force over area; for Force FΔ acting on a surface element of area SΔ, it is SF SΔΔ= →Δ 0)(limnt (3.3.1) and n denotes the normal to the surface element. An infinite number of traction vectors act at a point, each acting on different surf aces through the point, defined by different normals. 3.3.2 Cauchy’s Lemma Cauchy’s lemma states that traction vectors act ing on opposite sides of a surface are equal and opposite 1. This can be expressed in vector form: )( )( n nt t−−= Cauchy’s Lemma (3.3.2) This can be proved by applyi ng the principle of linear mo mentum to a collection of particles of mass mΔ instantaneously occupying a sma ll box with parallel surfaces of area sΔ, thickness δ and volume s vΔ=Δδ, Fig. 3.3.1. The resultant surface force acting on this matter is s sΔ+Δ−)( )( n nt t . Figure 3.3.1: traction acting on a small portion of material particles The total linear momentum of the matter is ∫∫Δ Δ= m Vdm dv v vρ . By the mean value theorem (see Appendix A to Chapter 1, §1.B.1), this equals mΔv , where v is the velocity at some interior point. Similarl y, the body force acting on the matter is v dv VΔ=∫Δb b , where b is the body force (per unit volume) acting at some interior point. The total mass 1 this is equivalent to Newton’s (third) law of action and reaction – it seems like a lot of work to prove this seemingly obvious result but, to be consistent, it is supposed that the only fundamental dynamic laws available here are the principles of linear and angular momentum, and not any of Newton’s laws )(nt )(nt−n n−sΔ thickness δ Section 3.3 Solid Mechanics Part III Kelly 331can also be written as v dv m VΔ==Δ∫Δρρ . From the principle of linear momentum, Eqn. 3.2.7, and since mΔ does not change with time, []dtdsdtdvdtdm mdtdv s sv v vv b t tn nΔ=Δ=Δ=Δ=Δ+Δ+Δ−δρρ)( )( (3.3.3) Dividing through by sΔ and taking the limit as 0→δ , one finds that )( )( n nt t−−= . Note that the values of )( )(,n nt t− acting on the box with finite thickness are not the same as the final values, but a pproach the final values at the surface as 0→δ . 3.3.3 Stress In Part I, the components of the traction v ector were called stress components, and it was illustrated how there were nine stress components associated with each material particle. Here, the stress is defined more formally, Cauchy’s Law Cauchy’s Law states that there exists a Cauchy stress tensor σ which maps the normal to a surface to the trac tion vector acting on that surface, according to jij i n tσ= = ,nσt Cauchy’s Law (3.3.4) or, in full, 3 33 2 32 131 33 23 2 22 121 23 13 2 12 111 1 n n n tn n n tn n n t σσσσσσσσσ ++=++=++= (3.3.5) Note: • many authors define the stress tensor as σnt= . This amounts to the definition used here since, as mentioned in Part I, and as will be (re -)proved below, the stress tensor is symmetric, ji ijσσ== ,Tσσ • the Cauchy stress refers to the current configuration, that is, it is a measure of force per unit area acting on a surface in the current configuration. Stress Components Taking Cauchy’s law to be true (it is proved below), the components of the stress tensor with respect to a Cartesian coordinate system are, from 1.9.4 and 3.3.4, ()j i j i ijete eσe ⋅==σ (3.3.6) which is the ith component of the traction vect or acting on a surface with normal je. Note that this definition is inconsistent with that given in Part I, §3.2 – there, the first Section 3.3 Solid Mechanics Part III Kelly 332subscript denoted the direction of the nor mal – but, again, the two definitions are equivalent because of the symmetry of the stress tensor. The three traction vectors acting on the surface elements whose outward normals point in the directions of the three base vectors je are jjeσ te=)(, () () () 333 223 13332 2 22 12331 221 11 32 e e e te e e te e e t 1e1e1e1 σσσσσσσσσ ++=++=++= (3.3.7) Eqns. 3.3.6-7 are illustrated in Fig. 3.3.2. Figure 3.3.2: traction acting on surfaces with normals in the coordinate directions; (a) traction vectors, (b) stress components Proof of Cauchy’s Law The proof of Cauchy’s law essentially follows the same method as used in the proof of Cauchy’s lemma. Consider a small tetrahedral free-body, with vertex at the origin, Fig. 3.3.3. It is required to determine the traction t in terms of the nine stress co mponents (which are all shown positive in the diagram). Let the area of the base of the tetrahedran, with normal n, be sΔ. The area 1ds is then αcossΔ , where α is the angle between the planes, as shown in Fig. 3.3.3b; this angle is the same as that between the vectors n and 1e, so () sns s Δ=Δ⋅=Δ1 1 1 en , and similarly for the other surfaces: sn sΔ=Δ2 2 and sn sΔ=Δ3 3 . 3x 2x2e3e 1e()1et()2et()3et 1x21σ 11σ31σ 12σ22σ32σ23σ33σ 13σ3x 2x 1x (a) (b) Section 3.3 Solid Mechanics Part III Kelly 333 Figure 3.3.3: free body diagram of a tetr ahedral portion of material; (a) traction acting on the material, (b) relations hip between surface areas and normal components The resultant surface force on the body, acting in the 1x direction, is sn sn sn st Δ−Δ−Δ−Δ3 13 2 12 111 1 σ σσ Again, the momentum is MΔv , the body force is vΔb and the mass is s h v m Δ=Δ=Δ )3/(ρρ , where h is the perpendicular distance from the origin (vertex) to the base. The principle of linear momentum then states that dtvds h s hbsn sn sn st1 1 3 13 2 12 111 1 )3/( )3/( Δ=Δ+Δ−Δ−Δ−Δ ρ σ σσ Again, the values of the traction and stress components on the faces will in general vary over the faces, so the values used in this equation are average values over the faces. Dividing through by sΔ, and taking the limit as 0→h , one finds that 3 13 2 12 111 1 n n n t σσσ ++= and now these quantities, 13 12 11 1 , ,, σσσt , are the values at the origin. The equations for the other two traction components ca n be derived in a similar way. Normal and Shear Stress The stress acting normal to a surface is given by )(ntn⋅=Nσ (3.3.8) The shear stress acting on the surf ace can then be obtained from 3x 2x 1xn()nt 23σ13σ 33σ12σ 22σ 32σ31σ21σ11σ1sΔ 3sΔ••n 2sΔ α 1e (a) (b) Section 3.3 Solid Mechanics Part III Kelly 334 22)( N S σ σ −=nt (3.3.9) Example The state of stress at a point is given in the matrix form [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −− = 12 32 213 12 ijσ Determine (a) the traction vector acting on a plane through the point whose unit normal is 3 2 1ˆ)3/2(ˆ)3/2(ˆ)3/1(ˆ e e e n − += (b) the component of this traction act ing perpendicular to the plane (c) the shear component of traction. Solution (a) The traction is ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −− = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −− = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ 392 31 221 12 32 213 12 31 321 33 32 3123 22 2113 12 11 )ˆ( 3)ˆ( 2)ˆ( 1 nnn ttt σσσσσσσσσ nnn or 3 2 1)ˆ(ˆˆ3ˆ)3/2( ee e tn−+−= . (b) The component normal to th e plane is the projection of )ˆ(nt in the direction of nˆ, i.e. .4.29/22)3/2()3/2(3)3/1)(3/2(ˆ)ˆ(≈=++ −=⋅= n tn Nσ (c) The shearing compone nt of traction is [] [ ] [ ][] []3 2 13 2 1)ˆ( ˆ)27/17(ˆ)27/37(ˆ)27/40(ˆ)27/44(1 ˆ)27/44(3 ˆ)27/22()3/2(ˆ)9/22( e e ee e en tn + + −=+−+ −+ −−=−=Sσ i.e. of magnitude 1.2 )27/17()27/37()27/40(2 2 2≈ + + − , which equals 22)ˆ(ˆ Nσ−nt . ■ Section 3.4 Solid Mechanics Part III Kelly 3353.4 Properties of the Stress Tensor 3.4.1 Stress Transformation Let the components of the Cauchy stress tensor in a coordinate system with base vectors ie be ijσ. The components in a second coordi nate system with base vectors je′, ijσ′, are given by the tensor transformation rule 1.10.5: pq qj pi ij QQσ σ=′ (3.4.1) where ijQ are the direction cosines, j i ijQ ee′⋅= . Isotropic State of Stress Suppose the state of stress in a body is [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = 000 0 00 00 0 σσσ σ One finds that the application of the tensor transformation rule yields the very same components no matter what the coordina te system. This is termed an isotropic state of stress, or a spherical state of stress (see §1.13.3). One ex ample of isotropi c stress is the stress arising in fluid at rest, which cannot support shear stress, in which case Iσ p−= (3.4.2) where the scalar p is the fluid hydrostatic pressure . For this reason, an isotropic state of stress is also referred to as a hydrostatic state of stress. A note on the Transformation Formula Using the vector transformation rule 1.5.5, the traction and normal transform according to [][][][ ][][]nQ ntQ tT T,=′ =′ . Also, Cauchy’s law transforms according to [] [] [] nσ t′′=′ which can be written as [][][][][]nQσ tQT T′= , so that, pre-multiplying by []Q, and since []Q is orthogonal, [][ ] [ ][]{}[]n QσQ tT′= , so [][][][]TQσQσ′= , which is the inverse tensor transformation rule 1.13.6a, showing the internal consistency of the theory. In Part I, Newton’s law was applied to a mate rial element to derive the two-dimensional stress transformation equations, Eqn. 3.4.7 of Part I. Cauchy’s law was proved in a similar way, using the principle of moment um. In fact, Cauchy’s law and the stress transformation equations are equivalent. Gi ven the stress components in one coordinate system, the stress transformation equations give the components in a new coordinate system; particularising this, they give the st ress components, and thus the traction vector, Section 3.4 Solid Mechanics Part III Kelly 336acting on new surfaces, oriented in some way wi th respect to the original axes, which is what Cauchy’s law does. 3.4.2 Principal Stresses Since the stress σ is a symmetric tensor, it has three real eigenvalues 3 2 1,,σσσ , called principal stresses , and three corresponding orth onormal eigenvectors called principal directions . The eigenvalue problem can be written as n nσ tnσ==)( (3.4.3) where n is a principal direction and σ is a scalar principal stress. Since the traction vector is a multiple of the unit normal, σ is a normal stress component. Thus a principal stress is a stress which acts on a plane of zero shear stress, Fig. 3.4.1. Figure 3.4.1: traction acting on a plane of zero shear stress The principal stresses are the root s of the characteristic equation 1.11.5, 03 22 13=−+− I I Iσσσ (3.4.4) where, Eqn. 1.11.6-7, 1.11.17, ()[] () [] 32132 23 122 12 332 31 222 23 11 33 22 113 21 2 23 3 31 313 32 212 312 232 12 11 33 33 22 22 112 2 21 23 2 133 22 111 2dettr trtr trtr trtr σσσσσσσσσσσσσσσσσσσσσσσσσσσσσσσσσσσσ =+−−− ==+ −=++=−−−++=−=++=++== σσσσσσσσ III (3.4.5) no shear stress – only a normal component to the traction n33 22 11)(e e e tnt t t ++= Section 3.4 Solid Mechanics Part III Kelly 337The principal stresses and principal directions are properties of the stress tensor, and do not depend on the particular axes chosen to describe the state of stress., and the stress invariants 3 2 1,, III are invariant under coordinate transformation. c.f. §1.11.1. If one chooses a coordinate system to coinci de with the three eigenvectors, one has the spectral decomposition 1.11.11 and the stress matrix takes the simple form 1.11.12, [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = ⊗=∑ = 3213 10 00 00 0 ,ˆ ˆ σσσ σ σ n nσ ii ii (3.4.6) Note that when two of the principal stresses are equal, one of the principal directions will be unique, but the other two will be arbitrary – one can choose any two principal directions in the plan e perpendicular to the uniquely de termined direction, so that the three form an orthonormal set. This stress state is called axi-symmetric . When all three principal stresses are equal, one has an isotro pic state of stress, and all directions are principal directions. 3.4.3 Maximum Stresses Directly from §1.11.3, the three principal stresses include the maximum and minimum normal stress components acting at a point. This result is re-derived here, together with results for the maximum shear stress Normal Stresses Let 3 2 1,,eee be unit vectors in the principal directions and consider an arbitrary unit normal vector 33 22 11 e e e n n n n ++= , Fig. 3.4.2. From 3.3.8 and Cauchy’s law, the normal stress acting on the plane with normal n is ()nnσ n tn⋅=⋅=)( Nσ (3.4.7) Figure 3.4.2: normal stress acting on a plane defined by the unit normal n 3 2 1n()ntNσ principal directions Section 3.4 Solid Mechanics Part III Kelly 338 With respect to the principal stresses, using 3.4.6, 333 222 111)(e e e nσ tnn n n σσσ ++== (3.4.8) and the normal stress is 2 332 222 11 n n nN σσσσ ++= (3.4.9) Since 12 32 22 1 =++ n n n and, without loss of generality, taking 3 2 1σσσ≥≥ , one has ( )N n n n n n n σσσσ σσ =++≥++=2 332 222 112 32 22 1 1 1 (3.4.10) Similarly, ( )32 32 22 1 32 332 222 11 σ σσσσσ ≥++≥++= n n n n n nN (3.4.11) Thus the maximum normal stress acting at a point is the maximum principal stress and the minimum normal stress acting at a poi nt is the minimum principal stress. Shear Stresses Next, it will be shown that the maximum sh earing stresses at a point act on planes oriented at 45 o to the principal planes and that they have magnitude equal to half the difference between the principal stresses. From 3.3.39, 3.4.8 and 3.4.9, the shear stress on the plane is () ( )22 332 222 112 32 32 22 22 12 12n n n n n nS σσσσσσσ ++−++= (3.4.12) Using the condition 12 32 22 1 =++ n n n to eliminate 3n leads to () () () ()[ ]2 32 2 3 22 1 3 12 32 22 32 22 12 32 12σσσσσσσσσσσ +−+−−+−+−= n n n nS (3.4.13) The stationary points are now obtained by equating the partial derivatives with respect to the two variables 1n and 2n to zero: ()() () () []{} ()() () () []{} 0 20 2 2 2 3 22 1 3 1 3 2 3 2 2 222 2 3 22 1 3 1 3 1 3 1 1 12 =−+−−−−=∂∂=−+−−−−=∂∂ n n nnn n nn SS σσσσσσσσσσσσσσσσσσ (3.4.14) One sees immediately that 02 1==n n (so that 13±=n ) is a solution; this is the principal direction 3e and the shear stress is by definition zer o on the plane with this normal. In Section 3.4 Solid Mechanics Part III Kelly 339this calculation, the component 3n was eliminated and 2 Sσ was treated as a function of the variables ),(2 1nn . Similarly, 1n can be eliminated with ) ,(3 2nn treated as the variables, leading to the solution 1en=, and 2n can be eliminated with ) ,(3 1nn treated as the variables, leading to the solution 2en=. Thus these solutions lead to the minimum shear stress value 02=Sσ . A second solution to Eqn. 3.4.14 can be seen to be 2/1 ,02 1 ±==n n (so that 2/13±=n ) with corresponding shear stress values ()2 3 2 41 2σσσ −=S . Two other solutions can be obtained as described earlier, by eliminating 1n and by eliminating 2n. The full solution is listed below, and these are evidently the maximum (absolute value of the) shear stresses acting at a point: 2 11 33 2 21,0, 21, 2121, 21,0, 2121, 21, 21,0 σσσσσσσσσ −=⎟ ⎠⎞⎜ ⎝⎛±±=−=⎟ ⎠⎞⎜ ⎝⎛±±=−=⎟ ⎠⎞⎜ ⎝⎛±±= SSS nnn (3.4.15) Taking 3 2 1σσσ≥≥ , the maximum shear stress at a point is ()3 1 max21σστ −= ( 3 . 4 . 1 6 ) and acts on a plane with normal oriented at 45o to the 1 and 3 principal directions. This is illustrated in Fig. 3.4.3. Figure 3.4.3: maximum shear stress at apoint Example (maximum shear stress) Consider the stress state 13 maxτ maxτprincipal directions Section 3.4 Solid Mechanics Part III Kelly 340[] ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ −−−= 1 12 012 6 00 0 5 ijσ This is the same tensor considered in the ex ample of §1.11.1. Using the results of that example, the principal stresses are 15 ,5 ,103 2 1 −=== σσσ and so the maximum shear stress at that point is ()225 21 3 1 max =−=σστ The planes and direction upon which th ey act are shown in Fig. 3.4.4. Figure 3.4.4: maximum shear stress ■ 3x 1x2x3ˆn1ˆn 2ˆno37maxτ Section 3.5 Solid Mechanics Part III Kelly 3413.5 Stress Measures for Large Deformations Thus far, the surface forces acting within a ma terial have been described in terms of the Cauchy stress tensor σ. The Cauchy stress is also called the true stress , to distinguish it from other stress tensors, some of which wi ll be discussed below. It is called the true stress because it is a true measure of the force per unit area in the current, deformed, configuration. When the defo rmations are small, there is no distinction to be made between this deformed configuration and some reference, or undeformed, configuration, and the Cauchy stress is the sensible way of describing the action of surface forces. When the deformations are large, however, one needs to refer to some reference configuration. In this case, there are a number of different possible ways of defining the action of surface forces; some of these stre ss measures often do not have as clear a physical meaning as the Cauchy stre ss, but are useful nonetheless. 3.5.1 The First Piola – Kirchhoff Stress Tensor Consider two configurations of a material, the reference and current configurations. Consider now a vector element of surface in the reference configuration, dSN , where dS is the area of the element and N is the unit normal. After deformation, the material particles making up this area element now occupy the element defined by dsn, where ds is the area and n is the normal in the current c onfiguration. Suppose that a force fd acts on the surface element (in the current confi guration). Then by definition of the Cauchy stress ds d nσf= (3.5.1) The first Piola-Kirchhoff stress tensor P (which will be called the PK1 stress for brevity) is defined by dS d NPf= (3.5.2) The PK1 stress relates the force acting in the current configuration to the surface element in the reference configuration. Since it relates to bot h configurations, it is a two-point tensor. The (Cauchy) traction vector was defined as dsdft= , nσt= (3.5.3) Similarly, one can introduce a PK1 traction vector T such that dSdfT= , NPT= (3.5.4) Whereas the Cauchy traction is the actual phys ical force per area on the element in the current configuration, the PK1 traction is a fictitious quantity – the force acting on an element in the current configuration divide d by the area of the corresponding element in Section 3.5 Solid Mechanics Part III Kelly 342the reference configura tion. Note that, since dS ds d T tf== , it follows that T and t act in the same direction (but have different magnitudes), Fig. 3.5.1. Figure 3.5.1: Traction vectors Uniaxial Tension Consider a uniaxial tensile test whereby a specimen is stretched uniformly by a constant force f, Fig. 3.5.2. The initial cross-sec tional area of the specimen is 0A and the cross- sectional area of the specimen at time t is )(tA. The Cauchy (true) stress is )()(tAtfσ= (3.5.5) and the PK1 stress is 0AfP= (3.5.6) This stress measure, force over area of the unde formed specimen, as used in the uniaxial tensile test, is also called the engineering stress . Figure 3.5.2: Uniaxial tension of a bar The Nominal Stress The PK1 stress tensor is also called the nominal stress tensor . Note that many authors use a different definition for the nominal stress, namely PNT= , and then define the PK1 stress to be the transpose of this P. Thus all authors use th e same definition for the PK1 stress, but a slightly different definition for the nominal stress. current configuration reference configuration dS dsN ntTdS ds d T tf== f current configuration Section 3.5 Solid Mechanics Part III Kelly 343 Relation between the Cauchy and PK1 Stresses From the above definitions, dS ds NP nσ= (3.5.7) Using Nanson’s formula, 2.2.59, dS J ds NF nT−= , T 1T FPσFσ P −− == JJ PK1 stress (3.5.8) The Cauchy stress is symmetric, but the defo rmation gradient is not. Hence the PK1 stress tensor is not symmetric , and this restricts its use as an alternative stress measure to the Cauchy stress measure. In fact, this la ck of symmetry and lack of a clear physical meaning makes it uncommon for the PK1 stress to be used in the modeling of materials. It is, however, useful in the description of the momentum ba lance laws in the material description, where P plays an analogous role to that played by the Cauchy stress σ in the equations of motion (see later). 3.5.2 The Second Piola – Kirchhoff Stress Tensor The second Piola – Kirchhoff stress tensor, or the PK2 stress , S, is defined by T 1−−= FσF SJ PK2 stress (3.5.9) Even though the PK2 does not admit a physical interpretation (except in the simplest of cases, but see the interpreta tion below), there are three good reasons for using it as a measure of the forces acting in a ma terial. First, one can see that () ()()T T1T1TTTT 1 −− − − −−= = FσF FσF σF F and since the Cauchy stress is symmetric, so is the PK2 stress: TSS= (3.5.10) A second reason for using the PK2 stress is that , together with the Euler-Lagrange strain E, it gives the power of a deforming material (s ee later). Third, it is parameterized by material coordinates only, that is, it is a ma terial tensor field, in the same way as the Cauchy stress is a spatial tensor field. Note that the PK1 and PK2 stresses are related through PFS FSP1,−= = (3.5.11) Section 3.5 Solid Mechanics Part III Kelly 344The PK2 stress can be interpreted as follows : take the force vector in the current configuration fd and locate a corresponding vector in the undeformed configuration according to fFf d d1−= . The PK2 stress tensor is th is fictitious force divided by the corresponding area element in the reference configuration: dS d SNf= , and 3.5.9 follows from 3.5.2, 3.5.8: dS J dS d NFσ NPfT−== 3.5.3 Alternative Stress Tensors Some other useful stress m easures are described here. The Kirchhoff Stress The Kirchhoff stress tensor τ is defined as στJ= Kirchhoff Stress (3.5.12) It is a spatial tensor field pa rameterized by spatial coordinates. One reason for its use is that, in many equations, the Cauchy stress appe ars together with the Jacobian and the use of τ simplifies formulae. Note that the Kirchhoff stress is the push forward of the PK2 stress; from 2.12.9b, 2.12.11b, () ()T 1 # 1 *T # * −− −= === τFFτ SFSF Sτ χχ (3.5.13) The Corotational Cauchy Stress The corotational stress σˆ is defined as σRRσTˆ= Corotational Stress (3.5.14) where R is the orthogonal rotation tensor. Wher eas the Cauchy stress is related to the PK2 stress through T 1SFFσ−=J , the corotational stress is related to the PK2 stress through (with F replaced by the right (s ymmetric) stretch tensor U): ()()()σRR UFσFU UσF FU SUUσT T 1 T 1 1 T 1ˆ = = = =− − −− − −J J J (3.5.15) The corotational stress is define d on the intermediate configuration of Fig. 2.10.8. It can be regarded as the push forward of the PK 2 stress from the reference configuration through the stretch U, scaled by 1−J (Eqn. 2.12.28b): ()() ( ) USU USU UG UG g g S σ GU1 T 1 1 1 # *1ˆ ˆ ˆ− − − − −= =⊗ =⊗ = = J J SJ SJ Jj iij j iijχ (3.5.16) Section 3.5 Solid Mechanics Part III Kelly 345or as the pull-back of the Cauchy stress with respect to R (Eqn. 2.12.27f): ()() σRR g g σσ gRT # 1 *ˆ ˆ ˆ =⊗= =− j iijσ χ (3.5.17) The Biot Stress The Biot (or Jaumann ) stress tensor BT is defined as USPR T ==T B Biot Stress (3.5.18) From 3.5.11, it is similar to the PK1 stress, only with F replaced by U. Example Consider a pre-stressed thin plate with 0 1 11σσ= , 0 2 22σσ= , that is, it has a non-zero stress although no forces are acting1, Fig. 3.5.3. In this initial state, IF= and, considering a two-dimens ional state of stress, ⎥ ⎦⎤ ⎢ ⎣⎡======0 20 1 B00ˆ σσTτσSPσ The material is now rotated as a rigid body o45 counterclockwise – the stress-state is “frozen” within the material and rotates with it. Then ⎥ ⎦⎤ ⎢ ⎣⎡−== 2/12/12/1 2/1RF The stress components with respect to the rotated * ix axes shown in Fig. 3.5.3b are 0 1* 11σσ= , etc.; the components with re spect to the spatial axes ix can be found from the stress transformation rule [][][][][][][]T * * TRσR QσQσ = = , and so ()() () ()⎥ ⎦⎤ ⎢ ⎣⎡ + −− +=0 20 1 21 0 20 1 210 20 1 21 0 20 1 21 σσσσσσσσσ Note that the Cauchy stress changes with this rigid body rotati on. Further, with 1=J , ⎥ ⎦⎤ ⎢ ⎣⎡===⎥ ⎦⎤ ⎢ ⎣⎡ −==0 20 1 B 0 20 10 20 1 00ˆ , 2/ 2/2/ 2/, σσ σ σσ σTσS Pστ Note that the PK1 stress is not symmetric. Now attach axes *x to the material and rotate these axes with the specimen as it rotates, as in Fig. 3.5.3b. The components with respect 1 for example a piece of metal can be deformed; when the load is removed it is often pre-stressed – there is a non-zero state of stress in the material Section 3.5 Solid Mechanics Part III Kelly 346to these rotated axes give the corotational stre ss; the corotational st ress is the stress in a body, taking out the stress changes caused by rigid body rotations – one says that the corotational stress (and PK2 st ress) “rotate” with the body. Figure 3.5.3: Pre-stressed material; (a) original position, (b) rotated configuration ■ 3.5.4 Small deformations From §2.7, when the deformations are sm all, neglecting terms involving products of displacement gradients, ) grad(O ) grad(O grad 2u I u u IF += ++= (3.5.19) Here, ) grad(O u means terms of the order of displacement gradients (and higher) have been neglected and 2) grad(O u means terms of the order of products of displacement gradients (and higher) have been neglected. Also, () ) grad(O1) grad(O div1 ) grad(O grad detdet 2 2u u u u u IF += ++= ++==J (3.5.20) From 3.5.8 and 3.5.9, using 3.5.19-20, one has ) grad(O ) grad(O) grad(O ) grad(O TT u S u σ FSFσu P u σ FPσ += +→=+= +→= JJ (3.5.21) In the linear theory then, with 0 ) grad(O→u , the stress measures encountered in this section are all equivalent. 1x2x 0 2σ0 1σ 1x2x 0 2σ0 1σ* 1x* 2x reference confi guration rotated confi guration (a) (b) Section 3.5 Solid Mechanics Part III Kelly 3473.5.5 Objective Stress Tensors In order to ascertain the objectivity of the stress tensors, first note that, by definition , force is an objective vector, and therefore so also is the traction vector. Similarly for the normal vector. The normal and traction vectors transform under an observer transformation according to 2.8.10, Qn n=* and Qtt=*. Then ()* T * *T *TnQQσ t nσQ tQ σn t =→ = →= (3.5.22) and so T *QQσσ= ; according to 2.8.12, the Cauchy st ress is objective. The PK2 stress S is objective, since it is a material tensor unaffected by an observer transformation. For the PK1 stress, using 2.8.23, () ()()T T TT* ** * − − −= = = σF Q QFQQσ Fσ P J J J (3.5.23) and so, according to 2.8.16, P is objective (transforming like a vector, being a two-point tensor). 3.5.6 Objective Stress Rates One needs to incorporate stress rates in m odels of materials wher e the response depends on the rate of stressing, for example with viscoe lastic materials. As discussed in §2.8.5, the rates of objective tensors ar e not necessarily objective. As discussed in §2.12.3, the Lie derivative of a spatial second order tensor is objective. For the Cauchy stress, there are a number of different objective rates one can use, based on the Lie derivative (see Eqns. 2.8.35-36, 2.12.41, 2.12.44): Cotter-Rivlin stress rate σlσlσ++T& σb vL= Jaumann stress rate σw wσσ+−& ()σσ# vb v 21L L+= (3.5.24) Oldroyd stress rate2 Tσl lσσ−−& σ# vL= Stress rates of other spatial st ress tensors can be defined in the same way, for example the Oldroyd rate of the Kirchhoff stress tensor is Tτl lττ−−& . The material derivative of the material PK2 stress tensor, S&, is objective. The push forward of S& is, from 2.12.9b, ()T# * FSF S&&=χ (3.5.25) 2 this is sometimes called the contravariant Oldroyd stress rate, to distinguish it from the Cotter-Rivlin rate, which is also sometimes called the covariant Oldroyd stress rate Section 3.5 Solid Mechanics Part III Kelly 348This push forward, scaled by the inverse of the Jacobian, T 1FSF&−J is called the Truesdell stress rate . This can be expressed in te rms of the Cauchy stress by using 3.5.9, and then 2.5.20, 2.5.5: () ()σdσl lσσF FσF FσF FσF FσFF FFσF F trTT. T 1 T 1 T. 1 T 1 1 T T 1 1 +−−=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ + + + =−− −− −− −− − −− − && & J J J J J JdtdJ (3.5.26) Thus far, objective rates have been constr ucted by pulling back, taking derivatives and pushing forward. One can construct object ive rates also by pulling back and pushing forward with the rotation tensor R only, since it is the rotation which causes the stress rates to be non-objective. For example, σ # vL, setting RF=, is, from 3.5.17 and 2.12.27b, ()()[] ()[]() () σΩσΩσRRσRRσRRσRRσ σ gR gRgR R Rdtd dtd −+=++ =⎟ ⎠⎞⎜ ⎝⎛=⎟ ⎠⎞⎜ ⎝⎛ − && & &T T T Tˆ# * ˆ# 1 * * ˆχ χχ (3.5.27) where TRRΩ&=R is the skew-symmetric angular ve locity tensor 2.6.3. The stress rate 3.5.27 is called the Green-Naghdi stress rate . From the above, the Green-Naghdi rate is the push forward of the time derivative of the corotational stress. Example Consider again the example di scussed at the end of §3.5.3, only let the plate rotate at constant angular velocity ω, so ()() () ()()( ) () ()⎥⎦⎤ ⎢⎣⎡ −− −==⎥⎦⎤ ⎢⎣⎡ −==t tt t t tt t ω ωω ωωω ωω ω sin coscos sin,cos sinsin cosRF RF && Again, using the stress transformation rule [][][][][][][]T * * TRσR QσQσ = = , () ()()()() ()()() () ()⎥ ⎦⎤ ⎢ ⎣⎡ + −− += t t t tt t t t ωσωσσσωωσσωω ωσωσ 2 0 22 0 10 20 10 20 12 0 22 0 1 cos sin sin cossin cos sin cosσ and, with 1=J , () () () ()⎥ ⎦⎤ ⎢ ⎣⎡===⎥ ⎦⎤ ⎢ ⎣⎡ −==0 20 1 B 0 20 10 20 1 00ˆ , cos sinsin cos, σσ σω σωσω σωTσS Pστ t tt t Also, Section 3.5 Solid Mechanics Part III Kelly 349⎥⎦⎤ ⎢⎣⎡−=====− 011 0T 1ωRΩ RR FFwl && Then () () () ()0 TσS P ===⎥ ⎦⎤ ⎢ ⎣⎡ −− −=⋅ B 0 20 10 20 1ˆ , sin coscos sin&& & σω σωσω σωω t tt t and () ()()()() ( )( ) () ()() ( ) () ()()⎥ ⎦⎤ ⎢ ⎣⎡ − +− −− − − −=0 20 10 20 12 20 20 12 2 0 20 1 cos sin2 sin cossin cos cos sin2 σσωω σσω ωσσω ω σσωωω t t t tt t t tσ& For a rigid body rotation, it can be seen that the definitions of the Cotter-Rivlin, Jaumann, Oldroyd, Truesdell and Green-Naghdi rates ar e equivalent, and they are all zero: 0σw wσσ =+−& This is as expected since objective stress ra tes for two configurations which differ by a rigid body rotation will, by definition, be equa l (the stress components will not change); they are zero in the reference configurat ion and so will be zero in the rotated configuration. ■ 3.5.7 Problems 1. Consider the case of uniaxial stress, wher e a material with initial dimensions length 0l, breadth 0w and height 0h deforms into a component with dimensions length l, breadth w and height h. The only non-zero Cauchy stress component is 11σ, acting in the direction of the length of the component. (a) write down the motion equations in the material description, ) (X xχ= (b) calculate the deformation gradient F and confirm that Fdet=J is the ratio of the volume in the current configuration to that in the initial configuration (c) Calculate the PK1 stress. How is it related to the Cauchy stress for this uniaxial stress-state? (d) calculate the PK2 stress 2. A material undergoes the deformation 3 3 2 1 2 1 1 , , 3 X x XtX xtX x = += = The Cauchy stress at a point in the material is [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −− = 0 0 00 202 t tt t σ (a) Calculate the PK1 and PK2 stresses at the point (check that PK2 is symmetric) Section 3.5 Solid Mechanics Part III Kelly 350(b) Calculate the expressions ESdσ FP & & :,: ,: J (for E&, use the expression 2.5.18b, dFFET=& ). In these expressions, d is the rate of deformation tensor. (You should get the same result for all three cases, since they a ll give the rate of internal work done by the stresses duri ng the deformation, per unit reference volume – see later) 3. Show that the Oldroyd rate of the Kirchhoff stress, Tτl lττ−−& , is equal to the Jacobian times the Truesdell stre ss rate of the Cauchy stress, 3.5.26. Section 3.6 Solid Mechanics Part III Kelly 3513.6 The Equations of Motion and Symmetry of Stress In Part II, §1.1, the Equations of Motion we re derived using Newton’s Law applied to a differential material element. Here, they are derived using the principle of linear momentum. 3.6.1 The Equations of Motion (Spatial Form) Application of Cauchy’s law σn t= and the divergence theorem 1.14.21 to 3.2.7 leads directly to the global form of the equations of motion [] ∫ ∫∫ ∫= ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ +∂∂=+ vi vi jij v vdvv dvbxdv dv & & ρσρ , div v bσ (3.6.1) The corresponding local form is then dtdvbx dtdi i jijρσρ =+∂∂=+ , divvbσ Equations of Motion (3.6.2) The term on the right is called the inertial, or kinetic, term, representing the change in momentum. The material time derivative of the spatial velocity field is () vvv vgrad+∂∂=t dtd so ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂+∂∂+∂∂=3 31 2 21 1 11 1 1vxvvxvvxv tv dtdv, etc. and it can be seen that the equations of motion are non-linear in the velocities. Equations of Equilibrium When the acceleration is zero, the equations reduce to the equations of equilibrium, 0 div=+bσ Equations of Equilibrium (3.6.3) Flows A flow is a set of quantities associated with the system of forces t and b, for example the quantities ρ,,σv . A flow is steady if the associated spatial quantities are independent of time. A potential flow is one for which the velocity fiel d can be written as the gradient of a scalar function, φgrad=v . An irrotational flow is one for which 0 curl=v . Section 3.6 Solid Mechanics Part III Kelly 3523.6.2 The Equations of Motion (Material Form) In the spatial form, the linear momentum of a mass element is dvvρ . In the material form it is dVV0ρ . Here, V is the same velocity as v, only it is now expressed in terms of the material coordinates X, and dV dv0ρρ= . The linear momentum of a collection of material particles o ccupying the volume v in the current configuration can thus be expressed in terms of an inte gral over the corresponding volume V in the reference configuration: ()( )∫= VdVt t , )(0 XVX Lρ Linear Momentum (Material Form) (3.6.4) and the principle of linear mo mentum is now, using 3.1.31, ()( ) )( ,0 0 t dVdtddVtdtd V VFVXVX ≡ =∫ ∫ρ ρ (3.6.5) The external forces F to be considered are those acting on the current configuration. Suppose that the surface for ce acting on a surface element ds in the current configuration is dS ds d T t f==surf , where t and T are, respectively, the Cauchy traction vector and the PK1 traction vector (Eqns. 3.5.3- 4). Also, just as the PK1 stress measures the actual force in the current configuration, but per unit surf ace area in the reference configuration, one can introduce the reference body force B: this is the actual body force acting in the current configuration, per unit volume in the reference configuration. Thus if the body force acting on a volume element dv in the current configuration is bodyfd , then dV dv d B b f ==body (3.6.6) The resultant force acting on the body is then dVB dST F dV dS t Vi Si i V S∫∫ ∫∫+= += , )( B T F (3.6.7) Using Cauchy’s law, PNT= , where P is the PK1 stress, and the divergence theorem 1.12.21, 3.6.5 and 3.6.7 lead to [] ∫ ∫=+ V VdVdtddVVBP0 Div ρ (3.6.8) and the corresponding local form is dtdVBXP dtdi i jij 0 0 , Div ρ ρ =+∂∂=+VBP Equations of Motion (Material Form) (3.6.9) Section 3.6 Solid Mechanics Part III Kelly 353Derivation from the Spatial Form The equations of motion can also be derived di rectly from the spatial equations. In order to do this, one must first show that ()TDiv−FJ is zero. One finds that (using the divergence theorem, Nanson’s formula 2.2.59 and the fact that 0 div=I ) () ()00 div Div 11T T =∂∂= == =∂∂= === = ∫∫∫∫∫∫∫∫∫ ∫ −−− − dvxdsn dsn dSN JF dVXJFdv ds ds dS J dV J v iij sj ij si Si ji V jjiv s s S V δδI In n NF F (3.6.10) This result is known as the Piola identity . Thus, with the PK1 stress related to the Cauchy stress through 3.5.8, T−=σF PJ , and using identity 1.14.16c, ()() () () TT TT : Grad: Grad DivDiv Div −− −− =+ == FσFσ FσFσ P JJ JJ (3.6.11) From 2.2.8c, σ P div Div J= (3.6.12) Then, with JdV dv= and 3.6.6, the equations of moti on in the spatial form can now be transformed according to [] []∫ ∫∫ ∫=+ → =+ V V v vdV dV dv dv V BP v bσ & &0 Div div ρ ρ as before. 3.6.3 Symmetry of the Cauchy Stress It will now be shown that the principle of angular momentum leads to the requirement that the Cauchy stress tensor is symmetric . Applying Cauchy’s law to 3.2.11, () dvvxdtddvbx dSn xdvdtddv ds vk j ijk vkj ijk sl klj ijkv v s ∫ ∫ ∫∫∫∫ = +×=×+× ρε ε σερv r br σn r (3.6.13) The surface integral can be converted into a volume integral using the divergence theorem. Using the index notation, and con centrating on the integrand of the resulting volume integral, one has, using 1.3.14 (the permutation symbol is a constant here, 0 /=∂∂l ijk xε ), Section 3.6 Solid Mechanics Part III Kelly 354()Tdiv σσ r :E+×≡ ⎭⎬⎫ ⎩⎨⎧+∂∂= ⎭⎬⎫ ⎩⎨⎧+∂∂=∂∂ kj lkl j ijk jl kl lkl j ijk lklj ijkxxxxxxσσεδσσεσε (3.6.14) where E is the third-order permut ation tensor, Eqn. 1.9.6, ( )k j i ijk e e e⊗⊗=εE . Thus, with the Reynold’s tran sport identity 3.1.31, {} ()∫∫ ∫× =×+ +× v v vdvdtddv dv vr br σσ r ρTdiv :E (3.6.15) The material derivative of this cross product is ()dtd dtd dtd dtd dtd vrvvvrvr vrvr ×=×+×=×+×=× (3.6.16) and so 0 divT= ⎭⎬⎫ ⎩⎨⎧−+×+∫∫ v vdvdtddvvbσ rσ ρ :E (3.6.17) From the equations of motion 2.6.2, the te rm inside the brackets is zero, so that 0 ,0T= =kj ijkσε σ:E (3.6.18) It follows, from expansion of this relation, that the matrix of stress components must be symmetric: ji ijσσ= = ,Tσσ Symmetry of Stress (3.6.19) 3.6.4 Consequences in the Material Form Here, the consequences of 3.6.19 on the PK1 and PK2 stresses is examined. Using the result Tσσ= and 3.5.8, T 1PFσ−=J , ()T 1TT 1 T 1FP PF PF− − −= = J J J (3.6.20) so that jk ik jk ik PF FP= = ,T TFP PF (3.6.21) These equations are trivial when ji=, not providing any constraint on P. On the other hand, when ji≠ one has the three equations Section 3.6 Solid Mechanics Part III Kelly 35533 23 32 22 31 21 33 23 32 22 31 2133 13 32 12 31 11 33 13 32 12 31 1123 13 22 12 21 11 23 13 22 12 21 11 PF PF PF FP FP FPPF PF PF FP FP FPPF PF PF FP FP FP ++=++++=++++=++ (3.6.22) Thus angular momentum considerations imposes these three constraints on the PK1 stress (as they imposed the three constraints 21 12σσ= , 31 13σσ= , 32 23σσ= on the Cauchy stress). It has already been seen that a consequence of the symmetry of th e Cauchy stress is the symmetry of the PK2 stress S; thus, formally, the symmetry of S is the result of the angular momentum principle. Section 3.7 Solid Mechanics Part III Kelly 3563.7 Boundary Conditions and The Boundary Value Problem In order to solve a mechanics problem, one must specify certain conditions around the boundary of the material under consideration. Such boundary conditions will be discussed here, together with the resulting boundary value problem (BVP ). (see Part I, 3.5.1, for a discussion of stress boundary conditions.) 3.7.1 Boundary Conditions There are two types of boundary condition, thos e on displacement and those on traction. Denote the body in the reference condition by 0B and in the curren t configuration by B. Denote the boundary of the body in the reference configuration by S and in the current configuration by s, Fig. 3.7.1. Displacement Boundary Conditions The position of particles may be specified over some portion of the boundary in the current configuration. That is, ()Xχx= is specified to be x say, over some portion us of s, Fig. 3.7.1, which corresponds to the portion uS of S. With )( )( xXx xu−= , or X Xx XU −= )( )( , this can be expressed as uu X XU XUx xu xu Ss ∈ =∈ = ),( )(),( )( (3.7.1) These are called displacement boundary conditions . The most commonly encountered displacement boundary condition is where some portion of the boundary is fixed, in which case () oxu=. Figure 3.7.1: Boundary conditions x XB0B sSus u0SUUuu == σSσS tt=TT= Section 3.7 Solid Mechanics Part III Kelly 357Traction Boundary Conditions Traction tt= can be specified over a portion σs of the boundary, Fig. 3.7.1. These traction boundary conditions are related to the PK1 traction TT= over the corresponding surface σS in the reference configura tion, through Eqns. 3.5.1-4, ds ds dS dS σn t PN T === (3.7.2) One usually knows the position of the boundary S and the normal ) (XN in the reference configuration. As deformation proceeds, the PK1 traction develops according to PNT= with, from 3.5.8, T−=σF P J . The PK1 stress will in general depend on the motion x and the deformation gradient F, so the traction boundary conditi on can be expressed in the general form ()FxXTT ,,= (3.7.3) Example: Fluid Pressure Consider the case of fluid pressure p around the boundary, n t p−= , Fig. 3.7.2. The Cauchy traction t depends through the normal n on the new position and geometry of the surface σs. Also, NF TT−−=pJ , which is of the general form 3.7.3. Figure 3.7.2: Fluid pressure on deforming material Consider a material under water with part of its surface deforming as shown in Fig. 3.7.2. Referring to the figure, 1E N−= , 2 1sin cos e e n θθ+−= , Iσ p−= , ()2xhg p−=ρ and 3 32 22 1 1 tan X xX xXa X x ==++= θ , ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = 1 0 00 1 00 tan1θ F , 1 det== F J The traction vectors and PK1 stress are θ Th 1 1,xX1 1,xX 1 1,eE2 2,eE an t p−= p Section 3.7 Solid Mechanics Part III Kelly 358() () ( ) ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −−−= ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡− −−= ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡− −−= 10 001 tan00 1 , 0tan1 , 0sincos 2 2 2 θ ρ θ ρ θθ ρ xhg Xhg xhg P T t with (note that θcos /=dsdS ) p=t and θcos/p=T . The traction vectors clearly depend on both position, and the deformation through θ. In this example, 2 11tan Grad grad e e FIU IFu ⊗=−==−=−θ and () u u u u grad: grad arctan grad arctan grad = =θ ■ Dead Loading A special case of loading is that of dead loading , where ()XTT= (3.7.4) Here, the PK1 stress on the boundary does not change with the deformation and an initially normal traction will not re main so as deformation proceeds. For example, if one considers again the geometry of Fig. 3.7.2, this time take () ( )() I XP N PN XT2 2 , 001 )( Xhg Xhg p −−= ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −=−== ρ ρ Then () ( ) () ( ) ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −−= ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ − = 10 001 tan00 1 , , 001 cos ,2 2 θ ρθ ρθθ xhg xhg xσ xt 3.7.2 The Boundary Value Problem The equations of motion 3.6.2, 3.6.9, are a se t of three differential equations. In the solution of any problem, one would have to supplement these equations with others, for example a constitutive equation expressing a relationship between the stress and the kinematic variables (see Part IV). This const itutive relation will typically relate the stress to the strains, or rates of strain, for example ) ,(deσ f = . Suppose then that the stresses are known in terms of the strains and hence the displacements u. The equations of motion are then a set of three second order differential equations in the three unknowns iu (assuming that the body force b is a prescribed function of the problem). They need to be subjected to certain boundary and initial conditions. Section 3.7 Solid Mechanics Part III Kelly 359 Assume that the boundary conditions are such that the displacements are specified over that part of the surface us and tractions are specified over that part σs, with the total surface σ us ss+= , with 0 =∩σ us s 1. Thus uss on ,on , uutnσt ===σ Boundary Conditions (3.7.5) where the overbar signifies quantities which ar e prescribed. Initial conditions are also required for the displaceme nt and velocity, so that 0 at ),( ),(0 at ),( ),( 00 = == = tt xutxuxutxu && Initial Conditions (3.7.6) and it is usually taken that Xx= at 0=t . Comparing 3.7.5 and 3.7.6, one also requires that u u=0 , ⋅ =u u0& over us, so that the boundary and init ial conditions are compatible. These equations together, the differential equations of motion and the boundary and initial conditions, are called the strong form of the initial bou ndary value problem (BVP): 0 at ),( ),(0 at ),( ),(on ,on ,div 00 = == ======+ ttss u xutxuxutxuuutσn tu v bσ &&&&& σρρ Strong form of the Initial BVP (3.7.7) When the problem is quasi-static, so the accele rations can be neglected, the equations of motion reduce to the equations of equilibrium 3.6. 3. In that case one does not need initial conditions and one has a boundary value problem involving 3.7.5 only. It is only in certain special cases and in certa in simple problems that an exact solution can be obtained to these equations. An altern ative solution strategy is to convert these equations into what is known as the weak form . The weak form, which is in the form of integrals rather than differential equations, can then be solved approximately using a numerical technique, for exam ple the Finite Element Method 2. The weak form is discussed in §3.9. 1 It is possible to specify both trac tion and displacement over the same portion of the boundary, but not the same components. For example, if one specified 11e tt= on a boundary, one could also specify 22e u u= , but not 11e u u= . In that case, one could imagine the boundary to consist of two separate boundaries, one with conditions with respect to 1e and one with respect to 2e, and still write 0=∩σ us s . 2 Further, it is often easier to prove results regarding the uniqueness and stability of solutions to the problem when it is cast in the weak form Section 3.7 Solid Mechanics Part III Kelly 360In the material form, the boundary conditions are uSS on ,on, UUT PNT ===σ Boundary Conditions (3.7.8) and the initial conditions are 0 at ),( ),(0 at ),( ),( 00 = == = tt XUtXUXUtXU & & Initial Conditions (3.7.9) and the initial vale problem is 0 at ),( ),(0 at ),( ),(on ,on ,Div 000 = == ======+ ttSS u XUtXUXUtXUUUT PNTU V BP & &&&& σρρ Strong form of the Initial BVP (3.7.10) Section 3.8 Solid Mechanics Part III Kelly 3613.8 Balance of Mechanical Energy 3.8.1 The Balance of Mechanical Energy First, from Part I, Chapter 5, recall wo rk and kinetic ener gy are related through K W W Δ=+int ext (3.8.1) where extW is the work of the external forces and intW is the work of the internal forces. The rate form is K P P &=+int ext (3.8.2) where the external and internal powers and rate of change of kinetic energy are KdtdK WdtdP WdtdP Δ= = = &, ,int int ext ext (3.8.3) This expresses the mechanical energy balance for a material. Eqn. 3.8.2 is equivalent to the equations of motion (see below). The total external force acting on the material is given by 3.2.6: dv ds v s∫∫+= b t Fext (3.8.4) The increment in work done dW when an element subjected to a body force (per unit volume) b undergoes a displacement ud is dvdub⋅ . The rate of working is () dvdtd dP /u b⋅= . Thus, and similarly for the traction, the power of the external forces is dv ds P v s∫∫⋅+⋅= vb vtext (3.8.5) where v is the velocity. Also, the total kinetic energy of the matter in the volume is ∫⋅= vdv K vvρ21 (3.8.6) Using Reynold’s transport theorem, ()∫ ∫⋅=⋅ = v vdvdtddvdtdKdtd vv vv ρ ρ21 (3.8.7) Thus the expression 3.8.2 becomes Section 3.8 Solid Mechanics Part III Kelly 362∫ ∫∫⋅=+⋅+⋅ v v sdvdtdP dv dsvv vb vt ρint (3.8.8) It can be seen that some of the power exer ted by the external forces alters the kinetic energy of the material and the remainde r changes its intern al energy state. Conservative Force System In the special case where the internal for ces are conservative, that is, no energy is dissipated as heat, but all energy is stored as internal energy, one can express the power of the internal forces in terms of a potential function u (see Part I, §5.1), and rewrite this equation as ∫∫∫∫⋅+ =⋅+⋅ v v v sdvdtddvdtdudv dsvv vb vt ρ ρ (3.8.9) Here, the rate of change of the internal energy has been written in the form dvdtdudvudtdUdtd v v∫∫= = ρρ (3.8.10) where u is the internal energy per unit mass, or the specific internal energy . 3.8.2 The Stress Power To express the power of the internal forces intP in terms of stresses and strain-rates, first re-write the rate of chan ge of kinetic energy using the equations of motion, ()∫∫+⋅=⋅= v vdv dvdtdKdtdbσ vvv div ρ (3.8.11) Also, using the product rule of differentiation, ()() ji ij jiji jij ixv xv xv∂∂−∂∂=∂∂−=⋅ σσσ,: div div lσ vσ σ v (3.8.12) rate of change of internal energy power of surface forces power of body forcesrate of change of kinetic energy power of internal forces Section 3.8 Solid Mechanics Part III Kelly 363where l is the spatial velocity gradient, j i ij xv l∂∂= / . Decomposing l into its symmetric part d, the rate of deformation, and its antisymmetric part w, the spin tensor, gives ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂=∂∂=+= ij ji ij ji ijxv xv xv 21,: : : : σσ dσwσdσlσ (3.8.13) since the double contraction of any symmetric tensor ( σ) with any skew-symmetric tensor ( w) is zero, 1.10.31c. Also, using Ca uchy’s law and the divergence theorem 1.14.21, () () ()dvxvdsvn dsvtdv ds ds ds v kiik sikik siiv s s s ∫∫∫∫∫∫∫ ∂∂= ==⋅=⋅=⋅ σσvσ n vσ vσn vt div (3.8.14) Thus, finally, from Eqn. 3.8.8, dvxv xvP dv P v ij ji ij v∫ ∫⎪⎭⎪⎬⎫ ⎪⎩⎪⎨⎧ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂−= −=21, :int int σ dσ Stress Power (3.8.15) The term dσ: is called the stress power ; the stress power is the (negative of the) rate of working of the internal forces, per unit volume. The complete equation for the conservation of mechanical energy is then ∫∫∫∫⋅+ =⋅+⋅ v v v sdvdtddv dv dsvv dσ vb vt ρ : Mechanical Energy Balance (3.8.16) The stress power is that part of the externa lly supplied power which is not converted into kinetic energy; it is converted into h eat and a change in internal energy. Note that, as with the law of conservation of mechanical energy for a particle, this equation does not express a separate law of continuum mechanics; it is merely a re- arrangement of the equations of motion (see below), which themselves follows from the principle of linear momentum (Newtons second law). Conservative Force System If the internal forces are conservative, one has dvdtduUdtddv v v∫ ∫== ρ dσ: (3.8.17) or, in local form, Section 3.8 Solid Mechanics Part III Kelly 364dtduρ=dσ: Mechanical Energy Balance (Conservative System) (3.8.18) This is the local form of the energy e quation for the case of a purely mechanical conservative process. 3.8.3 Derivation from the Equations of Motion As mentioned, the conservation of mechanical energy equation can be derived directly from the equations of motion. The derivation is similar to that us ed above (where the mechanical energy equations were used to de rive an expression for the stress power using the equations of motion). One has, multiplying the equations of motion by v and integrating, () ( ) {} (){} ∫∫∫∫∫ ∫∫ ⋅+⋅+−=⋅+−⋅=⋅+−⋅=+⋅=⋅ v s vvv v v dv ds dvdvdv dv dvdtd bv vt dσbvdσσvbvlσσv bσ vvv :: div: div div ρ (3.8.19) 3.8.4 Stress Power and the Continuum Element In the above, the stress power was derived usi ng a global (integral) form of the equations. The stress power can also be deduced by considering a differential mass element. For example, consider such an element whose boundary particles are m oving with velocity v and whose boundary is subjected to stresses σ, Fig. 3.8.1. Consider first the components of force and velocity acting in the 1x direction. The external forces act on the six sides. On thr ee of them (the ones that can be seen in the illustration) the stress and velocity act in the sa me direction, so the power is positive; on the other three they act in opposite direc tions, so there the power is negative. ),,(3 2 1 xxx 1xΔ2xΔ3xΔ 11σ1v 13σ12σ 11σ1v Section 3.8 Solid Mechanics Part III Kelly 365Figure 3.8.1: A differential mass element subjected to stresses As usual (see §1.6.6), the element is assume d to be small enough so that the product of stress and velocity varies lin early over the element, so that the average of this product over an element face can be taken to be repres entative of the power of the surface forces on that element. The power of the external surface forces acting on the three faces to the front is then () () () 3 32 21 21 21 13 21 32 21 21 1 3 21 32 21 21 1 , , 113 2 1, , 112 3 1 , , 111 3 2 surf x xx xx xx xx xx x x xx xx x v xxv xx v xx P Δ+Δ+Δ+Δ+Δ+Δ+ Δ+Δ+Δ+ ΔΔ+ΔΔ+ ΔΔ= σσ σ (3.8.20) Using a Taylor’s series expansion, and ne glecting higher order terms, then leads to ()()()() ()() () () ()() () () ⎭⎬⎫ ⎩⎨⎧ ∂∂Δ+∂∂Δ+∂∂Δ+ ΔΔ+⎭⎬⎫ ⎩⎨⎧ ∂∂Δ+∂∂Δ+∂∂Δ+ ΔΔ+⎭⎬⎫ ⎩⎨⎧ ∂∂Δ+∂∂Δ+∂∂Δ+ ΔΔ≈ 3113 3 2113 2 21 1113 1 21 ,, 113 2 13112 3 21 2112 2 1112 1 21 ,, 112 3 13111 3 21 2111 2 21 1111 1 ,, 111 3 2 surf 321321321 xvxxvxxvx v xxxvxxvxxvx v xxxvxxvxxvx v xx P xxxxxxxxx σ σ σσσ σ σσσ σ σσ (3.8.21) The net power per unit volume (subtracting the power of th e stresses on the other three surfaces and dividing through by the volume) is then ()()()() jj xv xv xv xvP∂∂=∂∂+∂∂+∂∂=11 3113 2112 1111 surfσ σ σ σ (3.8.22) Assume the body force b to act at the centre of the el ement. Neglecting higher order terms which vanish as the element size is allo wed to shrink towards zero, the power of the body force in the 1x direction, per unit volume, is simply 11vb. The total power of the external forces is then (including the ot her two components of stress and velocity), usin g the equations of motion, () ()() () dtddtdP dtvvd xv xvvbvdtdvbxvvbvx xvvbxvP ii ij ji ijii ii i ji ijii i jij ji ijii jiij vvdσvbvvb lσvbvσ lσvbvσ ⋅+=⋅+⋅ ⎭⎬⎫ ⎩⎨⎧+−+=⋅+⋅+=⋅+ = +⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂=+ ⎭⎬⎫ ⎩⎨⎧+−+∂∂=+∂∂+∂∂=+∂∂= 21::div:div 21 21T ext ext ρρ ρ σρ σσσσ (3.8.23) Section 3.8 Solid Mechanics Part III Kelly 366 which again equals the stress power term plus the change in kinetic energy. The power of the internal forces is dσ:− , a result of the forces acting inside the differential element, reacting to the applied forces σ and b. 3.8.5 The Balance of Mechanical Energy (Material form) The material form of the power of the external forces is wr itten as a function of the PK1 traction T and the reference body force B, 3.6.7, and the kinetic energy as a function of the velocity )(XV : ∫ ∫∫⋅=+⋅+⋅ V V SdVdtdP dV dSVV VB VT0 intρ (3.8.24) Next, using the identities 2.5.4, lFF=& and 1.10.3h, ()B AC BCA : )(:T= , gives ()()FσF FFσlσwσlσdσ & & : : : : : :T 1 − −= ==−= , (3.8.25) and so ()() ∫∫ ∫∫ == =− − VV v v dVJdV dv dv FPFσF FσF dσ && & :: : :T T (3.8.26) and ∫∫∫∫⋅+ =⋅+⋅ V V V SdVdtddV dV dSVV FP VB VT0 :ρ& Mechanical Energy Balance (Material Form) (3.8.27) For a conservative system, this can be written in terms of the internal energy ∫∫∫∫⋅+ =⋅+⋅ V V V SdVdtddVdtdudV dSVV VB VT0 0 ρ ρ (3.8.28) 3.8.6 Work Conjugate Variables Since the stress power is the double contra ction of the Cauchy stress and rate-of- deformation, one says that the Cauchy stress and rate of deformation are work conjugate (or power conjugate or energy conjugate ). Similarly, from 3.8.26, the PK1 stress P is power conjugate to F&. It can also be shown that the PK2 stress S is power conjugate to the rate of Euler-Lagrange strain, E& (and hence also the right Cauchy-Green strain) {▲Problem 1} : Section 3.8 Solid Mechanics Part III Kelly 367 CSESFPdσ &&& 21: : : : === J (3.8.29) Note that, for conservative systems, these quantities represent the rate of change of internal energy per unit reference volume. Using the polar decomp osition and the relation IRR=T, URFΩUR RURRURURF R&& &&&& +=+ =+= T (3.8.30) where RΩ is the angular velocity tensor 2.6.1. Then, using 1.11.3h, 1.10.31c, and the definitions 3.5.8, 3.5.12 and 3.5.18, UTUPRΩτUPRΩ PFURPFΩPFP RRR &&&& & :: :: :: : : BTT T =+=+ =+ = (3.8.31) so that the Biot stress is power conjug ate to the right stre tch tensor. Since U is symmetric, UT FP & & : sym :B= . Also, the Biot stress is conjuga te to the Biot strain tensor IUB−= introduced in §2.2.5. From 3.5.14 and 1.11.3h, dσd RσRdσ ˆ:ˆ : ˆ :T= = (3.8.32) so that the corotational stress is power conjugate to the rotated deformation rate , defined by dRRdT ˆ= (3.8.33) Pull Back and Push Forward From 2.12.12-13, the double contraction of two tensors can be expressed as push- forwards and pull-backs of those tensors. For example, the stress power (per unit reference volume) in the material description is ES&: . Then, using 3.5.13, 2.12.9a and 2.5.18b, dFFET=& , ()() dσ dτ E S ES : : : :*# * Jb== = & & χχ (3.8.34) This means that the material and spatial descriptions of the internal power can be transformed into each other using pu sh-forward and pull- back operations. Section 3.8 Solid Mechanics Part III Kelly 368Similarly, pulling back the corotational stress and rotated deformation rate to the intermediate configuration of Fi g. 2.10.8, using 2.12.13, 2.12.27, ()()()() dσ d σ dσ gR gRˆ:ˆ : :1 *# 1 * = =− − bχ χ (3.8.35) The stress power in terms of spatial tensors can also be expressed as a derivative of a tensor, using the Lie derivative. From 2.12. 42, the Lie derivative of the Euler-Almansi strain is the rate of deformation and hence (note that there is no universal function whose derivative is d), so eσ dσbJ JvL: := (3.8.36) 3.8.7 Problems 1. Show that the rate of intern al energy per unit reference volume dσ:J is equivalent to ES&: (without using push-forwards/pull-backs). Section 3.9 Solid Mechanics Part III Kelly 3693.9 The Principles of Virtual Work and Power The principle of virtual work was introduced an d discussed in Part I, §5.5. As mentioned there, it is yet another re-statement of the wo rk – energy principle, only it is couched in terms of virtual displacements, and the princi ple of virtual power to be introduced below is an equivalent statement based on virtual velocities . On the one hand, the principle of virtual work /power can be regarded as the fundamental law of dynamics for a continuum, and from it can be derived the equations of motion. On the other hand, one can regard the principle of linear momentum as the fundamental law, derive the equations of motion, and hence derive the principl e of virtual work. 3.9.1 Overview of The Principle of Virtual Work Consider a material under the actio n of external forces: body forces b and tractions t. The body undergoes a displacement )(xu due to these forces and now occupies its current configuration, Fig. 3.9.1. The problem is to find this displacement function u. Figure 3.9.1: a material displacing to its current configuration under the action of body forces and surface forces Imagine the material to undergo a small displacement uδ from the current configuration , Fig. 3.9.2, uδ not necessarily constant throughout the body; uδ is a virtual displacement, meaning that it is an imaginary displacement, and in no way is it related to the applied external forces – it does not actua lly occur physically . As each material particle moves through these virtual displacements, the external forces do virtual work Wδ. If the force b acts at position x and this point undergoes a virtual displacement )(xuδ , the virtual work is v W Δ⋅= ubδδ . Similarly for the surface tractions, and the total external virtual work is 0ext =⋅+⋅=∫∫ s vds dv W ut ub δ δ δ External Virtual Work (3.9.1) reference configuration current configuration u1b 2b3b1t Section 3.9 Solid Mechanics Part III Kelly 370 Figure 3.9.2: a virtual displacement field applied to a materi al in the current configuration There is also an internal virtual work intW due to the internal forces as they move through virtual displacements and a virtual kinetic energy Kδ. The principle of virtual work then says that K W W δδδ =+int ext (3.9.2) And this equation is then solved for the actual displacement u. Expressions for the internal virtual work and virtual ki netic energy will be derived below. 3.9.2 Derivation of The Principle of Virtual Work As mentioned above, one can simply write dow n the principle of virtual work, regarding it as the fundamental principle of mechanics, and then from it derive the equations of motion. This will be done further below. To begin, though, the starting point will be the equations of motion, and from it will be derived the principle of virtual work. Kinematically and Stati cally Admissible Fields A kinematically admissible displacement field is defined to be one which satisfies the displacement boundary condition 3.7.7c, usonuu= (see Part I, §5.5.1). Such a displacement field would induce some stress field within the body, but this resulting stress field might not satisfy the equations of motion 3.7.7a. In other words, it might not be the actual displacement field, but it does not violate the boundary conditions. A statically admissible stress field is one which satisfies the equations of motion 3.7.7a and the traction boundary conditions 3.7.7b, σson,tσn t== . Again, it might not be the actual stress field, since it is not specified how this stress fi eld should be related to the actual displacement field. Derivation from the Equations of Motion (Spatial Form) material moved an amount δu true position/solution u1b 3b1t 2b Section 3.9 Solid Mechanics Part III Kelly 371Let σ be a statically admissible stress field corresponding to a kinematically admissible displacement field u, so σsontσn= , usonuu= and u bσ &&ρ=+ div . Multiplying the equations of motion by u and integrating leads to ()∫∫⋅+=⋅ v vdv dv ubσ uu div &&ρ (3.9.3) Using the identity 1.14.16b, ) grad(tr div )(divTv A A v Av +⋅= , 1.10.10e, BA BA : ) (trT= , and the symmetry of stress, (){}∫∫⋅+ − =⋅ v vdv dv ubuσσu uu )(grad: div &&ρ (3.9.4) and the divergence theorem 1.14.22c and Cauchy’s law lead to ∫∫∫∫⋅+ =⋅+⋅ v v v sdv dv dv ds uu uσ ub ut &&ρ grad: (3.9.5) Splitting the surface inte gral into one over us and one over σs gives ∫∫∫∫∫⋅+ =⋅+⋅+⋅ v v v s sdv dv dv ds ds uu uσ ub ut ut σ u&&ρ grad: (3.9.6) Next, consider a second kinematica lly admissible displacement field *u, so uson*u u= , which is completely arbitrary, in the sense that it is unrelated to either σ or u. This time multiplying u bσ &&ρ=+ div across by *u, and following the same procedure, one arrives at ∫ ∫∫∫∫⋅+ =⋅+⋅+⋅ v v v s sdv dv dv ds ds* * * *grad: uu uσ ub ut ut σ u&&ρ (3.9.7) Let u uu−=*δ , so the difference between *u and u is infinitesimal, then subtracting 3.9.6 from 3.9.7 gives the principle of virtual work, ()∫ ∫∫∫⋅+ =⋅+⋅ v v v sdv dv dv ds uu uσ ub ut σδρδ δ δ && grad: Principle of Virtual Work (spatial form) (3.9.8) Note that since *,uu , are kinematically admissible, uson*ou uu =−=δ . If one considers the u in 3.9.8 to be the actual displacement of the body, then uδ can be considered to be a virtual displacement from th e current configuration, Fig. 3.9.2. Again, it is emphasized that this virtual displacem ent leaves the stress, body force and applied traction unchanged. One also has the transformed initial conditions: from 3.7.7d-e, Section 3.9 Solid Mechanics Part III Kelly 372 ∫ ∫∫ ∫ ⋅=⋅⋅=⋅ == v vtv vt dv dvdv dv uxu u txuuxu u txu δ δδ δ )( ),()( ),( 0 00 0 & & (3.9.9) Eqns. 3.9.8 and 3.9.9 together constitute the weak form of the initial BVP 3.7.7. The principle of virtual work can be grouped into three separate terms: the external virtual work: ∫∫⋅+⋅= v sdv ds W ub ut σδ δ δext External Virtual Work (3.9.10) the internal virtual work, ∫−= vdv W )(grad:int u σδ δ Internal Virtual Work (3.9.11) and the virtual kinetic energy, ∫⋅= vdv K uuδρδ&& Virtual Kinetic Energy (3.9.12) corresponding to the statement 3.9.2. Derivation from the Equations of Motion (Material Form) The derivation in the spatial form follows exac tly the same lines as for the spatial form. This time, let P be a statically admissible stress fi eld corresponding to a kinematically admissible displacement field U, so P T PN Son= , u UU Son= and U BP && 0 div ρ=+ . This time one arrives at ()∫ ∫∫∫⋅+ =⋅+⋅ V V V SdV dV dV dS UU U P UB UT Pδρ δ δ δ && 0 Grad: (3.9.13) Again, one can consider U to be the actual displacement of the body, so that Uδ represents a virtual displacement from the current configuration. With x X x UXxU δδδδ =−=−= (3.9.14) the virtual work equation can be expressed in terms of the motion ) (Xχx= , Section 3.9 Solid Mechanics Part III Kelly 373()∫ ∫∫∫⋅+ =⋅+⋅ V V V SdV dV dV dS χU χ PχBχT Pδρ δ δ δ && 0 Grad: Principle of Virtual Work (material form) (3.9.15) 3.9.3 Principle of Virtual Work in terms of Strain Tensors The principle of virtual work, in particular the internal virtual work term, can be expressed in terms of strain tensors. Spatial Form Using the commutative property of the variation 2.13.2, the term ()uδ grad in the internal virtual work expression 3. 9.8 can be written as ()() ()( ) ()() ()() Ωεu u u uu u u u u δδδ δδ δ δ δ δ +=− + + =− + + = T TT T grad grad21grad grad21)(grad)(grad21)(grad)(grad21)(grad (3.9.16) where ε is the (symmetric) small strain tensor and Ω is the (skew-symmetric) small rotation tensor, Eqn 2.7.2. Using the fact that the double contraction of a symmetric tensor (σ) and a skew-symmetric one ( Ω) is zero, 1.10.31c, one has ∫ ∫−= −= v vdv dv W εσ u σ δ δ δ : )(grad:int (3.9.17) Thus the stresses do internal virt ual work along the virtual strains εδ. One has ∫∫∫∫⋅+ =⋅+⋅ v v v sdv dv dv ds uuεσ ub ut σδρδ δ δ && : (3.9.18) Note that, although the small strain has been introduced here, this formulation is not restricted to small-strain theory. It is only the virtual strains that must be infinitesimal – there is no restriction on the magnitude of the actual strains. From 2.13.15, the Lie-variation of the Euler-Almansi strain e is ε eδδ=L , so the internal; virtual work can be expressed as ∫−= vdv W eσL int :δ δ (3.9.19) Material Form From Eqn. 3.9.15 and Eqn. 2.13.9, Section 3.9 Solid Mechanics Part III Kelly 374∫−= VdV W FPδ δ :int (3.9.20) so ∫∫∫∫⋅+ =⋅+⋅ V V V SdV dV dV dS χU FPχBχT Pδρδ δ δ && 0 : (3.9.21) Derivation of the Material Form directly from the Spatial Form To transform the spatial form of the virtual work equation into the material form, first note that, with 3.9.14b, ) grad(: ) grad(: xσ uσ δ δ= (3.9.22) Then, using 2.2.8b, 1Grad grad−= FV v , 2.13.9, ()u Fδ δ Grad= , 1.10.3h, ()B AC BCA : )(:T= , and 3.5.10, T−= Fσ PJ , ( ) () ()() F PFσFFFσFx σ xσ δδδδ δ :::) Grad(: ) grad(: 1T11 −−−− ==== J (3.9.23) which converts 3.9.17 into 3.9.120. Also, again comparing 3.9.17 and 3.9.120, using the trace properties 1.10.10, and Eqns. 3.5.9 and 2.13.11b, ()()()() ESεF FσFFεFσFεσεσ FP δδ δ δδδ : : tr tr : :T T 1 1= = = ==−− −J J J J (3.9.24) and so the internal work can also be expressed as an integral of ESδ: over the reference volume. The Internal Virtual Work and Work Conjugate Tensors The expressions for stress power 3.8.15, 3.8.29, and internal virtual work are very similar. For the material description, the time deriva tives in the former are simply replaced with the variation to get the latter: ESFP ESFP δδ : : : : = →=&& (3.9.25) For spatial tensors, the rate of strain tensor, e.g. d, is replaced with a Lie variation 2.13.14. For example, eσ dσbJ JvL: := (see 2.12.41-42) becomes: eσ eσ dσL v : L: : δJ J Jb→ = (3.9.26) Section 3.9 Solid Mechanics Part III Kelly 375 3.9.4 Derivation of the Strong Form from the Weak Form Just as the strong form (equations of moti on and boundary conditions) was converted into the weak form (principle of vi rtual work), the weak form can be converted back into the strong form. For example, () ( ){} (){} (){}∫∫∫∫∫∫ ∫∫∫ ⋅−−⋅=⋅−−⋅=⋅−−⋅=⋅+ =⋅+ v sv svv v v v dv dsdv dsdvdv dv dv dv uuσ utuuσ utuuσ uσuu uσ uuεσ σδρ δδρ δδρ δδρ δ δρδ &&&&&&&& && divdivdiv divgrad: : (3.9.27) and the last line follows from the fact that ou=δ on us. Thus the weak form now reads () ( ) 0 div =⋅−+−⋅−∫ ∫ v sdv ds uu bσ utt σδρ δ && (3.9.28) and, since uδ is arbitrary, one finds that the expr essions in the parentheses are zero, and so 3.7.7 is recovered. 3.9.5 Conservative Systems Thus far, no assumption has been made about th e nature of the intern al forces acting in the material. Indeed, the principle of virtual work applies to all types of materials. Now, however, attention is rest ricted to the special case where the system is conservative, in the sense that the work done by the external loads and the internal forces can be written in terms of potent ial energy functions 1. Further, for brevity, assume also that the material is in static equilibrium, i.e. the kinetic energy term is zero. In other words, it is assumed that the internal virtual work term can be expressed in the form of a virtual potential energy function: ∫ ∫= v vUdv dvδδ)(grad: u σ (3.9.29) Here, U is considered to be a function of u, and the variation is to be understood as in Eqn. 2.13.5, () ][ , u uuuδ δδ U U ∂≡ . If the loads can be regarded as functions of u only then, since they are conservative, they may be written as the gradient of a scalar potential: 1 The external loads being conservative would excl ude, for example, cases of frictional loading Section 3.9 Solid Mechanics Part III Kelly 376 utub∂∂−=∂∂−=t b U U, (3.9.30) Then, with uuuuδ δδ δ ⋅∂∂= ⋅∂∂=t tb bUUUU , (3.9.31) and using the commutative property 2.13.3 of the variational operato r, one arrives at () 0== ⎪⎭⎪⎬⎫ ⎪⎩⎪⎨⎧ ++∫∫∫u σU dvU dsU Udv vb st vδ δ (3.9.32) The quantity inside the brackets is the total pot ential energy of the system. This statement is the principle of stationary potential energy : the value of the quantity inside the parentheses, i.e. ()uU , is stationary at the true solution u. Eqn. 3.9.32 is an example of a Variational Principle , that is, a principl e expressed in the form of a variation of a functi onal. Note that the principle of virtual work in the form 3.9.8 is not a variational principle, since it is not expressed as the variation of one functional. Body Forces Body forces can usually be expressed in the form 3.9.30. For example, with gravity loading, g bρ= , where g is the constant accelera tion due to gravity. Then ug⋅−=ρbU ( og b==δδ and ()ub ub ⋅=⋅δδ , so dv dv∫∫⋅=⋅ ub ubδδ ). Material Form In the material form, one again has a stationary principle if one can write () U U B ∂ −∂= /BU , () U U T ∂ −∂= /TU (or, equivalently, replacing U with the motion χ). In the case of dead loading, §3.7.1, ()XTT= is independent of the motion so (similar to the case of gravity loading above) uT⋅−=TU with oT=δ and dV dV∫∫⋅=⋅ χT χT δδ . Deformation Dependent Traction In many practical cases, the traction will depend on not only the motion, but also the strain. In that case, one can write ()( ) ∫∫∫∫∫+⋅= = =⋅=⋅ v v s s sdv dv ds ds ds uσuσ uσ uσn unσ ut σδ δ δ δ δ δ grad: div div One might be able to then introduce a scalar function φ such that Section 3.9 Solid Mechanics Part III Kelly 377() εεuuεu δφδφφ : ,∂∂+⋅∂∂= with σεσu=∂∂=∂∂ φ φ, div (3.9.33) In the material form, one would have NPT= with ( ) ∫∫+⋅ =⋅ V SdV dS FPχP χT P: Divδ δ and then one might be able to introduce a scalar function ()Fχ,φ such that FFχχδφδφδφ :∂∂+⋅∂∂= with PFPχ=∂∂=∂∂ φ φ, Div (3.9.34) For example, considering again the fluid pressure example of §3.7.1, one can let pJ−=φ so that, using 1.15.7, T/−−=∂∂ F F pJφ . Then 1T/=−∂∂=−=Jp F F P φ , 2 Div E P gρ= and ∫∫−=⋅ pJdV dSδδχT . 3.9.6 The Principle of Virtual Power The principle of virtual power is similar to the principle of virtual work, the only difference between them being that a virtual velocity vδ is used in the former rather than a virtual displacement. To derive the virt ual power equation, multiply the equations of motion by the virtual velocity function, and integrate over the curre nt configuration, giving () ( ) () ∫∫∫∫∫ ∫∫ ⋅+ −⋅=⎭⎬⎫ ⎩⎨⎧⋅+∂∂− =⎭⎬⎫ ⎩⎨⎧⋅+∂∂− =⋅+=⋅ v v svv v v dv dv dsdvdv dv dvdtd vb dσ vtvbxvσvσvbxvσvσ vbσ vv δ δ δδδδδδδ δ δρ :: div)(: div div (3.9.35) These equations are identical to the mechanic al balance equations 3.8.16, except that the actual velocity is replaced with a virtual velocity. The term ∫− vdvdσδ: is called the internal virtual power . Note that here, unlike the virtual displacemen t function in the work equation, the virtual velocity does not have to be in finitesimal. This can be seen more clearly if one derives this equation directly from the virtual work equation. If the infinitesimal virtual displacement uδ occurs over an infinitesimal time interval tδ, the virtual velocity is the finite quantity tδδ/u , which here is labelled vδ. The virtual power equation can thus be obtained by dividing the virt ual work equation through by tδ. Section 3.9 Solid Mechanics Part III Kelly 378Again, supposing that the velocities are specified over that part of the surface vs and tractions over σs, the principle of virtual power can be written for the case of a kinematically admissible vi rtual velocity field: ∫∫∫∫⋅+ =⋅+⋅ v v v sdvdtddv dv ds vvdσ vb vt σδρδ δ δ : Principle of Virtual Power (3.9.36) In words, the principle of virtual power states that at any time t, the total virtual power of the external, internal and inertia forces is zero in any admissible virtual state of motion . 3.9.7 Linearisation of the Internal Virtual Work In order to solve the virtual work equations in anything but the mo st simple cases, one must apply some approximate numerical met hod. This will usually involve linearising the non-linear virtual work equations. To this end, the internal virtual work term will be linearised in what follows. Material Description In the material de scription, one has ()()()∫= VdV W uE uESδ δ :int (3.9.37) in which the Green-Lagrange strain is considered to be a function of the displacement, Eqn. 2.2.46, and the PK2 stress is a functi on of the Green-Lagrange strain; the precise functional dependence of S on E will depend on the material under study (see Part IV). The linearisation of the variation of a function is given by (see §2.13.2) () ()()uu u uu ΔΔ+ =Δ , , Lint int int W W W δ δ δ (3.9.38) where () () ()() ( ) ()() () ()() ( ) ()() ( ) ( ) () ( ) {}∫∫∫ ΔΔ+ΔΔ =⎭⎬⎫ ⎩⎨⎧Δ+ +Δ+ =Δ+ Δ+ =Δ+ =Δ∂=ΔΔ = === VVV dVdVdd dddVddWddW W uE uuES uuE uESuE u uES u uE uESu uE u uESu uu uuu δ δδεεεδεεδεεεδεδ δ ε εεε : , , :: ::][ , 0 00int 0int int (3.9.39) Section 3.9 Solid Mechanics Part III Kelly 379The linearization of the varia tion of the Green-Lagrange strain is given by 2.13.24, ()( )u u E δ δ Grad Grad symTΔ =Δ . With the PK2 stress symmetric, one has, with 1.10.3h, 1.10.31c, ()() ( ) ()()()( ) ()() ( ) () ( )()uESu uu u uESu u uES uuE uES Δ =Δ =Δ =ΔΔ Grad: GradGrad Grad:Grad Grad sym: , : TT δδδ δ (3.9.40) For the second term in 3.9.39, from 2.13.22, the variation of E is ()[ ] ()[] () u Fu F u Fu FFu E δδ δδ δ δ Grad symGrad GradGrad Grad TTTT 21T T 21 =+ =+ = (3.9.41) What remains is the calculation of the lineari sation of the PK2 stress. One has using the chain rule, ()() ()() ()()() ()()uuEEESu uEEESu uESuS uuESu ΔΔ∂∂=Δ+∂∂=Δ+ =Δ∂=ΔΔ == , ::][ , 00 εεεε εε dddd (3.9.42) Denote the fourth order tensor ()E ES∂∂ / by C and assume that it has the minor symmetries1.12.10. Then (see 3.9.41), with ()u F E Δ =Δ Grad symT (3.9.43) the linear increments in 3.9.38 become () ()()() { } () {}∫∫∫ ∂Δ∂+∂∂=⎭⎬⎫ ⎩⎨⎧ ∂Δ∂ ∂∂+∂Δ∂ ∂∂=ΔΔΔ +Δ =ΔΔ V dj abcd jc ia bdij biV dj jc abcd bk ka bd di biV dVXuCFF SXudVXuF CXuF SXu XuWdVW δδδ δδδδ δ uuu F u FuESu u uu ,Grad:: GradGrad: Grad , intT Tint C (3.9.44) Section 3.9 Solid Mechanics Part III Kelly 380The first term is due to the current stress and is called the ( initial ) stress contribution . The second term depends on the material properties and is called the material contribution . Solution formulations based on 3.9.44 are called total Lagrangian . Spatial Description The spatial description can be obtained by pushi ng forward the material description. First note that the linearization of the Kirchhoff stress is, from 3.5.13, () ()() ()() ( ) () () ( )T# *# *# * ,, ,, L , L FuuSFuτuuS uuSuuS uuτ ΔΔ+=ΔΔ+Δ=Δ =Δ χ χχ (3.9.45) so that, as in the derivation of the mate rial term in 3.9.44, and using 2.4.8, ()( ) () lk abcd ld kc jb iaxuCFFFF∂Δ∂=ΔΔΔ =ΔΔ uuτFuF F F uuτ ,grad: ,T TC (3.9.46) Define the fourth-order spatial tensor c through abcd ld kc jb ia ijkl CFFFFJ c1−= (3.9.47) so that () u uuτ Δ =ΔΔ grad: , cJ (3.9.48) Then, from 3.9.39, () ( ) ()()() { } ()() () {} (){} {} () {}∫∫∫∫∫∫ ∂Δ∂+∂∂=⎭⎬⎫ ⎩⎨⎧ ∂Δ∂ ∂∂+∂Δ∂ ∂∂=ΔΔΔ +Δ =Δ+ Δ =Δ + Δ =Δ+Δ =ΔΔ v dc abcd bd ac bav dc abcd ba bd da bavvVVb b dvxucxudvxucxu xu xuWdvdvdV JdV W σδδδσδδδ δδ δδ δδχχδχχ δ uuu u σu uu u u u σu u u u τE S E S uu ,grad:: grad grad: gradgrad: grad: grad grad:grad sym: grad: grad grad sym:: : , intTT*# * *# * int ccc (3.9.49) Section 3.9 Solid Mechanics Part III Kelly 381Solution formulations based on 3.9.49 are called updated-Lagrangian . Section 3.10 Solid Mechanics Part III Kelly 3823.10 Convected Coordinates Some of the important results from sections 3.1-3.9 are now re-expressed in terms of convected coordinates. As before, any relati ons expressed in symbolic form hold also in the convected coordinate system. 3.10.1 The Stress Tensors Traction and Stress Components Consider a differential parallelepiped elem ent in the current configuration bounded by the coordinate curves as in Fig. 3.10.1 (s ee Fig. 1.16.2). The bounding vectors are 22 11, g gΘΘ d d and 33gΘd . The surface area 1Sd of a face of the elemental parallelepiped on which 1Θ is constant, to which 1g is normal, is then given by Eqn. 1.16.35, 3 2 11 1 ΘΘ= ddgg Sd (3.10.1) and similarly for the other surfaces. Figure 3.10.1: vector elements bounding surface elements The positive side of a face is defined as that whose out ward normal is in the direction of the associated contravariant base vector. The unit normal in to a positive side is the same as the unit contrava riant base vector; as in Eqn. 1.16.14, ()sum no ˆ iii i i ggg n== (3.10.2) Let the force idF acting on the surface element with normal in be i iSdt (no sum over i), Fig. 3.10.2, so that it is the traction (force per unit area). x 1x2x3x 11gΘd33gΘd curve1−Θcurve2−Θcurve3−Θ 22gΘd Section 3.10 Solid Mechanics Part III Kelly 383 Figure 3.10.2: traction acti ng on a surface element The components of it along the unit covariant ba se vectors are denoted by jiσ : j jjji jji i gg g t1ˆσσ== (3.10.3) with no sum over the j in the jjg term; jiσ are called the physical stress components , Fig. 3.10.3. Figure 3.10.3: physical stress components Introduce now a new vector it defined by i ii igt t= (no sum over i) (3.10.4) It will be shown that this v ector is contravariant, that is, transforms between coordinate systems according to 1.17.3a (and so it does not satisfy the vector transformation rule, hence the superscript in pointed brackets). The components of it along the covariant base vectors are denoted by jiσ: jji ig tσ= (3.10.5) Comparing 3.10.3-5, 11σ31σ21σ 12σ22σ32σ13σ23σ33σ3Θ 2Θ 1Θ33gΘd22gΘd1g1 1ˆg n=1t Section 3.10 Solid Mechanics Part III Kelly 384 ji iijj ji ggσ σ= (no sum) (3.10.6) Cauchy’s Law and the Cauchy Stress Tensor Cauchy’s law can now be derived in the sa me way as in §3.3, by considering a small tetrahedral free-body, Fig. 3.10.4. Th e physical stress components ijσ shown act on the negative sides of the surfaces and so act in directions opposite that of the corresponding components on the positive sides (a consequence of Cauchy’s Lemma). It is required to determine the traction t in terms of the physical stress components and the unit normal n to the base area. Figure 3.10.4: free body diagram of a tetrahedral portion of material The normal to the base has components i i iin n g g n== (3.10.7) Consider the vector elements ad and bd shown in Fig. 3.10.5. Define the surface area element Sdto be the vector with magnitude equal to twice the area of the tetrahedron base and in the direction of the normal to the base, so () () ()() () ()3 2 1 211 31 3 3 23 2 2 12 1 2133 22 33 11 2121 21 21 S S Sgg gg ggg g g gba n S d d ddd dd ddd d d dd d dS d ++=×ΘΘ+×ΘΘ+×ΘΘ=Θ−Θ×Θ−Θ=×== (3.10.8) where 3 2 1 , , S S S d d d are the surface element areas of th e three coordinate sides of the parallelepiped of Fig. 3.10.1 (twice the area of the coordinate sides of the tetrahedron); from 3.10.2, 3Θ 2Θ 1Θnt 11σ 31σ21σ 12σ22σ 32σ 13σ 23σ 33σ Section 3.10 Solid Mechanics Part III Kelly 385i iiii i gSdSd dSd d d d gn nS S S S 13 2 1 ==++= (3.10.9) with no sum over the i in the iig term, or i ii ii i Sd gndS g g= (3.10.10) Figure 3.10.5: vector element of area for the base of the tetrahedron The principle of linear momentum, in vector fo rm, is then (cancelling out a factor of ½) 0=−iiSd dS t t (3.10.11) From 3.10.4, ii iiiidSn Sd gdS t t t = =1 (3.10.12) and so jiji iin n g ttσ== (3.10.13) Defining the (symmetric) Cauchy stress tensor σ through j iijg gσ⊗=σ Cauchy Stress Tensor (3.10.14) one arrives at Cauchy’s law nσt= . The Cauchy stress is naturally a contrava riant tensor because the normal vector upon which it operates to produce the traction is naturally represented in the form of a covariant vector (see 3.10.2). Note that the stress can also be expressed in the form adbdSd 22gΘd33gΘd 11gΘd Section 3.10 Solid Mechanics Part III Kelly 386jigtσ⊗= (3.10.15) Other Stress Tensors The PK1, PK2 and Kirchhoff stress tensors are j iijj iijj iij SP g gτG G SG G P ⊗=⊗=⊗= τ (3.10.16) By definition, στJ= , and so ij ijJστ= . By definition, T 1−−=σFF SJ , and so, from 2.9.8, j iij j iij j iij j iijJ J S G G G G gF gF G G ⊗=⊗=⊗ =⊗− −τ σ σ1 1 (3.10.17) Thus, as seen already, the Kirchhoff stress is the push-forward of the PK2 stress. Similarly, by definition T−=σF PJ and so () j ii kkjj km ii mkjj kkjj kkjj kkj j iij F JF JJJJ P G GG GG GG FGG gFg g G G ⊗ =⊗ ⊗ =⊗ =⊗=⊗=⊗ ⋅⋅− σσσσσ1 (3.10.18) 3.10.2 The Equations of Motion The Equations of motion have been given in the symbolic form by 3.6.2 and 3.6.9. To express these in curvilinear co ordinates, recall the definition of the divergence of a tensor, 1.18.28, ()i jij k j i kij k kg gg g gσσ | | div σ σ =⊗ =Θ∂∂= (3.10.19) The spatial and material descriptions of the equations of motion are then () ii ii i jijii ii i jij dtdVB Pdtvdb G G Ggg g 0 || ρρ σ =+=+ Equations of Motion (3.10.20) 3874 Fundamentals of Continuum Thermomechanics In this Chapter, the laws of thermodyna mics are reviewed and formulated for a continuum. The classical theory of thermodynamics, which is concerned with simple compressible systems, is discussed in secti ons 4.1-4.5, wherein are discussed the concepts of entropy, entropy production and entropy s upply, the second law, the notions of reversibility and irreversibility, the therm odynamic potential functi ons (internal energy, enthalpy and the Gibbs and He lmholtz free energies). C ontinuum thermomechanics is discussed in section 4.6. 388 Section 4.1 Solid Mechanics Part III Kelly 3894.1 Classical Thermodynamics: The First Law As an introduction to the thermomechanics of co ntinua, in this section particularly simple materials undergoing simple deformation and/ or heat-transfer processes are considered. 4.1.1 Properties and States First, here is some essential terminology used to describe thermodynamic processes. A property of a substance is a macroscopic charact eristic to which a numerical value can be assigned at a given time, w ithout knowledge of the history of the substance. Thus, for example, the mass, volume and energy of a mate rial, or the stress acting on a material, are properties. Work, on the other hand, is not a property, since it is in general history dependent and a material does not “have a certain amount of work”. The state of a material is the condition of the mate rial as described by its properties. For example a material which has properties volume 1V and temperature 1θ could be said to be in state ‘1’ whereas if at some later time it has different properties 2V and 2θ, it could be said to be in a different state, state ‘2’. 4.1.2 Thermal Equilibrium and Adiabatic Processes For the present purposes, a system can be defined to be a certain amount of matter which has fixed or movable boundaries. The state of a system can then be defined by assigning to it properties such as temperature, volume and so on. As will be seen, there are then two ways in which the state of the system can be changed, by interactions with its surroundings through heat or through work . The notion of heat, although familiar to us, will be defined precisely when the first la w of thermodynamics is introduced below. Before getting to the first law, it is helpful to consider the notions of thermal equilibrium and adiabatic processes. Thermal Equilibrium Consider the following experiment: two blocks of copper, one of which our senses tell us is “warmer than” the other, are brought into contact and isolated from their surroundings, Fig. 4.1.1a. A number of observation s would be made, for example: (1) the volume of the warmer body decreases with time whereas the volume of the colder body increases, until no further ch anges take place and the bodies feel equally warm (2) the electrical resistance of th e warmer block decreases with time whereas that of the colder block increases, until the electrical resistances would become constant also. When these and all such changes in observable properties cease, the interaction is at an end. One says that the two blocks are then in thermal equilibrium . In everyday language, one would say that the two blocks have the same temperature 1. 1 formally, temperature is defined through the zeroth law of thermodynamics , which states that if two systems are separately in thermal equilibrium with a th ird system, then they must be in thermal equilibrium Section 4.1 Solid Mechanics Part III Kelly 390 Figure 4.1.1: two blocks of copper brought in to contact; (a) no insulating wall, (b) insulating wall Adiabatic Conditions Suppose now that, before the bl ocks are brought together, an insulating wall is put in place to separate them, Fig. 4.1.1b. By this is meant that the volume, electrical resistance, etc. of one block does not affect those of the other block. Again, in everyday language, one would simply say that the temperature of one block does not affect the temperature of the other. The term adiabatic is used to describe this situation. 4.1.3 The First Law of Thermodynamics The Experiments of Joule Joule carried out experiments into the nature of work and heat transfer in materials in the 1840s. The essential features of hi s experiments were the following: (1) work was done on a known mass of fluid, cha nging the material from state “1” to state “2" (2) the fluid was thermally insulated so the process was adiabatic. (3) the work was performed in a variety of different ways (e.g. electrically or by stirring) (4) the same amount of work was required to go from state 1 to state 2, regardless of the method of work used The work done in taking a material from state 1 to state 2 by adiabatic processes therefore depends only on the initial and fi nal states, and is completely path independent , Fig. 4.1.2. Thus one can introduce a function U, a property of the system, such that 1 2U UU W −=Δ= (adiabatic process) (4.1.1) U is the internal energy , and the difference in internal en ergy between state 2 and state 1 is defined as equal to the work done in going from 1 to 2 by adiabatic means. with one another. This statement is tacitly assumed in every measurement of temperature – the third system being the thermometer (a)insulating wall (b) Section 4.1 Solid Mechanics Part III Kelly 391As an ideal example, one can think of a stiff metal spring. This “system” can be described by the property x, the extension of the spring fr om its equilibrium position. The work done in extending the spring depe nds only on its current “state”, that is x. It does not depend on how the spring may have be en extended and contracted in the past, assuming there has not been even the minutest te mperature change in the metal spring. In this example, the internal energy U is seen to be an el astic potential energy. Figure 4.1.2: a thermally insulated material taken from state 1 to state 2 through different “work paths” The First Law One can imagine now a careful experiment in which a material is thermally insulated from its surroundings and deformed through the wo rk of a set of forces. The material can be deformed into different stat es, Fig. 4.1.3. The internal energy U will in general be different in each state. U could be measured by carefully recording the work done on the material to reach a given state. Figure 4.1.3: a thermally insulated ma terial in three di fferent states Suppose that the internal energy of a material is known at va rious different states, through the conduction of the aforementioned experime nt, in particular one knows the internal energy for the material at two given states , 1 and 2. Relax now the condition that the changes are adiabatic. What this means is that if one now bri ngs the material into contact with another body, the properties of the material can be affected. Work is again done to take the material from state 1 to state 2 but it will now be found that, in general, 1 2U UU W −=Δ≠ (4.1.2) State 1 1U State 2 2U State 3 3U thermally insulatin g wallState 1 State 2 WW Section 4.1 Solid Mechanics Part III Kelly 392The difference between UΔ and W is defined as a measure of the heat Q which has entered the system in the change. Thus U Q WΔ=+ First Law of Thermodynamics (4.1.3) This is the first law of thermodynamics . In words, t he change in the internal energy is the sum of the work done plus the heat supplied . Note that the concept of heat Q (and internal energy) is in troduced and defined with the first law. Like work, heat is a form of energy transfer ; a body does not contain heat. Work is any means of changing the energy of a system other than heat. Sign Convention for Work and Energy The following sign convention will be used 2 0>Q - heat enters the system 0<Q - heat leaves the system (4.1.4) 0>W - work done on the system 0<W - work done by the system Other types of Energy When there are other energies involved, the first law must be amended. For a material moving with a certain velocity, one must also consider its kinetic energy, and the first law reads K U Q W Δ+Δ=+ (4.1.5) Other types of energy can be incorporated, fo r example gravitationa l potential energy and chemical energy3. All the different types of en ergy are often denoted simply by E, so the first law in general reads E Q WΔ=+ . Inside the Black Box In this continuum treatment of thermodynamics (or phenomenological thermodynamics ), it is not necessary to look in side and consider the billions of molecules inside the “black box” of a system. However, it is helpful to think of the molecules of a material as having certain micr o-velocities and it is the mean velocity of these micro-velocities which manifests itself as the macroscopic ve locity property, and the statistical fluctuations of the micro-veloci ties from the mean velocity are assumed to cancel out, Fig. 4.1.4. The micro-velocity fluctua tions give rise to an internal kinetic energy which manifests itself as the macroscopic temperature. Thus the internal energy will in general consist of both potential and kinetic energies. The inte raction between the elementary particles and 2 many authors use the exact opposite sign convention for work as used here 3 a potential energy which can be acce ssed when molecular bonds are broken Section 4.1 Solid Mechanics Part III Kelly 393the surroundings of the element causes energy to be transferred to the surroundings. This is the heat flow through the boundary of the system. This energy exchange can occur even when the shape of the element does not change, whereas a change in potential energy implies a deformation which will indu ce a re-arrangement of the molecules and change in shape or volume of the system. Figure 4.1.4: a system moving with velocity v Further, the property of pressu re or stress of the system is by definition determined by the forces exerted by the elementary particle s around the boundary. The fluctuations and micro-movement of the elementary particles wi ll cause stress fluctuations but again these are assumed to cancel out. 4.1.4 Simple Compressible Systems In order to demonstrate the meaning and use of the first law with examples and simple calculations, only simple systems will be considered. A simple system is one where there is only one possible work interaction. The clas sic example of a simple compressible system is that of a substance contained within a piston-cylinder apparatus, Fig. 4.1.5. The state of the material can be changed either by heat transfer or by the application of work, and the only work interaction possible is the application of a force to the piston head, compressing or expanding the material. More complex systems might include, for example, the possibility of doing work through electrical means – so here any effects due to surface, magnetic or electr ical effects, or due to mo tion or gravity, are ignored. Figure 4.1.5: A simple pi ston-cylinder system v Section 4.1 Solid Mechanics Part III Kelly 394A pure substance is one which has a uniform and inva riable chemical composition. In theory this could include different phases of the same substance (e.g. water and steam for H2O). In what follows, only pure substances in th e context of simple compressible systems will be considered. Now a general rule known as the state principle , based on experimental evidence, says that there is one independent property for each way a system’s energy can be varied independently. For a simple compressible sy stem, there are two wa ys of varying the energy and so the material has two independent properties. One can take any two of, for example, the temperature4 θ, pressure p, volume V or internal energy U. If p is the pressure at the piston face, and dV is a small change in volume of the material, then the work done in compre ssing/expanding the material is dVp W−=δ , (4.1.6) the minus sign because a positive work is done when the volume gets smaller. The total work done during a compression/expan sion of the mate rial is then ∫∫−==22 11, ,Vp VpdVp W Wδ (4.1.7) The symbol δ is used here to indicate th at the small amount of work Wδ is not a true differential5, ),( ),(1 1 2 2 VpW VpW dW W − =≠∫, since the work done is process/path dependent. The first law states that Q W dU δδ+= which can now be re-written as Q pdV dU δ+−= First Law for a Simple Compressible System (4.1.8) 4.1.5 Quasi-Static Processes A system is said to be in equilibrium when it experiences no change over time – it is in a steady state . Full thermodynamic equilibrium of a system requires thermal equilibrium with any surroundings and also mechanical equilibrium6. Much of the theory developed here requires th at the system be in a certain state with certain properties. If a property such as temperature is varying throughout the material, one cannot easily speak of its “state”. Thus when a material is undergoing some process, for example it is being deformed or heated, it is often necessary to assume that it is a quasi-static (or quasi-equilibrium ) process. This means that the process takes place so 4 the symbol θ denotes the absolute temperature , with 0>θ 5 but not to be confused with the use of this symbol to represent a variation, as in the context of the principle of virtual work 6 and also chemical equilibrium , where there are no net reactions taking place Section 4.1 Solid Mechanics Part III Kelly 395slowly that the rate of change of the proce ss is slow relative to the time taken for the properties to reach equilibrium. For example, if one heats water in the piston-cylinder arrangement of Fig. 4.1.5 by pu tting it directly over a hot flame, the water near the base will heat up first and cause convection current s and the water will not be anywhere near an equilibrium state. On th e other hand, one could imagin e heating the water extremely slowly with a low flame, so th at at any time instant the wate r temperature is very nearly constant throughout. To examine what this might mean in the case of the work performed, consider Fig. 4.1.6, which shows the system pressure p and the external pressure extp - the pressure exerted by the surroundings. Assuming thermal equilibrium, if extpp= then there is full equilibrium. If, however, there is an apprec iable difference between the two, for example if a large external pressure is suddenly applied, the piston head will depress rapidly and pressure will not remain uniform throughout th e system. However, if the pressures differ by an amount dp, the work done by the system is () dVp dVdp dVp dVdp p dVp Wext ext ext ∫∫∫ ∫∫−= −=±−=−= m (4.1.9) provided dp is extremely small. The smaller dp, the closer the system will be to mechanical equilibrium. As with the heat tr ansfer, this implies that quasi-equilibrium is maintained provided the piston is moved extr emely slowly by incrementally increasing the pressure by very small amounts. (It is of ten suggested that this might be achieved by repeatedly placing indi vidual grains of sand on the piston head.) Figure 4.1.6: pressures exert ed on a piston head Unless otherwise stated, it will be assumed that the material at any instance is in quasi- equilibrium. If the system is not in equilibrium, Eqn. 4.1.8, Q pdV dU δ+−= , does not make much sense, and one would have to use the more general version Q W dU δδ+= . Example A gas is contained in a rigid thermally insulate d container. It is th en allowed to expand into a similar container initially evacuated, Fi g. 4.1.7. There is no heat transfer and so 0=Q . Since a vacuum provides no resistance to an expanding gas, there is no pressure and hence no work done. Therefore there is no change in the internal energy of the gas. This is not a quasi-static process. pextp Section 4.1 Solid Mechanics Part III Kelly 396 Figure 4.1.7: a thermally insulated gas ex panding in an evacuated container ■ Example To illustrate that work is path dependent, consider the Vp− graph in Fig. 4.1.8, which shows three different proce ss paths between states 1 (1 1,Vp ) and 2 (2 2,Vp ). For path ABC, the work done is ) (1 2 2 V Vp− . For path CBA′, the work done is ) (1 2 1 V Vp− . The work for the third, curved, path requires an integration along AC and will in general be different from both the other results. Figure 4.1.8: a p-V diagram ■ Example Consider the cylinder arrangement of Fig. 4.1.9, which shows a gas contained by a weight. The gas is heated and this causes the weight to rise. The pressure is constant and so the work done is ()1 2V Vp Vp −=Δ . This example shows a system taking heat as input and performing work as output, w ith no necessary internal energy change. 2p1pA C Bp VB′ 1V2V Section 4.1 Solid Mechanics Part III Kelly 397 Figure 4.1.9: a heated gas causing a weight to move ■ The opposite process, whereby work is converted purely into heat is called dissipation (for example, as can occur in a frictional brake). 4.1.6 Equations of State The materials under considera tion have two independent pr operties. The remaining relations between the vari ous properties are called equations of state . For example, suppose that one takes the temperature and vol ume to be the independent properties. Then the relations ()()V UU Vpp , ,, θ θ = = (4.1.10) are equations of state. The first of these, re lating force variables (in this simple case, the pressure p) to kinematic variables (in this simple case, the volume V) and temperature, is called a thermal equation of state . The second, relating the in ternal energy to a thermal variable (here temperature) and a kinematic variable, is called a caloric equation of state . Beginning with the general cal oric equation of state ) ,(θVUU= , one can write the increment in internal energy as (the subscripts here indicate a variable which is held constant) dVVUdUdU V θθθ⎟ ⎠⎞⎜ ⎝⎛ ∂∂+⎟ ⎠⎞⎜ ⎝⎛ ∂∂= (4.1.11) This mathematical relation recognizes that U is a state function which depends only on the initial and final states and not on the path; thus one can first change θ with V constant and then change V with θ constant, and this will describe any arbitrary change dU. The internal energy and the othe r state variables can be expres sed in many different ways. For example, taking p and θ to be the independent variable s, the internal energy can be expressed as pweight pweight Section 4.1 Solid Mechanics Part III Kelly 398 θθθdUdppUdU p⎟ ⎠⎞⎜ ⎝⎛ ∂∂+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂= (4.1.12) 4.1.7 Specific Properties Specific properties are properties per unit mass . They are usually denoted by lower case letters. For example, the specific volume (r eciprocal of the density) and specific internal energy are mUumVv ==, (4.1.13) where m is the mass of the system. The properties V and U are extensive properties , meaning they depend on the amount of substance in the system. The specific properties on the other hand are intensive properties , meaning they do not depend on the amount of substance. Other intensive properties are the temperature θ and pressure p. One can also express the heat and work as per unit mass: mWwmQqδδδδ = = , (4.1.14) 4.1.8 Heat Capacity and In ternal Energy Measurements The internal energy of a material can seem qu ite an abstract concep t. To help quantify internal energy, the heat capacity is next introduced. Specific Heat and the Enthalpy The heat capacity is defined as the amount of heat required to raise the system by one unit of temperature . The higher the heat capacity, the more the heat required to increase the temperature. From Eqn. 4.1.8, the heat capacity at constant volume is then by definition V VVU dQC ⎟ ⎠⎞⎜ ⎝⎛ ∂∂=⎟ ⎠⎞⎜ ⎝⎛≡θθδ (4.1.15) In this case, all the supplied thermal energy goe s into raising the temp erature of the body. The heat capacity at constant pressure is by definition p p ppH VpU dQC ⎟ ⎠⎞⎜ ⎝⎛ ∂∂=⎟ ⎠⎞⎜ ⎝⎛ ∂∂+∂∂=⎟ ⎠⎞⎜ ⎝⎛≡θθθθδ (4.1.16) Section 4.1 Solid Mechanics Part III Kelly 399 where H is the enthalpy , defined by pV UH+= (4.1.17) In this case, some of the thermal en ergy is converted into work, and so V pC C≥ . The enthalpy is a property, since U, p and V are. The heat capacities ()V V U C θ∂∂= / and ()p p H C θ∂∂= / involve only properties and so th emselves must be properties. These expressions are thus valid for any quasi-static process, wh ether it be constant volume, constant pressure or neither. For example, consider a substance held at co nstant volume, heated using thermal energy Q so as to increase the internal energy by UΔ. The resulting temperature change is θΔ and θΔ=/Q CV . Consider, alternatively, the case wh ere the same substance, again held at constant volume, is impart ed a similar amount of energy W through work (e.g. by stirring). The internal en ergy change must again be UΔ. With ()V UU ,θ= , the temperature must again increase by θΔ and so now θΔ=/W CV . Note how ()V V U C θ∂∂= / is valid for both cases, even in the second case where no heat was actually supplied. Internal Energy Measurements Suppose now that the heat capaci ty at constant volume has been carefully measured over a given temperature range, by recording the heat required to effect increments in temperature. The internal energy ch anges within that range are then θθ θdC U UUV∫=−=Δ2 11 2 (constant volume) (4.1.18) Although this measurement technique requires co nstant volume proce sses, since internal energy is a property the results apply to all processes. Some values for the specific internal en ergy and enthalpy of steam for a range of temperatures, pressures and specific volumes are given in Table 4.1.1 below. The reference state for internal energy (where u is chosen to be zero) is for saturated water at 0.01 oC. The corresponding reference state fo r the enthalpy is obt ained from 4.1.177. θ (C0) v (m3/kg) u (kJ/kg) h (kJ/kg) 120 1.793 2537.3 2716.6 200 2.172 2658.1 2875.3 280 2.546 2779.6 3034.2 360 2.917 2904.2 3195.9 Table 4.1.1a: Properties for steam at pressure MPa1.0=p 7 note that u and h can take on negative values, depending on the reference state chosen Section 4.1 Solid Mechanics Part III Kelly 400 MPa)(p v (m3/kg) u (kJ/kg) h (kJ/kg) 0.035 6.228 2660.4 2878.4 0.100 2.172 2658.1 2875.3 0.300 0.716 2650.7 2865.5 0.500 0.425 2642.9 2855.4 Table 4.1.1b: Properties for steam at temperature C 2000=θ 4.1.9 The Ideal Gas A thermally perfect gas is one for which the ther mal equation of state is θmR pV= or θR pv= (4.1.19) where R is the universal gas constant. Further, an ideal gas is a thermally perfect gas whose internal energy depends on the temperat ure only, that is, its caloric equation of state is of the form )( θUU= (4.1.20) To see what this might mean from a physic al point of view, consider a gas at the microscopic level. Internal energy and pressure are related through intermolecular forces. If the pressure is very low, the internal energy is no longer affected by these forces, since the molecules are so far apart, but only by their kinetic energy of motion, i.e. the temperature. Moderate changes in volume will not bring the molecules of gas close enough together to alter this so le dependence on temperature. When the internal energy is a function of θ and V, one has dVVUdC dVVUdUdUV V θ θθ θθ⎟ ⎠⎞⎜ ⎝⎛ ∂∂+=⎟ ⎠⎞⎜ ⎝⎛ ∂∂+⎟ ⎠⎞⎜ ⎝⎛ ∂∂= (4.1.21) Thus for an ideal gas θdC dUV= (4.1.22) and this is valid for any process, not necessarily constant volume (although VC is measured at constant volume). Example Consider an ideal gas undergoing a volume change under isothermal , i.e. constant temperature, conditions. From 4.1.19, the quantity 22 11 Vp Vp pV == is a constant θmR . This constrains the process to lie on one particular path in a Vp− diagram. Also, from 4.1.22, 0=dU and so W Qδδ−= . If an ideal gas expands at constant Section 4.1 Solid Mechanics Part III Kelly 401temperature then the heat input exactly equa ls the work done against an incrementally changing external pressure. ■ Consider now a process involving wo rk and heat transfer. One has pdV dCQV+=θδ and the total heat input is ∫∫∫+ ==22 112 1, ,)(Vp VpV pdV d C Q Qθ θθθ δ (4.1.23) The second integral here cl early depends on the exact combination of pressure and volume during the process, so the heat input Q is path dependent, as expected. However, consider the following: ()1 2, , / ln)()()( 2 12 12 122 112 1 VV mR dCVdVmR dCpdVdC Q VV VVVp vpV + =+ =+ = ∫∫ ∫∫∫∫ θ θθ θθ θ θθθθθθθθθθ θδ (4.1.24) The quantity on the right is now path independent. In fact , for the simple case where VC is independent of θ, a good approximation for many “near-ideal” gases, one has () ()1 2 1 2 / ln / ln VV mR CQ V + =∫θθθδ (4.1.25) This expression means that, for an ideal gas undergoing a quasi-static process , although the quantity Q depends on the process, θδ/Q∫ does not and so is a property. It will be seen in the next section that this property is the entropy of the gas. 4.1.10 Problems 1. A gas is contained in a thermally insulated cy linder. It is quickly compressed so that its temperature rises sharply. Has there been a transfer of heat to the gas? Has work been done? Is the process quasi-static? 2. A gas expands from an initial state where kPa5001=p and 3 1 m1.0=V to a final state where kPa1002=p . The relationship between pr essure and volume during the particular process is k pV=, a constant. Sketch the process on a Vp− diagram and determine the work, in kJ. Interpre t the + or – sign on your result. 3. A system, whose equation of state depends only on the pressure p, volume V and temperature θ, is taken quasi-statically from state A to state B in the figure below Section 4.1 Solid Mechanics Part III Kelly 402along the path ACB at the pressures indicated. In this process 50J of heat enter the system and 20J of work are done by the system. (a) evaluate UΔ (b) how much heat enters the system along the path ADB ? (c) if the system goes from B to A by the curved path indicated schematically on the figure, the work done on the system is 25J. How much heat enters or leaves the system? (d) If the internal energy at A is denoted by AU, etc., suppose that J U UA D 25=− . What then is the heat transfer involved in the processes AD and DB? 4. Air is contained in a vertical piston-c ylinder assembly by a piston of mass kg 50 and having a face area of 2m01.0 . The mass of the air is 4 g, and initially the air occupies a volume of 0.005 m3. The atmosphere exerts a pressure of kPa 100 on the top of the piston. Heat transfer of magnitude kJ 1.41 occurs slowly fr om the air to the surroundings, and the volume of the air decreases to 3m 0025.0 . Neglecting friction between the piston and the cylinder wall, de termine the change in specific internal energy of the air, in kg /kJ . [Note that the force is constant on the piston-head.] 5. A closed system, i.e. a single mass of a substance, undergoes a thermodynamic cycle8 consisting of the following processes: Process 1-2: adiabatic compression with const.4.1= pV from kPa74.3441=p , 3 1 m 084951.0=V to 3/1 2V V= Process 2-3: constant volume Process 3-1: constant pressure, kJ 27317.493 1=−U U There are no significant changes in kine tic or gravitational potential energy. (a) sketch the cycle on a Vp− diagram (b) calculate the net work for the cycle (c) calculate the heat transfer for process 32− 6. How could you use the definition of the specif ic heat capacity at constant pressure to evaluate the internal energy of a material? 7. Show that for a system (not necessarily an ideal gas) undergoing a constant pressure process, the heat input is equal to the enthalpy. 8 meaning the substance is brought back to its initial st ate at the end of the process; state variables resume their initial values 1p15p A CD Bp V Section 4.1 Solid Mechanics Part III Kelly 403 8. Show that, for an ideal gas, V pC CR−= 9. Use the result of problem 8 to show that, when an ideal gas undergoes an adiabatic quasi-static change, const. =γpV where V pCC/=γ . 10. In Table 4.1.1: (a) Does the steam behave like an ideal gas? Nearly? (Note the internal energies in Table 4.1.1b) (b) The internal energy decreases as the steam is compressed. Is this what you would expect? Comment. Section 4.2 Solid Mechanics Part III Kelly 4044.2 Classical Thermodynamics: The Second Law 4.2.1 A Qualitative Sketch of the Second Law and Entropy The first law of thermodynamics is concer ned with the conservation of energy. The second law of thermodynamics is concerned w ith how that energy is transferred between systems. Its relevance to everyday experience can be seen from the following examples: • Ice is placed in a glass of water. It melts. • A hot metal tray is taken out of the oven and placed on a bench top. It cools. • A brittle plate is dropped from a height onto a hard floor. It smashes into small pieces. • A piece of iron is left outside. It rusts. • A bicycle tyre is pumped to high pressu re and punctured. The air rushes out. The common factor in all these exampl es is that energy is spreading out in a certain direction . • The energy in more rapidly moving warm air molecules disperses to the ice and breaks the intermolecular hydrogen bonds, allowing th e water molecules in the ice to move more freely. • The hot metal contains a relatively large am ount of energy due to its vibrating atoms and this energy is transferred to the surr ounding air molecules and thereby dispersed. • The potential energy in the plate disperses th rough a heating of th e surrounding air, the ground and the plate as it smashes. • The iron atoms and oxygen molecules in th e air have chemical (potential) energy stored in their bonds. When iron and oxyge n react, lower energy iron oxide bonds are formed and the energy difference is dispersed as heat1. • The relatively large energy of th e pressurized air in the tyre disperses when the tyre is punctured. Very qualitatively, the second law says that energy tends to spontaneously disperse unless hindered from doing so . If any of these processes were filmed and the tape accidentally played backwards, the mistake would immediately be evident. Howe ver, no physical law (apart from the second law) would be broken if the ev ents happened in reverse. For example, the plate falls because there is a gravitational force pulling it down. However, beginning at the end and working back, it is theoretically possible for the dispersed heat to flow back towards the broken pieces and so provide enough energy for the pieces to fly together and gain a kinetic energy to lift off the ground, rise up and eventually slow until it reaches its precise original position off the ground. The probability of this happening is to all intents and purposes zero. The second law says that en ergy simply does not spontaneously, that is without outside interference, gather togeth er and concentrate in a small locality. Entropy is closely associated with the sec ond law. Again, qualitatively, entropy is a measure of how dispersed energy is . Each system has a certain entropy and as energy 1 most spontaneous reactions of this type require a certain energy to get started, the activation energy , and this hinders the second law from wreaking havoc Section 4.2 Solid Mechanics Part III Kelly 405disperses, the entropy increases. When the air rushes out of the tyre, the entropy of the air and its surroundings increases. When the hot tray cools, the entropy of the tray and surrounding air increases. When the ir on and oxygen react, entropy increases. The second Law and Maximum work When heat is supplied to the confined gas of Fig. 4.1.9, work is done when the gas expands and raises the weight. However, if the flame is not placed under the apparatus but simply left to burn, the heat energy, according to the second law, will disperse into the air. It will not ever spontaneously gather back again in a sm all locality where it could again be used to do some work. The only way to get it back into a small locality again is to input even more energy. In this sense th e second law tells us that if we want to maximize the amount of work we can do, we n eed to use heat energy productively, and if any heat energy escapes it is not possible to use it again without expending more energy. A more formal and quantitative treatmen t of the second law will now be given. 4.2.2 Entropy and the Second Law Entropy The entropy S of a system is a property of th at system. The change in entropy dS is due to two quantities. First, define the entropy supply )(rSδ (an increment) through θδδQSr=)( (4.2.1) where Q is the heat supply. Define also the entropy production )(iSδ (also an increment) to be the difference between th e increment of entropy and the entropy supply: )( )( i rS S dS δδ+= (4.2.2) Thus the entropy change in a material is due to two components : the entropy supply, “carried” into the material w ith the heat supply, and the entropy production, which is produced within the material. (The reason for the “r” and “i” superscripts is given further below.) Note that, whereas the entropy S is a state function (a pr operty), the entropy supply and entropy production are not, since they depend on the particular process by which the state has changed, and hence the use of the symbol “ δ” for these functions. (Compare 4.2.2 with the first law, Q W dU δδ+= .) The Second Law The second law of thermodynamics states th at the entropy producti on is a non-negative quantity, 0)(≥iSδ The Second Law (4.2.3) Section 4.2 Solid Mechanics Part III Kelly 406 In terms of the entropy, the first law can be written as )()( ir S dS WS W dU θδθδθδδ −+=+= (4.2.4) or, including the second law, )(iS dS dU W θδθ δ +−= with 0)(≥iSδ . (4.2.5) A process is termed reversible if the equality holds, 0)(=iSδ , so that there is no entropy production, in which case )(rS dSδ= . Otherwise it is termed an irreversible process, in which case )( )( i rS S dS δδ+= . The superscript “ r” on the entropy supply is to indicate that the entropy supply is equivalent to the change of entropy in a reversible process. The superscript “ i” on the entropy production is to in dicate that entropy production is associated with irreversible processes. Alternative Statements of the Second Law There are many different statements of the second law and each can be “derived” from the others (there is no one agreed versi on). Another useful definition is that the heat input to the system in transforming from st ate A to state B is bounded from above , according to dS Qθδ≤ (4.2.6) The maximum possible heat input is dSθ, in which case the entropy change is due entirely to entropy supply, with no entropy prod uction – a reversible pr ocess. It can be seen that the statement dS Qθδ≤ is equivalent to the statement 0)(≥iSδ . A re-arrangement of Eqn. 4.2.6 gives the classic Clausius’s inequality : θδQdS≥ (4.2.7) 4.2.3 Reversibility Pure Heating As an example of a reversible process, consider a pure heating (or cooling) process, where the volume is held constant and so the work increment is zero, Fig. 4.2.1. Taking the two independent variables to be θ and V, it follows from 4.2.5 that the work increment can be expressed as )(iS dS UdVVS VUW θδθθθθθ δ +⎟ ⎠⎞⎜ ⎝⎛ ∂∂−∂∂+⎟ ⎠⎞⎜ ⎝⎛ ∂∂−∂∂= (4.2.8) Section 4.2 Solid Mechanics Part III Kelly 407 Figure 4.2.1: Pure heating With 0==dV Wδ , this reduces to 0)(=+⎟ ⎠⎞⎜ ⎝⎛ ∂∂−∂∂iS dS Uθδθθθθ (4.2.9) Now 0)(≥iSδ , 0>θ , and θd can be positive, negative or zero. Thus the equality in Eqn. 4.2.9 can only be satisfied in general if both 0 and0)(= =⎟ ⎠⎞⎜ ⎝⎛ ∂∂−∂∂i VSS Uδθθθ (4.2.10) The second equality shows that a quasi-static pure heating process is always reversible . The first equality is a relation betwee n state functions and hence holds for all processes, not just for pure heating. In reality there is never any such thing as a completely reversible process – in the case of pure heating, it is assumed that the temperat ure at any instant is uniform throughout the material, which will never be exactly true. It will be shown below that if there is any appreciably temperature gradient within a ma terial then there w ill be entropy production. Reversible Processes To be precise, a process is re versible when both the system and its surroundings can be returned to their original st ates. For example, if the ma terial in a piston-cylinder arrangement is compressed quasi-statically a nd there is no friction between the piston and cylinder walls, then the process is reversib le – the load can be reduced by very small amounts and the material will “push back” on the piston returning it to its original configuration, with no net work don e or heat supplied to the system. Irreversible Processes An irreversible process is one for which there is entropy production, 0)(>iSδ . In practice, irreversibilities ar e introduced into systems when ever there is spontaneity: ƒ unrestrained expansion of a gas/liquid to a lower pressure – for example when the lid is taken off a gas at high pressure, and is allowed to escape into the atmosphere ƒ heat transfer from one part of a material to another part at a lower temperature (except in the ideal case where the temperature difference is infinitesimal) ƒ friction (both the sliding friction of solid on solid and the friction that occurs between molecules in the flow of fluids) VSU ,, θQδ Section 4.2 Solid Mechanics Part III Kelly 408The common factor amongst all these is that the system and its surroundings cannot be returned to their original configurations. For example, with the piston-cylinder arrangement, friction between the piston head and cylinder walls means that further work needs to be expended on the return stroke so that, although the piston -cylinder is returned to its original state, a net amount of work needs to be done and so the “surroundings”, or whatever is producing the work, is not back at its original state. Similarly, if the piston was compressed very quickly to its final pos ition, the temperature, momentarily, might well be higher at the piston h ead than further down in the material. This would produce a spontaneous heat transfer from the upper part of the material to the lower part and it would not be possible to return the system a nd its surroundings to their original states. Irreversible Heat Transfer In the processes studied so far, it has been assumed that all the state functions were uniform throughout the material. In particular , it has been assumed that the temperature is uniform throughout. What if one now ha s a system whose parts are at different temperatures? Suppose that a quantity of heat Qδ flows from a body at temperature 1θ to a body at temperature 2θ, Fig. 4.2.2. One can imagine for the sake of argument that the heat capacities of both bodies are sufficiently large that their temperatures are effectively unchanged by the heat flow. The two bodi es are insulated fr om their surroundings. Figure 4.2.2: Heat flow from one body to another This is pure heating and the entropy change due to this heat tr ansfer are the entropy supplies 0 /1)( 1 <=θδδ Q Sr and 0 /2)( 2 >=θδδ Q Sr. Considering now the complete system (both bodies), there is no entropy s upply, so any entropy change must be an entropy production 1 2)( θδ θδδQ QSi−= (4.2.11) Since 0)(≥iSδ , it follows that 2 1θθ>, that is, heat flows from the warmer body to the colder body . In this example there is no work done, no heat transfer and no internal energy change, but there is an entropy change. If one wants the heat transfer to be very nearly reversible, one can make the entropy production very small. This can be achi eved by making the temperature difference between the two bodies very small: by letting θθθθθ Δ+==2 1 , , one has 2 2,Sθ1 1,SθQδ Section 4.2 Solid Mechanics Part III Kelly 409() () θθθδδ / /)(Δ−≈ Q Si. Keeping the entropy supply consta nt, this means that one must make θθ/Δ as small as possible. Thus heat transfer is reversible only if there is an “infinitely small” temperature difference betw een the two bodies. If, on the other hand, there is heat flow between bodies with an a ppreciable temperature difference, the process is irreversible, and a net am ount of energy will be require d to return the bodies and environment to their original states. Entropy supply is due to heat transfer, but the entropy production here is due to an adiabatic irreversible change. Entropy Measurements The entropy of a material can be meas ured as follows. First, since )(/iS Q dS δθδ+= , one has ())(/iS dC dS δθθ+ = where C is the specific heat cap acity. A re-arrangement shows that, for a reversible process, C is related to state variables through ()θθ ddS C /= (and in particular, ()V V ddS C θθ /= , ()p p ddS C θθ /= ). One can now use the expression ∫=Δ2 1/θ θθθdC S (reversible) (4.2.12) but one must ensure that the entropy production is zero. In practice, what one does is keep θθ/d small enough so that the entropy produc tion is sufficiently small for the accuracy required. Once the entropy cha nge is found, it of course applies to all processes, not just the reversible pro cess used in the experiment. Thermodynamic Equilibrium Thermodynamic equilibrium has already been mentioned – it occurs when no changes of the state variables can occur. Thus, one requires that 0 = == dUQ Wδδ . With 0 =Qδ , one has 0)(=rSδ and )(idS dS= . For full thermal equilibrium, one requires that 0)(=idS but, since entropy producti on tends always to incr ease the entropy, thermal equilibrium can only occur if the entropy has reached its maximum possible value . 4.2.4 Free Expansion of an Ideal Gas It was seen that, for an ideal gas undergoi ng a quasi-static process, (see Eqn. 4.1.25) ()()1 2 1 2 1 2 / ln / ln VV mR C S SSV + =−=Δ θθ (4.2.13) and the entropy production is zero. In other wo rds, any quasi-static process involving an ideal gas is reversible. Consider now a thermally insula ted container divided by a partit ion into two parts each of volume V. One of these contains an ideal gas an d the other is evacuated. The partition is taken away, so that the gas completely fills th e container (see Fig. 4.1.7). There is no heat Section 4.2 Solid Mechanics Part III Kelly 410supply and there is no work done and so the in ternal energy of the gas does not change. Since the internal energy is a function of temperature only, the temperature must be constant. Therefore the entropy change is 2lnmRS=Δ . Since there is no entropy supply, this must be entropy production. This example again illustrates that en tropy production occurs during spontaneous processes. During the spontaneous expans ion, there is a complex non-equilibrium turbulence. The gas eventually settles dow n and a new equilibrium position is reached. 4.2.5 Problems 1. A system undergoes a process in which work is done on the system and the heat transfer Q occurs at a constant temperature bθ. For each case, determine whether the entropy change of the system is pos itive, negative, zero or indeterminate: (a) reversible process, 0 >Q (b) reversible process, 0 =Q (c) reversible process, 0 <Q (d) irreversible process, 0 >Q (e) irreversible process, 0 =Q (f) irreversible process, 0 <Q 2. A block of lead at temperature K 200 has heat capacity 1KJ 1000−=C , which is independent of temperature in the range K 200 100− . It is to be cooled to K 100 in liquid baths, which are large enough that th eir temperatures do not change. What is the entropy supply for the lead and the liquid bath(s), a nd the net entropy production, during the following processes: the lead is (a) plunged straight into a liquid bath at K 100 (b) first cooled in a bath at K 150 and then in a second bath at K 100 (c) cooled using four baths at temperatures K 175 , K 150 , K 125 and K 100 (d) cooled in an infinite number of temper ature baths with a continuous range from K200 to K 100 [hint: no work is done; use the lead ’s heat capacity to evaluate Q] 3. Consider the freely expanded gas discussed in section 4.2.4. Suppose the gas is now quasi-statically (reversibly) compressed at constant temperature back to it original volume V. Is the gas back in its original state? Are the surroundings? 4. Consider an insulated piston- cylinder assembly which initially contains water as a saturated liquid at C 100 , as illustrated below. A pa ddle wheel acts on the water, which undergoes a process to the corres ponding saturated vapour state at the same temperature, during which the piston moves freely in the cy linder (no friction). Using the data below, determine (a) the net work per unit mass done – which is greater, the work done by the paddle wheel or that done by the expanding water? (b) the specific entropy supply ; the specific entropy production – why do you think it is non-zero? Section 4.2 Solid Mechanics Part III Kelly 411Next, consider the case where the initial and final states are the same as before, but the change is now brought about by the supply of heat only (with no paddle wheel). Determine (c) the work done per unit mass2 (d) the heat transfer per unit mass (e) the specific entropy supply and the sp ecific entropy production – is this a surprise? u (kJ / kg) v (m3 / kg) s (kJ / kg.K) p (MPa) Liquid 418.94 0.0010435 1.3069 0.1014 Gas 2506.50 1.673 7.3549 0.1014 5. A certain mass of an ideal gas for which 2 /3R CV= , independent of temperature, is taken reversibly from Pa 10 K,1005= = p θ to Pa 108 K,4005×= = p θ by two different paths (1) and (2): (1) consisting of (a) at constant volume from K400 100→=θ , (b) isothermally to the final pressure (2) consisting of (a) at constant pressure from K400 100→=θ , (b) isothermally to the final volume Calculate the entropy changes a nd show that the to tal entropy change is the same for both paths. Compare this with the heat ab sorbed or given out for each of paths (1) and (2) – they even turn out to be of opposite sign. [hint: use the ideal gas law and the fact that for a constant volume process, θδ dCQV= ; also, use Eqn. 4.2.13, the fact that dS QS S∫=2 1revθ , and the result of Q.8 from section 1.1] 2 the initial and final temper atures and pressures are C100 and 0.1014 MPa – th ese are the “end-points” for the initial and final states – in general, they may not necessarily be constant throughout the process – we do not know (and don’t have to know here) how the temperature and pressure changed during the process in parts (a-b); with the paddle wheel, the temperature an d pressure are unlikely to be uniform throughout the material. For parts (c-e), it is reasonable to assume that they are constant throughout Section 4.3 Solid Mechanics Part III Kelly 4124.3 Thermodynamic Functions Four important and useful thermodynamic func tions will be considered in this section (two of them have been encountered in the previous sections). These are the internal energy U, the enthalpy H, the Helmholtz free energy (or simply the free energy ) Ψ and the Gibbs free energy (or simply the Gibbs function ) G. These functions will be defined and examined below for both reve rsible and irrevers ible processes. 4.3.1 Reversible Processes Consider first a reversible process. The Internal Energy The internal energy is dS pdVQ W dU θδδ +−=+= (4.3.1) the second line being valid for quasi-static pr ocesses. The state variables for the pure compressible substance include S V,,θ and p. From 4.3.1, it is natural to take V and S as the independent variables: dSSUdVVUdU V S⎟ ⎠⎞⎜ ⎝⎛ ∂∂+⎟ ⎠⎞⎜ ⎝⎛ ∂∂= (4.3.2) so that V S SU VUp ⎟ ⎠⎞⎜ ⎝⎛ ∂∂=⎟ ⎠⎞⎜ ⎝⎛ ∂∂−= θ , (4.3.3) Thus ),(SVU contains all the thermodynamic in formation about the system; given V and S one has an expression for U and can evaluate p and θ through differentiation. U is a thermodynamic potential , meaning that it provides information through a differentiation. V and S are said to be the canonical (natural ) state variables for U. By contrast, expressing the internal energy as a function of the volume and temperature, for example, ),(θVUU= , is not so useful, since this cannot provide all the necessary information regarding the state of the ma terial. A new state function will be introduced below which has V and θ as canonical state variables. Similarly, the equation of state ) ,(pVθ does not contain all the thermodynamic information. For example, there is no information about U or S, and the equation of state must be supplemented by another, just as th e ideal gas law is supplemented by the caloric equation of state ) (θUU= . Section 4.3 Solid Mechanics Part III Kelly 413 Returning to the internal en ergy function, and taking the di fferential relations between θ,p and U, Eqns. 4.3.3, and differentiating them again, and using the fact that VS U SV U ∂∂∂=∂∂∂ / /2 2, one arrives at the Maxwell relation S V V Sp⎟ ⎠⎞⎜ ⎝⎛ ∂∂=⎟ ⎠⎞⎜ ⎝⎛ ∂∂−θ (4.3.4) The Helmholtz Free Energy If one wants to work with V and θ, one can use the (Helmholtz) free energy function , defined by S Uθ−=Ψ (4.3.5) One has, again for a reversible process, θδ/Q S= , θθδθθ Sd pdVdSQ dUdS dS dU d −−=−−=−−=Ψ (4.3.6) Now V and θ have emerged as the natural independent variables. Writing ),(θVΨ=Ψ , θθθd dVVd V⎟ ⎠⎞⎜ ⎝⎛ ∂Ψ∂+⎟ ⎠⎞⎜ ⎝⎛ ∂Ψ∂=Ψ (4.3.7) so that VSVp ⎟ ⎠⎞⎜ ⎝⎛ ∂Ψ∂−=⎟ ⎠⎞⎜ ⎝⎛ ∂Ψ∂−=θθ, (4.3.8) The Enthalpy and Gibbs Free Energy The enthalpy is defined by Eqn. 4.1.17, pV UH+= (4.3.9) To determine the canonical state variables, evaluate the increment: Vdp pdVQ WVdp pdV dU dH +++=++= δδ (4.3.10) For a quasi-static process, pdV W−=δ and so Vdp dS dH +=θ (4.3.11) Section 4.3 Solid Mechanics Part III Kelly 414 and the natural variables are p and S. Finally, the Gibbs free energy function is defined by pVS UG +−=θ (4.3.12) and the canonical state variables are p and θ. The definitions, canonical state variables and Ma xwell relations for all four functions are summarised in Table 4.3.1 below. Thermo- dynamic potential Symbol and appropriate variables Definition Differential relationship Maxwell relation Internal energy ),(VSU dS pdV dU θ+−= V SS V Sp VVUpSU ⎟ ⎠⎞⎜ ⎝⎛ ∂∂−=⎟ ⎠⎞⎜ ⎝⎛ ∂∂⎟ ⎠⎞⎜ ⎝⎛ ∂∂−=⎟ ⎠⎞⎜ ⎝⎛ ∂∂= θθ , Enthalpy ),(pSH pV UH+= dS Vdp dH θ+= p SS p SV ppHVSH ⎟ ⎠⎞⎜ ⎝⎛ ∂∂=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂=⎟ ⎠⎞⎜ ⎝⎛ ∂∂= θθ , Helmholtz free energy ),(VθΨ S Uθ−=Ψ θSd pdV d −−=Ψ VV p VSVp S ⎟ ⎠⎞⎜ ⎝⎛ ∂∂=⎟ ⎠⎞⎜ ⎝⎛ ∂∂⎟ ⎠⎞⎜ ⎝⎛ ∂Ψ∂−=⎟ ⎠⎞⎜ ⎝⎛ ∂Ψ∂−= θθ θθ, Gibbs free energy ),(p Gθ pVS UG +−=θ θSd Vdp dG−= pp V pSpGVGS ⎟ ⎠⎞⎜ ⎝⎛ ∂∂−=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂=⎟ ⎠⎞⎜ ⎝⎛ ∂∂−= θθ θθ, Table 4.3.1: Thermodynamic Potential Functions and Maxwell relations Mechanical variables: whereas the internal energy and the Helmholtz free energy are functions of a kinematic variable ( V), the enthalpy and the Gibbs function are functions of a force variable ( p). Thermal variables: whereas the internal en ergy and the enthalpy are functions of the entropy, the Helmholtz and Gibbs free energy fu nctions are functions of the temperature. If one is analyzing a process with, for example, constant temperature, it makes sense to use either the Helmholtz or Gibbs free energy fu nctions, so that there is only one variable to consider. Note that the temperature is an observable st ate variable and can be controlled to some extent. Values for the entropy, on the other hand, cannot be assigned arbitrary values in experiments. For this reason a description in terms of the free energy, for example, is often more useful than a descripti on in terms of the internal energy. Section 4.3 Solid Mechanics Part III Kelly 4154.3.2 Irreversible Processes Consider now an irreversible process. The Internal Energy One has )(iS dS dU W θδθ δ +−= and, with the internal en ergy again a function of the entropy and volume, )(i S VS dVVUdSSUW θδ θ δ +⎟ ⎠⎞⎜ ⎝⎛ ∂∂+⎥⎦⎤ ⎢⎣⎡−⎟ ⎠⎞⎜ ⎝⎛ ∂∂= (4.3.13) Consider the case of pure heating 0)(===iS dV W δ δ , so VSU⎟ ⎠⎞⎜ ⎝⎛ ∂∂=θ (4.3.14) as in the reversible case. This relation is of course valid for any process, not necessarily a pure heating one. Then )(i SS dVVUW θδ δ +⎟ ⎠⎞⎜ ⎝⎛ ∂∂= (4.3.15) Express the work in the form dVA dVAW dW W d qd q )( )()( )( +=+= δ δ (4.3.16) such that the quasi-conservative force )(qA is that associated with the work )(qW which is recoverable, whilst the dissipative force dA produces the work )(dW which is dissipated, i.e. associated with irreversibilities. From 4.3.15, Sq VUA ⎟ ⎠⎞⎜ ⎝⎛ ∂∂=)( (4.3.17) and the dissipative work )(dWδ is 0)( )( )(≥==i d dS dVA W θδ δ (4.3.18) The name quasi-conservative force for the )(qA (here, actually a force per area) is in recognition that the internal ener gy plays the role of a potential in 4.3.17, but it is also a function of the entropy. It can be seen from Eqn. 4.3.17 that the quasi-conservative force is a state function, and equals p− in a fully reversible process. Section 4.3 Solid Mechanics Part III Kelly 416In the isentropic case, 0=dS , one has )(iS dU W θδ δ+= (4.3.19) This shows that, in the isentropic case, the in ternal energy is that part of the work which is recoverable. The Free Energy Directly from S Uθ−=Ψ , with θ and V the independent variables, SS U−⎟ ⎠⎞⎜ ⎝⎛ ∂∂−∂∂=∂Ψ∂ θθθθ, VS VU V ∂∂−∂∂=∂Ψ∂θ (4.3.20) From the pure heating analysis give n earlier, Eqn. 4.2.9-10, the term θθθ∂∂−∂∂ / / S U is zero, so VS ⎟ ⎠⎞⎜ ⎝⎛ ∂Ψ∂−=θ (4.3.21) as in the reversible case and θ θSd dVVd −⎟ ⎠⎞⎜ ⎝⎛ ∂Ψ∂=Ψ (4.3.22) The work can now be written again as Eqn. 4.3.16, but now with the quasi-conservative force given by { ▲Problem 3} θ⎟ ⎠⎞⎜ ⎝⎛ ∂Ψ∂=VAq)( (4.3.23) The dissipative work is ag ain given by 4.3.18. Also, p Aq−=)( for a reversible process. In the isothermal case, 0=θd , )(iS d W θδ δ+Ψ= (4.3.24) This shows that, in the isothermal case, the fr ee energy is that part of the work which is recoverable. The quasi-conservative forces for the internal energy and free ener gy are listed in Table 4.3.2. Note that expressions for quasi-conserva tive forces are not available in the case of the Enthalpy and Gibbs free energy since they do not permit in their expression increments in volume dV, which are required for expre ssions of work increment. Section 4.3 Solid Mechanics Part III Kelly 417 Thermo-dynamic potential Differential relationship Relations ),(VSU )(iS dS pdV dU θδθ−+−= Sq V VUASU⎟ ⎠⎞⎜ ⎝⎛ ∂∂=⎟ ⎠⎞⎜ ⎝⎛ ∂∂=)(, θ S U Vθ θ−=Ψ ),( )(iS Sd pdV d θδθ−−−=Ψ θψ θψ⎟ ⎠⎞⎜ ⎝⎛ ∂∂=⎟ ⎠⎞⎜ ⎝⎛ ∂∂−=VA Sq v)(, Table 4.3.2: Quasi-Conservative Forces for Irreversible Processes , pdV S dVA Wi q−=+=)( )(θδ δ 4.3.3 The Legendre Transformation The thermodynamic functions can be transforme d into one another using a mathematical technique called the Legendre Transformation . The Legendre Transformation is discussed in detail in the Appendix to this Chapter, §4.A. For the present purposes, note that the Legendre transformation of a function ) ,(yxf is the function ) ,(βαg where ),( ),( yxfy x g −+=βαβα (4.3.25) and yf xf ∂∂=∂∂=βα , , βα∂∂=∂∂=gygx , (4.3.26) When only one of the two variables is switched, the transform reads ),( ),( yxfx y g −=αα (4.3.27) where xf ∂∂=α , α∂∂=gx (4.3.28) For example, if one has the function ) ,(VSU and wants to switch the independent variable from S to θ, Eqn. 4.3.27 leads one to consider the new function ),( ),( VSUS Vg −=θθ (4.3.29) and Eqns. 4.3.28 give VSU⎟ ⎠⎞⎜ ⎝⎛ ∂∂=θ and VgS⎟ ⎠⎞⎜ ⎝⎛ ∂∂=θ (4.3.30) It can be seen that ) ,(Vgθ is the negative of the Helmholtz free energy and the two differential relations in 4.3.30 are contained in Table 4.3.1. Section 4.3 Solid Mechanics Part III Kelly 418 4.3.4 Problems 1. By considering reversible processes, deri ve the differential relationships and the Maxwell relations given in Table 4.3.1 for (a) the enthalpy, (b) the Gibbs free energy 2. Let the two independent variables be V and θ. Consider the internal energy, ),(θVUU= . Use the pure heating example cons idered in §1.2 to show that the quasi-conservative force of Eqn. 4.3.17 can also be expressed as θθ⎟ ⎠⎞⎜ ⎝⎛ ∂∂−∂∂=VS VUAq)( 3. Show that Eqn. 4.3.22 leads to Eqn. 4.3.23. 4. Use the Legendre Transformation rule to transform the Helmholtz free energy ),(VθΨ into a function of the variables θ and σ. Derive also the two differential relations analogous to Eqns. 4.3.30. Show th at this new function is the negative of the Gibbs energy (use the relation S Uθ−=Ψ ), where p−=σ , and that the two differential relations correspond to two of the relations in Table 4.3.1. 5. Use the Legendre Transformation rule to transform the enthalpy ),(pSH into a function of the variables S and V. Show that this new func tion is the negative of the internal energy, and that the two differe ntial relations correspond to two of the relations in Table 4.3.1. Section 4.4 Solid Mechanics Part III Kelly 4194.4 Generalised Variables Here, the ideas of the last th ree section are generalised to the case of more complex processes. 4.4.1 Kinematical and Force Variables Consider an arbitrary continuum element, small enough so that, w ithin it, the state variables are uniform (although they differ from element to element). Let the state of the element, or system, be described by a set of independent kinematical variables ka, n kL,2,1= , and by its temperature θ. For a simple compressible system, there is only one kinematical variable, and one us ually takes this to be the volume V; for a more complex deforming element, the kinematical variables might include the six independent components of a strain tensor. If the system undergoes a change corre sponding to infinitesimal increments kda, the corresponding work done is of the form k kdaA W=δ (4.4.1) The coefficients kA are called the force variables corresponding to the kinematical variables ka (they are work-conjugate to the ka). For the simple compressible system, there is only one force variable, p A−= , the pressure. In a more complex system, they might include the 6 independent components of a stress tensor. This work expression can include magnetic eff ects, chemical effects, and so on. For example, if an amount of electrical charge dq flows into the system, then the work done is dqEW =δ , where E is the electrical poten tial difference driving the charge. In this context, E is regarded as a force variable wh ich drives the corresponding kinematical variable dq. The first law can now be written as Q daA dU k kδ+= (4.4.2) The results obtained thus far can be genera lised by replacing the (negative of the) pressure with the more general force vari ables, and by replacing the volume with the more general kinematical variables. The simple compressible system was completely described using two state variables. There are now 1+n independent variables; any of the sets, with n kL,2,1= , ()kaS, - internal energy description, ()kAS, - enthalpy description, ()ka,θ - Helmholtz free energy description or ()kA,θ - Gibbs free energy, can be used. Section 4.4 Solid Mechanics Part III Kelly 4204.4.2 The Internal Energy Taking ()kaS, to be the independent state variables, the results for the internal energy of the last section can be re -written and summarised as dSSUdaaUdUk k∂∂+∂∂= (4.4.3) kaSU⎟ ⎠⎞⎜ ⎝⎛ ∂∂=θ (4.4.4) Skq ki kq kkd k kq kd q aUA S daAdaA daAW W W ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂= +=+=+≡ )( )( )()( )()( )( ,θδδδδ (4.4.5) and in a reversible process, kq k A A=)(. 4.4.3 The Free Energy Taking ()ka,θ to be the independent state variables, the results for the internal energy of the last section can be re -written and summarised as S U ak θθ−=Ψ ),( (4.4.6) θθd daadk k∂Ψ∂+∂Ψ∂=Ψ (4.4.7) kaS ⎟ ⎠⎞⎜ ⎝⎛ ∂Ψ∂−=θ (4.4.8) θθδδδδ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂Ψ∂= +=+=+≡ kq ki kq kkd k kq kd q aA S daAdaA daAW W W )( )( )()( )()( )( , (4.4.9) and in a reversible process, kq k A A=)(. Section 4.5 Solid Mechanics Part III Kelly 4214.5 The Rate Equations The incremental equations of thermodynamics derived thus far can be written in rate form by dividing them through by dt. Thus for example the work increment Wδ becomes the rate of working or power *W - the superscript “*” is used here and in what follows to indicate rates of change of quantities whic h are not state functions. The rate form of the first law is * * *QaA Q WUkk+=+= & & (4.5.1) and the rate form of the second law is 0 )*(≥iSθ or *QS≥&θ (4.5.2) where )*( )*( i rS SS+=& and θ* )*( QSr= (4.5.3) *Q is called the thermal power , the non-mechanical power , or the rate of thermal work . 4.5.1 The Dissipation Function The rate of working can be expressed as )*( )( * i kq k S aA W θ+=& (4.5.4) with )(q kA expressed in any of a number of different ways (e.g. see Eqns. 4.4.5, 4.4.9). The quantity )*(iSθ is called the dissipation (or dissipation rate ) and is denoted by Φ. It is equivalent to the rate of working of the dissipative forces: 0)*( )( )*(≥===Φd kd kiW aA S & θ . (4.5.5) As with *Q, the dissipation is not a state function. The first law can now be written in the rate form * )(Q aA Ukq k+Φ+=&& (4.5.6) The isentropic case (see Eqn. 4.3.19) can be expressed as Φ+=U W&* (4.5.7) Section 4.5 Solid Mechanics Part III Kelly 422and the isothermal case (see Eqn. 4.3.24) can be expressed as Φ+Ψ=&*W (4.5.8) Section 4.6 Solid Mechanics Part III Kelly 4234.6 Continuum Thermomechanics The classical thermodynamics is now extended to the thermomechanics of a continuum. The state variables are allowed to vary thr oughout a material and processes are allowed to be irreversible and move far from thermal and mechanical equilibrium. Some schools of thought would question whether entropy is a st ate function at all unde r these conditions. Here, we simply accept the fact that it is. This approach is known as the rational thermodynamics and is generally accepted in the solid mechanics community. 4.6.1 The First Law The first law of thermodynamics is, in rate form, KU Q Pext&&+=+* (4.6.1) where extP is the power of the external forces, *Q is the rate at which heat is supplied, U& is the rate of change of the internal energy and K& is the rate of cha nge of kinetic energy. Recall from Part III, Eqn 3.7.2, th e mechanical energy balance, K P P &=+int ext (4.6.2) Eliminating extP and K& from these equa tions leads to U Q P &=+−* int (4.6.3) Heat supply It is convenient to write the to tal heat supply to a finite volum e of material as an integral over the volume. This is done by defining the heat flux q to be the rate at which heat is conducted from interior to exterior per unit area, Fig. 4.6.1. The rate of heat entering is thus dv s∫⋅− nq . Let there also be a source of h eat supply inside the material, for example a radiator of heat. Let dvr v∫ be the rate of such heat supply, where the scalar r is the heat source , the rate of heat generated per unit volume. Thus, with the divergence theorem, dvr dv Q v v∫∫+ −= qdiv* (4.6.4) q n ds Section 4.6 Solid Mechanics Part III Kelly 424Figure 4.6.1: heat flux vector and no rmal vector to a surface element Recall also from Part III, Eqns. 3.7.15, the stress power ∫−= vdv P dσ:int (4.6.5) Combining Eqns. 4.6.3-5, and expressing the strain energy rate in the form of an integral (see Part III, Eqn. 3.7.15) leads to ∫∫ ∫∫=+ − v v vvdv dvr dv dv u div : &ρ q dσ (4.6.6) Since this holds for all volumes v, one has the local form u div : &ρ=+− rq dσ The First Law (4.6.7) 4.6.2 The Second Law Entropy The entropy ) ,(tSx is defined as th e scalar property dvs S v∫=ρ ( 4 . 6 . 8 ) where s is the specific entropy or entropy density . The change in entropy is due to two quantities. First, very like the heat transfe rred into a body, Eqn. 4.6.4, define the entropy supply )*(rS to be the rate of entropy input, dvs ds S vr sr∫∫+⋅−= nsq)*( (4.6.9) where qs is the entropy flux through the element surface and rs is entropy supply due to sources within the element . Further, the en tropy flux is assumed to be proportional to the heat flux, and the proportionality factor is the reciprocal of the non-negative scalar absolute temperature θ (and similarly for the density rs and the heat supply density r) so that, using the divergence theorem, dvrdvdvrds S v vv sr ∫∫∫∫ +⎟ ⎠⎞⎜ ⎝⎛−=+⋅−= θθθ θ qnq div)*( (4.6.10) Define the entropy production )*(iS to be the difference between the rate of change of entropy and the entropy supply: Section 4.6 Solid Mechanics Part III Kelly 425 )*( )*( r iSS S−=& ( 4 . 6 . 1 1 ) The second law of thermodynamics states th at the entropy producti on is a non-negative quantity, 0)*(≥iS (4.6.12) The Clausius-Duhem Inequality Thus one has the Clausius-Duhem inequality : 0 div)*(≥−⎟ ⎠⎞⎜ ⎝⎛+ = ∫∫∫ v v vidvrdv dvsdtdSθθρq (4.6.13) In local form, the Clausius-Duhem inequality reads as (introducing a specific entropy production, *)(is) 0 div1*)(≥−⎟ ⎠⎞⎜ ⎝⎛+=ρθθρrs si q& (4.6.14) or, equivalently { ▲Problem 1}, 0) (1div1 2*)(≥∇⋅− +−= θρθ ρθρθq qrs si& The Second Law (4.6.15) This is the continuum statement of the Second Law. Note that, in the classical th eory, the temperature is assume d to be constant throughout, so that 0=∇θ , which leads to θ ρ /) (div*)(q−=rs which corresponds to the classical expression θδ/)(Q dSr= . 4.6.3 The Dissipation Inequality Eliminating qdiv (and r) from both the first and second laws leads to the dissipation inequality () 0) (1:1*)(≥∇⋅− +−= θρθ ρθθ q dσ us si&& Dissipation Inequality (4.6.16) The term *)(isθ is the specific dissipation (or internal dissipation ) and is denoted by the symbol φ. The Clausius-Duhem inequality can simply be written as 0)*(≥≡isθφ (4.6.17) Section 4.6 Solid Mechanics Part III Kelly 426 Multiplying Eqn. 4.6.16 across by the density leads to [] 0) (1:*)(≥⎥⎦⎤ ⎢⎣⎡∇⋅−++−= θθρρθρθ q dσu s si&& (4.6.18) Each term here has units of power per unit (current) volume. The term inside the first bracket is called the mechanical dissipation (per unit volume). The term inside the second bracket is the dissipation due to temperature gradients, i.e. heat flow, and is called the thermal dissipation (per unit volume). Note that the thermal dissipation is always positive since q and θ∇ are of opposite sign. Integrating over a volume v leads to {} () dv dvs u dvs v v vi∫ ∫∫⎭⎬⎫ ⎩⎨⎧⋅∇−++−= q dσθθρθρ ρθ1:)*(&& (4.6.19) The term *)(isρθ in Eqn. 4.6.18 is often denoted by the symbol γ and also termed the dissipation. This is a dissipation per unit vol ume. When the deformations are small, the volume changes are negligible. When the deformations are appreciable, however, the volume and density change, and it is better to work with specific quantities such as φ. The dissipation inequality 4.6.16 is in terms of the internal energy. In terms of the specific free energy θψ su−= , one has () 0) (1:1*)(≥∇⋅− +−−== θρθ ρψθθφ q dσ&&s si (4.6.20) 4.6.4 Special Thermodynamic Processes Reversible Processes In a reversible process, 0)*(==isφ and there are no temperat ure gradients (although the temperature may change), so ()dσ:1 ρθ+=s u&& or ()dσ:1 ρθψ+−=&& s (4.6.21) Isentropic Conditions For an isentropic process, the entropy is constant and remains constant, so 0=s& . In this case, the dissipation is mechanical dissipation thermal dissipation dissipation Section 4.6 Solid Mechanics Part III Kelly 427() 0) (1:1≥∇⋅− +−= θρθ ρφ q dσ u& (4.6.22) Isothermal Conditions In an isothermal process, the absolute temperature is constant, 0=∇θ , and remains constant, 0=θ& . This can be achieved, for example, by keeping the material’s surroundings at constant temperature, and load ing the material very slowly, so that any temperature differences which arise between the material and surroundings are allowed to disappear. One then has () 0 :1≥ +−= dσρθφ us&& or () 0 :1≥ +−= dσρψφ& (4.6.23) The second of these can be written as Φ+Ψ=+= &&ρφψρdσ: with 0≥Φ (4.6.24) where Ψ& is now the rate of change of the free energy per unit (current) volume , and Φ is the rate of dissipation per unit (current) volume . This equation is the most useful starting point for many applications. The free energy Ψ represents the energy that is stored and “free” to do more work or be recovered, whilst Φ is the rate at which energy is being dissipated and irreversibly lost from the “m acro-world”. The isothermal behaviour of different types of materials can be m odeled by choosing different free energy and dissipation functions. Equilibrium Conditions As mentioned in §4.2.3, a material which is unaffected by exte rnal conditions has no work done to it or heat supplie d and the first law then states that the internal energy is constant. In that case, when the entropy ha s reached a maximum and the dissipation is zero, there is no more change in any of th e state variables, and equilibrium has been reached. Adiabatic Conditions In an adiabatic process, oq=. This can be achieved, for example, by very rapid loading, so that there is no time for heat exchange with the surroundings. Under these conditions (and taking also 0=r ), the first law reads u :&ρ=dσ (recall that the internal energy change is equal to the work done in an adiabatic process). The dissipation is thus due solely to thermal effects. The dissipation inequality reduces to 0*)(≥=s si& (4.6.25) Section 4.6 Solid Mechanics Part III Kelly 428or 0≥=s&θφ (4.6.26) If the process is both adiaba tic and isentropic, then 0 ==s&φ . An adiabatic reversible process is equivalent to an isentropic reversible process. 4.6.5 The Clausius-Plank Inequality In many applications the thermal dissipation is very much smaller than the mechanical dissipation. If this is the case then the th ermal dissipation rate can be ignored, and one has the stronger form of the second law, in terms of internal energy and free energy, () () 0 :10 :1 ≥ +−−=≥ +−= dσdσ ρψθφρθφ &&&& sus Clausius-Plank inequality (4.6.27) which is known as the Clausius-Plank inequality . Note that the thermal dissipation is indeed zero under two of the commonest conditions, adiabatic and isothermal. Equivalently, one can argue that the processes of mechanical dissipati on and heat flow are independent, so that each are separately requ ired to be non-negative, again leading to Eqn. 4.6.27. Using the first law, Eqn. 4.6.27 can be rewritten in the alternative form qdiv1 1 ρρφθ −+= r s& (4.6.28) which is an evolution equation for s (showing how it evolves over time). 4.6.6 Small Strains When the strains are small, the rate of deformation is equivalent to the time rate of change of the small strain tensor: εd&=, ij ijdε&= (4.6.29) The dissipation inequalities are then 01 1 ,≥ −+−=ii ijij q us θρθεσρθφ &&& , 01:1≥∇⋅−+−= θρθρθφ qεσ&&&us or Section 4.6 Solid Mechanics Part III Kelly 42901 1 ,≥ −+−−=ii ijij q s θρθεσρψθφ &&& , 01:1≥∇⋅−+−−= θρθρψθφ qεσ&&&s (4.6.30) Reversible processes must involve vanish ing temperature gradients and lead to 01=+−ijij us εσρθ &&& or 01=+−−ijij s εσρψθ &&& (4.6.31) With ),(ijsuuε= and ) ,(ijεθψψ= , one has ij ijussuu εε&&& ∂∂+∂∂= and ij ijεεψθθψψ &&& ∂∂+∂∂= (4.6.32) Comparing with Eqn. 4.6.31 then leads to th e relations (compare these with 4.3.3 and 4.3.8) ijiju su ερσθ∂∂=∂∂= , and ijij sεψρσθψ ∂∂=∂∂−= , (4.6.33) Constitutive Relations for Small-strain Reversible Processes Note that the density (and volume) changes fo r small strains may be neglected, so that the density in 4.6.33 can be taken to be the curr ent density or the density in the undeformed configuration, 0ρ. 4.6.7 Thermomechanics in the Material Form The First Law In order to rewrite the energy balance equa tions in material form, first introduce the scalars (what follows is analogous to the defini tions of traction and st ress with respect to the current and reference configurations) NQnq Nn ⋅−=⋅−= )()( Qq (4.6.34) Here q is the Cauchy heat flux of Eqn. 4.6.4, defined per unit current surface area ds with outward normal n, and Q the Piola-Kirchhoff heat flux , defined per unit reference surface area dS and outward normal N. The rate of heat transfer into the material can now be written as either of dS ds S s∫∫⋅−=⋅− NQ nq (4.6.35) Section 4.6 Solid Mechanics Part III Kelly 430Using Nanson’s formula, Part III, Eqn. 2.2.59, dS J ds NF nT−= , the Cauchy and Piola- Kirchhoff heat flux vect ors are related through qF Q1−=J . The combination of the mechanical energy balance with the first law, i.e. Eqn. 4.6.3, then reads (see also Part III, Eqn. 3.7.26) ∫ ∫∫∫=+ − V VV VdV dVR dV dV u Div :0& & ρ Q FP (4.6.36) where dvr dVR v V∫∫= , or, in local form, u Div :0& & ρ=+− RQ FP ( 4 . 6 . 3 7 ) or u Div :0& & ρ=+− RQ ES (4.6.38) Note that, comparing the spatial and material forms, q Q q Q q Q div1Div1,div Div, div Div 0 ρ ρ= = =∫∫J dv dV v V (4.6.39) The Second Law Analogous to Eqn. 4.6.13, the second law can be expressed in material form as 0 Div0 ≥−⎟ ⎠⎞⎜ ⎝⎛+ ∫ ∫∫ V V VdVRdV dVsdtd θ θρQ (4.6.40) or, analogous to 4.6.16, one ha s the dissipation inequality () 0) Grad(1:1 0 0*)(≥⋅− +−= θθρ ρθθ Q FP& &&us si (4.6.41) Isothermal Conditions In an isothermal process, ()0 :1 0≥ +−= FP& && ρθφ us or ()0 :1 0≥ +−= FP& & ρψφ (4.6.42) The second of these can be written as Section 4.6 Solid Mechanics Part III Kelly 431Φ+Ψ=+= &&& φρψρ0 0 :FP with 0≥Φ (4.6.43) where Ψ& is now the rate of change of the free energy per unit (reference) volume , and Φ is the rate of dissipation per unit (reference) volume . 4.6.8 Objectivity By definition, the scalars heat Q, internal energy U, entropy S and temperature θ are objective, that is they remain unchanged unde r an observer transformation 2.8.7. It follows that the heat flux vector q is also objective, transfor ming according to 2.8.10. By definition, the vector entropy flux qs is objective, that is it transforms according to 2.8.10. 4.6.9 Problems 1. Show that ) (1div1div2θθθθ∇⋅−=⎟ ⎠⎞⎜ ⎝⎛q qq 2. Show that the relation qF Q1−=J is consistent with the relation 4.6.39, q Q div Div J= . Section 4.A Solid Mechanics Part III Kelly 4324.A Appendix to Chapter 1: The Legendre Transformation 4.A.1 One Dimensional Legendre Transformation Consider the curve Γ plotted in Fig. 4.A.1. This curv e can be described in a number of different ways. For example it can be expres sed in the conventional form, as a function of x: )(xf . One can also express it as a function of Y, where Y is the distance from the origin to the point where the tangent to the curve intersects the vertical axis, as illustrated. Only one independent variable is needed to describe the curve so x and Y are related. First, the slope of the curve is dxdfm= (4.A.1) From the construction of Fig. 4.A. 1, the slope is also given by () xxfYm+==θtan (4.A.2) and so ()xf xmY−= (4.A.3) Differentiating 4.A.3 with respect to m (considering x now to be a function of m) leads to dmdx dxdfmdmdxxdmdY−+= (4.A.4) Figure 4.A.1: A curve represented as a function of x or Y •pθ Y)(xfY+ xΓ o qθ Section 4.A Solid Mechanics Part III Kelly 433Using 4.A.1, dmdYx= (4.A.5) Equations 4.A.3, 4.A.1 and 4.A.5, constitu te a Legendre transformation between the function ()xf and the function ()mY :1 () () () ()()() dmmdYxdxxdfm mxfmmx mY = = −= , , Legendre Transformation (4.A.6) One says that Y is the Legendre dual of f and vice versa . Example To find the Legendre dual of ()23x xf= , note that x m6= and so () () () ()12 6362 2m mmmmxfmmx mY =⎟ ⎠⎞⎜ ⎝⎛−= −= and, as expected, xm dmdY==6 More generally, it can be shown that { ▲Problem 2} the Legendre dual of nx is ()( )11− ⎟ ⎠⎞⎜ ⎝⎛−=nn nmn mY Degenerate Case An interesting case is the function ()xxf=. The tangents at all points on this line go through the origin. Given x, one can determine m – it is always 1 tan==θ m – and () xxf= determines the complete curve. However, given 1=m , one cannot uniquely determine x; for the one value of m there are an infinite number of points x. This is a degenerate case of the Le gendre transform, called a singular transformation . In this case, again 1 /== dxdfm and the Legendre transform of ()xf is identically zero: () 0=mY . Since the transform can only be determ ined to within an arbitrary constant, instead of 4.A.3, introduce a new function Y such that ()() 0=−== f xm mY mYλ (4.A.7) 1 some authors use the definition xmf Y−= , i.e. the negative of this Section 4.A Solid Mechanics Part III Kelly 434 where λ is an arbitrary scalar. Then xdmYd dmdY==λ (4.A.8) Now for a given function ()mY , Eqn. 4.A.8 generates all values of x by assigning different values to λ. In summary ()() () () ()() () dmmYdxdxxdfm mxfmmx mY mY λ λ = = −== , , ) Singular Legendre Transformation (4.A.9) As an example, consider the function () 012=−=m mY . Then (taking the positive root) 1=m , () xxf= and the Legendre transform is λ2=x . Properties of the Legendre Transform Some useful properties of the Legendre transform follow: Let ()mY be the Legendre transform of ) (xf . Let ()qZ be the Legendre transform of )(pg . Then 1) () () xagxf= ⇒ ()amaZY /= scaling 1 2) () ( ) axgxf= ⇒ ()amZY /= scaling 2 3) () () axgxf += ⇒ ()a mZY−= translation 1 (4.A.10) 4) () ( ) axgxf += ⇒ ()am mZY−= translation 2 5) () () xgxf1−= ⇒ ()m mZ Y /1−= inversion 4.A.2 Multi-dimensional Legendre Transform Referring to Fig. 4.A.2, consider a function of two variables ()yxfz ,= and denote the partial derivatives by yf xf ∂∂=∂∂= β α , (4.A.11) The normal to the surface () 0 ,=−=Γ zyxf is 3 2 1 ee e n −+=Γ∇= βα (4.A.12) Let v be the vector joining q and p in the figure, so that Section 4.A Solid Mechanics Part III Kelly 435()3 2 1 e e e v Yf y x +++= (4.A.13) Since 0=⋅vn , one finds that () ()yxfy x Y , , −+=βαβα (4.A.14) Differentiating with respect to α and β (with ()βα,xx= , ()βα,yy= ) leads to βββββαβαααβααα ∂∂ ∂∂−∂∂ ∂∂−∂∂++∂∂=∂∂∂∂ ∂∂−∂∂ ∂∂−∂∂+∂∂+=∂∂ y yf x xf yyx Yy yf x xf y xxY (4.A.15) From 4.A.11, one has β α ∂∂=∂∂=YyYx , (4.A.16) Figure 4.A.2: A surface This can be generalised to higher di mensions. In summary, for vectors []TxxK,,2 1=x and []TwwK,,2 1=w , () ()() () ii i ii i i ii iwwYxxxfw xf xw wY∂∂=∂∂= −= , , Legendre Transformation (4.A.17) Example Consider the quadratic form jiji i xAx xf21)(= where A is symmetric; the two- dimensional version is •p YΓ o qx yn • •v Section 4.A Solid Mechanics Part III Kelly 436 [] ()2 2 22 21 122 111 21 21 22 1212 11 2 1 212 )( aaxxa xaxx a aa axx xfi + +=⎥⎦⎤ ⎢⎣⎡ ⎥⎦⎤ ⎢⎣⎡= Then ()nin mi m nin i i xA Ax xA xf w =+=∂∂=21/, s o m im i wA x1−= and () ()() j ijimp mp m imi q mqp mp m imiq nq mn p mp m imi n mn m ii i ii i wAwwwA wAw w wA wAwwAAwA wAw xAx xw xf xw wY 1 211 21 1 1 21 11 1 21 1 21)( −− − − −− − − =− = − =− = −=−= δ 4.A.3 Problems 1. Show that xexf=)( and mmm mY −=ln )( are Legendre duals 2. Show that ()nxxf= and ()( )11− ⎟ ⎠⎞⎜ ⎝⎛−=nn nmn mY are Legendre duals A1Answers to Selected Problems: Chapter 1 1.1 2. 3 1.3 1. -10 3. 9/19 4. 90o 6. 3 2 1 6 5 2 e e e++ 1.5 1. ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ − −− 2/1 2/12/12/12/1 2/12/1 0 2/1 1.6 2. 3 23 126 4 3 e e e t t t +− 3. (iii) x2, (iv) xx/ 5. 1 32 x xx+ 3 31 2 2 21 ) ( ) 1( e e xx x xx −+−− 1.7 1. 303 9. π2 1.8 1. No 3. when ba= 1.9 2. No 4. i jk ijkBA e or simply jk ijkBA . 6. 17 :5 15 15 123 3 2 3 3 2 2 2 3 1 =⊗+⊗−⊗−⊗+⊗=⋅ DFee ee eeee ee FD 7. 3 2 1 5 10 4 e e e++ A28. (a) a scalar, equals the trace of a second-order tensor (b) 3 functions of the 27 com ponents of a third-order tensor (c) 9 components of a second-order tensor (d) scalar 9. jiji dcba 1.10 1.11 1. The principal invariants are [] 0 det III2) tr()(tr II3 tr I 2 2 21 ===− === TT TT TTT and the eigenvalues are 2,1,0 7. (c) Spectral decomposition is ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ 100020008 Eigenvectors are ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ − 100 , 02/12/1 , 02/12/1 U is the square root of this. 1.12 1.13 1. (b) [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡− = 1 0 00 cos sin0 sin cos θθθθ Q (c) () ( )3 2 1 cos3 sin6 sin3 cos6 ee e u ′+′−+′−−= θθ θθ 1.14 1. (a) ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = 0 2 02 0 00 0 2 grad 231 xxx v (b) () ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ =⊗∇2 322 233 1 222 xxxxx vv 2. ou=∇2 A33. 1 3 grad eeu⊗= 1.15 7. (i) TA AT+ (ii)T T: : ATTA+ 1.16 12. Parabolic Cylindrical Coordinates (i) 1 , ,32 22 1 22 22 1 1 =Θ+Θ=Θ+Θ= h h h (ii) The Jacobian is =J2 22 1Θ+Θ (iii) ()()()2 32 22 22 12 12 22 12Θ+ΘΘ+Θ+ΘΘ+Θ=Δ d d d s ()2 12 22 1 33 12 22 1 23 22 22 1 1 ΔΘΔΘΘ+Θ=ΔΔΘΔΘΘ+Θ=ΔΔΘΔΘΘ+Θ=Δ SSS ()3 2 12 22 1 ΔΘΔΘΔΘΘ+Θ=ΔV 13. Elliptical Cylindrical Coordinates: (i) 1 , sin sinh , sin sinh3 22 12 2 22 12 1 =Θ+Θ=Θ+Θ= h h h (ii) The Jacobian is =J22 12sin sinh Θ+Θ (iii) ()( )( )2 32 2 22 12 2 1 22 12 2sin sinh sin sinh Θ+ΘΘ+Θ+ΘΘ+Θ=Δ d d d s ()2 1 22 12 33 1 22 12 23 2 22 12 1 sin sinhsin sinhsin sinh ΔΘΔΘΘ+Θ=ΔΔΘΔΘΘ+Θ=ΔΔΘΔΘΘ+Θ=Δ SSS ( )3 2 1 22 12sin sinh ΔΘΔΘΔΘΘ+Θ=ΔV 1.17 2. Jgg= 1.18 13. (a) ⎥ ⎦⎤ ⎢ ⎣⎡ −ΘΘ=⎥⎦⎤ ⎢⎣⎡ Θ∂∂ 021 2 jix A4 (b) ⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢ ⎣⎡ ΘΘ Θ− =⎥⎦⎤ ⎢⎣⎡ ∂Θ∂ 12 121210 ji x (c) () () ()() ()⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢ ⎣⎡ ΘΘ+ ΘΘΘ−ΘΘ− = ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ΘΘΘΘΘ+Θ= 2122 211212 21 2 12 122 41 44 41 ,4 ij ij g g (d) 12 212 122 222 111 221 211 121 111,0Θ=Γ=Γ=Γ=Γ=Γ=Γ=Γ=Γ (e) 2 1grad gg+=Φ