Kinematics. B r. Figure 1: Bodies, reference configuration B r and current configuration B t. κ : B (0, ) B. B r := κ(b, t 0 )

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1 1 Kinematics 1 escription of motion A body B is a set whose elements can be put into one-to-one correspondence with the points of a region B of three dimensional Euclidean three dimensional space. The region B is called the configuration of the body. Thus we have a invertible ^ E A mapping X P B r Figure 1: Bodies, reference configuration B r and current configuration B t. u κ : B (0, ) B which associate each element P of B at time t to a point κ(p, t) of B. The dependence on t implies that during the motion of the body, the body changes its configuration. In order to describe the motion of the body, we need to provide an arbitrary but fixed configuration of the body. This is caled reference configuration denoted by B r which refers to fixed (initial) time t 0. Thus ^ e i x P B t B r := κ(b, t 0 ) On the other hand the current configuration at time t is denoted by B t, thus B t := κ(b, t) Let P be an elements of B and X := κ(p, t 0 ) and x := κ(p, t)

2 2 We can eliminate P from these two relations and obtain a mapping χ from B r to B t : x = χ(x, t) The mapping is evidently bijective and it has the required differentiability properties. The mapping χ is called the motion of the body. For a specific particle χ defines the path line (trajectory) of the particle. Since χ(, t) is bijective, it has an inverse mapping χ 1 (, t) : B t B r so that X = χ 1 (x, t) Sometimes the above two mapping we write in the form x = x(x, t) (1) and X = X(x, t) (2) We choose a cartesian coordinate system with basis vectors {E A, E B, E C } for the reference configuration B r and refer position vector X in this coordinate system. On the other hand the position vector x is refrred with respect to a basis {ê 1, ê 2, ê 3 }. The coordinate system can be identical but often it may be advantageous to choose them to be different. 2 Material and spatial description The displacement is defined as the connecting vector between a given particle in B r and B t. If C denote the connecting vectors between the origins of the two bases, then u = x X + C Obviously the displacement will be different for different particles so we can write u = u(x, t) However this is not the only possibility. Since we can express X in terms of x, the above relation can also be written as u = u(x, t)

3 3 These two possibilities hold for other field quantities like scaler, tensor etc. also. Thus problems in continuum mechanics may be formulated using (X, t) as independent variable (Lagrangian or material description) or using (x, t) as independent variable (Eulerian or spatial description). To switch from material to spatial description we use X = X(x, t) and x = x(x, t) to change from spatial to material. 3 Velocity and acceleration efinition: The velocity field V (X, t) of a particle P B is defined as ( ) x(x, t) V (X, t) := X fixed where X is the position of P in B r. This is the rate of change of P in the material coordinate. The velocity in Eulerian description is given by ( ) x(x, t) v(x, t) := V (X(x, t), t) X=X(x,t) Note that X B r and x B t represents the same particle P B. efinition: The acceleration A(X, t) of a particle P B is defined as ( ) 2 x(x, t) A(X, t) := 2 X fixed where X is the position of P in B r. This is the rate of change of P in the material coordinate. The acceleration in Eulerian description is given by ( ) 2 x(x, t) a(x, t) := A(X(x, t), t) 2 X=X(x,t) 4 Material time derivative The time rate change of a quantity (scaler, vector or tensor) of a material particle is known as material time derivative. Thus velocity and acceleration we have defined earlier are material derivatives. There are two cases to consider. (i) The material description of the quantity is used i.e. the quantity is a function of X and t. We only have to take partial derivative with respect to t. Thus if φ(x, t) is a scaler filed, then its material derivative is given by φ := φ(x, t)

4 (ii) The spatial description of the quantity is used. If φ(x, t) is a scaler filed, then its material derivative is given by ( ) φ φ(x(x, t), t) := X fixed This could be simplified as follows: Since = φ(x(x, t), t) φ(x(x, t), t) = φ (x 1 (X, t), x 2 (X, t), x 3 (X, t), t) X=X(x,t) 4 we have φ(x(x, t), t) = + Using summation convention we can write φ(x(x, t), t) x 1 (X, t) φ(x(x, t), t) x 2 (X, t) + x 1 x 2 φ(x(x, t), t) x 3 (X, t) φ(x(x, t), t) + x 3 φ(x(x, t), t) φ(x(x, t), t) x i (X, t) = + x i φ(x(x, t), t) Now we come back to the spatial coordiantes by substituting X = X(x, t), t) to get φ(x(x, t), t) φ(x, t) φ(x, t) = v i (x, t) + = v ( φ) + X=X(x,t) x i φ(x, t) where v is the Eulerian velocity field. Thus for Eulerian field we have important formula φ φ(x, t) := v ( φ) + Next let u(x, t) be an Eulerian vector field. So we have ( ) u u(x(x, t), t) := u(x(x, t), t) = X fixed X=X(x,t) Now the components u i (x, t) of u(x, t) is a scaler field. Hence using the above result we write Thus u i := u i(x, t) + v ( u i ) = u i(x, t) + ( u) ij v j u := u(x, t) + ( u(x, t))v(x, t)

5 5 eformation gradient 5 The deformation gradient tensor F CT (2) is the material gradient (i.e. gradient w.r.t X) of the equations of motion (1) i.e F = Grad x(x, t) or in components form where F = F ia e i E A. F ia = x i X A = x i,a Using (2) we can define another tensor F 1 CT (2) by or in components form F 1 = grad X(x, t) F 1 Ai = X A = X A,i x i where F 1 = F A,i E A e i. We can verify that F 1 is the inverse tensor of F (i.e. F F 1 = F 1 F = I) easily. F F 1 = (F ia e i E A )(F 1 B,j E B e j ) = F ia F 1 B,j δ AB(e i e j ) = x i,a x A,j (e i e j ) = x i,j (e i e j ) = δ ij (e i e j ) = (e i e i ) = I Proceeding similarly we can show that F 1 F = I. The jacobian of the motion is defined by J := det(x i,a ) = det(f ) Since F is invertible, J 0. If the two cordinate systems are concident, then F = I at t = 0 i.e. at the reference configuration. Thus for concident coordinate systems J > 0 for t > 0. In describing motions and deformations, several measures of deformation are commonly used. First, let us consider that one based upon the change during the

6 6 deformation in the magnitude squared of the distance between the particles originally at P and Q i.e. dx 2 dx 2 = dx i dx i dx A dx A This we write as dx 2 dx 2 = x i,a dx A x i,b dx B δ AB dx A dx B = (x i,a x i,b δ AB )dx A dx B = (C AB δ AB )dx A dx B The symmetric tensor C = F T F = C AB E A E B is called Green s deformation tensor. It is a Lagrangian measure of deformation. Next define the Lagrangian finite strain tensor E as 2E = C I where the factor 2 is introduced for convenience in later calculations. Now we can write dx 2 dx 2 = dx 2EdX The difference dx 2 dx 2 may be expressed in terms of spatial variables similar way. We have dx 2 dx 2 = δ ij dx i dx j X A,i X A,j dx i dx j = (δ ij X A,i X A,j )dx i dx j = (δ ij c ij )dx i dx j The symmetric tensor c = (F 1 ) T F 1 is called Cauchy s deformation tensor. From this we define the Eulerian finite strain tensor e as 2e = I c and finally dx 2 dx 2 = dx 2edx Now from dx 2 dx 2 = 2e ij dx i dx j = 2E AB dx A dx B we derive that e ij = E AB X A,i X B,j and E AB = e ij x i,a x j,b

7 7 The Lagrangian and Eulerian finite strain tensor are expressed in terms of appropriate deformation gradient. These can also be expressed in terms of displacement gradient. We can write dispacement vector as Now we can write u = u i e i = u A E A u = x(x, t) X + = x X(x, t) + where is the distance between the origins of the two coordinate system. Taking Grad on the first equality gives Grad u = F I Similarly we have grad u = I F 1 Thus C = F T F = (I + (Grad u) T )(I + Grad u) = I + (Grad u) + (Grad u) T + (Grad u) T (Grad u) and c = (F 1 ) T F 1 = (I (grad u) T )(I grad u) = I (grad u) (grad u) T + (grad u) T (grad u) Thus Hence we also have 2E = F T F I = Grad u + (Grad u) T + (Grad u) T (Grad u) and 2e = I (F 1 ) T F 1 = grad u + (grad u) T (grad u) T (grad u) Properties: Let φ is a scaler field and u be a vector field. Then the following relations hold for gradient operator in the material and spatial coordinates. Grad φ = F T (grad φ) and Grad u = (grad u)f

8 8 To prove the first relation we proceed as follows Grad φ F T (grad φ) = φ,a E A F ia (E A e i )φ,j e j = φ,a E A F ia φ,j (E A e i )e j = φ,a E A F ia φ,j δ ij E A = φ,a E A φ,i F ia E A = φ,a E A φ,a E A = 0 Similarly Grad u (grad u)f = u A,B E A E B (u i,j e i e j )(F pq e p E Q ) = u A,B E A E B u i,j F pq δ jp (e i E Q ) = u A,B E A E B u i,p F pq (e i E Q ) = u A,B E A E B u i,q (e i E Q ) = (u A E A u i e i ),B E B = 0 since the quantity in the last line is the representation of the same vector field in two coordiantes. (Note: In the above proof we might have use Grad u = u i,a e i E A and grad u = u A,i E A e i. This does not effect the result.) From ɛ ijk J = ɛ ABC x i,a x j,b x k,c (3) By differentiating the above we get (J 1 x i,a ),i = 0 or div(j 1 F ) = 0 (4) To prove (4), we differentiate (3) by x i which gives ɛ ABC x j,b x k,c (J 1 x i,a ),i + ɛ ABC J 1 x i,a (x j,b x k,c ),i = 0 which gives ɛ ABC x j,b x k,c (J 1 x i,a ),i + ɛ ABC J 1 (x j,b x k,c ),A = 0 The second term on exapnding permutatio symbol is zero. Hence ɛ ABC x j,b x k,c (J 1 x i,a ),i = 0 Multiplying by X P,j X Q,k we get ɛ ABC δ P B δ QC (J 1 x i,a ),i = 0 or ɛ AP Q (J 1 x i,a ),i = 0

9 9 Multiplying the above by ɛ BP Q we get 2δ AB (J 1 x i,a ),i = 0 or (J 1 x i,b ),i = 0 Now from x i,a X A,j = δ ij we get J 1 x i,a JX A,j = δ ij Taking differentiation with respect to i we get Using the above result we get JX A,j (J 1 x i,a ),i + J 1 x i,a (JX A,j ),i = 0 x i,a (JX A,j ),i = 0 or (JX A,j ),A = 0 or iv(jf 1 ) = 0 Using the relation in (4) we can prove easily that for a vector field u and tensor field T iv u = J div(j 1 F u) and iv T = J div(j 1 F T ) To prove the first relation we proceed as follows iv u J div(j 1 F u) = u A J (J 1 F ib u B ) x A x i = u [ ] A J (J 1 F ib )u B + J 1 u B F ib x A x i x i = u A u B x i x A x i x B To prove the second relation we write = u A u B = 0 x A x B iv T J div(j 1 F T ) = T AB x A E B J This is due to the fact that = T AB x A E B J = T AB x A E B T Bj x B e j = x A (T AB E B T Aj e j ) = 0 (J 1 F ib T Bj )e j x [ i (J 1 F ib )T Bj e j + J 1 T Bj F ib e j x i x i ] T = T AB E A E B = T Ai E A e i

10 10 Thus we get E A (T AB E B T Ai e i ) = 0 or T AB E B T Ai e i = 0 6 Length and angle change Assume that N and n are unit vectors along dx and dx respectively. Thus dx = N dx and dx = n dx Using dx = F dx or n dx = F N dx we get (n dx ) (n dx ) = (F N dx ) (F N dx ) Simplifying the above expression we get Thus dx 2 = (N F T F N) dx 2 dx dx = N F T F N = N CN = F N : F N = F N Using the inverse relation X = F 1 x we arrive at Thus dx 2 = (n (F 1 ) T F 1 n) dx 2 = (n cn) dx 2 dx = F 1 n dx efinition: The ratio dx / dx is called stretch. Stretch may be expressed in terms of either N or n. To indicate these dependence we denote stretch either by Λ (N) or λ (n). They are of course the same physical quantity expressed differently i.e. Λ (N) = λ (n). Thus we have and Λ (N) := dx dx = N CN = C AB N A N B λ (n) := dx dx = 1 n cn = 1 cij n i n j Let us now investigate how angles bwtween two arbitrary lines in the reference configuration are changed during the deformation. Suppose dx 1 and dx 2 are two

11 11 vectors based at X with Θ be the angle between them. Now due to motion they changes to dx 1 = F dx 1 and dx 2 = F dx 2 Let θ be the angle between dx 1 and dx 2. If N 1, N 2 and n 1, n 2 are unit vectors in the dx 1, dx 2 and dx 1, dx 2 directions respectively. Hence cos Θ = N 1 N 2 cos θ = n 1 n 2 Now cos θ = dx 1 dx 1 dx 2 dx 2 = F dx 1 F dx 2 dx 1 dx 2 F N 1 = dx 1 / dx 1 F N 2 dx 2 / dx 2 = F N 1 F N 2 Λ (N 1 ) Λ (N 2 ) = N 1 (F T F N 2 ) Λ (N 1 )Λ (N 2 ) Thus cos θ = N 1 (CN 2 ) Λ (N 1 )Λ (N 2 ) The decrease in the angle Θ θ is called the angle of shear of the directions N 1 and N 2 in the plane of shear defined by N 1 and N 2. Using polar decomposition theorem, we have F = RU = V R where C = U 2 and b = V 2. The tensors U and V are called respectively the right and left stretch tensors. Now since U is symmetric and positive definite, the spectral representation theorem establishes the existence of an orthonormal basis consisting of eigen vectors of U i.e. there exists n A, n B, n C such that Un P = λ P n P, (P = A, B, C) and we write U as 3 U = λ P n P n P P =1

12 Here λ P are called the principal stretches of the deformation and n P are the principal Lagrangian directions. Let N = n P. Then Λ (np ) = n P Cn P = n P U 2 n P = Un P Un P = λ P n P λ P n P = λ P np n P = λ P 12 On the other hand if q p is the principal Eulerian directions then q p = Rn P and (q p ) = Rn P crn P = Rn P b 1 Rn P = Rn P (V 2 ) 1 Rn P = V 1 Rn P V 1 Rn P λ 1 Thus we have λ (qp ) = λ p = 1 λ p RnP Rn P = λ 1 7 Change of area and volume elements Let dx and dy be two material line elements based at X and dx and dy respectively are their images based at x under deformation. Let N and n are the unit normal to the surfaces. Then N = dx dy dx dy and p n = dx dy dx dz Now denoting da = dx dy and da = dx dz we get n da = dx dy = (F dx) (F dy ) Result: For any vector v, w we have for an invertible tensor A Av Aw = det(a)a T v w Proof: We have for any u V u Av Aw = A(A 1 u) Av Aw = [A(A 1 u), Av, Aw] = det(a)[a 1 u, v, w] which gives u Av Aw = det(a)a 1 u v w = det(a)u A T v w

13 13 Since u is arbitrary, we have the desired result. Using this result we get n da = det(f )F T (dx dy ) or n da = JF T NdA The volume dv at the reference configuration based at X is given by dv = dx (dy dz) [dx, dy, dz] Similarly for current configuration dv = [dx, dy, dz] = [F dx, F dy, F dz] = det(f )[dx, dy, dz] = JdV Thus J denotes the change in volume during deformation. efinition: If a motion x = x(x, t) is such that there is not change in volume, then the motion is called isochoric. Thus for a isochoric motion J(X, t) = det(f ) = 1, X B r, t > 0 8 Interpretation of right Green s deformation tensor Let N = E A. Now Λ (EA ) = E A CE A = C AA (no sum) or C AA = Λ 2 (E A ) Thus C P P (P = 1, 2, 3) is the square of the stretch undergone by material line element in B r at X and aligned in the directions of basis vector. Next let N 1 = E A and N 2 = E B (A B). Now cos θ = E A CE B Λ (EA )Λ (EB ) = C AB CAA CBB Thus C AB is the product of the stretches undergone by line elements along basis vectors in B r and the cosine of the angle between them in B t.

14 9 Interpretation of deformation gradient tensor 14 From the relation x i = x i (X, t) we get dx i = x i,a dx A = F ia dx A Thus dx = F dx so the deformation gradient tensor F shows how material line element are mapped into spatial line elements during motion. Thus it is mapping of infinitesimal vector dx of reference configuration into the infinitesimal vector dx of the current configuration. ue to the inverse mapping we also have dx = F 1 dx We know that det(f ) 0. Since F is invertible tensor, so it admits unique polar decomposition F = RU = V R, where R is orthogonal tensor and U and V are symmetric positive definite tensor. Since det(u), det(v ) > 0, we have det(r) has the same sign as det(f ). Since J = det(f ) > 0, we have det(r) = 1 i.e. R represents a pure rotation. Also we have proved that eigen values of U and V are identical. We already proved that the orthonormal eigen vectors n A, n B, n C is Lagrangian principal axes for U. Let n 1, n 2, n 3 be the corresponding Eulerian principal axes for V. We know that n 1 = Rn A, n 2 = Rn B, n 3 = Rn C The theorem of polar decomposition replaces the linear transfomation dx = F dx by two sequential transformation so that we can write dx = RU dx = R(UX) dx = V R dx = V (RX) Here dx is the vector connecting two neighbouring particles at X and X + dx of B r. Since U = F T F = (F ia E A e i )(F jb e j E B, U is a Lagrangian measure of deformation. Similarly V is a Eulerian measure of deformation. The deformation undergone by the body in the vicinity of a particle X in B r is regarded as a two step process

15 15 Figure 2: Polar decomposition of the deformation gradient. In this figure the edge of the cube in the reference configuration lies along the principal axes of U. a. Imposes the stretches λ A in the directions of n A (A = 1, 2, 3) of the eigen vectors of U. The result is going to be still in B r. b. Perform a rigid rotation of the output to B t. Because of the relation dx = V R dx = V (RX), the above operations can be thought of as in reverse order. a. Perform a rigid rotation which maps dx from B r to a new position in B t. b. Imposes the stretches λ i in the directions of n i (i = 1, 2, 3) of the eigen vectors n i of V. The result is going to be still in B t since V is a Eulerian measure of deformation. Since F, R, U and V are functions of X and t (or x and t), this decomposition is only local. In other words, an element at different position will experience different stretching and different rotation.

16 10 Infinitesimal deformation theory 16 Here we assume that the deformation is small. The small deformation corresponds to small displacement gradients i.e. Grad u (or grad u ) 1. Thus all first derivative of the dispacement with respect to coordinates are so small that linearization is justified. These are consequeces of the assumptions. enoting H = Grad u and h = grad u we have the following a. H = F I (exact) b. h = I F 1 (exact) c. Let δ = O( Grad u ). Now from the above two relations we have I = F F 1 = (H + I)(I h) = I + H h + O(δ 2 ) Hence for infinitesimal deformation theory, we have H h d. F 1 = I H + O(δ 2 ) This is obvious from the above relations e. det(f ) = 1 + tr(h) + O(δ 2 ) We have [using characteristic polynomial of H for λ = 1] det(f ) = det(i + H) = 1 + tr(h) + O(δ 2 ) ( d. C = I + H + H T + O(δ 2 ) E = ) 1 2 H + H T + O(δ 2 ) ( e. B = I H H T + O(δ 2 ) e = ) 1 2 h + h T + O(δ 2 ) f. Since H h the above two relations implies E e. Thus in infinitesimal theory the distinction between Lagrangian and Eulerian strain tensor disappears. Thus it is immaterial whether we differentiate the displacements with respect to the material or spatial coordiantes.

17 11 Velocity gradient 17 efinition: The velocity gradient tensor L CT (2) is defined as the spatial gradient of the velocity, i.e. L(x, t) := gradv(x, t) In component form we have L = L ij e i e j, L ij = v i,j This is an Eulerian tensor field and is defined without reference configuration. On the other hand the deformation gradient is a two-point tensor. We can decompose L as a sum of symmetric and skew-symmetric part i.e. L = + W Here = 1 2 (L + LT ) or ij = 1 2 (v i,j + v j,i ) is called stretching tensor or rate of deformation tensor or strain tensor. And W = 1 2 (L LT ) or W ij = 1 2 (v i,j v j,i ) is called vorticity or spin tensor. Properties: a. F = LF We have (x i,a) = ( ) xi (X, t) X ( ) A xi (X, t) = X A = v i,a = v i,j x j,a Since in the operation of / we have X A fixed, hence / and / X A commute i.e [ ] ( ) = [ ] ( ) X A X A On the other hand material time derivative of an Eulerian field does not commute i.e x i [ ] ( ) [ ] ( ) x i

18 18 From the above result we also have b. A corollary of the above result is L = F F 1 b. We have F 1 = F 1 L F F 1 = I Hence F F 1 + F Using the above result we have Hence F F 1 F 1 F 1 = 0 = LF F 1 = L = F 1 L J = J(divv) To prove this we use the result ( ) F (detf ) = (detf )tr F 1 Thus J = Jtr ( LF F 1) = Jtr(L) = J div(v) For an isochoric motion we have J 1, thus dj/dt = 0 and hence div v = Physical interpretation of and W Consider dx in B r and dx B t. Now let N and n are unit vectors. Then dx = N dx and dx = n dx Now let ds = dx and ds = dx. Then ds 2 ds 2 = dx dx dx dx = F dx F dx dx dx = dx (F T F I)X

19 19 Now (ds2 ds 2 ) = dx (F T F I)dX = dx (Ḟ T F + F T Ḟ )dx (P/ denoted by P ) = dx (F T L T F + F T LF )dx = F dx (L T + L)F dx = 2dx dx Now since (ds2 ) = 0 and dx = nds we get ds ds = n n If we choose n = e i (i = 1, 2, 3) we get the physical interpretation of the diagonal component of. Thus ii (no sum) is the rate of extension per unit length of a line element which in the current configuration B t is situtated at x and momentarily aligned with the direction defined by the base vector e i (i = 1, 2, 3). Next let n 1 and n 2 are unit vectors along dx 1 and dx 2 and θ is the angle between them. Thus dx 1 dx 2 = ds 1 ds 2 cos θ From the LHS we get after taking derivative with respect to time [dx 1 dx 2 ] = [dx 1 F T F dx 2 ] = dx 1 (Ḟ T F + F T Ḟ )dx 2 = dx 1 (F T L T F + F T LF )dx 2 = F dx 1 (L T + L)F dx 2 = 2dx 1 dx 2 On the other hand from the RHS we get [ds 1 ds 2 cos θ] = = ds1 ds 2 cos θ + ds2 ds 1 cos θ ds 1 ds 2 sin θ ( ) θ ds1 ds 2 cos θ + cos θ sin θ ds 1 ds θ ds 1 ds 2 2 Thus we get 2n 1 n 2 = [n 1 n 1 + n 2 n 2 ] (n 1 n 2 ) θ n 1 n 2

20 20 Figure 3: Interpretation of diagonal and non-diagonal element of or θ = [n 1 n 1 + n 2 n 2 ](n 1 n 2 ) 2n 1 n 2 n 1 n 2 which on substituting θ = π/2 γ becomes Thus if n 1 and n 2 are orthogonal we get γ 2 = n 1 n 2 Thus ij is the half rate of the decrease of the angle between a pair of line elements which in B t intersects at x and are momentarily in the directions of the unit vector e i and e j. For spin tensor W we note that (dx) = Ḟ dx = L dx = ( + W )dx Using dual vector w of W we can write the above as dv = W dx = dx + w dx Thus the relative velocity in the vicinity of a point p (the origin of dx) corresponds to the defomation part of the motion (the first term on the RHS) and a local rigidbody rotation about an axis through p. The dual vector w indicates the angular velocity, direction and the sense of rotation. Alternative interpretation of W : We have Also Λ = ds ds dx = F dx or nds = F NdS or Λn = F N

21 21 Taking material derivative we get Λ n + Λ ṅ = Ḟ N = LF N = LΛn = Λ( + W )n Since n is a unit vector it can only rotate and its rate of change is perpendicular to itself. n n = 1 From this we get ṅ n = 0 Taking dot product with n of the previous relation we get Λ = Λ(( + W )n) n = Λ(n n) where we have used the fact that n W n = 0 since W is symmetric. Thus we have ṅ = W n + [ (n n)i]n Thus rate of rotation of n is dependent on both and W. However if we chose n to be parallel to the principal direction, then n = (n n)n and hence ṅ = W n This shows that the spin tensor W gives the rate of rotation of the line element dx which in the current configuration is parallel to a principal direction of. Using dual vector w of W we get ṅ = W n = w n 12 Material derivative of line, area and volume elements We have already proved that [dx] = Ḟ dx = Ldx For area we use da = JF T da

22 22 Thus [da] = JF T da + J(Ḟ 1 ) T da = Jtr(L)F T da + JdA( F 1 L) T = tr(l)da L T da For volume we use dv = J dv Thus [dv] = J dv = Jtr(L) dv = tr(l) dv = v i,i dv Thus for an isochoric motion we have v i,i = 0. In summary, the deformation gradient determines the stretch of a line element, the change of an area element and the change of a volume element. On the otherhnad it is the velocity gradient L that determines the rate at which these changes occur. 13 Transport formulas We are interested in calculating the rates of change of integrals over material curve, surafces and volumes. Let C t, R t and Ω t denote a matrial curve, surface and volume respectively in the current configuration B t. If φ and u are continuously differen-

23 23 tiable scaler and vector fields respectively, then ( ) φ φdx = C t C t + φl dx [( ) ] φ φnda = S t S t + φ tr(l) n φl T n da ( ) φ φdv = Ω t Ω t + φ tr(l) dv ( ) u u dx = C t C t + LT u dx ( ) u u nda = + u tr(l) Lu nda S t S t ( ) u udv = Ω t + u tr(l) dv Ω t The main difficulty in the above formulas is the dependence on time in the integrand as well as on the C t, S t and Ω t. This is overcome by converting the left hand side to referential configuration. Once in the referential configuration the differentiation with respect to t and the integrals commute with each other. Next we transfer back the integral to current configuration. Let us prove ( ) u u nda = + u tr(l) Lu nda S t S t We transform the left hand side as u nda = u (JF T N) da S t S r = (JF 1 u) N da S r = S r (JF 1 u) N da [ J 1 ] = S r F 1 F 1 u u + J u + JF N da [ ] = J tr(l)f 1 u JF 1 1 u Lu + JF N da S r [ ] u = JF 1 + tr(l)u Lu N da S r [ ] u = + tr(l)u Lu (JF T N) da S r [ ] u = + tr(l)u Lu n da S t