Tensor Analysis. For orthogonal curvilinear coordinates, ˆ ˆ (98) Expanding the derivative, we have, ˆ. h q. . h q h q

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1 For orthogonal curvlnear coordnates, eˆ grad a a= ( aˆ ˆ e). h q (98) Expandng the dervatve, we have, eˆ aˆ ˆ e a= ˆ ˆ a h e + q q 1 aˆ ˆ ˆ a e = ee ˆˆ ˆ + e. h q h q Now expandng eˆ / q (some of the detals are left to the reader), 259

2 260 Tensor Analyss ˆ ˆ 1 ˆ ˆ 1 ˆ. k k k k k q q h h k h h q h k h h q δ = = = e e e e e No summaton on h thus must nsert to factor. ˆ k e

3 After much more work, employng eq. (18) and δ g = h δ, g =. (no summaton on h) 2 2 h one obtans, eˆ δ h δ h = q h q h q k k k eˆ k. (99) wth no summaton wth the h and h k and fnally, 1 aˆ aˆ h a h a= + δ h q h q hk q ee ˆˆ. k k k k k Agan wth no summaton of the scale factors. (100) 261

4 Example: Tensor gradent n cylndrcal and sphercal coordnates. For each case we wll frst wrte the terms n eq. (100), ncrementng the ndex then combne them for the fnal result. cylndrcal: q 1 = R, q 2 = φ, q 3 = z h 1 = 1, h 2 = R, h 3 = 1 262

5 = 1: 1 aˆ1 aˆ2 aˆ2 ˆˆ ˆˆ ˆˆ h ee + ee + ee 1 q q q aˆ1 h aˆ2 h1 aˆ 3 h h1 q h2 q h3 q ( δ ˆˆ ˆˆ ˆˆ 11ee δ12ee δ13ee 1 3) 0 0 aˆ ˆ ˆ 1 h 0 1 a1 h 0 1 a1 h 0 1 ee ˆˆ ˆˆ ˆˆ ee ee h1 q h2 q h3 q aˆ aˆ ˆ R φ az = ee ˆ ˆ R R + ee ˆ ˆ ˆ ˆ R φ + ee R z R R R 263

6 = 2: 1 aˆ1 aˆ2 aˆ 3 ˆˆ ˆˆ ˆˆ h ee + ee + ee 2 q q q aˆ1 h2 aˆ2 h 0 2 aˆ 3 h h1 q h2 q h3 q ( δ ˆˆ ˆˆ ˆˆ 21ee 2 1+ δ22ee 2 2+ δ23ee 2 3) 0 0 aˆ ˆ ˆ 2 h2 a2 h 0 2 a2 h0 2 ee ˆˆ ˆˆ ˆˆ ee ee h1 q h2 q h3 q 1 aˆ aˆ ˆ R φ az = ee ˆˆ φ R + ˆˆ ˆˆ ˆ ˆˆ ˆ ˆˆ R ee φ φ + ee φ + a ee φ φ aφee φ φ φ φ 1 aˆ 1 aˆ 1 ˆ R φ az = aˆ ˆˆ aˆ ˆˆ ˆˆ φ φ + + φ φ + φ R φ ee R φ ee ee R φ z R R R R z 264

7 = 3: 1 aˆ1 aˆ2 aˆ 3 ˆˆ ˆˆ ˆˆ h ee + ee + ee 3 q q q aˆ1 h 0 3 aˆ2 h 0 3 aˆ 3 h h1 q h2 q h3 q ( δ ˆˆ ˆˆ ˆˆ 31ee 3 1+ δ32ee 3 2+ δ33ee 3 3) 0 0 aˆ ˆ ˆ 3 h 0 3 a3 0 h3 a3 h 0 3 ee ˆˆ ˆˆ ˆˆ ee ee h1 q h2 q h3 q aˆ aˆ ˆ R φ az = ee ˆˆ z R + ee ˆˆ ˆˆ z φ + ee z z z z z 265

8 Combnng, we have the cylndrcal system, aˆ aˆ ˆ R φ az a= eˆ ˆ ˆ ˆ ˆ ˆ ReR + ereφ + erez R R R 1 aˆ 1 ˆ 1 ˆ R aφ az + aˆ ˆˆ ˆ ˆˆ ˆˆ φ φ R ar φ φ φ z R φ ee + R + φ ee + ee R φ aˆ aˆ ˆ R φ az + ee ˆˆ ˆˆ ˆˆ z R + ee z φ + ee z z z z z (101) 266

9 Followng ths example for the cylndrcal system, you can show that for the sphercal system, the tensor gradent s, aˆ ˆ ˆ r a a ˆˆ θ ˆˆ φ a= erer + ereθ + e ˆˆ reφ r r r 1 aˆ 1 ˆ 1 ˆ r a a ˆ ˆˆ θ φ + a ee ˆ ˆˆ ˆˆ r + + ar ee + ee r θ r θ r φ aˆ ˆ r aθ aˆ sn aˆ φ θ φ cosθ φ φ + ee ˆˆ ˆˆ φ r + ee φ θ rsnθ rsnθ aˆ φ aˆ sn ˆ r θ + aθ cosθ φ + ee ˆˆ φ φ r snθ θ θ θ θ θ φ (102) 267

10 Dvergence of a Second-Order Tensor We defne the dvergence operaton for a second-order tensor, k dv Φ Φ= e ( φ eek ) (103) q φ e e Φ = e e e + e + e q q q k k k k k φ k φ φ m n = e e e + e e + e e q k k k k k φ m k φ n. Now relabel the dummy ndces and factor, 268

11 φ r r Φ = e e e + e e + e e q rs s rk r s φ r s φ r s φ r r = + + rs s rk φ φ δr q e s. φ rk s φ e q k rs s Φ = + φ + In the Cartesan system ths s Φ φ = x ˆ. s. (104) (105) 269

12 Tensor Gradent Revsted There are two tensor operatons that we have not defned, but now wll because the result of these operatons may shed more lght on the physcal nterpretaton of the tensor gradent. The contracton of a second-order tensor or dyadc s the result of placng the dot operaton between the vectors of each dyad, φ φ e e = φ g s φ e e = φ g φ φ e e = φ s φ e e = φ Ths operaton reduces the order of the tensor by two. The vector of a second-order tensor or dyadc s the result of placng the cross product between the vectors of each dyad, 270

13 Φ φ e e = gφ ε e k V k φ e e = gφ g g ε e m n k mnk φ e e = gφ g ε e m k mk φ e e = gφ g ε e m k mk. Ths operaton reduces the order of the tensor by one. One can show the the followng two denttes, a Φ Φ Φ a T [ ] V, a b b b a T [ ( ) ] ( ). (106) (107) 271

14 Equaton (107) was brefly dscussed as eq. (97). Equaton (106) shows how the antsymmetrc part of Φ, whch has only three ndependent components, s related to the vector of Φ, Φ V. So for the gradent of vector a, a, k k ( a) S = a, e ek = a, k = dv a. (108) Note that dv a nvolves the dagonal terms a thus s assocated wth the symmetrc part of a (snce the dagonal terms of the antsymmetrc part are always zero). The vector of a, ( a) V s, ( a) = a e e = gg a ε e = curl a. k k m V, k, km (109) 272

15 Therefore, ( a) V and the antsymmetrc part of a are related to the rotaton of the vector feld as prevously presented. So we can conclude that the tensor gradent contans nformaton about the dltaton and rotaton of a vector feld. Ths nformaton s extracted wth the nternal operatons ust defned n eqs. (108) and (109) or when a s used to transform a vector nto another vector, (e.g., b n eq. (107) s used to transform a nto another vector). IN-CLASS EXAMPLES 273

16 Integral Theorems for Dyadcs and Second-Order Tensors The vector ntegral relatons (49)-(51) are ncreased by one order, R R R CS grad adτ = dv Φ dτ = S S na ˆ ds nˆ ΦdS curl Φdτ = nˆ ΦdS nˆ (curl Φ) ds = ds Φ S C (110) (111) (112) (113) 274

17 The nvarant forms can also be shown as before n eqs. (55)- (57) 1 grad a= lm ˆ ds τ 0 τ na S 1 dv = lm ˆ ds τ 0 τ Φ n Φ S 1 curl Φ = lm ˆ ds τ 0 τ n Φ S 1 nˆ curl Φ = lm S 0 S ds Φ C n (110) (111) (112) (113) Note that (113) and (117) are added for completeness. 275

18 Egenvalues and Egenvectors of Dyadcs We have shown how a dyadc can be wrtten n matrx notaton and some of the matrx operatons that apply to dyadcs. Let us now extend ths further by computng egenvalues and egenvectors of second-order tensors. The dyadc Φ s an operator that changes a vector nto another vector, e.g., Φ a= b. There s a specal case where only the magntude and/or sense of the vector s b changed,.e., a λa Φ a=λa where λ s the egenvalue that characterzes the change n a. 276

19 Φ a λa= 0 ( Φ λi) a= 0 Ths s the same relaton we developed for the homogeneous system n lnear algebra. We can wrte ths equaton n matrx form as, φ11 λ φ12 φ13 a1 0 φ φ λ φ a = φ31 φ32 φ33 λ a 3 0 From Cramer s theorem, we know that to obtan nontrval solutons, det( Φ λi) = φ λδ =

20 From Cramer s theorem, we know that to obtan nontrval solutons, det( Φ λi) = φ λδ = 0. Note:. Each egenvalue has an assocated egenvector.. Snce complex roots must appear n conugate pars, at least one λ s real for n = 3.. For symmetrc Φ, all λ are real. For most practcal engneerng problems Φ s symmetrc (Reddy & Rasmussen), so for the remander of the analyss we assume Φ s symmetrc. Furthermore, v. The egenvectors assocated wth dstnct egenvalues of a symmetrc dyadc are orthogonal. 278

21 Computng egenvectors (symmetrc dyadc): Snce only the drecton of the egenvectors s of mportance, we can choose to wrte the set of egenvectors as unt vectors,.e, f a s a gven egenvector, then the unt vector assocated wth a s * a aˆ 11 + a ˆ 22+ a ˆ 33 eˆ =± =. a a (119) 1 + a2 + a3 Then, a1 * λ1 ± = eˆ 1 a1 a2 * λ ˆ (120) 2 ± = e Choose ± to ensure a 2 a2 rght-handed system. a3 * λ ˆ 3 ± = e3 a 3 279

22 The represent the prncpal drectons of Φ. Axes of a coordnate system algned wth the prncpal drectons are the prncpal axes. The orthogonal set of egenvectors forms a bass. Thus we can use * * ( ˆ ˆ ˆ e) e and wrte, ˆˆ * * Φ = φ ˆ ˆ = φmnemen. Then, ( Φ λ I) a = 0 * * * ( φ ee ˆ ˆ λ I) a = 0(no summaton) mn m n 280

23 φ eˆ δ λ eˆ δ = 0 * * * mn m n k k φ eˆ λ eˆ = 0 * * * m m eˆ ( eˆ eˆ ) (no summaton) * * * * * φm m = λ φ = λδ or, λ λ3 1 * φ = λ 2 Ths result shows that a dyadc wrtten n prncpal axs coordnates reduces to only components along the dagonal that are the egenvalues. 281

24 Problems wth coordnate symmetry wll result n a repeated root, e.g., λ 1 dstnct from λ 2 = λ 3. In ths case, two drectons are ndependent and the thrd s arbtrary. So one can choose the thrd egenvalue such that the prncpal axes form a rghthanded coordnate system,.e., a 3 = a 1 a 2. Example: Fnd the egenvalues and egenvectors of

25 Taylor Seres Expanson From dfferental calculus, any analytc functon can be expanded as an nfnte power seres,.e., for a scalar functon of a scalar, u(x), δ x δ x x + δ 283

26 2 2 n n du δ d u δ d u ux ( + δ) = ux ( ) + δ dx 2! dx n! dx 2 n δ = 0 δ = 0 δ = 0 Ths s called the ncrement form of the Taylor seres. For a scalar functon of two scalar varables, ux ( + hy, + k) = uxy (, ) + h + k uxy (, ) + x y 1 + h k u( x, y) n! + x y n (121) (122) 284

27 u(x,y) x + h x y y + h Note, can you wrte aa sample term from the operator of eq. (122)? h + k u( x, y) =? x y 2 285

28 We now generalze these results to a scalar functon of a vector, φ(r), or a vector functon of a vector, a(r), φ(r +) r φ(r) 286

29 δ φ φ( r + δ) = φ( r) + δ φ + ( δ φ) + 2! δ a ar ( + δ) = ar ( ) + δ a + ( δ a) + 2! (123) (124) Recall the Taylor seres for the exponental functon, 2 n x x x e = 1 + x ! n! Notng the form of (121) - (124), we can defne a convenent operator form of the Taylor seres, d 2 2 n n δ d δ d δ d dx e = 1 + δ + + +, 2 n dx 2! dx n! dx (125) 287

30 Contnuum Velocty Feld: Relatve Moton Gven the velocty at a gven pont n a contnuum velocty feld, we can use the Taylor seres expanson to determne the velocty at a neghborng locaton. Ths moton can be decomposed nto a combnaton of translatonal, rotatonal and shearng (deformng) motons, v(r) vr ( + r) = vr ( ) + r v+ h.o.t. r r The gradent of the velocty s wrtten n ts symmetrc and antsymmetrc parts, 1 1 v= v+ v + v v 2 2 T T [ ( ) ] [ ( ) ], r+ r v(r+ r) 288

31 where, Thus, 1 [ ( ) T ] rate of stran tensor, v= v+ v ε = 2 1 T 1 r [ v ( v) ] = curl v r 2 2 = rate of rgd-body rotaton. 1 vr ( + r) = vr ( ) + r ε + curl v r. rate of stran 2 translaton rate of rotaton 289

32 The rate of stran tensor looks lke ε ε ε [ ε ] ε ε ε ε ε ε = Here s an llustraton of the combned stranng motons v x + v x v v v + = x x x extenson (dagonal terms) shear (off-dagonal terms) 290

33 Note: v v v trace[ ε ] = ε = + + = dv x1 x2 x3 = rate of change of volume per unt volume (dlataton). Ths concludes our dscusson of tensor analyss. v 291