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1.6.
Exercises
P1.1 Using Cartesian bases, show that (u Ä v) ⋅ (w Ä x) = (v ⋅ w)u Ä x where u, v, w, and x are rank 1 tensor. Solution: Using the Cartesian basis, (u Ä v)(w Ä x) = (ui ei Ä v j e j ) ⋅ (wk ek Ä xl el ) . Since the dot product occurs between adjacent bases, we have
(ui ei Ä v j e j ) ⋅ (wk ek Ä xl el ) = uiv j wk xl (e j ⋅ ek )(ei Ä el ) = uiv j wk xl djk (ei Ä el ) = uiv j w j xl (ei Ä el ) = v j w j (ui ei Ä xl el ) = (v ⋅ w)(u Ä x) In the above equation, we used the following properties: e j ⋅ ek = d jk , wk d jk = w j , and v jw j = v ⋅ w . P1.2 Any rank 2 tensor T can be decomposed by T = S + W, where S is the symmetric part of T and W is the skew part of T. Let A be a symmetric rank 2 tensor. Show A : W = 0 and A : T = A : S . Solution: Since A is symmetric and W is skew, we have A : W = AijWij = -AijW ji = -AjiW ji
Since in the above equation, the repeated indices i and j are dummy, the above equation can be rewritten as AijWij = -AijWij = 0
In addition, from the relation T = S + W,
A : T = A : (S + W) = A : S + A : W = A : S
P1.3 For a symmetric rank-two tensor E , using the index notation, show that I : E = E , where I = 21 [dik djl + dil djk ] is a symmetric unit tensor of rank-4. Solution: Using index notation, the contraction operator can be written as
(I : E)ij = 12 [dik djl + dil djk ]Ekl Since the Kronecker-delta symbol replaces indices, the above equation can be written as
(I : E)ij = 12 [Eij + E ji ] = Eij = (E)ij The symmetric property of E is used.
P1.4 The deviator of a symmetric rank-2 tensor is defined as Adev = A - Am 1 where Am = 13 (A11 + A22 + A33 ) . Find the rank-4 deviatoric identity tensor Idev that satisfies Adev = Idev : A . Solution: From Problem P1.3, it can be shown that I : A = A . In addition, Am can be written in the tensor notation as Am = 13 1 : A . Therefore, Adev = A - Am 1 and it can be written as
Adev = éêë I - 13 1 Ä 1ùúû : A = Idev : A The last equality defined the rank-4 deviatoric identity tensor Idev .
P1.5 The norm of a rank-2 tensor is defined as A = A : A . Calculate the following derivative ¶ A / ¶A . What is the rank of the derivative? Solution: From the definition
¶ A ¶A
=
¶ é A 1 (A : A)1/2 ùú = (A : A)-1/2(2A : I) = ê û 2 ¶A ë A
The result is a rank-2 tensor. Note that the property that ¶A / ¶A = I is used.
P1.6 A unit rank-2 tensor in the direction of rank-2 tensor A can be defined as N = A / A . Show that ¶N / ¶A = [I - N Ä N] / A . Solution: Using chain-rule of differentiation, the unit normal tensor can be differentiated as æ ö æ ¶ A ¶N ¶ çç A ÷÷ 1 çç ¶A = ÷÷ = A A Ä çç ç 2 ç ¶A ¶A çè A ÷ø÷ ¶A A çè ¶A
ö÷ ÷÷÷ ÷÷ø
It is straightforward to show that ¶A / ¶A = I . From Problem 1.5, we have
¶ A ¶A Therefore, we have
=
1 ¶ é A (A : A)1/2 ùú = (A : A)-1/2 (2A) = ê û 2 ¶A ë A
¶N 1 = ( I - N Ä N) ¶A A
P1.7 Through direct calculation of a rank-2 tensor, show that the following identity erst det[A] = eijk Air Ajs Akt is true Solution: In the index notation, (r, s, t) are real indices, while (i, j, k) are dummy indices. Since (r, s, t) only appears in the permutation symbol, it is enough to show the cases of even and odd permutation. Consider the following case of even permutation: (r, s, t) = (1, 2, 3). In such a case, non-zero components of the right-hand side can be written as eijk Ai 1Aj 2Ak 3 = e123A11A22A33 + e132A11A32A23 +e231A21A32A13 + e213A21A12A33 +e312A31A12A23 + e321A31A22A13
In the above equation, we have e123 = e231 = e312 = 1 and e132 = e213 = e 321 = -1 . Therefore, the above equation becomes
eijk Ai 1Aj 2Ak 3 = A11(A22A33 - A32A23 ) + A21(A32A13 - A12A33 ) + A31(A12A23 - A22A13 ) which is the definition of det[A ] . By following a similar appr