## A First Course in Continuum Mechanics by Professor Oscar Gonzalez, Professor Andrew M. Stuart

By Professor Oscar Gonzalez, Professor Andrew M. Stuart

A concise account of vintage theories of fluids and solids, for graduate and complicated undergraduate classes in continuum mechanics.

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**Extra info for A First Course in Continuum Mechanics**

**Example text**

15 that there are functions α0 , α1 , α2 : IR3 → IR such that C(H) = α0 (IH )I + α1 (IH )H + α2 (IH )H 2 for every symmetric H ∈ V 2 , where IH = tr H, 12 [(tr H)2 − tr(H 2 )], det H . Since C(H) is linear in H the only possibilities are α0 (IH ) = c0 tr H + c1 , α1 (IH ) = c2 , α2 (IH ) = 0, where c0 , c1 and c2 are scalar constants. Since C(O) = O it follows that c1 = 0. Thus, upon setting λ = c0 and µ = c2 /2, we have C(H) = λ(tr H)I + 2µH for every symmetric H. This together with the fact that C(W ) = O for every skew-symmetric W leads to the expression C(H) = λ(tr H)I + 2µ sym(H), ∀H ∈ V 2 .

11. 27 For any vectors a, b, c and d show (a ⊗ b) : (c ⊗ d) = (a · c)(b · d). 17). 13. 30 Let A, B and C be second-order tensors. Show that A : BC = AC T : B = B T A : C. 31 Let a and b be vectors and let A be a second-order tensor. Deﬁne C to be the fourth-order tensor given by C(S) = AS. Show that, if a is an eigenvector of A with eigenvalue α, then (C(a ⊗ b))v = α(b · v)a. 32 Find the components Cij k l of the fourth-order tensor deﬁned by C(S) = 12 (S − S T ). 14. 34 Suppose two symmetric second-order tensors S and E satisfy S = CE, where C is a fourth-order tensor with components Cij r s = λδij δr s + µ(δir δj s + δis δj r ) and λ and µ are scalar constants.

From the deﬁnitions of [S] and Sij we deduce the convenient expression [S] = ([Se1 ], [Se2 ], [Se3 ]), where [Sej ] is the 3 × 1 column matrix of components of Sej . 6) and the rules of matrix multiplication, the tensor equation v = Su can be written in various equivalent ways v = Su ⇔ vi = Sij uj ⇔ [v] = [S][u]. Thus the matrix representations of vectors and second-order tensors have the property [Su] = [S][u]. 1. In the respective coordinate frame of each example, the matrix representation of S is cos θ − sin θ 0 [S] = sin θ cos θ 0 , 0 0 1 and the matrix representation of T is −1 0 0 [T ] = 0 1 0 .