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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Additional resources for A First Course in Continuum Mechanics
Triple vector product. Let a = aq eq , b = bi ei , c = cj ej and d = dp ep . 5) where the last line follows from the fact that q k p = pq k . Thus, if we let d = a × (b × c), then dp = pq k ij k aq bi cj . Below we show that this expression can be used to derive a fundamental identity for the triple vector product. Remark: The geometrical deﬁnitions of the scalar and triple scalar products imply that these quantities are frame-independent. That is, the scalar and triple scalar products can be computed using components in any coordinate frame, with the same value obtained in each frame.
20), in particular S 3 − I1 (S)S 2 + I2 (S)S − I3 (S)I 3 λ3i − I1 (S)λ2i + I2 (S)λi − I3 (S) ei ⊗ ei = O. 27 Let v be an arbitrary vector. Since (a ⊗ b)T = (b ⊗ a) we have (a ⊗ b)T (c ⊗ d)v = (b ⊗ a)c(d · v) = b(a · c)(d · v) = (a · c)(b ⊗ d)v, which implies (a ⊗ b)T (c ⊗ d) = (a · c)(b ⊗ d). Using the fact that [b ⊗ d]ij = bi dj then yields (a ⊗ b) : (c ⊗ d) = tr((a ⊗ b)T (c ⊗ d)) = (a · c) tr(b ⊗ d) = (a · c)(b · d). 29 Since [S T D]ij = Sk i Dk j we have S : D = tr(S T D) = tr[S T D] = Sk i Dk i .
We usually write U = C. Proof See Exercise 26. 12 Polar Decomposition Theorem. Let F be a secondorder tensor with positive determinant. Then there exist right and left polar decompositions of F taking the form, respectively F = RU = V R, √ where U = F T F and V = F F T are symmetric, positive-deﬁnite tensors and R is a rotation. √ Proof Since F has positive determinant it is invertible. Thus F v = 0 for all v = 0. Similarly, the transpose F T is invertible and F T v = 0 for all v = 0. From these observations we deduce v · F T F v = (F v) · (F v) > 0 ∀v = 0, v · F F T v = (F T v) · (F T v) > 0 ∀v = 0.