Finite Element Analysis - Thermomechanics of Solids Part 2

2 Mathematical Foundations: Tensors 2.1 TENSORS We now consider two n ´ 1 vectors, v and w, and an n ´ n matrix, A, such that v = Aw. We now make the important assumption that the underlying information in this relation is preserved under rotation. In particular, simple manipulation furnishes that v¢ = Qv = QAw ∗M) = QAQTQw = QAQTw¢. (2.1) The square matrix A is now called a second-order tensor if and only if A¢ = QAQ . Let A and B be second-order n ´ n tensors. The manipulations that follow demonstrate that A , (A + B), AB, and A−1 are also tensors. (AT)¢ = (QAQT)T = QTTATQT (2.2) A¢B¢ = (QAQT )(QBQT ) = QA(QQT )BQT = QABQT (2.3) (A + B)¢ = A¢ + B¢ = QAQT + QBQT = Q(A + B)QT (2.4) A¢−1 = (QAQT)−1 = QT−1A−1Q−1 = QA−1QT. (2.5) 25 © 2003 by CRC CRC Press LLC 26 Finite Element Analysis: Thermomechanics of Solids Let x denote an n´ 1 vector. The outer product, xx , is a second-order tensor since (xxT )¢ = x¢x¢T = (Qx)(Qx)T = Q(xxT )QT (2.6) Next, d2j = dxTHdx  d T dj  dx  dx (2.7) However, dx¢TH¢dx¢ = (Qdx)T H¢Qdx = dxT(QTH¢Q)dx, (2.8) from which we conclude that the Hessian H is a second-order tensor. Finally, let u be a vector-valued function of x. Then, du = ¶x dx, from which duT = dxT ¶uT (2.9) and also duT = dxT ¶xT uT. (2.10) We conclude that  ¶uT ¶uT  ¶x ¶xT (2.11) Furthermore, if du¢ is a vector generated from du by rotation in the opposite sense from the coordinate axes, then du¢ = Qdu and dx = Qdx¢. Hence, Q is a tensor. Also, since du¢ = ¶x¢ dx¢ , it is apparent that ¶u¢ ¶u T ¶x¢ ¶x (2.12) from which we conclude that ¶u is a tensor. We can similarly show that I and 0 are tensors. © 2003 by CRC CRC Press LLC Mathematical Foundations: Tensors 27 2.2 DIVERGENCE, CURL, AND LAPLACIAN OF A TENSOR Suppose A is a tensor and b is an arbitrary, spatially constant vector of compatible dimension. The divergence and curl of a vector have already been defined. For later purposes, we need to extend the definition of the divergence and the curl to A. 2.2.1 DIVERGENCE Recall the divergence theorem òcTndS = òÑTcdV. Let c = ATb, in which b is an arbitrary constant vector. Now bT òAndS = òÑT(ATb)dV = òÑTATdVb = bT ò[ÑTAT]T dV. (2.13) Consequently, we must define the divergence of A such that *M) òAndS − ò[ÑTAT]T dV = 0. (2.14) In tensor-indicial notation, òbaijnjdS− òb[[ÑTAT]T]dV = 0. (2.15) Application of the divergence theorem to the vector cj = biaij furnishes b  ¶ aij −[[ÑTAT]T]i dV = 0. (2.16) j Since b is arbitrary, we conclude that [ÑTAT]i = ¶ aij = ¶ ajiT. (2.17) j j Thus, if we are to write Ñ×A as a (column) vector, mixing tensor- and matrix-vector notation, Ñ×A = [ÑTAT ]T. (2.18) © 2003 by CRC CRC Press LLC 28 Finite Element Analysis: Thermomechanics of Solids It should be evident that (Ñ×) has different meanings when applied to a tensor as opposed to a vector. Suppose A is written in the form aT A = aT, (2.19)  3  in which ai corresponds to the ith row of A:[ai ]j = aij. It is easily seen that ÑTAT = (ÑTa1 ÑTa2 ÑTa3). (2.20) 2.2.2 CURL AND LAPLACIAN The curl of vector c satisfies the curl theorem òÑ´ cdV = òn ´ cdS. Using tensor-indicial notation, òn ´ cdS = òeijknjaklb dS = òeijknjakl dSb = òcijnj dSb , cij = eijkakl l. (2.21) From the divergence theorem applied to the tensor cij = eijk akl bl, òn ´ AbdSi = ò¶¶j (eijkaklb )dV =  eijk ¶akl dV l j =  Ñ´ AT dV b , i if [Ñ´ A]il = eijk ¶¶j akl. (2.22) Let aT denote the row vector (array) corresponding to the lth row of A: [aT] = alk. It follows that Ñ´ A = [Ñ´a1 Ñ´ a2 Ñ´ a3 . (2.23) © 2003 by CRC CRC Press LLC Mathematical Foundations: Tensors 29 If bI is the array for the Ith column of A, then Ñ´ AT = [Ñ´ b1 Ñ´ b2 Ñ´ b3 . (2.24) The Laplacian applied to A is defined by [Ñ2A]ij = Ñ2aij. (2.25) It follows, therefore, that Ñ2A =[Ñ2b1 Ñ2b2 Ñ2b3]. (2.26) The vectors bi satisfy the Helmholtz decomposition Ñ2bi = Ñ(Ñ×bi )− Ñ´Ñ´ bi. (2.27) Observe from the following results that Ñ2A = Ñ(Ñ×AT )− Ñ´[Ñ´ AT]T. (2.28) An integral theorem for the Laplacian of a tensor is now found as òÑ2AdV = ò(nÑT )AdS− òn ´[Ñ´ AT]T dS. (2.29) 2.3 INVARIANTS Letting A denote a nonsingular, symmetric, 3 ´ 3 tensor, the equation det(A − ll) = 0 can be expanded as l3 − I1l2 + I2l − I3 = 0, (2.30) in which I1 = tr(A) I2 = 1[tr2(A)−tr(A2 )] I3 = det(A). (2.31) Here, tr(A) = dijaij denotes the trace of A. Equation 2.30 also implies the Cayley-Hamilton theorem: A3 − I1A2 + I2A − I3I = 0, (2.32) © 2003 by CRC CRC Press LLC ... - tailieumienphi.vn