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2.13 Strain rates 51 It is important to remember that the functional dependence is w ˙ [ d D = w ˙ [ d D ( { E >w) = (2.109b) Using the chain rule in Eq. (2.109a), w ˙ [ d D = w y d | o w [ o D = (2.109c) We define in the spatial configuration the velocity gradient tensor, w l = w y d | o w g d w g o > (2.110a) we can write the above as w l = w v u > (2.110b) w l W = u w v = (2.110c) Hence, we can write Eq.(2.109c) as w ˙ [ d D = w o d o w [ o D (2.111a) and therefore, w ˙ X = w l · w X = (2.111b) It is important to realize that the above is the material time derivate of the deformation gradient tensor, w ˙ X = G w X Gw . 2.13.2 The Eulerian strain rate tensor and the spin (vorticity) tensor We can decompose the velocity gradient tensor into its symmetric and skew- symmetric components: w l = w d + w $ (2.112a) where, w d = w d W = 1 2 ( w l + w l W ) (2.112b) is the Eulerian strain rate tensor (defined in the spatial configuration) and, w $ = w $ W = 1 2 ( w l w l W ) (2.112c) is the spin or vorticity tensor,alsodefined in the spatial configuration. Let us assume a deformation process referred to a fixed Cartesian system. The principal directions of w U form, in the reference configuration, a Cartesian system known as Lagrangian system. The principal directions of w V form, in the spatial configuration, a Cartesian system known as a Eulerian system (Hill 1978). 52 Nonlinear continua Fig. 2.5. Rotations We can go from one of the above-defined c oordinate systems to another one using the rotation tensors sketched in Fig. 2.5. From Fig. 2.5, we get w R H = w R · w R O = (2.113) For two consecutive rotations, w+w R = w+w w R · w R (2.114a) and therefore, w ˙ R = lim w $ 0 " w+w w R w g w # · w R = (2.114b) We can define a rotation rate w U = lim w $ 0 " w+w w R w g w # (2.114c) and using it in Eq. (2.114b) (Hill 1978), we get w ˙ R = w U · w R (2.115a) in the same way, w ˙ R O = w O · w R O > (2.115b) w ˙ R H = w H · w R H = (2.115c) 2.13 Strain rates 53 Since the rotation tensors are orthogonal we can write w R W · w R = g > (2.116a) taking the time derivative of the above equation and using Eq. (2.115a), we have w U + w W U = 0 (2.116b) in the same way, w O + w W O = 0 > (2.116c) w H + w W H = 0 = (2.116d) The above equations indicate that w U , w O and w H are skew- symmetric tensors. 2.13.3 Relations between dierent rate t ensors The time derivative of Eq. (2.113) leads to w H · w R H = w U · w R · w R O + w R · w O · w R O > (2.117a) and therefore, w R W · ( w H w U ) · w R = w O = (2.117b) Using Eqs. (2.111b) and (2.40), w l = w ˙ R · w R W + w R · w ˙ U · w U 1 · w R W > (2.118a) splitting the abov e equation into its symmetric and skew-symmetric compo- nen ts, we get w R W · w d · w R = 1 2 ( w ˙ U · w U 1 + w U 1 · w ˙ U ) (2.118b) and w R W · ( w $ w U ) · w R = 1 2 ( w ˙ U · w U 1 w U 1 · w ˙ U ) = (2.118c) It is very important to recognize that (Hill 1978): w U = g =, w $ = w U = (2.118d) An example of the above situation is the beginning of the deformation process (w =0). In the deformation process depicted in Fig. 2.6 (Truesdell & Noll 1965, Malvern 1969) we can write, using Eqs. (2.30a-2.30b): 54 Nonlinear continua Fig. 2.6. Relative deformation gradients X = w X · w X = (2.119a) For a fixed w-configuration, we can write d d X = d d w X · w X = (2.119b) Using Eq. (2.111b) and the polar decomposition in the abo ve equation, we get l · X =( d d w R · w U + w R · d d w U) · w X > (2.119c) for = w it is obvious that w U = w R = g , and since the above equation has to hold for any w X, d d w R | =w + d d w U | =w = w l = (2.119d) It is easy to show that the first tensor on the l.h.s. of the above equation is skew-symmetric and the second one is symmetric; hence, d d w R | =w = 1 2 ( w l w l W )= w $ (2.120a) d d w U | =w = 1 2 ( w l + w l W )= w d = (2.120b) We can obtain an interesting picture of the deformation process by refer- ring the Lagrangian tensors to the Lagrangian coordinate system (principal directions of w U) and the Eulerian tensors to the Eulerian coordinate system (principal directions of w V). Hence (Hill 1978): 2.13 Strain rates 55 • In the Lagrangian system the components of w U are w u (weassumethem to be dierent); the components of w ˙ U are w ˙ uv and the components of w O are w O uv . • In the Eulerian system the components of w V are of course also w u ;the components of w d are w g uv ; the components of w $ are w $ uv ; the components of w H are w H uv and the componen ts of w U are w U uv . From Eq. (2.117b) we have w H uv w U uv = w O uv > (2.121a) from Eq. (2.118b) we get w g uv = w ˙ uv w u + w v 2 w u w v (2.121b) and from Eq. (2.118c) we get w $ uv w U uv = w ˙ uv w u w v 2 w u w v ; (2.121c) (in Eqs. (2.121b) and (2.121c) we do not use the summation conven t ion). In the fixed Cartesian system the components of w U form the matrix [ w X]; hence, [ w X]=[ w U O ][ w ][ w U O ] W > (2.122a) where, [ w ]= 5 7 w 1 00 0 w 2 0 00 w 3 6 8 = (2.122b) Taking the time derivative of Eq. (2.122a), we obtain [ w ˙ X]=[ w U O ][ w ˙ ][ w U O ] W +[ w O ][ w U O ][ w ][ w U O ] W [ w U O ][ w ][ w U O ] W [ w O ] (2.122c) hence, [ w U O ] W [ w ˙ X][ w U O ]=[ w ˙ ]+[ w U O ] W [ w O ][ w U O ][ w ] [ w ][ w U O ] W [ w O ][ w U O ] = (2.122d) The above equation shows that: w ˙ uv = w ˙ u (u = v) w ˙ uv =( w v w u ) w O uv (u 6= v) = (2.123) From Eq. (2.121b) for the case u = v (diagonal components), we get w g uu = w ˙ u w u = d dw (ln w u ) = (2.124) 56 Nonlinear continua Example 2.16. JJJJJ Using Eqs. (2.65), (2.35), (2.111b) and (2.112b) we can show that: w ˙% = 1 2 w ˙ C = w X W · w d · w X = JJJJJ Example 2.17. JJJJJ The Hencky strain tensor components in the fixed Cartesian system are, [ w K]=[ w U O ]ln[ w ][ w U O ] W > hence, [ w ˙ K]=[ w U O ][ w ] 1 [ w ˙ ][ w U O ] W +[ w O ][ w U O ]ln[ w ][ w U O ] W [ w U O ]ln[ w ][ w U O ] W [ w O ] = JJJJJ 2.14 The Lie derivative In the deformation process represented in Fig. 2.2 we can define, for a Eulerian tensor w t,itsLie derivative associated to the flow of the spatial configuration (Simo 1988, Marsden & Hughes 1983): L w v ( w t)= w ! ½ d dw £ w ! ( w t) ¤ ¾ = (2.125) As we already kno w (see Sect. 2.9) the operation of pull-back is not a tensor operation since it operates on components. Hence, for calculating a Lie derivative using Eq. (2.125) it is important to identify the components of w t that we are using. The Lie derivative of a scalar is L w v = d dw = C Cw + C C w { d w y d = (2.126) The covariant components of the Lie derivative of a spatial vector w w are ¡ L w v w w ¢ l =( w [ 1 ) D l ½ d dw [( w [) m D w z m ] ¾ > (2.127a) after some algebra, 2.14 The Lie derivative 57 ¡ L w v w w ¢ l = C w z l Cw + C w z l C w { d w y d + C w y d C w { l w z d = (2.127b) Since, w z lw z l = w (2.128a) we can write ¡ L w v w w ¢ l w z l + w z l ¡ L w v w w ¢ l = µ d dw w z l ¶ w z l + w z l µ d dw w z l ¶ (2.128b) and from the above we get the contravariant components of the Lie derivative of a spatial vector w w, ¡ L w v w w ¢ l = C w z l Cw + C w z l C w { d w y d w z d C w y l C w { d = (2.128c) Followingtheaboveprocedurewecanshowthatthemixedcomponents of the Lie derivative of a general Eulerian tensor w t are ¡ L w v ( w t) ¢ d===e f===g = C w w d===e f===g Cw + C w w d===e f===g C w { s w y s (2.129) C w y d C w { s w w s===e f===g ··· C w y e C w { s w w d===s f===g + C w y s C w { f w w d===e s===g + ··· + C w y s C w { g w w d===e f===s Example 2.18. JJJJJ To calculate the Lie derivative of the spatial metric tensor w g we can directly use Eq. (2.125) ³ L w v w g ´ lm = w ! d dw h w ! ( w g) i LM ¸ > using now Eq. (2.93a), we get ³ L w v w g ´ lm = h w ! ( w ˙ C) i lm = Using the result in Example 2.16, we get ³ L w v w g ´ lm = ¡ 2 w d ¢ lm = JJJJJ 58 Nonlinear continua Example 2.19. JJJJJ To calculate the Lie derivative of the Almansi deformation tensor we use Eq. (2.125) and get ¡ L w v w e ¢ lm = w ! d dw £ w ! ( w e) ¤ LM ¸ and resorting to Eq.(2.94a), ¡ L w v w e ¢ lm = £ w ! ( w ˙%) ¤ lm = Taking into account the result obtained in Example 2.16 we can finally write ¡ L w v w e ¢ lm = ¡ w d ¢ lm = JJJJJ Example 2.20. JJJJJ The Lie derivative of the Finger deformation tensor is ¡ L w v w b ¢ lm = w ! d dw £ w ! ( w b) ¤ LM ¸ = Using Eq. (2.97b) we get £ w ! ( w b) ¤ LM = h w B ` i LM = j LM and since ˙g = 0 we get ¡ L w v w b ¢ lm =0= JJJJJ 2.14.1 Objective rates and Lie derivatives In this Section we will show that the Lie d erivative is the adequate mathe- matical tool for deriving covariant (objective) rates from covariant (objective) Eulerian tensors. Let us consider the deformation processes schematized in Fig. 2.7. It is ob vious that 2.14 The Lie derivative 59 Fig. 2.7. Deformation proc esses between three configurations ˆ w X = ˆ w w X · w X = (2.130a) For a covariant Eulerian tensor w t, that without losing generality we take as a second-order tensor: ˆ w w ˆ~ˆ = ˆ w w [ ˆ~ d ˆ w w [ ˆ e w w de > (2.130b) ¯ ¯ ¯ ˆ w ! ˆ w w ˆ~ˆ ¯ ¯ ¯ DE =( ˆ w [ 1 ) D ˆ~ ( ˆ w [ 1 ) E ˆ ˆ w w [ ˆ~ d ˆ w w [ ˆ e w w de = (2.130c) Using Eq. (2.130a) in the above, we get ¯ ¯ ¯ ˆ w ! ˆ w w ˆ~ˆ ¯ ¯ ¯ DE =( w [ 1 ) D o ( w [ 1 ) E n w w on > (2.130d) ¯ ¯ ¯ ˆ w ! ˆ w w ˆ~ˆ ¯ ¯ ¯ DE = ¯ ¯ w ! w w on ¯ ¯ DE = (2.130e) From the above and from Eq. (2.125) it follows that: L w ˆv ( ˆ w t) ˆd ˆ e = ˆ w w [ ˆd o ˆ w w [ ˆ e p h L w v ( w t) i op = ½ ˆ w w ! h L w v ( w t) i op ¾ ˆd ˆ e = (2.130f) The above equality shows that the Lie derivative of a covarian t Eulerian tensor is also a co variant Eulerian tensor. 60 Nonlinear continua Example 2.21. JJJJJ Considering again the case of a moving Cartesian frame and a fixed o ne from Example 2.15, we get w y = ˙f + ˙ T ( w } ) + T ( w y ) > therefore taking i nto accoun t that c > Q(w) and ˙ Q(w), for the case under consideration, are constant in space, w o = ˙ T C ( w } ) C w } + T ( w o ) C ( w } ) C w } hence, w o = ˙ T (T W ) + T ( w o ) (T W ) = Comparing with Eq. (2.101c) it is obvious that w l is not an objective tensor. Since the velocity gradient tensor is not objective in the classical sense we know that it is not a covariant tensor. JJJJJ Example 2.22. JJJJJ From Example 2.19, we know that w d is the result of a Lie derivative; hence we can assess that t he Eulerian strain rate tensor is a covariant (objective) tensor. JJJJJ Example 2.23. JJJJJ For a Eulerian tensor w t,wedefine in the reference configuration the tensor: w T ` = ¡ w [ 1 ¢ D d ¡ w [ 1 ¢ E e w w de g D g E which can be written as w T ` = w X 1 · w t · w X W = Since, w X 1 · w X = g we can derive that, d dw ¡ w X 1 ¢ = w X 1 · w l and d dw ³ w X W ´ = w l W · w X W we can write d dw ³ w T ` ´ = w X 1 · w ˙ t · w X W w X 1 · w l · w t · w X W w X 1 · w t · w l W · w X W = [...]... = + ããã + 2 1 4 à + ããã ảá = 0 If we now want to do a pull-back of Eq.(2.135b) the algebra can get quite lengthy, but we can use an analogy with the above tensor transformations: Ê Ê ( ) Ô ) ( Ô = = = = Using this formal analogy we can easily write: Ê ( à 1 4 à 1 4 Ô ) = 1 2 2 + + 2 Ă 2 1  2 + á Ă + ảà ảà 1  + + ả ảá JJJJJ 64 Nonlinear continua Hence, 2 1 2 2 2 á 2 + 1 4 1 4 à à ảà + ảà + +... (A.79a-A.79e), à 2 1 = + 2 Ă + 2 2 2  (2.1 34) ả (2.135a) where the are the Christo el symbols of the rst kind corresponding to the coordinate system { } Hence, using Eq (A.79b) ả à 2 2 2 2 1 = + 2 (2.135b) ảà ả à 1 + + + 4 ảà ảá à 1 + + =0 4 7 See Appendix 2.15 Compatibility Doing a pull-back operation on Eq (2.132) we obtain, Ă ÂÔ Ê = = 0 63 (2.136a) JJJJJ Example 2. 24 Equation (2.135b) represents the components... represent the spatial conguration of a body B corresponding to a time , and we identify a particle Fig 3.2 Internal forces at a point inside a continuum The external forces acting on B per unit mass are given by the vector eld b and the external forces per unit surface are given by the vector eld t 70 Nonlinear continua We now section the body B, in the -conguration, with a surface passing through... that: = 1 Ă 2  Introducing the above in Eq (2.135b) and linearizing (neglecting the higher powers of ) we get the compatibility equations corresponding to the assumption of linear kinematics: 66 Nonlinear continua 2 2 2 2 + = 0 The above represents a set of 6 equations, that proceeding as in Eqs (2.137), can be reduced to 3 independent compatibility conditions The above result was obtained for a Cartesian... Let us dene in the -conguration of a body a Cartesian system { = 1 2 3} with base vectors and an arbitrary curvilinear system e { = 1 2 3} with covariant base vectors g We consider the vector 68 Nonlinear continua Fig 3.1 External and internal forces in a bar eld b( x) to dene the forces per unit mass acting on the body at We can write b = = e g (3.1) The resultant of the forces per unit mass is:... tensor [( )] )] = d = Hence, d d = d d (2.131) From the above equation it is obvious that the covariant components of the Green deformation tensor are the covariant components of the metric tensor 62 Nonlinear continua of the pull-back space Note that the pull-back space is by no means coincident with the reference conguration, whose metric tensor has the covariant components Since we restrict our study... Tensor To deform a continuous body the exterior medium has to produce a loading on that body; therefore we get external forces acting on it Also, during the deformation of a continuous body, neighboring particles exert forces on each other, they are the internal forces in the body The study of the internal forces in a body leads to the notion of stresses, that we are going to develop in this chapter Some... tensor: ( ) = + Ă 1  1 2 à 2 2 2 ả 2 + ( ) = 0 (2.136d) is a metric of the Euclidean space The above equation indicates that Taking into account the Bianchi identities in Eq (A.82d) (Synge & Schild 1 949 ) and Eqs.(2.135a), we get 1212|3 + 1213|1 1313|2 + 1323|1 2323|1 1323|2 + = 0 1213|3 + 1213|2 = 0 1223|3 = 0 (2.137) and we reduce the number of independent compatibility conditions to 3 JJJJJ Example... acting on the region , a subset of the surface of the body at We can write 3.2 The Cauchy stress tensor t = e = (3.3) g The resultant of the forces per unit surface is: Z Z Z d = t d = e 69 g d (3 .4) It is important to point out that our description of the external forces acting on a body excludes the possibility of considering distributed torques per unit volume, mass or surface Therefore, the... of the reference conguration base vectors is zero and using the above together with the Lie derivative denition in Eq (2.125), we get Ê v Ô ( t) = | | The above equation is going to be used in Sect 3 .4 for deriving objective stress rates JJJJJ 2.15 Compatibility In our previous description of the kinematics of continuous media we went through the following path: Assume the existence of a regular mapping . w F UT C { S C { V á + Ă w F 1  PQ 1 4 à C w F VP C { U + C w F PU C { V C w F UV C { P ảà C w F TQ C { S + C w F QS C { T C w F ST C { Q ả 1 4 à C w F TP C { U + C w F PU C { T C w F UT C { P ảà C w F VQ C { S + C w F QS C { V C w F SV C { Q ảá JJJJJ 64 Nonlinear continua Hence, 1 2 C 2 w F ST C { U C { V + C 2 w F UV C { S C { T C 2 w F SV C { U C { T C 2 w F UT C { S C { V ¸ + ¡ w F 1 ¢ PQ (2.136b) 1 4 µ C w F VP C { U + C w F PU C { V C w F UV C { P ¶µ C w F TQ C { S + C w F QS C { T C w F ST C { Q ¶ 1 4 µ C w F TP C { U + C w F PU C { T C w F UT C { P ¶µ C w F VQ C { S + C w F QS C { V C w F SV C { Q ¶¸ =0= We. Fig. 2.6 (Truesdell & Noll 1965, Malvern 1969) we can write, using Eqs. (2.30a-2.30b): 54 Nonlinear continua Fig. 2.6. Relative deformation gradients X = w X · w X = (2.119a) For a fixed. case u = v (diagonal components), we get w g uu = w ˙ u w u = d dw (ln w u ) = (2.1 24) 56 Nonlinear continua Example 2.16. JJJJJ Using Eqs. (2.65), (2.35), (2.111b) and (2.112b) we can show